*1,211*
*127*
*2MB*

*English*
*Pages 481*
*Year 2009*

Engineering Thermodynamics Second Edition

M. ACHUTHAN Director MES College of Engineering Kuttipuram, Kerala Formerly Professor of Mechanical Engineering Indian Institute of Technology Bombay

New Delhi-110001 2009

ENGINEERING THERMODYNAMICS, 2nd ed. M. Achuthan © 2009 by PHI Learning 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-978-81-203-3845-6 The export rights of this book are vested solely with the publisher. Fourth Printing (Second Edition)

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…

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October, 2009

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Rajkamal Electric Press, Plot No. 2, Phase IV, HSIDC, Kundli-131028, Sonepat, Haryana.

Appendix F

Contents

Preface Chapter 1

ix

What is Thermodynamics? Introduction 1 Foundation of Thermodynamics 1 The Objective 3 Problem-solving 4 1.4.1 Numerical Calculations 5 1.4.2 Units 6 1.5 Some Systems 8 1.5.1 Internal Combustion (I.C.) Engine 1.5.2 Steam Turbine Power Plant 10 1.5.3 Heat Exchanger 11 1.5.4 Refrigerator 12 1.6 Conclusion 13 Review Questions 13

1–14

1.1 1.2 1.3 1.4

Chapter 2

8

Concepts of Thermodynamics 2.1 2.2

Introduction 15 Universally Primitive Concepts 15 2.2.1 System Description 15 2.2.2 System, Environment and Interactions 2.2.3 Classification of Systems 18 2.2.4 State and Properties 20 2.2.5 Thermodynamic Equilibrium 23 2.2.6 The Gibbs Phase Rule 24 2.2.7 Processes 24 iii

15–42

17

iv

Contents

2.3

Concepts from Properties of Matter 26 2.3.1 Empirical Temperature 27 2.3.2 Other Properties of Substances 30 2.4 Classical Thermodynamics 39 2.4.1 Models in Thermodynamics 39 2.4.2 Problems in Thermodynamics 40 2.5 Conclusion 40 Review Questions 41 Chapter 3

Work and Mechanical Energy

43–73

3.1 3.2

Introduction 43 Basic Concepts of Mechanics 43 3.2.1 Concepts from Geometry 43 3.2.2 Concepts from Kinematics 44 3.2.3 Concepts from Dynamics 44 3.3 Work 46 3.3.1 Different Modes of Work and Their Expressions 3.4 Evaluation of Work 52 3.4.1 Process Specification 52 3.5 Mechanical Energy 60 3.5.1 Kinetic Energy 60 3.5.2 Work–Energy Theorem 60 3.5.3 The Principle of Conservation of Energy 61 3.5.4 Conservative Force and Potential Energy 61 3.5.5 The Energy Equation 62 3.6 Expressions for Potential Energy 62 3.6.1 Gravitational Field 62 3.6.2 Electrostatic Field 63 3.6.3 Magnetic and Electromagnetic Fields 63 3.6.4 Total Energy 64 3.7 Adiabatic and Diathermal Walls 64 3.7.1 Adiabatic Wall 64 3.7.2 Diathermal (or Diathermic) Walls 65 3.7.3 Thermal Equilibrium and Heat 66 Review Questions 66 Exercises 67 Chapter 4

Temperature and Its Empirical Scales 4.1 4.2 4.3 4.4

Introduction 74 Thermal Equilibrium 75 Mathematical Development of Temperature Empirical Temperature 78

46

74–84

77

Contents

Ideal Gas Temperature 81 4.5.1 Experiments 81 4.5.2 Results and Their Interpretation 4.6 International Practical Temperature Scale Review Questions 84

v

4.5

Chapter 5

82 83

The First Law of Thermodynamics 5.1 Introduction 85 5.2 Born–Carathéodory Form of the First Law 85 5.3 Characteristics of the First Law 87 5.4 Measurement of Heat 87 5.5 Characteristics of Work, Heat and Energy 89 5.6 Independent State Variables 93 5.7 Thermodynamic Models of Working Substances 5.8 The First Law for Cycles 94 5.9 Equilibrium 95 5.10 Examples 95 Review Questions 102 Exercises 103

Chapter 6

93

Energy (First Law) Equation for Flow Systems 6.1 Introduction 107 6.2 Reference Quantities 110 6.3 Simplification of Energy (First Law) Equation 6.4 Examples 113 Review Questions 125 Exercises 126

Chapter 7

85–106

The Second Law of Thermodynamics 7.1 7.2 7.3

7.4

7.5 7.6 7.7 7.8 7.9 7.10 7.11

107–133

111

134–169

Introduction 134 Preliminary Definitions 135 The Need of the Second Law 136 7.3.1 Equivalence of Work and Heat for Processes 136 7.3.2 Equivalence of Work and Heat for Cycles 136 Different Forms of the Second Law of Thermodynamics 137 7.4.1 The Kelvin–Planck Form 138 7.4.2 The Clausius Form of the Second Law 139 Use of the Second Law of Thermodynamics 141 Reversible and Irreversible Processes 142 The Carnot Theorem 144 Thermodynamic Scale of Temperature 145 Relation to Ideal Gas Temperature 147 Negative Absolute Temperatures 148 Entropy 148 7.11.1 Evaluation of Entropy Change during a Process 149

vi

Contents

7.12

Clausius Inequality 150 7.12.1 The Principle of Increase of Entropy 151 7.12.2 Possible Processes 152 7.12.3 Possible Cyclic Devices 152 7.13 Adiabatic, Reversible and Isentropic Processes 153 7.14 Examples 154 7.15 Equilibrium 160 7.16 The Carathéodory Form of the Second Law 161 7.16.1 Accessible and Inaccessible States 161 7.17 The Second Law for Flow Systems 163 Review Questions 164 Exercises 165 Chapter 8

Auxiliary Functions

170–202

8.1 8.2

Introduction 170 Availability 171 8.2.1 Availability in Non-flow Processes 171 8.2.2 Availability in General Non-flow Processes 8.2.3 Availability in Flow Processes 176 8.2.4 Examples on Availability 177 8.3 Other Energy Functions 180 8.3.1 Internal Energy 181 8.3.2 Enthalpy 181 8.3.3 Gibbs Function 182 8.3.4 Helmholtz Function 182 8.3.5 Consequences 182 8.4 Property Relations 188 8.4.1 Maxwell Relations 191 8.4.2 Examples on Property Relations 191 8.5 Multicomponent Systems 193 Review Questions 196 Exercises 197 Chapter 9

Properties of Substances 9.1 9.2

9.3

Introduction 203 Data and Computational Procedure 203 9.2.1 Independent and Dependent Properties 9.2.2 Basic Data 204 9.2.3 Computational Procedure 206 Gases 210 9.3.1 Ideal Gas 210 9.3.2 Real Gases 214 9.3.3 Examples 223

175

203–247

203

Contents

Liquids 226 Mixture of Inert Ideal Gases 229 9.5.1 Psychrometry: Air-Water Vapour Mixtures Review Questions 245 Exercises 245

vii

9.4 9.5

234

Chapter 10 Thermodynamics of Combustion

248–301

10.1 10.2 10.3

Introduction 248 Major Energy Sources and Converters 248 Fuels 250 10.3.1 Coal 251 10.3.2 Liquid and Gaseous Fuels 252 10.4 Atmospheric Air 254 10.4.1 Interconversion of Mole and Mass Fractions 254 10.5 Combustion Process 258 10.6 Combustion Stoichiometry 261 10.6.1 Air Required for Combustion 263 10.6.2 Exhaust Gas Analysis 269 10.7 Reaction Equilibrium 276 10.7.1 Equilibrium Constant 276 10.7.2 Calculation of Equilibrium Composition 283 10.8 Heat Generated by Combustion 286 10.8.1 Heat of Formation 286 10.8.2 Heat of Reaction 289 10.8.3 The Calorific Value 291 10.9 Adiabatic Flame Temperature 294 Review Questions 297 Exercises 297 Chapter 11 Ideal Cycles 11.1 11.2 11.3 11.4

Introduction 302 Types of Cycles 303 Cycle Calculations 306 Cycle and Performance Parameters 309 11.4.1 Cycle Parameters 309 11.4.2 Performance Parameters 310 11.5 The Carnot Cycle 313 11.5.1 The Stirling and Ericsson Cycles 317 11.5.2 The Reversed Carnot Cycle 319 11.6 Ideal aAnd Real Cycles 322 Review Questions 322 Exercises 323

302–325

viii

Contents

Chapter 12 Internal Combustion (I.C.) Engine Cycles

326–344

12.1 12.2

Introduction 326 Operation of an I.C. Engine 327 12.2.1 Four-Stroke Engines 327 12.2.2 Two-Stroke Engines 328 12.3 Air-Standard Cycles 329 12.3.1 Air-Standard Otto Cycle 329 12.3.2 Air-Standard Diesel Cycle 331 12.4 Real Engines 332 12.4.1 Spark Ignition (S.I.) Engines 333 12.4.2 Compression Ignition (C.I.) Engines 335 12.4.3 The Air-Standard Dual Combustion Cycle 336 Review Questions 339 Exercises 339 Chapter 13 Brayton Cycle and Its Other Names 13.1 Introduction 345 13.2 The Brayton Cycle 346 13.2.1 Air-Standard Cycle 346 13.2.2 Basic Gas Turbine Plant 347 13.2.3 Modifications to the Basic Plant 353 13.3 Joule (Reversed Brayton) Cycle 356 13.4 The Rankine Cycle 357 13.4.1 The Basic Rankine Cycle 358 13.4.2 Modifications to the Basic Rankine Cycle 13.5 Vapour Compression Refrigeration Cycle— The Reversed Rankine Cycle 364 Review Questions 366 Exercises 367

345–371

361

Appendix A Additional Problems

373–390

Appendix B Explanatory Notes

391–409

Appendix C Problem-Solving

410–422

Appendix D Some Mathematical Principles

423–436

Appendix E Other Formulations of Thermodynamics

437–447

Appendix F Steam and R–12 Tables

448–485

Bibliography

487–488

Index

489–498

Appendix F

Preface

Why another book on thermodynamics? The answer to this question is simple. During every revamp of the thermodynamics course, a need is always felt for a textbook that logically and rigorously explains the subject from the first principles. Unfortunately, this need exists even today. Thermodynamics deals with energy which is a scalar quantity, while mechanics deals with forces and momenta which are vectors quantities. Hence, mathematically, thermodynamics is simpler than mechanics (even, classical (Newtonian) mechanics). In fact, all the equations of thermodynamics are algebraic and linear (mathematically, the simplest). Even then, thermodynamics is ‘judged’ to be ‘more difficult’ than mechanics. Since mechanics is taught first, it becomes ‘natural’. Thermodynamics is perceived to be more challenging as storage and flow of energy is basic to all natural processes and, therefore, the scope and variety of problems encountered become enormous. The main goal of this book is to help the reader develop ‘knowledge’ of thermodynamics. Knowledge is not something to be packed away in some corner of our brains, but what enters our being, colours our emotion, haunts our soul and is as close to us as life itself. It is overmastering power which through the intellect moulds the whole personality, trains the emotion, and disciplines the will. Dr. S. RADHAKRISHNAN On the Plaque in Convocation Hall of IIT Bombay

To provide the reader with a sound understanding of thermodynamics, the subject is developed from the first principles — clearly indicating what forms part of thermodynamics (i.e. showing what thermodynamics is and what it is not). Since this material was developed from courses offered to mechanical and electrical engineering students, equilibria (phase and reaction) are not elaborated. This material was also used as review for postgraduate students (with some additional topics, such as non-equilibrium thermodynamics, statistical mechanics, thermodynamics of combustion, direct energy conversion, which are not included here). A new Appendix containing problems from examinations of universities is added in the second edition. ix

x

Preface

And now the pleasant duty of acknowledging my indebtedness. First and foremost, the congenial environment in the Indian Institute of Technology Bombay and its Mechanical Engineering Department was excellent stimulants for ‘dreaming’ thermodynamics. The help of my former colleagues, (late) Prof. A. Jaganmohan and (late) Prof. B.S. Jagadish (both my teachers as well) in understanding the nuances of steam engineering, gas dynamics and jet propulsion was indispensable for a proper understanding of thermodynamics. (Late) Prof. M.N. Vartak of the Mathematics Department at IIT Bombay was instrumental in clarifying my fundamentals on mathematics (including analysis and mathematical logic) and statistics leading to my appreciation of the General Systems Theory and broadening my experience. The lectures of Prof. R.E. Bedford (of the Electrical Engineering Department at IIT Bombay) on Network Theory helped me understand the basic fact of science that “same things appear in different fields in different forms.” Numerous ‘awkward’ questions from my former students, both undergraduate and postgraduate, made me take another look at the concepts and methods of thermodynamics. Most of them did not know these, because they were satisfied with my answers, but I was not. My former colleague, Prof. U.N. Gaitonde (popularly referred to as UNG), was particularly helpful with numerous ‘discussions, debates and arguments’ on everything under the sun (in the garb of basics of thermodynamics), so much so that he should have been the co-author of this book. The author thanks all those who have commented on this book, especially Shri Jaspreet Singh, Lecturer of Mechanical Engineering, SUSCET, Tangori, Punjab. Thanks are also due to the management of MES College of Engineering, Kuttipuram, Kerala for the help rendered. I record my thanks to the publishers, PHI Learning, for the excellent work done by them in bringing out this book. Several persons have helped me produce this book. Errors that remain are solely due to me, and I shall be grateful if they are pointed out. Finally, I wish to express my appreciation to my wife for her unstinted support and encouragement during preparation of the manuscript. This book is written for and dedicated to Nachiketa, not the legendary boy-hero of Katha Upanishad, but an equally brilliant boy, who was frightened when he heard, “to understand thermodynamics you should master Jacobians, Legendre transformations, etc.” (The story and the name are true, but the name also represents numerous other students (boys and girls), who equally dread thermodynamics.) The book shows that, in most cases, these dreaded mathematical acrobatics are not needed. In fact, this book does not use them at all, except as additional pieces of information. I, therefore, will consider my efforts well rewarded if, after studying this book, the reader says, “hey! I have always known this and can now solve any problem in thermodynamics.” To conclude, this book is on thermodynamics, but the main point it tries to make is: “Do not accept anything (including this statement) as final till you have tested it yourself,” which is merely a restatement of: “Thus, …, having reflected upon it fully, you now act as you choose.” Srimad Bhagavath Geetha, 18.63 Central Chinmaya Mission Trust

M. ACHUTHAN

Chapter 1

What is Thermodynamics?

1.1

INTRODUCTION

The name ‘thermodynamics’ is derived from the Greek words thermē, meaning ‘heat’, and dynamikšs (or dynamīs) meaning ‘power’ or ‘motion’. Thus, thermodynamics essentially means ‘heat power’ or ‘heat-in-motion’. The Concise Oxford Dictionary defines thermodynamics as “the science of relation between heat and mechanical energy”. Thus, it is the science dealing with storage, inter-conversion and exchange of energy. In this book, the emphasis is on understanding fundamental concepts, and on systematic formulation and solution of problems from first principles. Appendix B titled “Explanatory Notes” contains additional material that further elaborates the material of main text. This and the next two chapters are merely reviews, and therefore, are necessarily concise.

1.2

FOUNDATION OF THERMODYNAMICS

From the time man created fire by striking flint stones, it was known that (a) fire is hot and (b) its hotness can change. These observations have given rise to the terms heat (hotness), temperature (the degree of hotness) and creation of heat by friction. Thermodynamics explains these observations on scientific basis by: ∑ Logically developing a quantitative relation between (frictional) work done by dissipation and energy stored (first law of thermodynamics). ∑ Defining and showing how to measure heat (first law of thermodynamics). ∑ Defining the concept of temperature (zeroth law of thermodynamics). ∑ Showing how to measure temperature (second law of thermodynamics). ∑ Logically developing criteria for inter-conversion of work and heat (second law of thermodynamics). 1

2

Engineering Thermodynamics

Historically, contributions from many branches of science have led to the development of thermodynamics (not necessarily under this name). Table 1.1 enumerates the history of development of thermodynamics. Appendix B gives a more elaborate account. Table 1.1 Year

History of development of heat and thermometry

Name

*

Homo sapiens**

50 B.C. 1592

Hero of Alexandria Galileo

1620 1631 1641

1680 1694

Francis Bacon Jean Rey Grand Duke Ferdinand II Robert Boyle and others Huygens Carlo Renaldini

1705 1766–99

Newcomen Joseph Black

1779

William Cleghorn

1798

Count Rumford

1824

Sadi Carnot

1842–43

Mayer and Joule

1847 1848 1850–55 1897 1907

Hemholtz Kelvin Clausius and Kelvin Maxwell Nernst

1665

Contribution Discovered that rubbing of two stones produces fire and found the properties of fire (Experiments of Joule, Mayer, … ?). Developed his ‘engine’ (which was really a turbine). Developed barothermoscope to measure pressure and temperature of atmosphere. Concluded that “heat is motion and nothing else”. Constructed the first liquid expansion thermometer. Developed the first sealed alcohol thermoscope. Independently proposed the one fixed point method of calibration. Developed the gunpowder engine (beginning of I.C. engines?) Proposed the two fixed point method of calibration using ice and steam points. Devised the first steam engine with cylinder and piston. Clarified concepts of ‘quantity of heat’ and ‘degree of hotness’ (temperature) and measured specific heats and latent heats. Proposed an extension to the material theory which was later named as ‘caloric’ by Lavoisier in 1787. Announced the results of the now famous cannon boring experiments (beginning of overthrow of caloric theory). Published (in imprecise form) the basics of the second law of thermodynamics. Published (the now well-known) experimental works on the mechanical equivalent of heat. Formulated the first law of thermodynamics. Developed the thermodynamic (Kelvin) scale of temperature. Stated their forms of the second law of thermodynamics Stated the zeroth law of thermodynamics.† Enunciated the third law of thermodynamics.

* This is the beginning of modern human beings regarded as a species. ** Anthropological name of the species of human race. † Fowler gave this name in 1930. Notes: 1. The years are only indicative and not exhaustive. 2. Count Rumford is Benjamin Thompson and Lord Kelvin is William Thomson. 3. Some other dates: 600 B.C.—Thales’ experiments on electricity and magnetism; 1600—Gilbert’s experiments on electricity and magnetism; 1686—Newton’s laws; 1785—Coulomb’s law.

Chapter 1: What is Thermodynamics?

3

Data of this table illustrate the following main points: 1. Concepts that are supposed to have been developed in thermodynamics were in existence long before thermodynamics itself came into existence. 2. Engineering has contributed significantly to the development of thermodynamics. 3. Temperature was measured using non-thermal effects, such as mechanical (mainly thermal expansion of liquids and gases), electrical or thermo-electrical. This is the reason why thermodynamic concepts (all, except entropy) are taken as primitives, i.e. those with ‘intuitive’ meanings requiring no formal definition. So, a student finds it difficult to appreciate the efforts to formalize the well-known concepts through innumerable definitions and hair splitting. Only thermodynamics alone defines them precisely through operational definitions and thereby shows how they should be used.

1.3

THE OBJECTIVE

The main objective of this book is to explain thermodynamics in the simplest possible way. This is done as follows: 1. Concepts and methods borrowed from other disciplines are clearly identified and their meanings are stated.1 These are primitives of thermodynamics. 2. The terminology of thermodynamics is logically developed in terms of these primitives using operational definitions.2 3. The laws that thermodynamics borrows from other disciplines are clearly stated. In mathematics, these are called axioms. These are also primitives of thermodynamics. 4. Consequences of the laws of thermodynamics are derived and proved. In mathematics, these are called theorems. 5. A systematic problem-solving methodology is used to illustrates the ‘patterns’ of examples and exercises.3 The emphasis is on systematically solving problems from first principles. The first four steps constitute the axiomatic method. These and the principle of operational definition are elaborated in Appendix B.

1

2 3

Thermodynamics operationally defines the concepts of heat and temperature, which have already been known but were assumed to be primitives and empirically measured (e.g. calories for heat). Indeed, internal energy and entropy are the only concepts developed first in thermodynamics. Out of these, internal energy posed no difficulty to the reader because it is another form of energy, which has been a primitive concept. Thus, only entropy was ‘accepted’ to be a difficult concept. Definition of a quantity in terms of the operations (including the paper and pencil operations, i.e. mathematical calculations) used for its measurement. The enormous and bewildering variety of possible examples and exercises (and, for the unfortunate student, questions in examination) can be classified into about one hundred types.

4

Engineering Thermodynamics

1.4

PROBLEM-SOLVING

Systematic method of problem-solving is essential (a) for identifying different aspects of the same problem, (b) for verifying the procedure of formulation and computations, and (c) for identifying necessary modifications in the case of different problems. Following method is generally found to be suitable. Problem statement.

In general, a problem is given as a set of verbal statements.

Identification of variables.

Making suitable assumptions, relevant variables are identified.

Formulation. In terms of the relevant variables and satisfying the basic conservation laws, the equation(s) describing the phenomenon (or the system) is (are) derived. The above steps form modelling. Solution. The equations are now solved. This is a purely mathematical exercise. This is called simulation, especially when computers are used. Verification. The solutions are verified using the results obtained from appropriate experiments. If the two do not agree within some predefined error, simulation is repeated with better solution techniques of a more accurate model.4 In this book, as far as possible (to avoid verbose), examples are solved using the above method.5 As science advanced, it demanded better results and, therefore, experimentation became more expensive (in terms of resources, namely materials, manpower, money and time). Hence, alternative methods of analysis were thought of. Most successful of these is modelling and simulation. Appendix B elaborates these concepts. Currently, mathematical models are extensively used in all disciplines of science (natural and social). Basically this consists of developing appropriate mathematical equations (called system equations) describing the phenomenon. Simulation6 is the process of solving these equations for different values of system parameters. In the next chapter, the basic models used in thermodynamics are described. All through this book, mathematical models (equations) are developed and used. Following philosophy is adopted in this book towards mathematics7:

4

5

6 7

A discipline is termed science only if its theories (predictions) can be confirmed (or denied) experimentally ([FEY], vol. I, p. 1–1). However, Born (Max., Experiments and Theory in Physics, Dover, 1956) traces how the development of science had many pitfalls and back-tracking. Systematic problem-solving is expected in all the exercises. The general scheme of marking of the exercises is: system and process diagrams = 1 mark; process specification (including initial and final states) = 1; answer = 1 (half for number and half for unit); assumptions = 3; numericals = 4. Thus, the emphasis is on approaching a problem from first principles and solving it systematically. For very large, or complex (or both) problems, computers are used for solving the system equations. This is why, these days, simulation automatically means computer simulation. The aim is not to dazzle others with mathematical gymnastics.

Chapter 1: What is Thermodynamics?

5

∑ “Suppose I apply myself to a complicated calculation and with much difficulty arrive at a result, I shall have gained nothing by my trouble if it has not enabled me to foresee the results of other analogous calculations, and to direct with certainty, avoiding the blind groping with which I had to be contended the first time”.8 ∑ Mathematics is used in the spirit of “Neither seeking nor avoiding mathematical exercitations we enter into problems solely with a view to possible usefulness to physical science”.9 In other words, in this book, mathematics is considered to be a language for compactly expressing the underlying physics as mathematical relations among different quantities. These relations are called equations. Each mathematical equation says something and, therefore, if any of them does not make sense, the statement it implies is physically meaningless.

1.4.1

Numerical Calculations

Numerical computations are an integral part of solving engineering problems. Hence, proper attention must be paid to them so that results are error free. A review of errors in calculations is presented in Appendix C titled “Problem-Solving”, which should be read for further explanation. The basic principle of all calculations is: the results are only as accurate as the least accurate data. For the purposes of this book following steps are recommended for obtaining maximum accuracies in calculations. ∑ In all calculations, the number of digits in the more accurate data should be one more than that in the least accurate one. After completing the calculations, round the result off to the number of significant digits of the least accurate data. For example, 987.6 + 5.432 = 987.6 + 5.43 = 993.03 = 993.0 (because the first number is accurate to four digits). ∑ Do all calculations at one stretch10 and finally round off the results. This avoids accumulation of round off errors. For example, 9.876 + (4.567 × 1.234 – 2.453)/6.1 = 10.39775 = 10 because the least accurate date (i.e. 6.1) has only 2 significant digits. ∑ During subtraction, especially of two nearly equal numbers, keep additional significant digits to compensate for their loss. For example, 987.65342 – 987.60211 = 0.05131. Thus, only four significant digits could be obtained by subtraction of two eight digit numbers. ∑ Use mathematical relations to transform equations such that the resulting operations are more accurate. For example, for small values of x, (1 – cos x) will result in subtraction of two nearly equal numbers. Hence, convert this expression to 2 sin2 (x/2). 8 9 10

Henri Poincaré, on ‘The Future of Mathematics’ in [POI2], p. 29. Attributed to Lord Kelvin and Peter Guthrie Tait by von Karman, T. and M.A. Biot, in ‘Mathematical Methods for Engineers’, McGraw-Hill, 1940, preface. This is important especially since internal accuracy of electronic calculators is more. However, calculation mistakes cannot be easily detected.

6

Engineering Thermodynamics

In this book, calculations are generally done to five significant digits and the results are rounded off to four digits.11

1.4.2

Units

The quantities that occur in engineering (and in sciences) are physical quantities. The characteristic that distinguishes them from others (say, mathematical quantities which are pure numbers) is that all of them have some dimensions and are expressed in some units.12 In order to emphasise the importance of units, in this book, they are written in brackets (e.g. 1 [kg]). SI units are mainly used in this book. However, FPS units can easily be converted to SI units by using following factors. Since they are definitions, these factors are exact. Therefore, it is good to memorize them. Metric or MKS units.

1 [kcal] = 4.1868 [kJ]; 1 [kgf] = 9.80665 [N].

British or FPS units. 1 [in] = 2.54 [cm]; 1 [lb] = 0.45359237 [kg]; 1 [°C] = (1/1.8) [°F]; and, 1 [Btu/(lb . °F)] = 1 [kcal/(kg . °C)]. It should be noted that the units of heat in British and metric systems, namely [Btu] and [kcal] are defined in such a way that the value of specific heat in both these systems is the same. The factor 1 with both [Btu] and [kcal] is merely a statement of this. In the following illustrative examples, the procedure is emphasized more than the result because a proper procedure always gives correct result. Appendix C contains a more elaborate review together with additional examples on interconversion. EXAMPLE 1.1 Convert a pressure of 2 [ft H2O] to SI units given that g = 9.81 [m/s2] and r = 1000 [kg/m3]. Solution

The SI unit of pressure is pascal defined as 1 [N/m2]. Then, ⎛ ⎛ ⎡ kg ⎤⎞ ⎛ ⎡ m ⎤⎞ ⎡ m ⎤⎞ p = rgh = ⎜ 1000 ⎢ 3 ⎥⎟ × ⎜ 981 ⎢ 2 ⎥ ⎟ × ( 2 [ ft ]) × ⎜ 0.3048 ⎢ ⎥⎟ = 5980 [Pa] ⎝ ⎝ ⎣ m ⎦⎠ ⎝ ⎣ s ⎦⎠ ⎣ ft ⎦ ⎠

EXAMPLE 1.2 Solution

Convert 2000 [°F] into [°C].

This well-known procedure is performed as follows: 2000 [°F] =

EXAMPLE 1.3

11 12

2000 − 32 = 1093 [°C] 1.8

Convert the density of 55 [lb/ft3] to SI units.

Do not write down all the digits that the calculator displays (1 mark penalty for doing this). CRC Handbook of Chemistry and Physics states: “A physical quantity is equivalent to the product of the numerical value, i.e. a pure number and a unit”, p. F-256, 68th ed., 1987–88.

Chapter 1: What is Thermodynamics?

Solution

This is done in one step as follows: 1 ⎡ 1b ⎤ ⎛ ⎡ 1b ⎤ ⎞ ⎛ ⎡ kg ⎤ ⎞ ⎛ 55 ⎢ 3 ⎥ = ⎜ 55 ⎢ 3 ⎥ ⎟ × ⎜ 0.45359237 ⎢ ⎥ ⎟ × ⎜ 3 ⎜ ⎣ ft ⎦ ⎝ ⎣ ft ⎦ ⎠ ⎝ ⎣ lb ⎦ ⎠ ⎝ (0.3048)

⎡ ft 3 ⎤ ⎞ ⎡ kg ⎤ ⎢ 3 ⎥ ⎟⎟ = 881.0 ⎢ 3 ⎥ ⎣m ⎦ ⎣m ⎦⎠

Convert 100 [psi] to SI units.

EXAMPLE 1.4 Solution

7

Now,

⎛ ⎡ lbf ⎤⎞ ⎛ ⎡ ft ⎤⎞ ⎛ ⎡ 1 ⎤⎞ 100 [psi] = ⎜ 100 ⎢ 2 ⎥⎟ = ⎜ 100[lb] × 32.174 ⎢ 2 ⎥⎟ × ⎜ 1 ⎢ 2 ⎥⎟ ⎝ ⎠ ⎝ ⎣ in ⎦ ⎣ s ⎦⎠ ⎝ ⎣ in ⎦⎠ ⎛ ⎡ kg ⎤⎞ ⎛ ⎡ ft ⎤ ⎡ m ⎤⎞ = ⎜ 100 [lb] × 0.45359237 ⎢ ⎥⎟ × ⎜ 32.174 ⎢ 2 ⎥ × 0.3048 ⎢ ⎥ ⎟ ⎝ ⎣ lb ⎦⎠ ⎝ ⎣s ⎦ ⎣ ft ⎦ ⎠ ⎛ ⎡ 1 ⎤ 1 × ⎜1 ⎢ 2 ⎥ × ⎜ ⎣ in ⎦ (0.0254)2 ⎝

EXAMPLE 1.5 Solution

⎡ in 2 ⎤ ⎞ ⎢ 2 ⎥ ⎟⎟ = 689500 [Pa] ⎣m ⎦⎠

Express 2500 [Btu] in SI units.

Since this is an illustration, 1 [Btu] is first converted to [kJ], the SI unit of energy.

⎛ ⎡ Btu ⎤ ⎞ ⎛ ⎡ kcal ⎤ ⎞ 1 [Btu] = ⎜ 1 ⎢ ⎟ × 1[lb] × 1[°F] = ⎜1 ⎢ ⎥ ⎥ ⎟ × 1[lb] × 1[°F] ⎝ ⎣ lb ⋅ °F ⎦ ⎠ ⎝ ⎣ kg ⋅ °C ⎦ ⎠ ⎛ 1[lb] ⎞ ⎛ 1[°F] ⎞ ⎛ ⎡ kJ ⎤⎞ = 1[kcal] × ⎜ = 1[kcal] × ⎜ 4.1868 ⎢ ⎥⎟ ⎝ 1[kg] ⎟⎠ ⎜⎝ 1[°C]⎟⎠ ⎝ ⎣ kcal ⎦⎠ ⎛ 1 ⎞ × 0.45359237 × ⎜ ⎟ = 1.055 [kJ] ⎝ 1.8 ⎠

Hence, 2500 [Btu] = 2638 [kJ]. EXAMPLE 1.6

Convert 2500 [Btu/h] to [kW].

Solution The answer is simply the answer to Example 1.5 divided by 3600 [s/h], but this is done again for clarity. ⎡ Btu ⎤ ⎛ ⎡ Btu ⎤ ⎞ ⎛ 1 ⎡ h ⎤ ⎞ ⎛ 2500 × 0.45359237 × 4.1868 ⎞ ⎡ kJ ⎤ 2500 ⎢ = ⎜ 2500 ⎢ ⎟⎢ ⎥ ⎥ ⎥⎟ × ⎜ ⎢ ⎥⎟ = ⎜ 1.8 × 3600 ⎣ h ⎦ ⎝ ⎣ h ⎦ ⎠ ⎝ 3600 ⎣ s ⎦ ⎠ ⎝ ⎠⎣ s ⎦

= 0.7327 [kW]. EXAMPLE 1.7 The FPS unit of power is horsepower, defined as 1 [hp] = 550 [lbf . ft/s]. How many [kW] is this?

8

Engineering Thermodynamics

Solution

By definition 1 [lbf] = 1 [lb] × g0; Then,

⎛ ⎡ ft ⎤⎞ ⎛ ⎡ ft ⎤⎞ ⎛ ⎡ kg ⎤⎞ 1 [hp] = ⎜ 550 [lb] × g0 ⎢ 2 ⎥⎟ × ⎜1 ⎢ ⎥⎟ = ⎜ 550[lb] × 0.45359237 ⎢ ⎥⎟ × (g0 ) ⎝ ⎣ s ⎦⎠ ⎝ ⎣ s ⎦⎠ ⎝ ⎣ lb ⎦⎠ ⎛ ⎡ ft ⎤ ⎡ m ⎤⎞ ⎛ ⎡ kg ⎤ ⎞ ⎛ ⎡ m ⎤⎞ × ⎜ 1 ⎢ ⎥ × 0.3048 ⎢ ⎥ ⎟ = ⎜ 550[lb] × 0.45359237 ⎢ ⎥ ⎟ × ⎜ 9.80665 ⎢ 2 ⎥ ⎟ ⎝ ⎣s⎦ ⎣ ft ⎦ ⎠ ⎝ ⎣ lb ⎦ ⎠ ⎝ ⎣ s ⎦⎠

⎛ ⎡ ft ⎤ ⎡ m ⎤⎞ × ⎜1 ⎢ ⎥ × 0.3048 ⎢ ⎥⎟ ⎝ ⎣s⎦ ⎣ ft ⎦⎠ Note that the value of g0 was substituted only at the end of the calculations. This ensures better accuracy because of the absence of round-off errors. Cancelling British units and collecting all numbers and units it is seen that, 1 [hp] = 550 × 9.80665 × 0.3048 × 0.45359237

[N ⋅ m] = 745.7 [W]. [s]

EXAMPLE 1.8 One metric horsepower (also called ‘Pferde Starke’ or ‘Cheval-vapeur’) (denoted as [Ps]) is defined as 75 [kgf . m/s]. How many [kW] is this? Solution

The calculation is carried out directly in one step.

⎛ ⎡ kgf ⋅ m ⎤ ⎞ ⎛ ⎡ kgf ⋅ m ⎤ ⎞ ⎛ ⎡ N ⎤⎞ 1[Ps] = ⎜ 75 ⎢ ⎟ = ⎜ 75 ⎢ ⎟ × ⎜ 9.80665 ⎢ ⎥ ⎟ = 735.5 [W] ⎥ ⎥ ⎝ ⎣ s ⎦⎠ ⎝ ⎣ s ⎦⎠ ⎝ ⎣ kgf ⎦ ⎠

1.5

SOME SYSTEMS

As seen from Table 1.1, thermal engineers who constructed and operated heat engines have contributed significantly to the development of thermodynamics. Moreover, thermal engineering forms an important (and, for us, the main) application of thermodynamics. Hence, this section presents descriptions of some typical thermal engineering systems and processes for which thermodynamic analysis is most commonly used. These descriptions are also aimed at introducing some commonly used terms for early acquaintance with them.

1.5.1

Internal Combustion (I.C.) Engine

Figure 1.1 is a schematic representation of an I.C. engine as used in automobiles. It is assumed to be a petrol13 engine and that all the processes are ideal. It operates as follows:

13

In general, a S.I. (spark ignition) engine.

Chapter 1: What is Thermodynamics?

9

Spark plug Inlet valve

Exhaust valve TDC

Cylinder head Gudgeon pin

Piston

Cylinder BDC

Axis

Connecting rod

Crank Crank shaft Figure 1.1 Schematic diagram of an I.C. engine.

1. At the beginning, let the piston be at the position denoted as TDC (Top Dead Centre). The inlet valve and the exhaust valve are closed. 2. The piston starts moving downwards. The inlet valve opens. A mixture of air and petrol is drawn (or sucked) into the cylinder because of the partial vacuum created. This suction stroke ends when the piston reaches the Bottom Dead Centre (BDC). At BDC the inlet valve closes. 3. Next, the piston starts moving upwards. Both the valves are closed so that the mixture is compressed. This compression stroke ends when the piston reaches TDC. 4. At TDC the mixture is ignited by an electrical spark across the electrodes of the spark plug. 5. Exhaust gases at high pressure and temperature produced by combustion of the mixture act on the piston and drive it downwards. This power stroke ends when the piston reaches BDC. At BDC the exhaust valve opens. 6. When the piston starts moving upwards again the product (exhaust) gases are expelled (exhausted) from the cylinder. This exhaust stroke ends at TDC when the exhaust valve closes. 7. The piston has now completed one cycle of operations and is ready to begin the next cycle of operations. The power produced in the cylinder is transmitted to the crank shaft through appropriate linkages to drive any load. Two distinguishing features of an I.C. engine14 are: (a) all the operations of the cycle take 14

In fact, the adjective ‘Internal Combustion’ denotes that the combustion of fuel takes place inside the cylinder itself.

10

Engineering Thermodynamics

place inside the cylinder–piston assembly, and (b) the basic motions of the operations are reciprocating, i.e. the I.C. engine is a reciprocating device. When such a cylinder–piston assembly is run backwards by driving it by some engine, it performs the task of compressing air or any other gas. Such a device is called a compressor. It may also be used to pump a liquid, in which case, it is called a pump. All these devices are called reciprocating engines or compressors or pumps because their basic movements are reciprocating i.e. up and down (or in and out; or to and fro). An I.C. engine produces power continuously. Such a system is called a cyclic device because it works in thermodynamic cycle. The work done is calculated from the first law. Then, its thermal efficiency is estimated from the second law.

1.5.2

Steam Turbine Power Plant

Figure 1.2 is schematic representation of a steam turbine power plant. This basic configuration is used for generating electricity for commercial and domestic applications. Hence it is also called electric utility plants. At the outset, note that unlike an I.C. engine, this plant contains more than one component. A steam power plant uses water as its working fluid. Water flows through all the components during which it evaporates and condenses, i.e. changes phase. The operation of this plant consists of the following processes, all of which are assumed to be ideal. 1. State 1 corresponds to the inlet to feed water pump. The working substance is in the form of saturated water at the condenser pressure. 2. The pump pressurizes the water to the high pressure existing in the boiler. This state is denoted as state 2, which is also the state at the inlet of the boiler. Q1, Heat absorbed 2

3 Boiler

WP, Pump work (input)

Pump

Turbine

Condenser

1

Wt, Turbine work (output)

4

Q2 , Heat rejected Figure 1.2

A steam turbine plant.

Dotted lines show the boundary of the equivalent closed system

Chapter 1: What is Thermodynamics?

11

3. In the boiler the feed water absorbs heat generated by combustion of fuel and, therefore, evaporates. This state is denoted by 3, which is also the inlet state of the turbine. 4. The high pressure steam enters the turbine where it expands to the condenser pressure. During this process energy is transferred from the steam to the turbine rotor as work, which is available to drive a load. The exit state of turbine is denoted as 4, which is also the inlet state of the condenser. 5. In the condenser the low pressure steam coming out of the turbine is cooled so that it becomes water. This state is denoted as 1, which is also the inlet state to the feed pump. 6. Now the working fluid has completed its cycle of states and is ready to begin the next cycle. The power produced during the expansion of steam in the turbine runs an alternator which produces electric power. From the above description, the following correspondence to the operations of an I.C. engine is evident. (a) The compression stroke takes place in the pump. (b) Combustion takes place in the furnace of the boiler. (c) The power stroke is carried out in the turbine. (d) The exhaust stroke is equivalent to the condensation of steam. Thus it is seen that in the steam plant, unlike an I.C. engine, each operation is carried out in one component; and that all the components taken together are equivalent to the operations in an I.C. engine. This is indicated by the dotted lines which denote the boundary of the equivalent cylinder–piston assembly, i.e. the closed system. A gas turbine plant is similar to the above except that its working fluid is a gas (typically air). Hence it does not change its phase. Moreover, with air as the working substance, the exhaust gases from the turbine are not recycled but are thrown into the atmosphere and fresh air is inducted by the air compressor or fan. It should be noted that, as in the case of I.C. engines, the turbine may be driven by an external power source so that it works as a compressor or fan, and that all these devices are rotary devices. Like an I.C. engine, a turbine plant also produces power continuously. Hence, it is a cyclic device because it works in a thermodynamic cycle. As before, the first law is used to calculate the work done. Then, the second law gives the thermal efficiency of the plant.

1.5.3

Heat Exchanger

This device is used to transfer heat from one fluid stream to another. Hence, it is one of the most important components of a thermal system. Boilers, combustion chambers (combustors), evaporators, condensers, etc. are some of the special types of heat exchangers. Figure 1.3 is the schematic diagram of the simplest type of heat exchanger. It has one tube placed concentrically inside another. As per this scheme, hot fluid flows inside the inner tube

12

Engineering Thermodynamics

and cold fluid flows in the annulus formed by the two tubes in a direction opposite to that of the hot fluid. Hence, this arrangement is called counterflow concentric tube heat exchanger. The hot fluid is cooled by the cold one. The overall heat exchanger is adiabatic. Hence, the energy given up by the hot fluid exactly equals the energy absorbed by the cold fluid. Cold fluid

Hot fluid

Cold fluid Figure 1.3

A concentric tube heat exchanger.

Like an I.C. engine or a steam turbine plant, the heat exchanger also works continuously. However, unlike an I.C. engine or a steam plant, the heat exchanger does not work in a thermodynamic cycle. In other words, the heat exchanger is not a cyclic device, but a continuous steady flow device. Moreover, unlike, I.C. engine or a turbine plant, the heat exchanger does not convert energy from heat to work, i.e. it is not an energy converter, but only an exchanger of heat. As usual, the first law is used to calculate the heat transferred, and the second law gives the effectiveness of the process of heat transfer.

1.5.4

Refrigerator

A refrigerator is the basic device for removing heat from a cold body. It is a very important device both commercially and industrially. Figure 1.4 shows the schematic diagram of a domestic refrigerator. As in the case of the steam turbine plant, this figure also shows the components in which each process is carried out. The same scheme also represents a heat pump. Atmosphere 3

2

condenser

Capillary

4

Q1, Heat rejected

Compressor

Evaporator

Wc , Work (input)

1

Dotted lines show the boundary of the equivalent closed system

Q2, Heat absorbed Cold storage

Figure 1.4

A domestic refrigerator.

Chapter 1: What is Thermodynamics?

13

Whether a refrigerating system represents a refrigerator or a heat pump depends on its specifications. Thermodynamically, they are the same. The working fluid of a refrigerating system is called a refrigerant. Domestic refrigerators use Freon which is a trade name. The processes (assumed to be ideal) in the operation of a refrigerator are: 1. Low pressure refrigerant in the saturated vapour state, denoted as state 1 enters the inlet valve of the compressor. 2. This vapour is compressed to a high pressure and reaches the state 2 at the exit of the compressor. 3. This refrigerant vapour at high pressure and temperature enters the condenser, where it is condensed to liquid at high pressure by rejecting heat to the surroundings (e.g. air, river or sea water, etc.). This state is denoted as state 3. 4. The high pressure liquid refrigerant is then throttled to a low pressure by passing it through a throttle valve, capillary, etc., during which it also cools and reaches the state denoted as 4. 5. The low pressure liquid refrigerant at low temperature (i.e. at state 4) is then passed through the cold storage where the product to be cooled is kept. During this process the refrigerant absorbs heat from the product and reaches state 1 (saturated vapour state) by evaporation. Like an I.C. engine, a refrigerating system is also a cyclic device and works in a thermodynamic cycle. As before, the first law gives the work required. Then, the second law is used to calculate its COP (Coefficient of Performance).

1.6

CONCLUSION

In this chapter, some thermal systems are described which can be analyzed by thermodynamics. Some basic concepts regarding accuracy of calculations and units used in thermodynamics are also stated. In the next chapter, the concepts and methods primitive to thermodynamics will be stated. In terms of these, the basic concepts of thermodynamics will also be developed.

REVIEW QUESTIONS 1. How did the word thermodynamics originate? 2. How did thermodynamics attempt to put the ancient concepts on scientific basis? 3. Name the difficulties that arose because so many disciplines contributed to the development of thermodynamics. 4. What is an operational definition? 5. Why is modelling needed in scientific study? Give examples. 6. What is simulation?

14

Engineering Thermodynamics

7. 8. 9. 10.

According to Poincaré, what is the purpose of mathematical calculation? How did Kelvin and Tait approach mathematics? What is the basic principle of accurate numerical calculations? How does one avoid loss of significant digits when two nearly equal numbers are subtracted? 11. What attributes does a number have when used to represent a physical quantity? 12. Explain, with neat sketches, working of the following thermal engineering systems. (a) I.C. engine (b) Steam turbine plant (c) Heat exchanger (d) Refrigerator.

Chapter 2

Concepts of Thermodynamics

2.1

INTRODUCTION

It was mentioned in the preceding chapter that the primitives of a discipline are concepts and methods that are either intuitive (undefined) or those borrowed from other disciplines. Thermodynamics was the last part of classical physics to be formalized. Hence, it does not contain any intuitive quantities. Its primitives are those borrowed from other disciplines (mainly physics and chemistry). In this chapter, these are used to define1 the basic terminology and concepts of thermodynamics.2 Appendix B contains further explanations on all these concepts (and methods). By necessity, these explanation have been kept brief. Standard textbooks on physics, chemistry, etc. may be consulted for further details.

2.2

UNIVERSALLY PRIMITIVE CONCEPTS

The analytical approach of Newtonian mechanics was the first to meet with complete success. Hence its methods (and concepts) were also adopted by other disciplines of science. Presently, all scientific analyses use the methodology of mechanics in one form or the other. Thus, they are universal, i.e. common to all disciplines. In this section, they are stated in the form appropriate to thermodynamics.

2.2.1

System Description

The mode of defining a system, its environment, interactions, etc. depends upon the way it is described. 1 2

In this book, from now on, unless specifically stated, the word ‘definition’ means operational definition (including those defined by mathematical equations and formulae). In fact, these are the primitives modified to suit the purposes of thermodynamics. 15

16

Engineering Thermodynamics

A system can be described from microscopic or macroscopic3 points of view. They are also called molecular and continuum descriptions, respectively. Following are their main characteristics: Microscopic or molecular description The microscopic description looks at the particles (atoms, molecules, ions, photons, …) of a system. Even the macroscopic ‘zero’ volume at laboratory conditions contains a large number of these particles. Hence, this approach requires a very large number of variables and equations (one set for each particle). The results are then averaged in order to obtain the macroscopic properties, which results in loss of information.4 This disadvantage is compensated by the fact that, with the techniques currently available for measurements and computations, only the small-sized samples are needed for testing. Following are the two approaches included under this category. Kinetic theories. The states are defined by the position and velocities of each particle. The interactions are momentum (force) and energy. The relations are integro-differential equations expressing conservations of mass, momentum and energy of each particle and the overall system. Both thermodynamics and transport properties [viscosity (m), thermal conductivity (k) and diffusion coefficient (D)] can be calculated using this approach. Behaviours of gases and liquids can be described using this approach. Statistical mechanics (thermodynamics). Here, the state is defined by energy. Interactions are also energy. The relations are algebraic equations describing particle distribution in different energy levels and averaging rules. This approach can predict only the thermodynamic properties. This approach can be used to describe behaviours of gases, liquids and solids. Macroscopic or continuum approach The number of equations required is less. The equations are generally algebraic. The state is defined by primitive properties. All the quantities are real (i.e. they are measured as real numbers). Consequently, all relations are real functions (perhaps, with finite number of discontinuities). All quantities can be measured directly, perhaps with less accuracy. Figure 2.1 shows the molecular and continuum regimes with respect to variation of density. It is seen that the density fluctuates as the size of the macroscopic volume decreases. This fluctuation is due to non-uniformity in the mass enclosed by the macroscopic volume. As the volume decreases and becomes infinitesimal, the density becomes constant. Note that there is some uncertainty in specifying the macroscopic ‘zero’ volume since there is a range of it for which density is constant.

3 4

From Greek words, skopeen = ‘to look at’, micros = ‘little things’, and macros = ‘long or great’. Appendix B contains an interesting explanation, originally proposed by Callen ([CAL]), on this.

Chapter 2: Concepts of Thermodynamics

17

r = dm/dV

dV Figure 2.1

Molecular and continuum regimes.

As the volume decreases still further, molecular effects are felt and the density fluctuates because of random entry and exit of molecules into the volume. In the limit, when the volume equals that of one molecule the density fluctuates randomly between zero (corresponding to no molecule in the volume) and infinity (corresponding to one molecule in the volume). Note that, the microscopic zero volume is the volume of one molecule. When the mean free path of molecular motion becomes of the order of the size of a system,5 the microscopic effects become significant.

2.2.2

System, Environment and Interactions

This book deals mainly with macroscopic description. Hence all definitions presented below pertain to this approach. The universe. It is the totality of all systems which interact among themselves by exchanging mass and energy. Thermodynamic system. That part of the universe which is isolated (at least in imagination6) and idealized for the purpose of study. Figure. 2.2 shows how a system is represented. A system may contain mass (our usual systems) or it may contain only energy (e.g. electromagnetic field in vacuum). A system containing mass will automatically contain energy, although the reference state may be chosen to make this zero.

5 6

That is, the Knudsen number, defined as Kn = l/L (where, l is the mean free path and L is the characteristic dimension of the vessel) becomes unity or more. So that an observer can study it ‘impartially’, ‘objectively’ etc.

18

Engineering Thermodynamics

Figure 2.2

Representing a system.

Boundaries. These are the surfaces that isolate a system from its universe and thereby operationally define it. They are specified either verbally or by figures (generally, in both ways). Free-body diagrams of mechanics are examples of pictorial representation. A boundary may coincide with real surfaces (e.g. walls, piston head of a cylinder–piston assembly) or may be imaginary (e.g. inlet section of a turbine). It may be fixed (e.g. cylinder walls) or it may be moving (e.g. piston head). The importance of the boundaries will be understood from the fact that mass or energy (or both) become interactions only when they cross the system boundaries. Environment. The immediate part of universe with which a system directly interacts. There is no special name for the rest (i.e. the ‘non-environment’ part) of the universe. A system interacts with the rest of the universe only through its environment.7 Interactions. Interactions are flow of mass and energy between a system and its environment across the system boundaries. It should be emphasised that only the mass and energy that cross the boundaries of a system constitute interactions and not those that flow within the system. In other words, interactions occur only when a system and its environment are in contact.

2.2.3

Classification of Systems

Depending upon the interactions, a system is classified as follows: Isolated system. There are no interactions with the environment. If all the interactions between a system and its environment can be controlled then it is equivalent to isolating it since the effects of these interactions can always be subtracted. Closed system or control mass. There are no mass interactions. Hence the mass of the system remains constant. Only energy interactions are allowed. The Lagrangian equations in fluid mechanics are derived using this approach. 7

Hence the environment may be considered to be another system. Moreover, the combination of a system with its environment forms a bigger system.

Chapter 2: Concepts of Thermodynamics

19

Open system or control volume. Interactions in the forms of flow of mass (bulk flow and/ or diffusion) as well as of energy flow are allowed. Flow systems are those in which bulk flow takes place. The Eulerian equation in fluid mechanics are obtained using this approach. The above definitions mean that experiments will have to be performed on the system boundaries to determine whether mass and/or energy flows across them. An equivalent method of definition of system is based on measurements performed inside the system. Then, the classification is: Isolated system. Closed system.

An isolated system is one in which both energy and mass are constant. A closed system is one in which mass is constant but energy can vary.

Open system. An open system is one in which both mass and energy can vary. The mass inside a system can vary because of (a) chemical reactions which produce (or consume) some species; (b) mass diffusion of chemical species; and, (c) flow of chemical species (also called mass convection). In thermal engineering open systems generally mean flow systems. However, in this book, these types are distinguished by specific mention. It should be noted that the above classes are mutually exclusive, i.e. a system can either be isolated or closed or open. In other words, an isolated systems can neither be closed nor open; or, a closed system can neither be isolated nor open, an open system can neither be isolated nor closed. All the concepts, laws and other results of thermodynamics are developed with respect to closed systems. However, all systems in real life are open. Hence, to make thermodynamics useful, the results of closed systems should be extended to open systems. Otherwise it will remain just another theory, however good it may be. In this book, unless stated otherwise the term system will denote a closed system. The standard symbol for it will be the cylinder–piston assembly shown in Figure 2.2(a). Relation between closed and open systems The relation between these two types of representations is now determined. Let subscripts i and e denote the quantities at the inlet and exit, respectively of a flow system [see Figure 2.2(b)]. This system is also called a control volume. The subscript CV indicates quantities corresponding to it. Since the classification of systems into closed and open is based on the mass flow, it seems reasonable to use the principle of conservation of mass8 to establish a relation between them. For this purpose this principle is written as: The rate of increase of mass inside a control volume equals the net (i.e. the algebraic sum of) mass flowing into it.

8

Only Newtonian dynamics is considered here for which this law is stated as mass can neither be created nor be destroyed, i.e. the total mass of the system is constant. Of course, in chemical reactions, chemical species can be created and used up. This law is also a primitive of thermodynamics.

20

Engineering Thermodynamics

The mathematical form of this principle, called the continuity equation, is written as m CV = m i − m e

Now, consider the time interval dt between t and t + dt. Multiplying both sides of the continuity equation by dt gives9 dmCV = dmi – dme Using the definition of differential this becomes mCV, t + dt – mCV, t = dmi – dme

or

mCV, t + dt + dme = mCV, t + dmi

This shows that, if the system is considered as (a) the control volume plus the entering mass at time t, and, (b) the control volume plus the leaving mass at time t + dt, then an open system is transformed into an equivalent closed system. Since the limit of dt Æ 0 is finally considered, this is a valid procedure.10

2.2.4

State and Properties

A process is one of the basic units11 of thermodynamic analysis. It is discussed in a later subsection. But a process is defined as a sequence of change of states. Hence, the concept of state becomes fundamental. This subsection deals with the definition of the state. State The quantity that determines the (inside) condition of a system is called a state. It is defined by fixed values of a set of independent primitive properties (e.g. 1 [kg], 1 [bar] and 1 [m3]). Two questions arise immediately. The first is: How many independent properties are needed to define a state uniquely? The second is: What are these independent properties? For most of the commonly used substances, the Gibbs phase rule (presented in Subsection 2.2.6) provides the answers. However, in a general case, the answers are obtained using the state principle stated as follows: State principle. The stable state of a system which is bounded by a fixed surface and subjected to prescribed fields at its boundaries is fully determined by its energy. This is a modification by Hatsopoulos and Keenan ([HAT], p. xxvii) who attribute this to Kline and Köenig. The original form12 is a modification of the Gibbs phase rule. However, this form seems to be a logical extension from quantum mechanics where a state is defined by the wave

9 10 11 12

Note that dmi and dme mean infinitesimal mass flows while dmCV means the difference of the mass of control volume between times t + dt and t. Our apologies for this messy nomenclature. It should be pointed out that this equation is also valid for mass diffusion if the net mass inflow, viz. mi – me, is expressed in appropriate relations (e.g. Fick’s law). Others are the systems that execute processes and the working substances the systems use. Kline S.J. and F.O. Köenig, Journal of Applied Mechanics, Vol. 24, p. 29, (1957).

Chapter 2: Concepts of Thermodynamics

21

function y, which, in turn is determined by total energy E and potential energy V through Schrödinger’s wave equation. In Chapter 5, using the state principle, the answers to the above two questions about state variables are presented. The following definitions are a direct result of the above definition of the state. Change of state. state.

It is change in the value of at least one property used in the definition of the

Equilibrium. A state is an equilibrium state if it remains unchanged with time and the interactions are zero (i.e. change of state and interactions are both zero). Steady state. The state remains unchanged with time but the interactions are non-zero (i.e. change of state is zero but interactions are non-zero). Properties A property is any quantity whose value depends only on the state. Hence, it is also called point function, the adjective point referring to one in state space. It can (mathematically) operationally be defined as the quantity F which is a property if and only if 2

∫

1, A

2

dF =

∫ dF

or

v∫ dF = 0

1, B

Note that this definition becomes a tautology (same statement in another set of words) if F is used to define the state. This is the reason why only primitive properties are used in the definition of state. However, if we already know that a quantity is property, then it can be used to define the state for this test. Properties can be classified in the following two ways: Based on the defining characteristics. properties are known as:

Depending on the manner of definition, these

Primitive properties. These are not defined within thermodynamics. The properties of relevance are pressure (p, defined in mechanics), volume (V, defined in geometry) and mass (m, defined in mechanics) since they are known to determine the energy of the system. Chapter 3 contains brief operational definitions of these properties. However, technically, properties such as refractive index (defined in optics) etc. may also be used but these are not directly useful for determining the state since they do not determine the energy of a system directly. Thermodynamically defined properties. These are energy (U, defined by the first law), entropy (S, defined by the second law), and, T, the temperature.13 However, in Chapter 4, the concept of 13

As explained in Section 2.3.1 below, the temperature has been assumed to be a primitive property because (a) the concept of temperature was in existence long before the science of thermodynamics was formulated, and (b) temperature measurement uses non-thermal characteristics (e.g. expansion (mechanical), change in resistance (electrical), thermal emf (electrical), etc.).

22

Engineering Thermodynamics

temperature is developed rigorously, on the basis of the experimental results including the zeroth law of thermodynamics. Mathematically derived properties. These are mathematical combinations of the above properties. Density (r = m/V), enthalpy (H = U + pV), Gibbs function (G = H – TS) and Helmholtz function (A = U – TS) are the most common ones. In this book, they will be considered as convenient short-hand symbols. Based on the size of the system.

These properties are known as:

Extensive properties. These properties depend upon the mass (size, extent) of the system and are operationally defined as follows: When the system is divided into two halves, the values of all extensive properties will be halved. The upper-case letters are used to denote these properties. The most common extensive properties are V, U, S, G and A. Intensive properties. These properties do not depend upon the mass of the system. The operational definition of these properties is similar to that of extensive properties. The two intensive properties of our interest are p and T. Specific properties. These are extensive properties expressed per unit mass of the system. Lower-case letters corresponding to those of the extensive properties denote them u, e, h, g and a. Although their values are independent of the mass of the system, they are not true intensive properties in the sense of the above definition. The adjective specific generally denotes mass basis while molar indicates mole basis. For example, specific volume will be volume per unit mass and molar volume will be the volume per unit mole. They are related through the molecular weight (RMM, the relative molecular mass). It is important to note that these two classifications of properties are not mutually exclusive. For example, a thermodynamically defined extensive property is the complete specification of energy E. Similarly, p is a primitive intensive property. However, within each classification, the categories are mutually exclusive (i.e. a property can be either intensive or extensive but not both). State space A geometric space constructed with the independent properties as axes is known as the state space. Consequently, a state will be represented by a point in state space. It is a real space since all the properties are real variables. Due to historical reasons, the practice of using only the primitive properties for state space is not followed. For example, long before it was known that only primitive properties could be used in the definition of state, thermal engineers have been using others, e.g. the Mollier (h–s) chart is most commonly used in steam turbine design. Other common state spaces used in classical thermodynamics are p–v, p–T, v–T, T–s, and p–h.

Chapter 2: Concepts of Thermodynamics

2.2.5

23

Thermodynamic Equilibrium

A system is in thermodynamic equilibrium when its pressure, temperature and composition are same throughout its size (generally, volume). This operational definition requires almost an infinite number of measurements (i.e. at all points in the system). Consequently, it is convenient to define thermodynamic equilibrium in the following equivalent form, comprising three criteria. Mechanical equilibrium. (i.e. pS = pE). Thermal equilibrium. (i.e. TS = TE).

The pressure of the system is equal to that of the environment

The temperature of the system is equal to that of the environment

Chemical equilibrium. The concentration of each species in the system is equal to that in the environment. The absence of chemical reactions and mass diffusion ensure this. This means that mi,S = mi,E, where mi is the chemical potential of the ith species. This relation is proved in Chapter 8. After the discussion on non-equilibrium presented below, it will be evident that the above two definitions of thermodynamic equilibrium are equivalent. The only difference between equilibrium and steady state is that in the former the interactions are also zero. Consequently, to distinguish the two, measurements should be made inside the system (to determine change of state) and on the boundaries (to determine the fluxes) simultaneously. This is where the second definition is superior since it uniquely tests for equilibrium. In other words, pS = pE uniquely means that no energy flow (as work) exists; TS = TE uniquely means that no energy flow (as heat) exists; mi,S = mi,E, uniquely means that no mass flow exists. They also imply that pS, TS and species concentration remain uniform and constant. To understand the effects of non-equilibrium, consider a closed system (i.e. a cylinder– piston assembly) containing an ideal gas at pressure pS, 0. Let the environment pressure be pE,0. Then at mechanical equilibrium pS,0 = pE, 0. Now, suppose pE is suddenly decreased to a new value (say, pE,1). Then the gas in the system will undergo a sudden expansion. The pressure in the bulk of the gas will still be pS, 0. However, the gas at the piston will have attained the value pS,1, that will be equal to pE,1. Hence any value of pressure between pS,1 to pS, 0 can be found at some point within the system. Therefore, a unique value of pressure to the system cannot be assigned. The same argument in terms of temperature or concentration is also valid. Hence, it can said that: In the case of non-equilibrium, pressure, temperature and composition are not defined for a system as a whole. It has already been mentioned that only a few variables are needed for macroscopic description of systems. The advantage of thermodynamic equilibrium is that simple functions define their behaviour. This advantage is so great that it is retained in the case of nonequilibrium, as well through one of the following stratagems. 1. A thermodynamic system in non-equilibrium wherein pressure, temperature, composition, etc. vary with space coordinates is divided into small volumes so that

24

Engineering Thermodynamics

they are uniform within these. When these small volumes become infinitesimal, differential equations are obtained in terms of species concentrations, pressure, temperature, etc. 2. In a thermodynamic system whose condition of non-equilibrium arises out of finite rate of the processes, fast and slow processes are assumed to have reached equilibria; the former reaching the next state and the latter remaining in the current state.

2.2.6

The Gibbs Phase Rule

This rule was derived by Gibbs and determines the number of independent properties (called the degrees of freedom) when different species exist in equilibrium in different phases. Here it is derived to answer the question regarding the number of independent properties needed to determine a thermodynamic state uniquely. Mathematics asserts that this number be equal to the difference between the number of variables and the number of independent equations among them. This difference is called the degrees of freedom (of choice of independent variables) of the set of mathematical equations. Consider a system consisting of C number of distinct chemical species present in P phases (all species are assumed to be present in all phases) in thermodynamic equilibrium.14 Then: 1. The number of variables for each phase is C + 1 (i.e. C – 1 mole fractions, p and T). 2. Therefore, the total number of variables is P(C + 1). 3. The number of equations is counted as follows. Let the subscripts of alphabets denote the phases and let the numerical subscripts denote species. Then, Mechanical equilibrium implies, pa = pb = … = pP Thermal equilibrium implies, Ta = Tb = … = TP Chemical equilibrium implies, ma,i = mi,b = … = mi,P (i = 1, 2, … , C). 4. There are P – 1 equalities each of pressure, temperature and chemical potential of each species. Thus, there are C + 2 such equalities. 5. Therefore, the total number of equations is (C + 2)(P – 1). 6. Thus, the number of variables which can be independently chosen is the difference between the number of variables and the number of independent equations connecting them, which equals P(C + 1) – (C + 2)(P – 1) = C + 2 – P. 7. This difference gives the degrees of freedom and is denoted by F. 8. Thus, the Gibbs phase rule is F = C + 2 – P. For example, for a pure substance (C = 1) in single phase (P = 1), the number of variables which can be independently chosen is, F = 2.

2.2.7

Processes

A process is defined as a sequence of change of states. It characterizes the interaction between a system and its environment. It is basic unit of the behaviour of a thermodynamic system. 14

Suitable numbers of species reservoirs connecting the system through semi-permeable membranes (an arrangement called van’t Hoff box) are assumed.

Chapter 2: Concepts of Thermodynamics

25

Flow processes are executed by open systems and non-flow processes are executed by closed systems. Only non-flow processes are considered now. Quasi-static processes A system can change its state only by interacting with its environment. However, the definition of thermodynamic equilibrium given above means that such a system cannot change its state. Thus, thermodynamic equilibrium specifies only static states. Consequently, they do not have any practical use. However, their greatest advantage is that they can be specified by very few variables. Hence, in order to retain the advantage of the condition of thermodynamic equilibrium (namely, simple relations and very few variables), quasi-static processes are cooked-up. As the name implies, a quasi-static process is one which is almost15 static. Thus, it is operationally defined as one during which the system will always be in thermodynamic equilibrium. Another way of stating this is: A quasi-static process is succession of equilibrium states. To understand these ideas, consider the process of reducing the system pressure from p1 to p2 quasi-statically. Since the system is in equilibrium at the initial state, its pressure will be equal to that of the environment (i.e. pS = pE). Thus, the experiment will be done as follows: Step 1. Reduce pE infinitesimally from p1 to p1 – dp. Step 2. Wait till the system attains the new pressure (i.e. till pS = p1 – dp). This requires infinite time. Step 3. Reduce pE to p1 – 2dp. Step 4. Wait till pS = p1 – 2dp. Step 5. Wait till pS = p2. Note that there are an infinite number of steps, each of which requires infinite time for reaching equilibrium. Thus, quasi-static processes cannot be executed in practice. They are similar to the slow motion sequences in motion pictures.16 Heating a system quasi-statically from T1 to T2 is also similar to the above situation and requires an infinite number of reservoirs. Since all the states of a quasi-static process are well and uniquely defined, it is represented by a single continuous curve in the state space (e.g. in Figure 2.3, 1 is a quasi-static process). To understand a non-quasi-static process, consider the sudden expansion of an ideal gas contained in a closed system. Let the initial pressure of the gas be pS, 0. Let the environment pressure be pE, 0. Since the initial state is one at mechanical equilibrium, pS, 0 = pE, 0. For the sudden expansion to take place, the pressure pE should be suddenly decreased to a new value (say, pE,1). Then, the gas will undergo unrestricted expansion with the pressure of the gas remaining undetermined, though lying between the bulk pressure pS, 0 and the new environment 15 16

Concise Oxford Dictionary defines quasi as: seemingly, not real, half, almost. The author is indebted to Dr. C.N. Chablani, former Research Scholar, Mech. Engg. Dept. for this suggestion.

26

Engineering Thermodynamics

pressure pE,1. This process will continue till the system pressure becomes equal to pE,1. Thus, in a non-quasi-static (non-equilibrium) process the intermediate states are not defined. Hence, in state space, it is represented by non-unique broken lines (e.g. curves 2 and 3 of Figure 2.3 represent the same non-quasi-static process). These conclusions follow directly from the definitions of state, state space and quasi-static process. In this book, unless specified otherwise, all processes are assumed to be quasi-static.

Figure 2.3

Quasi-static and non-quasi-static processes.

Common processes The most common processes (all assumed to be quasi-static) are: Isochoric (constant volume) where the system volume is constant. Isobaric (constant pressure) where the system pressure is constant. Isothermal (constant temperature) where the system temperature is constant. Adiabatic where no heat is exchanged. It should be noted that since the processes are quasi-static, when constant values are specified for them, these remain the same throughout the system. For example, during a constant pressure (isobaric) process, the pressure will remain constant at the specified value. Moreover, since the process is quasi-static, the system will be in mechanical equilibrium. Similarly, in an isothermal process the system will always be in thermal equilibrium. Cycle A cycle is a sequence of non-flow processes such that the end state of the last process is identical with the beginning state of the first one.

2.3 CONCEPTS FROM PROPERTIES OF MATTER The definition of a thermodynamic system and its classification shows that the characteristics

Chapter 2: Concepts of Thermodynamics

27

of the working substance are not taken into account. For example, though each of the following systems is a closed system, each has a different working substance. I.C. engine.

In the ideal case, the same mass of gas executes a cycle.

Tensile/Compression test. A specimen is subjected to stress till it is fractured. The system contains a solid which undergoes an expansion/compression. Furnace heat transfer. Heat transfer is mainly by thermal radiation. Since the furnace volume is constant, it is taken as a closed system with radiation field acting as the working substance. These examples show that the properties of the working substance do not form part of thermodynamics. They are borrowed from the discipline called Properties of Matter. At macroscopic level, solid mechanics, fluid mechanics, electricity, magnetism, optics etc. deal with properties appropriate to themselves as part of their theories. At microscopic level, kinetic theory of gases and liquids, solid state physics, etc. are used. In fact, these are kinetic theories since they deal with dynamical aspects of behaviour. Hence, they are not part of thermodynamics. Only the following aspects of the properties of matter are directly useful in thermodynamics. 1. Thermal expansion and other effects of heat (e.g. thermal resistance and thermoelectricity), which are used in measuring temperature. 2. The fact that not all properties of a substance can be changed independently which give rise to equations of state. 3. When a substance is heated, that energy is stored which gives rise to the concept of specific heats. 4. Some substances (called fuels) burn, thereby releasing heat. In this section, only the first three aspects are discussed. Traditionally, the last aspect has been part of chemistry under the name thermochemistry. In this book, this will be discussed in Chapter 10 as an application of thermodynamics. The Appendix A contains further details.17

2.3.1

Empirical Temperature

A look at the dates of discoveries in thermometry, heat and thermodynamics presented in Table 1.1 shows that thermometry, uses non-thermal effects, i.e. mechanical (mainly thermal expansion), electrical (resistance) or thermo-electrical (thermal emf). The most common of these are expansion thermometers which make use of thermal expansion liquids (and gases). Hence, thermometry predated heat and thermodynamics. This also has contributed to the mistaken belief that temperature is a primitive property. 17

Standard textbooks on physics and chemistry (as appropriate) should be consulted for a more elaborate explanation.

28

Engineering Thermodynamics

Consequently, these temperatures and their scales will be called empirical temperatures and empirical scales of temperature. Saha and Srivastava ([SAH], Ch. I) give an excellent account of the modern temperature measuring techniques. Here, only the basic ideas will be discussed. Ideal gas temperature scale This is the most important empirical scale. An ideal gas is defined to be one which obeys the Boyle’s law and the Joule’s law. For the present purposes, these may be stated as follow: Boyle’s law.

At constant temperature, the volume of a gas varies inversely with pressure.

Joule’s law.

The internal energy of an ideal gas is a function of temperature only.18

The main consequence of the Boyle’s law (proposed in 1662) is the definition of the ideal gas temperature and ideal gas temperature scale. In this section, this aspect will be discussed. In Chapter 7 on the second law, the Joule’s law is used to show that the ideal gas temperature is identical to the Kelvin of temperature (and, evidently these scales are also identical). At low pressures, all gases are found to behave ideally. Then, temperatures (and their scales) defined in terms of the properties of ideal gas will be independent of the behaviour of the gas used as the thermometric substance. Hence, it is appropriate to call these as the ideal gas temperatures (and, scales). Mathematically, the Boyle’s law is written as: at constant temperature, p μ (1/v) where v is the specific volume. This can be rewritten as pv = C, where the constant C = f(q) with f(q) being some function of the empirical temperature q. Since temperature is empirical,19 it is adequate to choose the simplest form for the function f(q), namely f(q) = Cq where C is a constant. Thus, the Boyle’s law may be written as pv = Cq, or q = (pv)/C. This is the definition of the empirical ideal gas temperature q. The scale of temperature constructed by this method is also empirical. Hence, it is adequate to choose: 1. The simplest substance for measuring temperature (called the thermometric substance). 2. The simplest property that varies with temperature (called the thermometric property). 3. The simplest function for defining the scale. Boyle’s experiments (reported in 1662) showed20 that all these criteria can be met by choosing an ideal gas as the thermometric substance and choosing its volume (at constant pressure) or its pressure (at constant volume) as the thermometric property. Then, the Boyle’s law stated above shows that the scale would be linear.

18 19 20

Based on the Joule’s experiment on the free expansion of gases. This will be shown in Chapter 7. Concise Oxford Dictionary defines, empirical as “based or acting on observation or experiment, not theory; deriving knowledge from experience alone”. A look at Table 1.1 on the important dates of discovery in thermometry and heat shows that Galileo (in 1592) had used the principle of liquid expansion in his barothermoscope.

Chapter 2: Concepts of Thermodynamics

29

Thus, the general form of the scale becomes q = a + bX, where the thermometric property X is either the pressure p or the specific volume v and, a and b are constants to be determined experimentally. The constant a (intercept of the line) determines the zero of the scale and the constant b (slope of the line) fixes its interval i.e. number of degrees. To evaluate these constants, two conditions are needed. These are called the primary21 fixed-points. If X1 and X2 are the values of the thermometric property at the fixed-points q1 and q2, then q1 = a + bX1 and q2 = a + bX2. Simultaneous solution of these equations gives b = (q2 – q1)/(X2 – X1);

a = (X2q1 – X1q2)/(X2 – X1)

Prior to 1954, the primary fixed-points were the ice-point—the temperature at which ice and water in equilibrium at 1 [atm]—and the steam point—the temperature of equilibrium between water and steam (its vapour) at 1 [atm]. In the Celsius scale the temperature of the ice-point and the steam-point are fixed as 0 [°C] and 100 [°C] respectively. This means that the zero of the scale is the ice-point and there are 100 ‘degrees’ between the ice-point and the steam-point. In the Fahrenheit scale the temperatures of ice and steam points are 32 [°F] and 212 [°F], respectively, which gives the ice-point as 32 [°F] and 180 degrees between the ice and steam points. This results in the well-known temperature conversion rule, namely, q [°F] = (1.8) × q [°C] + 32.0. These definitions show that the Celsius and Fahrenheit scales are ideal gas temperature scale, i.e. scale in which temperatures are defined in terms of the properties of ideal gas. In this book, ideal gas temperature will mean temperature in degree Celsius and ideal gas temperature scale will mean Celsius scale of temperature. Conversions to Fahrenheit scale will be made (if needed) using the above rule. This is the method used in S.I. units as well. Once the temperature scale is defined, other properties can also be used to measure temperature even when the relations are nonlinear. Some typical examples are vapour pressure of a liquid, electrical resistance and thermal emf. It has always been difficult to obtain, experimentally, the values of ice and steam points because of presence of air and other gases in steam alter their values. Hence, after 1954 only one fixed-point was used—corresponding to the triple point of H2O substance taken as 273.16 [K]—since it is constant.22 The second fixed-point was not needed since the interval of the Celsius scale was retained. In 1787, Charles found that at constant pressure, specific volumes of hydrogen, air, carbon dioxide and oxygen increase linearly with temperature according to the formula v = v0 (1 + bq) where q is the temperature (in [°C]) and v is the specific volume at that temperature. Then, it is evident that v0 should be the specific volume at ice-point, i.e. 0 [°C]. Gay-Lussac (in 1802) found that all gases obey this rule. Hence, this law is now known as law of Charles and Gay-Lussac. Ideal gases obey it exactly while real gases approximate it. It should be noted that b is the coefficient of volume expansion and has the dimension of inverse of temperature.

21 22

Secondary and auxiliary fixed-points have been agreed upon for industrial purposes. As per the Gibbs phase rule, this point has zero degrees of freedom (C = 1 and P = 3).

30

Engineering Thermodynamics

Experiments on gases showed that b = 1/273.15. Therefore, the law of Charles and GayLussac can be rewritten as v q q + 273.15 T =1+ = = v0 273.15 273.15 T0 where T, defined as T = q + 273.15, is called the absolute Celsius temperature. In Section 7.9, it is shown that this temperature is identical to the thermodynamic (i.e. Kelvin) temperature. In anticipation of that result, the unit of T is denoted as [K]. Since v0 is the volume at ice-point, it follows that 0 [°C] = 273.15 [K]. Now, volume v should always be positive, which means that its minimum value is zero. The above relation then also shows that T should be positive and T = 0 corresponds to q = – 273.15 [°C]. This temperature is called the absolute zero. Since the Celsius and Kelvin scales differ only in their zeroes, the latter can as well be used in the Boyle’s law. Then, the Boyle’s law23 can be written as pv = RT, where the proportionality constant is now written as R. This constant is called the gas constant. This equation is recognized to be the well-known ideal gas equation. Rewriting this equation as T = pv/R shows that the ideal gas equation is really the definition of the ideal gas temperature T.

2.3.2

Other Properties of Substances

The empirical scale of temperature discussed above makes use of the fact that a gas expands either when its pressure is reduced or when it is heated. These are the general characteristics of all substances. During the discussions on the property relations it will be shown that these are thermodynamic requirements. These are also part of the general relations between pressure, volume and temperature of a gas. In this subsection, this aspect will be briefly discussed. Substances and their phases Based on its chemical properties, a working substance can be classified as an element if it consists entirely of atoms of one type, or as a compound if it is chemically formed from one or more elements combined in a definite proportion by moles, or as a mixture if (i) its constituents can be separated by suitable physical or mechanical means, (ii) its constituents can occur in all proportions, (iii) its heat of formation is zero, and (iv) the overall properties are aggregate (e.g. weighted average) properties of the constituents. A mixture is homogeneous if its composition is same throughout. Otherwise, it is called heterogeneous. A homogeneous molecular mixture of two or more substances of dissimilar molecular structures (generally, solids in liquid) is called solution. Although a solution has all the properties of a mixture, its solubility (which determines the proportion of combining) has restricted range. The term phase means solid, liquid or gaseous (vapour) states. A substance (even as component of a mixture) may exist in any one or more of the phases. 23

Note that this includes the law of Charles and Gay-Lussac.

Chapter 2: Concepts of Thermodynamics

31

Another convenient way to classify a working substance is as pure substance, i.e. one chemical constituent (an element or a compound) in one or more (solid, liquid or gas) phases or as a mixture of two or more pure substances in one or more phases occurring as heterogeneous or homogeneous mixture. In the following subsections, specifications of a working substance are described qualitatively using (state) diagrams. Quantitative specifications of a working substance are described in terms of the p–v–T data, as well as specific heat and other data. These are described in later subsections. The p–v–T diagram All substances (especially those of direct interest here) are pure substances and may exist in solid, liquid or gas phases. Figure 2.4 shows the states of a typical substance that expands during melting.24

Figure 2.4

Pressure–volume–temperature diagram.

For a pure substance (C = 1) in a single phase (i.e. P = 1), the Gibbs phase rule (F = C + 2 – P) shows that the degrees of freedom F = 2. This means that two intensive properties can be chosen independently. All others can be expressed in terms of these two. Hence, it is convenient to consider the processes in two dimensions (i.e. planes). Three possible combinations of these processes are p–T, T–v and p–v. In the following paragraphs, these curves (shown in Figure 2.5 to Figure 2.7) are explained.

24

Water is the typical example of a substance which contracts on melting. For water, the solid + liquid phase will be folded back in the figure.

32

Engineering Thermodynamics

p–T diagram Figure 2.5 shows the pressure–temperature diagram (the p–T curve) of a pure substance. This is obtained by projecting the p–v–T surface on the p–T plane. The legend mentions its salient features.

Figure 2.5

Pressure–temperature diagram.

The horizontal lines represent isobars since p is constant along them. For each isobar, T increases from left to right. Thus, horizontal lines from left to right are the loci of states corresponding to isobaric heating25; the line 1–2–3–4 being a typical one. Similar arguments show that vertical lines from top to bottom are loci of states corresponding to isothermal expansion and line 5–6–7 is a typical one. Consider an experiment of isobaric heating the substance, that is initially at some low temperature. This state is shown as 1 in Figure 2.5. The substance exists in the subcooled solid phase. Since it is a pure substance in a single phase, the number of components, C = 1 and the number of phases, P = 1. Then the Gibbs phase rule shows that number of degrees of freedom, F = 1 + 2 – 1 = 2, which means that p and T can vary independently. Since the process is isobaric heating, only the temperature increases. Thus the point representing its state moves to the right along a horizontal line towards the point 2. The heat absorbed is called the sensible heat of solid. When the temperature of the substance reaches its fusion point (represented by the point 2) it begins to melt so that the substance exists in two phases, namely the solid and its liquid are in phase equilibrium. Then, the Gibbs phase rule shows that F = 1 + 2 – 2 = 1 which means that only p or T can vary independently. Since the process of heating is isobaric, p remains constant. Hence, the temperature also remains constant till all the solid has melted and the state point remains at 2. The heat absorbed during melting is called the latent heat (more correctly, enthalpy) of fusion. 25

Evidently, the opposite direction, i.e. right to left corresponds to isobaric cooling.

Chapter 2: Concepts of Thermodynamics

33

Once the substance has become completely liquid, the number of phases becomes P = 1. Then the Gibbs phase rule shows that the number of degrees of freedom, F = 2 so that during the isobaric heating the temperature of the substance begins to rise again and the state point begins to move towards the point 3. These states are called the subcooled liquid states. The heat absorbed during this part of the heating process is called the sensible heat of the liquid. When the liquid temperature reaches the boiling point, it begins to evaporate so that the liquid co-exists in phase equilibrium with its vapour. Now the number of phases has become, P = 2 so that the number of degrees of freedom, F = 1. Once again, since the process is isobaric, the temperature remains constant at the boiling point (represented by point 3) till all the liquid has evaporated. The heat absorbed during evaporation is called latent heat (more correctly, enthalpy) of evaporation. When all the liquid has evaporated, once again the number of phases reduces to one so that the number of degrees of freedom becomes, F = 2. Thus, during the isobaric heating the temperature of the vapour begins to rise again and the state point begins to move towards the point 4. These states are called superheated vapour states. By repeating the experiment for different pressures, variation of the fusion temperature with pressure is obtained. The locus of the fusion temperature is called fusion curve and it is represented as BD in Figure 2.5. Note that this curve has a positive slope. This is the shape for all liquid which shrinks on solidifying. Water is an important exception since the specific volume of ice is more than that of water,26 for which the fusion curve will have a negative slope. These experiments also generate data on the variation of boiling point with pressure. This locus is represented as BC in Figure 2.5. This is called the vaporization curve. At low pressures, the solid does not melt but vaporizes. This phenomenon is called sublimation. The curve AB, called sublimation curve, represents the locus of sublimation temperatures as pressure varies. The Clausius–Clapeyron equation describes these (sublimation, fusion and vaporization) curves. These represent state of equilibrium of two phases (called the phase equilibrium) i.e. the sublimation curve represents the solid–vapour phase equilibrium, the fusion curve is the locus of the equilibrium states between the solid–liquid phases, and the vaporization curve consists of liquid–vapour equilibrium states. The triple point is the state when the three phases (solid, liquid and vapour) are in equilibrium. Thus, it is the meeting point of all the curves. It is labelled B in Figure 2.5. At the triple point the number of phases, P = 3 so that by the Gibbs phase rule the number of degrees of freedom is F = 1 + 2 – 3 = 0. Thus, both p and T remains constant.27 It should be noted that at the triple point only the two intensive properties, namely p and T are constant. However, extensive properties, such as volume, energy, entropy, etc. change during the phase change. Hence, the triple point is a point only in the p–T plane. It will be a line in the p–v and T–v planes. It should also be noted that no liquid phase exists28 below the triple-point pressure. Above it, a solid can reach vapour phase only through the liquid phase. 26 27 28

This is why the icebergs float and the ship TITANIC sank and the movie became a hit. This is why the triple point is chosen as the primary fixed-point in thermometry. However, above the triple point a substance can exist in liquid phase even when the temperature is below the triple point. This can be verified by drawing a vertical line through the triple point.

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Engineering Thermodynamics

As the experiments are conducted at higher and higher pressures, the differences between the states of solid and liquid phases become less and less.29 The state at which the liquid phase vanishes is called the critical point (or state). It is labelled C in Figure 2.5. The adjective critical is used to indicate quantities at the critical state. For example, the pressure, temperature and volume at triple point are called critical pressure, critical temperature and critical volume, respectively. Similarly, the isobar and isotherm passing through the critical point are called critical isobar and critical isotherm, respectively. T–v and p–v diagrams Figure 2.6 is the projection of the p–v–T surface on the T–v plane. As before, the legends explain the salient features. The main point to be noted is that since the abscissa (the x-axis) is v, an extensive property, it does not remain constant during phase change. Consequently, the triple point becomes a line and the two phase regions (solid–liquid, solid–vapour and liquid– vapour) are explicitly seen.

C: BG : S: S+L: L: L+V: V: S +V : XX : 1–2–3–4–5–6 : 7–8–9–10 : AB : BD : FE : FC : CG :

Figure 2.6

Critical point Triple point Solid phase Solid phase + liquid phase Liquid phase Liquid phase + vapour phase Vapour phase Solid phase + vapour phase Critical isobar Typical isobar Sublimation isobar Sublimation curve Fusion curve Fusion curve Sat. liquid curve Sat. vapour curve

Temperature–volume diagram.

This diagram is also explained considering an isobaric heating used above. A typical curve of isobaric heating is 1–2–3–4–5–6. The point 1 represents the initial state of solid phase of the substance. On heating its temperature increases and, therefore, the state point moves along the line 1–2 towards the point 2. This process continues till the melting point 2 is reached. Then the temperature remains constant since the solid is melting. As more heat is added, more and more of the solid melts. Thus the state point moves along the line 2–3 towards the point 3. The state 3 corresponds to the completion of the melting process.

29

This will be evident from the T–v and p –v diagrams described in Figures 2.6. and 2.7, respectively.

Chapter 2: Concepts of Thermodynamics

35

Further heating increases the temperature of the liquid till its boiling point 4 is reached. Then the temperature remains constant till all the liquid evaporates and point 5 is reached. Still further heating increases the temperature of the vapour indefinitely. All the processes explained in connection with the p–T diagram can be seen in Figure 2.6 as well. In thermal engineering, only the regions of the liquid, liquid + vapour and vapour are of interest. Hence only these curves are usually drawn. The projection of the p–v–T surface on the p–v plane is shown in Figure 2.7. This diagram is similar to the T–v diagram except that the isotherms run rightwards and downwards. Hence, it will not be elaborated further.

XX : 1–2–3–4–5–6 : 7–8–9–10 : S: S+L: L: L+V: V: S +V : C: BG : AB : BD : FE : FC : CG :

Figure 2.7

Critical isotherm Typical isotherm Sublimation isotherm Solid phase Solid phase + liquid phase Liquid phase Liquid phase + vapour phase Vapour phase Solid phase + vapour phase Critical point Triple point Sublimation curve Fusion curve Fusion curve Sat. liquid curve Sat. vapour curve

Pressure–volume diagram.

Equations of state Mathematically, the p–v–T data is represented in general by the implicit relation f(p, v, T) = 0 or by the explicit relations of the type, v = v(p, T), p = p(v, T), etc. These are also known as equations of state. For liquid and vapours the form of the function f (.) is complex. Hence, the data on liquids and vapours are presented in the form of tables or graphs. The tables of properties of steam and refrigerant R–12 presented in the Appendix E are examples of these.30 All these will be called equations of state in whatever form (equations, tables or charts) they are. In the following subsections, the most common equations of state for a gas are described. Ideal gas equation This is the best known of the equations of state of a gas. It is written as pv = RT 30

or

V p ⎛⎜ ⎞⎟ = RT ⎝ m⎠

or

pV = mRT

The Mollier (h–s) diagram for steam, p–h charts for refrigerants are diagrammatic representations of thermodynamic properties of substances. These are not used in this book.

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Engineering Thermodynamics

where m is the mass of the gas. In this equation, pressure p is measured in31 [Pa], volume v is measured in [m3], mass m is measured in [kg], and temperature T is measured in [K]. Consequently, the unit of R will be [J/kg·K]. This is called the specific gas constant. An alternative form is pv = RT

or

⎛V⎞ p ⎜ ⎟ = RT ⎝ n⎠

or

pV = nRT

where n is the number of moles (mole number) of the gas. In this equation, pressure p is measured in [Pa], volume v is measured in [m3], mole number n is measured in [mol], and temperature T is measured in [K]. Consequently, the unit of R will be [J/mol·K]. This is called the molar gas constant and it is denoted as R. Now, the Avogadro’s law states that at the same pressure and temperature equal volumes of all gases contain equal number of molecules. This number is defined as the Avogadro’s Number. Moreover, this amount of substance (containing the Avogadro number of molecules) is defined as the mole. Its value is 6.022045 × 1023 and its unit is, obviously, [1/mol]. Then, the above equation shows that R will be same for all gases. Hence, it is called the universal gas constant. It should be remembered that mass and mole are different. A mole is the unit of quantity of matter (particles, i.e. atoms, molecules, ions, electrons, etc.) equal in number to the Avogadro number. The mass of a mole of substance is its molecular (formula, etc.) weight (also called the Relative Molecular Mass, abbreviated RMM), M. Note that in engineering, the unit of the molecular weight M is [kg/kmol]. Hence, the specific and universal gas constants are related by the equation, R = MR. In this book, for simplicity, the ideal gas equation is generally written as pv = RT. Whether R is the specific gas constant or universal gas constant depends on the specification of the specific volume v or molar volume since it is the only term that contains mass (or mole) units. Thus, if v is the molar volume then it will be the universal gas constant and if v is the specific volume then R will be the specific gas constant. van der Waal equation This equation is given by a⎞ ⎛ ⎜⎝ p + 2 ⎟⎠ ( v − b) = RT v

In the ideal gas, the molecules are assumed to be point masses which interact only on collision. Thus, the molecular volume and the molecular attraction are neglected. The van der Waal equation considers both. The constant a accounts for the molecular attraction and the constant b is the correction for the finite molecular volume. These constants are determined experimentally. The importance of the van der Waal equation lies in the fact that it was the first equation which represented the phenomenon of phase change. These details are discussed in Chapter 9. 31

Remember that S.I. units are used in this book.

Chapter 2: Concepts of Thermodynamics

37

Other data In addition to the p–v–T data [in the form of tables, charts or equations (of state)], some other properties of matter are also inducted into thermodynamics. These are explained below. Specific heat data. The specific heat at constant volume, cv and the specific heat at constant pressure cp were originally defined as cv = (dQ/dT)v and cp = (dQ/dT)p. They are currently defined as cv = (∂u/dT)v and cp = (∂h/dT)p and are used in this book. The equivalence of these two definitions is shown in Chapter 8. Experiments show that variation of specific heats with temperature can be expressed as a polynomial in temperature, i.e. cp (or, cv) = a + bT + cT2 + …, or, more generally, cp +n

(or, cv) =

∑a T j

j

.

j =– n

Coefficient of volume expansion. This is the volume analogue of the coefficient of linear expansion. It is defined as b = (1/v)(∂v/∂T)p. Rewriting the equation of state as v = v(p,T) gives an expression for calculating the variation of b with p or v or T. For example, the equation of state of an ideal gas can be written as v = RT/p, so that R 1 ⎛ 1⎞ ⎛ ∂v ⎞ = b =⎜ ⎟⎜ ⎟ = ⎝ v ⎠ ⎝ ∂ T ⎠ p pv T

This shows that for an ideal gas, b depends only on temperature.32 Temperature measurement by expansion thermometers uses this property. Isothermal bulk modulus. This is also the volume analogue of the Young’s modulus and defined as BT = – v(∂p/∂v)T. The equation of state is used to express its variation with p or v or T. For example, the equation of state of an ideal gas can be written as p = RT/v, so that

RT ⎛ ∂ p⎞ BT = − v ⎜ ⎟ = =p ⎝ ∂v ⎠T v This shows that the isothermal bulk modulus is only function of pressure. Hence, at high pressures it almost remains constant. The inverse of the isothermal bulk modulus is called the isothermal compressibility and denoted by k (the Greek letter ‘kappa’). The actual values of the constants and the form of these relations can only be obtained from measurements (either macroscopic or microscopic) made on the substance. However, thermodynamics imposes restrictions on the relations between the properties. For example: 1. Thermodynamics cannot predict what the values of b or BT would be, but it shows that b and BT should be positive.

32

This is not surprising since, by definition, the ideal gas (also thermodynamic) temperature scale is linear in v.

38

Engineering Thermodynamics

2. The nature of the function f(.) in equation of state and the values of the constants a, b, …, in specific heat equation can only be determined experimentally. However, once these two are known, thermodynamics shows how to calculate the variation of specific heats with p, V, etc. 3. Thermodynamics also demands that, for all substances 2

⎛ ∂v ⎞ ⎛ ∂ p⎞ c p − cv = − T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂ v ⎠ T from which it is easy to show cp > cv. These (and many other) results are presented in Chapter 8. For liquid and vapours (and, of course for solids), the forms of equations of state as well as the equations for the properties are complex. Hence, they are presented in the form of tables or graphs. Appendix F presents these for water substance and the refrigerant R–12. Explanations and examples on the use of these tables are presented therein. These should be read now, since they will be used from the next chapter onwards. Here is an illustrative example on the use of steam tables. Even though the properties such as energy u, enthalpy h and entropy s are not defined so far, at this stage they are considered simply as some properties which are tabulated.33 Property Tables of R–12 are used similarly. EXAMPLE A closed system contains 1 [kg] of steam at 5 [bar] and 0.3 [m3]. Calculate changes in h, s, v, T and x when it executes a constant volume process till the pressure reduces to 2 [bar]. Solution The section ‘Steam Tables’ in Appendix E contains these tables together with examples on their usage. It is assumed that these are read. Then, the problem is solved in the following steps. 1. Since the state is specified in terms of p and v, pressure based saturation tables are used to determine the initial state. 2. This table shows that at 5 [bar], vf = 0.0010928 [m3/kg] and, vg = 0.37468 [m3/kg]. 3. Now, as per the given data v1 = 0.3 [m3/kg] which means that vf < v1 < vg. Hence the steam is wet. 4. Then the temperature can be directly read off from the tables as T1 = Tsat = 151.9 [°C]. 5. Since the steam is wet, the dryness fraction, x, of the steam should be determined. The specified volume is used for this purpose. The formula given in Appendix E shows that v = vf + (x) · (vg – vf), which can be solved for x = (v – vf)/(vg – vf). Substituting the above values gives x1 = 0.8. 6. Similarly, h and s are calculated from the formulae h = hf + x · (hg – hf), and s = sf + (x) · (sg – sf). 33

Just like you used log tables in school and learnt its theory only in college.

Chapter 2: Concepts of Thermodynamics

39

7. Reading the values from the table and substituting in these formulae gives h1 = 640.22 + (0.8) . (2747.5 – 640.12) = 2326.04 [kJ/kg], and s1 = 1.8604 + (0.8) · (6.8192 – 1.8604) = 5.8274 [kJ/kg . K]. 8. The process is given to be isochoric (i.e. constant volume). This means that v2 = v1 and by given data v1 = 0.3 [m3/kg]. 9. Now, at p = 2 [bar], vf = 0.001068 [m3/kg] and vg = 0.88544 [m3/kg]. 10. Once again, since vf < v2 < vg the steam is wet in this state as well. 11. Using the formula in step 5, x2 = (0.3 – 0.001068)/(0.88544 – 0.001068) = 0.33802 ª 0.34. 12. The tables then give T2 = 120.2 [°C]. 13. Substituting the appropriate values hf, hg, sf and sg, read off from the tables in the formulae in steps 6 and 7, gives h2 = 504.70 + (0.34) · (2706.3 – 504.70) = 1253.2 [kJ/kg], and s2 = 1.5301 + (0.34) · (7.1268 – 1.5301) = 3.4330 [kJ/kg·K]. 14. Hence, Dh = h2 – h1 = – 1073.8 [kJ/kg], Ds = s2 – s1 = – 2.3944 [kJ/kg·K], and DT = T2 – T1 = – 31.61 [K].

2.4 CLASSICAL THERMODYNAMICS Classical thermodynamics uses the macroscopic description of a system in thermodynamic equilibrium at all times. Consequently, all the processes are quasi-static. This book essentially deals with classical thermodynamics. Exceptions are indicated as appropriate. In classical thermodynamics the basic equation of the behaviour of a system may be written symbolically as Change of state = f (interactions) It should be noted that time does not enter into this equation since all processes in classical thermodynamics require infinite time. In other words, they deal with only changes and not rates of changes. Consequently, there have been some proposals to call this as thermostatics and the non-equilibrium (or irreversible) thermodynamics is then called the thermodynamics.34 This terminology is not used in this book since time becomes a natural coordinate only in the equations of mechanics so that the state space of mechanical systems requires time to be explicitly (as an axis) or implicitly accounted for. However, in thermodynamics, the motion (i.e. the dynamics) takes place in thermodynamic state space. Hence the terminology is appropriate. Because of its general nature, thermodynamics is applicable in almost all branches of science. This wide range of coverage requires some codification so that it can be applied with confidence. This is attempted in the following subsections.

2.4.1

Models in Thermodynamics

The models used in thermodynamics are of the following types: 34

Tribus, M., Thermostatics and Thermodynamics, D. Van Nostrand Co. Inc., 1961, p. 383

40

Engineering Thermodynamics

System models.

Isolated, closed and open systems.

Process models.

Quasi-static, reversible and irreversible processes.

Models of working substance. Pure, simple, compressible and incompressible substances. Simple compressible substances are further classified as ideal gas, van der Waal gas, real gas, etc. In this book, these subclasses will be used as types of specifications for classifying the problem.

2.4.2

Problems in Thermodynamics

Due to its wide applicability, a large number of problems (examples and exercises) can be generated in thermodynamics. This gives the impression that thermodynamics is a complex subject. One way out of this difficulty is to classify the problems, so that same problem appearing in different guises can be recognized. Problems in thermodynamics may be classified as follows. Fundamental problems.

These are:

1. Obtain relation(s) between the energy stored and the energy flows (this is based on the energy equation). 2. Determine what processes are possible (the second law does this). Application problems. These are: Energy conversion and heat pumping problems. In these, the thermodynamic relations between the different energy flows are obtained. Energy conversion, refrigeration and heat pumps are the most common ones under this category. Heat transfer problems. These are similar to the above but the basic interest is in the relation between energy transferred as heat. Property evaluation. Using the thermodynamic laws and principles, an attempt is made to obtain relations between the properties of substances. It should be again emphasized that thermodynamics does not state anything about the details of these relations but only puts some restrictions on their forms. Equilibria. Conditions of equilibria are derived using the principles and laws of thermodynamics.

2.5

CONCLUSION

Now that all the terminologies of thermodynamics have been developed, the attention is turned

Chapter 2: Concepts of Thermodynamics

41

to development of the concepts of thermodynamics. It should again be emphasized that thermodynamics only tries to show how the quantities called (thermal) internal energy (U), temperature and heat are measured, i.e. it tries to define these quantities operationally. It also shows that some types of processes are not possible. This law (called the second law) is the real law of thermodynamics.

REVIEW QUESTIONS 1. What is a microscopic description of a system? Give examples. 2. What is a macroscopic description of a system? Show by diagram the continuum region of macroscopic approach to system description. 3. What is the importance of boundaries of a system? 4. How does a system interact with its environment? 5. How are systems classified? What is the importance of each type? 6. State the state principle. What is a state space? Name the common state spaces used in classical thermodynamics. 7. What is a property? What are the methods of classifying properties? Give examples. 8. How is a process of a thermodynamic system defined? 9. What is a quasi-static process? What are the common quasi-static processes? Explain how to execute a quasi-static expansion from pressure p1 to pressure p2. 10. What is a cycle? 11. Define thermodynamic equilibrium. How is it different from steady state? 12. State and prove the Gibbs’ phase rule. How is it used? 13. How is a system in non-equilibrium state analyzed? 14. What is classical thermodynamics? 15. How are working substances classified? 16. What is the difference between a compound and a mixture? 17. What is an equation of state? Give the most common ones. 18. Define B, b, cp and cv. 19. Show how the ideal gas temperature scale is developed. State its connection with the thermodynamic (Kelvin) scale. 20. What are primary fixed-points? What are their values in Fahrenheit and Celsius scales? 21. Why were the two fixed points of temperature used prior to 1954? Why was a single fixed-point found adequate after 1954? 22. Why does Tribus propose that classical thermodynamics be called thermostatics? Then, what, according to him, is called thermodynamics? 23. What are the types of models used in thermodynamics?

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Engineering Thermodynamics

24. What are the types of problems encountered in thermodynamics? 25. For the following processes, determine changes in h, s, v, T and x (if appropriate). Assume p1 = 5 [bar]. (a) constant volume, v1 = 0.3 [m3/kg], p2 = 3.0 [bar]; (b) constant entropy, s1 = 6.3 [kJ/kg.K], p2 = 1.5 [bar]; (c) constant volume, h1 = 2500 [kJ/kg], p2 = 2.0 [bar]; (d) constant enthalpy, s1 = 6.4 [kJ/kg.K], p2 = 2.0 [bar]; (e) constant temperature, T1 = 200 [°C], p2 = 2.0 [bar]; (f) constant pressure, x1 = 0.5, T2 = 400 [°C].

Chapter 3

Work and Mechanical Energy

3.1

INTRODUCTION

In Chapter 2, the universal primitives (i.e. those common to all disciplines of science) were stated in terms suitable for thermodynamics. A brief review of the properties of working substance (borrowed from ‘Properties of Matter’) was also presented. It was also mentioned that thermodynamics deals with storage and flow of energy. Work is one of the two forms of energy flow. Similarly, kinetic and potential energies are the two forms of energy stored. These terms are defined in mechanics. Elastic, electrical, magnetic, …, are different forms of potential energy. Hence, all these are called mechanical energies. Since thermodynamics deals with energy storage and energy interaction (energy flow), it is logical to use the above concepts as the basis for developing thermodynamics. Consequently, in this chapter these concepts are discussed in some detail. The objective is to understand the basic assumptions involved in them. Hence, only the simplest case (i.e. the linear motion of a Newtonian particle in one dimension) will be considered. Essentially, this is a review of elementary mechanics presented in the first course of college physics. Appendix B contains further elaborations.

3.2 BASIC CONCEPTS OF MECHANICS In this section, the basic concepts of mechanics are defined operationally. These also serve as illustrations of the procedure of operational definition.

3.2.1

Concepts from Geometry

Mechanics uses geometrical concepts like position, length, etc. which are defined in this subsection. Only Euclidean (plane) geometry is considered. The concepts of a point, a (straight) line, a coordinate system, its origin, etc. are not defined in geometry. Hence, their intuitive meanings are assumed. Length is also not defined 43

44

Engineering Thermodynamics

in geometry. Hence, as per the exemplar (model, pattern) method, it is defined1 as (1/299792458) of the distance travelled by light in vacuum in 1 [s]. Notice that this definition requires time (i.e. second [s]) which is defined in the next subsection. Once the length is defined, the area and volume can be defined by geometric formulae. For example, geometry defines the area of a circle of diameter d as pd2/4. Then, the operational definition of the area of a circle involves the following steps: (a) draw a circle, (b) measure its diameter,2 and (c) calculate (a paper-and-pencil operation) the area as pd2/4. Similarly, geometry defines the areas and volumes of different shapes. The main idea of this discussion is that, a mathematical formula for a quantity is its operational definition. The verbal statements of the operational definitions are only explanations of the formula. This is the general nature of all formulae. Hence, from now onwards, mathematical equation defining a quantity will be considered as its operational definition.

3.2.2

Concepts from Kinematics

Kinematics deals with the geometry of motion. Hence, it uses concepts like velocity, acceleration, etc. It does not use the concept of force and those derived from it. Time is also an undefined concept in mechanics.3 Hence, it (i.e. second [s], the S.I. unit of time) is defined as the duration of 91926331770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the Caesium-133 atom. Then, the concepts of velocity and acceleration are operationally defined as: V(t) = dx(t)/dt and a = dV(t)/dt; where dx is the infinitesimal distance4 covered in infinitesimal time dt.

3.2.3

Concepts from Dynamics

The concept of force and those related to it are defined in dynamics. Since the main aim is to develop these concepts, only Newtonian [classical (non-relativistic, non-quantum)] mechanics is dealt with. Mass is a concept developed basically by chemists—through weighing of chemicals for reactions. Thus the gravitational measure of mass (or, in short, gravitational mass) becomes its accepted intuitive meaning.5 Chemists also discovered the principle of conservation of mass (really, of chemical atoms) which states that mass (of atoms) can neither be created nor be destroyed. Thus, this law is also a primitive to mechanics. Newton’s first law may be stated as follows. In the absence of any external influence a body continues to be in the state of rest or of uniform motion along a straight line. The external influences are called forces. This law implies the existence of a property which ensures that, in the absence of forces, a body continues to be in its state (of rest or of uniform motion along a straight line). This 1 2 3 4 5

This is the definition of metre, the S.I. unit of length. Defined as the smallest length between opposite points on the circumference of the circle. In fact, Einstein arrived at the special theory of relativity by trying to define time operationally. Defined as the length from an arbitrary origin. For example, the mass of the international prototype of 1 [kg] kept in Paris.

Chapter 3: Work and Mechanical Energy

45

property is called (inertial) mass. It only states that mass exists but does not show how to measure it6. Hence, mechanics assumes that the inertial mass of a body is identical with its gravitational mass. In terms of the concept of mass, the other concepts of dynamics (dealing with forces) are defined as shown in Table. 3.1. Table 3.1

Concepts in dynamics

Name and symbol

Defined as

Density (r) Kinetic Energy (Ek) Momentum (P) Force (F) Pressure (p)

(dm/dV) (1/2)(mV2(t)) mV(t) μ (dP/dt) (second law of motion) (dF/dA)

Note that the operational definition of force given in Table 3.1 is the Newton’s second law of motion, i.e. the rate of change momentum of a body is proportional to the applied force.7 The concept of gravitational mass used above can now be shown to be equal to the inertial mass. Consider the process of weighing masses by spring balance8. A spring balance really shows the force acting on a mass due to the action of gravity. Assuming the spring to be linear, the expression for force F exerted by it, is given by F = kx, where k is called the spring constant and x is the displacement of the end of the spring (where the hook is attached). For this case, the Newton’s second law can be written as kx = mg, where m is the inertial mass of the body and g is the acceleration due to gravity (assumed to be constant). This equation can be rewritten as m = (k/g)x, which shows that m μ x thereby implying that mass m can be ordered according to the displacement x it produces on weighing. This is why this m is called gravitational mass. This shows that inertial mass is identical to gravitational mass since the proportionality constant is assumed to be unity. It is important to note the difference between quantity of a substance and its mass. The former is measured in mole,9 defined as the amount of substance in a system which contains as many elementary entities (particles, atoms, molecules, ions, electrons, photons, …) as there are atoms in 0.012 [kg] of carbon–12. This number, called the Avogadro number, is denoted as NA and equals 6.022045 × 1023. The mass of 1 [mol] of a substance is its molecular weight10 in [g]. It is convenient to view the Avogadro number as a factor which magnifies the quantity of a substance from microscopic to macroscopic level. For example, 2 molecules of hydrogen 6 7 8 9 10

Hence, this is a non-constructive definition. Unlike the first law of motion, this law not only defines the existence of the quantity called force, but also shows how to measure it. Hence this is a constructive definition. A pan balance compares the weight of a body with that of a standard body. Hence the mass indicated by the pan balance is also the gravitational mass. Also called the gram mole. Only for material particles. For example, 1 [mol] of photons at 1014 [Hz] has 39.90 [kJ] energy—which is its characterization. Molecular weight is also called relative molecular mass (RMM), formula weight, ….

46

Engineering Thermodynamics

react with 1 molecule of oxygen to produce 1 molecule of steam. Then 2NA molecules of hydrogen will react with NA molecules of oxygen to produce NA molecules of steam, i.e. 2 [mol] of hydrogen will react with 1 [mol] of oxygen to produce 1 [mol] of steam. Thus, the Avogadro number is simply a scale factor magnifies the quantity of the substances from microscopic to macroscopic level.11

3.3 WORK The concept of work is defined in mechanics (see Appendix A) and is used in all other disciplines of physics and chemistry, including thermodynamics. It is defined as follows: Definition of work When the point of application of a force F moves from point 1 to point 2, the work done by F is W = ∫12 F · ds. The following properties of work are the consequence of the mathematical properties of the integral and do not part of mechanics (i.e. properties of mass, force, distance, velocity, etc.). Well-definedness. The integral exists only when F is well-defined function12 so that the sum which represents the integral (in the sense of Riemann integration) exists. Thermodynamically, this means that the process if quasi-static. Path dependence. The value of the integral equals the area under the curve of F–s. Hence it depends on the nature of the function F(s), i.e. the path of the point of application of the force F in the F–s plane. Consequently, W is a path function. This is indicated by writing, d−W = Fds for an infinitesimal process (i.e. with a bar on d). Mathematically, this means that d−W is not an exact differential, i.e. W = ∫12 d−W π (W2 – W1). Additivity over processes. Since, ∫ ba y dx = ∫ ca y dx + ∫ bc y dx the work done in a series of processes equals the algebraic sum of the work done in the individual processes. Thus, work can be evaluated as the ∫12 F · ds only when the process is quasi-static. Unit of work. Being a product of force ([N]) and displacement ([m]), work is measured in joules [J], but the engineering unit [kJ] is used in this book. Sign convention for work. Sign conventions are so chosen such that the negative sign does not commonly occur. In thermal engineering, systems are required to produce work. Thus, the sign convention is: work done by a system is positive.

3.3.1

Different Modes of Work and Their Expressions

The definitions of work given above have been directly used in deriving expressions for work in other fields. These expressions are described below. In all cases except in the case of fields 11 12

The Avogadro number also magnifies mass in [amu] to [g]. Continuous function with a finite number of essential singularities or one with bounded fluctuations.

Chapter 3: Work and Mechanical Energy

47

(gravitational, electrical, magnetic, electromagnetic) the force and the point of its application can be directly identified. Hence, such expressions can be derived straightaway. Expansion work (Wx) This is the most common (and, therefore, the most important) mode of work in thermodynamics. Consider a gas contained in a closed system [i.e. a cylinder–piston assembly of Figure 3.1(a)]. Let p be the pressure it exerts on the piston whose area is A. The cylinder walls are rigid and stationary and only the piston moves. If the piston undergoes an infinitesimal displacement dx, then the displacement of the point of application of the force will also be dx. W

p 1

Piston Gas p

Process Cylinder

(a) Figure 3.1

2 (b)

V

Expansion work: (a) the system and (b) the process diagram.

Then the definition of work done gives, d−Wx = Fdx. Now the pressure is defined as p = F/A, so the expression for the work done becomes d−Wx = (pA)dx. This expression can also be written as d−Wx = p(Adx). But, by the definition of volume, we have dV = Adx, where dV is the infinitesimal change in volume. Thus the final expression for the work done becomes d−Wx = p · dV

(3.1)

Equation (3.1) shows that the work d−W done by the gas is positive when it expands, i.e. when dV is positive. Similarly when the gas is compressed, the work d−W done on the gas is negative, i.e. when dV is negative. It should be noted that expansion work is the area under the curve of the process in the p–V plane. Such a diagram [Figure 3.1(b)] is called the indicator diagram13 and is used for determining the work done (called the indicator work) by internal combustion engines and the work done on reciprocating compressors (which use cylinder–piston assembly).

13

This used to be measured using a drum which was screwed on to the engine cylinder-head. The cylinder pressure drove a small piston to which a pen was connected to record the change in pressure. The drum was rotated back and forth by the crank shaft and indicated the piston displacement. This was called indicator since it indicated the variation of pressure with crank angle (which is proportional to the volume displaced by the piston) in a cycle. With sophisticated computer-based DAS (Data Acquisition System), indicators have become obsolete.

48

Engineering Thermodynamics

Elastic work (Wel ) This mode of work is similar to that of expansion work. Imagine that the closed system of Figure 3.1(a) contains a solid which is being pulled by an external force14 (see Figure 3.2). This is equivalent to the expansion of the solid except that the force is in the opposite direction. Let F be the tension and dL the deformation. Then, using the previous definition, the work done becomes d−Wel = – F · dL

(3.2)

The negative sign is included since work must be done on the solid. The d−W is negative when the solid elongates because of the applied tension, i.e. when dL is positive. Similarly work must be done on the solid, when it is compressed. Here too, d−W is negative even though both F and dL are negative. F F 1 ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

Moving head of UTM ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ Solid ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

2 L

Fixed head of UTM (b)

(a) Figure 3.2

Elastic work: (a) the system and (b) the process diagram.

Surface tension work (Ws ) This mode of work is also similar to that of expansion work with one exception. A method to confine a liquid film is needed. Now, if the cylinder–piston assembly is cut by two planes perpendicular to the piston then the resulting shape is equivalent to a wire bent in the form of U with a cross-wire connecting the two open limbs (in the place of piston) and the liquid film spread in between them (see Figure 3.3).

14

Better still, remove (in imagination) the sides of the cylinder: fix one end of the solid (in the form of a rod) to the cylinder head and the other to the piston. Now keeping the cylinder stationary, pull the piston. This is the principle of working of the UTM (Universal Testing Machine).

Chapter 3: Work and Mechanical Energy

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

49

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

Figure 3.3 Surface tension work: (a) the system and (b) the process diagram.

Now, surface tension s is defined as the force per unit length of the liquid film. Then, if surface of the liquid is of length dL and it distorts by dx the work done will be d−Ws = – (s · dL) · dx = –s · (dL · dx) = –s · dA

(3.3)

where dA is the change in the surface area. The negative sign is included since work must be done on the film. The d−W is negative when the liquid expands i.e. when dA is positive. Similarly work must be done on the film, when it is contracts, i.e. when both s and dL are negative. Here too, d−W is negative. Stirrer (or Paddle or Shaft) work (Ws) This is also a case where the force and its point of application can be identified. Imagine that the closed system (i.e. a cylinder–piston assembly) (Figure 3.4) now contains a fluid. Let it be stirred with a torque t through an angle dq. Now, torque t exerted by a force F is defined as the product of the force and the radius r from the axis of rotation, at which it is acting (called the torque arm), i.e. t = F · r. But the definition of the angle gives ds = r · dq, where ds is the displacement along the curve. W

τ 1 Cylinder

Piston

Fluid

τ, dθ Stirrer

2

(a) (b) Figure 3.4 Stirrer work: (a) the system and (b) the process diagram.

θ

50

Engineering Thermodynamics

Then, applying the above definition, the work done becomes d−W = –F · ds = –F · (r · dq) = – (F · r) · dq = –t · dq

(3.4)

The negative sign is included since work must be done on the fluid, i.e. d−W must be negative when it is stirred. Note that neither t nor dq can be negative. So, work cannot be done by the fluid. Electrical work (We) In all the cases discussed above, it was possible to identify a force and the displacement of its point of application. In the case of fields (e.g. gravitational, electrical, and magnetic) this identification becomes difficult. Hence, alternative quantities are defined for evaluating work. This is now illustrated in connection with the electrostatic field. In this case, the Coulomb’s inverse square law gives the expression for the force. This law is written mathematically as F = (q1q2)/(4pe0r2), where the two charges are denoted as q1 and q2 and r denotes the separating distance between them. The constant of proportionality has been written as 1/(4pe0), where e0 is called the permittivity (of vacuum). Now, the work done for a displacement dr is

⎛ qq W = F ⋅ dr = ⎜⎜ 1 2 2 ⎝ 4pe 0 r

⎞ ⎟⎟ ⋅ ( dr ) ⎠

Integrating this equation, ⎛ q q ⎞ 2 dr ⎛ q1q2 ⎞ ⎛ 1 1 ⎞ F ⋅ dr = ⎜ 1 2 ⎟ =⎜ ⎟⎜ − ⎟ 2 1 ⎝ 4pe 0 ⎠ 1 r ⎝ 4pe 0 ⎠ ⎝ r1 r2 ⎠ The voltage (or potential difference) DV is defined as the work done per unit charge. Let q1 be the standard charge (called the test charge assumed to be vanishingly small) used for testing, then W=

∫

2

∫

⎛ W ⎞ ⎛ q ⎞⎛ 1 1 ⎞ DV = lim ⎜ ⎟ = ⎜ 2 ⎟ ⎜ − ⎟ q1 → 0 q ⎝ 1 ⎠ ⎝ 4pe 0 ⎠ ⎝ r1 r2 ⎠ so that the work done becomes We = DV q1 (finite charges) and d− We = DV dq (infinitesimal charges) The point to be noted is that, now work done by a charge moving in an electrical field is specified indirectly—through voltage difference. Since gravitational force governed by the Newton’s law of gravity is similar to electrostatic force, the above arguments are also applicable to the gravitational field. The mathematical form of the Newton’s law of gravitational attraction between two masses m1 and m2 separated by a distance r is F = (G) · (m1m2/r2), where G is the gravitational constant. Then, the work done is obtained as

W=

∫

2

1

F ⋅ ds = (Gm1m2 ) ⋅

∫

2

1

⎛1 1⎞ ⎛ Gm2 ⎞ dr = (Gm1m2 ) ⎜ − ⎟ = (m1 ) ⎜ ⎟ (r1 − r2 ) 2 r ⎝ r1 r2 ⎠ ⎝ r1r2 ⎠

Chapter 3: Work and Mechanical Energy

51

Comparing with the usual formula, F = mgz, for the potential energy of a body in gravitational field shows that the term in the second parenthesis is g, the acceleration due to gravity. These arguments also show that the gravitational potential difference may be written as

⎛W ⎞ ⎛1 1⎞ lim ⎜ ⎟ = (G ⋅ m2 ) ⎜ − ⎟ m1 → 0 m ⎝ 1⎠ ⎝ r1 r2 ⎠ Electrical battery. The definition of voltage difference can now be applied to all electrical systems. One of the most common (and hence, important) systems is the electrolytic cell (see Figure 3.5). (+)

(–)

q ∆V

V

1 Electrodes

Electrolyte

Battery 2 q

(a)

(b)

Figure 3.5 Electrical work: (a) the system and (b) the process diagram.

Reversible work modes All the work modes mentioned above, except the stirrer work, are reversible in the sense that the reversed process exists and is physically meaningful. For example, the reverse of an expansion is a compression and the reverse of charging a battery is discharging it. However, the stirrer work is basically irreversible in this sense, since a fluid at uniform pressure and temperature can be stirred thereby transferring work to it. But, it cannot run the strirrer as a turbine to extract work out of the fluid at uniform pressure and temperature (as we shall see in Chapter 7, this is the essence of the Kelvin–Planck form of the second law of thermodynamics). This mode is taken into account only because, historically, in all experiments involving heat transfer the fluid was stirred to keep its temperature and composition uniform and that it is a factor to be subtracted from the total work. Generalized expressions All the expressions for work derived so far can be generalized as (for infinitesimal processes) d−Wi = Xi dYi Wi = ∫12 Xi dYi

(for finite processes)

52

Engineering Thermodynamics

where Xi and Yi are called the generalized force and displacement, respectively for the ith mode. Table 3.2 gives the quantities Xi and Yi for different modes. Table 3.2

Generalized forces (Xi) and displacements (Yi) for different modes of work Mode of work Expansion (Wx)

Stirrer (Ws)

Elastic (Wel)

Electrical (We)

Surface (Ws)

p V

t q

F L

e q

s A

Generalized force (Xi) Displacement (Yi)

Note: Ws = shaft/paddle work; Xi are intensive/specific properties and Yi are extensive properties.

3.4 EVALUATION OF WORK The need for a systematic problem-solving procedure has already been emphasized in Chapter 1 and is illustrated in detail in Appendix B. In this section, some well-known formulae are derived.

3.4.1

Process Specification

During the definition of work, it was explained that the integral representing work done cannot be evaluated unless the process is specified. This aspect is discussed in this subsection. It should be noted that, sometimes, for simplicity, an expression for work will be obtained using specific volume v, instead of total volume, V. Then, the expressions give work done per unit mass since V = mv and total work done obtained by multiplying by the system mass m. Depending on the specification of the process [Xi(Yi), i.e. Xi as a function of Yi], the technique for evaluation of work may be divided into two categories. For simplicity, only examples on expansion work are considered. However, these ideas are applicable to other work modes as well. Direct specification In this case the process is specified in terms of the variables which enter into the expression of work [i.e. Xi(Yi)], so that the integral ∫12 Xi dYi can be evaluated directly. Some common examples are: Isobaric15 process.

Constant pressure implying Wx = p(V2 – V1).

Isochoric process.

Constant volume implying Wx = 0.

Polytropic process. C = p1V1 = p2V2. 15

pV n = C, implying Wx = (p1V1 – p2V2)/(n – 1) since the constant

In Greek: isos = equal, baros = weight, and chōrā = space; Chambers Twentieth Century Dictionary, New Edition, 1972, Allied Publishers, pp. 696 & 697.

Chapter 3: Work and Mechanical Energy

53

It may be noted that the expression for the polytropic expansion is a general one. It reduces to isobaric (constant pressure) process for n = 0 and to isochoric (constant volume) process for n = •. It is indeterminate for n = 1. This condition corresponds to either an isothermal process of an ideal gas or an adiabatic process of saturated steam. The former follows from the definition of an ideal gas and that of an isothermal process. This is illustrated in Example 3.2, below. The latter is an experimental approximation (cp ª cv). This procedure for evaluation of work is illustrated by the following examples. Even though different methods are employed for this purpose, the basic emphasis is still on problem solving from first principles. During the discussion on units in Chapter 1, it was shown how to write the unit of each quantity adjacent to it. In the initial stages of solving problems in thermodynamics (or, in any other area), it is advisable to write the units of each quantity adjacent to it, each time it occurs. This practice has the following two advantages: (a) It gives a feel for various quantities and relations between them. (b) It helps in avoiding the common errors (such as, for example, dividing instead of multiplying by conversion factors). However, the main disadvantage of this procedure is that it is cumbersome since it involves additional writing effort and space (and hence, extra time) during problem solving. Therefore, it is recommended that this procedure be strictly followed (a) during practice sessions problem solving, (b) during the examinations, unless specifically stated, for writing units for the intermediate and final results, and (c) during checking when units need to be written. Despite these suggestions, in many chapters in this book, in order to save space, units are not written adjacent to each quantity. EXAMPLE 3.1 A fluid expands according to the law pVn = C. What is the work done? Use subscripts 1 and 2 to denote the initial and final states. Solution The system is closed. The system and its process diagram are shown in Figure 3.6. In this example, a tabular form as shown below is used for calculations. p

W

1

Piston

Gas

n

Process (pV = C)

Cylinder

2 V

(a)

(b)

Figure 3.6 Fluid expansion: (a) the system and (b) the process diagram.

54

Engineering Thermodynamics

Step

Governing equation

Reason

1. 2. 3.

Wx = ∫12 p · dV p = C/Vn Wx = C · ∫12 dV/V n

Expression derived earlier. Direct process specification. Mathematical substitution.

4.

Wx = C ⎡⎣V − n +1 ⎤⎦ ( − n + 1) 1

Result of integration.

5.

− n +1 − V2− n +1 ⎤⎦ ( n − 1) Wx = C ⎡⎣V1

Substituting the limits.

6.

Wx = [p1V1 – p2V2]/(n – 1)

C = p1V1n = p2V2n

2

Indirect specification In this case, the variable used in the process specification is not Xi(Yi). Hence, this specification should be converted into a relation of the form Xi(Yi). Some common examples are: Isothermal process. The temperature,16 T, is constant. This should be converted into a specification in p and V. The equation of state of the working substance is used for this purpose. Adiabatic process. No heat is transferred, i.e. Q = 0. Then the first law of thermodynamics is used since it relates W and Q. Examples 3.2 and 3.5 illustrate these concepts. EXAMPLE 3.2 done?

An ideal gas expands isothermally from state 1 to state 2. What is the work

Solution Figure 3.7 shows the system and the process diagrams for this case. The tabular form is used for calculations in this example also. p

W

1

Piston

Gas

Process (pV = C)

Cylinder

2 V

(a)

(b)

Figure 3.7 Gas expansion: (a) the system and (b) the process diagram. 16

In Greek: isos = equal and therme = heat.

Chapter 3: Work and Mechanical Energy

Step

Governing equation

Reason

1.

Expression derived earlier

2. 3. 4.

Wx = ∫12 p · dV T = const. = T0 (say) pV = mRT pV = mRT0 = C (say)

Indirect process specification Ideal gas equation Mathematical substitution

5.

Wx = C · ∫12 dV/V

p = C/V

6.

Wx = C [ ln V ]1

Result of integration

7.

Wx = C[ln (V2/V1)]

Substituting the limits

2

55

The equation for an isothermal process of an ideal gas is pV = C. Applying this for the initial and final states gives C = p1V1 = p2V2. Now, the equation of state becomes pV = mRT. Comparing these two equations shows that C = mRT0, where T0 is the specified value. Thus, the expression for work done can be written in several alternate (and convenient) forms that can be written compactly as Wx = C [ln (V2/V1)] = C [ln (p1/p2)], where C = p1V1 = p2V2 = mRT0. The above procedure is convenient only for ideal gases or those with simple equations of state. In the general case, other properties which indirectly represent the equation of state should be used to express the dV in the integral of work. The most common ones are the coefficient of volumetric expansion17 defined (in Chapter 2) as b = (1/V)(∂ V/∂T) p, and isothermal compressibility18 defined as k = –(1/V)(∂V/∂p)T. Since the process is specified in terms of T, we write V = V(T,p) so that partial differentiation gives ⎛ ∂V ⎞ ⎛ ∂V ⎞ dV = ⎜ dT + ⎜ dp = bVdT − k Vdp ⎝ ∂ T ⎟⎠ p ⎝ ∂ p ⎟⎠ T

Then, the expression for work can be written as W = ∫12 pdV = ∫12 pV(bdT – kdp). The main advantage of these expressions is that for liquids and solids, the average values of volume may be used in computations since b and k do not vary significantly over large ranges of pressure and temperature. For example, when mercury is pressurized from 0 to 1000 [atm] at 0 [°C] it was found19 that b changed by 4% and k by 2%. Note that, for an ideal gas b =1/T and k = 1/p. Following two examples illustrate this procedure. EXAMPLE 3.3 Rework Example 3.2, i.e. find the work done when an ideal gas expands isothermally from state 1 to state 2. Solution Since this is a repetition of Example 3.2, the same system and process diagrams are applicable. Thus, only the steps of the solution are indicated here. As a variation, this example is solved in text form. 17 18 19

3D analogue of linear expansion. Inverse of isothermal bulk modulus (B), the 3D analogue of Young’s modulus. See [ZEM], pp. 246 and 248.

56

Engineering Thermodynamics

Step 1.

As Eq. (3.1) shows, the expression for expansion work is, Wx = ∫12 p · dV.

Step 2.

As shown above, dV = V(b dT – kdp).

Step 3. Since T is constant, the first term on the RHS in step 2 is zero. Therefore, dV = – Vkdp and hence, Wx = – ∫12 p · V · k · dp. Step 4. For an ideal gas the equation of state, V = (mRT/p), shows that k = (1/p). Therefore, Wx = – ∫12 p · V · (1/p) · dp = – ∫12 m · R · T · (dp/p), since, for an ideal gas, pV = mRT. Step 5. For an isothermal process the temperature remains constant, (say, at T0). Substituting this in the integral gives, Wx = (mRT0) ∫12 (dp/p). Step 6. Evaluating the integral gives W = mRT0 ln (p1/p2), which is the same result as that of Example 3.2. EXAMPLE 3.4 To understand the real advantage of the technique given above, consider the case of isothermal compression of mercury from 0 [bar] to 1000 [bar]. Solution Since this example is similar to Example 3.3 except that the working substance is a liquid, the same procedure can be used. In fact, the first three steps of both the examples are common. So only steps from 3 onwards are given below. Step 3. Since T is constant, the first term on the RHS in step 2 is zero. Therefore, dV = – Vkdp and hence, Wx = – ∫12 p · V · k · dp. Step 4. As discussed above for liquids, k is constant over large ranges of pressure and temperature. Note that this is an experimental fact and hence is an approximation. Thus, the integral reduces to Wx = –k · ∫12 p · V · dp. Step 5. The volume of mercury does not change significantly with pressure in this range, i.e. it is incompressible. This is also an experimental fact and hence is an approximation. Then, the integral becomes Wx = – k V + ∫12 pdp. 2

Step 6.

Evaluating the integral gives, Wx = –k V

⎡ p2 ⎤ 2 2 ⎢ ⎥ = –(k V /2)(p1 – p2 ). 2 ⎣ ⎦1

Step 7. The values from tables are: v = 14.7 [cc/mol], and k = 3.84 × 10–12 [cm2/dyne]. Also, 1 [atm] = 1.013 × 106 [dyne/cm2]. Substituting the values, gives W x = 2.876 [J/mol]. EXAMPLE 3.5 What will be the work done by an ideal gas in expanding adiabatically from state 1 to state 2?

Chapter 3: Work and Mechanical Energy

57

Solution Figure 3.8 shows the system and the process diagrams for this case. In this example the equation of this process will be found in terms of p and v so that the expressions developed in Example 3.1 can be used. For this, the first law is used since the process is specified in terms of heat transferred. Once again the text form is used for this purpose. p

W

1

Piston

Gas

N

Process (pV = C)

Cylinder

2 V

(a) Figure 3.8

(b) Adiabatic expansion: (a) the system and (b) the process diagram.

Step 1. For a closed system, the first law for an infinitesimal process becomes (see Chapter 5) d−Q = dE + d−W. Step 2. Since the process is adiabatic, i.e. Q = 0, this relation becomes 0 = dE + d−W. Step 3. Since only expansion work is specified, d−W = pdV = mpdv, where, by definition, v = V/m. Step 4. Now, E = U (since no other energy modes are specified); U = U(T) (ideal gas); and, dU = mcvdT (definition of cv). Step 5. Substituting these relations in the first law equation gives, 0 = m(cvdT + pdv); or, 0 = cvdT + pdv. Step 6. To express T in terms of p and v, the equation of state is used. For this, it is written as T = pv/R. Therefore, dT = (pdv + vdp)/R. Step 7. Substituting for dT in the equation in step 5 and collecting terms gives 0 = (cv + R) pdv + cvvdp; or, 0 = cppdv + cvvdp (since, by definition, for an ideal gas, cp = cv + R); or, 0 = g pdv + vdp (dividing by cv and using the definition g = cp/cv); or, 0 = g (dv/v) + (dp/p); or, g (dv/v) = – (dp/p). Step 8. This equation is integrated to give g ln (v) = ln (C) – ln (p), where the constant of integration is written as ln (C). On rearrangement, we get pvg = C. Step 9.

Now the result of Example 3.1 can be directly used to get Wx = (p1V1 – p2V2)/(g – 1)

58

Engineering Thermodynamics

An important mathematical property of an integral is the additivity over limits, i.e. ∫12 ydx = ∫13 ydx + ∫13 ydx. Thermodynamically this means that work is additive over processes, i.e. the net work done in a series of processes is the algebraic sum of the work done in each of the processes. This is illustrated by the following example. EXAMPLE 3.6 One kilogram of an ideal gas (M = 30 [kg/kmol] and g = 1.4) at 1 [bar] and 300 [K] is first compressed at constant pressure till the volume becomes one-tenth of the initial. It is then heated at constant volume till the pressure becomes 15 [bar]. Finally, it is expanded till the initial conditions are reached according to the law pV n = C; where C is a constant. Determine the net work done. Solution In this problem only one work mode (i.e. expansion work) exists, but there are three processes. Hence the net work done is the algebraic sum of the work done in each of the processes. The system and the processes are shown in Figure 3.9. W

p 3

Piston

pV

n

=C

Cylinder

Gas

1

2 (a)

(b)

Figure 3.9

V

(a) The system and (b) the processes.

The expression for the work done in each process is: constant pressure compression (1–2): W12 = p(V2 – V1) constant volume heating (2–3): W23 = 0, since V is constant and expansion is the only work mode. polytropic expansion (3–1): W31 = (p3V3 – p1V1)/(n – 1). Hence, the exponent n of the process should be determined before evaluating this expression. The following calculations are straightforward. 1. Since the gas behaves ideally, V1 = (mRT/p) = (1)(8.3143)(300)/(100)(30) = 0.83143 [m3]. 2. From the given data, V2 = V1/10 = 0.083143 [m3].

Chapter 3: Work and Mechanical Energy

3. 4. 5. 6. 7.

59

W12 = (100)(0.083143 – 0.83143) = – 74.829 [kJ]. W23 = 0. Since p3V3n = p1V1n, n = ln (p3/p1)/ln (V1/V3), or, n = 1.1761. W31 = [(1500)(0.083143) – (100)(0.83143)]/0.1761 = 236.07 [kJ]. Hence, the net work done W = 236.07 + 0 – 74.829 = 161.2 [kJ].

In mechanics, as a consequence of the principle of conservation of energy it can be shown (see Section 3.5) that work is an energy flow. As a consequence of this and the principle of energy conservation, it follows that the net work done by the different modes, simultaneously present, equals the algebraic sum of the work done in each of the modes. The following two examples illustrate this property. EXAMPLE 3.7 During an experiment of 1 [h] duration, 1 [kg] of air (assumed an ideal gas) initially at 1 [bar] and 300 [K], is stirred with a constant torque of 10 [Nm] at 100 [rpm]. Meanwhile, the gas expands at constant pressure till the volume is doubled. What is the net work done? Solution In this case there are two modes of work done, namely expansion work Wx and stirrer work, Ws. Hence the total work done is, W = Wx + Ws. This example can be solved in the following steps. Step 1. Since the pressure is constant and the volume is doubled, Wx = ∫12 pdV = p(V2 – V1) = pV1. Step 2.

Similarly, since torque and speed are constant, Ws = – ∫12 t dq = -t (q2 – q1) = –(t)(2p)(N)(1)(60).

Step 3.

To evaluate Wx, the initial volume should be determined first.

Step 4.

Since the gas is ideal, V1 = (mRT)/(p1) = [(1)(8.3143)(300)]/[(29)(100)] = 0.8601 [m3].

Step 5.

Substituting the values, Wx = (100)(0.8601) = 86.01.

Step 6.

Similarly, Ws = –[(10)(2p)(100)(60)]/1000 = –377 [kJ].

Step 7.

So the net work done is (–377 + 86) = –291 [kJ].

EXAMPLE 3.8 As another example, consider an electric battery being charged at 6 [V] and 0.1 [A] for 10 [h]. During this process, 1 [mol] of H2 gas was liberated at constant pressure of 1 [bar] and temperature of 300 [K]. Assuming that the gas behaves ideally, what is the net work done? Solution Part (a) of Figure 3.5 gives the system diagram in this case. The process diagram will be in e–q plane as shown in part (b) of Figure 3.5 (corresponding to We the electrical work done) and as in part (b) of Figure 3.1 (corresponding to Wx the expansion work done). The following steps are used to solve this example:

60

Engineering Thermodynamics

Step 1. By definition, Wx = ∫ pdV. Step 2. Since the pressure is constant, Wx = pDV; or, Wx = D(pV). Step 3. Since the gas is ideal, by its equation of state, the expression for Wx reduces to Wx = D(nRT). Step 4. For an isothermal process the above relation becomes, Wx = RTDn. Step 5. Substituting the values, Wx = (8.3143)(300)(0.001) = 2.4943 [kJ]. Step 6. The electrical work done during charging is given by We = – (6)(0.1)(10)(3600)/(1000) = –21.600 [kJ]. Step 7. Hence, the net work done is, W = Wx – We = – 19.10 [kJ].

3.5 MECHANICAL ENERGY Energy is a concept that is basically defined in mechanics but applied to all disciplines such as electromagnetism, elastic deformation of solids, waves (e.g. sound, ocean) etc. Therefore, all these types of energies will be called mechanical energies.20 Here, mechanical energy is reviewed briefly with the aim of demonstrating the procedure since it is also used in thermodynamics. For the present purpose it is adequate to consider one-dimensional (assumed as the x direction) non-relativistic motion of a particle of mass m at velocity V(t).

3.5.1

Kinetic Energy

Kinetic energy Ek of a body of mass m moving with an instantaneous velocity V(t) is defined as Ek = m · V2(t)/2. The following properties of this definition should be noted. 1. This is a constructive definition since it not only defines the existence of kinetic energy but also shows how it is measured. 2. The kinetic energy is only a function of the velocity V(t) at the time instant t and not how that velocity was reached, i.e. it is not a path function. In other words, in the terminology of thermodynamics,21 it is property. 3. The kinetic energy depends upon the mass of the body. Hence, in the terminology of thermodynamics, it is an extensive property. 4. Only changes in the kinetic energy (i.e. values relative to a datum assumed to possess zero velocity) are significant and not their absolute values.

3.5.2

Work–Energy Theorem

For the motion of a body of mass m the Newton’s second law can be written as, F = m(dV(t)/dt), where all the symbols have their usual meanings. Now, let dx be the infinitesimal distance 20 21

Meaning all types of energies defined in terms of the concept of energy in mechanics. In other words, they are non-thermal energies. See the discussion on properties in Chapter 2.

Chapter 3: Work and Mechanical Energy

61

travelled in an infinitesimal time interval dt. Then, for this time interval the above equation can be written as F · dx = m · (dV(t)/dt) · dx = m · (dV(t)/dx) · (dx/dt) · dx = m · V(t) · (dV(t)/dx) · dx = d(m · V2(t)/2) Now, by definition, the work done is, d−W = F · dx. Using the definition of kinetic energy, namely dEk = m · V2(t), the equation for the work done can be written as d−W = dEk (infinitesimal motion)

and

W = DEk (finite motion)

In mechanics, this result is known as the work–energy theorem. It can be stated in words as follows. The work done on a body is stored as its kinetic energy.

3.5.3

The Principle of Conservation of Energy

In Chapter 2, the principle of conservation of mass, a primitive to mechanics (and hence, thermodynamics) was stated as follows. The rate of change of mass inside a control volume (i.e. an open system) equals the net (i.e. algebraic sum of) mass flowing into it. This principle was also used to show the relation between the closed and the open system. In other words, this principle implies that any mass that flows into an open system should be stored in it. Allowing for creation of chemical compounds through chemical reactions, this principle can be restated as follows. The rate of increase of mass of a chemical species in a system should be equal to the sum of the net rate of its mass flow into the system and the rate of its creation within the system. Using this as an analogy, the work–energy theorem can be restated as: the increase in the kinetic energy of a body equals the net work flow into it. This means that the work–energy theorem is really the principle of conservation of energy stated as follows. Energy can neither be created nor be destroyed; or, equivalently, the total energy of a system is constant. This principle is derived below (Section 3.5.5) for a more general case. The work–energy theorem further shows that (a) the work is an energy flow; (b) the magnitude of work done depends on the mass of the body;22 and (c) it is meaningless to talk of work stored23 in a body.

3.5.4

Conservative Force and Potential Energy

A force is defined to be conservative if the work done by it around a closed loop is zero.24 By the definition of a property presented in Chapter 2, this implies that during a motion the work

22 23

24

This is already clear from the definitions of force and work. The analogy of clouds and rain (due to UNG) can be cited. Rain existing as water vapour in the clouds condenses and flows down to earth. Thus it is meaningless to talk of rain stored in clouds. The term rain cloud, an abbreviation of ‘rain bearing cloud’, is a misnomer. Similarly, a river exists only when water flows and ‘dry’ river is a misnomer. Since dissipative effects (like friction, hysteresis, etc.) always oppose the motion, the work done by them cannot be zero around a closed loop. Hence, in a conservative force field these must be absent. As a consequence of Newton’s laws, all the fundamental forces of nature are conservative ([FEY], pp. 14-6, vol. I.).

62

Engineering Thermodynamics

done by a conservative force equals the change in a property. This property is called the potential energy since the force (i.e. the potential for motion) can be expressed as its derivative. This principle can also be stated as follows. The work done by a conservative force during a finite motion (process) depends only on the end states and not on the path of motion. As an illustration, consider the infinitesimal one-dimensional motion25 discussed above. For this case, the work done becomes F · dx, which, by the reason that it is a property, equals dEp, where Ep denotes the property, i.e. F · dx = – dEp. This relation can be rearranged as F = – (dEp/dx), i.e. the force is the derivative of a potential. For a finite motion (process), this becomes

∫12 F ⋅ dx = − ∫12 d E p = − DE p = E p1 − E p2 As the consequence of the work–energy theorem given above, i.e. that work is an energy flow, this equation shows that Ep is an energy. It is called the potential energy because it is the potential of a conservative force. Now, since work is an energy flow, the principle of conservation of energy demands that the potential energy should decrease when the force does work; i.e. work is done at the expense of the potential energy. Hence the negative sign is essential in the definition. This equation is called the potential energy theorem, which states that there exists a property called the potential energy whose decrease between two states equals the work done by a conservative force.

3.5.5

The Energy Equation

Next, consider an infinitesimal one-dimensional motion of a body in a conservative force field. Now, the work–energy theorem is d−W = dEk. Since the force is conservative it has a potential and the potential–energy theorem gives, d−W = – dEp. Combining these two relations gives, dEk = –dEp, or dEk + dEp = 0 or Ek + Ep = C, where the constant C is called the total energy and denoted by E. Thus, the mathematical statement of the principle of conservation of energy is, Ek + Ep = E. This equation is called the energy conservation equation or, more concisely, the energy equation.

3.6

EXPRESSIONS FOR POTENTIAL ENERGY

In this section, the above definitions are used to derive expressions for the potential energy of different fields.

3.6.1

Gravitational Field

The Newton’s law of gravitation states that the gravitational force between two masses m1 and m2 is directly proportional to the product of the masses and inversely proportional to the square of the distance r separating them. The constant of proportionality is called the gravitational constant and is denoted as G. Mathematically, this is written as F = G(m1m2)/r2. Now, the 25

In Appendix A, this is shown for the general case using the Stoke’s theorem of vector analysis.

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strength of the gravitational field is defined as the force per unit test mass, i.e. F/m1; where, m1 is assumed to be the test mass.26 Then, the expression for the strength of the gravitational field produced by the mass m2 is Gm2/r2. In elementary physics, the gravitational force on a mass m1 is usually written as F = m1g, where g is called the acceleration due to gravity. Comparing this with the Newton’s law of gravitation, it shows that the acceleration due to gravity represents the strength of gravitational field, i.e. g = Gm2/r2. Note that force is a vector while mass is a scalar. Hence, the strength of the gravitational field, i.e. the acceleration due to gravity, should be a vector since it incorporates the sense of direction of the force. Gravitational field is a conservative force field. As mentioned above, this implies that the work done during a motion should be equal to the change of the potential energy.27 Mathematically, this means that F = – (dEp/dr), where Ep is known as the gravitational potential energy. In terms of the definition of the acceleration due to gravity given above, this can be written as mg = – (dEp/dr). Assuming that the acceleration due to gravity g is constant, this equation can be integrated to give DEp = – mgz, where the distance is denoted as z instead of r. Note that the distance z is measured from some arbitrary position assumed to have zero potential energy and called the datum.

3.6.2

Electrostatic Field

The same approach as for the gravitational field is used since the Coloumb’s law is also an inverse square law. In the usual symbols, the Coloumb’s law of electrostatics, is F = q1q2/ 4pe0r2. Then, the strength of the electrostatic field produced by the charge q1 is defined as e = F/q1 and equals q2/4pe0r2. Since electrostatic field is also conservative, arguments on the lines given above show that F = –dEp,e/dr, where Ep,e is the potential energy of the electrostatic field. This equation can be separated to give, dEp,e = –F · dr = – (q1q2/4pe0r2) · dr. Assuming that the potential energy is measured from infinite distance, i.e. at infinite distance it is zero, this equation can be integrated to give, Ep,e = q1q2/4pe0r. The electric potential, (also called the voltage), denoted as V, is defined as the work done per unit charge, i.e. V = d−W/q1 = q2/4pe0r. The electrical potential energy of the system of charges is defined as the work done in assembling them. Thus, for example, the electrical potential energy of charges q1 and q2 separated by the distance r equals q1q2/4pe0r, which is precisely the relation obtained above. The study of physics shows that the typical electric field is the one that exists between the plates of a parallel plate capacitor, with Ee = (ke0) V 2/2, where V is the electric field strength (voltage).

3.6.3

Magnetic and Electromagnetic Fields

It is well known that magnetism arises because of the movement of electric charges. Thus, it is not strictly correct to deal with magneto-statics separately. Therefore, these aspects will be mentioned in passing. 26

Ideally, the test should be done with a vanishingly small mass, so that the field strength is defined as lim (F/m1)

27

For fields, especially, electrical, magnetic, etc. potential is the most commonly used function.

m1 → 0

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Energy is required to produce a magnetic field such as that produced by an induction coil and this is given by, Em = m0B2/2 where B is the magnetic induction. This is also known as the magnetic energy. An electromagnetic field arises because of moving charges. Therefore, it is not possible to identify electrical and magnetic effects separately. It is shown in the study of physics that the power transmitted by an electromagnetic wave is given by the Poynting vector, S = (1/m0) (E × B), where B and E are the vectors of magnetic induction and the electric field strength, respectively. The preceding three forms of energy (electrostatic, magnetic, and electromagnetic) can also exist in the absence of mass inside a system, i.e. even if the system is perfectly evacuated.

3.6.4

Total Energy

The total energy E is given by E = Ek + Ep + Ee + Em + … It should be noted that all the formulae for different types of energies require that the quantities which define the respective energies be uniform within the system (e.g. uniform electric field). Now, the principle of conservation of energy can be restated as follows: In the absence of nuclear reactions, the total energy of a system remains constant. Hatsopoulos and Keenan present the following definition (and attribute to Gibbs) and explain the need for it. Thermodynamic definition of work. The energy flow across the system boundaries that can be completely reduced to lifting of a weight using devices that are idealizations of the existing ones (i.e. one whose working principles are known). As a consequence of the fact that work is energy flow, this definition is superfluous.

3.7

ADIABATIC AND DIATHERMAL WALLS

An important consequence of the definition of work is that it permits the definition of two important classes of boundary surfaces. They are the adiabatic (also called adiathermal) surfaces and its complement (the logical negation) the diathermic (or diathermal) surface. Either of these may be taken as basic (or primary) so that the other is defined as a logical negation. In order to illustrate this symmetry, in the following sections both approaches are presented. However, in this book, work and adiabatic surfaces are assumed as the basic concepts.

3.7.1

Adiabatic Wall

Assuming the adiabatic wall as the primary (basic) concept, a diathermic wall is defined as its logical negation. Thus, the following definitions are self-evident. Adiabatic wall (or boundary or surface). One that permits energy flow only in the form of work. Such a surface is an ideal thermal insulator. Then, the definition of diathermic wall

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follows as logical negation of this statement, i.e. a diathermic wall is one which is not adiabatic. Evidently, a diathermic surface should be a perfect thermal conductor.

3.7.2

Diathermal (or Diathermic) Walls

In contrast to the approach in Subsection 3.7.1, a diathermic surface can be assumed to be as basic and adiabatic may be defined as its logical negation. This is done in this subsection. The generalized definition of work given above (for the ith work mode) is, dWi = Xi · dYi, where Xi and Yi are the generalized force and the generalized displacement associated with the ith work mode. Table 3.2 shows that for all work modes except electrical (and hence, magnetic and electromagnetic) Yi is constant when the boundary remains fixed. In other words, if the system boundaries remain fixed, no work of the non-electrical type will be done. This table also shows that in the absence of current flow no electrical work is done. Moreover, it is known from electromagnetic theory that no magnetic (and hence, electromagnetic) work will be done when the applied fields are steady. Thus, no energy is transferred across a system boundary as work if (a) the system has fixed boundaries, (b) the boundaries are impermeable to electric current, and (c) the system is subjected to steady fields at its bounding surfaces. The second condition ensures that no energy flows as electrical work and the third one takes care of induced effects. Based on this idea, a perfect thermal conductor is defined as one that transmits only non-work type of energy. Definition of diathermal wall. One that is impermeable to electrical force fields as well as electric currents and which allows energy interaction in the absence of any displacement. Note that this definition essentially states, in many words, that such a wall will permit only non-work type of interactions. Note also that in the case of non-equilibrium (when the work interaction cannot be evaluated as ∫12 XdY), it is experimentally easier to verify only if the non-work type of interactions have taken place. Hence, the non-work type of interaction seems more general.28 From the definitions cited above, the following ones are directly obtained. Adiabatic system. Diathermal system. Adiabatic process. Diathermal process.

The system that is completely enclosed by adiabatic walls. The system that is completely enclosed by diathermic walls. The process that is executed by an adiabatic system. The process that is executed by diathermic system.

Two important consequences of the definition of diathermal walls are the existence of the quantity called heat and the condition called thermal equilibrium. These are formally defined now. 28

Strictly speaking, this argument is deceptive because, if someone asks us, “how do you know that keeping the boundaries fixed and impermeable to electric current and fields ensures only non-work type of interactions?” we can only say that we know that the energy transferred by displaced boundaries and changing electrical fields is work and the non-work type of interaction is its negation.

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Definition of heat.

Interactions that take place in a diathermal process.

Definition of thermal equilibrium. It is the condition of equilibrium between systems in contact through diathermal walls and with the whole assembly isolated. Note that the non-work interaction has been called heat and that it is the form of energy exchanged during the process of reaching thermal equilibrium.

3.7.3

Thermal Equilibrium and Heat

In order to emphasize the symmetry in the development of the concepts developed so far and that one is not superior to the other, they are summarised below. ∑ Energy is exchanged as work when the boundary of a system distorts or electric current flows or the applied force fields change. ∑ Energy is exchanged as heat (non-work) when the boundary of a system is fixed, is impermeable to electric current, and subjected to steady applied force fields. ∑ Adiabatic wall permits energy flow only as work. ∑ Diathermal wall permits energy flow only as heat. ∑ Thermal equilibrium is attained by exchange of energy only as heat. Note that all of these definitions are operational definitions. Without experimental evidence to support the existence of such quantities, the definitions of this type are only of theoretical interest since any number of them can always be generated using mathematico-logical rules. However, the importance of the above definitions lie in the fact that experiments show the existence of both. It should be clearly understood that this logically symmetric nature of the relation arises due to the fact that (a) the total energy flow can be only work or heat and (b) heat is defined as the non-work type energy flow. It should be noted that in this book work is taken as the basic mode of energy flow and heat is defined as the non-work type of energy flow.

REVIEW QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8.

Define, operationally, the volume of a sphere of diameter D. How are inertial and gravitational masses defined? Why are they identical? What is the difference between the quantity of a substance and its mass? Explain how the unit [mol] magnifies the quantity of the substances from the microscopic level to the macroscopic level? What are the properties of work which arise from its mathematical nature (i.e. work is an integral)? What is an indicator diagram? What does it represent and how? What are reversible work modes? Define kinetic energy and list its characteristics.

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9. Derive the work–energy theorem. 10. State the principle of conservation of energy. 11. List the characteristics of work which arise from the principle of conservation of energy and the work–energy theorem. 12. Define a conservative force. 13. Define potential energy and show how it arises. 14. Derive the relations for the potential energy of gravitational and electrical fields. 15. Define adiabatic and diathermal walls. 16. Define adiabatic system and adiabatic process. 17. Define thermal equilibrium. 18. Explain the symmetry in the definitions of adiabatic and diathermal walls.

EXERCISES Instructions A thermodynamic cycle. A series of non-flow processes such that the end state of the last process is the same as the beginning state of the first one. In state space a cycle forms a closed curve. Indicator diagram.

A cycle drawn on the p–V plane.

Sign convention for W and Q. The work done by a system and the heat absorbed by a system are termed positive. Net work and heat. Net work done and heat absorbed in a series of processes equal the algebraic sum of those in the individual processes. Assumptions. The following are the default assumptions: 1. 2. 3. 4. 5.

A closed system is the cylinder–piston assembly. Non-flow processes are those that are executed by closed systems. All processes are quasi-static. Subscripts ‘1’ and ‘2’ denote initial and final states, respectively. Air behaves like an ideal gas with molecular weight (RMM) = 29 [kg/kmol] and g = 1.4. Methodology. Solve each exercise systematically from first principles. Use the methodology mentioned in Chapter 1, namely first draw the system and process diagrams; indicate each assumption each time it is used; systematically formulate the equations and solve them; and state units of each quantity at each stage. Write the unit of each quantity adjacent to it (see the examples in Chapter 1). 1. List the advantages and disadvantages of defining units in terms of the fundamental physical constants.

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2. Assuming the Bohr radius as the basic unit of length and electron mass as the basic unit of mass, develop the rest of the S.I. units in terms of these two. 3. Measurement of mass by the spring balance is a direct method and that by the pan balance is a comparative method. Classify the measurements of other units into direct and comparative methods. 4. We can also define Wx = – ∫12 pdV, where p = pE, the pressure exerted by the environment. Taking this as the basic definition, rederive the relations for Wel, Ws, Ws,.… 5. Derive the relation for the energy stored in an electromagnetic field. 6. Determine the work done when 2 [kg] of a gaseous substance undergoes the following non-flow processes from p1 = 5 [bar], V1 = 0.1 [m3], to p2 = 2 [bar], V2 = 0.25 [m3]. (a) p varies linearly with V. (b) pV = constant. (c) p remains constant till the volume becomes 0.15 [m3] and pV n = constant after that till V2 = 0.25 [m3]. [Hints: (a) p μ V ﬁ p = CV, so that W = ∫12 pdV = C ∫12 VdV = (C/2)[V22 – V12]. But C = p1/V1 = p2/V2. Substituting, the above relation simplifies to W = (1/2) [p2V2 – p1V1]. (This could be also obtained from the formula for polytropic expansion with n = –1). From the above relation, V2 = V1 · (p2/p1) = 0.04 [m3], so that W = –21 [kJ].] (b) pV = C, so that W = ∫12 pdV = C ∫12 dV/V = C ln (V2/V1) = C ln (p1/p2), where C = p1V1 = p2V2. Substituting the numerical values gives, W = (500) (0.1) ln (5/2) = 45.81 [kJ]. (c) In this case there are two processes and the net work done is their algebraic sum. Let the isobaric and polytropic processes be denoted by 1–3 and 2–3, respectively, so that the net work done is W = W13 + W32. Since 1–3 is an isobaric process, W13 = p · (V3 – V1) = 25 [kJ]. The work done in the polytropic process has been shown in the text to be W32 = (p3V3 – p2V2)/(n – 1). The index of expansion n is determined from the process relation, namely p3V3 n = p2V2n. Substituting the values gives n = 1.7937, so that W32 = 31.50 [kJ]. Then, W = 56.5 [kJ]. 7. We know that ∫12 pdV, the work done during a quasi-static process, equals the area of the curve of the process in the p–V plane enclosed by the curve and V-axis between coordinates V = V1 and V = V2. Similarly, ∫12 Vdp, the area enclosed by the curve and the p-axis between coordinates p = p1 and p = p2, is also another form of work done (because it has the unit of work). Show that for the polytropic (pV n = C) expansion, the relation becomes W = ∫12 Vdp = [n/(n – 1)](p1V1 – p2V2). [Hints: The process equation is pVn = C. Differentiating, (p) (nVn–1) dV + dpVn = 0, or, npdV = – Vdp. Then, the work done is W = ∫12 Vdp = –n ∫12 pdV. Evaluation of this integral gives the result.] 8. A compressor is used for compressing a gas from a low pressure (pressure of the source, called the suction pressure) to a high pressure (called the delivery pressure,

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i.e. the pressure of the load). If the fluid being compressed is a liquid, then the device is called a pump. It consists of a cylinder in which a tight-fitting piston reciprocates (moves back and forth). Hence the name. The gas is drawn (or sucked or inducted) into the cylinder volume through the inlet valve. It is discharged (or delivered) from the cylinder volume through the delivery valve. The compressor can be operated by hand (e.g. the hand pump used for filling air in bicycle tyres) or driven by an electric motor (e.g. one used for filling air in bicycle and automobile tyres) or by an I.C. engine (like the ones used in compressed air drills (for example, those employed for breaking concrete road surface)). (See Figure 1.1 of the I.C. engine.) For convenience of description, assume that the axis of cylinder is horizontal. Let p1 and p2 denote the suction and delivery pressures. Let V1 denote the cylinder volume. The operation of the system consists of the following processes: (a) Initially, let the piston be near the cylinder head (a position known as IDC, the inner dead centre). The cylinder volume enclosed by the piston when it is at IDC is called the clearance volume. (b) When the piston moves outwards, the inlet valve opens and because of the partial vacuum created, the gas from the source flows into the cylinder at constant pressure equal to p1. This is called the suction stroke. (c) When the piston reaches the end of its motion (called ODC, the outer dead centre) the inlet valve closes so that the cylinder is completely filled with gas at the suction pressure. (d) During the subsequent motion of the piston from ODC to IDC, both valves are closed. Hence, the gas trapped in the cylinder is compressed continuously. When its pressure reaches p2, the delivery valve opens and the gas is discharged at constant pressure of p2. Thus, the first part of the return stroke is the compression stroke and the rest is the discharge stroke. Let V2 be the cylinder volume when the pressure of the gas inside the cylinder reaches p2. Assuming that all the above processes are ideal and that the clearance volume is zero: (a) Draw all the processes on the p–V plane. (b) Calculate the mass of the gas inducted (sucked-in) if the gas behaves ideally. (c) Calculate the work done in each process and the net work done if the process of compression follows the law pV n = C, where C is a constant. [Hints: (a) Since the suction process is at constant pressure, it is a horizontal line at p = p1, from V = 0 to V = V1. Similarly, the process of discharge is also represented by a horizontal line at p = p2 from V = V2 to V = 0. The compression process is a polytropic that connects the points (p1, V1) and (p2, V2). (b) Using the ideal gas law, m = (p1V1)/(RT1). (c) W = p2V2 + (p2V2 – p1V1)/(n – 1) – p1V1 = [n/(n – 1)] × (p2V2 – p1V1) = [n/(n – 1)] × (p1V1) × [(p2/p1)(n – 1)/n – 1] = [n/(n – 1)] × (mRT1) × [(p2/p1)(n – 1)/n – 1] = p2

∫ p1 Vdp].

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9. Arrange the magnitudes of work done in increasing order when an ideal gas is compressed (a) isothermally, (b) polytropically, and (c) isentropically from p1 to p2. [Hints: In the p–V plane, the work done is the area of the polygon enclosed by the line of the compression process, the vertical line from p2 and the horizontal lines, p = p1 and p = p2. The slope of the polytropic process pV n = C at any point the p–V plane is obtained as follows: By log differentiation we get (dp/p) + n (dV/V) = 0, or dp/dV = – n(p/V). This shows that isentropic process is steepest (n = 1.4). Therefore, we have Wis < Wpoly < Wisen.] 10. A two-stage reciprocating compressor compresses a gas from source pressure p1 to delivery pressure p2 in two stages. The first stage compression is from p1 to some intermediate pressure (say, p3). The second stage then compresses the gas from p3 to p2. Generally between the first and second stages of compression the gas is cooled by passing it through an inter-cooler. With perfect inter-cooling, the inlet temperature to the second stage (say, T3) is the same as T1, the temperature of the first stage. Show that for minimum work input p3 = p1 p2 . [Hints: For two-stage compression

⎡⎛ p ⎞(n −1) / n ⎤ ⎛ n ⎞ ⎡⎛ p ⎞(n −1) / n ⎤ ⎛ n ⎞ 3 2 W =⎜ − 1⎥ + ⎜ − 1⎥ ⎟ ( p1V1 ) ⎢⎜ ⎟ ⎟ ( p3V3 ) ⎢⎜ ⎟ ⎢⎝ p1 ⎠ ⎥ ⎝ n − 1⎠ ⎢ p ⎥ ⎝ n −1⎠ ⎣ ⎦ ⎣⎝ 3 ⎠ ⎦ With no inter-cooling, the gas temperature at the inlet to the second stage will be the exit temperature of the first stage. Let this be denoted by T3. Then, since the gas behaves ideally

⎡⎛ p ⎞(n −1) / n ⎤ ⎛ n ⎞ ⎡⎛ p ⎞( n−1) / n ⎤ ⎛ n ⎞ 3 2 ⎢ ⎥ ⎢ W =⎜ −1 + ⎜ − 1⎥ ⎟ (mRT1 ) ⎜ ⎟ ⎟ ( mRT3 ) ⎜ ⎟ ⎢⎝ p1 ⎠ ⎥ ⎝ n − 1⎠ ⎢ p ⎥ ⎝ n −1⎠ ⎣ ⎦ ⎣⎝ 3 ⎠ ⎦ With perfect inter-cooling, T3 = T1 so that ⎡⎛ p ⎞( n−1) / n ⎛ p ⎞( n−1) / n ⎤ ⎛ n ⎞ 3 2 ⎢ ⎥ ( ) 2 W =⎜ mRT + − ⎜ ⎟ ⎜ ⎟ ⎟ 1 ⎢⎣⎝ p1 ⎠ ⎥⎦ ⎝ n −1⎠ ⎝ p3 ⎠ Now, W will be minimum when (∂W/∂p3) = 0; or, (∂/∂p3)[(p3/p1)m + (p2/p3)m – 2] = 0, where m = (n – 1)/n; or, (∂/∂p3)[(p3)m · (p1)–m] = – (∂/∂p3)[(p2)m · (p3)–m]; or, m · (p3)(m–1) · (p1)–m = (p2)m · m · (p3)(–m–1), which simplifies to (p3)2m = (p1 · p2)m; or, p3 = p1 ⋅ p2 .] 11. Draw each of the following cycles on the p–V plane and calculate work done in each process and the net work done in each cycle. Assume that (a) the air is the working substance (hence these cycles are called air standard cycles); (b) the system contains 1 [kg] of air, and (c) in all cycles the initial state (state ‘1’) of air is p1 and V1. Carnot cycle. The processes are: (a) Air is compressed isentropically to p2 and V2 (isentropic compression to state ‘2’); (b) it is then ‘heated’ at constant temperature to p3 and V3 (isothermal ‘heating’ to state ‘3’); (c) it is then isentropically expanded to p4 and V4 (isentropic ‘expansion’ to state ‘4’); and (d) it is finally ‘cooled’ at constant temperature to the initial state (isothermal ‘cooling’ to state ‘1’).

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Otto cycle. This cycle is the same as the Carnot cycle but with isothermals being replaced with constant volume processes (isochorics). Thus, this cycle comprises the following processes: (a) isentropic ‘compression’ of air to state ‘2’; (b) isochoric ‘heating’ to state ‘3’; (c) isentropic ‘expansion’ to state ‘4’; and (d) isochoric ‘cooling’ to state ‘1’. Diesel cycle. This cycle is same as the Otto cycle but with ‘heat’ addition at constant pressure instead of constant volume. Hence the following processes constitute this cycle: (a) isentropic ‘compression’ of air to state ‘2’; (b) isobaric ‘heating’ to state ‘3’; (c) isentropic ‘expansion’ to state ‘4’; and (d) isochoric ‘cooling’ to state ‘1’. Dual-combustion cycle. In this cycle ‘heat’ is added at constant pressure as well as constant volume. Hence, this cycle is a combination of Otto and Diesel cycles. The constituent processes of this cycle are: (a) isentropic ‘compression’ to state ‘2’; (b) isochoric ‘heating’ to state ‘3’; (c) isobaric heating to state ‘4’; (d) isentropic ‘expansion’ to state ‘5’; and (e) isochoric ‘cooling’ to state ‘1’. Brayton cycle. This cycle is the same as Otto cycle but with constant pressure ‘heat’ absorption and rejection. Hence, it comprises of the following processes: (a) isentropic ‘compression’ to state ‘2’; (b) isobaric ‘heating’ to state ‘3’; (c) isentropic ‘expansion’ to state ‘4’; and, (d) isobaric ‘cooling’ to state ‘1’. Stirling cycle. isentropics.

This cycle is the same as the Otto cycle with isothermals replacing

Ericsson cycle. This cycle is the same as the Brayton cycle with isothermals replacing isentropics. 12. Central electric utility power stations operate on the Rankine cycle. Basically it is a Brayton cycle but with steam as the working substance. The initial state ‘1’ is assumed to be saturated water at the low pressure p2. A pump pressurises the water to light pressure, p1, so that state ‘2’ becomes subcooled water at the higher pressure p1. In the basic Rankine cycle, the state ‘3’ is the dry saturated steam (and, in the modified Rankine cycle it is the superheated steam). The state ‘4’ is generally wet steam at the low pressure, p2. Draw the basic Rankine cycle on the p–V plane and calculate the work done in individual processes and the net work of the cycle. 13. A vessel contains a fluid at a pressure p. It has an inlet pipe of area A. Show that the work required to pump unit mass of the fluid is pv, where v is the specific volume of the fluid corresponding to the conditions in the vessel. This work is known as the flow work. [Hints: Consider a small mass dm being pushed-in. Let it occupy length dL in the inlet pipe. Work done to push it in, is given by – p × A × dL. The mass dm = r × A × dL. Then the work done per unit mass is – (p × A × dL)/(r × A × dL), which simplifies as – p/r = – pv.]

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14. An electric water heater with a resistance of 40 [W] is connected across a power source of 240 [V] for a period of 1 [h]. (a) Determine the work done by the power source on the heater. (b) How many units of electricity are consumed? V. V ), the work done is We = DV [Hints: (a) By the definition of voltage difference (DV q, where q is the charge which flows during time Dt. The definition of current i gives V/R, where R is the resistance) this relation q = i · Dt. Using the Ohm’s law (i.e. i = DV V )2 · Dt/R. Then, the relation for the work done reduces We = becomes q = (DV V )2 · Dt/R. Substituting the values, We = (2402/40) · (3600/1000) = 5184 [kJ]. (DV (b) One unit of electricity is defined to be equal to 1 [kWh], i.e. 3600 [kJ]. Then, the quantity of electricity consumed becomes 5184/3600 = 1.44 [units].] 15. A closed system containing 5 [kg] of a substance is stirred with a torque of 1.0 [Nm] at a speed of 500 [rpm] for 24 [h]. Meanwhile, it expands from V1 = 1.5 [m3] to V2 = 2.0 [m3], against a constant pressure of 5.0 [bar]. Determine the net work done by the system. [Hints: There are two modes of work, namely expansion and stirrer. So that the net work is W = Wx + Ws. Since the expansion is isobaric, Wx = p ¥ DV = (500) (2.0 – 1.5) = 250 [kJ]; and, Ws = – (t)(N)(Dt)(2p) = – [(1.0)(500)(60)(24)(2p)]/1000 = – 4523.9 [kJ]. Therefore, W = –4274 [kJ]. 16. One kilogram of air contained in a closed system at 1 [bar] and 300 [K] is compressed isothermally till the volume halves. During this process it is also stirred with a torque of 1 [N.m] at 500 [rpm] for 1 [h]. Calculate (a) Wx, (b) Ws, and (c) W. [Hints: Similar to Exercise 15, but now the air is compressed. Hence, the answers are: (a) Wx = –59.62 [kJ], (b) Ws = –188.5 [kJ], and (c) W = –248.1 [kJ].] 17. One kilogram of air contained in a closed system at 1 [bar] and 300 [K] is compressed at constant pressure till the volume halves. During this process it is also stirred with a torque of 1 [N.m] at 500 [rpm] for 1 [h]. Calculate (a) Wx, (b) Ws, and (c) W. [Hints: This is similar to Exercise 16. Ans: (a) Wx = – 50.0 [kJ], (b) Ws = – 188.5 [kJ], and (c) W = – 238.5 [kJ].] 18. An electrical battery is charged at 12 [V] and 1 [A] for 24 [h]. During this period, 5 [mol] of hydrogen is liberated at 1 [bar] and 300 [K]. Calculate the net work done. [Hints: There are two modes of work, namely expansion work Wx and electrical work We. Using the standard formulae, We = – (12)(1)(24)(3.6) [kJ] and Wx = pDV (since the process is at constant pressure) = DnRT (since H2 behaves like an ideal gas) = (0.005)(8.3143)(300). Then, the net work done being the algebraic sum of these two is equal to – 1024.3 [kJ].] 19. A storage battery is being charged with a current of 1 [A] at 12 [V] for 5 [h] at T = 300 [K]. During this period 5, [mol] of H2 (assumed an ideal gas) was liberated against constant atmospheric pressure of 1 [bar]. Calculate (a) We, (b) Wx, and (c) W. [Hints: Same as Exercise 18. Ans: Wx = 12.5 [kJ] and We = –216 [kJ].]

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20. Calculate the work done when 1 [kg] of steam is compressed from 5 [bar] to 250 [bar] isothermally at 600 °C. [Hints: In this case, the variation of p with v is obtained from steam properties. Reading the volumes for different pressures from the steam tables (Appendix F, Tables F.2.7) and numerically integrating (say, the trapezoidal rule) gives, Wx = –649 [kJ].] 21. One kilogram copper block at 1 [bar], 300 [K] is compressed isothermally to 1000 [bar]. Calculate the work done. Assume the following constant values: v = 1.1 × 10–4 [m3/kg], and k = 7.76 × 10–12 [1/Pa]. [Hints: The work done in this case is expansion work, i.e. W = ∫12 pdV. Let V = V(T, p). Then, dV = (∂V/∂T)pdT + (∂V/∂p)T dp. In terms of the coefficient of volume expansion, defined as b = (1/V) (∂V/∂T)p, and the isothermal compressibility defined as k = – (1/V) (∂V/∂p)T, the expression for dV becomes dV = bVdT – kVdp. For the isothermal process, the first term vanishes. Substituting the relation for dV in the integral and evaluating it, the relation for the expansion work becomes Wx = –(k)(V)(p22 – p12)/2. The volume does not change as copper is a solid and hence incompressible. Substituting the numerical values gives, Wx = –0.3841 [J].]

Chapter 4

Temperature and Its Empirical Scales

4.1

INTRODUCTION

Temperature is generally considered a primitive (i.e. non-thermal) property because of the following reasons. 1. The concept of temperature makes only qualitative use of the concept of heat. 2. It has been in use long before thermodynamics was formalized as a part of physics. 3. It is measured through non-thermal effects of heat on bodies (e.g. expansion, change of phase, change in resistance, thermal emf, etc.). These measurement techniques were also in existence long before thermodynamics was formulated. Consequently, it was not recognized that the quantity being measured is a thermal property. Some important years in the progress of thermometry, heat and heat engines presented in Table 1.1 justify these statements. Saha and Srivastava ([SAH], Ch. I) present an excellent discussion on temperature measurement. In this chapter the concept of temperature and its scales of measurement are systematically developed based on experimental results including the zeroth law of thermodynamics. It will be seen from later discussions that non-constructive definitions of concepts of heat, thermal equilibrium, … , are adequate1 for this purpose. Consequently, these scales of measurement of temperature are called empirical2 scales. In order to emphasize the empirical nature of temperature, this chapter is deliberately located here. For the same reason, it is brief.

1

2

Some authors call these ‘qualitative’ (meaning non-mathematical) definitions, but it is clear that they are operational and hence as quantitative as any other. It is best to consider them as short-hand terms for the detailed definitions. Meaning ‘based on experiment and not on theory’ (derived from the Greek words ‘empeiria’ meaning experience and ‘empireikos’ meaning skilled. 74

Chapter 4:

4.2

Temperature and Its Empirical Scales

75

THERMAL EQUILIBRIUM

Discussions in Chapter 3 on work and mechanical energy show that work is the energy transfer owing to displacement of boundaries (corresponding to Wx, Wel, and Ws) or owing to flow of electric charge (i.e. electric current) and/or time varying fields (corresponding to We, Wm and Wem). Then, it follows that: Adiabatic wall is one that transfers energy only by displacement or by flow of electric charges (i.e. electric current) through itself or owing to time varying fields (induction effect). In Chapter 3, heat was defined as the non-work type of energy interaction. Then, it follows once again, that: Heat is the energy transferred through a wall that is fixed (absence of displacement) and that does not allow flow of electric charges when the applied fields are steady. Now, the following definition is a direct consequence of the above definition of heat. Diathermal wall is one that permits energy flow only as heat. It should be noted that the above definitions of heat and diathermal wall are nonconstructive (existential type) definitions. Experimental evidence that heat transferred through a wall depends directly on its thermal conductivity may be used for an alternative definition as follows: Diathermal wall is a perfect thermal conductor. And its negation is defined as: Adiabatic wall is a perfect thermal insulator. The following is a direct consequence of the above definitions. Definition of isolating a system. Enclosing a system in adiabatic walls which are fixed and ensuring that electromagnetic fields are steady and electrical currents are absent. Now, consider the case in which a system at some state is brought in contact with another through common diathermal walls, the assembly consisting of both the systems being isolated. Experiments show that these two systems reach a state of equilibrium. Such an equilibrium is called thermal equilibrium. Thus, its operational definition is: Definition of thermal equilibrium. The state of equilibrium reached by a system (or a number of systems) when in mutual contact through diathermal walls, the assembly being isolated. Figure 4.1 illustrates the concept of thermal equilibrium. The definition of thermal equilibrium shows that any system will be in thermal equilibrium with (a duplicate copy of) itself. Mathematically, this property of thermal equilibrium is known as reflexivity. Similarly, if system A is in thermal equilibrium with system

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B then system B will be in thermal equilibrium with system A. This property is known, in mathematics, as symmetry. Diathermal wall

Systems A and B are in contact through a diathermal wall, but together they are isolated.

System A

System B

Hatched portions are isolating walls Figure 4.1

Thermal equilibrium.

When these experiments are repeated with three systems A, B and C, it is found that if system B is in thermal equilibrium with A and C independently, then A and C are in thermal equilibrium. This result is generalized as the: Zeroth law of thermodynamics. Two systems in thermal equilibrium independently with a third, are in thermal equilibrium with each other. Figure 4.2 illustrates this zeroth law schematically. This law essentially states the following mathematical property, called transitivity.

⇒

Figure 4.2 Schema of the zeroth law of thermodynamics.

Transitivity. If system A is in thermal equilibrium with system B and system B is in thermal equilibrium with system C, then system A will be in thermal equilibrium with system C. The transitivity property allows comparison of any number of systems by testing them pairwise. Note that the properties of reflexivity, symmetry and transitivity are due to the fact that nature does not show preference to any of the systems under experimentation, i.e. they are all same. This characteristics is common to all types of equilibrium.3 3

In contrast, heat transfer takes place always from higher temperatures to lower ones, thereby exhibiting directional asymmetry (the second law of thermodynamics arises out of this).

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Temperature and Its Empirical Scales

77

The above arguments show that systems in thermal equilibrium have something common among them. The property that establishes this commonality is named the temperature. In other words, they show that the existence of temperature and its equality are the basic characteristics of thermal equilibrium.

4.3

MATHEMATICAL DEVELOPMENT OF TEMPERATURE

In mathematics (analysis) any relation which possesses the characteristics of reflexivity, symmetry and transitivity is called an equivalence relation. The two properties of equivalence relations4 used here are: Existence of common features. All the elements possess some features (at least one) which are common to all systems in the class. These common features have equal values in all systems. Choice of a representative system. investigating the common features.

Any one system can be chosen as representative for

From the discussion in Section 4.2 it is clear that the concept of being in thermal equilibrium is the equivalence relation which possesses the common feature of equality of temperatures and the system used to represent the class of systems in thermal equilibrium is the thermometer. In this section the concept of temperature is developed using functional analysis as the consequence of the experimental results. In the next section, a physical approach is presented. To simplify arguments, consider a simple compressible system5 whose states are defined by pressure p and volume v. 1. Then, the condition of thermal equilibrium between systems A and B may be symbolically written as F1(pA, vA, pB, vB) = 0, which may be functionally solved as pB = G1(pA, vA, vB). 2. Similarly, the condition of thermal equilibrium between systems B and C may be written as F2(pB, vB, pC, vC) = 0, which may be solved as pB = G2(pC, vC, vB). 3. Thus, the condition of thermal equilibrium of the body B independently with A and C respectively, may be expressed as G1(pA, vA, vB) = pB = G2(pC, vC, vB) 4. Since this is a functional relation it can also be written as F3(pA, vA, pC, vC, vB) = 0. 5. Now, the condition of thermal equilibrium between systems A and C may be written as F4(pA, vA, pC, vC) = 0. 4 5

A simple example is the geometric concept of similar triangles. Essentially, we have the gas thermometer in mind which is the primary standard for temperature measurement.

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6. Then, the zeroth law implies that F3(pA, vA, pC, vC, vB) = F4(pA, vA, pC, vC). 7. Mathematically this requires that vB in F(.) (or, G(.)) should appear in such a way that it cancels out. For example, G1(pA, vA, vB) = f1(pA, vA)g(vB) + h(vB). 8. Thus, the earlier equation can be written as f1(pA, vA) = f2(pB, vB). 9. These arguments may be extended to arrive at the relation f1(pA, vA) = f2(pB, vB) = f3(pC, vC) = … . Now, if the temperature is defined as q = f(p,v) then the zeroth law gives

q = f(pA, vA) = f(pB, vB) = f(pC, vC) = … Any of these equations can be rewritten as f(p,v,q) = 0, which is seen to be the equation of state. In other words, what is traditionally named as the equation of state is really a definition of temperature. It should be noted that the above arguments did not use, anywhere, the facts (a) that thermal equilibrium can be represented functionally in terms of four variables (e.g. f (pA, vA, pB, vB) = 0) and (b) that these are the pressures and volumes. This means that the arguments are valid for systems containing any substance and not necessarily simple compressible substances for which two state variables may be independent and which can be chosen as p and v.

4.4 EMPIRICAL TEMPERATURE In Section 4.3, based on mathematical arguments it was shown that temperature exists and it will be the same for all systems that are in thermal equilibrium. This temperature was denoted as q since it was only shown to exist and not how it is measured.6 Hence, empirical methods need to be devised to measure temperature. Consequently, it will be called the empirical temperature. This is done in this section. Now, consider the following experiments. 1. Let system A be in a fixed state (say, (pA,1, vA,1)). 2. Let system B be brought in contact with A and be allowed to attain thermal equilibrium. Let this state of equilibrium of B be (pB,1, vB,1). 3. Plot the state of B on the pB–vB plane with pB as x-axis (abscissa) and vB as y-axis (ordinate). 4. Remove system B and change its state. 5. Let system B again attain thermal equilibrium with A in its fixed state. 6. It is found that this will be a new state (say (pB,2, vB,2)). 7. Plot this point also on the pB–vB plane. 8. Repeat the experiments for different states of B. 6

In this respect this law is less powerful than the first law that defines the existence of the property called internal energy and also shows how to measure it (see Chapter 5).

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79

The main result of these experiments is that all these points (representing the thermal equilibrium states of B) are found to lie on one curve. This curve is called an isotherm.7 Repeating the above experiments for different states of system A, different isotherms of the system B can be generated. Since there are infinity of states of system A, infinity of isotherms of system B can be generated. These results may be summarized as: Existence of isotherms. Isotherms are continuous curves in the state space and they completely cover it. When similar experiments are done but with state of system B held fixed at any state on its isotherm (say, pB,1, vB,1) then, and by changing the states of system A, an isotherm of A is generated. This pair of isotherms is called the corresponding isotherms. The zeroth law allows this concept to be extended (pairwise) to any number of systems so that it becomes meaningful to consider the corresponding isotherms of systems A, B, C, … . The important property of corresponding isotherms is that, when systems are in states represented by points on their respective corresponding isotherms they will be in thermal equilibrium among themselves. Existence of corresponding isotherms proves the symmetric nature of the state of thermal equilibrium. The measurement of temperature and construction of its scale begin with the following facts: (a) temperature is property to check whether two systems are in thermal equilibrium, (b) all systems in mutual thermal equilibrium will have the same temperature, and (c) an isotherm is the locus of states of thermal equilibrium. These facts lead to the following definitions: (a) temperature is a label for an isotherm and (b) the temperature scale is a rule for labelling all the isotherms. The experiments also show that systems whose states are on the corresponding isotherms are in thermal equilibrium with one another. Then, the concept of temperature stated above implies that all corresponding isotherms should have the same temperature. It follows from these results that all corresponding isotherms should have the same label. It is adequate to label the isotherms of the thermometer (the standard system used in testing for thermal equilibrium) since the zeroth law can be used to extend the results to all corresponding isotherms. Hence, in further discussions, only the isotherms of the thermometer are considered. All the above requirements based on thermodynamics are satisfied if each isotherm is assigned a unique label, i.e. each isotherm has one and only one label and two isotherms do not have the same label. Thermodynamics does not put any restriction on the types of the labels allowed for marking isotherms. However, for practical reasons8 the following rules are adopted in construction of the temperature scale. For consistency with other physical usage: (a) the labels are numbers and (b) the isotherms corresponding to physiologically hotter states are assigned larger numbers.

7 8

Derived from the Greek words isos meaning equal and therme meaning heat, i.e. states of equal heat. Remember temperature was being measured long before the zeroth law was enunciated. Hence, our rules should be consistent with these long-established practices.

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For convenience of experimentation, during measurements, only one property (called the thermometric property) is varied. Figure 4.3 shows this. y The x is the thermometric property

θ1 θ2 θ3 θ4

x( θ 1)

x( θ 2 )

Figure 4.3

x( θ3 )

x(θ 4)

x

Thermometric property.

The common thermometric properties are shown in Table 4.1. Table 4.1

Common thermometric properties

Type of thermometer

Constant property

Thermometric property

Mode of operation

Gas expansion Gas expansion Liquid expansion Resistance Thermocouple

V (Gas volume) p (Gas pressure) p (Liquid pressure) p, t (Wire tension) p, t (Wire tension)

p (Gas pressure) V (Gas volume) L (Liquid col., length) R (Electric resistance) E (Thermal emf)

Constant volume Constant pressure — — —

For convenience of calculations including interpolation: (a) labels are real numbers9 and (b) the scale is linear. Note that not all thermometric properties vary linearly. For example, the thermal emf (the emf of thermocouples) and the electrical resistance are quadratic functions of temperature. However, for defining the scale, a property that varies linearly [e.g. volume of an ideal gas or liquid (like mercury)] is used. Properties that vary non-linearly are used essentially in small range where linear approximation is adequate. Combining all these requirements, the temperature scale may be written as q = ax + b, where q is the temperature, x is the thermometric property and a and b are constants. Note that the constant a determines the interval between two degrees of the scale and the constant b determines the zero of the scale. 9

Since experiments show that isotherms are continuous and they cover the entire state space the temperature (i.e. f(.)) is a real function. Then the conclusion that the labels are real numbers, follows automatically.

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Temperature and Its Empirical Scales

81

Since this equation contains two constants, the values of x and q at two specified states are needed to evaluate them. These states, called the primary fixed points, are: (a) the ice point [the temperature at which ice melts at atmospheric pressure, i.e. the state of equilibrium between ice and (air saturated) water at atmospheric pressure] taken as the lower fixed point and (b) the steam point (the temperature at which water boils at atmospheric pressure, i.e. the state of equilibrium between water and steam at atmospheric pressure) taken as the upper fixed point. The two common scales are: Fahrenheit. Celsius.

The lower fixed point is –32 [°F] and the upper fixed point is 212 [°F].

The lower fixed point is 0 [°C] and the upper fixed point is 100 [°C].

These scales are well known and will not be discussed further. In 1954, the Kelvin scale was adopted for universal use. It was assumed that this will be a centigrade scale whose interval coincides with that of the Celsius scale, i.e. dT of 1 [K] = dT of 1 [°C]. Consequently, it will have only one constant. Hence, it needs only one fixed point for evaluation of this constant. This fixed point is taken as the triple point of water. The constant of this scale is then evaluated from the relation, “triple point of water = 273.16 [K]”.

4.5

IDEAL GAS TEMPERATURE

In the empirical scale of temperature defined above, it is difficult to separate the contribution of the working substance. Consequently it is important10 to know whether it is possible to construct a temperature scale that is independent of the working substance. In Section 7.8, the thermodynamic temperature and its scale (the Kelvin scale) constructed as a consequence of the Carnot theorem are shown to do this. The precursor to this was the ideal gas scale, which is described now.

4.5.1

Experiments

Consider the following experiments conducted with a gas thermometer operating under constant volume mode.11 1. 2. 3. 4. 5. 10 11 12 13

Fill the bulb with H2. Put the thermometer in a triple point cell.12 Measure the pressure indicated (say p3,1). Now put the thermometer in a steam cell.13 Measure the indicated pressure (say ps,1).

Both theoretically and practically. This is known to involve least errors. A bath maintained at the temperature of the triple point of water. A bath maintained at steam point, i.e. the boiling point of water at 1[atm].

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6. Calculate q (ps) = 273.16° (ps/p3). 7. Plot this point on the plane with p3 as the abscissa (i.e. x-axis) and q (ps) as the ordinate (i.e. y-axis). 8. Remove some mass of the gas from the bulb so that p3 is reduced. 9. Repeat the above steps to obtain a second point on the graph. 10. Repeat the experiments till the p3 reaches as small a value as possible. 11. Repeat the experiments with different gases, such as N2, O2, etc. 12. Repeat the experiments for different fixed points (e.g. sulphur-point, silver-point, goldpoint, etc. called for convenience, the upper fixed points) in place of the steam point.

4.5.2

Results and Their Interpretation

The results of the above series of experiments show the following important characteristics: (a) For any gas, at a fixed point, all the points on the graph lie on a straight line and (b) the lines of all gases, when extrapolated, intersect the y-axis (i.e. zeroth value of p3) at the same point. Figure 4.4 shows these for the steam point. O2 p/p 3

Air

Scales are exaggerated N2

H2

0

p3

0 Figure 4.4

Ideal gas temperature of steam point.

Only the second result is needed here. It shows that when the pressure tends to zero, the value becomes independent of the gases. Since it is known that at zero pressure, all gases behave ideally, the temperature corresponding to zero pressure is defined as the ideal gas temperature of that fixed point. This is written mathematically as

q = 273.16° lim

p3 → 0

p ; p3

V = constant

Repeating the experiments by operating the gas thermometer in constant pressure mode gives V ; p3 → 0 V 3

q = 273.16° lim

p = constant

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Temperature and Its Empirical Scales

83

The above two expressions can be combined as

q = 273.16° lim

p3 → 0

= 273.16° plim →0 3

pV/n ( pV/n)3 pV ( pV )3

Since ideal gases (i.e. gases in the limit of zero pressure) are being dealt with, the limit is changed from p3 to p. Then, this equation becomes the definition of the ideal gas temperature. Denoting it as qig gives

qig = 273.16°

lim p → 0 ( pv) lim p→ 0 ( pv)3

The universal gas constant is defined as R=

lim p→0 ( pv)3 273.16°

so that the above definition reduces to

qig = which can be rewritten as

lim p → 0 ( pv) R pv = Rq

and is recognized as the ideal gas equation. Thus, it is seen that the ideal gas equation is really a definition of ideal gas scale of temperature. In Section 7.9 it is shown that, as a consequence of the Carnot theorem, the ideal gas temperature is the same as the thermodynamic (Kelvin scale) temperature. Anticipating this result, the ideal gas temperature qig is denoted as T. Then the ideal gas equation can be put in the more familiar form, i.e. pv = RT.

4.6 INTERNATIONAL PRACTICAL TEMPERATURE SCALE The temperature scales defined in this chapter and those to be developed later are not convenient for use in industrial environments. Consequently, the International Practical Temperature Scale (IPTS) was developed. Essentially this is nothing other than the usual scale with a large number of temperatures at convenient intervals accepted as fixed points. It also specifies the thermometer to be used as standard as well as the equation(s) to be used for calculating the temperature. This is not elaborated here since it is a part of the course on measurements.

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REVIEW QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

13. 14. 15. 16.

Why has temperature been considered for long as a primitive property? Why are temperature scales called empirical scales? Define the concept of isolating a system. Define thermal equilibrium. State the zeroth law of thermodynamics. What are the properties of the relation ‘being in thermal equilibrium’? Which of these properties is the result of the zeroth law of thermodynamics? What is an equivalence relation? What are its characteristics? Develop the concept of temperature from functional analysis. What is an isotherm? Explain an experiment to generate it. What are corresponding isotherms? What is temperature and its scale in isotherm approach? What are the rules of constructing a temperature scale? Which of these are thermodynamic requirements which of these are assumed for convenience of measurements and of calculations? What are fixed points? Why are they needed? What were the fixed points before and after 1954? Why? Describe an experiment to develop the ideal gas temperature scale. How is it constructed? Develop the concept of temperature from physical arguments using an ideal gas as the working substance. What is IPTS? Enumerate its details. (Hint: Look up a standard textbook on measurements or instrumentation).

Chapter 5

The First Law of Thermodynamics

5.1

INTRODUCTION

In this chapter, the first law of thermodynamics and its consequences are discussed. These discussions begin by noting that this law is needed to explain the following observed facts: (a) measurement of heat as a form of energy and (b) interrelation of heat and work, especially, the observation as to why friction seems to convert work into heat. There is another way to look at the first law of thermodynamics. Since thermodynamics deals with energy storage and energy exchange, the following two basic questions arise: 1. How is the energy stored within a system related to that it exchanges with its environment when it executes a process (or a cycle)? 2. Which are the possible processes between two given states? The first law answers the first question. The second law provides the answer to the second question. In this chapter, the first law is developed for a closed system. It is then used to show how heat is measured on the basis of the principle of conservation of energy. In Chapter 6, these concepts are extended to flow1 systems.

5.2

BORN–CARATHÉODORY FORM OF THE FIRST LAW

Of the several forms of the first law of thermodynamics, the one developed by Born and Carathéodory2 is used here. This form of the law is a generalization of the results of experiments of Mayer, Joule and others conducted for determining the ‘mechanical equivalent of heat’. 1

2

It should be noted that a closed system is defined as one in which the mass remains constant and an open system is defined to be one which is not a closed system. This means that the mass contained in an open system is not constant. Now, the mass inside a system can change by (a) diffusion, (b) reaction, and (c) flow. Thus, while all flow systems are open systems but the converse (i.e. all open systems are flow systems) is not true. See [ZEM], Ch. 4, or [KES1], Vol. I., p. 150. 85

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Basically all of the experiments consisted of heating a fixed mass of water at atmospheric pressure, contained in an insulated vessel, from 14.5 [°C] to 15.5 [°C] by one of the following methods: (a) stirring by paddles, (b) rubbing two bodies immersed in the water, and (c) electricity, etc. The main result was that, in all cases, within experimental errors, the ‘mechanical equivalent of heat’ was constant at 1 [cal] = 4.1858 [J] (this is known as the 15° calorie). In order to understand the implications of this result as well as those of the experiments, they are restated below in tabular form in the terminology of thermodynamics (i.e. the terms used so far in this book). Then, Statement

Thermodynamic meaning

Fixed mass Atm. pr. and 14.5 [°C] Atm. pr. and 15.5 [°C] Insulated vessel Various types of ‘heating’ Result

Closed system Fixed initial state Fixed final state Adiabatic system Different work modes (processes) Work transferred is independent of processes

Thus the above statements can be rephrased as: the work transferred to an adiabatic system between a fixed initial state and a fixed final state is the same for all processes (and is equal to 4.1858 [J/cal]). In most of the discussions of the experiments of Joule and others, the emphasis is on the value of the mechanical equivalent, i.e. that it is 4.1858 [J/cal]. However, as the first law of thermodynamics, Born and Carathéodory emphasized the fact that the mechanical equivalent was constant. In other words, their generalization will still hold if, a million years later, someone found that this value was 1234.56 × 10789 (or any other number), provided it is same for all processes. In the chapter on primitive concepts (see Chapter 2), a quantity that is independent of processes is defined as a property. This means that the work done in the above experiments (i.e. in adiabatic experiments between fixed states) should equal the change in a property. Thus, the Born–Carathéodory form of the first law of thermodynamics is: The work transferred in an adiabatic process equals the change in a property. This property is called the internal energy3 12 and is denoted as U. The mathematical form of the first law is, therefore, given by DU = – Wad . The negative sign is introduced for consistency between the sign conventions of D and of work. According to mathematics, D is positive when the final value is greater than the initial one. The principle of energy conservation demands that the energy of a system should increase only when energy flows into it. The only form of energy flow involved in these experiments is work for which the sign convention adopted is: work outflow is positive. These two requirements can be met only when the negative sign is present.

3

Strictly, thermal internal energy. This is a generalization of the definition of a conservative force field.

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The First Law of Thermodynamics

87

5.3 CHARACTERISTICS OF THE FIRST LAW By defining the existence of a new form of energy called the internal energy, the first law exerts an influence on all aspects in which energy is involved. These aspects are now taken up. The following characteristics of internal energy are directly deduced from the first law: (a) It changes when work is done on the system. (b) It is stored inside the system. (c) Only its changes are defined so that any convenient datum may be chosen. (d) It is measured in the same units as work. (e) Energy states may be arranged according to some order, since for any two adiabatic processes ‘1’ and ‘2’, U1 > U2, i.e. (U1 – U0) > (U2 – U0) if and only if – W1 > – W2, where U0 is the energy at some datum. (f) It is an extensive property. This may be understood by assuming that the work is done by adiabatic expansion, for which – DU = W = ∫12 pdV = m ∫12 pdv, where v is the specific volume, V/m. With the inclusion of this new form of energy, namely, the internal energy, U, it is clear that following are the different forms in which a system can store energy. (a) Kinetic energy (Ek) defined as Ek = m · v2/2. (b) Potential energy (gravitational) (Ep) defined as Ep = m · g · z. (c) Electric energy (Ee) calculated as Ee = (k · e0) · E2/2, where E is the electric field strength. (d) Magnetic energy (Em) calculated as Em = B2/2m0, where B is the magnetic induction. (e) Electromagnetic energy (Eem) given by the Poynting vector, S = (E ¥ B)/m0, where B and E are the vectors of magnetic induction and the electric field strength, respectively. These expressions were presented in Chapter 3 on work. The last three forms of energy (i.e. electrostatic, magnetic and electromagnetic) can exist in the absence of mass inside a system, i.e. even if it is perfectly evacuated. It should be noted that all these formulae require that the quantities that define the respective energies be uniform within the system (e.g. uniform electrical field). There is an important difference between U and all other form of energies. The latter all are non-thermal energies since U arises because of the random motion of molecules caused by thermal agitation.4 Consequently, energy stored as U cannot be completely converted into work. This is the underlying idea of the second law of thermodynamics. In other words, if work is converted by dissipation into U then it cannot be completely recovered. Some authors emphasize this fact by calling U the thermal energy (implying that it has something to do with heat) while terming all others mechanical energies. Now, the total energy E is obtained as E = U + Ee + Em + Ek + Ep + …

5.4

MEASUREMENT OF HEAT

Towards the end of Chapter 3 on work and in the beginning of Chapter 4 on empirical temperature, it was shown that the existence of heat as non-work form of energy follows from the definition of work and from the experiments on determination of isotherms. However, they do not show how to measure it.5 Only the principle of conservation of energy together with the 4 5

Appendix B contains an interesting example originally proposed by Callen ([CAL]). In this respect it is a non-constructive (existential) definition like temperature.

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first law prescribes how heat should be measured. In this section, this aspect is discussed. Other consequences are dealt with in later sections. The principle of conservation of energy states that energy can neither be created nor be destroyed. This was shown in Chapter 3 during the discussions on work and other ‘mechanical energies’. Hence, the total (‘mechanical’) energy in all (‘mechanical’) forms is constant. This means that the increase in ‘mechanical’ energy in a system equals the net ‘mechanical’ energy flowing into it and the ‘mechanical’ energy flow is work (in all forms). Now, the first law shows the existence of a ‘non-mechanical’ energy. Taking this energy into account, the preceding relation can be written as the increase in total energy in a system equals the net total energy flowing into it, where the total energy flow now includes the ‘nonmechanical’ energy flow (i.e. non-work) as well. For further arguments it is convenient to write these relations in symbolic form as far as possible. Denoting the total energy as E and the total work done as W (= Si Wi), the said relation can be compactly written as DE = –W + Net ‘non-mechanical energy’ (‘non-work’) inflow where for mathematical consistency the negative sign is used for the work term. This equation can be rearranged as Net ‘non-mechanical’ energy inflow = DE + W Assuming that only one type of ‘non-mechanical’ energy, called heat (denoted as Q) exists, this equation becomes Q = DE + W dQ = dE + dW

Finite process Infinitesimal process

(5.1)

where the inflow of heat is assumed to be positive. The above arguments can be better understood by developing these concepts from experimental results. Consider the following experiments with a closed system between fixed initial and final states. (a) In the first experimently, the system undergoes an adiabatic process 12 absorbing work Wad . (b) In the second experiment, the adiabatic walls are removed and the system is allowed to exchange energy by ‘non-work’ mode as well. Let the work done by the system be W. In both of the above experiments, the end states are the same and therefore, the changes of energy in both the processes are equal. Moreover, in both processes, the work done (as 12 . Now, heat energy flow) is stored as the energy of the system. Then, it follows that W £ Wad is defined as (all of) the energy other than work. In other words, the energy flow can be work or heat (i.e. all other than work). It is then evident that the difference in energy between these two processes should be due to the flow of heat. This argument is written mathematically as 12 Q = Wad – W12

(definition of heat)

Using the first law this becomes Q = –DU – W = – (DU + W) or

–Q = DU + W

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or considering all other modes of energy storage as well, –Q = DE + W It is convenient to absorb the negative sign in Q by adopting the sign convention for Q as heat absorbed by the system is positive. Thus, the above equation may be written as Q = DE + W which is the same as Eq. (5.1). This equation is called the mathematical form of the first law of thermodynamics. But it should be clear that this equation is only the definition of Q, the energy flow as heat. A device that creates or destroys energy and thereby violates the principle of energy conservation (also called the first law of thermodynamics) is called a Perpetual Motion Machine of the First Kind and is denoted as PMM1. Then, the principle of conservation of energy may also be stated as follows: It is impossible to construct a PMM1. The above discussions show that heat has the same characteristics as work because of the definition. In other words, heat is measured in the same units (i.e. [kJ]) as work and it is a path function (i.e. mathematically, it is an inexact differential). The quantity and direction of heat flow are determined from the energy (first law) equation as follows. In the absence of work transfer and of all ‘mechanical’ modes of energy storages, the energy (first law) equation becomes Q = DU. Hence, the following steps are involved in the measurement of the magnitude and direction of heat flow: ∑ Take a system with walls impermeable to electric currents and to electromagnetic fields (so that Ee, Em, Eem are constant and We = 0 = Wm = Wem). ∑ Fix its boundaries so that they do not deform (so that Wx = 0 = Wel = Ws). ∑ The discussions on work and thermal equilibrium show that the preceding two steps ensure that only heat interactions exist. ∑ Ensure that the system is at rest (so that Ek = 0). ∑ Ensure that the system is at the reference with respect to all potentials (so that Ep = 0). ∑ The above two steps together with the first step ensure that E = U. ∑ Measure the change in internal energy. ∑ The change in the magnitude of internal energy equals the quantity of heat transferred. ∑ If the internal energy has increased, then heat has flown into the system. Note that all the above steps are only the verbal explanation of the energy (first law) equation (i.e. definition of heat) since the general equation, namely Q = DE + W = D(U + Ek + Ep + Ee N + Em + … ) + ∑i= Wi reduces to Q = DU, when Ek = Ep = Ee = Em = … = const., and, 0 Wi = 0 for all i.

5.5 CHARACTERISTICS OF WORK, HEAT AND ENERGY It is essential to understand the relations between work, heat and energy clearly, since many doubts can arise on account of misunderstanding them. Hence, this section is devoted to a critical review of these relations.

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All the quantities dealt with so far are basically energy. However, work and heat are the energy flows, i.e. they exist only when the energy flows across the system boundaries. The analogy to rain6 is illustrative. When moisture is stored in the clouds, it is not called rain. After a rain, when the water collects as a puddle it is still not called rain. Only during the process of condensation of the steam in clouds as water and the falling down as water, it is called rain. Thus, rain may be considered as water in transit (or motion or in flow). Similarly, work and heat are energy in transit or motion or flow. Instead of rain, a river may also be used for the analogy because a river exists only when water flows through it.7 These arguments are summarized as: (a) Energy contained in a system is the energy stored. (b) Only when energy flows across the boundaries of the system, it is either work or heat. The energy equation [i.e. the first law, DE = Q + (–W)] expresses the relation quantitatively between work, heat and energy stored. Mathematics asserts that this equation can be used to evaluate any one of DE, Q and W in terms of the other two. This equation is also an identity in the sense that if all the three quantities can be independently evaluated, then on substitution in the equation the left-hand side should be equal to the right-hand side. Since this is a frequent source of confusion, it is worthwhile to look at it critically. The three quantities, DE, Q, and W can be independently calculated as follows: Mechanical energy (Emech). The individual ‘mechanical’ energies (Emech, i for i = 1, ..., M) are calculated as discussed in Section 5.2, and then Emech = ∑ i Emech, i. Internal energy (U). This is calculated as du = (∂u/∂T)v dT + (∂u/∂v)T dv, which may be written as du = cV dT – [T(∂p/∂T)v – p]dv, where the first term arises because of the definition of cV and the second is obtained using the Maxwell’s relation (Section 8.4). If the specific heat data in the form cV = cV (T) and the equation of state of the substance in the form p = p(T, V) are known, then the equation for dU can be integrated to obtain the values of U at all states. Total energy (E). This is calculated as E = Emech + U. Work (W). The net work done is W = ∑ i Wi even if any (and all) of the processes is non-quasistatic. However, Wi = ∫12 Xi · dYi, if (and only if) the ith process is quasi-static. In other words, only when all the processes are quasi-static can the total work done be calculated as W = ∑ i ∫12 Xi · dYi. Heat (Q).

This can be evaluated as follows:

Combustion. As Q = mf . Qcal, where Q is the heat released when mass mf of the fuel with calorific value Qcal burns. 6 7

Prof. U.N. Gaitonde, Mech. Engg. Dept., IIT Bombay, Private Communication. These analogies are interesting since they also indicate the improper usages like ‘rain clouds’ and ‘dry river’. Compare this with the usage like “heat stored, heat capacity”, etc.

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Conduction. Using the Fourier’s law, i.e. Q = –(k) . (A) . (DT/Dx), where the heat flows at the rate of Q across area A when the temperature gradient is (DT/Dx) and the thermal conductivity k is a property of the substance. Convection. As Q = hADT, where the symbols have the same meanings as given above. The heat transfer coefficient h is correlated to the flow conditions and the fluid properties. However, it should be noted that this equation really defines the quantity h. Radiation. Using the Stefan–Boltzmann law, i.e. Q = (s) . (A) . (T14 − T24 ) , where s is the Stefan–Boltzmann constant, T1 and T2 are the temperatures of the radiation source and sink respectively. Other symbols have the same meaning as above. It should be noted that all the quantities mentioned above are not part of thermodynamics, i.e. they are primitives to it. The data and expressions for ‘mechanical energies’ and different work modes are borrowed from mechanics, electromagnetics, etc. The data and expressions for calculating the internal energy (i.e. specific heats and equation of state) are borrowed from properties of matter. The calorific value is also a property of matter. The expressions for calculating the heat transfer rate are empirical. In other words, extra information is needed if any of Q, W and E are to be calculated. To understand these ideas, consider a closed system under the following experiments. Case 1. Experiments showed that there was a change in one of the properties used for definition of state. No other information is available. The only conclusion that can be drawn is that some process has taken place since the state has changed. Since no other information is available, DE cannot be calculated and hence the energy (first law) equation, namely Q = DE + W is useless since none of the terms can be calculated. Case 2. The same experiment as above but now let the system be thermally insulated, which means that Q = 0. The only conclusion possible now is that energy is transferred as work, because the energy (first law) equation becomes DE = – W. Case 3. The same experiment as above, but now let the boundaries of the system be rigid. This means that no expansion work is present. Case 4. The same experiment as above, but now let all other modes of work be absent. Then the energy (first law) equation shows that the energy of the system is constant (because DE = 0). Case 5. The same experiment as above, but now let all forms of energy except U be zero. Then, it shows that the internal energy of the system is constant. Case 6. The same experiment as above, but now let the system contain an ideal gas. Then, its temperature remains constant since the internal energy of an ideal gas depends only on temperature.8 The main point to be noted in all these experiments is that as more and more information is available, better quantitative (numerical) answers can be provided. Conversely, better answers require more information. This is true for all other disciplines as well. 8

This is the Joule’s law which is illustrated in Example 5.7 in this chapter.

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As an illustration, consider a rigid insulated vessel divided into two parts (say A and B) by a frictionless perfectly heat-conducting piston held fixed initially at some place as shown in Figure 5.1. Let all other work modes (e.g. Ws, We, …) be absent so that W = Wx. Let all ‘mechanical’ energy modes (e.g. Ek, Ep, …) be also absent so that E = U. Let both the parts contain some gas at different pressures and volumes. Let the piston be released and let it come to equilibrium. Basically, the processes of expansion of the gases in the compartments are non-quasi-static. Hence, the first law can be applied only to the end states which are the equilibrium states. Heat-conducting piston

Rigid insulated walls

Part B

Part A

Stoppers System = Part A + Part B

Figure 5.1

The system.

Since the piston can move freely, mechanical equilibrium implies that at the final state the pressures must be equal, i.e. pA, f = pB, f. Moreover, since the piston is a perfect conductor TA,i = TB,i and TA,f = TB,f. Now for the system consisting of both parts, Wx = 0 because it is rigid, and Q = 0 because it is insulated. The energy (first law) equation then shows that DU = 0. Since energy is additive over systems, this means DUA + DUB = 0. The energy (first law) equation for part A is DUA = QA – Wx,A. Similarly, for the part B it is DUB = QB – Wx,B. Consequently, the above relation becomes QA – Wx,A + QB – Wx,B = 0; or QA + QB = Wx,A + Wx,B. In other words, all that can be said is that the total energy of the combined system is constant, and that it is not possible to evaluate the heat transferred and work done for each of the gas compartments separately. A careful analysis of the statement like “expansion work is first done which is converted into internal energy and is transferred back as heat”, etc. will reveal that essentially it is a description of the non-equilibrium process being executed with the conclusion that the algebraic sum of the work done and the heat (nothing other than the energy stored) is constant. These ideas may be summarized as follows: ∑ If the working substance is known, DU can be calculated from the measured change of state. ∑ In addition, if it is known that the processes are quasi-static then work can be calculated using the appropriate forces and displacements. Then the energy (first law) equation gives the value, Q = DU + W. ∑ If the process is known to be non-quasi-static and Q can be calculated in some way, then W can be evaluated using the energy (first law) equation since W = Q – DE, where DE can be calculated from the initial and final states.

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∑ If the processes are known to be non-quasi-static and there is no data to calculate Q, then only Q – W can be calculated; since the energy (first law) equation gives DE = Q –W and DE can be calculated from the initial and final states. The above discussion shows that the only way to measure heat (other than using the empirical laws like the Fourier’s law) is by using the first law. Now, no work is done when the system boundaries are fixed, also no work is done in the absence of flow of electric charges and when the fields are steady. Then, the energy (first law) equation shows that DE = Q. Consequently, when experiments are conducted under these conditions, the change in the system energy is solely because of heat flow.

5.6

INDEPENDENT STATE VARIABLES

During the discussion in Chapter 2 on primitive concepts and state of a system, it was mentioned that independent state variables and their numbers are chosen on the basis of the state principle and the energy (first law) equation. This is described now. The state principle basically asserts that the state of a system is uniquely defined by its energy. However, energy is determined by the energy (first law) equation, which can be written as n

n

i

i

dE = dQ – dW = dQ – ∑ dWi = dQ – ∑ Xi dYi This shows that the value of E depends on (n + 1) variables, since it is the algebraic sum of n work modes plus heat. Thus, the number of independent properties required to determine E uniquely is one more than the number of work modes. The property corresponding to dQ is always chosen as temperature T. Then, this equation also shows that the (n + 1) variables can be chosen out of the nXi’s, nYi’s and T. These ideas are summarized in the following rule: Independent state variables. The number of independent state variables (properties) required for uniquely determining the state of a system admitting n types of reversible work mode is n + 1, out of nXi’s, nYi’s and T. Two comments on the above rule are relevant here: (a) Ideal gas is a special case since its equation of state, written as T = (1/R) · (pv), is really the definition of temperature T. (b) Stirrer work is not included since it is basically irreversible9 (and t as well as q are not properties of the system).

5.7

THERMODYNAMIC MODELS OF WORKING SUBSTANCES

Based on the state principle, the following important thermodynamic models of working substance may be defined. 9

The Carathéodory’s formulation of the second law of thermodynamics discussed in Chapter 7 uses this fact as the basis.

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Pure substance. This is a thermodynamic model of a homogeneous substance of uniform composition of one chemical species. It is defined as one which stores energy only as internal energy, i.e. E = U. Simple substance. properties.

One that permits only one work mode. Hence its state is defined by two

Simple compressible substance. A simple compressible substance is one that permits only expansion work, dWx = pdv. Then, the state of this substance is defined by any two of p, v, T, e, h, s, a, and g. This is the thermodynamic model of a gas and a vapour. Simple incompressible substance. A simple incompressible substance is one that does not permit any expansion work. Then, the number of independent properties is one, which cannot be p or v (since the term incompressible means dv = 0 and dp Æ •). This is a degenerate case of the above. Its state is defined by one of T, u and s. This is the thermodynamic model of a liquid (and a solid).

5.8

THE FIRST LAW FOR CYCLES

In Chapter 2, a cycle was defined as a series of processes such that the final state of the last process coincides with the initial state of the first one. Applying this definition to the first law,

v∫

v∫

v∫

v∫

dU = 0 . Hence Eq. (5.1) gives, dQ = dU + dW. However, since U is a property, dQ = dW. Conventionally the net heat transferred to the system, Q, is divided into, Q1, the total heat absorbed and, Q2, the total heat rejected. Hence the final form of the energy (first law) equation for a cycle is

v∫

v∫

v∫ dQ = v∫ dW

(integral form)

Q1 – Q2 = W

(finite form)

(5.2)

Many authors use the integral stated above as the first law of thermodynamics.10 It is clear that this has the disadvantage that Q, the energy flux in the form of heat, will have to be operationally defined first. However, if it is defined either explicitly (as is done in this and some other books) or implicitly (as is done in many other books) as ‘all the energy flux other than work’, then this equation is simply the principle of energy conservation. Then, the first law is superfluous. Alternatively, if the conservation of energy principle is assumed, then this equation simply becomes the operational definition of heat, namely, as ‘all the energy other than work’. This is done in this book. Then, the Born–Carathéodory form of the first law, namely, dE = dQ – dW, follows as a corollary of the energy conservation principle. 10

Traditionally, this is known as the Planck-Poincare formulation.

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The First Law of Thermodynamics

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EQUILIBRIUM

The first law gives a criterion for testing whether a system will be in equilibrium or not. Rewriting the Born–Carathéodory form of first law as Wad = –DU, it is seen that (a) during an adiabatic process, work is done by the system at the expense of internal energy and (b) the internal energy decreases (negative sign) when the system delivers work (positive sign). Now, as the system executes an adiabatic process and spontaneously delivers work, its energy keeps decreasing. The process goes on till the energy reaches a minimum value. This can be formally stated as: The condition of equilibrium. An adiabatic system reaches equilibrium when its energy becomes minimum. It is usual to assume that the process is reversible and write this mathematically as (DU)S = 0 Generally the test for the second derivative is mathematically more involved. Hence, only the first derivative is tested and the above physical argument is used to confirm that the energy is minimum.

5.10

EXAMPLES

This section presents some numerical examples as illustrations of the method of energy analysis of thermodynamic systems [i.e. application of the energy (first law) equation to systems]. Once again, the emphasis is on systematic problem-solving. In order to save space, units are not written adjacent to each quantity. In actual problem-solving this practice is recommended. Remember that closed and open systems are only models of real systems. The separation made here is artificial and is meant for convenience of learning only. EXAMPLE 5.1 Air at 2 [bar], 300 [K] contained in a rigid vessel of 0.5 [m3] volume is heated till the pressure rises to 5 [bar]. Calculate (a) DE and (b) Q. Solution

The system and the process diagram are shown in Figure 5.2. W p 2

Piston

Gas

Process V = constant Cylinder 1

(a)

Figure 5.2

(b)

V

Heating and stirring gas: (a) the system and (b) the process diagram.

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This problem is solved using the following steps: The system. Type.

The rigid vessel.

Closed.

Working substance. Initial state. Process.

Air assumed to be an ideal gas with M = 29 [kg/kmol] and g = 1.4.

p1 = 2 [bar], V1 = 0.5 [m3], and T1 = 300 [K].

Heating at constant volume.

Final state.

p2 = 5 [bar] and V2 = V1 (since the vessel is rigid).

Formulation of equations and their solution.

This is done in the following convenient steps:

1. Since air behaves like an ideal gas, the equation of state gives m = pV/RT = (200)(0.5)/ [(8.3143/29)(300)] = 1.1626 [kg]. 2. Also, for an ideal gas, cv = R/(g – 1) = (8.3143/29)/(1.4 – 1) = 0.71675 [kJ/kg·K]. 3. Since the volume is constant, the ideal gas equation gives, p2/T2 = p1/T1; or, T2 = T1 (p2/p1) = (300) (5/2) = 750 [K]. 4. Since U for an ideal gas is only a function of temperature, using the definition of cv gives DU = mcv DT = (1.1626) (0.71675) (750 – 300) = 374.98 ª 375.0 [kJ] [Ans. (a)]. 5. Energy (first law) equation for a closed system is Q = DU + W; but since the vessel is rigid (i.e. V is constant) Wx = 0 and no other work modes are mentioned. Hence, W = Wx = 0; so that the energy equation becomes, Q = DU = 375 [kJ] [Ans. (b)] EXAMPLE 5.2 A closed system contains saturated liquid water at 30 [°C]. Heat is added to it at constant pressure. Determine the final state (dryness fraction, enthalpy) and the change in internal energy, if the amount of heat added is 2000 [kJ/kg]. Solution The system and the process diagram are shown in Figure 5.3. The problem is solved using the following steps: The system. A cylinder–piston assembly (the typical closed system). Type. Closed. Working substance. Water. So properties are obtained from steam tables. Initial state. Saturated water at T1 = 30 [°C]. Process. Heating at constant pressure with heat, Q = 2000 [kJ/kg] added. Final state. p2 = 1 [bar] and h2 = h1 + 2000 [kJ/kg] (as shown below). Formulation of equations and its solution. It is convenient to use the following steps: 1. The energy equation for a closed system is Q = DE + W; steam is a pure substance and Ek = Ep = 0 and hence E = U; W = Wx (no other work mode is specified); Wx = ∫12 pdV (by definition); Wx = pDV = D(pV) (pressure is constant during evaporation). Thus, Q = DU + D(pV) = D(U + pV) = DH (by definition of H)

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W p

Saturated liquid line Piston 1

Water

(a) Figure 5.3

2. 3. 4. 5. 6. 7. 8. 9.

Saturated vapour line 2

Cylinder

(b) Evaporation of water: (a) the system and (b) the process diagram.

V

= H2 – H1; so that H2 = H1 + Q; or, h2 = h1 + (Q/m) = hf + (Q/m) (since the initial state of water is saturated). For T = 30 [°C], the T-based saturated tables gives hf = 125.66 [kJ/kg] so that h2 = 2000 + 125.66 = 2125.7 [kJ/kg] [Ans.] The above tables also show that hf < h2 < hg so that the steam in the final state is wet. Then, the standard formula (see Appendix F on steam tables) h2 = hf + x2(hg – hf) can be solved for x2 = (h2 – hf)/(hg – hf). Substituting the values from steam tables, x2 = (2125.7 – 125.7)/(2556.4 – 125.7) = 0.82281 [Ans.] Using the similar formula, v2 = vf + x2(vg – vf) = 0.0010043 + (0.82281)(32.928 – 0.0010043) = 27.0937 [m3/kg]. Then, by definition, u2 = h2 – p2v2 = 2125.7 – (0.042415)(100)(27.0937) = 2010.8 [kJ/kg]. Similarly, u1 = 125.66 – (0.042415)(100)(0.0010043) = 127.7 [kJ/kg]. Hence, Du = 1885.1 [kJ/kg] [Ans.]

EXAMPLE 5.3 Three kilograms of air in a rigid container changes its state from 5.0 [bar] and 60 [°C] to 15 [bar] while being stirred. The heat absorbed is 200 [kJ]. Determine the final temperature, change in internal energy, and work done. Solution The system and the process diagram are shown in Figure 5.4. This problem is solved using the following steps: The system. Type.

The rigid vessel.

Closed.

Working substance.

Air.

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W

p 2 Cylinder

Piston Process (V = constant) Gas

, d Stirrer 1

(a)

(b)

V

Figure 5.4 Heating a gas: (a) the system and (b) the process diagram.

Initial state.

m = 3 [kg], p1 = 5 [bar] and T1 = 60 [°C] = 333 [K].

Process. Stirring at constant volume (because the container is rigid) with heat Q = 200 [kJ/kg] being added. Final state.

p2 = 15 [bar] and V2 = V1 (rigid vessel).

Formulation of equations and their solution.

It is convenient to use the following steps:

1. The energy (first law) equation for a closed system is Q = DE + W. Air is a pure substance and Ek = Ep = 0 and hence E = U. Now, W = Wx + Ws (no other work mode is specified); Wx = ∫12 pdV (by definition) = 0 (volume is constant). Then, Q = DU + Ws; or, Ws = Q – DU. 2. Since air is assumed to behave like an ideal gas, its equation of state for constant volume reduces to p/T = C, or, p1/T1 = p2/T2; or, T2 = (p2/p1)T1 = (15/5) (333) = 999 [K] [Ans.] 3. Since air is an ideal gas, cv = R/(g – 1) = (8.3143/29)(1.4–1) = 0.71675 [kJ/kg·K]. 4. Again, since air behaves like an ideal gas DU = mcvDT = (3)(0.71675)(999 – 333) = 1432.1 [kJ] [Ans.] 5. Then, energy (first law) equation gives Ws = Q – DU = 200 – 1432.1 = –1232.1 [kJ] [Ans.] It was mentioned earlier that work done by a force can be evaluated as W = ∫12 XdY only for a quasi-static process. The work done in non-quasi-static processes can be evaluated only by using the energy (first law) equation. This is illustrated in the following example. EXAMPLE 5.4 A closed system executes a quasi-static process absorbing 80 [kJ] of heat and expanding from 2 [m3] to 2.25 [m3] against a constant pressure of 2 [bar]. A non-quasi-static process is used to bring the system to its initial state during which it absorbs 120 [kJ] of heat. Calculate the work done in the second process.

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The system and the process diagrams are shown in Figure 5.5. p

W

Quasi-static process, A 1

2

Piston

Unspecified substance

Cylinder

Non-quasi-static process, B

V (a)

(b)

Figure 5.5 Gas expansion: (a) the system and (b) the process diagram.

The problem is solved using the following steps: The system. Type.

A cylinder–piston assembly (the typical closed system).

Closed.

Working substance. Initial state.

Not mentioned.

p1 = 2 [bar] and V1 = 2 [m3].

Processes. Forward quasi-static constant pressure expansion (labelled A) with heat absorption QA = 80 [kJ]. A reverse non-quasi-static process (labelled B) with heat absorption QB = 120 [kJ]. Final state.

p2 = 2 [bar] (isobaric forward process) and V2 = 2.25 [m3].

Formulation of equations and their solution.

It is convenient to use following steps:

1. The process ‘1A2’ and the process ‘2B1’ form a cycle. Then, the energy (first law) equation becomes

v∫ dQ = v∫ dW (since v∫ dE = 0 , because E is a property); so that

QA + QB = WA + WB; or, WB = QA + QB – WA. 2. For the first process (i.e. process A), W = Wx (no other work modes are given) = ∫12 pdV (quasi-static process) = pDV (isobaric process) = (200)(2.25 – 2.00) = 50 [kJ]. 3. Substituting in the energy (first law) equation WB = 80 + 120 – 50 = 150 [kJ] [Ans.] EXAMPLE 5.5 The energy (in[J]) of a closed system can be expressed as E = 100 + 50T + 0.04T2. The heat absorbed (in [J]) is given by Q = 5000 + 20T. The temperature in these relations is in [K]. Calculate the work done during a process when the temperature rises from 500 [K] to 1000 [K].

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Solution In this case, the system is a closed one. So it can be represented as shown in the preceding example. However, since the process is not specified, the process diagram cannot be drawn. Hence, only the solution steps are indicated below. The system. Type.

A cylinder–piston assembly (the typical closed system).

Closed.

Working substance. is in [K]. Initial state.

Unknown, but with E = 100 + 50T + 0.04T2, where E is in [J] and T

T1 = 500 [K]. Unspecified; but with Q = 5000 + 20T, where Q is in [J] and T is in [K].

Processes. Final state.

T2 = 1000 [K].

Formulation of equations and their solution. Since the details of the process are not specified, W will have to be evaluated from the energy (first law) equation as W = Q – DE = =

∫

1000

∫

1000

500

500

[5000 + 20T – 100 – 50T – 0.04T2]dT [4900 – 30T – 0.04T2] dT = –20,467 [kJ] [Ans.]

EXAMPLE 5.6 Two kilograms of steam at critical conditions is sealed in a rigid container. It is then placed inside an oven which is maintained at 350 [°C] and allowed to reach thermal equilibrium. Determine the masses of water and steam in the final state, and the heat transferred. Solution

The system and the process diagram are shown in Figure 5.6. W

1

p Saturated liquid line

Saturated vapour line

Piston 1 Steam

(a)

2

Cylinder

(b)

Figure 5.6 Condensation of steam: (a) the system and (b) the process diagram.

V

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101

The problem is solved using the following steps: The system. Type.

The sealed rigid vessel.

Closed.

Working substance. Initial state. Process.

Steam. Its properties are obtained from the steam tables.

Steam at critical conditions.

Cooling at constant volume (since the vessel is rigid).

Final state. system)].

T2 = 350 [°C] and v2 = v1 [(since V (rigid container) and m are constant (closed

Formulation of equations and its solution. It is convenient to use the following steps: 1. The energy (first law) equation for a closed system is Q = DE + W; steam is a pure substance and Ek = Ep = 0 and hence E = U; W = Wx (no other work mode is specified); Wx = ∫ 12 pdV (by definition) = 0 (volume is constant). Then, Q = DU. 2. Using the p-based saturation tables, v1 = 0.0031751 [m3] = v2. 3. The tables (and, in fact the process diagram too) show that the steam at the final state will be wet. 4. Since the tables do not have an entry corresponding to T = 350 [°C], the values will have to be interpolated (the linear rule will do). 5. Then, p2 = 166.7 [bar], vf = 0.0017495 [m3/kg], hf = 1677.7 [kJ/kg], vg = 0.0086746 [m3/kg]; and, hg = 2562.9 [kJ/kg]. 6. Then, the final dryness fraction is x2 = (0.0031751 – 0.0017495)/(0.0086746 – 0.0017495) = 0.2058. Now, using the definition of x, the mass of steam is, ms = m · x2 = (2)(0.2058) = 0.4116 [kg] [Ans.]; and, mw = m – ms = 1.5884 [kg] [Ans.] 7. Then, h2 = hf,2 + x2(hg,2 – hf,2) = 1677.7 + (0.2058)(2562.9 – 1677.7) = 1859.9 [kJ/ kg]; and, by definition, u2 = h2 – p2v2 = 1859.9 – (166.7) (100) (0.0031751) = 1807.0 [kJ/kg] [Ans.] 8. Similarly, u1 = 2108.3 – (221.2)(100)(0.0031751) = 2038.1 [kJ]. 9. Then, Q = DU = (2)(1807.0 – 2038.1) = – 462.2 [kJ] [Ans.] EXAMPLE 5.7 A rigid insulated vessel is divided into two halves by a membrane. One compartment initially contains an ideal gas at p and T. The other compartment is evacuated. The membrane is punctured and the gas is allowed to expand into the evacuated compartment. Calculate the final state of the gas. Solution This is known as free expansion since a restraining force does not resist the expansion of the gas. Figure 5.7 shows the system diagram.

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Engineering Thermodynamics Membrane

Rigid insulated walls

Ideal gas Part A

Evacuated Part B

System = Part A + Part B

Figure 5.7

The system.

It is convenient to do the calculations in tabular form as shown below. The system consists of both compartments as shown in Figure 5.7. No.

Governing equation

1. 2. 3. 4. 5. 6. 7.

Q = Q = E = W= Wx = DU = DT =

DE + W 0 U Wx 0 0 0

Reason First law for closed systems Insulated system Pure substance All other work modes are absent Rigid vessel By substituting in the first law The gas is ideal (Joule’s law)

This means that the temperature of an ideal gas does not change during a free expansion. Joule obtained this result experimentally. From this, he proposed that at constant temperature, the internal energy of an ideal gas does not depend on volume. This is now known as the Joule’s law. It should be remembered that the gas expands unrestricted into the evacuated chamber. This is basically a non-quasi-static process.11 Note that free expansion is different from the expansion against a resisting force (resisted expansion). Free expansion may be made quasistatic by choosing the resisting forces suitably.

REVIEW QUESTIONS 1. Why is the first law of thermodynamics needed? 2. What basic question does the first law answer? 3. What are the experiments results whose results form the basis of the Born– Carathéodory form of the first law? State these experiments and their results and show how Carathéodory reinterpreted them. What is the advantage of this interpretation? 4. State the mathematical form of the first law of thermodynamics. Explain why the negative sign is needed in it. 11

During the discussions on entropy, in Chapter 7, it is shown that this process is also irreversible.

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103

5. What are the characteristics of energy that follow directly from the first law? 6. What is the difference between internal energy and other forms of energy? 7. State the principle of conservation of energy and explain how it was modified because of the first law. 8. Show how heat can be defined on the basis of the principle of conservation of energy. What are the characteristics of heat? 9. What is PMM1 (i.e. Perpetual Motion Machine of the First Kind)? 10. Explain the difference between the physical and mathematical characteristics of energy and work/heat. 11. Use the energy (first law) equation and state the principle to obtain the number of independent variables needed to uniquely define the state of a substance. 12. What are the thermodynamic models of working substances? 13. Derive the energy (first law) equation for a cycle. 14. Based on the energy (first law) equation, derive the expression for the state of equilibrium.

EXERCISES Notes: 1. 2. 3. 4. 5.

The following are the default assumptions: A closed system is the cylinder–piston assembly. Non-flow processes are those which are executed by closed systems. All processes are quasi-static. Subscripts ‘1’ and ‘2’ denote the initial and final states, respectively. Air behaves like an ideal gas with molecular weight (RMM), M = 29 [kg/kmol] and g = 1.4.

Sign convention for W and Q. positive.

Work done by a system and heat absorbed by a system are

Net work and heat. Net work done and heat absorbed in a series of processes equal the algebraic sum of those in the individual processes. Solve each exercise systematically from first principles as illustrated in the examples. Write the unit of each quantity adjacent to it (see the examples in Chapter 1). Remember.

Each assumption and each data gives rise to one equation.

1. A mass of 2.5 [kg] of air at 2 [bar], and 26 [°C] in a closed system executes a constant pressure process with heat absorption of 650 [kJ]. Calculate (a) the final temperature, (b) the change in enthalpy, (c) the work done, and (d) the change in internal energy.

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2.

3.

4.

5.

6.

[Hints: Energy equation (first law) for closed systems gives dQ = dE + dW = dU + dWx = dU + pdV = dU + d(pV) = d(U + pV) = dH = mcpdT. Integrating, Q = mcpDT; or, T2 = T1 + (Q/mcp). For air, cp = [g R/(g –1)] = 1.003 [kJ/kg·K]. Substituting the values gives. (a) T2 = 285.2 [°C]. (b) DH = Q = 650 [kJ]. (c) Now, W = Wx = ∫12 pdV = p(V2 – V1). Since air is an ideal gas, V = mRT/p. Substituting the values, V1 = 1.07154 [m3] and V2 = 2.00045 [m3], then Wx = 185.782 = 185.8 [kJ]. (d) DU = DH – D(pV) = DH – pDV = Q – Wx = 464.2 [kJ].] Saturated water at 5 [bar] and 100 [°C] is supplied 1.5 [MJ/kg] of heat at constant pressure. Calculate h2. [Hints: Since water evaporates, the pressure is constant and only expansion work is done. Then, as shown in Exercise 1, the energy equation becomes Q = DH so that H2 = H1 + Q; or h2 = h1 + (Q/m). Since the working substance is water, steam tables should be used. Since at 5 [bar], Tsat = 151.84 [°C], state ‘1’ is subcooled water. So using the approximation illustrated in Example E.3 (Appendix E), we get h1 = 419.48 [kJ/kg]. Then, evaluating the above expression gives, h2 = 1919.5 [kJ]. Hence, the steam is wet.] A closed system containing 10 [kg] of liquid water at 1 [bar] and 30 [°C] absorbs 25 [MJ] of heat at constant pressure. Calculate U2. [Hints: Same as Exercise 2, so that h2 = h1 + (Q/m). Now, the tables for subcooled water do not have an entry for 1 [bar] and 30 [°C]. So h1 should be calculated by the approximate method (see Example F.3 (Appendix F)). Then, h1 = 125.66 + (1 – 0.042415)(100)(0.0010043) = 125.76 [kJ/kg], so that h2 = 125.76 + (25,000/10) = 2625.8 [kJ/kg]. The p-based saturation tables show that for p = 1 [bar], hf < h2 < hg so that the steam is wet. Then, the dryness fraction is evaluated from the expression h2 = hf + x2(hg – hf) as x2 = (h2 – hf)/(hg – hf) = 0.978. Then, v2 = vf + x2 (vg – vf) = 1.6565 [m3/kg]. Then, U2 = mu2 = m(h2 – p2v2) = 24.602 [MJ].] Determine DH, Q, W and DU when 1 [kg] of saturated water at 1 [bar] is completely evaporated into steam at constant pressure. [Hints: The energy equation (first law) for the closed system gives Q = Du + Wx = Du + pDv = Du + D(pv) = D(u + pv) = Dh. Substituting values from the p-based tables gives Dh = hg – hf = 2258 [kJ/kg] = Q. Now, W = Wx = pDv = p(vg – vf) = 169.3 [kJ]; and, Du = Dh – D(pv) = Du – pDv = Dh – Wx = 2089 [kJ].] A mass of 1 [kg] of air contained in a closed system at 1 [bar], 300 [K] is stirred with a constant torque of 1 [Nm] at a speed of 100 [rpm] till the volume doubles at constant pressure. The initial and final temperatures were found to be the same. If 10 [kJ] of heat is absorbed during the experiment, calculate its duration. [Hints: The energy equation (first law) gives Q = DU + Wx + Ws; or, Ws = Q – DU – pDV. Since air behaves ideally, DU = 0 as T1 = T2. Moreover, by the ideal gas law V1 = mRT1/p1 = 0.8601 [m3]. Then, Wx = pDV = 86.01 [kJ]. Now, Q = 10 [kJ] so that Ws = – 76.01 [kJ]. Since, by definition, Ws = t q Dt, substitution gives Dt = 121 [min].] A perfectly insulated vessel containing 0.05 [m3] of hydrogen at 25 [°C] and 5 [bar] is stirred at constant pressure till the temperature reaches 50 [°C]. Determine (a) the heat transferred, (b) the change in internal energy, (c) the net work done, and (d) the stirring work done. Assume hydrogen to be an ideal gas with M = 2 [kg/kmol] and g = 1.4.

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105

[Hints: Same as Exercise 5, since the system is insulated, put Q = 0 [Ans. (a)] (instead of DU) so that the energy equation (first law) gives W = –DU; and, Ws = W – Wx. Since H2 is an ideal gas, m = p1V1/RT1 = 0.02018 [kg]. Moreover, cv = [R/(g – 1)] = 10.393 [kJ/kg·K]. Then, DU = mcvDT = 5.2433 [kJ] [Ans. (b)]; and hence, W = – 5.2433 [kJ] [Ans. (c)]. Since p is constant, V2 = V1(T2/p1) = 0.054195 [m3]. Then, Wx = pDV = 2.0975 [kJ], and, Ws = 7.341 [kJ] [Ans.(d)].] 7. Three kilograms of air in a rigid vessel at 5 [bar] and 100 [°C] is stirred till the pressure reaches 15 [bar]. Meanwhile the air absorbs 200 [kJ] of heat. Calculate (a) T2, (b) DU, and (c) Ws. [Hints: Same as Exercise 6. Since the vessel is rigid, V is constant. Then, the ideal gas equation gives T2 = T1(p1/p2) = 1119 [K] [Ans. (a)]. Also, cv = [R/(g –1)] = 0.71675 [kJ/kg·K]. Then, DU = mcvDT = 1604.1 [kJ] [Ans. (b)]. Finally, the energy equation (first law) gives, W = Ws = Q – DU = –1404.1 [kJ] [Ans. (c)].]. 8. The work and heat exchange during a non-flow process are given by dQ = 80dT [J/K] and dW = 60dT [J/K]; where T is in [K]. Calculate the change of energy of the system during a process in which the temperature rises from 60 [°C] to 120 [°C]. [Hints: For closed systems, the energy equation gives DE = Q – W = ∫ 393 (dQ – dW)dT = ∫ 393 20dT = 1.2 [kJ].] 333 333 9. Half a kilogram of water (c = 4.18 [kJ/kg.K]) is stirred in an insulated vessel by a mass of 50 [kg] falling through 20 [m]. Determine the temperature rise of water. Assume g = 9.81 [m/s2]. [Hints: Consider the water as the system. It is closed. From the above data, Q = 0 (insulated); E = U (pure substance; Ek = Ep = … = 0); and, W = mgh. Hence solving the equation of the first law gives DT = 4.69 [K].] 10. Five kilograms of steam at 5 [bar], 0.5 dry, contained in a closed rigid system is adiabatically stirred till the steam becomes dry and saturated. If the stirrer delivers a torque of 1 [N.m] and its speed is 1000 [rpm], calculate the time for which the stirrer should be on. [Hints: The system is closed, rigid (Wx = 0), adiabatic (Q = 0), and for steam E = U. Hence, the energy (first law) equation becomes DU = – Ws = t q Dt. Moreover, since the water is evaporating, p remains constant at 5 [bar]. Since the steam is initially wet, the dryness fraction x1 is used to obtain h1 = 1693.8 [kJ/kg] and v1 = 0.18789 [m3/kg], so that u1 = h1 – pv1 = 1599.8 [kJ/kg]. Similarly, u2 = ug = hg – pvg = 2560.2 [kJ/kg]. Then, DU = m(u2 – u1) = 4802.0 [kJ] = –Ws. Using the above expression for Ws gives Dt = 764.3 [min].] 11. A bullet (m = 2 [g]) flying horizontally at 1 [km/s] strikes a fixed wooden block (m = 5 [kg], c = 0.14 [kJ/kg·K]) and is embedded in it. Assume that there is no heat loss from the block and that block does not change its volume. What will be the change in the temperature of the block? [Hints: Consider the wooden block and the bullet as the system. The initial state is when the bullet is about to strike the block. The final is the equilibrium state reached after the impact. This is a closed system. So, the first law is, Q = DE + W. For the above data, Q = 0 = W and E = U + Ek. By the definition of specific heat, DU = (m)(c)(DT). Substituting the values, DT = 1.43 [K].]

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12. A steel cylinder (V = 10 [cc]) contains an ideal gas at 25 [bar] and 300 [K]. It is placed inside another insulated evacuated cylinder of 2 [lit] volume. The gas from the smaller cylinder leaks and fills the larger one. At the final equilibrium state, determine the temperature and pressure of the gas. [Hints: Considering the larger insulated cylinder as the system, the first law gives DU = 0. Since the gas is an ideal one, this means DT = 0; so that T2 = T1 = 300 [K]. Then, the Boyle’s law gives p2 = (10/2000) × 25 = 0.125 [bar].] 13. An open rigid container of 0.1 [m3] volume (remember it will contain air) is heated to 427 [°C]. It is then sealed and cooled to 27 [°C]. Compute the heat to be extracted during the cooling process. [Hints: By the first law, Q = DE (because W = Wx = 0, there are no other work modes and the container is rigid). Since air is an ideal gas, DE = DU, and it is only a function of T. Using the definition of cv, DU = (m)(c v)(DT). For an ideal gas, cv = R/(g – 1) where R = R/M = 0.717 [kJ/kg·K]. From the ideal gas law, m = pV/(RT) = 0.04983 [kg]. Substituting in the energy (first law) equation gives Q = 14.29 [kJ].] 14. Two-and-a-half kilograms of saturated liquid water at 10 [bar] is mixed with 1 [kg] superheated steam at 10 [bar], 250 [°C]. The mixing is adiabatic and at constant pressure. Determine the dryness fraction of the final state, the change in volume, the work done and the change in internal energy. [Hints: Let A denote the saturated water at 10 [bar] and B denote the superheated steam at 10 [bar] and 250 [°C]. Now, making the same assumption as earlier the energy (first law) equation becomes 0 = Q = DU + Wx = DU + pDV = DU + D(pV) = DH, so that H2 = H1; or, mAhA + mBhB = (mA + mB)h2. Substituting the values from steam tables and solving gives h2 = 1385.6 [kJ/kg]. Since the steam is wet, x2 = (h2 – hf)/(hg – hf) = 0.30939, so that v2 = 0.060890 [m3/kg]; and, V2 = 3.5 v2 = 0.213115 [m3]. Now, V1 = (2.5)(0.0011274) + (1) (0.23275) = 0.235681 [m3]. Then, Wx = pDV = – 22.57 [kJ]; and, DU = – Wx (since DH = 0).] 15. One kilogram of saturated liquid water at 1 [bar] in a rigid adiabatic container is mixed with steam at 1 [bar] and 300 [°C]. Determine (a) p2, (b) DU, and (c) W. [Hints: Same as Exercise 14, but now the total volume remains constant because the vessel is rigid so that mAvA + mBvB = (mA + mB)v2, which may be rewritten as v2 = (mAvA + mBvB)/(mA + mB). Since the mixture is likely to be wet (this should be verified later) x2 = (v2 – vf)/(vg – vf) (Eq. I). A second equation is obtained from the energy (first law) equation which reduces to DU = 0; or, U2 = U1; or, (mA + mB)u2 = mAuA + mBuB, which may be rewritten as u2 = (mAuA + mBuB)/(mA + mB). Once again, since the mixture is likely to be wet, x2 = (u2 – uf)/(ug – uf) (Eq. II). Then, Eqs. I and II should be solved simultaneously (by trial and error). The data from the steam tables gives, v2 = 1.3198 [m3/kg] and u2 = 1614.0 [kJ/kg], such that trial and error gives p2 = 0.73 [bar] [Ans. (a)]. Then, by energy (first law) equation, DU = 0 [Ans. (b)]; and by data W = 0 (rigid vessel) [Ans. (c)].]

Chapter 6

Energy (First Law) Equation for Flow Systems

6.1

INTRODUCTION

In this chapter, the energy (first law) equation is extended to flow1 systems. A typical flow system (consisting of one inlet and one exit) is shown in Figure 6.1. Wx

Piston Cylinder

System W

Q Inlet i

Exit e

Figure 6.1 A flow system.

The subscripts i and e are used to denote the quantities at the inlet and exit, respectively. The system itself is called control volume. The subscript CV indicates quantities corresponding to it. If the inlet and exit valves of a flow system are closed, it becomes a closed system. Consider the flow system executing a general process (called the flow process). Let it exchange heat at the rate Q . Let the rate of work done be W ¢. Let the rates of inflow and outflow of mass from the system be m i and m e , respectively. Let the rate of change of energy of the system be E . 1

Also called bulk flow. As explained in Chapter 4, all flow systems are open systems but all open systems need not be flow systems because the mass in the system can change (definition of an open system) owing to mass diffusion and chemical reaction. 107

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In Chapter 2 on concepts, it was shown that if the system is taken as the control volume plus the inlet mass at t and as the control volume plus the exit mass at time t + dt, then the original flow system is equivalent to a closed one. In that chapter, this was proved using the continuity equation (i.e. the equation of conservation of mass), namely m CV = m i − m e To use this principle for writing the energy (first law) equation, note that DE = ECV,t+dt + dEe – ECV,t – dEi where DE is the net change of energy of the equivalent closed system (defined above). Then, the energy (first law) equation for this equivalent closed system, namely dE = dQ – dW¢ becomes

ECV,t+dt + dEe – ECV,t – dEi = dQ – dW¢

which can be rewritten as ECV,t+dt – ECV,t = dQ – dW¢ + dEi – dEe Dividing both sides by dt and taking the limit d t Æ 0, this equation becomes

ECV = Q − W ′ + E i − E e which is rewritten in terms of the specific energies as

ECV = Q − W ′ + m i ei − m e ee In thermal systems of our prime interest, the most common forms of energy associated with a flowing element are internal energy, kinetic energy and potential energy2, so that e = u + V2/2 + gz. The W¢ is the algebraic sum of all the modes of work done (such as shaft work Ws, expansion work Wx, etc.). Out of these, one type of work depends directly on the mass flow rate. It is called the flow work, and is denoted as Wf l. It is the work done in pushing the fluid into the control volume. The relation for flow work is obtained using the following arguments. Let p be the pressure inside the control volume. Let d m be the mass being pushed in. If v is the specific volume of the fluid then the volume occupied by this mass element will be (dm)v. If A is the area of cross-section of the inlet, then the length of this mass element will be dx = (dm)(v/A). The work done in pushing the fluid in is dWfl = –Fdx = – (pA)dx = – (pA)(dm)(v/A) = –(pv)(dm) Dividing this by d m, the work done in pushing unit mass of the fluid into the control volume is obtained as (Wf l/m) = –pv. 2

For example, in the analysis of MHD (Magneto Hydro Dynamic) systems the energy stored and the work done by the electromagnetic fields should also be included.

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Energy (First Law) Equation for Flow Systems

109

It should be noted that the flow work pivi per unit mass is done (by the pump) on the fluid element at the inlet when it is pumped in. Then, when it flows out of the system at the exit, it can do work peve per unit mass on the environment (e.g. on the piston of an engine, blades of a turbine, etc.). Let W denote the net work (i.e. algebraic sum of all work modes) done by the system, except the flow work Wf l, i.e. W = W¢ – Wf l. Then, W¢ = W + Wf l = W – (mpv)i + (mpv)e Now, substituting the expressions derived above for the specific energy of a fluid element and the flow work done by it, the energy (first law) equation becomes

⎛ ⎞ ⎛ ⎞ V2 V2 E CV = Q − W + m i ⎜⎜ ui + pi vi + i + gzi ⎟⎟ − m e ⎜⎜ ue + pe ve + e + gze ⎟⎟ 2 2 ⎝ ⎠ ⎝ ⎠ In terms of enthalpy (defined as h = u + pv), this equation becomes

⎛ ⎞ ⎛ ⎞ V2 V2 ECV = Q − W + m i ⎜⎜ hi + i + gzi ⎟⎟ − m e ⎜⎜ he + e + gze ⎟⎟ 2 2 ⎝ ⎠ ⎝ ⎠ To understand this equation better as general form of the energy conservation equation, rewrite it as ⎡ E CV = [Q + (−W )] + ⎢ m i ⎣⎢

⎛ ⎞ ⎛ ⎞⎤ Vi2 V2 + gzi ⎟⎟ − m e ⎜⎜ he + e + gze ⎟⎟ ⎥ ⎜⎜ hi + 2 2 ⎝ ⎠ ⎝ ⎠ ⎦⎥

Then, the left most term on the left-hand side is seen to be the rate of increase of the energy stored in the system (i.e. the control volume). The terms in the first brackets on the right-hand side represent the net energy flow into the system that is independent of the mass flow.3 Note that a negative sign is attached to the work term, since by convention the work done by the system (i.e. work flow out of the system) is taken as positive. The term in the second brackets is the net energy flowing into the system because of the mass (bulk) flow.4 Thus, the right-hand side represents the net flow of energy into the system by flow as well as non-flow processes. Then, this equation can be written, in words as Rate of increase of energy of a system = Net energy flowing into it This is the form of the energy conservation equation stated in Chapters 2 and 3 as the form derived in mechanics (and hence primitive to thermodynamics). Notice that the continuity equation, namely m CV = m i − m e

relates the mass flows m i and m e . 3 4

In heat transfer, this mode of energy flow (the heat part) is known as heat conduction (and as heat radiation). In heat transfer, this mode of energy transfer is known as heat convection. Thermodynamics (the nonequilibrium part) does not use this term.

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Thus, it is seen that the behaviour of a flow system is determined by the pair of equations, namely the continuity equation (abbreviated as ‘ce’) m CV = m i − m e

(6.1)

and the energy equation (abbreviated as ‘ee’)

⎛ ⎞ ⎛ ⎞ V2 V2 ECV = Q − W + m i ⎜⎜ hi + i + gzi ⎟⎟ − m e ⎜⎜ he + e + gze ⎟⎟ 2 2 ⎝ ⎠ ⎝ ⎠

(6.2)

6.2 REFERENCE QUANTITIES It should be noted that the energy (first law) equation contains DU. Consequently, when it is extended to flow systems the effect of the energy at reference state should be determined. Following discussion deals with this aspect. The derivation for flow work presented above does not involve the reference (say, ambient) pressure since the same pressure pi is assumed to act on both sides of the piston pushing the fluid in (the assumption of local equilibrium). Thus, these terms are not affected by the choice of reference. The fluxes (heat and work) also do not contain any reference terms since it is their actual values (and not differences) that are taken. Then, using the subscript 0 (zero) to denote the reference quantities, the energy (first law) equation becomes ⎡ ⎤ V2 d ( mCV eCV ) = Q − W + m i ⎢(ui − u0 ) + i + g(zi − z0) ⎥ − dt 2 ⎣ ⎦

⎡ ⎤ V2 m e ⎢(ue − u0 ) + e + g( ze − z0 ) ⎥ 2 ⎣ ⎦ and differentiating the left-hand side this beomes ⎡ ⎤ V2 ( mCV )eCV + m CV (eCV ) = Q − W + m i ⎢(ui − u0 ) + i + g( zi − z0 ) ⎥ − 2 ⎣ ⎦ 2 ⎡ ⎤ V m e ⎢(ue − u0 ) + e + g( ze − z0 ) ⎥ 2 ⎣ ⎦ Since reference quantities are constant, they do not affect the first term on the left-hand side. Similarly, since Q and W are energy fluxes they are also not affected by the reference quantities. It was argued that they do not affect the expression for flow work. Thus, for further analysis, the above equation can be simplified as ⎡ ⎤ ⎡ ⎤ V2 V2 m CV ⎢(u − u0 ) + + g( z − z0 ) ⎥ = Q − W + m i ⎢(ui − u0 ) + i + g( zi − z0 ) ⎥ − 2 2 ⎣ ⎦CV ⎣ ⎦ ⎡ ⎤ V2 m e ⎢(ue − u0 ) + e + g( ze − z0 ) ⎥ 2 ⎣ ⎦

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Energy (First Law) Equation for Flow Systems

111

Consideration, for clarity, of only the terms involving the reference quantities5 on both sides gives m CV (u0 + gz0 )CV = m i (u0 + gz0 ) − m e (u0 + gz0 )

where the common negative sign has been cancelled out. Using the continuity equation (Eq. 6.1), this is seen to be an identity since the left-hand side is always equal to the right-hand side. This means that the choice of reference quantities do not affect the form of energy (first law) equation for a flow system.

6.3 SIMPLIFICATION OF ENERGY (FIRST LAW) EQUATION In actual applications, it is a usual practice to simplify the energy equation. This simplification is based on the following assumptions. Steady flow. It is defined as m CV = 0. Then, the continuity (conservation of mass) equation (Eq. 6.1) becomes m i = m e = m

(say)

Steady state. It is defined (as per the state principle) as E CV = 0, so that the energy (first law) equation (Eq. 6.2) reduces to

⎡ ⎤ V 2 − Ve2 0 = Q − W + m ⎢(hi − he ) + i + g( zi − ze )⎥ 2 ⎣ ⎦ This is known as the steady flow energy equation6 (abbreviated SFEE). Adiabatic flow (or process).

It is defined as Q = 0.

It should be pointed out that the conditions of steady flow and steady state are independent of each other. This can be easily understood if the following physical meanings of the various terms of the energy equation are recognized. (a) E CV represents the energy stored in the system. (b) Q represents the energy inflow in the absence of fluid flow (called bulk flow) and of work being done. (c) – W represents the energy inflow in the absence of fluid flow (called bulk flow) under adiabatic conditions. (d) As a consequence of (b) and (c), Q + (– W ) represents the energy inflow in the absence of bulk flow. 5 6

The other terms satisfy the energy (first law) equation anyway. However, it should properly be called the steady flow steady state energy equation (SFSSEE), but this terminology is cumbersome.

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⎛ ⎞ V2 (e) m i ⎜ hi + i + gzi ⎟ represents the energy inflow owing to bulk inflow of the fluid. ⎜ ⎟ 2 ⎝ ⎠ ⎛ ⎞ V2 (f) m e ⎜⎜ he + e + gze ⎟⎟ represents the energy outflow due to bulk outflow of the fluid. 2 ⎝ ⎠ Thus, the energy equation shows that the energy of the system can be changed by one (or more) of the following: (a) Inflow of heat Q. (b) Inflow of work –W.

⎛ ⎞ V2 (c) Changing the energy ⎜ hi + i + gzi ⎟ of the incoming fluid ⎜ ⎟ 2 ⎝ ⎠ ⎛ ⎞ V2 (d) Changing the energy ⎜ he + e + gze ⎟ of the outgoing fluid. ⎜ ⎟ 2 ⎝ ⎠ For thermodynamic (energy) analysis, these equations can be simplified further. They are presented here only as illustrations. In solving any problem, it is always convenient to begin with the main equation and simplify it according to its requirements rather than remember a dozen equations (which are really the same). Hence a tabular form (see Table 6.1) is used for this purpose. Table 6.1 Device Turbines, Compressors

2

Boilers, Heat exchangers Nozzles, Diffusers4 5

Throttle Valves

3

Simplified forms of SFEE

Assumptions1

Equation

Q = 0 = Dek = Dep

W = m (–Dh)

W = 0 = Dek = Dep

Q = m (Dh)

Q=0= W

0 = Dh + Dek + Dep

Q = 0 = W = Dek = Dep 2

where: Dh = he – hi; Dep = gDz = g(ze – zi); Dek = D(V /2) =

hi = he (Ve2

– Vi2)/2

1. These are default (unless stated otherwise) assumptions. 2. These absorb or deliver work and, therefore, are called work transfer devices. 3. These absorb or deliver heat and, therefore, are called heat transfer devices. 4. These convert enthalpy into kinetic and potential energies. 5. Also capillaries used in refrigerators. These only reduce the pressure.

Finally, consider a steady adiabatic flow of a simple incompressible fluid with zero work transfer. For this, Q = 0, W = 0, and DU = 0. Then, the energy equation reduces to 0 = D(pv) + Dek + Dep. This is recognized as the Bernoulli’s equation, since by definition v = 1/r.

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113

6.4 EXAMPLES This section presents some numerical examples as illustrations of the method of energy analysis of thermodynamic systems. Once again, the emphasis is on systematic problem-solving. In order to save space, units are not written adjacent to each quantity. In actual problem-solving, this practice is recommended. Remember that closed and open systems are only the models of real systems. The separation made here is artificial and is meant for convenience of learning only. EXAMPLE 6.1 A fluid passes through a turbine with hi = 2.8 [MJ/kg] and he = 2.2 [MJ/kg]. The entrance and exit velocities are 180 [m/s], and 300 [m/s], respectively. If the heat loss is 20 [kJ/kg], what is the work done? Solution

The system and the process diagram are shown in Figure 6.2. Exit e

Turbine Inlet i

Shaft work, Ws

h i p = pi e

p = pe s

(a)

(b)

Figure 6.2 A turbine: (a) the system and (b) the process diagram.

It is convenient to solve this problem in the following steps: 1. By the default assumption, the flow is steady meaning m CV = 0. Then, the continuity equation gives m i = m e . Let m denote this common mass flow rate. 2. Again the default assumption of steady state implies E CV = 0 so that the energy equation (ee) [together with the continuity equation (ce)] reduces to ⎡⎛ ⎞ ⎛ ⎞⎤ V2 V2 0 = Q − W + m ⎢⎜⎜ hi + i + gzi ⎟⎟ − ⎜⎜ he + e + gze ⎟⎟ ⎥ 2 2 ⎠ ⎝ ⎠ ⎦⎥ ⎣⎢⎝

which can be rewritten as

⎡ ⎤ V 2 − Ve2 0 = Q − W + m ⎢( hi − he ) + i + g( zi – ze ) ⎥ 2 ⎣ ⎦

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or as [since Dy = ye – yi (by definition of D)] ⎡ ⎛V2 0 = Q − W − m ⎢ Dh + D ⎜⎜ ⎝ 2 ⎣⎢

⎤ ⎞ ⎟⎟ + D( gz) ⎥ ⎠ ⎦⎥

or as [since Dx = xe – xi (by definition)] ⎡ ⎛ V2 0 = Q − W − m D ⎢ h + ⎜⎜ ⎝ 2 ⎣⎢

⎤ ⎞ ⎟⎟ + gz ⎥ ⎠ ⎦⎥

or, using the definitions of kinetic energy (ek) and potential energy (ep), as 0 = Q – W – m D(h + ek + ep) or, in terms of total quantities, namely, H, Ek and Ep, as 0 = Q – W – ( H + E k + E p ) which can be solved for power as W = Q – ( H + E k + E p )

3. Substituting the given data, this equation becomes 300 2 − 180 2 = 551.2 [kJ/kg] W = (–20) – (2200 – 2800) – 2000 where the kinetic energy term is divided by 1000 since the unit [m2/s2] is the same as [kg·m2/kg·s2], or [kg·m/s2] · [m/kg], or [N·m/kg], or [J/kg]. The same argument is also applicable to the potential energy term Ep. EXAMPLE 6.2 Saturated liquid water at 0.1 [bar] is pumped to a pressure of 20 [bar]. The pumping is isentropic. Determine the work done per kg of water pumped. Solution The system and the process diagram are shown in Figure 6.3. This example is also solved the same way as the above example (obviously, after all it is the same energy equation), but to save space the steps are combined. 1. The default assumption that the flow is steady (i.e. m CV = 0) reduces the continuity equation to m i = m e = m . 2. Again the default assumption of steady state implies E CV = 0 so that the energy and continuity equations reduce to ⎡⎛ ⎞ ⎛ ⎞⎤ V2 V2 0 = Q − W + m ⎢⎜⎜ hi + i + gzi ⎟⎟ − ⎜⎜ he + e + gze ⎟⎟ ⎥ 2 2 ⎠ ⎝ ⎠ ⎦⎥ ⎣⎢⎝

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115

Saturated liquid line h

Inlet i

p = 20 [bar] Shaft work, Ws

Pump

Exit e

e

i

p = 0.1 [bar]

(b)

(a)

s

Figure 6.3 A pump: (a) the system and (b) the process diagram.

3. Using the default assumptions for work transfer devices [i.e. adiabatic ( Q = 0) with negligible changes in ek and ep], this gives

W = hi − he m 4. The inlet state is saturated water at 0.1 [bar]. The steam tables give hi = hf = 191.83 [kJ/kg]. 5. Since the compression is isentropic (default assumption) se = si; and since water is an incompressible fluid (default assumption) this means that Te = Ti = 45.83 [°C]. 6. Then, the exit state is 20 [bar] and 45.83 [°C] which means that it is subcooled. 7. Since the tables do not have entry for this state, the enthalpy should be estimated using the approximate formula (see Section F.1.2, Appendix F). 8. Thus, he = ue + peve (definition of h); but since the fluid is incompressible, ue and ve will correspond to those of the saturated liquid at the specified temperature. Since Ti = Te = 45.83 [°C], this means that ue = ui and ve = vi, so that he = ui + pevi = hi – pivi + pevi = hi + (pe – pi)vi. 9. Then, the energy equation becomes

W = hi − he = − ( pe − pi )vi m 10. Reading off the values from the p-based saturation tables (since p is specified) and substituting in this equation gives

W = (20 – 0.1) (100) (0.0010102) = 2.0103 [kJ/kg] m EXAMPLE 6.3 An air compressor has an inlet state of 1 [bar], 300 [K]. It compresses 10 [kg/s] of air to a pressure of 10 [bar], with an isentropic efficiency of 0.75. For the actual process, determine (a) the power consumed and (b) the exit temperature.

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Solution

The system and the process diagram are shown in Figure 6.4. p = 10 [bar]

h Inlet i

Compressor

e¢ e

Shaft work, Ws

i

p = 1 [bar]

Exit e

s (a)

Figure 6.4

(b)

A compressor: (a) the system and (b) the process diagram.

Once again this example is solved the same way as the above one (including combining the steps). 1. The default assumption of steady adiabatic flow with negligible changes in kinetic and potential energies reduces the energy equation (together with the continuity equation) to

W = hi – he m 2. For an ideal gas the enthalpy is only a function of temperature. Then, using the definition of cp the energy equation becomes

W = hi – he = cp(Ti – Te) m 3. Since air behaves like an ideal gas, using the expression for cp of an ideal gas gives

⎛ g ⎞ ⎛ 1.4 ⎞ ⎛ 8.3143 ⎞ R=⎜ = 1.003 [kJ/kg.K] cp = ⎜ ⎝ 0.4 ⎟⎠ ⎜⎝ 29 ⎟⎠ ⎝ g − 1⎟⎠ 4. Since compression is an isentropic process of an ideal gas, this gives ⎛p ⎞ Te = Ti ⎜ e ⎟ ⎝ pi ⎠

(g −1) / g

= (300) × (10)0.4/1.4 = (300) (10)(1/3.5) = 579.21 [K]

5. Substituting the values, the energy equation in step 2 gives

W = (1.003) (300 – 579.21) = –280.05 [kJ/kg] m where the negative sign indicates that work is done on the system.

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6. The isentropic efficiency is defined as hs = Ws/W, where Ws denotes the work done during isentropic compression and W denotes the work done in actual compression. Then,

W W s /m 280.05 = =− = – 373.33 [kJ/kg] [Ans. (a)] hs m 0.75 7. Now, the energy equation shows that for the actual compression

W = –373.33 = cp(Ti – Te¢) m which can be solved for Te¢ as W /m Te′ = Ti − = 300 − ( −373.33 /1.003) = 672.2 [K] [Ans. (b)] cp EXAMPLE 6.4 Steam enters the nozzle of a steam turbine with a velocity of 3 [m/s] at 40 [bar], 300 [°C]. The nozzle exit was measured to be 12 [bar], 200 [oC]. Determine (a) the exit velocity and (b) isentropic efficiency of the nozzle. Solution The system and the process diagram are shown in Figure 6.5. This example is also solved the same way as the above one (including combining the steps). h

p = 40 [bar]

i

Nozzle Inlet i

Exit e

e

p = 12 [bar] e¢ s

(a)

(b)

Figure 6.5 A nozzle: (a) the system and (b) the process diagram.

1. The default assumptions for a nozzle are: steady flow ( m CV = 0), steady state ( E CV = 0), adiabatic flow ( Q = 0), no work interaction ( W = 0) and negligible potential energy change (Dep = 0). This reduces the energy and continuity equations to Ve2 − Vi2 = hi – he or Ve = Vi2 + 2(hi − he ) 2 2. For the actual inlet and exit states, the steam tables show that the steam is superheated. Hence reading off these values and substituting them values step 1 gives

Ve′ = (3)2 + (2)(2962.0 − 2814.4)(1000) = 543.33 [m 2 ][Ans. (a)]

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3. For the ideal case of isentropic flow, se = si. From steam tables, si = 6.3642 [kJ/kg·K]. 4. Now, saturated steam tables show that at 12 [bar], sg = 6.5194 [kJ/kg.K] so that the steam is wet. Then, using the standard formula (see notes in Appendix E) se = s f + x e (sg − s f )

xe =

or

se − s f sg − s f

5. Substituting the values, xe =

6.3642 − 2.2161 = 0.964 6.5194 − 2.2161

6. Then, using the standard formula he = hf + xe(hg – hf) = 798.43 + 0.964(2782.7 – 798.43) = 2711.3 [kJ/kg] 7. Now, using the definition of the isentropic efficiency, i.e. hs = (hi – he¢)/(hi – he) and substituting the values

hs =

2962.0 − 2814.4 = 0.5888 = 58.9% [Ans. (b)] 2962.0 − 2711.3

EXAMPLE 6.5 In a throttling calorimeter, the pressure of the mains is 7 [bar]. The state after throttling is 0.5 [bar], 100 [°C]. Determine the quality of steam in the mains. Solution Throttling is one of the important processes in thermal engineering. It is used for reducing the pressure without decreasing the energy of the fluid. Thermodynamically it is important because it is an inherently irreversible process. In other words, other processes can be made to approach reversible conditions by making them more and more quasi-static.7 However, throttling effect cannot exist when a process is quasi-static. To understand this, consider the throttling process taking place when a fluid (in this example, steam) flows through a valve. The system and the process diagram are shown in Figure 6.6. Outlet e

Inlet i

h

p = 7 [bar]

(i) Throttle valve

p = 0.5 [bar] i

Porous plug Outlet e

Inlet i (ii) Porous plug experiment (a)

Figure 6.6 7

e

s (b)

The throttling process: (a) the system and (b) the process diagram.

Of course, for simplicity, we assume that a quasi-static process is reversible.

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119

1. Basically, in a throttling process the fluid (steam in this case) flows through a valve across a finite pressure drop that increases the velocity of the fluid. The kinetic energy so produced is dissipated by fluid friction, so the exit temperature of the fluid rises (or remains constant—in a normal expansion it decreases). The porous plug is a solid (e.g. ceramic) manufactured with a large number of fine holes (micro-pores). Thus, it is equivalent to a number of micro-throttle valves. Because of the small diameters (with large surface to cross-sectional area), the porus plug offers very large frictional resistance to flow. Consequently, the velocity practically remains zero because the kinetic energy is dissipated very fast. 2. The inlet and exit of a throttle valve (or a porous plug) are generally connected to large sources so that Vi = Ve = 0, or Dek = 0. Moreover, they are generally horizontal, i.e. zi = ze = 0, or Dep = 0. 3. These devices operate at steady state ( E CV = 0), under adiabatic ( Q = 0) steady flow ( m CV = 0) with zero power output ( W = 0). 4. Under the conditions in step 3, the continuity and energy equations reduce to hi = he. 5. It should be noted that the equation in step 4 only shows that, in a throttling process, the enthalpies in the initial and final states are the same. It does not say anything about the intermediate states (which are non-equilibrium states). Thus, the throttling effect exists only owing to the process of viscous dissipation (which is an irreversible process). 6. A throttling calorimeter uses this principle to measure the quality of steam in a vessel (mains, boiler, …). If xi is the dryness fraction in the vessel then, by the usual formula, hi = hf + xi (hg – hf). 7. It should be noted that if xe is the quality of steam at exit, then the usual formula he = hf + xe(hg – hf) and the energy equation (viz. hi = he) will contain the two unknowns xi and xe and, therefore, these equations cannot be solved for xi. Hence, the exit state of the steam should at least be dry saturated. 8. The calorimeter is so operated that the exit state of the steam is generally superheated. Then, this state can be determined by measuring the pressure and temperature of the steam at the exit and he can be read off from the steam tables. 9. In the present example, the exit state of the steam is superheated since, pe = 0.5 [bar] (data) and Te = 100 [°C] (data) is greater than Tsat for this pressure. Then, the steam tables gives hi = 2682.6 [kJ/kg]. 10. Then, the above equations give he = hi = h f + xi (hg − h f )

or

xi =

he − h f hg − h f

11. Substituting the values read off from the steam tables xe =

2682.6 − 697.06 = 0.962 2762.0 − 697.06

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EXAMPLE 6.6 In a heat exchanger, 150 [kg/h] of saturated water enters at 7 [bar], and leaves at 5 [bar], 250 [°C]. Hot air enters at 2 [bar], 500 [°C] and leaves at 2 [bar], 250 [°C]. Determine (a) the flow rate of air and (b) the heat transfer rate from air to water. Solution

The system and the process diagram are shown in Figure 6.7. Hot fluid exit h,e

T

Cold fluid exit c,e

Th,i Hot fluid Tc,e

T h,e Cold fluid

Cold fluid inlet c,i

Tc,i Hot fluid inlet h, i

(a)

0 (b)

x

L

Figure 6.7 A heat exchanger: (a) the system and (b) the process diagram.

For the flow directions shown in Figure 6.7, the heat exchanger is called counter (current) flow heat exchanger. But, for thermodynamical analysis this is not relevant. 1. As per the default assumptions, the heat exchanger is an adiabatic device ( Q = 0) that operates at steady state ( E CV = 0) under steady flow ( m CV = 0) with no work output ( W = 0) and with negligible kinetic and potential energies (Dek = 0 = Dep). 2. Then, the continuity equation reduces to m i – m e = 0, which can be expanded as

(m c,i + m h,i ) − (m c,e + m h,e ) = 0 3. Similarly, the energy equation, H i − H e = 0 can be expanded as ( H c,i + H h ,i ) – ( H c,e + H h,e ) = 0 4. Since the individual streams flow steadily, their continuity equations are

m c,i = m c,e

and

m h,i = m h,e

5. The individual streams are non-adiabatic steady flow at steady state with no work output and with negligible kinetic and potential energies. Thus, their energy equations reduce to

Q h + H h ,i – H h ,e = 0

and

Q c + H c,i – H c,e = 0

Adding these equations and rearranging yields (Q h + Q c ) + ( H h ,i + H c,i ) − ( H h ,e + H c,e ) = 0 6. Comparison with the energy equation of the overall heat exchanger shows that

Q h + Q c = 0,

or

Q c = − Q h = Q (say)

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121

7. Thus, the energy and continuity equations for the overall heat exchanger can be written as H − H = H − H c ,e

c ,i

h, i

h ,e

8. In terms of the mass flow rates, this becomes

m c (hc,e − hc,i ) = m h (hh,i − hh, e ) 9. In the present example, air rejects heat (and hence is the hot fluid) and water absorbs heat (and hence is the cold fluid). Since the air flow rate is to be determined, the above equation can be rearranged as

m h = m c

hc,e − hc,i hh ,i − hh ,e

10. Since air behaves like an ideal gas its, h = cpT, where cp = Rg/(g – 1) = 1.003 [kJ/kg.K]. 11. The enthalpies of water and steam are read from steam tables as hw = 697.06 [kJ/kg] and hs = 2961.1 [kJ/kg]. 12. Substitution of these values and the given data in the relation in step 9 gives m a = 150

2961.1 − 697.06 = 1354.4 [kg/h] [Ans. (a)] 1.003(500 − 250)

13. To evaluate the rate of heat transferred, appropriate values are substituted in the energy equation of air or water. Then, using the water stream 150 Q = (2961.1 − 697.06) = 94.34 [kW] [Ans. (b)] 3600

EXAMPLE 6.7 A device for mixing steam has two inlets and two exits. The stream at the first inlet is steam at 5 [bar], 400 [°C] flowing at the rate of 0.01 [kg/s] with a velocity of 20 [m/s]. The second stream is steam at 1 [bar], 150 [°C] flowing at the rate of 0.1 [kg/s] with a velocity of 120 [m/s]. The stream at the first exit is 0.001 [kg/s] of saturated water at 80 [°C]. Heat loss from the device is 25 [kW]. Neglecting the exit velocities and differences in the heights of the inlets and exits of all streams, calculate the exit enthalpy of the second stream. Solution This device is a heat exchanger because it transfers heat from one fluid to another. But it is of a mixing type so that the continuity and energy equations cannot be written for each stream separately. Except for this, the same equations and procedure as those for Example 6.6 are applicable. The system diagram will only be a rectangular box with two inlets and two exits. The process diagram is complex since the states of two different streams are involved. Hence, to save space, they are not drawn. The following are the solution steps. Let subscripts ‘1’ and ‘2’ denote the two streams.

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1. The device is of mixing type. Hence the continuity equation, written for the exchanger as a whole, becomes

m 1,i + m 2,i = m 1,e + m 2, e which can be solved for the exit mass flow rate of the second stream as

m 2,e = m 1,i + m 2,i − m 1,e = 0.01 + 0.1 − 0.001 = 0.109 [kg/s] 2. Similarly, the energy equation becomes 0 = Q + H 1,i + H 2,i − H 1,e − H 2,e which can be solved for the exit enthalpy of the second stream as

⎛ 1 ⎞ h2,e = ⎜⎜ ⎟⎟ (Q + H 1,i + H 2,i − H 1, e ) m ⎝ 2,e ⎠ 3. Substituting the values reading off from the steam tables yields H 2,e = Q + H 1,i + H 2,i − H 1,e

or

⎛ 1 ⎞ h2,e = ⎜ × [( −25) + (0.01)(3272.1) + (0.1)(2776.3) − (0.001)(334.92)] ⎝ 0.109 ⎟⎠

= 2615.0 [kJ/kg] In Section 6.2, it was argued that steady flow does not imply steady state and vice versa. The following example, which is a typical everyday problem in the operation of any electrical power generation station, illustrates this. EXAMPLE 6.8 40 [°C].

Consider a steam turbine with inlet steam at 100 [bar], 400 [°C] and exit at

1. Since the objective here is to show the difference between the steady state and the steady flow conditions, the system and process diagrams are not drawn. 2. For operation under all the default assumptions, in terms of the usual notations, the continuity and energy equations give, W = m (hi – he). 3. From the steam tables, hi = 3099.9 [kJ/kg] and si = 6.2182 [kJ/kg.K]. 4. For isentropic flow, se = si. The T-based steam tables show that at Tsat = 40 [°C], sf = 0.57212 [kJ/kg.K] and sg = 8.2583 [kJ/kg.K]. Hence the steam is wet, so that the usual formula gives xe =

se − s f sg − s f

=

6.2188 − 0.57212 = 0.734 8.2583 − 0.57212

5. Then, substituting values from steam tables in another usual formula yields he = hf + (xe) (hg – hf) = 167.45 + (0.734) (2574.4 – 167.45) = 1934.2 [kJ/kg].

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123

6. Thus, the work output under steady flow, steady state operation is

W = hi − he = 3099.9 − 1934.2 = 1165.7 [kJ] m

7.

8. 9. 10.

11.

which means that steam flowing at the rate of 1 [kg/s] will produce 1.1657 [MW] of power. The conventional method of stating this is to say that the plant requires (1/1.1657) = 0.8578 [kg/s] of steam to produce 1 [MW] of power; or, the plant’s steam rate is 0.88 [kg/MW]. In view of the inefficiencies involved at various stages, this number is generally taken as 1 [kg/MWe], where [MWe] denotes [MW] of electrical energy produced. Now, let the electrical power demand be increased by 10% (due, for example, to lights being switched on in all houses, shops, … in the evening). This demand first loads the terminals of the alternator (electrical generator) and is then transmitted to the shaft of the turbine coupled to the alternator. Thus, the power output at the shaft of the turbine to meet this load should be 1.1 × 1165.7 = 1282.3 [kJ/kg]. Now, the mass of the steam flowing through the turbine is controlled by appropriate flow valves, and hence, it is not directly affected by the increase in power demand. The inlet state of the steam is determined by the conditions in the steam generator which are also not directly affected by the increase in load. Similarly, the state of the exit steam also remains unchanged since the conditions at the steam condenser have not changed. Thus, the power delivered by the steam, namely, m (hi – he) remains unchanged. Then, the energy equation for the new condition yields E CV W + (hi – he) = – 1282.3 + (3099.9 – 1934.2) =− m m = – 1282.3 + 1165.7 = – 116.6 [kJ/kg]

which shows that –116.6 [kJ/kg] as work output (i.e. the 10% increase in the load) is met at the expense of the energy stored in the control volume (which decreases as shown by the negative sign). 12. Increasing the input energy by 10% may restore the steady operating state at the new conditions. This can be achieved by several means. For example, by increasing the mass flow by 10%, or by increasing the temperature on the inlet steam so that hi is increasedº. The above example is an illustration of the unsteady state conditions occurring during a steady flow. The following two examples, on the other hand, illustrate the processes of unsteady state with unsteady flow. EXAMPLE 6.9 A rigid evacuated, insulated vessel is being adiabatically filled with steam from mains at 5 [bar], 200 [°C]. Determine the final state. Solution

The following are the steps involved in solving this example.

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1. The system (the rigid adiabatic vessel) is modelled as a flow system with one inlet and no outlet (because it is being filled). 2. The system diagram is trivial since it is simply a vessel (say, a box) with one inlet and no exit. The process diagram cannot be drawn since the mass inside the system varies. Hence the system and the process diagram are not drawn. 3. This is an unsteady process where only the initial and final states are defined since they are the equilibrium states. Let Dt denote the time taken for filling the vessel completely. 4. The continuity (mass conservation) equation for this case is, m CV = m i [no exit, i.e. m e = 0 (data)]. Integrating both sides with respect to time for the period Dt, this becomes DmCV = mi

or

m2 – m1 = mi

or

m2 = mi

where m2 and m1 are the respective mass contained in the vessel at the initial and final states. The last step (viz. m1 = 0) arises because of the vessel being in an evacuated state initially (data). 5. Similarly, the energy equation written for the initial and final states becomes [W = Wx = 0, rigid vessel (data)]; Q = 0 [insulated vessel (data)]; e1 = 0 [evacuated vessel (data)] and no radiation energy stored (default assumption); ek = 0 = ep (default assumptions); and e = u [pure substance (default assumption)] DUCV = Hi

or

U2 – U1 = Hi

or

U2 = Hi

or

u2 = h i

where the last step arises because m2 = mi (by continuity equation). 6. Steam tables show that at inlet (and in the mains) the steam (at 5 [bar], 200 [°C]) is superheated and the superheat tables give, hi = 2855.1 [kJ/kg]. 7. The condition of mechanical equilibrium shows that the flow will stop when the steam pressure in the vessel equals the pressure in the mains, i.e. p2 = p1 = 5 [bar]. 8. Substituting the values from steam tables in the definition of u reduces the energy equation to u2 = h2 – p2v2 = h2 – (500)v2 = h1 = 2855.1 [kJ/kg] which shows that the steam is still superheated at 5 [bar]. 9. Solving this equation by trial and error gives, T2 = 332 [°C]. 10. Note that the given data is insufficient for determining mi since it cancels out of the energy equation. Example 6.10, which illustrates another common industrial process, is a slight modification of this example. EXAMPLE 6.10 A balloon, initially evacuated, is being filled, adiabatically, with air from a cylinder. The air in the cylinder is at 1.5 [bar] and 300 [°C]. When the volume of the balloon becomes 0.015 [m3], the process of filling stops. At this condition, the pressure of the air inside the balloon is found to be 1.1 [bar]. Determine the final mass of the air in the balloon assuming that the atmospheric pressure is 1 [bar].

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125

Solution Since the procedure is similar to Example 6.9, only the main steps will be indicated. 1. The system, being the balloon, is modelled as a flow system with one inlet and no outlet (because it is being filled). 2. As earlier, the system and the process diagram are not drawn. 3. This is an unsteady process where only the initial and final states are defined since they are the equilibrium states. Let Dt denote the time taken for filling the balloon completely. CV = m i [no exit, i.e. 4. The continuity (mass conservation) equation for this case is m m e = 0 (data)]. Integrating both sides with respect to time for the period Dt, this becomes or m2 – m1 = mi or m2 = mi DmCV = mi where m2 and m1 are the respective mass contained in the baloon at the initial and final states. The last step (viz. m1 = 0) arises because of the balloon being in an evacuated state initially (data). 5. But the energy equation is different in the sense that because it takes into account the work done when the balloon expands against the atmospheric pressure. Thus, written for the initial and final states, it becomes [W = Wx = paDV = paV2 (V1 = 0, being evacuated balloon (data))]; Q = 0 [insulated vessel (data)]; e1 = 0 [evacuated vessel (data) and no radiation energy stored (default assumption)]; ek = 0 = ep (default assumptions); and, e = u (pure substance (default assumption)). DUCV = Hi – paV2

or

U 2 – U 1 = H i – pa V 2

or

U2 = Hi – paV2

or in terms of specific quantities m2u2 = mihi – paV2 6. Since air is an ideal gas, u2 = cvT2, hi = cpTi, V2 = (m2RT2)/p2, cv = R/(g – 1) and g = cp/cv. 7. Then, the energy equation given above becomes m2T2 = g m2Ti –

pa (g – 1)(m2T2) p2

where the continuity equation, namely, m2 = mi has been used. 8. This equation can be solved for T2 and then substituting the data gives T2 =

g Ti (1.4)(300) = = 308.0 [°C] 1 + (g − 1)( pa /p2 ) 1 + (0.4)(1.0/1.1)

9. Then, using the equation of state, m2 =

p2V2 (110)(0.015) = 0.01869 [kg] = RT2 (8.3143 / 29)308

REVIEW QUESTIONS 1. What is the difference between an open system and a flow system? 2. What is the closed system that is equivalent to a flow system? Why?

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3. Obtain an expression for the flow work. 4. Show how the reference quantities affect the equations of a flow system. 5. What are the simplified forms of the equations of the flow systems? Where are they applicable?

EXERCISES Notes:

The following are the default assumptions:

1. Flow processes are those processes that are executed by flow systems and non-flow processes are those that are executed by closed systems. 2. For this chapter, the terms ‘system’ ‘flow system’ and ‘control volume’ are synonymous. 3. The abbreviations ‘ce’ and ‘ee’ mean continuity equation and energy equation. 4. All processes are quasi-static. 5. Subscripts ‘1’ and ‘2’ denote, respectively, the initial and final states of the system. 6. Subscripts ‘i’ and ‘e’ denote the inlet and exit states of the system. 7. Air behaves like an ideal gas with molecular weight (RMM) of 29 [kg/kmol] and g = 1.4. 8. The value of the acceleration due to gravity is g = 9.81 [m/s2]. Write the unit of each quantity adjacent to it (see the examples in Chapter 1). 1. A steady flow system receives 45 [kg/min] of a fluid with negligible velocity and discharges it at a point 25 [m] above the inlet with a velocity of 2500 [m/min]. During the process it absorbs 15 [kW] of heat and increases its enthalpy by 8.5 [kJ/kg]. Determine the work done. [Hints: Since the flow is steady (i.e. m CV = 0), the ce becomes m i = m e = m (say) = (45/60) [kg/s] (data). For steady state operation (default assumption) ( E CV = 0), the ee becomes 0 = Q – W + m [(hi – he) + (Vi2 – Ve2)/2 + g(zi – ze)], which can be solved for W = Q + m [(hi – he) + (Vi2 – Ve2)/2 + g(zi – ze)]. Substituting the values and evaluating gives, W = 7.882 [kW].] 2. A PC cabinet dissipates 100 [W] of electric power. It is kept cool by circulating air through it. Determine the flow rate of air (cp = 1.0 [kJ/kg.K]) so that the temperature rise of air does not exceed 10 [°C]. [Hints: Using default assumptions (including air to be an ideal gas), the ce and ee become Q = m cp(Te – Ti), or m = Q /[cp(Te – Ti)] = 0.01 [kg/s].] 3. In a steady flow system, 50 [kJ] of work is done per [kg] of the fluid. The values of v, p and V at the inlet and exit sections are 0.4 [m3/kg], 6 [bar], 15 [m/s] and 0.6 [m3/kg], 1 [bar], 250 [m/s] respectively. The inlet is 30 [m] above the exit. The heat loss from the system is 8 [kJ/kg]. Calculate Du.

Chapter 6:

4.

5.

6.

7.

Energy (First Law) Equation for Flow Systems

127

[Hints: Since the flow is steady ( E CV = 0), the ce become m i = m e = m (say). The ee for steady state ( E CV = 0), i.e. 0 = Q – W + m [(ui – ue) + (pivi – peve) + (Vi2 – Ve2)/2 + g(zi – ze)] can be solved for Du = ue – ui = ( Q / m ) – ( W / m ) + (pivi – peue) + (Vi2 – Ve2)/2 + g(zi – ze), which gives Du = 91.01 [kJ/kg].] In a refrigerator with R–12 as the working fluid, saturated liquid at 35 [°C] is throttled through a capillary to 5 [°C]. Determine the quality of refrigerant at the exit of the capillary. [Hints: The ce is m i = m e (steady flow, i.e. m CV = 0, default assumption). The ee reduces to hi = he (steady state, i.e. E CV = 0, default assumption); adiabatic process (i.e. Q = 0, default assumption); no work done (i.e. W = 0, default assumption); and negligible changes in kinetic and potential energies (i.e. Dek = 0 = Dep, default assumption). Since the exit vapour is wet, use the usual formula, he = hf,e + xe(hg,e – hf,e). Substituting the values from the saturated R–12 tables in ee (i.e. hi = he) gives, hi = 68.988 = he = 40.304 + xe (189.27 – 40.304), or xe = 0.1926.] A centrifugal compressor sucks 6 [m3/min] of air and compresses it from 0.8 [bar] to 6.4 [bar]. The initial and final specific volumes are 0.8 [m3/kg] and 0.2 [m3/kg], respectively. The diameter of the duct is 10 [cm] at the inlet and 5 [cm] at the exit. Determine (a) m , (b) DV, and (c) net flow work done on the fluid. [Hints: By definition of v, m i = [(6/60)/0.8] = 0.125 [kg/s]. [Ans. (a)] By the ce for steady flow (i.e. m CV = 0), this becomes = m e . Then, Vi = mivi/Ai = 40/p, and similarly, Ve = meve/Ae = 40/p therefore, DV = 0, [Ans. (b)]. By definition, – 80 × 0.8) = – 8 [kW]. [Ans. (c)].] Wf l = m (peve – pivi) = (0.125) (640 ×m 0.2 i Water at the rate of 1500 [kg/min] flows through a horizontal venturimeter with inlet and exit diameters of 3.5 [cm] and 7 [cm] respectively. If the density of water remains constant at 1000 [kg/m3], calculate the pressure drop between the inlet and the throat. [Hints: The default assumptions for this venturimeter are steady flow (i.e. m CV = 0 implying m i = m e ) at steady state E CV = 0, adiabatic flow (i.e. Q = 0), and no work output (i.e. W = 0). Being a horizontal venturimeter zi = ze. Thus, the ee and the ce reduce to (Vi2 – Ve2)/2 = he – hi. The ce also gives, m = riAiVi = reAeVe. The definition of h means that, he – hi = (ue – ui) + (pivi – peve) and since water is incompressible (i.e. ui = ue, adiabatic pumping), this becomes he – hi = peve – pivi = (pe – pi)/r where r = 1/v. Then the ee gives, (pe – pi) = (r/2) (Vi2 – Ve2) = 1.899 [bar].] Petrol (n-octane, C8H18) vapour at 25 [°C] is burnt with 500% air in the combustor of a gas turbine engine. The combustion air is at 500 [K]. If the products leave at 1000 [K], what will be heat loss from the combustor per kg of the fuel? Assume that the properties of petrol vapour and exhaust gases are the same as those of air. = m e (the default assumption of steady flow (i.e. [Hints: The ce reduces to m CV = 0)). Since the input streams are air and fuel and the exit stream is exhaust (product) gases, this becomes m g = m e = m i = m a + m f , where the subscripts a, f and g denote air, fuel and product gases, respectively.

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The ee reduces to Q = m e he – m i hi (the default assumptions of steady state ( E CV = 0), of no work output ( W = 0), and of negligible kinetic and potential energy changes ( Dek = 0 = Dep)); or using the ce becomes the ee Q = mghg – mf hf – maha. Since the fuel, air and products behave like ideal gases, the ee further becomes Q = mgcp,gTg – macp, aTa – mf cp, f Tf = mgcpTg – macpTa – mf cpTf (identical specific heats (data)). Combustion stoichiometry shows that stoichiometric air fuel ratio is 15.26 [kg A/kg F]. Assuming that m f = 1 [kg/s], the actual air supplied is ma = 76.3 [kg/s] (500% of stoichiometric air (data)), and mg = 77.3 [kg/s] (ce). Substituting the values in the ee gives, Q = 38,969 [kW].] 8. An ideal gas (g = 1.4) expands reversibly in a turbine from 10 [bar] to 1 [bar]. Assume that the process law is p = 12 – 5v, where p is in [bar] and v in [m3/kg]. If the heat loss from the turbine is 200 [kJ/kg], calculate the shaft work done. [Hints: Since the flow is steady, the ce reduces to m i = m e = m . Since the turbine operates at steady state ( E CV = 0) with negligible Dek and Dep (default assumptions), the ee reduces to ( W / m ) = ( / m ) + (hi – he) = ( Q / m ) + cp(Ti – Te) = ( / m ) + [g /(g – 1)] ¥ (pivi – peve). The process law gives, vi = 0.4 [m3/kg] and ve = 2.2 [m3/ / m = 430 [kJ/kg].] kg]. Substituting all the values, the ee gives, 9. Water at a rate of 1 [kg/s] enters a pump at 1 [bar], 40 [°C]. The pump power is 60 [kW] and it raises the pressure to 5 [bar]. The water then passes through a boiler where 2000 [kJ/kg] of heat is absorbed. Determine (a) state at the exit of the boiler and (b) the exit velocity if the diameter of the exit duct is 20 [cm]. [Hints: There are two processes that W take place here. In the first process, water is Q pumped to a high pressure and in the second one, it is heated at constant pressure. At the inlet to the pump, thewater is subcooled and this data is not listed in the tables. So using the approximation, h1 = hf,40°C + (p – psat,40°C) vf,40°C = 167.54 [kJ/kg]. Then, from the ee h2 = h1 – W = 167.54 – (–60) = 227.54 [kJ/kg]. For the second process, the ee gives, h3 = h2 + Q = 2227.5 [kJ]. The steam tables show that at 5 [bar], h3 < hg so that the steam is wet. Then from the usual formula, x2 = (2227.5 – 640.12)/ (2747.5 – 640.12) = 0.7532. Thus the steam at the exit of the boiler is, p = 5 [bar] and x = 0.7532 [Ans. (a)]. Then from the usual formula, v3 = 0.28248 [m3/kg] so that V3 = 8.992 [m/s] [Ans. (b)].] 10. A six cylinder four stroke petrol engine of 10 [cm] bore and 12.5 [cm] stroke was run on full throttle at a constant speed of 1500 [rpm] from which the following data were obtained. Determine the engine performance for each case. pm,i [bar]

7.82

8.32

8.14

8.02

8.32

8.06

6.91

6.06

pm,b [bar]

6.27

6.62

6.76

6.72

6.69

6.34

5.79

5.14

sfc [g/kWh BP]

547

486

432

337

346

334

346

401

Other data: compr. ratio = 5; and fuel cal. value = 44.5 [MJ/kg]. pm,i = IMEP (Indicated Mean Effective Pressure); pm,b = BMEP (Brake Mean Effective Pressure); IP = Indicated Power, and BP = Brake Power.

Chapter 6:

Energy (First Law) Equation for Flow Systems

129

[Hints: The computations are explained for the first case and the results are tabulated for others. Let Ncyl and Ncyc denote the number of cylinders and the number of cycles per second. Then, BP = (BP/cycle/cylinder) × Ncyc × Ncyl, and by the definition of BMEP (i.e. pm,b), BP = (pm,b · Acyl · L) × [(rpm/2)/60] × Ncyl (where, the term in brackets arises because the engine is four stroke) = [6.27 · p (0.1)2/4· (0.125)] × (1500/120) × 6 = 46.1667 [kW]. Similarly, using the definition of IMEP (i.e. pm,i), the indicated power is, IP = 57.5795 [kW]. Now, m f = (0.547)(46.1667) = 25.2532 [kg/h], and Q1 = ( m f ) × (Qcal) = (25.2532/ 3600) 44,500 = 312.157 [kW]. Then, the definition of brake thermal efficiency (hb,th) gives, hb,th = BP/ Q1 = 14.79%. Similarly, the definition of the indicated thermal efficiency gives, hi,th IP/ Q1 = 18.44%. The mechanical efficiency defined as hm = BP/IP gives, hm = 80.12%. Results for other cases are tabulated below. IP [kW]

61.26

59.94

50.05

61.26

59.35

50.88

44.62

BP [kW]

48.74

49.77

49.48

49.26

46.68

42.63

37.85

Q1 [kW]

292.83

265.80

206.12

210.68

192.73

182.34

187.60

hi, th

0.2092

0.2222

0.2428

0.2908

0.3079

0.2790

0.2378

hb, th

0.1664

0.1872

0.2400

0.2338

0.2422

0.2338

0.2018

hm

0.7956

0.8303

0.9886

0.8041

0.7865

0.8378

0.8483

Note: The hb,th and hm values of the 3rd column are too high and so data are erroneous.

11. From the following test data on a Q turbine: (a) enthalpy at the inlet section = comb 3373.7 [kJ/kg], (b) specific volume at the inlet section = 3.279 [m3/kg], (c) enthalpy at the exit section = 2676.0 [kJ/kg], (d) specific volume at the exit section = 1.673 [m3/ kg], (e) area of the inlet section = 0.01 [m2], (f) area of the exit section = 1.0 [m2], and (g) heat loss from the turbine = 20 [kJ/kg], determine the work done per [kg] of the working substance. [Hints: Since the flow is steady ( m CV = 0), the ce gives m i = m e = m . Then, the ee gives W / m = (hi – he) + (Vi2 – Ve2)/2. But, Vi = m i vi/Ai = 327.9 [m/s]. Similarly, Ve = 1.673 [m/s]. Substituting in the ee gives, W / m = 731.4 [kJ/kg].] 12. A petrol engine operates steadily with a shaft output of 100 [kW]. Heat transferred to the cooling water and other surroundings is 400 [MJ/h]. The air-fuel mixture enters the engine at 30 [°C] and the exhaust gases leave at 800 [°C]. The flow rate of the mixture is 100 [kg/s]. For the exhaust gases, assume cp = 1.05 [kJ/kg.K]. Determine the heat released by combustion of the fuel. [Hints: The ce is m i = m e ( m CV = 0, default assumption of steady flow) = m (say) – Qloss – W + m (hi – he) (default assumption = 100 [kg/s] (data). The ee is 0 = of steady state ( E CV = 0), and of Dek = 0 = Dep). The ee can be solved as Q comb = Q loss + W + m (he – hi) = Q loss + W + m cp(Te – Ti) (for an ideal gas and by definition of cp). Substituting the values gives, Q comb = 81061 [kW].] 13. Air flowing at 1 [m/s] through pipe is heated with constant heat flux. At the inlet V, p and T are uniform across the cross-section with p = 1 [bar], and T = 30 [°C]. At the exit, p is uniform at 1 [bar], but the variations in V and T are, V = 150[1 – (r2/R2)] [m/s], and T = 250[1 + (r2/R2)] [°C]. Calculate Q.

130

14.

15.

16.

17.

Engineering Thermodynamics

[Hints: This exercise is similar to the earlier ones, but now the velocities and temperatures are not uniform. So, their average values will have to be obtained first. R Now, the mass average velocity V is defined by the relation (pR2) · (rV) = ∫ 0 (2prdr) 2 (rV(r)). Similarly, the average temperature T is defined by the relation (pR ) (rVcpT) = ∫ 0R (2prdr) · [rV(r)cpT(r)]. Substituting the data and evaluating these expressions give, Ve = 75 [m/s] and Te = 333 [°C]. Then, the ee gives Q / m = 306.7 [kJ/kg].] An air turbine expands 25 [kg/s] of air from 10 [bar], 1600 [K] to 2 [bar]. If the isentropic efficiency is 0.8, determine (a) the exit temperature and (b) the power output. [Hints: With default assumptions, the ee reduces to W = m (hi – he) = m cp(Ti – Te) (since air is an ideal gas with constant properties). For the isentropic process, Te = Ti (pe/pi)g/(g – 1) = 1010.2 [K]. By definition hs = (hi – he¢)/(hi – he) = (Ti – Te¢)/(Ti – Te) (since air is an ideal gas with constant properties). This equation can be rearranged as Te¢ = Ti – hs(Ti – Te) = 1028.2 [K] [Ans. (a)]. Substituting in the ee gives, W = 11,830 [kW]. [Ans. (b)].] An adiabatic steam turbine handles 10 [kg/s] of steam. The inlet state is 15 [bar], dry saturated steam. The exit pressure is 1 [bar]. If the isentropic efficiency is 0.8, for the actual process, determine (a) the exit state and (b) the power output. [Hints: With the default assumptions, the ce and ee give, W = m (hi – he). Now for the isentropic process se = si = 6.4406 [kJ/kg.K]. Then the steam tables show that, steam at the exit is wet, so that xe = 0.8482 (from the usual formula). Then, another usual formula gives, he = 2332.6 [kJ/kg]. The definition of hs gives, he¢ = hi – hs (hi – he) = 2424.1 [kJ/kg]. Thus, xe¢ = 0.8887. Thus the exit state is p = 1 [bar] and CV W = 3658 [Ans. (b)].] x = 0.8887 [Ans. (a)]. Then from the E ee, It is proposed to use R–12 as the working fluid in a solar heat pump. The fluid is to be heated from 10 [bar], 30 [°C] to 10 [bar], 80 [°C] at the rate of 10 [kg/h]. The efficiency of the evaporator is 60%. If the absorbed solar heat flux is 700 [W/m2], calculate the required surface area. CV = 0), [Hints: This is a heat exchanger. So the default assumptions (steady flow, ( m = 0), no work output ( W = 0) and Dek = 0 = Dep) reduce the ee steady state ( with ce to Q = m (he – hi). Since the R–12 tables show that the inlet liquid is subcooled, using the approximation hi = 64.2516 [kJ/kg]. The R–12 tables also show that the exit vapour is superheated. Interpolation shows that the exit vapour may be taken as Tsat = 40 [°C] with a degree of superheat of 40 [°C], so the superheat tables gives, he = 230.81 [kJ/kg]. Then, the required area is, A = (10/3600)[(230.81 – 64.25)/ (0.6 · 0.7)] = 1.102 [m2].] Water is pumped at the rate of 10,000 [lit/h] from 1 [bar], 25 [°C] to 200 [bar]. If the rating of the pump is 70 [kW], calculate its efficiency. [Hints: Using the default assumptions of steady adiabatic flow, steady state operation, with negligible kinetic and potential energy changes the ee along with the ce gives, W = m (hi – he), and, by definition of the efficiency, h = W / P , where P is the input power. Since the water is subcooled at the inlet as well as at the exit, using the usual formula gives, hi = 104.87 [kJ/kg] and he = 124.82 [kJ/kg]. Substituting in the formulae above gives, h = 79.17%.]

Chapter 6:

Energy (First Law) Equation for Flow Systems

131

18. Air at 30 [°C] enters a cooling tower at a height of 1 [m] above the ground level with a velocity of 20 [m/s] and leaves at 70 [°C] at a height of 7 [m] with a velocity of 30 [m/s]. Water enters the tower at 80 [°C] at a height of 8 [m] with a velocity of 3 [m/s] and leaves at 50 [°C] at a height of 0.8 [m] with a velocity of 1 [m/s]. Determine the mass flow rate of air required to cool 1 [kg/s] of water. [Hints: This is really a mixing type heat exchanger. Using the default assumptions, the a,i [ha,i + (V2a,i/2) + gza,i] = ( m w,e ) [hw,e ee becomes ( m w,i )[hw,i + (V2w,i/2) + gzw,i] + m 2 2 + (V w,e/2) + gzw,e] + m a,e [ha,e + (V a,e/2) + gza,e]. Since the flow of air and water are steady this can be rewritten as m a [(ha,e + (V2a,e/2) + gza,e) – (ha,i + (V2a,i/2) + gza,i)] = m w [(hw,i + (V2w,i/2) + gzw,i) – (hw,e + (V2w,e/2) + gzw,e]. Substituting the values and solving gives, m a = 3.110 [kg].] 19. Superheated steam at 7 [bar] and 250 [°C] flowing at the rate of 2 [kg/s] is mixed with wet steam at 7 [bar] and 0.5 quality flowing at the rate of 4 [kg/s] in a steady flow device. Calculate the quality of mixture [Hints: Using the default assumptions, the ee becomes m i hi = m e he, or H i = H e , or m s hs + m w hw = ( m s + m w )he. Substituting the values and solving gives, h2 = 2137.7 [kJ/kg], and using the usual formula, xe = 0.698.] 20. A rigid tank of 0.304 [m3] volume initially contains steam at 15 [bar] and 250 [°C]. Saturated water 0.0035 [m3] in volume at 15 [bar] is pumped into the tank. The water is then cooled. When equilibrium is re-established, the pressure is found to be 3 [bar]. Calculate the heat transferred during the process. [Hints: The process is non-quasi-static with only the initial and final states being equilibrium ones. Then the unsteady flow ce becomes, m2 – m1 = mi. Using properties from steam tables, m1 = 0.304/0.152 = 2.0 [kg], and mi = 0.0035/0.0011539 = 3.0332 [kg], so that m2 = 5.0332 [kg]. The unsteady ee becomes, m2u2 – m1u1 = Q + mihi, or Q = m2u2 – m1u1 – mihi. Now state 2 is, p2 = 3 [bar] and v2 = 0.304/5.0332 = 0.0604 [m3/kg], and the steam tables show that it is wet so that x2 = 0.09814. Then the properties are evaluated as, u2 = 755.61 [kJ/kg], u1 = 2695.5 [kJ/kg], and hi = 844.66 [kJ/kg]. Substituting these values in ee gives, Q = – 4150 [kJ].] 21. A 250 [litre] rigid vessel containing air at 1 [bar] and 300 [K] is being filled adiabatically by connecting it to a pipe supplying air at 5 [bar] and at unknown temperature T2. Once the vessel is completely filled, it is disconnected from the supply mains and cooled to 27 [°C]. Assume that the temperature and pressure of the air inside the vessel when it is full is T2 and 5 [bar], respectively. Determine (a) T2 and (b) the heat loss from the vessel during the process of cooling it. [Hints: There are two processes involved here. First one is the adiabatic filling of the tank and the second one is the tank losing heat to the surrounding at constant volume (rigid vessel). The first one is a flow process and the second a non-flow process. For the filling process, m2 – m1 = mi (ce, unsteady flow). The ee becomes m2u2 – m1u1 = mihi (default assumptions); or, since air behaves like an ideal gas, m2T2 – m1T1 = mi g Ti = (m2 – m1)(g Ti) (using the ce); or, m2(T2 – g Ti) = m1(T1 – g Ti); or, since T2 = Ti, m2T2(g – 1) = m1T1 [g (T2/T1) – 1]; or, using ideal gas equation, p2V2(g – 1) = p1V1 [g (T2/T1) – 1]; or, since V2 = V1 (rigid vessel), p2(g – 1) = p1[g (T2/T1) – 1], which can be solved as T2 = (T1/g )[1 + (g – 1)(p2/p1)]. Substituting the values gives,

132

22.

23.

24.

25.

26.

Engineering Thermodynamics

T2 = 642.8 [K] [Ans. (a)]. Now the equation of the state gives, m2 = 0.6783 [kg]. Then during the cooling process at constant volume (rigid vessel), Q = m2cv(300 – 642.9) = –166.7 [kJ] [Ans. (b)].] An evacuated and well insulated tank of 0.3 [m3] capacity is slowly being filled with oxygen (M = 32 [kg/kmol] and g = 1.4) from a mains. The oxygen in the mains is at 6.5 [bar] 27 [°C]. The flow stops when the tank pressure reaches 5.5 [bar]. If the ambient temperature is 15 [°C], calculate (a) the pressure of oxygen in the tank when it reaches the ambient temperature and (b) the heat loss from the tank. [Hints: As in Exercise 21, there are two processes involved here also. The first one is the adiabatic filling of the tank and the second one is the tank losing heat to the surroundings at constant volume (rigid vessel). The first one is a flow process and the second a non-flow process. For the first process, since the tank is initially evacuated, m1 = 0. Hence, m2 = mi (by ce). Then, the ee reduces to u2 = hi, or, T2 = g Ti (by definitions of cv and cp for an ideal gas) = 420 [K]. Then, m2 = 1.512 [kg] (by the ideal gas law). Since the second process (non-flow one) is one of heat loss at constant volume (rigid container), p3 = 3.771 [bar] (isochoric process of an ideal gas) [Ans. (a)] and Q = DU = –129.6 [kJ] [Ans. (b)].] Find the answers to Exercise 22 if the tank initially contains oxygen at 1 [bar], and 15 [°C]. [Hints: The gas law gives, m1 = 0.4009 [kg]. Now the ee gives, m2cvT2 – m1cvT1 = micpTi or, mi = (m2T2 – m1T1)/(gTi) = (V/R)[(p2 – p1)/(gTi)] = 1.2371 [kg]. Then the ce gives, m2 = m1 + mi = 1.638 [kg]; and the ideal gas equation gives, T2 = 387.71 [°C]. Now, for the non-flow process, p3 = 4.086 [bar] [Ans. (a)] and Q = DU = –106.1 [kJ] (means heat loss) [Ans. (b)].] An evacuated, insulated rigid tank is filled with steam from mains at 7 [bar] and 250 [°C]. Calculate the final temperature of the steam in the tank. [Hints: The ce gives m2 = mi (evacuated tank); and, then the ee gives, u2 = hi = 2954.0 [kJ/kg] (from steam tables); or, h2 – (700)v2 = 2954.0 [kJ/kg] (by definition of u) and p will be 7 [bar]. Solving by trial and error, T2 = 395.6 [°C] at 7 [bar].] A rigid, insulated tank initially containing 10 [kg] of air at 2 [bar] and 40 [°C] is being filled with air from mains at 16 [bar] and 60 [°C]. Calculate (a) the mass of air flowing into the tank and (b) the final temperature of the air in the tank. [Hints: Same as Exercise 24, but now the tank is not evacuated. The condition of mechanical equilibrium demands that when the flow stops, the p2 = pi = 16 [bar]. The ideal gas equation gives, V1 = 4.4869 [m3] = V2 (rigid tank). The ce gives, m2 = m1 + mi; and, since air behaves ideally, the ee reduces to m2T2 – m1T1 = migTi; or, (p2V2/R) – (p1V1/T1) = mig Ti; or, (V/R)(p2 – p1) = mig Ti; which can be solved as mi = 46.998 [kg]. Substituting in ce, m2 = 56.998 [kg] [Ans. (a)], and from the ideal gas equation, p2 = 439.3 [K] [Ans. (b)].] A pneumatic press is supplied with an ideal gas (g = 4/3) from a source of 7 [bar] and 0.3 [m3/kg]. The control valve between the source and the cylinder of the press is initially closed. It is then opened and closed after some time. The initial and final states of the gas in the cylinder are 1 [bar], 0.005 [m3] and 5.5 [bar], 0.02 [m3], respectively. The work done by the gas on the piston is 2.5 [kJ]. Assuming adiabatic conditions, calculate the mass of the gas that entered the cylinder.

Chapter 6:

27.

28.

29.

30.

Energy (First Law) Equation for Flow Systems

133

[Hints: This is also an unsteady process whose end states are equilibrium states. Then, writing the equations for the initial and end states, the ce: m2 – m1 = mi, gives, the ee gives, m2u2 – m1u1 = mihi – Wx; and, for an ideal gas, m2RT2 – m1RT1 = migRTi – (g – 1)Wx; or, p2V2 – p1V1 = mig pivi – (g – 1)Wx. Solving for mi and substituting the values gives, mi = 0.0398 [kg].] A steam press is supplied with steam at 10 [bar], 200 [°C] from mains. The cylinder of the press (V = 0.01 [m3]) initially contains 0.002 [m3] of dry saturated steam at 9.6 [bar]. When the flow stops, the steam in the cylinder is found to be dry saturated at 9.4 [bar]. Calculate the work done by the press. [Hints: The working fluid here is steam. Then, using the (saturated) steam tables, m1 = 0.0099 [kg], and m2 = 0.04852 [kg]. So the ce gives mi = 0.03862 [kg]. The ee gives, m2u2 – m1u1 = mihi – Wx, which can be solved as Wx = 9.535 [kJ].] One kilogram of steam at 1 [bar], 300 [°C] in a rigid adiabatic container is mixed with saturated liquid water at 1 [bar] and then stirred till the steam becomes dry and saturated at 1 [bar]. Determine (a) DU and (b) W. [Hints: Only the equilibrium end states are considered. Then the ce is m2 – m1 = mi (no exit), and the ee becomes, m2u2 – m1u1 = mihi – Ws (no exit), W = Ws,Wx = 0 (rigid container, data), Q = 0 (adiabatic process, data), Dek = 0 = Dep (default assumptions); the ee can thus be rewritten as Ws = mihi + m1u1 – m2u2. Also, V1 = m1v1 = 2.6387 [m3] (using data from steam tables) = V2 (rigid container) = m2v2 (definition of v), so that m2 = 1.55795 [kg], and mi = 0.55795 [kg] (by ce). Substituting the values in the ee gives, Ws = –861.1 [kJ].] At a specific time at the inlet and exitm iduring transient operations of a turbine, the following conditions were observed: pi = 5 [bar], Ti = 1000 [°C], Vi = 200 [m/s], Ai = 20 [cm2], pe = 1 [bar], Te = 800 [°C], Ve = 300 [m/s], and, Ae = 30 [cm2]. The power output from the turbine was 50 [kW]. Assuming that the working substance can be approximated to air, calculate E CV of the system. = riAiVi = 0.548 [Hints: This is an unsteady operation. Using the usual formulae, [kg/s]. Similarly, m e = 0.29259 [kg/s], so that the ce gives, m CV = 0.2554 [kg/s]. The turbine is an adiabatic device with Dep = Dek = 0 (default assumptions). Then substituting the data in the ee gives, E CV = 315.0 [kW]] The following readings were obtained during an interval of 1 [min] in the operation of a gas turbine, pi = 10 [bar], Ti = 800 [°C], Vi = 8 [m/s], Ai = 40 [cm2], pe = 2 [bar], Te = 400 [°C], Ve = 5[m/s], and, Ae = 80 [cm2.]. The power output was 1 [kW]. The heat loss was 8 [MJ] Assuming that the working substance can be approximated to air, calculate (a) DmCV and (b) DECV of the system. [Hints: Similar to Exercise 29. The ce gives DmCV = 0.06256 [kg] [Ans. (a)] and DECV = –2193 [kJ] [Ans. (b)].]

Chapter 7

The Second Law of Thermodynamics

7.1 INTRODUCTION It was shown during the discussion on work and mechanical energies (those which can be derived from some force) that these concepts are part of mechanics. It was also shown that the principle of conservation of energy is postulated using the analogy of the principle of conservation of mass which is a primitive even to mechanics. Thus, these concepts and laws are primitives to thermodynamics. The first law of thermodynamics only defines the existence of another type of energy called the internal energy (denote as U), so that the total energy E stored by a system becomes, E = U + Ek + Ep + Eem +
. In addition, one more form of energy flow different from work also exists. Retaining the principle of conservation of energy, it logically follows that this form of energy flow (called heat and denoted by Q) should be defined as Q = DE (W), where the second negative sign indicates that the positive directions of work and heat are opposite to each other. These arguments make it clear that what is known as the first law of thermodynamics is really the modified form of the principle of conservation of energy.1 Thus, the second law of thermodynamics is the real law of thermodynamics. This law can be interpreted in different ways. The more important of these (namely as efficiency of interconversion of work and heat2 as well as the possible direction in which the processes can take place (including heat flow)) are discussed here. This second law is one of the most interesting laws in physics. It has generated adequate controversy and confusion. However, its nature has been responsible for developing the statistical (microscopic) thermodynamics. It should be noted that this is the only law that remained least affected when extended to microscopic analysis (i.e. quantum and statistical mechanics3). 1 2 3

This will be evident from the discussion on flow systems presented in the last chapter. This can also be viewed as the difference between the characteristics of internal energy U and all other forms of mechanical energies. This should not be surprising since quantum mechanics (and statistical mechanics) basically modified the characteristics of energy at microscopic level. 134

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In this chapter the second law of thermodynamics is developed following the classical sequence. Later, another formulation is mentioned.

7.2

PRELIMINARY DEFINITIONS

It can rightly be stated that this law is really the outcome of human inability to extract energy from atmosphere, ocean and other large bodies, whose temperature remains uniform, and to convert it continuously to work.4 Such a machine will produce work continuously and for ever, i.e. it will be a perpetually working (in motion) machine. Hence, such a machine is named a perpetual motion machine of the second kind (PMM2). Some definitions needed to formalise the above result are stated below. A reservoir is a system of infinite size (mass, volume,
) which can exchange finite quantities of energy as heat without changing its state. Note that this is the thermodynamic model for our atmosphere, oceans, rivers, lakes, etc. which act as a source or sink for our engines. The temperature of a reservoir defines its state. Note also that a work reservoir need not be defined since none exists in nature. Thus, the word reservoir will mean only the heat reservoir. Reservoirs can be arranged in an order in terms of their temperatures. This is based on the basic law that heat flows spontaneously from a body at a higher temperature to one at a lower temperature. Then, if a reservoir at temperature T1 is brought in contact with another at T2, and the heat flows out of the first one and into the second, then T1 is said to be higher than T2. The thermodynamic model for an engine, which exchanges heat from reservoirs and produces work continuously, will be a cyclic device (i.e. one which executes a thermodynamic cycle) that exchanges heat with reservoirs and produces work.5 Since, it absorbs heat (energy in one form) from one (or more) reservoir (system) and converts it into work and deliver to some load (transfer it in another form), it is also called a heat engine.6 The following definitions are also used later. An nT engine is an engine which exchanges heat with n reservoirs and produces work. A heat pump (and a refrigerator) is a cyclic device that absorbs heat from a cold reservoir (one system) and delivers it to a hot reservoir7 (another system). In this chapter, these ideas are developed systematically. However, before this is attempted, another way of looking at the second law is presented first. The first law (the BornCarathéodory form) presented earlier is for a process. The formulation of the second law is also for a process. However, it has an unfamiliar form since it is purely based on mathematical arguments. This is explained in Section 7.16 of this chapter. 4 5 6 7

In other words, this means that the experiments of Joule, Mayer,
for measuring the mechanical equivalent of heat cannot be reversed. These are energy transducers. A transducer is a device that transfers power from one system to another in the same or another form (from transacross and ducereto lead). The modern utility power plants, aircraft jet propulsion systems, etc. are really heat engines in the thermodynamic sense (of this definition) even though they are not really named so. As per the definition given above, these are also transducers since they pump (transfer) heat (one form of energy) from the cold reservoir (one system) into the hot reservoir (another system).

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Here the second law is presented for a cycle first. Its form for a process is derived in a later section. This has two advantages: (a) This is the way in which this law was developed originally. (b) Engines (and refrigerators and heat pumps) are basic devices in thermal engineering, and therefore, have more physical appeal.

7.3

THE NEED OF THE SECOND LAW

In Chapter 2, it was mentioned that a problem in applied thermodynamics may be (a) a process problem typically stated as: bring about the specified change by transferring energy as heat or work or (b) an energy conversion problem typically formulated as: determine the heat required to produce specified work and the efficiency of the process, or (c) a heat pumping problem with the typical statement of: determine the work required for pumping a given quantity of heat between specified temperature levels. However, the discussion presented above shows that, from the thermodynamic point of view, the last two problems could be combined since engines, heat pumps and refrigerators are basically cyclic devices. Thus, these three categories reduce to two, (a) process problems and (b) cycle problems. In this section, the work and heat are examined for these two cases to determine if these two forms of energy flow are identical in the sense that both of them produce identical results.

7.3.1

Equivalence of Work and Heat for Processes

First consider the process problems. These pertain to estimating changes that occur in the state of the system. The state principle asserts that the state of a system is determined by its energy. Moreover, change in energy is related to the work and heat through the energy (first law) equation,8 i.e. DE = Q + (W). It is thus seen that the same change of state can be brought about by transferring energy either as work or as heat. For example, the energy increase of DE = 100 [kJ] may be produced by Q = x [kJ] and W = x100 (since by the energy (first law) equation, W = Q DE), for all values of x such that 0 £ x £ 100. Thus, since heat and work produce the same change of state during a process, they are equivalent (identical) and therefore need not be distinguished or named separately. This should not be surprising since, during the discussion on energy it was emphasized that only when energy flows can it be classified as work or heat; once they are stored it is simply energy.

7.3.2

Equivalence of Work and Heat for Cycles

Next, consider thermodynamic cycles (i.e. engines (energy converters) and heat pumps (and refrigerators)). For clarity, the discussion here is restricted to engines. Performance criteria Since an engine works in a cycle exchanging heat with some reservoirs and producing work, in this case, the equivalence of heat and work is examined from this point of view. To be useful 8

Recall that the negative sign of W only means that its positive direction is opposite to that for Q.

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an engine should work continuously9 and, therefore, continuousness should be chosen as one criterion in this examination. Purely economic consideration shows that the conversion should be as complete as possible. Hence completeness is chosen as the second criterion. In other words, for an engine (cycle), continuousness and completeness of interconversion should be the criteria for examining the equivalence of heat and work. Conversion of work into heat Consider an experiment in which an electric resistor is immersed in a river. When the heater is switched on, it converts the electrical work into heat. Since the heater is immersed in the river, the water absorbs the heat completely and the resistor does not store any energy. Thus, the electrical work is completely converted into heat and transferred to the water in the river. This process continues till the heater is switched off. In place of the electrical resistance, one can also use a rigid vessel containing an ideal gas being stirred continuously.10 This shows that work can be converted into heat completely and continuously. Conversion of heat into work At the outset it should be noted that the above two devices cannot be operated in the reverse direction,11 since (a) an electric resistor always dissipates electrical work, and (b) a turbine cannot be operated to produce work in an environment where the pressure is uniform. Hence, a cylinderpiston assembly should be used for converting heat into work. Since the working substance should convert all heat into work, it should be an ideal gas undergoing an isothermal expansion. However, this expansion stops when the system pressure equals the external pressure. Thus, when heat is completely converted into work it cannot be continuous. To obtain continuous conversion of heat into work, a cyclic device (one which works in a cycle) is needed which may absorb heat and convert it into work.12 The classical form (KelvinPlanck form) of the second law of thermodynamics asserts that such devices cannot convert all the heat into work. The arguments presented so far can be summarized as follows: the conversion of heat into work cannot be both complete and continuous. Hence, they should be distinguished. In the rest of this chapter, the second law is developed first. The later sections contain the consequences of this law.

7.4 DIFFERENT FORMS OF THE SECOND LAW OF THERMODYNAMICS The second law has been stated in different ways. The most important of them are the Kelvin Planck form13, the Clausius form and the Carathéodory form. They are equivalent in the sense 9 10 11 12 13

So that it can replace human and animal powerthe basis of industrialization. These devices are called dissipators. They exhibit friction (solid/fluid), Joule heating (electrical friction), hysteresis (magnetic friction), etc. The operating principle of these devices are based on friction (in some form or other) which is unilateral in the sense that it always opposes motion. In the discussions so far, these devices have been named engines. For simplicity, this will be referred to as the KP form.

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that on the basis of any one of them all other forms may be derived. Here the KP form is assumed to be the basic one and all other forms are shown to be equivalent to it. In this section, the KelvinPlanck and the Clausius forms are discussed. The Carathéodory form is presented in Section 7.16.

7.4.1

The KelvinPlanck Form

The KelvinPlanck (KP) form of the second law of thermodynamics may be stated ([ZEM], p. 147) as follows: It is impossible to construct an engine that, operating in a cycle, will produce no effect other than extraction of heat from one reservoir and the performance of an equivalent amount of work. In Section 7.2, an nT engine was defined as one which exchanges heat with n reservoirs, with at least one of the exchanges being a heat absorption. In terms of this, the KP form of the second law simply means that it is impossible to construct an 1T engine. A Perpetual Motion Machine of Second Kind (PMM2) is defined as a device that violates the second law of thermodynamics. Then, following the KP form of second law, it can be defined as an engine which absorbs heat from one reservoir and completely converts it into work, and in terms of the definition of an nT engine PMM2 is an 1T engine. The above statement shows that an engine should exchange heat with at least two reservoirs, and in order to satisfy the principle of conservation of energy, one of these should be a heat absorption reservoir. The other heat exchange should be a heat rejection. To understand this, consider the following arguments. (a) Let an engine exchange heat with two reservoirs R1 and R2. (b) Since temperature characterizes a reservoir, their temperatures should be different; otherwise both of them can be combined into a single bigger reservoir. Let the temperature of reservoir R1 be T1 and that of R2 be T2 with T1 > T2 (arbitrarily). (c) Now, contrary to the above proposition, assume that the engine absorbs heat Q1 from R1 and Q2 from R2. (d) Bring both the reservoirs in contact (as explained later, in this section) so that heat flows spontaneously (the natural phenomenon) from reservoir R1 to R2. (e) Adjust the contact such that heat Q2 flows from reservoir R1 to R2 so that whatever energy R2 gives to the engine is received from R1. (f) Now, the combined system, consisting of R1, in contact with R2 and the engine, is logically equivalent to the engine absorbing heat (Q1 + Q2) from the reservoir at T1, i.e. it is a 1T engine. (g) This shows that the above proposition should be true, i.e. Q1 or Q2 should be a heat rejection reservoir. Traditionally, the heat rejection is denoted by Q2, so that Q1 denotes the heat absorbed. A 2T engine is schematically represented as shown in Figure 7.1(a). In Section 7.3, it was argued that conversion of heat into work can either be complete or continuous, and that to be useful, it should be continuous (so that it cannot be complete). In thermal engineering (and hence in thermodynamics), thermal efficiency measures the degree of completeness. It is defined an h = W/Q1 where W is net work done and Q1 is the total heat absorbed14. Now, the energy conservation principle implies that, W, the work done cannot be

14

Note that h has the dimension of [kJ W/kJ Q], where W and Q denote work and heat, respectively.

Chapter 7:

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6

139

61

31

3

'

9

32

Figure 7.1

4

9

3

6

6

C

D

(a) 2T engine and (b) 2T refrigerator/heat pump.

more than Q1, the total heat absorbed, so that, h £ 1. Moreover, this definition together with that of an 1T engine (i.e. W = Q1) implies that h1T engine = 1. But, the KP form of the second law, which asserts that 1T engines cannot be constructed, implies that h1T engine ¹ 1. Taken together all these results simply means that h < 1. The term thermal efficiency is meaningful only to engines (which convert heat into work), and since engines are expected to deliver work (to drive a load), W ³ 0 so that h ³ 0. Combining this result with the earlier ones finally gives, 0 £ h < 1. Since engines are cyclic devices, the energy (first law) equation becomes ± F3 = F9 ,

v

v

v

v±

and being a property, ± F' = 0. Traditionally, the net work done ± F9 denoted as W itself. However, the net heat exchanged ± F3 is divided into Q1, the total heat absorbed, and Q2, the

v

total heat rejected. Then, the energy (first law) equation becomes W = Q1 Q2. Note that the KP form of the second law implies that Q2 cannot be zero.15 The above arguments may be summarized as:

h = W/Q1

(by the definition of thermal efficiency) (using the energy (first law) equation for a cycle) = 1 Q2/Q1

An engine is said to idle when it does not produce any work output16 to drive the load, i.e. W = 0. Then, the energy (first law) equation given above shows that Q1 = Q2. This means that an idling engine is the perfect heat conductor between a hot reservoir and a cold one.

7.4.2

The Clausius Form of the Second Law

The KP form of the second law deals with the process of continuous conversion of heat into work by an engine. The Clausius form pertains to the process of heat transfer. 15 16

This is another way to state the second law. Real engines need finite power during idling to overcome friction and to run auxiliary systems, so that for them W > 0 even during idling.

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It was shown above that an idling engine can be used to transfer heat from a high temperature reservoir to a low temperature one, where the terms high and low temperatures have been chosen to be consistent with the basic experimental fact that heat flows spontaneously from hotter bodies to colder ones. The Clausius form (C form) of the second law asserts another aspect of this process, i.e. that heat cannot be transferred spontaneously in the reverse direction (from a cold reservoir to the hot one). The Clausius form of the second law of thermodynamics may be formally stated as follows ([ZEM], p. 148): it is impossible to construct a device that, operating in a cycle, will produce no effect other than the transfer of heat from a cooler to a hotter body. Refrigerators and heat pumps are the cyclic devices used for pumping heat from a cold reservoir to a hot one. They are schematically represented as in Figure 7.1(b). Then, the energy (first law) equation becomes, Q1 = Q2 + W, where Q2 is the heat absorbed from the cold reservoir, W is the work absorbed, and Q1 is the heat rejected to the hot reservoir. The coefficient of performance (denoted as COP) is the parameter used for estimating their performance of refrigerators. The purpose of a refrigerator is to keep a room (a low temperature reservoir) cool. Hence, the main quantity is the heat extracted from the low temperature source. The C form of the second law asserts that this cannot be done without expenditure of work.17 Then, it is logical to define its COP as COPref = Q2/W

(by definition)

= Q2/(Q1 Q2)

(by the energy (first law) equation for cycle)

A transducer as defined earlier show that refrigerator is indeed a transducer. On the other hand, the main function of a heat pump is to pump heat into a higher temperature reservoir with some expenditure of work so that, COPhp = Q1/W

(by definition)

= Q1/(Q2 Q2)

(by the energy (first law) equation for a cycle)

In terms of the COPs defined above, the C form of the second law becomes as follows: The COP of refrigerators and heat pumps cannot be infinity. There is another way to look at the Clausius form of the second law. During the discussion on the KP form of the second law it was mentioned that the perfect heat conductor is an idling engine, because the energy (first law) equation shows that Q1, the heat absorbed from the hot reservoir equals Q2, the heat rejected to the cold reservoir. The Clausius form of second law shows that this idling engine is a unilateral device (like the diode in electronics, the rack-andpinion in mechanisms, etc.) since it cannot be run in the reverse direction. This idea provides a method to rigorously define the concept of high and low temperatures as follows. Given two reservoirs, take an engine unconnected to any load and connect it to the reservoirs arbitrarily. If the engine runs, then the reservoir to which the input is connected is the high temperature reservoir. 17

For which, we have to pay, e.g. the electricity bill.

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7.5 USE OF THE SECOND LAW OF THERMODYNAMICS The well-known method of reductio ad absurdum18 is the general strategy of proving all propositions under the second law of thermodynamics and its corollaries. Following example illustrates this strategy. EXAMPLE 7.1 Show that the thermal radiation inside an enclosure with constant wall temperature is isotropic. Solution For simplicity (a) the enclosure is assumed to be evacuated so that conduction and convection of heat are absent and (b) the enclosure is assumed to be isolated so that other interactions can be neglected. The proposition to be proved is that the radiation is isotropic, i.e. it is the same in all directions. This is done as follows: (a) Assume that the radiation is not isotropic. (b) Then, there must be at least one direction in which the radiations is more that in others. (c) Run an engine such that it absorbs heat from this direction, converts part of it into work, and rejects the rest in the directions of lower radiation. (d) Such an engine will run indefinitely and, therefore, be a PMM2. (e) This means that the proposition is false. Sometimes, this procedure requires modifications as illustrated in the following example. EXAMPLE 7.2 Show that the KelvinPlanck (KP) form of the second law of thermodynamics is equivalent to the Clausius (C) form. Solution Mathematical logic shows that the logical relation A is equivalent to B is the same as A implies B and B implies A. This relation is true when both A and B are true or when both of them are false.19 However, for natural phenomena it is impossible to show that if A is true then so is B, because this requires infinite number of experimental observations. Hence, it is only possible to show that if A is false so is B. This uses the well-known principle of the excluded middle.20 This is followed here. First, it is shown that if the KP form is false then so is the C form. For convenience, it is shown step-by-step as follows: (a) Assume that the KP form is false. This means that an engine can be constructed that absorbs heat from one reservoir and completely converts it into work. (b) Now consider a refrigerator extracting heat Q2 from a cold reservoir. Let it absorb W of work. Note that this does not violate the C form of the second law. Then, the energy (first law) equation shows that the heat rejected by it to the hot reservoir is Q1 = Q2 + W. (c) Next, as per the assumption in the first step (which violates the KP form of the second law), construct an engine which absorbs heat from the hot reservoir and completely converts it into 18 19 20

In Latin, reduction to absurdity (COD), i.e. absurd conclusion logically follows if the proposition is false. This can be verified by writing the truth table for the relation (A ® B) AND (B ® A). However, some do not accept this strategy since, if the premise that the object sought does not exist results in a contradiction, then this, in the opinion of intutionists, cannot be considered as a proof of its existence, a note on Mathematical Logic, in Encyclopedia of Mathematics, Kluwer Academic Publishers, 1990. See also [BRI1], p. 38 for another opinion on the applicability of the principle of excluded middle to physical problems.

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work. Let the work produced by this engine be W, i.e. the work absorbed by the refrigerator. Then, by the energy (first law) equation, Q1 = W, where Q1 is the heat absorbed from the hot reservoir. (d) Now, make a composite system by directly coupling the refrigerator to the engine. This composite system is a refrigerator that does not absorb any work from external sources because the engine that produces the work is a part of it, but it pumps heat Q2 from the cold to the hot reservoir. Thus, this composite system violates the C form of the second law. (e) This means that if the KP form of the second law is false, then, so is the C form. Next, it is shown in the following steps that if the Clausius form of the second law is false, then so is the KP form. (a) Assume that the Clausius form is false. This means that a heat pump (or, a refrigerator) can be constructed that can pump heat from a low temperature reservoir to one at a higher temperature without absorbing work from external sources. (b) Let an engine work between these two reservoirs, absorbing heat Q1 from the hot reservoir, producing work W and rejecting heat Q2 to the cold reservoir. Note that this engine does not violate the KP form of the second law. Then, by the energy (first law) equation, Q2 = Q1 W. (c) Next, construct the heat pump of the first step. Let it pump heat Q2 (that is rejected by the engine) from the cold reservoir to the hot one. (d) Now, consider the composite system of the engine and the heat pump. This system does not require the cold reservoir since, Q2 the heat rejected by the engine can be fed directly into the heat pump. Hence, it becomes an 1T engine, which produces work W by absorbing heat only from the hot reservoir. Thus, it violates the KP form of the second law. (e) This proves that if the Clausius form is false, then so is the KP form. Thus the given proposition, namely that the KelvinPlanck form of the second law is equivalent to the Clausius form is proved. In the rest of the chapter, many of the results which follow as corollaries of the second law are presented.

7.6

REVERSIBLE AND IRREVERSIBLE PROCESSES

Consider a process executed by a system exchanging heat and work with its environment. Let the process be reversed so that the system comes back to its original state. Then if, during the execution of the reversed process, the environment and all the rest of the universe also return to their original states, the process is called a reversible process. Thus, a reversible process can be operationally defined as follows: A reversible process is one which, when reversed, reaches the initial state such that there are no changes in the environment and the rest of the universe. An equivalent definition is as follows: A reversible process is one which, when executed in the reverse direction, leaves no trace of it on the environment and the universe. Operationally, the above definitions require a large number of measurements since the state of the system and that of all others that constitute the environment should be measured. A more convenient operational definition tests a reversible process in two components, each of which is designed check the violation of one of the above forms of the second law of thermodynamics.

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First consider an electric resistor in contact with a reservoir as a system. In this system, the electrical work input is dissipated into heat and absorbed by the reservoir. Now, to reverse this process the heat supplied to the reservoir should be extracted, completely converted into work and supplied to the source from which it was originally drawn. This process violates the KP form of the second law. This shows that, any process that involves dissipative forces (any type of friction, hysteresis
) is irreversible. Next, consider the spontaneous heat transfer across a finite temperature difference from a hot reservoir to a cold one. To reverse this process the heat that flowed spontaneously into the cold reservoir should be pumped back to the hot reservoir without taking work from any source. This violates the Clausius form of the second law. This shows that any process that involves spontaneous heat transfer across a temperature difference is irreversible.21 A process that is not irreversible is called a reversible process. The above discussion shows that a reversible process can be operationally defined as follows: A reversible process is one that does not contain dissipative effects or involve heat transfer across a finite temperature difference. Zemansky ([ZEM], pp. 132134) classifies irreversibilities as follows: (a) Isothermal external mechanical irreversibilities such as irregular stirring of a liquid; coming to rest of a rotating/vibrating liquid; transfer of electricity through a resistor and magnetic hysteresis. In all these, the system is in contact with a reservoir so that the processes are isothermal which convert work into heat. (b) Adiabatic external mechanical irreversibilities that constitute all of the above processes but are now insulated. Thus, the work done is stored as the internal energy of the system. (c) Internal mechanical irreversibilities caused by free expansion of a gas, throttling of a fluid, snapping of a stretched wire, and collapse of a soap bubble. In these, the internal energy is used to do work which is then converted into internal energy again. (d) External thermal irreversibilities caused by heat transferred by conduction and radiation. (e) Internal thermal irreversibility that is similar to (d). (f) Chemical irreversibilities arising out of formation of new species by chemical reactions; mixing of two different substances (e.g. diffusion of gases, mixing of alcohol and water); and, sudden phase change (e.g. freezing of supersaturated liquid and condensing of supersaturated vapour). Note that the cases in (a), (b) and (c) above are essentially those of dissipation, while the irreversibilities in (d) and (e) are due to heat transfer across finite temperature differences. Example 7.8 shows that the diffusion processes in (f) are irreversible. Most of the known dissipative forces vanish when the forces tend to zero.22 In such cases, a quasi-static process is also a reversible one. Exceptions occur when the process curve has essential singularities (e.g. backlash). 21 22

Example 7.5 shows that, by definition, heat transferred across an infinitesimal temperature difference is reversible. The irreversible (or non-equilibrium) thermodynamics deals with the cases where the fluxes vary linearly with potentials. The observed physical phenomena support this assumption (at least as a first approximation). In these cases, this assumption is a good one.

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Following definitions can be directly obtained from the preceding discussion. Reversible cycle.

One in that all the processes are reversible.

Irreversible cycle.

One that contains at least one irreversible process.

Reversible engine.

An engine that works in a reversible cycle.

Irreversible engine.

7.7

An engine that works in an irreversible cycle

THE CARNOT THEOREM

The second law of thermodynamics states that the efficiency of a 2T engine cannot be cent per cent. This is not a very effective criterion to use. The Carnot theorem gives a stronger condition than this. In this section, this theorem is stated and proved. This theorem and other propositions are proved by showing that, if they are false then a PMM2 can be constructed. Thus, logically, the Carnot theorem becomes a corollary of the second law. However, Table 1.1 shows that Sadi Carnot proposed this theorem long before the second law was formulated. The Carnot theorem deals with 2T engines (i.e. those which work between two reservoirs). It states that out of all 2T engines working between the two given reservoirs, the reversible 2T engine is most efficient. Mathematically, this means hR ³ h, where the subscript R refers to a reversible 2T engine.23 For proving the Carnot theorem statement, it is convenient to split its statement into two parts, namely, (a) the efficiency of an irreversible 2T engine is less than that of a reversible one working between the same reservoirs and (b) the efficiencies of all reversible 2T engines working between two reservoirs are the same. The first part of the above statement is proved as follows: (a) Assume that this statement is not true. This means that an irreversible 2T engine can be constructed which is more efficient than a reversible 2T engine when working between the two given reservoirs, i.e. such that h > hR. (b) Let the heat absorbed, work done and heat rejected by the irreversible engine be Q1, W, and Q2, respectively. (c) Choose the size of a reversible 2T engine such that it absorbs heat Q1,R from the hot reservoir, does work W (i.e. the same quantity as that done by the irreversible engine), and rejects heat Q2,R to the cold reservoir. (d) Since, by definition, h = W/Q, the condition h > hR means that Q1,R > Q1 (since both engines do the same work). (e) Reverse the reversible engine so that it runs as a heat pump24 absorbing work W and delivering heat Q1,R to the hot reservoir. (f) Now couple the irreversible engine to 23

24

It should be noted that this theorem puts a condition only on the thermal efficiency. Different 2T engines may absorb different quantities of heat and may do different amounts of work, but their efficiencies (i.e. the ratios of W to Q1) will be the same. In other words, the work done by each engine is proportional to the heat it absorbs. By the way it is defined, when a reversible engine is operated in the reverse direction so that it becomes a reversible refrigerator or heat pump, only the directions of Q1, W and Q2 get reversed and their magnitudes do not change.

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the heat pump so that the work required by the heat pump is supplied by the irreversible engine. (g) This combined system does not absorb work from any external sources. (h) The net heat pumped (from the cold reservoir) by this combined system into the hot reservoir is Q1,R Q1, which is positive. (i) Therefore, the combined system violates the Clausius from the second law, thereby making it a PMM2. (j) Hence the proposition is false. These arguments form the basic structure of most proofs under the second law as corollaries of the Carnot theorem. For example, by replacing the irreversible engine with another reversible engine we can prove that the efficiencies of all reversible 2T engines working between the same reservoirs are the same. Similarly, we can show that the efficiency of a Carnot engine does not depend upon its size, working substance, etc. Consequently, from the point of view of thermal efficiencies all reversible engines are the same and can be generically called Carnot engine. The above results can be summarized as follows: The efficiency of a reversible 2T engine (and the COP of a reversible 2T refrigerator) working between two reservoirs depends only on the temperatures of the reservoirs. This means that any reversible engine (refrigerator/heat pump) is a Carnot engine (refrigerator/heat pump). It should be noted that the Carnot theorem is applicable only to any cyclic device (engine or heat pump or reservoir) exchanging heat with two reservoirs. For example, if an engine absorbs heat from two reservoirs and rejects heat to a third one, then the Clausius inequality is used. If a device works at constant temperature (e.g. fuel cells, electrical storage batteries) then also the Carnot theorem is not applicable and, therefore, a suitable modification of it is used. The Carnot theorem, (namely, hR ³ h) has an equality part and an inequality part. The two important consequences of the equality part are the definition of the thermodynamic scale of the temperature and the definition of entropy. The inequality part gives the Clausius inequality. These aspects are discussed in rest of this chapter.

7.8 THERMODYNAMIC SCALE OF TEMPERATURE This is the first important concept that arises as a consequence of the equality part of the Carnot theorem. The thermal efficiency of a reversible engine working between two reservoirs depends only on their temperatures. Let these temperatures be q1 and q2, respectively.25 Then, the Carnot theorem says that h = y (q1, q2). But, h = W/Q1 (by definition); and h = 1 (Q2/Q1) (by the energy (first law) equation), where Q1 and Q2 are, respectively, the total heat absorbed and rejected by the engine. Combining this relation with the Carnot theorem gives, Q1/Q2 = f (q1, q2). Next, let the reversible engine work between the reservoirs at q1 and q3, absorbing heat Q1 at q1 and rejecting heat Q3 at q3. For this, we write Q1/Q3 = f (q1, q3). Finally, let the reversible engine work between the reservoirs at q3 and q2, absorbing heat Q2 at q2, and rejecting heat Q3 at q3, so that Q2/Q3 = f (q2, q3). Now, 3 3 25

3 3 3 3

UQ VJCV

H R R

H R R H R R

We shall use q to denote these temperatures and reserve T for the temperature on the thermodynamic scale.

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Then mathematical arguments show that26 H R R

H R H R

and combining this with the preceding relation, H R 3 H R 3

The last relation simply means that the heat exchanged by a reversible engine with a reservoir at temperature q is proportional to f(q). In the thermodynamic (Kelvin) scale27 of temperature, this function f(.) is simply chosen as the temperature itself, i.e. f(q) = q. T indicates these temperatures. Thus the heat exchanged (Q) by a reversible engine with a reservoir at T is proportional to T, i.e. Q = aT, where a is the constant of proportionality. The constant of proportionality is chosen from the condition that the temperature of the triple point of water is 273.16 [K]. This makes 0 [K] = 273.16 [°C]. Combining this with the earlier definitions gives 36 6 3 6 3 3 6

or 6

3 3

where Q3 is the heat that a Carnot engine exchanges with the reservoir at 273.16 [K]. In terms of the definition of the thermodynamic scale of temperature, the relations for the efficiency of a heat engine get modified as shown below.

I

9 3

I

26

27

3 3

(the basic definition applicable to all engines) (application of the energy (first law) equation for a cycle)

Out of the several arguments, the most direct seems to be that presented by Denbigh ([DEN], p. 30). This is summarized here. Consider the relation f(x, y) = f(x, z)/f(y, z), which can be written as f(x, y) × f(y, z) = f(x, z). Partially differentiating this expression with respect to z gives, f(x, y) × (¶f(y, z)/¶z) = (¶f(x, z)/¶z). Eliminating f(x, y) using the original expression, this relation becomes [¶ÿ ln f(y, z)/¶z] = [¶ ln f(x, z)/¶z] = f(z) (say). Integration of these equations give, f(x, z) = f(z) + ln f(x) and, f(y, z) = f(z) + ln f(y), which can then be transformed easily as f(x, z)/f(y, z) = f(x)/f(y). In other words, the dummy variable z drops out. The original choice of Kelvin was f(q) = ln (q) so that temperatures could vary from ¥ to +¥.

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6 (by definition of Kelvin scale (reversible engines only)) 6 It is seen that the efficiency of a reversible engine becomes unity at absolute zero, which violates the KP form of the second law. Thus, it may be stated that it is not possible to reach the absolute zero on the thermodynamic scale. This law is known as the third law of thermodynamics which states it is impossible, by any procedure, no matter how ideal, to reduce any system to absolute zero temperature, in a finite number of operations ([ZEE], p. 397). The following example shows the application of the Carnot theorem (together with the definition of Kelvin scale of temperature).

I

EXAMPLE 7.3 An inventor claims to have developed an engine that absorbs 100 [kW] of heat from a reservoir at 1000 [K], produces 60 [kW] of work and rejects heat to a reservoir at 500 [K]. Will you advise investment in its development? Solution Since the engine is supposed to produce 60 [kW] of work while absorbing 100 [kW] of heat, its efficiency is 60/100 = 0.6. Now, a reversible engine working between the same reservoirs will have an efficiency of h = 1 (T2/T1) = 1 (500/1000) = 0.5. Thus, it is clear that the proposed engine is more efficient than a reversible one. Since, by the Carnot theorem this is impossible we note that this engine cannot be constructed and investment in its development will be a waste of resources.

7.9

RELATION TO IDEAL GAS TEMPERATURE

During the discussion on working substances of a system (in Chapter 2) it was pointed out that an ideal gas is defined to be one which follows (a) the Boyles law and (b) the Joules law. In Chapter 4 on empirical temperature, it was shown that the Boyles law leads to the definition of the ideal gas temperature q, as q = pv/R, where p is the pressure, v is the specific volume and R is a constant called the gas constant. It was also pointed out that this relation is conventionally written as pv = Rq and is called the equation of state of an ideal gas (but it really is the definition of temperature written in another way). In this section, the Joules law is used to show that the ideal gas temperature q is identical to the Kelvin scale of temperature T. 1. The Joules law (based on the experiments on the free expansion of ideal gases, as explained in Example 7.8), states that at constant temperature, the internal energy of an ideal gas does not depend on its volume. 2. Since an ideal gas is a simple compressible substance, its state (and, therefore, all properties) is determined by two independent properties. 3. Choosing them as T and v, the internal energy is written as u = u(T,v); so that by differentiation, du = (¶u/¶T)vdT + (¶u/¶v)Tdv. 4. Then, the Joules law means that (¶u/¶v)T = 0. 5. Now, for a simple compressible substance, the basic property relation can be written as du = Tds pdv, so that (¶u/¶v)T = T(¶s/¶v)T p. Using one of the Maxwells relations (derived in Chapter 8), this becomes (¶u/¶v)T = T(¶p/¶T)v p.

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6. Now, imposing the Joules law this becomes (¶p/¶T)v = p/T. 7. Assuming that q and T are only functions of each other and using the chain rule, this can be written as (¶p/¶q)v × (dq/dT)= p/T. 8. For an ideal gas, p = Rq/v, so that (¶p/¶q)v = (R/v) = (p/q). 9. Substituting relation in step 8 the equation in step 7 and simplifying gives, dq/dT = q/T. 10. Separating the variables and integrating gives, T = C × q, where C is a constant. 11. It is customary to assume the constant to be unity, so that T = q. This shows that the ideal gas temperature is identical to the thermodynamic (Kelvin) temperature.

7.10 NEGATIVE ABSOLUTE TEMPERATURES Ramsey shows28 that: · Thermodynamics has to admit the existence of negative absolute temperatures because of the definition of T = (¶U/¶S)V, since transitions when S decreases with U are observed. · The ordering is + 0 [K],
, + 300 [K],
, + ¥ [K],
, 0 [K],
, 300 [K],
, ¥ [K]. · The only change required in thermodynamics is in the KelvinPlanck statement of the second law. This should be modified as follows: It is impossible to construct a cyclic device which absorbs (rejects) heat with only one reservoir at positive (negative) temperature and produces (absorbs) equivalent work. · However, it is not possible to construct Carnot engines that operates between positive and negative temperatures.

7.11

ENTROPY

In Section 7.8, the first important consequence of the equality part of the Carnot theorem was used to construct the thermodynamic (Kelvin) scale of temperature. In this section, the second important consequence of the equality part of the Carnot theorem, namely the definition of entropy, is presented. In that section, it was shown that due to the definitions of the Carnot engine and of the Kelvin scale of temperature, the heat exchanged by a reversible engine with a reservoir is proportional to its temperature. Now, consider an infinitesimal reversible engine absorbing heat F Q1 from reservoir at temperature T1 and rejecting heat F Q2 to another reservoir at T2. Then, from Section 7.8, 28

F 3 6 F 3 6

Ramsey, N.F., Thermodynamics and Statistical Mechanics at Negative Absolute Temperatures, in Temperature, Vol. 3, Part 1, F.G. Brickwedde (Ed.), Robert. E. Krieger Publishing Co., 1972, pp. 1520.

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or F 3 F 3 F 3 F 3 or 6 6 6 6 Since F 32 is the heat rejected by the engine and hence is negative, the preceding equation can be written as F 3 F 3 6 6 or

v±

F3 6

(cyclic device)

This means that F 3 /T equals the change of a property. This property is called the entropy and is denoted by S. Since the engine is reversible, all the processes (including the heat transfer) are reversible. This gives the following definitions: F5

F 34 6

'5 ±

(infinitesimal process)

F 34 6

(finite process)

Since the temperature of a reservoir is constant (due to its definition), its entropy change is given as

F5 4

¦ F 3µ § ¶ ¨ 6 ·4

'5 4

¦ 3µ § ¶ ¨6 ·4

(infinitesimal process) (finite process)

where T is the temperature of the reservoir.

7.11.1

Evaluation of Entropy Change during a Process

Entropy is a property. Consequently, its value is fixed once the state is fixed. Then, the change of entropy is evaluated as the difference between its value at the final and initial states. The actual procedure used depends upon the type of available data: (a) If the property tables of the substance (e.g. steam tables) are available, then the steps are: (i) determine the final state from the actual process, (ii) read the values of entropy for the initial and final states, and (iii) subtract the initial entropy from the final one. (b) If the equation of state and the equations of properties of the substance is available (e.g. air assumed to be an ideal gas) then the change of entropy is calculated from the defining equation given in Section 7.11.

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The entropy algorithm consisting of the following steps is recommended for calculation of entropy: (a) Given the initial state and a process. (b) From the actual process calculate the final state. (c) Replace29 the actual process by an equivalent reversible process30. (d) Calculate the heat transferred using the energy (first law) equation assuming that the processes are reversible and that only expansion work is done, i.e. FQR = dU + pdV. (e) Evaluate entropy as DS = ± FQR/T = ± (dU + pdV)/T. These procedures are illustrated by examples in Section 7.14.

7.12 CLAUSIUS INEQUALITY The inequality part of the Carnot theorem leads to (a) the Clausius inequality and (b) the principle of increase of entropy. This section deals with the first. The principle of increase of entropy is developed in the next section. Consider an infinitesimal irreversible engine absorbing heat FQ1 from a reservoir at T1 and rejecting heat FQ2 to a reservoir at T2. (a) The inequality part of the Carnot theorem is h < hR; (b) by definition, for any engine, h = 1 ( FQ2/ FQ1); (c) but for a reversible engine, in terms of the Kelvin scale of temperature, hR = 1 (T2/T1); (d) substituting in the inequality and rearranging, FQ2/FQ1 > T2/T1. (e) since T1 and T2 are positive and non-zero, FQ2/T2 > FQ1/T1); (f) since FQ2 is the heat rejected, this becomes ( FQ1/T1) + ( FQ2/T1) < 0; (g) the left-hand of this inequality is the net change of the quantity ( FQ/T) during the cycle so that this inequality can be written as ± F36 < 0; (h) during the derivation of the relation for entropy, it was shown that, for a reversible cycle,

v

v

v

this relation becomes ± F36 ; (i) combining these two results gives ± F36 £ 0. This is called the Clausius inequality. The equality part is used for the reversible processes. One of the most useful applications of the Clausius inequality is to decide whether a cyclic device (engine or heat pump or refrigerator) exchanging heat with more than two reservoirs is possible or not. The Carnot theorem cannot be applied to this case. The following example illustrates this. EXAMPLE 7.4 An inventor claims to have developed an engine which absorbs 100 [kW] of heat from a reservoir at 1000 [K], 80 [kW] of heat from another reservoir at 1600 [K], produces 130 [kW] of work and rejects heat to reservoir at 500 [K]. Will you advise investment in its development? Solution The basic idea is to check if the engine is thermodynamically possible using the second law of thermodynamics. Since, the engine absorbs heat from two reservoirs and rejects heat to one, i.e. it is a 3T engine, the Carnot theorem cannot be applied. So the Clausius inequality is used as follows: 29 30

This can be done because entropy depends only on the state. This should be a convenient process for evaluation of the work done and change of internal energy.

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(a) Let Q1 = 100 [kW], Q2 = 80 [kW], and W = 130 [kW]. (b) Then, the energy (first law) equation gives Q3 = 50 [kW]. (c) The Clausius inequality (second law) demands that (Q1/T1) + (Q2/T2) + (Q3/T3) £ 0. (d) But, (100/1000) + (80/1600) (50/500) = 0.05 (> 0). (e) Hence, this engine violates the second law and so it cannot be constructed. Therefore, investment in it is not justified.

7.12.1

The Principle of Increase of Entropy

Consider an irreversible cycle consisting of an irreversible forward process from 1 to 2 and a reversible return process from 2 to 1. The Clausius inequality for this cycle is F3 F3 4 ± 6 ± 6 c

Transposing the second term to the right and changing its limits by using the minus sign,31 this relation becomes F3 F3 4 c ± 6 ± 6

Since, the second term is the definition of DS, the rearrangement of this relation gives F3 DS ³ ± 6

(finite process)

F3 (infinitesimal process) 6 An adiabatic system is defined as one that does not permit heat transfer. Thus, for an adiabatic system, the above two relations become

dS ³

(DS)ad ³ 0

(finite process)

(dS)ad ³ 0

(infinitesimal process)

A thermodynamic system may exchange heat with a number of reservoirs, but the extended system consisting of the system and all the reservoirs form an adiabatic system. In thermodynamics, this equivalent adiabatic system is called the universe.32 The subscript U denotes the quantities pertaining to it, so that the above relations become (DS)U ³ 0

(finite process)

(dS)U ³ 0

(infinitesimal process)

These relations are the mathematical form of the second law of thermodynamics. They show that, for all the processes occurring inside the universe, the entropy cannot decrease. In particular, for the irreversible processes occurring inside the universe, the entropy keeps 31 32

This can be done because it is a reversible process. Mathematically, this can be done for the other process too, but physically it is meaningless. The most unfortunate term that generated great controversies.

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increasing. This is the well-known principle of increase of entropy stated as follows: the entropy of thermodynamic universe (i.e. the equivalent adiabatic system) cannot decrease (i.e. either it increases or remains constant).

7.12.2

Possible Processes

The following discussion deals with only the finite processes since they are the practical ones. The principle of increase of entropy can be used to determine as to which of the processes can possibly be executed. This test can be written as DS ³ ± ( FQ/T) and applied in the following steps. (a) The initial state and the process are given. (b) Use the energy (first law) equation to calculate the end states. (c) If the property tables of the substance (e.g. steam tables) are available, then read (or estimate) the properties corresponding to the end state and go to step (e). (d) If the equation of state and the equations for properties are given (e.g. ideal gas), then choose a convenient reversible (non-degenerate) process and calculate the entropy

change as '5 ± F7 RF8 6 .

(e) For the actual process, evaluate ± F 3 6 .

(f) Check that '5 s ± F 3 6 ).

An alternate procedure is to check if (DS)U ³ 0. The steps of this procedure are also the same as above except that the last step is replaced by: (a) Calculate the entropy change of each reservoir as Q/T. (b) Form the universe by combining the system with all the reservoirs. (c) Obtain the entropy change of the universe as the algebraic sum entropy changes of the system and all reservoirs (because, being a property, entropy is additive over system, i.e. it is an extensive property), therefore, (DS)U = Sproc (DS)proc + Sres (Q/T)res. (d) Verify that this is positive, i.e. (DS)U ³ 0. Examples in Section 7.14 illustrate the application of both of these methods.

7.12.3

Possible Cyclic Devices

For cyclic devices the more convenient criteria are Carnot theorem for 2T cyclic devices (engines, heat pumps, refrigerators), i.e. only those 2T cyclic devices which satisfy the Carnot theorem are possible. In other words, for possible cycles, h £ hR, if the device is an engine and COP £ COPR if the device is a refrigerator or heat pump. The computational procedure (for an engine) is: (a) For the actual cycle calculate W, Q1 and hence, h = W/Q1. (b) Taking the highest and the lowest temperatures in the cycle as T1 and T2, respectively, calculate the corresponding Carnot efficiency, i.e. hR = 1 (T2/T1).

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(c) Verify that h £ hR. The calculations for a refrigerator or heat pump are similar. For cyclic devices which exchange heat with more than two reservoirs, the Clausius inequality is used, i.e. 6 [( FQ/T)j] £ 0. L

7.13

ADIABATIC, REVERSIBLE AND ISENTROPIC PROCESSES

Since these processes differ from one another, it is necessary to distinguish between them. This is done now. For an infinitesimal process, the Clausius inequality becomes F5 s

or

F3 6

F3 F5R 6 where dSp is a non-negative (i.e. zero or positive) quantity added to convert the inequality to an equality.33 Since the above step is done purely for mathematical convenience, it is necessary to interpret the terms to see if they have any physical meaning. (a) Now, the term dS represents the change of entropy. It can be positive, zero or negative. (b) The term FQ/T is called the entropy flow, because it is associated with the heat flow. It can be positive, zero or negative depending upon the sign of FQ. (c) The term dSp is called the entropy produced. It is positive for an irreversible process and zero for a reversible one. This leads to the following definitions: F5

(a) An isentropic process is one for which dS = 0.

(b) An adiabatic process is one for which FQ = 0. (c) A reversible process is one for which dSp = 0. Now the above equation shows that a term will be zero if and only if other two terms are zero which means that (a) an adiabatic reversible process will be isentropic, (b) a reversible isentropic process has to be adiabatic, and (c) an adiabatic isentropic process will be reversible. In terms of the concept of entropy produced, the second law can be written as dSp = dS dSp ³ 0

33

F3 6

(by definition) (the second law)

This is a standard procedure used in solving a system of inequalities by converting the inequalities into equalities. For example, in the LP (Linear Programming) method of optimization, such quantities are called slack variables.

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for an infinitesimal process. And as F3 Sp = DS ± 6 Sp ³ 0

(by definition) (the second law)

for a finite process. Note that for an adiabatic system (and, therefore, the thermodynamic universe) (Sp)ad = (DS)ad (Sp)U = (DS)U

(since universe @ adiabatic system)

In the beginning of this chapter, a (heat) reservoir has been defined as one that can exchange (absorb or reject) finite quantities of heat without changing its temperature. Then, for the case of a reservoir at temperature T exchanging Q of heat with a system (or, more systems), the definition of entropy reduces to (DS)R = (Q/T)R. One of the most important consequences of the distinction between quasi-static and reversible processes is the commonly employed statement, namely the area under the curve in the TS plane equals the heat transferred reversibly. This can be easily understood from the following arguments: (a) A quasi-static process is defined as one with infinitesimal (vanishingly small) differences of potentials (i.e. dp, dT and dmi all tend to zero). (b) Consequently, it is a infinite succession of equilibrium states. (c) Since all the properties are uniquely defined at equilibrium states, a quasi-static process is represented by a unique (well-defined) curve in the state space, including the TS plane. (d) Then, the definition of the Riemann integral implies that ± TdS exists and it represents the area under a curve in the TS plane.

(e) It was shown above that the second law can be written as TdS ³ FQ, where the equality sign is valid for the reversible processes. (f) During the discussion or reversible processes it was shown that dissipation (including all types of friction) and heat transfer across a finite temperature difference makes a process irreversible. (g) However, when dissipations do not vanish when the potentials become zero (i.e. they are nonlinear functions of potential), a process may become quasi-static but not reversible.

7.14 EXAMPLES This section contains examples to illustrate the concepts introduced above. To save space, the system and process diagrams are omitted. They are the same as in Chapters 5 and 6.

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EXAMPLE 7.5 A thermal conductor connects two thermal reservoirs at temperatures of 1000 [K] and 400 [K]. The steady-state heat flow rate from the hot to the cold reservoir is 1000 [W]. Determine the rate of entropy production by the conductor. Solution

This example illustrates why the irreversible processes are said to produce entropy.

1. The system is the thermal conductor which is a closed one since its mass remains constant. 2. This system absorbs heat (say, 3 1) from the hot reservoir (subscript R1) and rejects heat (say, 3 2) to the cold reservoir (subscript R2). The quantities pertaining to the system are unsubscripted. 3. Since the system operates at steady state, 7 5 (since properties are only functions of state). 4. The energy (first law) equation for a closed system reduces to 3 = 0 (since 9 = 0, (no expansion and no other work modes are present)); or 3 3 3 (say). 5. For this case, the second law is written as 57 ³ 0. 6. The universe of a system has been defined as the equivalent adiabatic system obtained by adding, to the system, all the reservoirs with which it exchanges heat. Hence, here the universe will consist of the two reservoirs and the conductor. 7. Since entropy is an extensive property, the second law becomes, 57 54 5 54 8. Since entropy change of a reservoir is defined as (DS)R = Q/T, where Q is the heat absorbed from the reservoir at temperature T, this relation becomes 57

§ 3 · § 3 · ¨ ¸ ¨ ¸ © 6 ¹ © 6 ¹

or 57

§ ·

3 ¨ ¸ © 6 6 ¹

9. Since universe is an equivalent adiabatic system, the expression for the entropy produced reduces to DSU = DSp, and substituting the values gives, 5 R 57 = (1000)[(1/400) (1/1000)] = 1.5 [W/K]. The main objective of this example is to show the origin of the term entropy produced. Now, entropy flows into the system at the rate of 3 /T1 and it flows out at the rate of 3 /T2. Since T2 < T1, this means that more entropy flows out of the system than into it. Retaining the conservation principle34 implies that the extra entropy, namely 3 /T2 3 /T1 must have been produced by the conductor owing to heat transfer across the temperature difference (T1 T2). This is the reason for the statement that heat transfer across a finite temperature difference (and, any other irreversible process) produces entropy. 34

This has served us well (in conservation of mass, momentum, energy, electric charge,
).

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Consider the special case, with T1 = T and T2 = T dT. Let dQ be the heat transferred. Then, the above formula reduces to dSp = [(dQ × dT)/T2], which shows that dSp, the entropy produced will become zero when the temperature differential dT vanishes. This is the basis for the assertion that heat transfer at a vanishing temperature differential is reversible. EXAMPLE 7.6 An electric water heater has a resistance of 40 [ohm]. It is connected across a power source of 240 [V] for a period of 1 [h]. During the process, the temperature of the heater remains constant at 90 [°C]. Determine (a) the work done by the heater, (b) the heat transferred to the heater, (c) the change in entropy of the heater and the water, and (d) the entropy change of the universe. Solution

The following steps illustrate the procedure:

1. The resistor is the system. It is closed since its mass is constant. 2. In terms of the usual notation, the electrical work done by system is, We = i2RDt = (V 2/R)Dt = (2402/40) × (3600/1000) = 5184 [kJ], where the negative sign indicates that work is done on the resistor [Ans. (a)]. 3. Since the system is closed, the energy (first law) equation reduces to Q = W. Since DE = DU = 0 [the resistor is a simple electrical system (default assumption) and its temperature is constant]. Thus, Q = 5184 [kJ], which means that heat is lost from the system (resistor) [Ans. (b)]. 4. Since the temperature of the resistor (the system) is constant, its state is constant and hence its DS = 0 [Ans. (c)]. 5. Since the mass of the water is large compared to that of the resistor (default assumption), water behaves like a reservoir at 90 [°C] = 363 [K]. Hence, its entropy change is (DS)R = Q/T = 5184/363 = 14.28 [kJ/K] (since water absorbs heat) [Ans. (c)]. 6. Since the universe will consist of the resistor and the water, the second law is (DS)U = (DS) + (DS)R = 0 + 14.28 = 14.28 [kJ/K] [Ans. (d)]. EXAMPLE 7.7 Three kilograms of air in a rigid insulated container changes its state from 5 bar, 300 [K] to 15 [bar] while it is stirred. Determine (a) the change in entropy of the system, and (b) the entropy produced. Solution

The solution consists of the following steps:

1. The system is the ideal gas. The rigid container means that Wx = 0 and it is insulated means that Q = 0. 2. Since it is a closed system, the energy (first law) equation becomes 0 = DE + Ws = DU + Ws (since air is a pure substance). 3. Since air behaves like an ideal gas, the equation of state gives, T2 = T1(p2/p1) = (300) × (15/5) = 900 [K] [since volume is constant (rigid container)]. 4. It has been stated that the stirrer work is irreversible. Hence, for evaluating entropy, it should be replaced by an equivalent reversible process. 5. A suitable process is the reversible heating of air at constant volume.

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6. The entropy change of the system is defined as '5 ±

F3

6

4

or

'5 ±

F7 RF8 6

where the second relation is obtained by using the energy (first law) equation for a simple compressible substance (i.e. air). 7. Since the reversible process is chosen as a constant volume one, this equation for the entropy change of the system becomes OE F6 F7 X '5 ± or 6 6 where the last equation is obtained from the definition of an ideal gas and that of cv. 8. Integrating this equation and substituting the values gives

'5 ±

'5 OEX NP

6 t t NP = 2.363 [kJ/K] [Ans. (a)]. 6

9. Since the system is insulated, its entropy change is also its entropy produced, so that Sp = 2.363 [kJ/K] [Ans. (b)]. 10. Since Sp > 0, this process is irreversible. Note that this is the typical case of the experiments of Joule, Mayer and others used for determination of the mechanical equivalent of heat. This result justifies the statement that it is impossible to perform the experiments of Joule (and others) in the reverse. EXAMPLE 7.8 An insulated chamber of volume 2V1 is divided by a thin, rigid partition into two halves. One chamber contains an ideal gas at a temperature of T1. The other chamber is evacuated. The partition is suddenly removed and the equilibrium is allowed to be reestablished. Determine (a) the final temperature and (b) the change in entropy. Solution

The solution is worked out as per the following steps:

1. The system is rigid (i.e. DV = 0) and insulated (i.e. Q = 0). 2. Since the mass contained in the system is constant, it is a closed system. Note that one half of the system contains an ideal gas while the other half is evacuated is a detail that is internal to the (i.e. within the boundaries of) system and, therefore, has no influence on the analysis. 3. Then, the energy (first law) equation becomes, DU = 0. 4. Since air is assumed to be an ideal gas, its internal energy, U is only a function of the temperature. Then, the above equation will mean that DT = 0, or T1 = T2 [Ans. (a)]. 5. This expansion of the air from (T1, V1) to (T2, V2) is called a free expansion, since there is no resisting pressure to ensure that the process is quasi-static. 6. Since the actual process is irreversible (as shown below), a reversible process is chosen to evaluate the entropy change of the gas.

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7. Since both T and V of the gas change, a general process from (T1, V1) to (T2, V2) is chosen as the equivalent reversible process. 8. Then, the entropy change in an infinitesimal process can be written as F5

F34 6

or

F5

F7 RF8 6

where the first equation is the definition of entropy change in a process, and the second one is the result that air is a pure substance (i.e. E = U) and only expansion work is done. 9. Using the definitions of an ideal gas and of cv, this equation becomes F5

OEX F6 O46 8 F8 6

or

¦ F8 µ ¶ ¨ 8 ·

F5 O4 §

where the second step arises because the energy (first law) equation above shows that T is constant. 10. Integrating this equation and substituting the data gives '5

O4 NP

8 8

O4 NP =#PU D?

11. Since DS > 0, this result means that free expansion of an ideal gas is irreversible. Joule conducted these famous experiments on free expansion of gases first. He discovered that the temperature does not change during free expansion. This is known as the Joules law which states that the internal energy of an ideal gas at constant temperature is not a function of its volume. EXAMPLE 7.9 A metal block (m = 5 [kg], c = 0.4 [kJ/kg.K]) at 40 [°C] is kept in a room at 20 [°C]. It is cooled in the following ways: (a) using a Carnot engine (executing integral number of cycles) with the room itself as the cold reservoir and (b) naturally. In each case calculate the change of entropy of (i) the block, (ii) the air of the room, and (iii) the universe. Solution

The solution is worked out as per the following steps:

1. The system is the metal block with constant specific heats and which does not change volume (default assumptions). 2. Then, the energy (first law) equation for a reversible infinitesimal process reduces to dQR = mcdT. 3. The defining relation of entropy then gives DS = ± ( FQ/T) = mc ln (T2/T1), where the last step is obtained by evaluating the integral and substituting the limits. 4. Substituting the values, this result reduces to

'5 t t NP

[kJ/K] [Ans. (a)(i)].

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5. Since the first process [i.e. process (a)] is reversible, the entropy change of the air in the room is equal to that of the metal block, i.e. (DS)air = + 0.1321 [kJ/K] [Ans. (a)(ii)], and (DS)U = 0 [kJ/K] [Ans. (a)(iii)]. 6. For the second process [process (b)] too, the entropy change of the metal block is the same, i.e. DS = 0.1321 [kJ/K] [Ans. (b)(i)]. 7. But since the air directly absorbs the heat from the block, it acts like a reservoir so that its entropy change is (DS)air = mcDT/T = + 0.1365 [kJ/K] [Ans. (b)(ii)]. 8. The entropy change of the universe is then, (DS)U = + 0.0044 [KJ/K] [Ans. (b)(iii)]. This example gives a quantitative measure of the maximum work that can possibly be obtained during a process in which a body exchanges heat with one reservoir. This can be understood by evaluating the work done by the Carnot engine of case. (a) above. For clarity, the temperature of air is denoted as T0. For simplicity, the Carnot cycle is assumed to be infinitesimal. At some arbitrary time, during the cooling of the metal block, let its temperature be T. Now, consider one cycle executed by the engine. Let it absorb FQ from the metal block, thereby cooling it by dT. Then, the energy (first law) equation for the metal block becomes FQ = DU = mcdT. The efficiency of the Carnot engine at this time will be hR = 1 (T0/T). Then, the work output of the engine at that time will be FW = [1 (T0/T)](mcdT). The total work output from the engine when the block cools from T1 to T0 is obtained by integrating this expression, i.e. 6

W = ± [1 (T0/T)](mcdT) 6

or

W = (mc)(T0 T1) (T0)[mc ln (T0/T1)]

or, using the results derived above, this becomes W = (DU)block T0( DS)block

or

W = D(U T0S)block

Since the Carnot theorem asserts that the work output by a Carnot (reversible) engine is maximum (and, of course, all others are less than this), this result can be generalized as follows: The maximum work done by a system executing a process by exchanging heat with a reservoir at TR is, W = D(U TRS). This is, an important function since the availability function discussed in Chapter 8 is a modification of this. The examples so far discusses here have dealt with substances with simple equations for state and properties so that DS can be evaluated by appropriate equations. The following example shows the procedure of employing property tables for this purpose. EXAMPLE 7.10 Ten kilograms of dry saturated steam at 30 [bar], contained in a closed system, is brought into thermal contact with a heat sink at 200 [°C]. The steam rejects 16,200 [kJ] of heat during a constant pressure process. Determine (a) the final state of the steam, (b) the change in entropy of the steam, and (c) the change in entropy of the universe. Solution This process of steam condensing involves heat transfer across a finite temperature difference, since at p = 30 [bar], Tsat = 233.84 [°C], while the temperature of the cold reservoir (heat sink) is 200 [°C].

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1. Since the initial state is saturated vapour at 30 [bar], the steam tables give, h1 = 2802.3 [kJ/kg] and s1 = 6.1837 [kJ/kg.K]. 2. It is known that when steam condenses the pressure remains constant. 3. Hence with the default assumptions, the energy equation becomes, Q = DE + W = DU + Wx = DU + pDV = D(U + pV) = DH = m(h2 h1). 4. Solving for h2 and substituting the data gives, h2 = h1 + (Q/m) = 2802.3 (16200/10) = 1182.3 [kJ/kg]. 5. Since the steam is wet, using the standard formula to evaluate the dryness fraction gives, x2 = 0.097 [Ans. (a)]. 6. Since the properties of the working substance (water substance) are available, it is not necessary to assume an equivalent reversible process.35 7. Again using the standard formula, the entropy of the wet vapour is obtained as s2 = sf + x2(sg sf) = 2.9887 [kJ/kg.K]. 8. Hence, DS = m(s2 s1) = 31.95 [kJ/K] [Ans. (b)]. 9. Using the definition, the entropy change of the reservoir is, DSR = QR/T = + 16200/ 473 = 34.249 [KJ/K]. 10. The universe is constructed by combining the system with the cold reservoir. Then, (DS)U = DS + DSR = + 2.300 [kJ/K] [Ans. (c)]. This shows that the process of condensation of a vapour is irreversible. Using similar arguments one can show that the process of evaporation of a liquid is also irreversible.

7.15

EQUILIBRIUM

It is shown in mechanics that a system reaches equilibrium when its energy is minimum. Since it does not consider heat transfer or friction, this means that the processes in mechanics are isentropic (i.e. reversible adiabatic). Thus the condition of equilibrium in mechanics can be written as (DE)S = 0 (see also Chapter 5.). For thermodynamic processes which involve work and heat transfers, the second law provides the criterion for testing whether a system will be in equilibrium or not. The principle of increase of entropy implies that as the system executes an adiabatic process spontaneously, its entropy either remains constant or increases. If the entropy increases, the process goes on till the entropy reaches a maximum value. Thus, at equilibrium (DS)E = 0. Thus, the condition of equilibrium is usually written as

35

(dS)E = 0

(infinitesimal process), and

(DS)E = 0

(finite process)

This has been done by those who prepared the tables.

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Generally, the test for the second derivative is mathematically more involved, and, therefore, only the first derivative is used to evaluate these expressions. The physical arguments are used to confirm that the entropy is maximum at constant energy or energy is minimum at constant entropy (whichever is more convenient).

7.16 THE CARATHÉODORY FORM OF THE SECOND LAW Unlike the KelvinPlanck and Clausius statements of the second law, in this formulation the second law is stated as the impossibility of executing some adiabatic processes,36 i.e. the impossibility of reaching some states through adiabatic processes from a given state. Hence, it is known as the reachability form (or, statement) of the second law. The Carathéodorys formulation of the second law is very simple and profound. It is slightly unconventional in the sense that it uses only mathematical arguments,37 and, therefore, has been regarded as non-physical and thus (automatically) difficult. In this section, the Carathéodory formulation is developed with emphasis on the basic ideas. For clarity, the discussion is confined to simple compressible systems. Appendix D presents the details.

7.16.1

Accessible and Inaccessible States

The only physical assumption made in the Carathéodorys formulation is the existence of inaccessible (and hence, accessible) states. Therefore, these are reviewed below. For example, when two liquids are mixed in an adiabatic vessel, the final temperature should lie between the temperature of the colder liquid and that of the hotter one. All other states are inaccessible. For ease of understanding, consider an ideal gas expanding from an initial state i to the final state f. In terms of T and v, the relation for change of entropy is

'U E NP X

6

H

6

X

4 NP

H

X

K

K

As per the second law, only those adiabatic processes are possible for which

'U s

'U E NP

KG

X

6

H

6

4 NP

X

H

X

K

s

K

For an ideal gas, cv = R/(g 1), the above relation, therefore, simplifies to 6

H

§X · NP NP ¨ ¸ 6 ©X ¹ 36 37

H

H

K

K

t

or as

6 XH s 6 XH H

H

K

K

Recall that adiabatic processes have been defined as those taking place inside an adiabatic enclosure (i.e. processes executed by adiabatic systems). This should not be surprising since Carathéodory was a mathematician and is famous for his basic contributions to the measure theory and the theory of integration.

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This relation can be explicitly written as H

H §6 · §X · or as vf ³ vi ¨¨ ¸¸ Tf ³ Ti ¨ ¸ ¨X ¸ ©6 ¹ © ¹ Note that this is a special case of the isentropic expansion of an ideal gas which may be written as K

K

H

H

pvg = C

Tv(g 1) = C

or

6

or

R

(g 1)/g

1

H

H

%

where C = pivi = Tivi = Ti/(pi) . This result may be summarized as: · The accessible states of an ideal gas for adiabatic processes are those for which p, V and T are related through pvg ³ C; or, Tvg 1 ³ C; or, 6R H H s %, where, the value of the constant C is evaluated from the initial state as shown above. · When the adiabatic process is also reversible (i.e. isentropic), only states on the curve defined by the equality are accessible. These states are shown in Figures 7.2(a) and (b). g

R R

g

6 6

R H X H H CEEGUUKDNG t R X H K

K

6 X H

H H

CEEGUUKDNG t 6 X K

H

K

K

K

#EEGUUKDNG \QPG

#EEGUUKDNG \QPG RXH

6X

RK XK H

+PCEEGUUKDNG \QPG

H

6 X H K

K

+PCEEGUUKDNG \QPG X

X K

X

X K

C Figure 7.2

D Accessible states for adiabatic processes.

The main purpose of this discussion was to illustrate the fact that near any (arbitrary) point in the state space of a system, there are other points that cannot be reached from it through an adiabatic process.38 Carathéodory uses this experimental fact to prove his theorem, namely, if in the neighbourhood of an arbitrary point in the domain of a Pfaffian differential equation, there are points that are inaccessible to it along the solution surface of the differential 38

But, we already know this because the Joules experiment cannot be reversed (or, equivalently, the KP form of the second law forbids this).

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equation, then this equation has an integrating denominator. Explanation of these terms as well as proof of this theorem is presented in Appendix C. It can be shown (see Appendix D) that this integrating denominator can be written as U

=%G

±

J R F R

?= ) G ?

where q is the empirical temperature, h(q) is an arbitrary function of q, f is some state variable, F(q) is some arbitrary function of state, and C is a constant. Then, the heat transferred dQ is written as dQ = t df and comparing with the second law (i.e. dQ = TdS) it is seen that T is defined as T = Ceòh(q)dq, and dS should be defined as dS = F(f). The constant C is chosen positive so that the principle of increase of entropy will be satisfied, and its value is chosen such that Ceòh(q )dq = 273.16 [K], where q3 is the empirical temperature at the triple point of water. 3

7.17

THE SECOND LAW FOR FLOW SYSTEMS

The extension of the second law to flow systems is direct. As was done for the energy (first law) equation, the entropy change of the control volume owing to the mass flows should be taken into account so that 5%8 s

¥

3 6 O K UK O G UG

TGUGTXQKTU

or in terms of the entropy produced, as 5 R 5%8

¥

3 6 O K UK O G UG (definition) and 5 R s (second law)

TGUGTXQKTU

The following example illustrates the application of this equation. EXAMPLE 7.11 Steam at 100 [bar] and 500 [°C] is expanded through a turbine to 1 [bar] with hs = 0.90. During the process the turbine loses 20 [kJ/kg] of heat to the atmosphere at 300 [K]. Calculate (a) the final state of the steam, (b) the entropy change of the steam, (c) the entropy change of the atmosphere, and (d) the net entropy produced. Solution Since no explicit assumptions are specified, the default ones for a turbine mentioned in Chapter 6 are applicable. 1. Then, the energy (first law) equation together with the continuity equation gives, 9 O 3 O J J . 2. For the isentropic expansion, se = si = 6.5994 [kJ/kg.K] (from steam tables). 3. Using the usual formula this gives, xe = (6.5994 1.3027)/(7.3598 1.3027) = 0.8745. 4. Then, another standard formula gives, he = 417.51 + (0.8745)(2675.4 417.51) = 2392.0 [kJ/kg]. 5. Then, the work done during this isentropic process becomes, Ws = ( 20) + (3374.6 2392.0) = 962.6 [kJ/kg], where, hi = 3374.6 [kJ/kg] (from steam tables). K

G

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6. 7. 8. 9. 10. 11. 12.

The definition, hs = W/Ws implies that W = hsWs = (0.9)(962.6) = 866.3 [kJ/kg]. Then,39 he¢ = hi W = 3374.6 866.3 = 2508.3 [kJ/kg]. Then, as earlier, xe¢ = (2508.3 417.51)/(2675.4 417.51) = 0.9260 [Ans. (a)]. Then, se¢ = 1.3027 + (0.9260) (7.3598 1.3027) = 6.9116 [kJ/kg.K]. Then, the entropy change of the system in the actual process is, 6.9116 6.5994 = 0.2122 [kJ/kg.K] [Ans. (b)]. Since the atmosphere acts as a reservoir, (DS)atm = 20/300 = 0.0667 [kJ/kg.K] [Ans. (c)]. Since the turbine and the atmosphere to which the turbine loses heat form the universe, the net entropy produced equals the change of entropy produced, i.e. Sp = (DS)U = (DS)sys + (DS)atm = 0.2789 [kJ/kg.K] [Ans. (d)].

REVIEW QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 39

Define a reservoir. What is an nT engine? What is a heat pump? What is a refrigerator? What is the difference between them? What are the typical statements of (a) a process problem, (b) an energy conversion problem, and (c) a heat pumping problem? Show that work and heat are equivalent to one another for process problems. What are the performance criteria used for testing the equivalence of work and heat for a cycle? Why are they chosen? Show that the conversion of work into heat is complete and continuous. State the KelvinPlanck (KP) form of the second law of thermodynamics. What is PMM2? State the second law in terms of this device. Show that if an engine exchanges heat with more than one reservoir, then at least one of the reservoirs should be a heat rejection one. Define thermal efficiency. What is its significance? State the Clausius form of the second law of thermodynamics. How are the performance parameters of a heat pump and a refrigerator defined? Show that the KP form and the C form of the second law are equivalent. Define reversible and irreversible processes. Classify irreversibilities. Define reversible and irreversible cycles. State and prove the Carnot theorem for an engine. Develop the thermodynamic (Kelvin) scale of temperature. Using the Joules law show that it is identical to the ideal gas scale.

Note that heat loss has already been accounted for in the earlier step. Hence it should not be included here. This relates only to the flow irreversibilities.

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21. 22. 23. 24. 25. 26.

Define negative thermodynamic temperatures. What are their characteristics? Derive an expression for the change of entropy of a non-flow process. Derive the Clausius inequality. Derive the principle of increase of entropy. What is the test of possibility of execution of a process and a cycle? Define adiabatic, reversible and isentropic processes. What are the differences between them? 27. Express the condition of equilibrium according to the second law. 28. State the Carathéodorys form of the second law of thermodynamics. 29. Derive the second law equation for a flow system.

EXERCISES Notes:

The following are the default assumptions.

1. All processes are quasi-static. 2. Subscripts 1 and 2 denote the initial and final states, respectively. 3. Air behaves like an ideal gas with molecular weight (RMM) M = 29 [kg/kmol] and g = 1.4. 1. Prove that in an enclosure at uniform wall temperature (a) the radiation is uniform, i.e. it is the same at all points and (b) the temperature within the enclosure is uniform. [Hints: Assume that these propositions are false, run an engine between the high and the low points and directions, and show that it becomes a PMM2.] 2. Prove the Carnot theorem by assuming that the heat rejected by both the reservoirs is the same. [Hints: Directly feed the heat rejected by the engine to the heat pump and eliminate the cold reservoir. Show that it is a 1T engine (i.e. W > 0).] 3. Prove that the efficiencies of all the reversible engines are the same. 4. Prove that the efficiency of a reversible engine depends only on the temperature of the two reservoirs (i.e. it does not depend on the size of engines, working substance,
.) 5. Show that the COP of a reversible refrigerator is greater than that of any other refrigerator working between the same two reservoirs. [Hints: (For exercises 35): (a) assume the contrary, (b) reverse the reversible device so that it works as a heat pump, (c) equalize W or Q2 or Q1, and (d) show that the combined system is a PMM2.] 6. An inventor claims that his engine absorbs 100 [kW] of heat from a reservoir at 2000 [K], produces 90 [kW] of work and rejects heat to the atmosphere at 300 [K]. As an entrepreneur, will you invest in this project? Why?

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7.

8.

9.

10.

[Hints: The maximum possible work output from an engine under the given conditions is (100) × [1 (300/2000)] = 85 [kW]. So the engine cannot be constructed. The inventor is making wrong claims, and money should not be invested in this project.] A Carnot engine working between 227 [°C] and 27 [°C] drives a Carnot heat pump working between T [°C] and 27 [°C]. The ratio of the net heat transferred to the reservoir at 27 [°C] to the heat absorbed by the engine is 4.6. What is the value of T? [Hints: Let Q1 and Q2 be heat absorbed and heat rejected by the engine, respectively. Let Q3 and Q4 be heat absorbed and heat rejected by the heat pump, respectively. Since they are Carnot devices, (Q2/Q1) = (300/500) and (Q3/Q4) = (T/300). Since the engine drives the heat pump, Q1 Q2 = Q4 Q3. Finally, the given data implies (Q2 + Q4)/Q1 = 4.6. Solving all these equations gives, T = 330 [K].] A Carnot engine absorbs heat Q1 from a reservoir at T1 and rejects heat Q2 to a reservoir at T2. However, due to heat conduction effect, the temperatures drop by DT1 and DT2 on the hot and cold sides, respectively, where DT1 = CQ1 and DT2 = CQ2 with C as a constant (it is called the thermal conductance). Show that the efficiency of the engine is h = 1 [T2/(T1 2CQ1)]. [Hints: Let q1 and q2 be the actual temperatures as seen by the engine. Then, q1 = T1 DT1 = T1 CQ1 and q2 = T2 + DT2 = T2 + CQ2. Since the engine is reversible, by the definition of Kelvin temperature, q2/q1 = Q2/Q1, and for C ¹ 0, this gives q2/q1 = CQ2/CQ1. Substituting the above expressions for q1 and q2, gives q2/q1 = (T2 + CQ2)/(T1 CQ1). Now, using one of the well-known (dividendo) rule of ratios this can be written as: q2/q1 = (T2 + CQ2 CQ2)/(T1 CQ1 CQ1). Since the efficiency of a Carnot engine is h = 1 (q2/q1), the result follows immediately.] A reversible heat engine absorbs heat from reservoirs at 800 [K] and 600 [K] and rejects heat to one at 300 [K]. It develops 500 [kJ] of work. The heat received by the engine from the 800 [K] reservoir equals double of that received from the 600 [K] reservoir. Determine the magnitudes of the heat flows and the efficiency of the engine. [Hints: Since the engine is reversible, the Clausius inequality gives Q/T = 0. Let the heat absorbed from the reservoirs at 800 [K] and 600 [K] be Q11 and Q12. Let Q2 denote the heat rejected. Let W be the work done. Then, the energy (first law) equation gives: Q11 + Q12 = Q2 + W. From the given data, Q11 = 2Q12. The Clausius inequality may be written as: (Q2/300) = (Q11/800) + (Q12/600). Solving these equations gives Q11 = (4000/7), Q12 = (2000/7), Q2 = (2500/7) and h = 7/12.] Five kilograms of ice (c = 2 [kJ/kg.K]) at 10 [°C] is placed in an insulated tank containing 50 [kg] of water (c = 4.186 [kJ/kg.K]) at 30 [°C]. The ice melts (latent heat = 335 [kJ/kg]) and cools the water till equilibrium is reached. Calculate the entropy production. [Hints: During melting of ice, the pressure remains constant. Since the tank is insulated (Q = 0), with the default assumptions, the energy (first law) equation gives DH = 0; or, H2 = H1; or, (DH) ice = (DH)water. Let T be the final temperature of the mixture. Since ice and water are incompressible, cp = cv = c. Substituting the values gives, (5)(2)[0 (10)] + (5)(335) + (5)(4.186)(T 0) = (50)(4.186)(30 T); or, T = 19.56 [°C] = 292.56 [K]. Using the usual formulae, (DS)ice = (5)(2) ln (273/263) + (5)(335/273) + (5)(4.186) ln (292.56/273) = 7.9570 [kJ/K]; and, (DS)water = (50)(4.186) ln (292.56/ 303) = 7.3387 [kJ/K]; so that Sp = (DS)U = (DS)ice + (DS)water = 0.6183 [kJ/K].]

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11. During heat treatment of a steel product (m = 10 [kg]; c = 0.5 [kJ/kg.K]) at 800 [K], it is quenched in oil (m = 50 [kg]; c = 2 [kJ/kg.K]) at 400 [K]. Assume that the overall process is adiabatic. Calculate (a) the final temperature of the steel, and (b) the entropy production of the process. [Hints: With default assumptions, the energy (first law) equation gives, [mc(Ti Tf)]steel = [mc(Tf Ti)]oil. So Tf = 419 [K]. Since the overall process is adiabatic 5 R = (DS)U = (DS)sys = (DS)steel + (DS)oil = (10)(0.5)(ln (419/800)) + (50)(2)(ln (419/400)) = 1.407 [kJ/K].] 12. Five hundred grams of water (c = 4.18 [kJ/kg.K]) at 300 [K] is stirred in an insulated vessel. The energy required for the stirring is obtained from a mass of 50 [kg] falling through a height of 20 [m]. Assuming g = 9.81 [m/s2], compute the entropy change of water. [Hints: Consider the water as the system. It is closed. From the above data Q = 0, E = U, and W = mgh. Then, the energy (first law) equation gives, (DU)water = mgh, or substituting the values, (0.5)(4.18)(T 300) = (50)(9.81)(20)/1000, or, T = 304.69 [K]. Then, the definition of entropy gives, (DS)water = (0.5)(4.18) ln (304.69/ 300) = 0.03242 [kJ/K].] 13. Five kilograms of water at 1 [bar], 50 [°C] contained in a closed system is heated and compressed to 50 [bar], 100 [°C]. Calculate the change in entropy of the water, assuming it to be incompressible. [Hints: Both initial and final states are the sub-cooled water. The steam tables give, s1 = 0.70347 [kJ/kg.K] and s2 = 1.3030 [kJ/kg.K], so that DS = m(s2 s1) = 2.9976 [kJ/K]. By the approximation of simple incompressible fluid, s1 = sf,50°C = 0.70351 [kJ/ kg.K], and s2 = sf,100°C = 1.3069 [kJ/kg.K], so that DS = 3.0171 [kJ/K]. Hence, the approximation is good.] 14. When an ideal gas (cv = 0.748 [kJ/kg.K]) at 0.06 [m3/kg] and 540 [K] was expanded reversibly and adiabatically to 0.19 [m3/kg], its temperature was found to fall by 170 [K]. However, when it was expanded adiabatically but irreversibly, its temperature was found to fall only by 30 [K]. Determine the change of entropies in both cases. [Hints: The value of R should be found first so that the standard formula can be applied. Now, for the reversible process, Ds = 0 = cv ln (T2/T1) + R ln (V2/V1). Substituting the values, R = 0.2453 [kJ/kg.K]. Using this value of R for the irreversible process gives, Ds = 0.240 [kJ/kg.K].] 15. A gaseous mixture of 1 [kg] O2 and 2 [kg] N2 (assumed to be ideal with g = 1.4) at 7 [bar] and 400 [K] is expanded reversibly according to the process law pV1.2 = C, till the temperature reaches 300 [K]. Calculate (a) specific gas constant and (b) the change of specific entropy. [Hints: From definition, xO2 = 0.30435 and xN2 = 0.69565. Then the molecular weight is, M = (0.30435)(32) + (0.69565 (28) = 29.2174 [kg/kmol] [Ans. (a)]. Now the ideal gas law gives, V1 = 0.48783 [m3]. Then the process law gives, V2 = (V1) · (T1/T2)(1/0.2) = 2.056 [m3]. Substituting these values in the standard formula gives, Ds = 0.1833 [kJ/K][Ans. (b)].]

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16. A rigid, well-insulated vessel is divided into two parts by a perfectly heat conducting piston of negligible heat capacity. The piston is initially clamped at some place. One part contains 1 [kg] of air at 5 [bar] and 350 [°C] and the other 3 [kg] of CO2 at 2 [bar] and 450 [°C]. The piston is released and the system is allowed to attain equilibrium. Assuming that the two gases behave ideally with constant specific heats, determine (a) the final state of gases in both compartments, (b) the changes in entropy of both gases, and (c) whether this process is possible or not. [Hints: Since the piston can move freely and is perfectly heat conducting, p and T should be same on both sides of the piston. So the air goes from 5 [bar] and 350 [°C] to p and T, and CO2 goes from 2 [bar] and 450 [°C] to p and T. So DS for each gas can be computed from the standard formula. Let A and B denote the two compartments. Since, the total system is adiabatic and rigid, the energy (first law) equation, with the other default assumptions, gives DU = 0. Since the working substances are ideal gases, this gives T = (mA . cv,A . TA,i + mB . cv,B . TB,i)/(mA . cv,A + mB . cv,B = 689.4 [K] [Ans. (a)]. Since the vessel is rigid, the total volume is constant, i.e. VA,i + VB,i = VA,f + VB,f. Using the ideal gas equation to convert this relation in terms of pressures and solving for the common pressure gives, p = 1.363 [bar]. [Ans.(a)]. Finally, the standard formula gives DSA = 0.4743 [kJ/kg.K] [Ans. (b)]. DSB = 0.1231 [kJ/kg.K] [Ans. (b)]. Since, (DS)U = 0.5974 [kJ/kg.K] which is (+), this process is possible [Ans. (c)].] 17. A rigid vessel of 2 [m3] volume is initially divided into two halves. Initially, one part (say A) contains air at 10 [bar] and 400 [K] and the other (say B) contains air at 5 [bar] and 500 [K]. The partition is removed and the gases are allowed to mix with each other. Assume that during mixing 1 [MJ] of heat is lost to the atmosphere at 300 [K]. Calculate (a) DSA, (b) DSB, and (c) DSU. [Hints: Since air behaves like an ideal gas (default assumption), its equation of state gives, mA = 8.7199 [kg] and mB = 3.4880 [kg]. The energy (first law) equation becomes, DU = Q, since W = Wx (default assumption) = 0 (rigid vessel, i.e. DV = 0). Since U is an extensive property, this reduces to (DU)A + (DU)B = Q, or (8.7199)(0.717) (400 T) + (3.488)(0.717)(500 T) = 1000, or T = 314.33 [K]. Then, by the usual formula, DSA = (8.7199)[(0.717) ln (314.33/400) + (0.2867) ln (2)]= 0.2259 [kJ/K] [Ans. (a)]. Similarly, DSB = (3.488)[(0.717) ln (314.33/500) + (0.2867) ln (2)] = 0.4677 [kJ/K] [Ans. (b)]. Since atmosphere is a reservoir, (DS)R = (1000/300) = 3.3333 [kJ/K]. Then, (DS)U = (DS)A + (DS)B + (DS)R = 3.0915 [kJ/K] [Ans. (c)].] 18. Saturated water of mass 2.45 [kg] at 10 [bar] is mixed at constant pressure with 1 [kg] of superheated steam at 10 [bar], 250 [°C]. During the process, the mixture is stirred with a constant torque of 1 [N.m] at 500 [rpm] for 1 [h] while absorbing 100 [kJ] of heat. Determine (a) the change of energy, (b) the net work done, and (c) the change in entropy of the system. [Hints: Considering this as a process of mixing at constant pressure in a closed system, the default assumptions reduce the energy (first law) equation to Q = DH + Ws; or, 100 = (3.45)(h2) [(2.45)(762.61) + (1)(2943.0)] (1)(2p)(500)(60)/1000; or, h2 = 1478.2 [kJ/kg]. Then, the usual formula gives, x2 = 0.3554. With the usual formulae,

Chapter 7:

19.

20.

21.

22.

The Second Law of Thermodynamics

169

v2 = 0.06978 [m3/kg] and s2 = 3.7178 [kJ/kg.K], then, DU = 283.17 [kJ] [Ans.(a)]. Then, Wx = 5.2289 [kJ], Ws = (60p) = 188.4956 [kJ], so that the net work done is W = 183.3 [kJ] [Ans.(b)]. Then, DS = (3.45)(3.7178) 6.9259 (2.45)(2.1382) + (100/452.88) = 0.8827 [kJ/K] [Ans.(c)].] Air may be taken as an ideal gas mixture of mole fractions of nitrogen and oxygen of 0.79 and 0.21, respectively, each component assumed to be an ideal gas as well. If air is separated at 1 [bar] and 300 [K] to its components at the same pressure and temperature, what will be the change in entropy? [Hints: The entropy of separation = negative of that of mixing, since the end states of these processes are the same. Now, for mixing of ideal gases, DS = R Sni ln yi where yi = (pi/Pi) with pi and Pi being the partial pressures of the component i after and before mixing ([DEN], p. 118). If all the components are in pure state, all Pi = P(say). Moreover, if P is the same as the total pressure of the mixture, then yi = xi then, DS = nR S xi ln xi. Consequently, for separation, DS = nR S xi ln xi. Substituting the values, DS = (8.3143)(1)[(0.21) ln (0.21) + (0.79) ln (0.79)] = 4.2732[kJ/K].] An evacuated, rigid and insulated bottle is filled with steam from mains at 5 [bar] and 250 [°C]. After the flow has stopped, 1 [kg] of steam was found to have entered the bottle. Calculate the entropy produced during the process of filling. [Hints: 5 R = (DS)U = (DS)steam = 7.2721 0 = 7.2721 [kJ/K].] Steam at 100 [bar] and 600 [°C] is expanded through a turbine to 1 [bar] with hs = 0.95. During the process the turbine loses 20 [kJ/kg] of heat to the atmosphere at 300 [K]. Calculate (a) the final state of the steam, (b) the entropy change of the steam, (c) the entropy change of the atmosphere and (d) the net entropy produced. [Hints: This is a standard exercise. But, the entropy is produced because of the flow irreversibility in turbine and because of heat loss to atmosphere. The data gives, xe = 0.9243 and then, the energy equation gives, he = 2504.5 [kJ/kg]. Then for hs = 0.95, he¢ = 2560.4 [kJ/kg], so that xe¢ = 0.9491 [Ans. (a)]. Then, the usual formula gives, se¢ = 7.0515 [kJ/kg.K], so that (DS)sys = 0.2011 [kJ/kg.K] [Ans. (b)]. Then, (DS)atm = (20/300) = 0.0667 [kJ/kg.K] [Ans. (c)], so that 5 R = 0.2169, kJ/kg.K] [Ans. (d)].] Saturated steam at 15 [bar] enters an adiabatic turbine at the rate of 10 [kg/s] and expands to 1 [bar]. Assuming that hs = 0.9, calculate (a) the actual power developed and (b) the rate of entropy produced. Assume the following data from the steam tables in the usual units p [bar]

hf

hg

sf

sg

15 1

844.7 417.5

2790.0 2675.4

2.314 1.303

6.441 7.360

[Hints: Standard procedure. xe = 0.8483 and he = 2332.9 [kJ/kg]. Since hs = 0.9, we get he¢ = 2355.7 [kJ/kg], so that xe¢ = 0.8584. Then, the energy equation gives, 9 = 434.3 [kW] [Ans. (a)]. Now, using the usual formulae, se¢ = 6.50623 [kJ/kg.K]. Since the turbine is adiabatic, D 5 = 0.613 [kW/kg.K] = Sp [Ans. (b)].]

Chapter 8

Auxiliary Functions

8.1

INTRODUCTION

All the chapters till now dealt with the basic concepts of thermodynamics. The rest of this book, from this chapter onwards, presents the application of thermodynamics to processes and systems important in thermal engineering. The discussion presented so far may be summarized as follows. Thermodynamics deals with storage and exchange (flow) of energy. These forms are: (a) energy flow as work (W) and energy storage as any one (or more) of the ‘mechanical energies’ (including flow work) (borrowed from mechanics); (b) storage of energy as (thermal) internal energy (U) as a consequence of the first law; (c) heat as an energy flow as a consequence of the principle of conservation of energy and the fact that energy flow can either be heat or work; and (d) T DS as an energy flow from the definition of entropy (DS = ÚdQ/T). This chapter deals with some other energy functions in addition to the above. They are energy functions since they are mathematical combinations of energy functions defined above (mathematically defined properties). All these are energy functions since they have the dimensions of energy. They are called auxiliary functions.1 These auxiliary functions are: (a) availability function defined as F = E + p0V – T0S, where p0 and T0 are the pressure and temperature of the environment, (b) enthalpy (H) defined as H = U + pV, (c) Gibbs function2 (G) defined as G = H – TS, and (d) Helmholtz function3 (A) defined as A = U – TS. Auxiliary functions are true energy functions in the sense of the work-energy theorem of mechanics, since work is done by a system at their expense, i.e. the work done equals the decrease in their values. In thermodynamics, they are employed in three ways namely (a) as energy functions, when they are used to calculate the maximum possible work available from 1 2 3

helping, subsidiary; from Latin auxili a ris = help; aug e re = to increase; Chambers 20th Century Dictionary, New ed., 1972. Gibbs function is also called free energy. Helmholtz function is also called work function. 170

Chapter 8: Auxiliary Functions

171

a process, (b) as energy functions, when they are also used to express the condition of equilibrium, and (c) as properties, when they give rise to some relations which are purely mathematical and act as constraints to all property relations. This chapter deals with these uses. The second application is presented in Section 10.7 as well.

8.2

AVAILABILITY

In Chapter 7 it was shown that the second law of thermodynamics (the Kelvin–Planck form) asserts that it is impossible to absorb heat from one reservoir and completely and continuously convert it into work using a cyclic device, i.e. W1T–engine d £ 0. In other words, the maximum possible work output by an engine absorbing heat from one reservoir is zero. Now, the question that naturally arises is: “Is there a similar upper limit for conversion of heat into work when a system executes a process exchanging heat with only one reservoir?” The answer to this question is in the affirmative and this is shown in this section. For convenience in future discussions, the following definitions ([KEE], p.290) are used. Atmosphere.

The environment of the system that acts as the reservoir.

Availability. The maximum useful work produced by a system executing a process exchanging heat with its atmosphere. It should be noted that the coldest reservoir available for heat interactions is chosen as the atmosphere. All quantities pertaining to the atmosphere are denoted with the subscript ‘0’ (zero). For example, its temperature is T0, pressure p0, …, etc. The quantities pertaining to the system are unsubscripted. In this section, the relation for availability for a non-flow process is developed first. In the general case, the work done can only be evaluated using the energy (first law) equation, i.e. as W = Q – DE. However, the second law, i.e. DS ≥ ∫ 12 dQ/T, constrains the heat transferred. Thus, to calculate the maximum possible work output from a system, the energy (first law) equation combined with the second law is used. These concepts are then extended to flow processes. Finally, they are illustrated by examples. For simplicity, the discussions are confined to engines but with appropriate changes in the terminology and signs they can also be applied to refrigerators/heat pumps.

8.2.1

Availability in Non-flow Processes

Non-flow processes are executed by closed systems. Here the relation for availability is derived in two different, though equivalent ways. Then, for any situation, the more convenient of the two methods can be used. System interacting with its atmosphere Consider a closed system executing an infinitesimal process within an atmosphere as shown in Figure 8.1.

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Atmosphere

T0

dQ

System Dotted lines = Universe Figure 8.1

A system and its atmosphere.

The principle of increase of entropy is

(dS)U ≥ 0

In this case, the universe consists of the system and its atmosphere. Since entropy is an extensive property, (dS)U = (dS)0 + dS so that the above relation becomes dS0 + dS ≥ 0 Using the definition of entropy this relation can be converted as

−

dQ + dS ≥ 0 T0

Now, the heat d−Q leaves the reservoir and enters the system so that the energy (first law) equation for the system becomes, d−Q = dE + d−W. Then, the above equation can be written as –

dE + dW + dS ≥ 0 T0

which can be rearranged (since T0 > 0) as d−W £ – (dE – T0dS)

When the boundaries of the system are displaced against the atmospheric pressure, work to be done on the atmosphere is p0dV. This work is not available for other uses. Hence, the useful work done by the system is d−Wu = d−W – p0dV

or

d− Wu £ – (dE + p0dV – T0dS)

or, since p0 and T0 are constants, (infinitesimal process) d−Wu £ – d(E + p0V – T0S) (finite process) Wu £ – D(E + p0V – T0S) where the extension to the finite process is self-evident.

(8.1)

Chapter 8: Auxiliary Functions

173

This relation shows that the useful work is maximum when the processes are reversible (when the equality part of the second law of thermodynamics is applicable). At the beginning of this section, this maximum useful work has been defined as the availability. Hatsopoulos and Keenan ([HAT], p.165) use the symbol L (the Greek capital letter ‘lambda’) to denote the availability. In this book, the symbol4 A is used. Then, dA = d−Wu, max = – d(E + p0V – T0S) = dF (infinitesimal process) A = Wu, max = – D(E + p0V – T0S) = DF (finite process) where F is the availability function defined as F = E + p0V – T0S. The above discussion shows the following important points. (a) The availability function is a property because it is a mathematical combination of properties.5 (b) It is an extensive property since it essentially is the work output (which was shown, in Chapter 3 on work, to depend upon the mass of the system). (c) When a system does work spontaneously (i.e. W > 0), the value of the availability function decreases.6 (d) When the system comes to an equilibrium with its environment, its state will be the same as that of the environment. This state is called the dead state. The value of the availability function at the dead state is, F0 = E0 + p0V0 – T0S0. (e) For a finite process between states ‘1’ and ‘2’, the relation for availability can be written as A = Wu, max = – DF = F1 – F2 = (F1 – F0) – (F2 – F0) Following the analogy to motion in gravitational field, this relation can be interpreted as the sum of the work done by the system when it goes from the initial state (state ‘1’) to the dead state (state ‘0’) and the work done on the system to take it from the dead state to the final state (state ‘2’). Irreversibility Now the relation for availability will be derived by another method. Even though this method is equivalent to that described above, it has the advantage of showing how the relation for irreversibility arises naturally. Since the system executes an infinitesimal process absorbing heat d−Q from the atmosphere at T0 and producing work d−W, the energy (first law) equation gives d−W = d−Q – dE In terms of the entropy produced, the second law is

dS = 4 5 6

dQ + d−Sp T0

Calligraphic (from Greek: kallos—beauty and graphein—to write) (or script) form of the alphabet A. However, it is not a true (blue-blooded) property because p0 and T0 are constants. In this respect it is similar to a body in the earth’s gravitational field that does work when its potential energy (height from a datum) decreases.

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Multiplying this equation7 by T0 and subtracting from the energy (first law) equation gives d−W = T0dS – T0 d−Sp – dE

or

d− W = – d(E – T0dS) – T0 d−Sp

Using the definition of useful work this becomes, d−Wu = – d(E + p0V – T0dS) – T0 d−Sp

or

d− Wu = – dF – T0 d−Sp

Comparing with the definition of the availability function, F, given above, this relation becomes d−Wu = d−Wu, max – T0 d−Sp During the discussion on the earlier method it was pointed out that the maximum possible useful work, d−Wu, max, occurs when all the processes are reversible (corresponding to the equality part of the second law). Then, it follows that the difference between this maximum useful work and the actual useful work d−Wu is the work lost because of irreversibility. Thus, it is convenient8 to define irreversibility (I) through the relation dI = dA – d−Wu = d−Wu, max – d−Wu = T0 d−Sp I = A – Wu = Wu, max – Wu = T0Sp

(infinitesimal process) (finite process)

System interacting with one reservoir as well In order to develop a methodology for extending the above analysis to any number of reservoirs, consider a system exchanging heat with one reservoir in addition to the atmosphere. Let it absorb d−Q1 from the reservoir at T1 and reject d−Q0 to the atmosphere at T0. Then the energy (first law) equation for the system becomes d−W = d−Q1 – d−Q0 – dE

which can be rewritten as d−Q0 = d−Q1 – d−W – dE

The second law and the entropy principle give

( DS )U = −

dQ1 d Q0 + + dS = dSp T1 T0

First multiply the above relation by T0, then substitute for d−Q0 from the energy (first law) equation and finally rearrange the result. This gives

⎛ T ⎞ dW = dQ1 ⎜ 1 − 0 ⎟ − d ( E − T0 S) − T0 d Sp T1 ⎠ ⎝ 7 8

This converts the terms of the equation (which have the units of entropy) into energy units. Mathematically, any monotonic function will do but an identity function is generally chosen.

Chapter 8: Auxiliary Functions

175

which then gives

⎛ T ⎞ d−Wu = ⎜ 1 − 0 ⎟ d−Q1 – d(E + p0V – T0S) – dI T1 ⎠ ⎝

(infinitesimal process)

⎛ T ⎞ Wu = ⎜ 1 − 0 ⎟ Q1 – D(E + p0V – T0S) – I (finite process) T1 ⎠ ⎝ This relation can be rewritten in terms of the availability function F as ⎛ T dA = dWu, max = d−Wu + d−I = ⎜ 1 − 0 T1 ⎝ ⎛ T A = Wu, max = Wu + I = ⎜ 1 − 0 T1 ⎝

⎞ − − ⎟ d Q1 – d F ⎠

⎞ ⎟ Q1 – DF ⎠

(infinitesimal process)

(finite process)

The definition of availability, A, and the relation for it in terms of the availability function, F, derived above shows that it is logical to define the availability of the heat flux dq (or Q) absorbed from the reservoir at the temperature T1 as ⎛ T ⎞ – dFQ = d−Wu, max = ⎜ 1 − 0 ⎟ d−Q T1 ⎠ ⎝

(infinitesimal process)

⎛ T –DFQ = Wu, max = ⎜ 1 − 0 T1 ⎝

(finite process)

⎞ ⎟Q ⎠

It should be noted that this definition follows directly from that of the availability since the maximum work obtainable from a heat flux Q absorbed from a reservoir at T equals the work done by a Carnot engine.9 Then, the above relations can be rewritten as d−A = d−W = d−W + d−I = – dF – dF = – d(F + F) (infinitesimal process) u, max

u

Q

Q

A = Wu, max = Wu + I = – DFQ – DF = – D(FQ + F)

(finite process)

The last relation shows that availability can be calculated as the (algebraic) sum of the availability of the heat fluxes and that caused by the change of state. This principle of additivity is verified below.

8.2.2

Availability in General Non-flow Processes

The above arguments can be extended directly to a general case in which a system absorbs heat from a number of reservoirs (reservoir j being at temperature Tj for j = 1, 2, … , N) in addition to its atmosphere (of course at T0). For simplicity, only the finite processes are discussed.

9

It is, of course, assumed that the work output from the Carnot engine is completely utilized.

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Engineering Thermodynamics

For a non-flow finite process, the energy (first law) equation and the second law are

W=

∑Q

j

− DE

DS =

and

∑ (Q

j

/ Tj ) + S p

j

j

which can be manipulated (as usual) to get T0 ⎞ ⎟⎟ Qj – D(E + p0V – T0S) – T0Sp = – j ⎠ j ⎝ and rewritten in the usual symbols as

Wu =

⎛

∑ ⎜⎜ 1 − T

A = Wu, max = Wu + I

or

to confirm the principle of additivity stated above.

8.2.3

⎛ A=– ⎜ ⎜ ⎝

∑ DF

∑ DF

Qj

j

Qj

− DF – I

j

⎞ + DF ⎟ ⎟ ⎠

Availability in Flow Processes

In this section, the relation for availability for a flow process (a process executed by a flow system) is obtained exactly in the same manner as above. Consider a flow system executing a finite process exchanging heat Qj with the reservoir j at the temperature of Tj for j = 1, 2, …, N. Then, the energy (first law) equation is written as W = − E CV =

∑ Q j

and the second law as

j

⎛ ⎞ ⎛ ⎞ V2 V2 + m i ⎜ hi + i + gzi ⎟ − m e ⎜ he + e + gze ⎟ ⎜ ⎟ ⎜ ⎟ 2VCV 2 W ⎝ ⎠ ⎝ ⎠ W u = W − p0VCV = − ( E + p0V − T0 S )CV +

⎛

j

0 = − SCV +

Q j

∑T j

⎝

+ m i si − m e se + S p

j

As before, multiply the second law equation by T0 and subtract it from the energy (first law) equation and rearrange the results to get W = − ( E − T0 S )CV +

∑ j

⎛ T ⎜⎜ 1 − 0 Tj ⎝

⎞ ⎡ ⎤ V2 ⎟⎟ Q j + m i ⎢( hi − T0 si ) + i + gzi ⎥ − 2 ⎣ ⎦ ⎠

⎡ ⎤ V2 m e ⎢(he − T0 se ) + e + gze ⎥ − T0 S p 2 ⎣ ⎦ or, using the definition of the useful work (i.e. Wu =

– p0

), it becomes

⎡ ⎤ ⎡ ⎤ V2 V2 m i ⎢(hi − T0 si ) + i + gzi ⎥ − m e ⎢(he − T0 se ) + e + gze ⎥ − T0 S p 2 2 ⎣ ⎦ ⎣ ⎦

T0 ⎞ ⎟⎟ Q j + j ⎠

∑ ⎜⎜ 1 − T

Chapter 8: Auxiliary Functions

177

Based on the analogy of expressions of availability defined so far, the (bulk) flow availability and the (bulk) flow availability function10 can be defined as

⎡ ⎤ ⎡ ⎤ Vi2 Ve2 ( ) ( ) A flow = − F = m h − T s + + gz − m h − T s + + gze ⎥ ⎢ ⎥ ⎢ i i i e e flow 0 i 0 e 2 2 ⎣ ⎦ ⎣ ⎦ or, as

⎛ ⎞ ⎛ ⎞ V2 V2 i ⎜ bi + i + gzi ⎟ − m e ⎜ be + e + gze ⎟ A flow = − F flow = m ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ where the new function b, defined (following Keenan) as b = h – T0s, is called exergy.11 Then, the above relation for useful work becomes ⎛ + W u = − ⎜ F ⎜ CV ⎝

∑ F

Qj

j

⎞ +F flow ⎟ − I ⎟ ⎠

which shows that the principle of additivity of availability functions is valid for flow processes as well. For the work transfer devices like turbines, compressors, pumps, etc. with the default assumptions (steady state operation at steady adiabatic flow with negligible kinetic and potential energy changes) the above relation reduces to

Wu = Bi − B e − I which shows that exergy is the availability of the bulk flow part of a flow process.

8.2.4

Examples on Availability

This section deals with examples that illustrate the procedures for calculating availability and irreversibility. These procedures are based on the following relations.12 Availability (A A ). It is defined as the maximum useful work output when a system executes a process while interacting with any number of reservoirs and an environment [called the atmosphere (of the system)], i.e. A = Wu, max. The energy (first law) and second law equations show that Wu, max = – (DF + DFQ), where F is the availability function for change of state defined as F = E + p0V – T0S, with p0 and T0 as the pressure and temperature of the atmosphere, and FQ is the availability of heat flux defined as DFQ = Sj (1 – T0/Tj)Qj, with Qj as the heat absorbed from the reservoir at temperature Tj. The definition and causes of irreversibility together with the energy equation show that the irreversibility (I) should be defined as I = Wu,max – Wu, where Wu is the actual useful work. 10 11 12

This is not a standard terminology. Originated in Continental and Russian literatures. These are written for finite processes, but the extension to infinitesimal ones is obvious.

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The above definition together with that of entropy produced and the energy and second law equation also show that I = T 0Sp, where Sp is the entropy produced owing to the irreversibility. The actual useful work, Wu, is calculated from the specification of the load to be driven. These results can be summarized as A = Wu, max = – D(F + FQ) = Wu + I where by definition, F = E + p0V – T0S, FQ = Sj (1 – T0/Tj)Qj, and I = T0Sp. Then, it is logical to use the simplest set of equations for a given problem. This is done here. As before, to save space, the system and process diagrams are not drawn. For the same reason, units are also not written alongside the numbers. EXAMPLE 8.1 Calculate the availability of 1 [kg] of flue gas (considered ideal with cp = 1.1 [kJ/kg.K] and g = 1.41) at 1.5 [bar] and 1500 [K]. Solution Since this is a change of state, it is convenient to calculate the availability function. Moreover, since only one state is given, its availability is calculated with respect to the dead state. This is done in the following steps. 1. In terms of the availability function, F, the availability is defined as A = – DF = F1 – F0, where F0 is the value of the availability function at the dead state. 2. By the definition of the availability function, it becomes A = (E1 – E0) + p0(V1 – V0) – T0(S1 – S0). As the gas is ideal, this relation reduces to, A = (U1 – U0) + p0(V1 – V0) – T0 (S1 – S0). 3. Using the usual formulae for the equation of state of an ideal gas and for those of its properties, the availability function becomes A = mcv(T1 – T0) + (p0).(mR).[(T1/p1) – (T0/p0)] – (T0).[(mcp).ln (T1/T0) – (mR).ln (p1/p0)] 4. From an ideal gas, cv = cp/g and R = cp.(g – 1)/g. 5. Substituting the given data, m = 1 [kg], cp = 1.1 [kJ/kg.K], g = 1.41, T1 = 1500 [K], T0 = 300 [K], p0 = 100 [kPa], and p1 = 150 [kPa], in the above relation for availability and evaluating it gives, A = 1652.0 [kJ]. EXAMPLE 8.2

Calculate the availability of 1 [kg] of steam at 2 [bar], 150 [°C].

Solution This example is similar to Example 8.1 except that steam tables are used to determine the properties. Once again, it is conveniently done in the following steps: 1. For 2 [bar], 150 [°C], the steam tables give, h = 2768.5 [kJ/kg], v = 0.95954 [m3/kg], and s = 7.2794 [kJ/kg.K]. 2. Then, F = U + p0V – T0S = (H – pV) + p0V – T0S = H – (p – p0)V – T0S, and substituting the values gives, F = 488.73 [kJ]. 3. It is convenient to calculate F0 once and for all, so that it can be used in all the examples involving steam.

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4. Now, at 1 [bar] and 300 [K] (i.e. 27 [°C]), water is subcooled and its values are not tabulated. Hence, it is calculated using the standard approximation (of an incompressible fluid), i.e. h0 = (hf, T = 27°C) + (p0 – (psat, T = 27°C))(vf,T = 27°C) = 113.13 + (1 – 0.035636) . (100) . (0.0010034) = 113.23 [kJ/kg]. 5. Now, F0 = U0 + p0V0 – T0S0 = (H0 – p0V0) + p0V0 – T0S0 = H0 – T0S0 = 113.23 – (300) . (0.39495) = – 5.255 [kJ/kg]. 6. Substituting these values in the relation for availability, A = 488.73 – (– 5.355) = 494.0 [kJ] A typical practice followed in industries using steam at low pressures (for heating, processes, …) is to throttle steam generated at higher pressure (depending upon the available boiler). In this case, Wu = 0 and hence, I = A = – DF. Note the double penalty involved in this practice. First, the available work (= A) is not used and hence is lost. Secondly, the expenditure is incurred for the energy required for running the auxiliaries. EXAMPLE 8.3 Heat at the rate of 100 [kW] is to be transferred from a reservoir at 500 [K] to another at 400 [K]. What will be the availability of this heat? Solution In this case, it is convenient to use the definition of availability, i.e. it is the maximum useful work produced directly. By the Carnot theorem this useful work equals the work output of a Carnot engine running between two given temperatures. Thus, A = Wu, max = (1 – (400/500)) . (100) = 20 [kW]. It is assumed that all the work output of the Carnot engine is used (and none is lost). EXAMPLE 8.4 What is the availability when 10 [kg] of copper (c = 0.4 [kJ/kg.K]) is cooled from 600 [K] to 400 [K]? Solution In this case too, a change of state is specified. Hence, it is convenient to use the availability function. 1. By definition A = – DF = (E1 + p0V1 – T0S1) – (E2 + p0V2 – T0S2), and since a solid is incompressible (V1 = V2) pure substance (i.e. E = U), this relation reduces to A = (U1 – U2) – T0(S1 – S2). 2. Using the standard formulae for the terms involved in the relation for availability, it becomes A = (mc) . [(T1 – T2) – T0 . ln (T1/T2)]. 3. Substituting the values, this relation becomes, A = [(10) · (0.4)] . [(600 – 400) – (300) . ln (600/400)] = 313.4 [kJ]. EXAMPLE 8.5 Calculate the availability of an evacuated insulated rigid vessel of volume 1 [m3] at 0 [K]. Solution In this case too, it is convenient to use the definition of availability directly. This case will be recognized to be the same as the case of filling a bottle from the mains. The maximum work done will correspond to the work of reversibly filling the vessel, i.e. equal to the flow work. Then, A = p0V = (100) . (1) = 100 [kJ].

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Alternatively, the availability can also be calculated from the availability function since DE = 0 = DS (the vessel is evacuated) and V is constant (it is rigid). Then, A = – DF = (p0 – 0)V = (100) . (1) = 100 [kJ]. EXAMPLE 8.6 A turbine gets a supply of 5 [kg/s] of steam at 7 [bar], 250 [°C] and discharges it at 1 [bar]. Calculate the availability. Solution A turbine is a flow system with the default assumptions of steady state operation ( ECV = 0 = SCV ), steady adiabatic flow ( m CV = 0 = Q ), and negligible changes in the kinetic and potential energies (Dek = 0 = Dep). 1. Then, the relation for availability is

A = Bi − Be = m ⋅ [(hi − T0 si ) − (he − T0 se )] = m ⋅ [(hi − he ) − T0 ⋅ (si − se ) ] 2. Since the flow is isentropic, si = se and the relation for availability reduces to, A = m . [(hi – he)]. 3. From the steam tables, at 7 [bar] and 250 [°C], hi = 2954.0 [kJ/kg] and si = 7.1066 [kJ/kg.K] = se (since the process is isentropic). Then from the standard formulae, xe = 0.9582 and he = 2581.0 [kJ/kg]. 4. Substituting the values in the above formula for availability and evaluating it gives, A = (5).(2954.0 – 2581.0) = 1865 [kW]. EXAMPLE 8.7 Air at the rate of 2.5 [kg/s] at 1000 [K] and 1 [MPa] enters a turbine (hs = 0.8) with expansion ratio of 10. Determine the actual power and the irreversibility. Solution This example is the same as Example 8.6 except that the working substance is air (assumed to be an ideal gas) and the turbine is not ideal. Hence, this example is also solved following the same steps but with appropriate changes. 1. For the isentropic flow, Te = (Ti) . (pe/pi)(g – 1)/g = (1000) . (0.1)0.4/1.4 = 517.9 [K]. 2. By the definition of hs, for the actual process, Te¢ = Ti – (hs) . (Ti – Te) = 614.3 [K]. 3. Then the actual work done is, W = m . cp . (Ti – Te¢) = (2) . (1.003) . (1000 – 614.3) = 773.7 [kW]. 4. Also, A = Bi − Be = m . [cp(Ti – Te) – (T0) . {cp ln (Ti/Te¢) – R ln (pi/pe)}]. 5. Substituting the values, A = 1069.9 [kW], so that the irreversibility is I = A − W = 96.2 [kW].

8.3

OTHER ENERGY FUNCTIONS

In this section the other functions mentioned at the beginning of the chapter, namely (a) internal energy (U) defined by the first law, (b) enthalpy defined as H = U + PV, (c) Gibbs function defined as G = H – TS, and (d) Helmholtz function defined as A = U – TS, are shown to be energy functions in the sense that work output equals the decrease in their values.

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Note that: (a) for an isentropic non-flow process (DS = 0) taking place in a rigid vessel (DV = 0), F = U; (b) for an isentropic non-flow process in which the system pressure p is the same as its atmospheric pressure for every state (this requires an infinite number of atmospheres), F = H; (c) for a general non-flow isochoric (DV = 0) process in which the temperature of the system T is the same as that of its atmosphere at each of its states, F = A; and (d) for a general non-flow process in which the pressure p and temperature T of the system are the same as its atmosphere for all states of the system, F = G. Thus, it is evident that these are energy functions since they are special cases of the availability function. However, this is now shown from the first principles. The same procedure as Section13 8.2 is used for this purpose. For clarity, only the infinitesimal processes are considered.

8.3.1

Internal Energy

In the usual symbols, the energy (first law) equation shows that d−W = d−Q – dU

and the second law shows that d−Q £ TdS

Substituting the second law in the equation for the first law gives the expression for work as d−W £ TdS – dU

d− WS = – dUS

or

This last relation shows that work is done by the system (i.e. d−W > 0) in an isentropic process (when the equality sign is valid) when the energy decreases and that this work done equals the decrease in the internal energy. This result has already been mentioned in Chapter 5 on the first law of thermodynamics.

8.3.2

Enthalpy

Enthalpy is defined as H = U + pV, so that dH = dU + pdV + Vdp

or

– dU = – dH + pdV + Vdp

Substituting for dU in the above relation for work gives d−W £ TdS – dH + pdV + Vdp

or

( d−W – pdV)S, p = – dH

The left-hand side of the last relation is the work done in an isentropic isobaric process14 by all modes except by expansion. It is seen that this work is done [i.e. ( d−W – pdV) > 0] at the expense of enthalpy (i.e. dH < 0).

13 14

This should not be surprising since the same arguments are presented in different forms. In the ideal case a fluid is assumed to flow steadily at constant pressure, because being ideal there is no friction and, therefore, no pressure drop.

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8.3.3

Gibbs Function

The Gibbs function is defined as G = H – TS = U + pV – TS, so that or

dG = dU + pdV + Vdp – TdS – SdT

– dU = – dG + pdV + Vdp – TdS – SdT

Substituting for dU in the above relation for d−W and simplifying gives d−W £ – dG + pdV + Vdp – SdT

or

( d−W – pdV)T,p £ –dG

Here also it is seen that the work done over and above the expansion work (i.e. d−W – pdV) in an isobaric and isothermal process15 is positive when the Gibbs function decreases.

8.3.4

Helmholtz Function

This is defined as A = U – TS, so that dA = dU – TdS – SdT

or

– dU = – dA – TdS – SdT

Again, substituting for dU in the expression for d−W and simplifying gives d−W £ – dA – SdT

or

d−WT £ – dA

Here again the work output for an isothermal process is seen to be positive when the Helmholtz function decreases. In order to bring out the basic patterns in these functions, they are arranged as follows. For convenience, they are written for a finite process. (W – Wx)S,V = – DU

(W – Wx)S,p = – DH

(W – Wx)T,V £ – DA

(W – Wx)T,p £ – DG

By reading horizontally, this arrangement shows that what the function U does for an isochoric isentropic process the function H does for isobaric isentropic process. Similarly, by reading vertically, it shows that what U does for an isochoric isentropic process, A does for an isochoric isothermal process. Similar arguments can also be used to arrive at the pattern between H and G.

8.3.5

Consequences

Now, the two most important consequences of the above results (of expressing work output in terms of energy functions) are discussed. Efficiencies In all of the above results (including that of availability), the maximum work output possible under appropriate conditions corresponds to the processes being reversible (when, as the 15

Typically, all flows take place at constant pressure. In addition, phase change and chemical reactions take place at constant pressure and temperature. Hence, in engineering (e.g. chemical and metallurgical engineering) this function plays a fundamental role.

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consequence of the second law of thermodynamics, the equality sign becomes valid). Then, it is possible to define efficiency as, h = W/Wmax, where W is the actual work output in the process (which may be zero) and Wmax is the maximum possible work output in that process. The expression for the maximum work depends upon the process. These relations have been derived above and are reproduced below. General process within an atmosphere. When the system executes a non-flow process exchanging heat with its atmosphere, then Wu, max = – DF. Isentropic isochoric process. For a system executing an isentropic non-flow process in a rigid vessel, (Wmax)S,V = – DU. Isentropic isobaric process. Isothermal isochoric process. [(W – Wx)max]T,V = – DA.

For a system executing this process, [(W – Wx)max]S,p = – DH. In such a process the relation for the maximum work done is,

Isothermal isobaric process. In this case the maximum work becomes [(W – Wx)max]T,p = – DG. Now, by taking each one of these expressions for maximum work by turns in the denominator of the definition for efficiency, four efficiencies can be defined. However, only the following efficiencies have been found to be useful. Exergy efficiency. In this definition, the denominator is the exergy (i.e. Wu,max = Bi – Be). This efficiency has been found useful in the analysis of refrigerating and heat pumping system.16 Chemical efficiency. Since chemical reactions are assumed to occur at constant pressure and temperature, the Carnot efficiency is not applicable. Hence, for chemical reactions, efficiency should be defined as h = W/DG. For example, the efficiency of an electrolytic cell (battery) is V ) . (i) . (Dt)/DG, where V is the voltage at which current i flows for the time Dt. defined as h = (V Since these efficiencies are not unique, in this book, these are rarely used. They are mentioned here just as information of the possibilities. Equilibrium It was mentioned that in (classical) thermodynamics, a system is assumed to be in mechanical, thermal and chemical equilibria simultaneously. It was shown, as a consequence of the energy (first law) equation that a system is in equilibrium if its energy is minimum under adiabatic conditions. It was also shown that the condition of equilibrium in terms of the entropy maximum principle follows from the second law. It is shown here that this principle implies mechanical and thermal equilibria. 16

Some authors call this the second law efficiency. However, it is clear from Chapter 7 on the second law of thermodynamics that the concept of thermal efficiency arises naturally from it. Hence, the thermal efficiency (including the Carnot efficiency) is the real second law efficiency.

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Let a multicomponent system of simple compressible substances (described below in Section 8.5) be in a state of thermodynamic equilibrium. Let the boundaries of the system be rigid and insulated. Let it be divided by an internal diathermal tightly fitting free piston, permeable to the ith species, into two parts A and B. Taking U, V and the mole numbers n1, n2, …, as the independent variables, the entropy becomes S = S(U, V, n1, …). For an infinitesimal virtual displacement from equilibrium, the change in entropy becomes ⎛ ∂S ⎞ ⎛ ∂S ⎞ dS = ⎜ ⎟ dU + ⎜ ⎟ dV + ⎝ ∂U ⎠ ⎝ ∂V ⎠

⎛ ∂S ⎞

∑ ⎜⎝ ∂n ⎟⎠ dn

i

i

i

⎛ dU ⎞ ⎛ p ⎞ =⎜ ⎟ + ⎜ ⎟ dV + ⎝ T ⎠ ⎝T ⎠

⎛ mi ⎞

∑ ⎜⎝ T ⎟⎠ dn

i

i

where to simplify the writing, subscripts on the partial derivatives are omitted on the understanding that all variables are constant, except the one involved in the derivative as the independent variable. The symbol mi, which represents the chemical potential of the ith component is explained in Section 8.5 of this chapter. This is one of the functional equations of thermodynamics. Now, the principle of maximum entropy asserts (DS)U = 0. Since, the system is insulated, the system itself becomes the universe. Then, this relation reduces to dS = 0. 1. For simple compressible substances, the only admissible work mode is the expansion work, which is zero since the walls are rigid. 2. Since the system is insulated, Q = 0. 3. Then the energy (first law) equation reduces to DU = 0. 4. Since U is an additive property, dUA + dUB = 0. 5. Since the system boundaries are rigid, V is constant. 6. Since V is an additive property, dVA = – dVB. 7. Since entropy is additive, dSA = – dSB. 8. Since the total mole numbers are constant, but the piston is permeable only to the ith species, dnAi = – dnBi. 9. Using these assumptions, the fundamental equation becomes

⎡ P A PB ⎤ 1 ⎤ ⎡ 1 0 = dS = ⎢ A − B ⎥ dU A + ⎢ A − B ⎥ dV A − T ⎦ T ⎦ ⎣T ⎣T

∑ i

⎡ miA miB ⎤ A ⎢ A − B ⎥ dni T ⎦ ⎣T

10. Since the variations dUA, dVA and dniA are arbitrary, this implies that the quantities in the brackets should vanish. Then, TA = TB, pA = pB and miA = miB, which are the conditions of thermal, mechanical and chemical equilibria respectively. The conditions of equilibrium as energy minimization and entropy maximization are equivalent. This is illustrated below for the thermal equilibrium part. 1. Let the piston in the above illustrative example be fixed. 2. Then U = UA(SA, VA) + UB(SB, VB)

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3. Since the piston is fixed and the system is rigid, the volumes remain constant. 4. Consequently,

⎛ ∂U A ⎞ ⎛ ∂U B ⎞ dU = ⎜⎜ A ⎟⎟ dS A + ⎜⎜ B ⎟⎟ dS B ⎝ ∂S ⎠ ⎝ ∂S ⎠ 5. Using the definition of temperature, this relation becomes dU = TAdSA + TBdSB 6. 7. 8. 9. 10.

Now the energy minimum principle is, (dU)S = 0 Since the entropy is additive, S = SA + SB = constant. Hence, dSA + dSB = 0, or dSA = – dSB. Substituting all these gives, (TA – TB) dSA = 0. This implies that T A = T B, which is the condition of thermal equilibrium.

By assuming that the volumes also change, the energy minimum principle can be shown to involve mechanical equilibrium. Next, consider the condition of membrane equilibrium, i.e. equilibrium of a species (denoted by the subscript i) present on both sides of a membrane which is permeable to it. Let the system described in the above illustrations be in mechanical and thermal equilibria. Then: 1. Since the vessel is rigid and the piston is fixed, dV = dVA = dVB = 0. 2. Since the system is rigid and insulated, W = Wx = 0 = Q, so that the energy (first law) equation becomes dU = 0. 3. Since the system is in thermal equilibrium, TA = TB. 4. Under all these conditions, the fundamental equation, namely, dU = TdS – pdV + Simidni reduces to

⎡ m A dn A m B dnB ⎤ dS = − ⎢ i i + i i ⎥ T ⎦ ⎣ T 5. Since the total mole numbers of all the species are constant, dniA = – dniB (say), where dni is the infinitesimal quantity of species i exchanged between parts A and B. Then this equation becomes

(

)

A B dS = – mi − mi . (dni/T)

miA

miB ,

6. This equation shows that if then dS > 0 only if dni is negative, i.e. dn moles > of ith species leave part A. 7. This means that species diffuse across membranes permeable to them from parts of higher chemical potential to those of lower ones. In other words, the chemical potential is the driving force for mass transfers. 8. At equilibrium, when dS = 0, this equation shows that miA = miB . This means that for membrane equilibrium, the chemical potential of each species should be the same on both sides of the membrane.

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The concept of equilibrium and its conditions are borrowed from mechanics, wherein a system is defined to be in (stable) equilibrium if it cannot deliver work spontaneously, i.e. (W π 0). Now, the relations for work output from a system, obtained earlier in this section, show that when a system delivers work spontaneously, these energy functions decrease. Hence, it follows that the processes will stop when these energy functions reach their minimum value. This can be expressed mathematically as DF = (DU)S, V = (DG)S, p = (DH)T, p = (DA)T, V = 0 This condition is valid for maximum and minimum (and inflexion) points. Hence for a mathematically rigorous test, the second derivatives should be verified. Because of the complex nature of dependence of energy on the properties of substances, it is difficult to test the second derivatives mathematically. Hence, physical arguments are used to satisfy that a point is minimum. Even though the energy minimum and the entropy maximum principles derived from the first and second laws are the basic criteria for equilibrium, many a times it is simpler and more convenient to use the auxiliary functions. However, following example illustrates one of its interesting applications. EXAMPLE 8.8 The above discussion shows that the condition of equilibrium can be stated as the state of minimum energy for isochoric isentropic change of state, i.e. U(S,V) is minimum. This means that (∂2U/∂S2)V > 0 and (∂2U/∂V2)S > 0. Now, consider the first of the derivatives given above. It can be written as (∂2U/∂S2)V = [∂/∂S(∂U/∂S)]V. The property relation for U (i.e. dU = TdS – pdV) implies that (∂U/∂S)V = T so that (∂2U/∂S2)V = (∂T/∂S)V = T/cv (since by definition cv = T(∂S/∂T)V). Substituting these in the original expression gives, (∂2U/∂S2)V = T/cv. Now the condition of equilibrium means that T/cv > 0, or since, T > 0, then cv > 0. This implies that for any substance in stable equilibrium17 the specific heat at constant volume should be positive. Similar arguments gives (∂2U/∂V2)S = [(∂/∂V)(∂U/∂V)]S = – (∂p/∂V)S. This shows that for all stable substances, volume decreases as the applied pressure is increased. This is why the isentropic bulk modulus18 is defined inclusive of the negative sign, i.e. as BS = – (1/V) (∂p/∂V)S. Similarly, since U is a property (i.e. an exact differential), the order of differentiation does not matter. This gives (∂2U/∂S∂V) = (∂2U/∂V∂S) = [∂/∂V(∂U/∂S)] = (∂T/∂V) = 1/(b × V) which shows that b, the coefficient of volume expansion should be positive, i.e. all stable substances should expand when heated. Similarly, starting with the Helmholtz function, A(T, V), it can be shown that the isothermal bulk modulus, defined as BT = – (1/V)(∂p/∂V)T should be positive. 17 18

All the substances we deal with in our daily lives are stable, otherwise we will not be able to observe them in normal way. Volume analogue of Young’s modulus.

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Phase equilibrium During a change of phase, some mass leaves one of the phases and enters the second. For example, when water vaporizes, H2O substance leaves water (the liquid) phase and enters steam (the vapour) phase. During the discussion on membrane equilibrium presented above, it was shown that: ∑ The mass of the ith species is transferred, across a membrane permeable to it, from the region of higher chemical potential to that of the lower one. ∑ The transfer of mass stops when its chemical potential is the same in both the regions. If the interface between the phases is considered as a membrane, then it is clear that these rules are also applicable to phase equilibrium. However, to understand this result better, it is derived independently as follows: 1. Phase changes occur at constant pressure (mechanical equilibrium) and constant temperature (thermal equilibrium). 2. Hence it is convenient to express the condition of phase equilibrium as, (dG)p,T = 0. 3. By its definition, dG = Vdp – SdT + Smidni, which for the above conditions, reduces to Smidni = 0. 4. Let dni moles of the ith species leave phase B and enter phase A, so that dniA = – dniB = dni (say). 5. Then this equation becomes (miB – miA)dniA = 0, which reduces to miA = miB. The Clausius–Clapeyron equation governs the condition of the commonly encountered phase equilibrium19 of simple compressible substances. This is derived now. 1. Consider a simple compressible substance in equilibrium in phases A and B. 2. Then the condition of phase equilibrium asserts that, mA = mB. 3. Since for a single component system m equals the molar Gibbs function, this becomes gA = gB . 4. Now consider a virtual displacement from the equilibrium, so that dgA = dgB. 5. Since dg = vdp – sdT, this equation becomes vAdp – sAdT = vBdp – sBdT, which can be rearranged as dp s A − s B = dT v A − v B

6. Using the relation between entropy and enthalpy (i.e. latent heat) for phase change, denoted as lAB, this is rewritten as

l AB dp = dT T ( v A − v B ) 19

Technically, this is called the first order phase change.

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7. Clapeyron derived this equation first. Hence, it is called the Clapeyron equation. 8. If the volume of one of the phases (say, A) is large, then this equation reduces to dp l AB = dT Tv A 9. Further, if this phase behaves like an ideal gas, so that vA = RT/p, then the Clapeyron equation becomes

d ln p l AB = dT RT 2 Clausius proposed this modification. Hence, this equation is called the Clausius– Clapeyron equation. The Clapeyron equation written as

lAB = (T) . (vA – vB) . (dp/dT) may be used to estimate the latent heat. For example, consider saturated steam at 30[°C] (= 303.15 [K]). 1. Then the preceding equation can be written as hfg = (Tsat) . (vg – vf) . (dp/dT)sat 2. The temperature-based steam tables give, vf = 0.0010043 [m3/kg], vg = 32.928 [m3/kg]. 3. Also at T = 29 [°C], psat = 0.040040 [bar], and at T = 31 [°C], psat = 0.044911 [bar]. 4. Then, (dp/dT)sat, T = 30°C = (Dp/DT) = 100(0.044911 – 0.040040)/(31 – 29) = 0.24355 [kN/m2 . K]. 5. Substituting in the above equation gives, hfg = 2431.07 [kJ/kg]. The steam tables give 2430.74 [kJ/kg].

8.4

PROPERTY RELATIONS

In this section another application of the auxiliary functions (i.e. as relations imposing constraints on the properties) is presented. Only simple compressible fluids which are the working fluids in most thermal systems are only considered. During the discussions on evaluation of entropy, it was mentioned that since properties are functions of only states, it is possible to choose any process that is convenient for evaluating the energy change and work flow. Thus, it is assumed20 that (a) the process is reversible so that the second law reduces to d−Q = TdS, (b) only internal energy is present, i.e. E = U, and (c) the only work mode present is the expansion work, i.e. d−W = d−Wx, and since the process is quasi-static d−Wx = pdV. This idea is best illustrated by the older and the current definitions of specific heats. 20

Verify see Section 7.11.1 that the same assumptions were also made in the evaluation of entropy that is a property too.

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EXAMPLE 8.9 The older definition of specific heat at constant volume is presented as cv = (1/m) . ( d−Q/dT)v, where all the symbols have the usual meanings. The older definitions show the remnants of caloric theory. Since the process is assumed to be reversible, the second law reduces to d−Q = TdS. Then, the above definition becomes ⎛ 1 ⎞ ⎡ ⎛ ∂ S ⎞ ⎤ ⎡ ∂( S / m) ⎤ ⎛ ∂s ⎞ cv = ⎜ ⎟ ⋅ ⎢T ⎜ ⎟ ⎥ = ⎢ = T⎜ ⎟ ⎝ m ⎠ ⎣ ⎝ ∂ T ⎠ v ⎦ ⎣ ∂ T ⎥⎦ v ⎝ ∂T ⎠ v

The same arguments show that the older definition of specific heat at constant pressure presented as cp = (1/m) . ( d−Q/dT)p becomes

⎛ 1 ⎞ ⎡ ⎛ ∂S ⎞ ⎤ ⎡ ∂ ( S / m) ⎤ ⎛ ∂s ⎞ c p = ⎜ ⎟ ⋅ ⎢T ⎜ ⎟ ⎥ = T ⎢ ⎥ = T ⎜⎝ ∂ T ⎟⎠ ⎝ m ⎠ ⎢ ⎝ ∂T ⎠ p ⎥ ∂Τ ⎣ ⎦p p ⎣ ⎦ Now, the energy (first law) equation becomes d−Q = dU + pdV, and ( d−Q)v = (dU)v. Then, the definition of cv becomes

⎛ 1 ⎞ ⎛ d Q ⎞ ⎛ 1 ⎞ ⎛ ∂ U ⎞ ⎡ ∂ (U/m) ⎤ ⎛ ∂u ⎞ cv = ⎜ ⎟ ⋅ ⎜ = ⎜ ⎟ = ⎜ ⎟⋅⎜ ⎟ =⎢ ⎟ ⎥ ⎝ m ⎠ ⎝ dT ⎠v ⎝ m ⎠ ⎝ ∂ T ⎠v ⎣ ∂ T ⎦ v ⎝ ∂ T ⎠v Now, for a constant pressure process, d−Wx = pdV = d(pV), so that the energy (first law) equation becomes d−Q = dU + d(pV) = d(U + pV) = dH (by the definition of enthalpy). Then, the definition of specific heat at constant pressure becomes

⎛ 1 ⎞ ⎛ d Q⎞ ⎛ 1 ⎞ ⎛ ∂H ⎞ ⎡ ∂ (H/m) ⎤ ⎛ ∂h ⎞ cp = ⎜ ⎟ ⋅⎜ = ⎜ ⎟ ⋅⎜ =⎢ = ⎜ ⎟ ⎟ ⎟ ⎥ ⎝ m ⎠ ⎝ dT ⎠ p ⎝ m ⎠ ⎝ ∂T ⎠ p ⎣ ∂ T ⎦ p ⎝ ∂ T ⎠ p The boxes show the modern definitions of cv and cp. Now, the energy (first law) equation becomes d−Q = dU + pdV, which can be written as dU = d−Q – pdV. Since the process is reversible, the second law becomes d−Q = TdS. Thus, the combined energy (first law) and second law equation reduces to dU = TdS – pdV. Now, enthalpy is defined as H = U + pV, so that dH = dU + pdV + Vdp. Using the combined energy (first law) and second law equation derived above, this relation becomes dH = TdS + pdV. Similarly, the Gibbs function is defined as G = H – TS = U + pV – TS, or, as dG = dU + pdV + Vdp – TdS – SdT, and by the combined energy (first law) and second equations, this reduces to dG = Vdp – SdT. Analogously, the Helmholtz function defined as A = U – TS, or, as dA = dU – TdS – SdT together with the combined energy (first law) and second law equations gives, dA = – pdV – SdT. These equations are identical to those used in the earlier section. Thus, it is evident that these functions are also relations among properties of simple compressible substances.

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It should be emphasized that as property relations these equations are valid for all processes since change of a property is independent of a process. However, as process relations they are valid only for reversible processes (i.e. d−Q = TdS) of a simple compressible substance (i.e. W = Wx) and E = U. These functions are very basic to the development of property relations. Hence, they are compiled below. dU = TdS – pdV (8.2) dH = TdS + Vdp

(8.3)

dA = – pdV – SdT

(8.4)

dG = Vdp – SdT

(8.5)

Equations (8.2)–(8.5) show that it will be convenient to choose S and V as independent properties when variations in U are considered. Thus, (S, V) may be called the natural coordinates for calculations involving U. Similarly, the natural coordinates for H, A and G are (S, p), (V, T), and (p, T), respectively. In the analysis of the relations among properties of simple substances, it is mathematically convenient to choose natural coordinates.21 All the above relations can be put in the general form, dz = Mdx + Ndy. Such relations are called Pfaffian22 differential equations (or, more loosely Pfaffians). Appendix C contains a brief review of them. A quantity is defined to be a property if and only if its variation is independent of a process. In mathematics, such Pfaffians are termed exact. In mathematics if the Pfaffian dz = Mdx + Ndy is exact, then (a) M = (∂z/∂x)y and N = (∂z/∂y)x, and (b) (∂M/∂y)x = (∂N/∂x)y. It has been shown that the second characteristic [i.e. (b) here] is necessary and sufficient for a Pfaffian to be exact. Applying the first characteristic to the above property relations gives ⎛ ∂U ⎞ ⎛ ∂H ⎞ = T =⎜ ⎝ ∂ S ⎟⎠ V ⎜⎝ ∂ S ⎟⎠ p

⎛ ∂U ⎞ ⎛ ∂ A⎞ p= −⎜ = −⎜ ⎟ ⎟ ⎝ ∂V ⎠ S ⎝ ∂V ⎠ T ⎛∂H ⎞ ⎛∂G⎞ V =⎜ ⎟ =⎜ ⎟ ⎝ ∂ p ⎠ S ⎝ ∂ p ⎠T ⎛ ∂ A⎞ ⎛ ∂G⎞ S =−⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ V ⎝ ∂T ⎠ p

The above relations are called thermodynamic definitions of temperature, pressure, volume and entropy, respectively, because the dependent variables in these equations are energy functions. 21 22

Callen’s axiomatic method (see Appendix D and [CAL]) uses them. In honour of J.F. Pfaff (pronounced ‘phuf’), who first investigated this form of functions and correctly interpreted them ([FOR], pp. 80–83, or, [INC], § 2.83).

Chapter 8: Auxiliary Functions

8.4.1

191

Maxwell Relations

Due to its definition, the change in the value of a property between two states is independent of the path. Mathematically, this means that the order in taking the second derivative does not matter.23 This is the characteristic (b) of an exact Pfaffian mentioned above. Applying this condition to Eq. (8.2) gives

⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎡ ∂ ⎛ ∂U ⎞ ⎤ ⎡ ∂T ⎤ ⎛ ∂ p⎞ ⎢ ∂V ⎥ = ⎢ ∂V ⎜⎝ ∂ S ⎟⎠ ⎥ = ⎢ ∂ S ⎜⎝ ∂V ⎟⎠ ⎥ = − ⎜⎝ ∂ S ⎟⎠ ⎣ ⎦S ⎣ V ⎦S S ⎦V V ⎣ Usually the above second derivatives are more loosely written as (∂ 2 U/∂V∂S) and (∂ 2U/∂S∂V). Similar relations can be obtained from Eqs. (8.3), (8.4) and (8.5). The complete list of these equations, called Maxwell’s relations, is given below:

⎛ ∂T ⎞ ⎛ ∂ p⎞ ⎜⎝ ∂V ⎟⎠ = − ⎜⎝ ∂ S ⎟⎠ S V

(8.6)

⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜⎝ ∂ p ⎟⎠ = ⎜⎝ ∂ S ⎟⎠ p S

(8.7)

⎛ ∂ p⎞ ⎛ ∂S ⎞ ⎜⎝ ∂ T ⎟⎠ = − ⎜⎝ ∂V ⎟⎠ V T

(8.8)

⎛ ∂S ⎞ ⎛ ∂V ⎞ ⎜⎝ ⎟⎠ = − ⎜ ⎟ ∂T p ⎝ ∂ p⎠T

(8.9)

These equations are simple and can be derived easily. Hence, there is no necessity for any mnemonics.24 However, when several types of variables are needed, systematic procedures for deriving them are essential. All the methods are based on (a) Legendre transformation and (b) use of Jacobians. These are explained in Appendix C.

8.4.2

Examples on Property Relations

In this section, the above methods are illustrated by some examples. EXAMPLE 8.10 Solution 23 24

Obtain the Gibbs–Helmholtz equation, namely, [∂/∂T(G/T)]p = – (H/T2).

This equation is obtained in the following steps:

Recall that, in calculus, (∂2z/∂x∂y) means increment first along the x-direction by dx, then along the y-direction by dy, and finally taking the limit as dx and dy tend to zero. However, for the sake of completeness Born diagram is described in Appendix A.

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1. By definition, G = H – TS, or, dG = Vdp – SdT. 2. Since G is a property, S = – (∂G/∂T)p. 3. Substituting for S in the defining relation for G gives, G = H + T .(∂G/∂T)p, and transposing this gives, (G/T) – (∂G/∂T)p = (H/T). 4. Now, [∂/∂T(G/T)]p = (1/T) . (∂G/∂T)p – (G/T2). 5. Hence, (G/T) – (∂G/∂T)p = (– T) . [∂/∂T(G/T)]p. 6. Combining with step (3) above, step (5) becomes (– T) . [∂/∂T(G/T)]p = (H/T). Transposing this gives the result. Sometimes the Gibbs–Helmholtz equation is also written as H = – T2 . [d/dT(G/T)]p. EXAMPLE 8.11 Solution

Show that: (∂U/∂p)T = – T(∂V/∂T)p – p(∂V/∂p)T.

This relation is obtained in the following steps:

1. Differentiating both sides of the property relation dU = TdS – pdV with respect to p gives (∂U/∂p)T = T . (∂S/∂p)T – p . (∂V/∂p)T. 2. The property relation dG = Vdp – SdT gives the Maxwell relation (∂V/∂T)p = – (∂S/∂p)T. 3. Substituting the Maxwell relation in the equation in step (1) gives the result. EXAMPLE 8.12 Solution

Show that: cv = – T (∂ 2A/∂T 2)V.

The following steps are suitable for this example.

1. By definition, cv = T (∂S/∂T)V. 2. But the property relation dA = – pdV – SdT shows that S = – (∂A/∂T)V. 3. Substituting in the above equation for cv gives, cv = T [∂/∂T(–∂A/∂T)V]V = – T (∂2A/∂T2)V. EXAMPLE 8.13

Show that: cp/cv = (∂p/∂V)S/(∂p/∂V)T.

Solution This example is solved in the following steps. Since cv and cp occur in the two TdS equations the solution begins with them. 1. The property relation, TdS = mcvdT + T (∂p/∂T) VdV, for an isentropic process (dS = 0), becomes mcv = – T . (∂p/∂T)V . (∂V/∂T)S. 2. Similarly, for an isentropic process, the relation TdS = mcpdT – T(∂V/∂T)p dp gives mcp = T . (∂V/∂T)p . (∂p/∂T)S. 3. Dividing one by the other gives, (cp/cv) = – (∂V/∂T)p . (∂T/∂p)V . (∂p/∂V)S. 4. But the cyclic relation (∂ V/∂T) p . (∂ T/∂p)V . (∂p/∂ V)T = – 1 shows that (∂V/∂ T) p . (∂T/∂p)V = – 1/(∂p/∂V)T.

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193

5. Substituting in the the relation in step (3) gives cp/cv = (∂p/∂V)S/(∂p/∂V)T = BS/BT, where BS and BT are the isentropic and isothermal bulk moduli. 6. The following is an alternative way to derive the given relation. By definition, cp = T(∂s/∂T)p and cv = T(∂s/∂T)v. Thus cp/cv = (∂s/∂T)p/(∂s/∂T)v. By the cyclic relation this becomes (∂p/∂T)s . (∂s/∂p)T/(∂v/∂T)s . (∂s/∂v)T and, by the usual rule of partial differentiation this reduces to, [(∂p/∂v)s/(∂p/∂v)T] = Bs/BT.

8.5

MULTICOMPONENT SYSTEMS

So far the working substance was assumed to be pure one in a single phase, where a pure substance was assumed to be a single distinct chemical constituent. Even when the same substance was present in two phases (e.g. wet steam), its state was represented by an overall property, which was generally calculated as the weighted average of the individual phases. In short, single mass or mole number represented the system. In this section, the concepts derived so far are extended to multicomponent systems. Consider a system containing a number of chemical species in different phases in thermodynamic equilibrium. The number of components of a mixture is the minimum number of substances from which all the phases of the system (i.e. the complete mixture) could be prepared ([DEN], p.171). A component can occur more than in one phase [e.g. water substance in liquid phase (i.e. as water) and vapour phase (i.e. as steam)]. Such a system is called a multicomponent system. Here, the components and the mixture are assumed to be simple compressible substances. For a single-component, simple-compressible substance, U = U(S, V). Then, for an infinitesimal process ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ dS + ⎜ dV = TdS − pdV ⎝ ∂ S ⎟⎠ V ⎝ ∂V ⎟⎠ S where the second equality arises out of the definitions of temperature and pressure presented in Section 8.4. In a multicomponent system, the complete state of the system is represented by specifying the composition of the working substance in addition to the two-state variables specified for a simple-compressible substance. Then, for a mixture of N-component simple-compressible substances U = U(S, V, n1, n2, …, nN) where ni is the number of moles (more commonly called mole number) of the ith component. Partial differentiation of the above relation gives

⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ dV + ⎟ dS + ⎜ ⎟ ⎝ ∂ S ⎠V , ni ⎝ ∂ V ⎠S , ni

N

⎛ ∂U ⎞

∑ ⎜⎝ ∂ n ⎟⎠ i =1

i

dni S ,V , n j

Comparing this equation with that for the single-component system shows ⎛ ∂U ⎞ =T ⎜⎝ ∂ S ⎟⎠ V , n i

and

⎛ ∂U ⎞ =−p ⎜⎝ ∂ V ⎟⎠ S , n i

(8.10)

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Engineering Thermodynamics

The third term of Eq. (8.10) represents the additional change in energy owing to change in the mole numbers at constant volume and entropy. These derivatives are called the chemical potential. For the ith species, the derivative is denoted by the letter mi and defined as

⎛ ∂U ⎞ ⎟ ⎝ ∂ ni ⎠S ,V , n j

mi = ⎜ Then, Eq. (8.10) can be written as

N

dU = TdS − pdV +

∑ m dn i

i

i =1

This argument can be extended to other properties as well to obtain similar relations for dH, dG and dA, all of which are compiled below for easy reference. N

dU = TdS – pdV +

∑ m dn

i

(8.11)

i

(8.12)

i

(8.13)

i

i =1 N

dH = TdS + Vdp +

∑ m dn i

i =1 N

dG = Vdp – SdT +

∑ m dn i

i =1

N

dA = – pdV – SdT +

∑ m dn i

i

(8.14)

i =1

where the chemical potential mi is defined as

⎛ ∂U ⎞ ⎛∂H ⎞ ⎛∂G⎞ ⎛∂A⎞ =⎜ =⎜ =⎜ ⎟ ⎟ ⎟ ⎟ ⎝ ∂ ni ⎠S ,V ,n j ⎝ ∂ ni ⎠S , p,n j ⎝ ∂ ni ⎠ p,T , n j ⎝ ∂ ni ⎠T ,V ,n j

mi = ⎜

(8.15)

In engineering, the most convenient independent variables are p and T. Hence, the definition of chemical potential in terms of G [Eq. (8.13)] is used from now onwards. A large number of Maxwell type relations can also be developed. However, the following two relations obtained from the expression for dG [i.e. Eq. (8.13)] have been found to be most useful.

⎛ ∂S ⎞ ⎛ ∂ mi ⎞ = −⎜ ⎟ ⎜ ∂Τ ⎟ ⎝ ⎠ p, ni , n j ⎝ ∂ ni ⎠ p,T , n j and

⎛ ∂V ⎞ ⎛ ∂ mi ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ∂ p ⎠T , ni ,n j ⎝ ∂ ni ⎠ p,T , n j

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195

The partial derivative on the right-hand side of these equations, viz. (∂ /∂ ni)p,T,nj occurs frequently in analyzing multicomponent systems. It is indicated by the adjective partial molar. For example, the derivative (∂X/∂ni)p,T,n is called the partial molar X and is denoted as X i . j Thus, the above equations for partial molar entropy and partial molar volume can be written as

⎛ ∂S ⎞ ⎛ ∂ mi ⎞ = −⎜ = Si ⎟ ⎜ ∂T ⎟ ⎝ ⎠ p ,ni , n j ⎝ ∂ ni ⎠ p,T ,n j

(8.16)

⎛ ∂V ⎞ ⎛ ∂ mi ⎞ =⎜ = Vi ⎟ ⎜ ⎟ ⎝ ∂ p ⎠T , ni , n j ⎝ ∂ ni ⎠ p,T , n j

(8.17)

and

For a single-component system (i.e. N = 1) containing n number of moles of the substance, the above definitions reduce to: (a) the chemical potential equals the molar Gibbs function, i.e. m = (∂G/∂n)p,T; (b) variation of m with temperature equals the molar entropy, i.e. (∂m/∂T)p = – (∂S/∂n)p,T; and (c) similarly, the variation of m with pressure equals the molar volume, i.e. (∂m/∂p)T = (∂V/∂n)p,T. Following examples serve as illustrations. EXAMPLE 8.14 For n moles of an ideal gas, the second of the above equations for a singlecomponent system gives (∂m/∂p)T = v; and by using the ideal gas equation this becomes (∂m/∂p)T = (RT/p). Integration of this partial differential equation gives, m = m0 + RT ln p, where m0 is only a function of temperature. Thus, it is seen that the equation of state of an ideal gas can also be written as

m = m0 + RT ln p

(8.18)

where m is only a function of temperature. In Chapter 10, Eq. (8.18) is used during the discussion on reaction equilibrium of ideal gases. 0

EXAMPLE 8.15 of Eq. (8.18)]. Solution

Obtain the expression for m0 in the equation of state of an ideal gas [i.e.

This is done indirectly, by obtaining an expression for m of an ideal gas as follows:

(a) For a single-component, simple-compressible substance, m = g. Hence, molar quantities are used in this analysis. (b) Choose p and T as independent variables since they are the natural variables for g. (c) By definition, g = h – Ts. (d) Since h = h(T, p), dh = (∂h/∂T) pdT + (∂ h/∂ p) Tdp. (e) For an ideal gas, h is only a function of temperature. (f) Moreover, by definition, cp = (∂ h/∂T)p.

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Hence, dh = cpdT; or, h = Ú∂cp dT + C0. Now, TdS = dh – vdp; and hence, ds = (dh/T) – (vdp/T). For an ideal gas, this becomes ds = (cpdT/T) – (Rdp/p). Integrating this expression gives s = Ú(cpdT/T) – R ln p + C1, where C1 is another integration constant. (k) Substituting for h and s in the relation of g gives

(g) (h) (i) (j)

⎡ g = ⎡ c p dT + C0 ⎤ − T ⎢ ⎣ ⎦ ⎣

∫

∫

cp dT T

⎤ − R ln p + C1 ⎥ ⎦

(l) This above relation for g can be rewritten as c p dT ⎡ ⎤ g = ⎢ c p dT + C0 − T − C1T ⎥ + RT ln p T ⎣ ⎦ (m) By comparing with Eq. (8.18) it is seen that the expression in step (l) in the brackets is only a function of temperature. Hence,

∫

∫

∫

m0 = c p dT − T

∫

cp dT T

− C1T + C0

REVIEW QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

What are auxiliary functions? What are three ways in which auxiliary functions are used in thermodynamics? Define the terms: (a) atmosphere and (b) availability. Derive the relation for availability when a closed system interacts (a) with its atmosphere and (b) with a reservoir in addition to its atmosphere. Define irreversibility and derive relations for it. Obtain the relation for availability of a general flow system. Obtain relations for the maximum work output in terms of the energy functions, U, H, G and A. What are the different types of efficiencies? Show that the condition of maximum entropy implies mechanical, thermal and chemical equilibria. Taking thermal equilibrium as an example, show that minimum energy condition is equivalent to the maximum entropy condition. Obtain an relations for (a) membrane equilibrium and (b) phase equilibrium. Derive the Clausius–Clapeyron equation of phase equilibrium. What are the necessary and sufficient conditions for a Pfaffian to be exact? Obtain the thermodynamic definitions of p, V, T and S. Derive the Maxwell’s relations.

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197

EXERCISES Notes: Assume air to be an ideal gas with molecular weight M = 29 [kg/kmol] and specific heat ratio g = 1.4. Assume that the atmosphere is at 1 [bar] and 300 [K]. Calculate the maximum possible work in the following cases (Exercises 1–11): 1. One kilogram of air at the following states: (a) p = 50 [bar], T = 1000 [K]; and (b) p = 0.5 [bar], T = 200 [K]. [Hints: Since the states are specified, the maximum work is calculated as the change in the availability function. Then, A = – DF = F1 – F0 = (u1 – u0) + p0 . (v1 – v0) – T0 . (s1 – s0) (since air is an ideal gas, and m = 1 [kg]). Using the equation of state and property relations of ideal gas, the availability function becomes: A = cv . (T1 – T0) + (p0) . (R) . [(T1/p1) – (T0/p0)] – (T0) . [cp ln (T1/T0) – (R) ln (p1/p0)]. For air, R = 0.2867 [kJ/kg.K], cv = 0.717 [kJ/kg.K], and cp = 1.0 [kJ/kg.K]. Substituting these values as well as the given data in the above relation and evaluating it gives, A = 396.9 [kJ] [Ans. (a)] and A = 18.99 [kJ] [Ans. (b)].] 2. Steam/water at: (a) 5 [bar], 0.5 dry; (b) 10 [bar], 250 [°C]; (c) 2 [bar], 50 [°C]. [Hints: Same as above but steam tables are used now. Then, A = F1 – F0, where F0 = – 5.26 [kJ/kg] (calculated in Example 8.2). Evaluating F1 and substituting in the above relation gives A as (a) 320.9 [kJ], (b) 661.0 [kJ], and (c) 3.55 [kJ].] 3. One kilogram of steam at: (a) saturated vapour at 1 [bar]; (b) 10 [bar], 100 [°C]. [Hints: Since a state is specified, calculate the change in the availability function. Then, F1= u1 + p0v1 – T0s1 = h1 – (p1 – p0)v1 – T0s1 (since m = 1 [kg]). Substituting the values from the steam tables and evaluating this relation gives, F1 = h1 – T0s1 = 476.46 [kJ] (for case (a) since p1 = 1 [bar]); and, F1 = 26.94 [kJ] (case (b)). In Example 8.2, it was shown that for steam, F0= – 5.26 [kJ/kg]. Thus, A = F1 – F0 = 472.7 [kg] [Ans. (a)] and A = 32.20 [kJ] [Ans. (b)].] 4. Ten kilograms of a liquid (c = 4 [kJ/kg.K]) is cooled from 800 [K] to 400 [K]. [Hints: Calculate the change in the availability function since change of state is specified. Then, A = – DF = (U1 – U2) – T0(S1 – S2) (since liquid is incompressible) = (mc)[(T1 – T2)] – T0 . ln (T1/T2)] = 7682 [kJ].] 5. Heat of 100 [kJ] available from reservoirs at (a) 1000 [K], (b) 1200 [K], (c) 1500 [K], and (d) 2000 [K]. [Hints: Since heat transferred from a reservoir is specified, the definition of availability (i.e. maximum possible work output) is directly used. Since this corresponds to the work output of a Carnot engine, A = (1 – T0/T1) . Q. Substituting the values, A = 70 [kJ] [Ans. (a)], A = 75 [kJ] [Ans. (b)], A = 80 [kJ] [Ans. (c)], and A = 85 [kJ] [Ans. (d)].] 6. One hundred kilograms of water cooled from 5 [bar] and 100 [°C] by rejecting heat to a reservoir at triple point of water. [Hints: This can be calculated as for the above exercise, but as a variation the change of availability function is used here. The A = F1 – F2. Using the values from steam tables, this becomes A = (100) . (29.963 + 0.1) = 3006 [kJ].]

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7. Sixty kilowatts of heat available from a reservoir at a temperature of 800 [K]. The cold reservoir is at 400 [K]. [Hints: Since heat transferred from a reservoir is given, proceeding as above A = (1 – 400/800) . (60) = 30 [kW].] 8. 4.5 [MJ] of heat available from a reservoir at 600 [K]. [Hints: Same as above. A = (1 – 300/600) . (4.5) = 2.25 [MJ].] 9. A refrigerator cooling 10 [kg] of a solid (c = 3 [kJ/kg.K]) to – 10 [°C] from (a) 7 [°C], and (b) 300 [K]. [Hints: Same as above. A = – DF = (U1 – U2) – T0 . (S1 – S2) (since solid is incompressible), or A = (mc)[(T1 – T2) – T0 . ln (T1/T2)] = – 53.72 [kJ] [Ans. (a)] and A = –74.66 [kJ] [Ans. (b)]. Note that the values are negative since work is done for cooling the solid (i.e. in refrigerating it).] 10. Exhaust gas (assumed ideal with cp = 1.2 [kJ/kg.K] and g = 1.41) at 1.2 [bar] and 1000 [K]. [Hints: Since state is given, availability is evaluated as the change of the availability function. For an ideal gas it has already been derived in Exercise 1 as, A = – DF = (U1 – U0) + p0(V1 – V0) – T0(S1 – S0) = (m)(cv)(T1 – T0) + (p0)(mR)[(T1/p1) – (T0/p0)] – (T0)[(mcp) ln (T1/T0) – (mR) ln (p1/p0)]. For an ideal gas, cv = cp/g and R = cp(g – 1)/g. Using these relations and substituting the given data shows that, A = 739.7 [kJ/kg].] 11. Water (assumed incompressible with c = 4.2 [kJ/kg.K]) available from a flat plate collector at 90 [°C] at the rate of 1 [kg/s] being cooled to (i) 310 [K], and (ii) 300 [K]. [Hints: Flow system and hence A = Bi – Be = m [(hi – T0si) – (he – T0se)] = m [(hi – he) – T0(si – se) = ( m c)[(Ti – Te) – T0 ln (Ti/Te)]. Substituting the values, A = 23.23 [kW] [Ans. (a)] and A = 24.42 [kW] [Ans. (b)].] 12. A copper (c = 0.39 [kJ/kg.K]) block of 10 [kg] at 300 [°C] is dropped into a large tank containing water at 300 [K]. Calculate the availability due to cooling of the copper block. [Hints: Since this is a change of state, availability function is used. Then, A = – DU + T0DS (since solid is incompressible), or A = (mc)[(T1 – T2) – T0 ln (T1/T2)] = 307.6 [kJ].] 13. A steel (c = 0.5 [kJ/kg.K]) bar of mass 10 [kg] at 600 [°C] is quenched in a large tank of oil (c = 2 [kJ/kg.K]) at 300 [K]. Assuming no phase change of oil, calculate the irreversibility. [Hints: Same as Exercise 12. I = 1262.8 [kJ].] 14. Two identical bodies (of constant properties) at T1 and T2 (T1 > T2) enclosed in an adiabatic container interact without phase change. Show that the maximum possible work obtainable is mc(T1 + T2 – 2Tf), where Tf =

T1 T2 .

[Hints: By the energy (first law) equation, W = – DU = (mc)(T1 – Tf) + (mc) (T2 – Tf) = (mc)[T1 + T2 – 2Tf]. Now, W will be maximum when Tf becomes minimum. By the second law, (DS)U ≥ 0. Since S is a property, (DS)U = (DS)1 + (DS)2 =

Chapter 8: Auxiliary Functions

15.

16.

17.

18.

19.

199

(mc) ln (Tf /T1) + (mc) ln (Tf /T2) = (mc) ln (Tf2/T1T2) ≥ 0. The work will be maximum when the process is reversible, i.e. when the equality sign is valid. Then, this expression shows Tf, min = T1T2 corresponding to the logarithmic term becoming unity.] One kilogram of steam at 5 [bar] and 200 [°C] in a closed system is first mixed adiabatically with 1 [kg] of saturated water at 5 [bar]. The mixture is then cooled at constant volume by heat loss to atmosphere at 300 [K] till its final state is 1 [bar], 0.55 dry. Calculate the irreversibility. [Hints: The first process (1–2) is adiabatic mixing at constant pressure for which the energy (first law) equation gives 0 = DU + pDV = DH, or H2 = H1. Then, from the data of steam tables, h2 = 1747.6 [kJ], and s2 = 4.4662 [kJ/kg.K]. The second process (2–3) is constant volume cooling for which the energy (first law) equation becomes Q = DU = U3 – U2. Now, x2 = 0.5255, v2 = 0.19741 [m3/kg], so that u2 = 1648.9 [kJ/kg]. From the given data and steam tables, u3 = 1261.2 [kJ/kg]. Then, Q = – 775.4 [kJ], (negative because heat is lost). Then, s3 = 3.348 [kJ/kg.K], so that the second law gives, Sp = (DS)U = DS + (DS)R = 0.34107 [kJ/kg.K] and I = T0Sp = 102.3 [kJ].] During a process, a closed system absorbs heat Q from a reservoir at TR and rejects heat QL to atmosphere at T0. The only work done is against the atmosphere. Express the irreversibility in terms of DV, DU and DS of the system. [Hints: The energy (first law) equation gives, Q – QL = DU + p0DV, or QL = Q – DU – p0DV. The second law is (DS)U = Sp = DS + (DS)R + (DS)0 = DS – (Q/TR) + (QL/T0), or QL = T0Sp + (T0/TR)Q – T0DS. Since the left-hand side of the second law equation is same as left-hand side of the energy equation, equating their right-hand sides gives Q – DU – p0DV = I + (T0/TR)Q – T0DS, which can be arranged as I = (1 – T0/TR)Q – DU – p0DV + T0DS = – DFQ – DF.] An evacuated rigid bottle is filled with steam from mains at 40 [bar] and 400 [°C]. It was found that 1 [kg] of steam had entered the bottle when the flow stopped. Calculate the availability if the process of filling is adiabatic. [Hints: This is also a change of state. Hence, A = – DF. Since the mass of steam is 1 [kg], this equals A = – Df = – D(u + p0v – T0s). Since the bottle is rigid, this reduces to A = – D(u – T0s) = – Du + T0Ds = (u1 – u2) – T0(s1 – s2). Since the bottle was initially evacuated, u1 and s1 are zero. Thus, Df = – (u2 – T0s2). Substituting the values from steam tables gives, A = – 890.2 [kJ].] Steam at 20 [bar], 300 [°C] enters a steam turbine and exhausts at 0.5 [bar]. Determine the irreversibility if hs = 0.9. [Hints: For the isentropic expansion, si = se gives xe = 0.8731. Then, he = 2353.4 [kJ/kg]. Then, by the definition of hs, he¢ = hi – hs(hi – he) = 2420.6 [kJ/kg], so that xe¢ = 0.9022 and se¢ = 6.9242 [kJ/kg.K]. Then, I = T0DS = 56.7 [kJ].] Exhaust gas at 400 [°C] (assumed to be large, i.e. a reservoir) is used to generate saturated steam at 10 [bar] from saturated water at 10 [bar]. Calculate the irreversibility per [kg] of steam. [Hints: Using the entropy principle, Sp = (DS)U = DS + (DS)R. Taking data from the steam tables, Ds = s2 – s1 = 4.4446 [kJ/kg.K], and (DS)R = – (h2 – h1)/TR = – 2.9925 [kJ/kg.K], so that I = T0Sp = 435.8 [kJ/kg].]

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20. Saturated steam at 5 [bar] is generated from saturated water at 5 [bar] using exhaust gas at 500 [°C]. Determine the irreversibility. [Hints: Same procedure as above. I = T0Sp = 300 [(6.8192 – 1.8604) – (2747.5 – 640.12)/773] = 669.8 [kJ].] 21. Five kilograms per second of air at 1.5 [bar] and 320 [K] enters a compressor with a velocity of 50 [m/s]. Its exit conditions are 3 [bar], 450 [K] and 100 [m/s]. Calculate (a) the actual work done, (b) minimum work required, and (c) the irreversibility. [Hints: This is a flow system. Hence the actual work done may be calculated from the SFEE as – 668.75 [kW]. The minimum work is calculated as the change of availability, W min = m [(bi – be) + (Vi 2 – Ve2)/2], where b = h – Ts. Substituting the values, W min = – 455.45 [kW]. Finally, calculate the irreversibility as the difference. Then I = 213.3 [kW].] 22. What will be the maximum discharge pressure of steam in a diffuser with inlet at 4.5 [bar], 350 [m/s]? [Hints: This corresponds the reversible flow. Then under the default assumptions the energy equation reduces to hi + Vi2/2 = he, and the second law gives si = se. The final state will have to be solved for this state. Ans: pe = 4.55 [bar] and Te = 148.7 [°C] (by linear interpolation).] 23. Dry saturated steam at 4 [bar] is irreversibly expanded to 1 [bar] through a steam turbine. The work done is 80 per cent of that for reversible expansion between the same pressures. During the expansion, heat is extracted from the steam so that its initial and final entropies are equal. Calculate the irreversibility. [Hints: For the reversible process, s1 = s2. From the p-based saturation tables at 4 [bar], h1 = 2737.6 [kJ/kg] and s1 = 6.8943 [kJ/kg.K]. For p = 1 [bar], sf < s2 < sg so that the steam is wet at state ‘2’. Then x2 = (s2 – sf)/(sg – sf) = 0.9231 and h2 = 2501.8 [kJ/kg] so that Wirr = (0.8) (2737.6 – 2501.8) = 188.64 [kJ/kg]. Then, h2' = 2737.6 – 188.6 = 2549.6 [kJ/kg] so that x2¢ = 0.9442 and s2¢ = 7.0218 [kJ/kg.K]. Then, by definition I = T0DS = (300) (7.0218 – 6.8943) = 38.25 [kJ/kg].] 24. Show that [∂/∂T(A/T)]V = – (U/T 2) (i.e. U = – T 2[∂/∂T(A/T)]V). [Hints: Same procedure as that for Example 10. By definition A = U – TS, so that dA = – pdV – SdT; or, S = – (∂A/∂T)V. Substitute this relation for S in the definition of A, divide both sides by T2, and rearrange.] 25. Show that (∂U/∂V)T = T(∂p/∂T)V – p. [Hints: Use the relation dU = TdS – pdV. Hence, (∂U/∂V)T = T(∂S/∂V)T – p. Use one of the Maxwell’s relations to convert the first term on the right-hand side to the required form.] 26. Show that Tds = cvdT + T(∂p/∂T)v dv. [Hints: Let s = s(T, v). Then, by partial differentiation, ds = (∂s/∂T)vdT + (∂s/∂v)T dv. Multiplying both sides of this equation by T gives, Tds = T(∂ s/∂ t)v dT + T(∂s/∂v)T dv. Now, by definition cv = T (∂s/∂T)v. Use a Maxwell relation to convert the second term on the right-hand side to the required form.]

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201

27. Show that Tds = cpdT – T(∂v/∂T)pdp. [Hints: Same procedure as above, but consider s = s(T, p) and use the definition, cp = T(∂s/∂T)p.] 28. Show that Tds = cv(∂T/∂p)v dp + cp(∂T/∂v)p dv. [Hints: S = S(p, v). Then, by partial differentiation ds = (∂s/∂p)v dp + (∂s/∂v)p dv. Multiplying both sides by T gives, Tds = T(∂s/∂v)pdv. Now, (∂s/∂p)v = (∂s/∂t)v(∂T/∂p)v. Similarly, (∂s/∂v)p = (∂s/∂t)p(∂T/∂v)p. Use the definitions of cv and cp given earlier and rearrange these to get the result.] 29. Show that: cp = – T(∂2g/∂T2)p. [Hints: Same procedure as that used for Example 12. By definition cp = T(∂s/∂T)p, and since G is a property: cp = T[∂/∂T(–∂G/∂T)p]p = – T(∂2G/∂T2)p.] 30. Show that: (∂cv/∂v)T = T(∂2p/∂T2)v. [Hints: (∂cv/∂v) T = [∂ /∂ v(T ∂s/∂T) v]T = (by definition of cv); or (∂c v/∂v) T = T [∂/∂ T(∂ s/∂ v)T]v (by interchanging the order of differentiation); or, (∂ cv/∂v) T = T [∂p/∂T(∂/∂T)v]T = T(∂2p/∂T2)v (Using a Maxwell’s relation).] 31. Show that: (∂cp/∂p)T = T(∂2v/∂T2)p. [Hints: Same procedure as above, but start with the definition of cp and use the appropriate Maxwell’s relation.] 32. Show that: cp – cv = – T(∂v/∂T)2p (∂p/∂v)T. [Hints: The two TdS relations are Tds = cvdT + T((∂p/∂T)vdv and Tds = cpdT – T(∂v/∂T)p dp. Since the left-hand sides of these equations are same, so will the righthand sides be. Then, by transposition, (cp – cv)dT = T[(∂p/∂T)vdv + (∂v/∂T)p dp]; or, dT = [T/(cp – cv)][(∂p/∂T)vdv + (∂v/∂T)pdp]. Considering T = v(v, p) gives dT = (∂T/∂v)pdv + (∂T/∂p)vdp. Comparing the coefficient of dv shows that [T/(cp – cv)]. (∂p/∂T)v = (∂T/∂v)p; or, (cp – cv) = T(∂p/∂T)v(∂v/∂T)p. The cyclic relation between these properties, namely, (∂p/∂T)v( ∂T/∂ v)p( ∂v/∂p) T = – 1 gives (∂ p/∂ T)v = –(∂v/∂ T) p (∂p/∂v)T. Substituting this in the above relation for cp – cv gives the answer. It should be verified that equating the coefficient of p also gives the same answer.] 33. Show that cp – cv = Tvb2/k ; where the coefficient of volume expansion, b and the isothermal compressibility k are defined as: b = (1/v)(∂v/∂T)p and k = – (1/v) (∂v/∂p)T, respectively. [Hints: Substitute the parameters in the relation for cp – cv derived in Exercise 32.] 34. Show that c dT ⎡ ⎤ a = ⎢ cv dT + C0 − T v − CiT ⎥ + RT ln v T ⎣ ⎦ [Hints: Same procedure as that in Example 8.15 for the Gibbs function g.] 35. The condition of phase equilibrium demands that chemical potential (Gibbs function) should be the same for all the phases. Verify this for the water–steam equilibrium. [Hints: This means that gf = gg, where g = h – Ts. Let T = 97 [°C]. Then, substituting the data from the steam tables in the above relation gives, gf = – 64.553 [kJ/kg], and gg = – 63.703 [kJ/kg].]

∫

∫

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Engineering Thermodynamics

36. Suppose saturated steam at 1 [bar] is brought in contact with water at 1 [bar] and 300 [K]. By calculating the Gibbs function, show that the steam will condense. [Hints: The system is in mechanical equilibrium since the pressures of both phases are the same. When the steam is brought in contact with water, thermal equilibrium will prevail so that both phases will be at the same temperature. For water at 1 [bar] and 300 [K], using the data from T-based saturation tables shows that gw = 5.258 [kJ/kg] and gs = 68.19 [kJ/kg]. Since gs > gw, the mass transfer will be from steam phase to water phase, i.e. the steam will condense.] 37. (a) Assuming that the isothermal compressibilitykT is constant, obtain a relation for the change in Gibbs function when a liquid is compressed isothermally from p1 to p2. (b) Calculate Dg when water is compressed from 1 [bar] to 20 [bar] at 300 [K]. [Hints: (a) dg = vdp – sdT (by definition) and dg = vdp (isothermal process). Now, kT = – (1/v)(∂v/∂p)T = – (∂ ln v/∂p)T; or, d(ln v) = – kT dp, which can be integrated as v = v0 e–kT(p – p0). Substituting in the relation for the Gibbs function and integrating gives, g – g0 = v0 ∫ p e–kT(p – p0)dp. (b) Writing kT = (1/v) (Dv/Dp) and using the data 0 from the T-based steam tables gives, kT = 0.1191 [bar–1]. Evaluating the above integral and substituting this value of kT gives, g = 7.548 [J/kg].] 38. The densities of saturated liquid and vapour of a substance at 350 [K] are 828 [kg/m3] and 3.23 [kg/m3], respectively. The rate of change of pressure with temperature is 23 [torr/K]. Compute the heat of vaporization using the Clapeyron equation. [Hints: Writing the Clapeyron equation as hfg = T(vg – vf)(dp/dT). Substituting the values and remembering that 760 [torr] = 101.352 [kN/m2], the Clapeyron equation gives, hfg = 331.0 [kJ/kg].] 39. The vapour pressure (in [torr]) of a substance is given by log p = 7.2621 – (1402.46/T) – (51387.5/T2), where T is in [K]. Calculate (a) its normal boiling point (NBP) (i.e. at p = 760 [torr]); (b) heat of vaporization at this pressure, if the specific volumes of vapour and liquid are 0.356 [m3/kg] and 0.0012 [m3/kg], respectively; and (c) its boiling point at a pressure of 800 [torr]. (Note: log x = log10 x.) [Hints: (a) Solving the given vapour pressure equation gives, T = 353.3 [K]; (b) obtaining (dp/dT) from this equation and substituting in the Clapeyron equation along with other data gives, hfg = 172.5 [kJ/kg]; and (c) same as (a) above, T = 355 [K].] p

Chapter 9

Properties of Substances

9.1

INTRODUCTION

It was pointed out in the early chapters that the characteristics of working substances of a thermodynamic system do not form part of thermodynamics. They are borrowed from the properties of matter (an area common to physics and chemistry). Some preliminary ideas were presented in Chapter 2. In this chapter, these are elaborated. The discussion here is restricted to properties of gases (including vapours) and liquids since they form the most common working fluids of thermal systems. In the ensuing discussions, the following two primitive concepts are used. The phases of a substance are solid, liquid, and gas, and a pure substance is one that contains only one chemical species irrespective of the number of phases. In this chapter pure substances are dealt with, since two of the most common working substances of thermal systems [namely, gases (e.g. air) and liquids (e.g. water)] are of this type.1 The major part of the discussion here pertains to gases since (a) their properties are expressible by equations of reasonable complexity, (b) vapours can be well approximated as gases, and (c) properties of liquids are directly determined by experiments.

9.2

DATA AND COMPUTATIONAL PROCEDURE

In this section, the basic data required and the method of computing the properties of substances are discussed.

9.2.1

Independent and Dependent Properties

The thermodynamic properties of primary interest are pressure (p), volume (v), temperature (T), internal energy (u), enthalpy (h), entropy (s), Gibbs function (g), and Helmholtz function (a). It is seen that there are eight properties (i.e. variables) whose values are needed. 1

For historical reasons, air is treated as a pure substance although it is a mixture. 203

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Since only simple compressible substances are considered, two of these properties should be taken as independent ones as, according to the state principle, others are functions of these independent properties. During the discussion on the state principle it was shown that, for simple compressible substances, the independent properties should be chosen from among p, v and T. It was also mentioned therein that it is usual to choose T as one of the independent properties since the techniques of measuring temperature accurately (particularly by resistance thermometers and thermocouples) have been well developed. Then, the other independent property can be either p, or v. Now, the values of the six dependent properties, namely, p (or v), u, h, s, g and a are to be calculated in terms of the properties chosen as independent ones. According to mathematics, this requires six equations. The following defining relations, namely (a) h = u + pv, (b) g = h – Ts, and (c) a = u – Ts provide three (algebraic) equations. Then, three other equations are needed. These are (a) the equation of state, (b) the equation for internal energy u, and (c) the equation for entropy s. Since u and s are defined in terms of energy, both of them can be calculated once the specific heats are known. Thus, only the equation of state and the values of specific heats are required to calculate all the properties. It would be ideal if the values of the properties are measured at all states. However, due to several constraints on the resources (i.e. money, manpower, time, …, etc.) this is not possible. The states are so chosen that the data obtained has maximum accuracy with the available resources.

9.2.2

Basic Data

The basic data are generally obtained from macroscopic measurements. However, for vapours and gases at low pressures, where they approach ideal gas behaviour, spectroscopic (microscopic) measurements and methods of statistical mechanics are used. This is mentioned at the end of this section. The p–v–T data The states chosen for obtaining the p–v–T data should cover the entire range of desired p, v, and T. The statistical methods are then used to smoothen the data and obtain the equation of state (e.g. by the methods of least squares). The implicit form of the equation of state, namely f(p, v, T) = 0 is not convenient for calculations. Hence, it is generally expressed in the explicit form such as p = p(T, v), or as v = v(T, p). For some substances either of these forms is convenient. For example, the ideal gas equation, pv – RT = 0, can be written as p = RT/v, or as v = RT/p. However, the equation of state for a van der Waal gas can be solved directly for p as p = [RT/(v – b)] – a/v2. Since it is a cubic equation in v, solving for v is more difficult. There are other criteria that are also to be considered in choosing the second independent variable (the first one is T). These are mentioned below. The following quantities also contain part of the p–v–T data. (a) The volume expansivity defined as b = (1/v)(∂v/∂T)p. Note that this is the coefficient of volumetric expansion defined analogous to the coefficient of linear expansion. (b) The isothermal bulk modulus defined as

Chapter 9:

Properties of Substances

205

BT = – v(∂p/∂v)T analogous to the Young’s modulus. The inverse of this is defined as the isothermal compressibility (i.e. kT = 1/BT). (c) The isentropic bulk modulus defined as BS = – v(∂p/∂v)S, and its inverse the isentropic compressibility defined as kS = 1/BS. The importance of these quantities lies in the fact that some of them can be more easily (or accurately) determined and that they vary only slowly with p and T. Specific heat data This terminology is remnant of the caloric theory. The two specific heats are cv and cp. Currently, they are defined as cv = (∂u/∂T)v and cp = (∂h/∂T)p. Generally, it is adequate to measure the values of specific heats at different temperatures at a fixed pressure, (generally atmospheric) (for cp), or at a convenient volume (for cv). These data are then expressed as a polynomial in temperature. For example, the variation of cv can be expressed as cv = f(T), where f(T) = a0 + a1T + a2T2 + …, or more generally, f(T) =

∑

+∞ j = −∞

a j T j . Similarly, the variation of cp can be expressed as cp = g(T), where g(T) = b0 +

b1T + b2T2 + …, or more generally, g(T) =

∑

+∞ j = −∞

bj T j .

Then, using the thermodynamic relations developed in Chapter 8 (see property relations) their values at all other states are calculated. For example, since cv is also a property, cv = cv(T, v)

or

⎛∂2p⎞ ⎛∂c ⎞ ⎛∂c ⎞ ⎛∂c ⎞ dcv = ⎜ v ⎟ dT + ⎜ v ⎟ dv = ⎜ v ⎟ dT + T ⎜⎜ 2 ⎟⎟ dv ⎝ ∂ T ⎠v ⎝ ∂ v ⎠T ⎝ ∂ T ⎠v ⎝ ∂ T ⎠v where the second term in the last part of the equality is the transformation given (as Exercise 21) in Chapter 8. It is seen that since the variation of cv with T can be expressed as a polynomial in T from measurement, the first term is known. The equation of state, written as p = p(T, v), is used for evaluating (∂2p/∂T2)v. Thus, the partial differential equation of cv becomes ⎛∂2p⎞ dcv = f (T )dT + T ⎜⎜ 2 ⎟ ⎟ dv ⎝ ∂Τ ⎠v or

cv =

∫

T

T0

f (T ) dT + T

⎡⎛ ∂ 2 p ⎞ ⎤ ⎢⎜ ⎟ ⎥ dv v0 ⎢ ⎜ ∂ T 2 ⎟ ⎥ ⎠v ⎦ ⎣⎝

∫

v

By integrating this partial differential equation, the values of cv at all states are obtained. Using the same arguments, it can be shown that ⎛ ∂ 2v dc p = g(T ) dT + T ⎜⎜ 2 ⎝ ∂Τ

⎞ ⎟⎟ dp ⎠p

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or

cp =

∫

T

T0

g(T ) dT + T

∫

⎡⎛ ∂ 2 v ⎞ ⎤ ⎢⎜ ⎟ ⎥ dp p0 ⎢⎜ ∂ T 2 ⎟ ⎥ ⎝ ⎠v ⎦ ⎣ p

By integrating this partial differential equation, the values of cp at all states can be obtained. This confirms the earlier argument that it is adequate to measure cv (and cp) at all temperatures for one volume (and pressure for cp). The values at all other states can be computed using this value and the equation of state. It should also be noted that it is not necessary to measure both cp and cv since they are related by the expression 2

TV b 2 ⎛ ∂V ⎞ ⎛ ∂ p ⎞ c p − cv = − T ⎜ ⎟ ⎜ ⎟ = ⎝ ∂ T ⎠ p ⎝ ∂V ⎠ T kT or, through the ratio of the specific heats as

g =

⎛ ∂ p⎞ =⎜ ⎟ cv ⎝ ∂V ⎠ S

cp

BS ⎛ ∂ p⎞ ⎜⎝ ⎟⎠ = ∂V T BT

where BS and BT are the isentropic and isothermal bulk moduli, respectively. These were presented in Chapter 8 (as Example 8.13). Microscopic data The data mentioned above are obtained by macroscopic measurements. However, in some cases (e.g. ideal gas, solids, …), it is convenient to make measurements at microscopic levels by spectroscopic methods. In this, the emission or absorption spectra of a (small) sample is obtained from experiments. From this, the distribution of particles (atoms, molecules, …) in various energy levels and the partition function are calculated using the models of statistical mechanics (the Maxwell– Boltzmann (or classical), the Bose–Einstein, the Fermi–Dirac, …) as appropriate. As shown in the next subsection, from the partition function all the thermodynamic functions can be calculated.

9.2.3

Computational Procedure

Experience has shown that some combinations are more suitable for computing the dependent properties since they involve less complicated calculations.2 These combinations are listed below: 1. If the cv data is available, since cv is defined as (∂u/∂T)v, it is convenient to choose T and v as the independent variables (properties). In this case, one term (each) of the (partial differential) equations for u and s involves the cv data which is directly measured. 2. By the same arguments, if the cp data is available, then T and p are chosen as the independent properties. 2

The formulation based on the fundamental functions is explained in the notes to the steam tables in Appendix F.

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207

3. If cv and cp are both available, the more accurate of the two is chosen. 4. If both of cv and cp are equally accurate, then T and v are chosen if the equation of state in the form p = p(T, v) is more accurate or simpler (or, both). 5. If cp and cv are equally accurate, then T and p are chosen if the equation of state in the form v = v(T, p) is more accurate or simpler (or, both). 6. If cp, cv, v(T, p) and p(T, v) are all known with equal accuracy and convenience, then either approach can be chosen. The discussion given below will clarify these ideas further. v and T as independent properties In this case, the following steps form the computational procedure. 1. The pressure p is calculated from the equation of state written as, p = p(v, T). 2. The appropriate specific heat data is, cv = a + bT + cT2 + … . 3. Since T and v are the natural coordinates of u, the relation for u is written as, u = u(T, v). Now, partial differentiation gives du = (∂u/∂T)vdT + (∂u/∂v)Tdv. The definition of cv shows that the first term is cvdT. The second term is converted as follows. The basic property relation, namely, du = Tds – pdv, gives (∂u/∂v) T = T(∂s/∂v)T – p. Using the Maxwell’s relation (∂s/∂v)T = (∂p/∂T)v, the equation for u becomes

⎡ ⎛ ∂ p⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − p⎥ dv ⎣ ⎝ ∂Τ ⎠ v ⎦ or ⎡ ⎛ ∂ p⎞ ⎤ T ⎜ ⎟ − p⎥ dv ⎢ T0 v0 ⎣ ⎝ ∂T ⎠ v ⎦ By evaluating the integrals, the values of u at all states can be calculated. u − u0 =

∫

T

cv dT +

∫

v

4. Now the enthalpy is calculated from its definition, namely, h = u + pv, where the pressure p is obtained from the equation of state. Similar to the equation for u, an equation (partial differential) can be set up for h too, as follows. The property relation dh = Tds + vdp gives

⎛ ∂h⎞ ⎛ ∂s ⎞ ⎛ ∂ p⎞ ⎛ ∂ p⎞ ⎛ ∂ p⎞ ⎜⎝ ∂ v ⎟⎠ = T ⎜⎝ ∂ v ⎟⎠ + v ⎜⎝ ∂ v ⎟⎠ = T ⎜⎝ ∂ T ⎟⎠ + v ⎜⎝ ∂ v ⎟⎠ T T T v T where the first term of the last equality is obtained through a Maxwell’s relation. In addition, the following equation for the variation of h can also be derived.

⎛ ∂h ⎞ ⎛ ∂s ⎞ ⎛ ∂ p⎞ ⎛ ∂ p⎞ ⎜⎝ ⎟⎠ = T ⎜⎝ ⎟⎠ + v ⎜⎝ ⎟⎠ = cv + v ⎜⎝ ⎟⎠ ∂T v ∂T v ∂T v ∂T v

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Note that this equation involves more number of derivatives than the above procedure of computing u first and then h through an algebraic equation. 5. As for u, the (partial differential) equation for s is

⎛ ∂s ⎞ ⎛ ∂ p⎞ ds = ⎜ ⎟ dT + ⎜ ⎟ dv ⎝ ∂T ⎠ v ⎝ ∂T ⎠ v or ⎡⎛ ∂ p ⎞ ⎤ ⎢⎜ ⎟ ⎥ dv v0 ⎝ ∂ T ⎠ ⎣ v⎦ By evaluating the integrals, the values of s can be calculated for all states. s − s0 =

⎛ cv ⎞ ⎜ ⎟ dT + T0 ⎝ T ⎠

∫

T

∫

v

6. Then, Gibbs and Helmholtz functions are calculated from their defining equations, namely, g = h – Ts and a = u – Ts, respectively. p and T as independent properties In this case too, the procedure is similar to the one given above. Hence, the following steps need no explanation. 1. The specific volume v is calculated from the equation of state written as v = v(p, T). 2. The appropriate specific heat data is, cp = a + bT + cT2 + … . 3. Since T and p are the natural coordinates of h, the relation for h is written as h = h(T, p). Now, partial differentiation gives dh = (∂h/∂T)p dT + (∂h/∂p)T dp. The definition of cp shows that the first term is cp dT. The second term is converted as follows. The basic property relation, namely, dh = Tds + vdp, gives (∂h/∂p)T = T(∂s/∂p)T + v. Using the Maxwell’s relation (∂s/∂p)T = – (∂v/∂T)p the equation for h becomes ⎡ ⎛ ∂v ⎞ ⎤ dh = c p dT − ⎢T ⎜ ⎟ − v ⎥ dp ⎝ ⎠ ⎥⎦ ⎣⎢ ∂ T p

or h − h0 =

∫

T

T0

c p dT −

∫

⎡ ⎛ ∂v ⎞ ⎤ ⎢T ⎜ ⎟ − v ⎥ dp p0 ⎝ ⎠ ⎥⎦ ⎣⎢ ∂ T p p

By evaluating the integrals, the values of h at all states can be calculated. 4. Now, the internal energy is calculated from its definition, namely, u = h – pv, where the v is obtained from the equation of state. Similar to equation for h, an equation (partial differential) can be set up for u too, as follows. The property relation du = Tds – pdv gives ⎛ ∂u ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜⎝ ∂ p ⎠⎟ = T ⎝⎜ ∂ p ⎠⎟ − p ⎝⎜ ∂ p ⎠⎟ = − T ⎝⎜ ∂ T ⎠⎟ − p ⎝⎜ ∂ p ⎠⎟ p T T T T

where the first term of the last equality is obtained through a Maxwell’s relation.

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In addition, the following equation for the variation of u can also be derived. ⎛ ∂u ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜⎝ ⎟⎠ = T ⎜⎝ ⎟⎠ − p ⎜⎝ ⎟⎠ = c p − p ⎜⎝ ∂T p ∂T p ∂T p ∂ T ⎟⎠ p

Note that this equation involves more number of derivatives than the above procedure of computing h first and then u through an algebraic equation. 5. Using the same arguments as for u, the (partial differential) equation for s is obtained as ⎛ ∂s ⎞ ⎛ ∂v ⎞ ds = ⎜ ⎟ dT − ⎜ ⎟ dp ⎝ ∂T ⎠ p ⎝ ∂T ⎠ p

or s − s0 =

T

cp

T0

T

∫

dT −

∫

⎛ ∂v ⎞ ⎜ ⎟ dp p0 ⎝ ∂ T ⎠ p p

As before, by evaluating these integrals the values of s can be calculated for all states. 6. Then, Gibbs and Helmholtz functions are calculated from their defining equations, namely, g = h – Ts and a = u – Ts, respectively. Computations for microscopic data The data and computational procedure presented so far pertain to macroscopic measurements. It is also possible to calculate the thermodynamic properties based on the spectroscopic data (microscopic measurements). This approach is now described. In this approach, all thermodynamic properties are expressed in terms of the partition function, denoted as Z, as follows: The Helmholtz function is written as A = – NkBT ln Z, where N is the total number of particles in the system at temperature T, and kB is the Boltzmann constant (equals universal gas constant divided by the Avogadro number). The internal energy is written as U = [NkBT2][d(ln Z)/dT]. The expression for entropy is written as S = NkBT ln Z + (U/T), which is the same as S = (A/T) + (U/T). Basically, the following three types of distribution functions have been found to be most suitable. The Maxwell–Boltzmann distribution function for which Z = Si exp(ei/kBT). This assumes that all the particles are distinguishable and only posses kinetic energy. Since these are the assumptions applicable in classical mechanics, this is also called the classical statistics. The Bose–Einstein statistics considers all the rules of quantum mechanics except the exclusion principle. Here, the expression for partition function is, Z = Si 1/[exp(ei/kBT) – 1]. The particles that obey this statistics are called bosons. Photons are the most famous bosons. The distribution function which includes the exclusion principle is called the Fermi– Dirac statistics for which, Z = Si 1/[exp(ei/kBT) + 1]. The particles which obey this statistics are called fermions. Electrons are the most famous fermions.

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9.3

GASES

The interest in gases is primarily due to the fact that (a) their behaviour can, in most cases, be expressed by simple equations, (b) they are the most common working substances in thermal engineering, and (c) in many conditions, the behaviour of vapours can be approximated by those of gases.

9.3.1

Ideal Gas

Among gases, the ideal gas3 is the most important because (a) its behaviour can be expressed by very simple equations and (b) for the working conditions of thermal systems, most gases (e.g. air, combustion products, steam at atmospheric temperature, …) behave ideally, and (c) its behaviour can be estimated from the experiments on real gases by extrapolating to zero pressure [in fact, this (i.e. all gases behave ideally at zero pressure) is generally taken as the definition of an ideal gas]. Equation of state An ideal gas is defined to be one that follows the equation pv = RT at all pressures, where R is the gas constant. Whether it is specific or universal (molar) gas constant depends on whether v is the specific or molar volume. This is determined as follows. For the sake of arguments, let R and R denote the specific and universal gas constants. Similarly, let v and v¢ denote the specific and molar volumes. Let M denote the molecular weight. Let m and n denote the mass and moles of the gas. The relations between these quantities are: m = nM; R = R/M and v = v¢/M. To begin with, assume that v in the above equation of state is the specific volume. Then, using the relations given above the following transformations are directly obtained: pv = RT; or, p(V/m) = RT; or, pV = mRT; or, pV = n(M)RT; or, pV = nRT; or, p(V/n) = RT; or, pv¢ = RT. This shows that: ∑ The ideal gas equation can simply be written as pv = RT provided that — if v is the specific volume v, then R should be the specific gas constant (i.e. if v = V/m, then R = R); and, — if v is the molar volume v¢, then R should be the universal gas constant (i.e. if v = V/n, then R = R). ∑ Only the molar (universal) gas constant (= 8.3143 [kJ/kmol.K]) is needed since it can be converted to mass unit when needed by dividing the universal gas constant by the molecular weight. In this book, the ideal gas equation is written only as pv = RT. The value of the gas constant is chosen as 8.3143 [kJ/kmol.K] only if the volume v is the molar volume. Otherwise R is taken as that specific gas constant (= 8.3413/M) [kJ/kg.K], where M is the molecular weight. 3

It will be recalled that ideal gases, van der Waal gas (to be described below), .... are only models of real gases and hence are approximate. In real applications, for each case, the most appropriate model should be chosen.

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In Chapter 2, the Boyle’s law (proposed4 in 1662) was used to define an empirical temperature q = pv/R, where all the symbols have the usual meaning. In the same chapter, it was also shown that the law of Charles (1787) and Gay-Lussacs (1802) implies that the absolute Celsius temperature can be defined as q¢ = q + 273.15, so that the laws of Boyle and of Charles and Gay-Lussacs can be combined to give q¢ = pv/R. In Chapter 4, during the discussion on empirical temperature, it was shown that the ideal gas temperature can be obtained by extrapolation to (i.e. as the limit of) zero pressure of the data on real gases. During the discussion on the second law (in Chapter 7) it was shown that the consequence of the Joule’s law (i.e. U of an ideal gas is independent of V) is q¢ = T, where T is the thermodynamic temperature. It will now be clear that the ideal gas equation written in the above form incorporates all these results. Equations for properties The Joule’s law states that for an ideal gas (u)T is independent of volume. Then, it was assumed that u will be a function of T only. Mathematically, this is correct only if (u)T is independent of p too. This is determined as follows. Internal energy.

The basic property equation, du = Tds – pdv, gives ⎛ ∂u ⎞ ⎛ ∂s ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎜⎝ ∂ p ⎟⎠ = T ⎜⎝ ∂ p ⎟⎠ − p ⎜⎝ ∂ p ⎟⎠ = − T ⎜⎝ ∂ T ⎟⎠ − p ⎜⎝ ∂ p ⎟⎠ p T T T T

where the first term in the second equality is obtained from a Maxwell’s relation. Writing the ideal gas equation as v = RT/p, simple differentiation gives (∂v/∂T)p = R/p, and (∂v/∂p)T = – (RT/p2). Substituting these in the above expression shows that (∂u/∂p)T = 0, i.e. u is independent of p. This proves the statement that the internal energy of an ideal gas depends only on temperature. The specific heat at constant volume is defined as cv = (∂u/∂T)v. For ideal gases, this becomes cv = du/dT. This shows that the cv of an ideal gas is only the function of temperature. Then, the equations for evaluating u are

⎛ ∂u ⎞ ⎛ ∂u⎞ du = ⎜ ⎟ dT + ⎜ ⎟ dv = cv dT ⎝ ∂T ⎠ v ⎝ ∂v⎠ T

and

u = u0 +

∫

T

T0

cv dT

Enthalpy. Now, enthalpy is defined as h = u + pv, which for an ideal gas, becomes h = u + RT. Since, u is only function of T (as shown above), this implies that the enthalpy of an ideal gas is also a function of only temperature. 4

See Conant, J.B., “Robert Boyle’s Experiments in Pneumatics”, Case 1, in [CON], Vol. 1. But, the first statements of a detailed atomic hypothesis in a form approximating that which prevails today was made by William Higgins in 1789. In 1803, Dalton made his first announcement of his atomic theory (see Nash, L.K., “The Atomic-Molecular Theory”, Case 4, ibid).

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The specific heat at constant pressure is defined as cp = (∂h/∂T)p, which, for ideal gases becomes cp = dh/dT. This implies that cp of an ideal gas is only the function of temperature. Then, the equations for evaluating h become ⎛ ∂u ⎞ ⎛ ∂h ⎞ dh = ⎜ ⎟ dT + ⎜ ⎟ dp = c p dT ⎝ ∂T ⎠ p ⎝ ∂ p⎠T

h = h0 +

and

∫

T

T0

c p dT

Specific heats. Differentiating the definition of h for ideal gases (i.e. h = u + pV = u + RT) with respect to T given above and substituting the definitions of cp and cv gives, cp = cv + R, or, cp – cv = R. This shows that the difference between the specific heats of ideal gases is constant. The ratio of specific heats cp/cv is denoted by g. Then, dividing the above relation throughout by cv gives, [(cp/cv) – 1] = (g – 1) = R/cv, which can be rewritten as cv = R/(g – 1). Then, cp = gcv = Rg/(g – 1). This shows that both the specific heats of an ideal gas can be evaluated if its gas constant and specific heat ratio are known. Table 9.1 gives the values of cp for some common gases (assumed ideal) which form combustion products. Table 9.1 Species H2 N2 O2 CO H2O CO2 SO2

Specific heats of some gases

a

b

14.54 0.9643 0.7974 0.9483 1.679 0.6086 0.4198

c

–0.4187 0.2113 0.4256 0.2745 0.5522 0.9611 0.9056

– – – – –

d

1.006 0.1207 0.1331 0.04188 0.06210 0.3240 0.5955

— — — — — — 0.1346

cp = a + bq + cq2 + dq3, where q = T/1000, with T in [K] and cp in [kJ/kg.K]. Entropy. The equations for entropy change of an ideal gas are obtained as follows. By definition of entropy, ds = (dQR/T), and by energy (first law) equation, ds = (du + pdv)/T; or, using the definition of cv, ds = cv(dT/T) + (p/T)dv. By the ideal gas equation, this becomes ds = cv

dT dv +R T v

(infinitesimal process)

and s = s0 +

and

⎛ dT ⎞ cv ⎜ ⎟ + R T0 ⎝ T ⎠

∫

T

⎛T s = s0 + cv ln ⎜ ⎝ T0

∫

v

v0

dv v

⎞ ⎛ v ⎞ ⎟ + R ln ⎜ ⎟ ⎠ ⎝ v0 ⎠

(finite process)

(finite process, constant cv)

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This last equation can be rewritten in two different forms by using the ideal gas equation and the relation between cp and cv. These are given below for a finite process with constant cp and cv. By the equation of state, v/v0 = (T/T0)(p0/p). Substituting this in the above equation and using the relation cp = cv + R, the entropy equation reduces to ⎛T s = s0 + c p ln ⎜ ⎝ T0

⎞ ⎛ p ⎞ ⎟ − R ln ⎜ ⎟ ⎠ ⎝ p0 ⎠

(finite process, constant cp)

Similarly, eliminating the temperature by T/T0 = (p/p0)(v/v0), and using the relation cp = cv + R gives ⎛ v ⎞ ⎛ p⎞ s = s0 + c p ln ⎜ ⎟ + cv ln ⎜ ⎟ ⎝ v0 ⎠ ⎝ p0 ⎠

Gibbs and Helmholtz functions.

(finite process, constant cp and cv)

The equations for the Gibbs and Helmholtz functions are cp

∫

∫ T dT − n R T ln p − c T + c

∫

∫ T dT − n R T ln V − c T + c

G = c p dT − T

A = cv dT − T

3

cv

1

4

2

These equations have been presented in Chapter 8. Processes It is necessary to consider only the relations for isothermal and isentropic processes since these were shown (in Chapter 3) to be indirect specifications. The isothermal process is defined by the equation T = constant = T0 (say). Then, the equation of state of an ideal gas reduces to pv = RT0 = C, where C is a constant. This is the equation for an isothermal process of an ideal gas. The relation for an isentropic process of an ideal gas can be derived directly from the above equations for entropy change. An isentropic process is defined to be one for which Ds = 0. Imposing this condition on the first form of the entropy equation and using the various expressions for specific heats gives

⎛T ⎞ ⎛R⎞ ⎛ v ⎞ 0 = ln ⎜ ⎟ + ⎜ ⎟ ln ⎜ ⎟ ⎝ T0 ⎠ ⎝ cv ⎠ ⎝ v0 ⎠

or

or

⎡⎛ T ⎞⎛ v ⎞g −1 ⎤ 1 = ⎢⎜ ⎟⎜ ⎟ ⎥ ⎢⎣⎝ T0 ⎠⎝ v0 ⎠ ⎥⎦

or

⎡⎛ T ⎞⎛ v ⎞g −1 ⎤ 0 = ln ⎢⎜ ⎟⎜ ⎟ ⎥ ⎢⎣⎝ T0 ⎠⎝ v0 ⎠ ⎥⎦

T ( v)g −1 = T0 (v0 )g −1

which can be compactly written as Tv(g –1) = C, where C is a constant (= T0v0(g –1)).

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Similarly, starting with the other forms of the equations derived above the following alternative forms, namely, pvg = C, and, T/(p)(g –1)/g = C can be derived. These forms can also be obtained by transforming the first equation using the equation of state. All these are equivalent forms of the equation for an isentropic process of an ideal gas. Since they are all equivalent, in any given situation, the most convenient form can be used.

9.3.2

Real Gases

There are two ways to approach the equation of state of real gases. Microscopic description. According to this, the atoms of an ideal gas are point masses with no interaction other than collision. Then, modifications should account for the atomic volume and for other types of interactions (e.g. through potential energy). This is known as ‘the kinetic theory of dense gases’. Macroscopic description. According to this, an ideal gas equation is one that contains no arbitrary constants5 (or parameters). Then, we can have equations with increasing number of arbitrary constants. Here, the equation of state is simply the best fit equation of the p–v–T data obtained from experiments through statistical techniques such as the method of least squares. Both the above approaches are equivalent in the sense that they lead to the same types of equations. In this section, only a few of the more common equations of state are mentioned. As in the case of an ideal gas, here too, the polynomial functions in temperature have been found to be adequate to represent variation of specific heats with temperature. Their variation with p (or v) is evaluated from the equation of state through appropriate formulae mentioned (as examples or exercises) in Chapter 8. van der Waal gas Logically, the first step in extending the ideal gas equation is to assume an equation with one (arbitrary) constant. The physical arguments suggest that this should be a correction factor that accounts for the finite volumes of the molecules. This gives the Clausius equation of state that has the form p(v – b) = RT where b is the constant which accounts for the finite volume of the atoms. This constant is determined from the p–v–T data obtained experimentally. The next logical step is to assume an equation with two (arbitrary) constants. Once again the physical arguments suggest that this additional factor should account for the interaction owing to attraction between the atoms. This is done by the van der Waal equation of state that has the form a⎞ ⎛ ⎜⎝ p + 2 ⎟⎠ ( v − b) = RT v 5

Generally, the equations are written in terms of molar volume so that the gas constant R is the universal gas constant and, therefore, is the same for all gases.

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215

where a and b are constants to be determined from experiments. A gas that follows the van der Waal equation is called a van der Waal gas. The following are the two reasons for the importance of the van der Waal equation of state. First, it is based on the arguments of the kinetic theory of gases.6 Since the molecules of a gas have non-zero volumes, the actual (free) volume occupied will be less. The constant b accounts for this reduction in volume. This was the argument used by Clausius while proposing his equation of state (given above). The kinetic theory of ideal (dilute) gases shows that the pressure exerted by a gas is due to the collision of its molecules with the walls of the vessel. The collision changes the momenta of the molecules since the walls reflect them.7 Now, the next step is to account for the potential energy interactions between the molecules and the wall. The molecules in the bulk gas attract those moving towards the wall and thereby reduce their momenta. This attraction is directly proportional to the number of molecules in the bulk gas and is inversely proportional to the distance between them. The number of molecules of the gas is directly proportional to the density8 of the gas. The distance between the molecules decreases as its number (per unit volume, i.e. density) increases. In other words, the distance between the molecules is inversely proportional to the density. Thus, the attraction is proportional to the square of the gas density, i.e. it is inversely proportional to the specific volume. The constant of proportionality was assumed to be ‘a’ by van der Waal. The second reason for the importance of the van der Waal equation of state is that it predicts the phase changes (i.e. evaporation, condensation, sublimation and solidification) of a gas. The isotherms of an ideal gas (i.e. pv = constant) are rectangular hyperbolas in the p–v plane. Then, each isobar [constant p (i.e. horizontal) line] intersects an isotherm at only one point. The reason for this is as follows. The equation pv = C is really the relation p = C/v from which p can be calculated for any given value of v. Now, if the value of v is to be determined for a given value of pressure, say p = p1, then the equation to be solved is v = C/p1, which is equivalent to the pair v = C/p and p = p1. Geometrically, this is the point of intersection of the isotherm and the isobar. This shows that ideal gases cannot undergo any phase change.9 The importance of the van der Waal equation of state is that a van der Waal gas can condense and evaporate. To understand this, consider the nature of the isotherms of a van der Waal gas as shown in Figure 9.1. At the outset, note that for v >> a/p and v >> b the van der Waal equation reduces to the ideal gas equation. In other words, for any given temperature, when the volume is far greater than the values given above, a van der Waal gas behaves like an ideal gas. At low pressures, this volume is determined by the first criterion (i.e., v >> a/p ), and at high pressures, this term becomes small so that the second criterion, namely, v >> b becomes applicable. 6 7 8 9

Other theories simply reuse these assumptions in more refined forms. Since momentum has both magnitude and direction, it can change when either of thses changes. Because of its definition, density equals the number of molecules in a vessel multiplied by the mass of one molecule divided by the volume of the gas. This is why they are also called permanent (or perfect) gases.

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The equation for the isotherms of a van der Waal gas is (p + a/v2)(v – b) = C, where the constant C = RT1 = RT2 = …. In Figure 9.1 the isotherms corresponding to low and high pressures are indicated by G0–G0 and by G1–G1, respectively. Their temperatures (namely, T0 and T1 respectively) are also indicated therein. G1 p

5

p = p1 C

p = pc

1

T1

Tc 6 V

L+V

L

G0

G1

B 3

p = p2 2

2¢

T0

T2

A 4 G0 v Figure 9.1 Isotherms of a van der Waal gas.

Note that G1–G1 is also a typical supercritical isotherm since T1 > Tc. Note also that, the typical supercritical isobar, indicated as p1 intersects the typical supercritical isotherm only at one point. Similarly, the critical isobar that has been indicated as pc intersects the critical isotherm (one passing through the critical point C) indicated by 5–6 critical isotherms only at one point. In Figure 9.1, the curve 1–2–A–2¢–B–3–4 represents a typical subcritical isotherm. Since its temperature T2 is less than the critical temperature Tc, it is called a subcritical isotherm. Note that subcritical isotherms go through a maximum and a minimum. Similarly, the isobar denoted as p = p2 is a typical subcritical isobar since p2 < pc. The figure shows that they intersect each other at three points denoted as 2, 2¢ and 3, respectively. For specified values of p and of T, the van der Waal equation is a cubic equation in v. Mathematically, this means that the equation should have three roots. Moreover, since the coefficients of this equation are real, the complex roots should occur in conjugate pairs. Then, the possible distribution of roots is (a) one real and two complex roots, (b) three real but equal roots, and (c) three real and distinct roots. The above discussion shows that these three cases correspond to the supercritical, critical and subcritical states. It was shown in Chapter 8 that for stable substances v should decrease when p increases,10 i.e. dp/dv < 0. In other words, the part A–2¢–B of the subcritical isotherm is unstable since it 10

In common terminology, this means that if you compress a substance its volume should decrease.

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217

has a positive slope. Thus, the horizontal line 2–2–3 represents the stable isothermal states. This represents a phase change. The states corresponding to the part 3–B are called the supersaturated states. For example, they occur during the last stages of expansion steam turbine when the steam flows so fast that it does not have adequate time to condense. However, after some time the steam condenses and the state moves on to the line 2–3. Similarly, the part 2–A occurs when a fluid without any impurities boils off from surface without any irregularities to act as nucleation sites. This part is also unstable since such ideal conditions get disturbed soon. van der Waal constants The constants of the van der Waal gas a and b are related to pc, vc and Tc. These relations are derived as follow: The typical subcritical isotherm in Figure 9.1 (whose temperature is denoted as T2) shows a maximum and a minimum. Now, as T2 increases, these approach each other. In the limit, when T2 = Tc, they coincide and, the isotherm becomes the critical isotherm. In mathematics, such a point is called an inflexion point. Thus, the critical point C in the p–v plane is an inflexion point on the critical isotherm. Mathematically, this means that ⎛∂ p⎞ ⎜ ⎟ =0 ⎝ ∂ v ⎠Tc

and

⎛∂ 2p⎞ ⎜⎜ 2 ⎟⎟ = 0 ⎝ ∂ v ⎠Tc

The van der Waal equation is first written as p = RT/(v – b) – (a/v2). Then, ⎛ ∂ p⎞ ⎜⎝ ⎟⎠ = 0 ∂v T

gives

⎛∂ 2p⎞ ⎜⎜ 2 ⎟⎟ = 0 ⎝ ∂ v ⎠Tc

gives

c

2a vc3

=

RTc ( vc − b)2

and 2 RTc 6a = 4 vc (vc − b)3

Solving these given equations and then substituting in the van der Waal equation gives vc = 3b,

Tc =

8a , 27 Rb

and

pc =

a 27b2

Now, the two constants can be evaluated from any two of these equations. Generally, the critical volume vc is known with less accuracy since it is a difficult parameter to measure. Thus, the van der Waal constants are expressed in terms of the pc and Tc as a=

27 R 2 Tc2 64 pc

and

b=

RTc 8 pc

The critical pressure and critical temperature of some common gases are shown in Table 9.2. These have been computed from the data of CRC Handbook ([WEA], pp. F-64–66).

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Engineering Thermodynamics

Table 9.2 Species

CO2

Critical constants of some gases CO

H2

304 133 Tc [K] pc [MPa] 7.39 3.50 0.12801 0.11955 vc [l/mol] Zc (dimensionless) 0.374 0.378

N2

O2

H2O

33.1 126 154.6 647.1 1.30 3.39 5.08 22.12 0.07983 0.11739 0.09549 0.09147 0.377 0.380 0.377 0.376

Corresponding States Principle (CSP) This important principle, also called the Principle of Corresponding States (PCS), arises as a consequence of the van der Waal equation. To understand how this arises, first define the dimensionless properties as pr = p/pc, vr = v/vc and Tr = T/Tc, so that p = pr pc, v = vrvc and T = TrTc. These properties, namely, pr, vr and Tr are called reduced properties (or, coordinates) since the actual p, v and T are expressed as fractions of the critical constants, i.e. they are reduced with respect to the critical values. Then, the van der Waal equation becomes ⎛ a ⎜ pr pc + vr vc ⎝

⎞ ⎟ (vrvc – b) = RTrTc ⎠

Substituting appropriate expressions for pc, vc and Tc from above and simplifying, the van der Waal equation becomes

⎛ 3 ⎜⎜ pr + 2 vr ⎝

⎞ ⎟⎟ (3vr – 1) = 8Tr ⎠

This equation is called the reduced equation of state. The most important feature of this equation is that the van der Waal constants ‘a’ and ‘b’ which characterize a gas do not appear in it. Hence, it is a universal equation applicable to all gases. Since gases are simple compressible substances, only two properties can be independently chosen. All other properties are functions of these two independent properties. Now, this equation shows that if the independent properties are chosen as any two of pr, vr and Tr, then the third property11 has a fixed value irrespective of the gas. This state i.e. the state defined by two reduced properties, is called the corresponding state.12 Then, the reduced equation shows that the reduced (in the above sense) properties will be the same for all gases in their corresponding states. This is called the Principle of Corresponding States (PCS). For example, consider CO2 at 36.45 [bar], –121 [°C] (152 [K]) and N2 at 16.97 [bar], –108.5 [°C] (64.5 [K]). It is seen that for both gases pr = 0.5 = Tr. Thus, they are in corresponding states. For these values, the reduced equation gives vr = 0.4, i.e. rr = 2.5. Then, rCO2 = 1150 [kg/m3] and rN2 = 777.5 [kg/m3]. 11 12

And also all other properties appropriately reduced. The adjective corresponding is used here in the same sense as in corresponding isotherms.

Chapter 9:

Properties of Substances

219

The behaviour of most gases approximate that of van der Waal gas only near the critical point.13 Consequently, the Principle of Corresponding States (PCS) is also valid (at best) only in this range. However, as an empirical principle, it is often used to estimate the transport properties ([PER], p. 278) [i.e. viscosity (m), thermal conductivity (k) and diffusion coefficient, (D)], to addition to the estimation of thermodynamic properties (ibid., p. 268). Compressibility factor The compressibility factor is denoted by Z and defined as Z = pv/RT. The ideal gases equation shows that for it, Z = 1. Hence, Z can be used as a parameter to estimate the departure in the behaviour of any gas from that of an ideal gas. The more Z deviates from unity, the less ideal is the behaviour of the gas. The equation for the compressibility of a van der Waal gas is presented in the next section. However, its critical compressibility factor of a van der Waal gas becomes Zc = pcvc/RTc. Substituting for pc, vc and Tc, this equation becomes Zc = 3/8. Thus, the departure in the behaviour of a gas from the van der Waal gas can be estimated from the deviation of its Zc from this value. The values of Zc in Table 9.2 shows that, to a good degree of accuracy these common gases behave like the van der Waal gas. Virial equation To understand this concept, rewrite the van der Waal equation as follows. First solve for p as p=

RT a − 2 v−b v

Then, multiply both sides by v to get pv =

RTv a − v−b v

or ⎛ pv = ( RT ) ⎜ 1 − ⎝

b⎞ ⎟ v⎠

−1

−

a v

or pv ⎛ b⎞ = ⎜1 − ⎟ RT ⎝ v⎠

−1

⎛ a ⎞ ⎛ 1⎞ −⎜ ⎝ RT ⎟⎠ ⎜⎝ v ⎟⎠

or expanding the first term on the right-hand side

⎞ ⎛ a ⎞⎛ 1 ⎞ pv ⎛ b b2 = ⎜⎜ 1 − + 2 + " ⎟⎟ − ⎜ ⎟⎜ ⎟ RT ⎝ v v ⎠ ⎝ RT ⎠⎝ v ⎠ 13

It is interesting to plot the van der Waal equation using one of the computer plotter packages. For this, the reduced form is adequate since it does not contain any unknown constants. It will be seen that the general subcritical isotherms are meaningful only for Tr > 0.85.

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Engineering Thermodynamics

or

pv a ⎞ ⎛ 1 ⎞ b2 ⎛ = 1 + ⎜b − +" ⎟⎜ ⎟ + ⎝ RT RT ⎠ ⎝ v ⎠ v 2 Generalizing these arguments, the equation of state for a real gas may be written as the infinite series A A pv = 1 + 1 + 22 + " RT v v Sometimes this is also written as pv B B = 1 + 1 + 22 + " RT p p

i.e. as polynomials of pressure. The coefficients are polynomials in temperature. Such equations are called the virial equations of state. The coefficients A1, A2, … , B1, B2, … , are called the virial coefficients. They are related to the molecular properties. The coefficient A1, called the second virial coefficient, has been investigated (both theoretically and experimentally) exhaustively. Note that the left-hand side of the virial equations is the compressibility factor. Then, these equations can be symbolically written as Z = f(v, T)

and

Z = g(p, T)

where the functions f(∑∑) and g(∑∑) are polynomials with coefficients which are specific to each gas (as shown above). Similar to the reduced equation of state which is independent of any specific gas, the compressibility factor can also be made to be independent of a gas as follows. Now, the compressibility factor is defined as Z = pv/RT. By the definition of the reduced properties, this equation becomes Z=

pr vr pv pv × c c = r r × Zc RTc Tr Tr

where Zc is the compressibility factor at the critical point (or critical compressibility factor), i.e. Zc =

pc vc RTc

Transposing Z c and using the definition of the reduced compressibility factor as Zr = Z/Zc, the above relation is equivalent to Zr = prvr/Tr. Then, the virial equations become Zr = f(vr, Tr) and

Zr = g(pr, Tr)

These generalized compressibility factors are graphically presented as generalized compressibility charts, schematically represented in Figure 9.2.

Chapter 9:

Properties of Substances

221

Z

Increasing Tr Increasing Tr 1.0

pr Figure 9.2 Generalized compressibility charts.

Fugacity This is another form of representing the behaviour of real gases. The concept of equations of state (including the compressibility factor) measures the deviation from an ideal gas in terms of one (or more) of p, v or T. Fugacity measures the deviation from the ideal gas behaviour in terms of the chemical potential (the molar Gibbs function). It was shown in Chapter 8 that the chemical potential for an ideal gas can be written as

m = m0 + RT ln p where m0 is only a function of temperature. Then, the fugacity, denoted as f, is defined by the relation m = m0 + RT ln f The following features of fugacity are evident from the above definition, namely (a) it has the dimension of pressure,14 and (b) since all gases behave ideally as zero pressure is approached it tends to the pressure p, i.e. limpÆ0 (f/p) Æ 1. The differential equation for calculating the variation of fugacity with temperature is obtained as follows. It was shown in Chapter 8 that the chemical potential for pure substances is identical to the molar Gibbs function. Then, using the definition of the latter gives (∂m)T = (∂g)T = vdp. Substituting for (∂m)T from the above equation gives RT[∂ (ln f )]T = vdp 14

It is pseudo pressure that a real gas exerts in order to have the same chemical potential.

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Engineering Thermodynamics

Integrating the preceding equation gives the fugacity at all temperatures. Out of the several methods of solving this equation, the following are the more important ones. In the first method, define the quantity15 a by a = (RT/p) – v. Substituting for v from this equation gives ⎛ RT ⎞ − a ⎟ dp RT [∂ (ln f)]T = ⎜ ⎝ p ⎠

[∂ (ln f)]T =

or

dp a dp − p RT

Integration (of the second form) of the above equation gives 1 RT

ln f = ln p –

∫

p

0

a dp

since by definition f Æ p at the lower limit of p = 0. The integral on the right-hand side can be evaluated, perhaps numerically, from the p–v–T data. The second method consists in integrating the defining equation directly. This gives ⎛ f ⎞ 1 ln ⎜ ⎟ = ⎝ f0 ⎠ RT

∫

p p0

vdp

where f0 is the fugacity at the lower pressure of p0. When the equation of state is available in the form of v = v(T, p), the integral on the right-hand side can be evaluated (perhaps, numerically). Some equations (e.g. the van der Waal equation) can be solved more easily as p = p(T,v). In this case the integral is transformed first as follows. Integration by parts gives

∫

p p0

vdp = ( pv − p0 v0 ) −

∫

v

pdv

v0

which can be substituted in the above equation to get ⎛ f ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ln ⎜ ⎟ = ⎜ ( pv − p0 v0 ) − ⎜ ⎟ ⎟ ⎝ RT ⎠ ⎝ f0 ⎠ ⎝ RT ⎠

∫

v

pdv

v0

wherein the integral can now be evaluated for all values of volume. The third method uses the compressibility factor Z. Now, the deviation of volume a of the first method can be written as a = (RT/p) – v = (RT/p)[1 – (pv/RT)] = (RT/p)(1 – Z). Substituting this in the original equation and integrating, it gives ln f = ln p +

1 RT

∫

p p0

( Z − 1)(dp/p)

wherein the integral can be evaluated from the data of compressibility factor. It should be noted that all these methods are equivalent to one another since they use the p–v–T data in different forms. These equations give the values of fugacity at all pressures for a given temperature. 15

Note that this represents the deviation of volume of the real gas from that of the ideal one.

Chapter 9:

Properties of Substances

223

To obtain an equation expressing the variation of f with T at constant p, we proceed as follows. The defining equation is written as RT ln f = m – m0 . Since the chemical potential m is identical to the (molar) Gibbs function g, this equation can be written as ln f = (g – g0)/(RT). Then, [∂ ln f/∂T]p = (1/R) . [∂(g/T)/∂T]p = –h/T2, where the last step has been proved in Chapter 8. This equation can be integrated to obtain the values of f at all temperatures for any given pressure. Then, combining the integrals of the preceding two equations of f gives values of f at all points in the state space (with T and p as coordinates). Generalized fugacity charts (similar to generalized compressibility charts) with pr and Tr as coordinates (see, for example, [GLA], p. 257) are prepared like this. These concepts are mostly used in the analysis (and design) of processes in chemical engineering. In thermal engineering they are used only in the analysis of refrigerating systems with working substances (e.g. CO2) whose operating ranges of pressure and temperature are near pc and Tc. Since the main objective of this chapter is to explain the concepts, these charts along with many well-known and well-used equations such as the Beattie–Bridgeman equation, the Benedict–Webb–Rubin (BWR) equation, etc., have not been mentioned. Standard references (for example, see [PER], p. 3-267ff) should be consulted for this purpose. Moreover, for the ranges of pressure and temperature encountered in thermal power engineering, the working substances (air and common flue gases such as CO2, N2, … ,) may be assumed to be ideal ones. The above discussions are meant as an illustration of the method of approach and the results are not used later.

9.3.3

Examples

EXAMPLE 9.1 Steam can be modelled as an ideal gas. Determine the range of pressure where this model holds good. Solution Testing such as this is meaningful only when the criteria of the test are specified. For the purpose of this example, assume that the model is to be validated for the prediction of the specific volume v. Let vg and vs denote the specific volumes of steam as an ideal gas and as a vapour. The latter is obtained from steam tables. Then, the following are the steps. 1. Consider a very low pressure of p = 0.01 [bar] but a high temperature of 500 [°C]. 2. Then, from the superheat tables vs = 356.81 [m3/kg]. 3. Assuming steam to be an ideal gas (with molecular weight of 18 [kg/kmol], the ideal gas equation gives, vg = RT/p = (8.3143/18)(773/1) = 357.05 [m3/kg]. 4. Then, the error defined as, e = (vg – vs)/vg becomes 0.067%. 5. Similar calculations give the following results: (a) for saturated steam at 1 [bar], e = 1.63%; (b) for saturated steam at 100 [bar], e = 33.1%; and (c) for superheated steam at 100 [bar] and 500 [°C], e = 8.27%. 6. This shows that superheated steam approximates an ideal gas reasonably well even at high pressures, but steam at all states is well approximated by an ideal gas at low pressures. EXAMPLE 9.2 It was stated that when b > b and 1/v0 Æ 0. Applying these conditions, the above equation reduces to ln f =

pv a ⎛ v − b⎞ − 1 − ln ⎜ − ⎟ ⎝ RT ⎠ RTv RT

5. The van der Waal equation shows that pv v a b a −1= −1− = − RT v−b RTv v − b RTv 6. Substituting this in the equation in step 4 and simplifying gives

ln f =

⎛ RT ⎞ 2a b + ln ⎜ ⎟− v−b ⎝ v − b ⎠ RTv

This is the required equation for fugacity of a van der Waal gas.

9.4 LIQUIDS Working fluids in common thermal systems are gases and liquids. Section 9.3 dealt with the properties of gases. In this section, the properties of liquids are presented. The main difference between these fluids is that liquids undergo phase changes during the operation of thermal systems. Thus, the properties of vapours should also be computed in addition to those of liquids. Vapours behave like gases (ideal or real), therefore, the methods of Section 9.3 are also applicable to them. The relations among properties imposed by thermodynamics presented in Chapter 8 (and applied to gases in Section 9.3) are valid for liquids as well. It is more common to measure (and predict) liquid density as a function of pressure and temperature. Similarly, the specific heats are also measured as polynomials. These functions are quite complex. Hence, in actual applications, tables of properties are used. In this section, a brief description of the procedure of computing these tables (with the refrigerant R–12 as an illustration) is presented. Appendix F elaborates the steam tables16, too. These tables show that the following properties are to be computed. General. Saturation pressure psat for a given temperature (or, equivalently, saturation temperature Tsat for a given pressure). 16

It should be noted that since the mass of the working fluid varies from system to system, all the extensive properties are always calculated for unit mass, i.e. they are specific properties.

Chapter 9:

For saturated liquid. For saturated vapour.

227

Volume (vf), enthalpy (hf) and entropy (sf). Volume (vg), enthalpy (hg) and entropy (sg).

For superheated vapour. Basic data.

Properties of Substances

Volume (v), enthalpy (h) and entropy (s).

The following experimental data17 are available.

1. Equation for the saturated pressure in the form ps = ps(T). Note that this is the standard form of the vapour pressure equation [the integrated (the Kirchoff’s) form of the Clausius–Clapeyron equation]. 2. The p–v–T data of vapour is represented by the Martin–Hou equation in the form p=

∑

Si ( Ai + BiT i + Ci e DiT ) ( v − b) j

j

where some of the coefficients may be zero. Note that this equation has the general form p = p(T, v). 3. For vapour values of cv, is written as a polynomial in the form

cv (T ) =

∑A

j

⋅ T ( j −3)

j

4. The density of the saturated liquid is written as

r f (T ) =

∑ A (T − a) j

j

j/3

+

∑ B (T − a)

k /2

k

k

Note that, since the equation of state has T and v as the independent properties and the data of cv = cv(T) are available, it is convenient to compute u = u(T, v) and s = s(T, v) first. Then, the defining relation, h = u + pv is used to calculate enthalpy. Computational procedure. With the above data, the following basic equations as well as the computational procedure are suitable. For computational convenience, all the equations are made dimensionless with appropriate constants (including the critical constants). 1. A value of temperature is chosen. 2. Assuming that this temperature is Tsat, the pressure of the saturated vapour, psat, is calculated from the vapour pressure equation. 3. For this temperature, the specific volume of saturated liquid is calculated since vf = 1/rf. 4. For this temperature, the specific volume of the saturated vapour, vg, is calculated from the Martin–Hou equation of state by iteration. 17

McHarness, R.C., B.J. Eiseman (Jr) and J.J. Martin, Refrigeration Engg., Vol. 63, No. 9, (1955), pp. 32–44.

228

Engineering Thermodynamics

5. The enthalpy of vaporization (hfg) is calculated from the Clausius–Clapeyron equation, namely hfg dp = dT T ( vg − v f )

as hfg =

dp T (vg − vf ) dT

where the slope of the vapour pressure curve on the right-hand side is obtained from the vapour pressure equation given above. 6. The internal energy of vaporization ufg is calculated as ufg = hfg – psat (vg – vf) 7. The entropy of vaporization is obtained as sfg =

h fg Tsat

8. The internal energy of the saturated vapour, ug, is computed by integrating the equation ⎡ ⎛ ∂ p⎞ ⎤ du = cv dT + ⎢T ⎜ ⎟ − p ⎥ dv ⎝ ⎠ ⎣ ∂T v ⎦

9. The enthalpy of saturated vapour for all states is then obtained from the defining relation hg = ug + pgvg. 10. The enthalpy of saturated liquid is then calculated as hf = hg – hfg. 11. The entropy of saturated vapour, sg, is calculated similarly, by integrating the equation

⎛c ⎞ ⎛ ∂ p⎞ ds = ⎜ v ⎟ dT + ⎜ ⎟ dv ⎝ ∂Τ ⎠ v ⎝T⎠ 12. The entropy of saturated liquid is obtained as sf = sg – sfg 13. A new temperature is assumed and the steps 2–13 are repeated. Once the saturation tables are generated, the appropriate equations for the properties of vapour are solved for different values of p and T. This generates the superheat tables. It should be pointed out that in the refrigeration literature, the following procedure have been adopted.18 18

For example, see the reference at the footnote 17 as well as Chan, C.Y. and G.G. Haselden, Int. Journal of Refrigeration, Vol. 4, No. 1, 1981, pp. 7–12.

Chapter 9:

Properties of Substances

229

1. Writing h = h(T, v), gives dh = (∂h/∂T)v dT + (∂h/∂v)T dv. 2. Since h = u + pv; (∂h/∂T)v = (∂u/∂T)v + v (∂p/∂T)v = cv + v (∂p/∂T)v. Using the cv data and the equation of state, the right-hand side of this equation can be evaluated. 3. The relation dh = Tds + vdp gives, (∂h/∂v)T = T (∂s/∂v)T + v (∂p/∂v)T = T (∂p/∂T)v + v (∂p/∂v)T where the Maxwell relation (∂s/∂v)T = (∂p/∂T)v has been used once again. 4. Substituting for (∂h/∂T)v and (∂h/∂v)T from steps 2 and 3, respectively, in the original equation for dh in step 1 gives the differential equation for calculating h at all states. It will be noticed that this procedure involves considerable additional computations. Hence, this method is not used to generate the table of properties of R–12 presented in this book.

9.5 MIXTURE OF INERT IDEAL GASES Consider a system of volume V, at a temperature T containing a mixture of non-reacting (inert) ideal gases at a pressure p. Let n1 be the mole number of the component gas ‘1’, n2 the mole number of gas ‘2’, etc., that is let ni be the number of moles of gas i for i = 1, 2, … , N. Now, each component gas is inert (i.e. it does not react with others). Then, the principle of conservation of mass19 asserts that n T = n 1 + n 2 + … + nN =

N

∑n

i

i =1

where nT is the total number of moles in the system. Since the mixture is in thermodynamic equilibrium, assuming20 that it behaves like an ideal gas, its equation of state becomes

⎛ N ⎞ pV = nTRT = ⎜ ni ⎟ RT = (n1 + n2 + … + nN) RT ⎜ i =1 ⎟ ⎝ ⎠ which can be rewritten as

∑

N

p = (n1 + n2 + … + nN)(RT/V) =

∑ i =1

ni RT V

(9.1)

(9.2)

Now, the ratio (niRT/V) is the pressure exerted by the gas i if it behaves like an ideal gas occupying the whole volume V as if other gases were absent. Denoting this pressure as pi, Eq. (9.2) becomes N

p = p1 + p2 + … + p N =

∑p

i

i =1

Hence pi is called the partial pressure of the component gas i. 19 20

The principle of conservation of mass is really the principle of conservation of matter. This assumption is verified below

(9.3)

230

Engineering Thermodynamics

A mixture of ideal gases obeys the Dalton’s law (of partial pressures) which states that the total pressure exerted by a mixture of ideal gases equals the sum of the individual pressures which each gas would exert as if it alone were present. Equation (9.3) is the mathematical form of the Dalton’s law. Equations (9.1–9.3) show that the Dalton’s law implies the following: (a) each gas behaves like an ideal gas and (b) the mixture too behaves like an ideal gas. Now, the first equality of Eq. (9.1) shows that (RT/V) = p/nT. Substituting this in the first equality of Eq. (9.2) gives ⎛n ⎞ ⎛n ⎞ ⎛n ⎞ p = ⎜ 1 ⎟ p + ⎜ 2 ⎟ p +" +⎜ N ⎟ p = ⎝ nT ⎠ ⎝ nT ⎠ ⎝ nT ⎠

⎛ ni ⎞ ⎟p T ⎠ i =1 N

∑ ⎜⎝ n

which on simplification gives 1=

n n1 n2 + + " + N = x1 + x2 + " + x N nT nT nT

where the mole fractions are defined21 as xi = ni/nT with i = 1, 2, … , N. Since the component gases and the mixture behave like ideal gases, using their equations of state gives xi =

ni p V/RT pi = i = nT pV/RT p

or

pi = xi p

Currently this last relation is used as the definition of the partial pressure that the component gas i in a mixture of ideal gases exerts. Now, since each component gas behaves ideally, we have piV = niRT

or

(xi p)V = niRT

or p(xiV) = niRT

or

pVi = niRT

where, Vi, defined as xiV, is the volume occupied by the component gas i if it exerts the pressure p (i.e. the total pressure) as if the other gases are absent. Then, summing this relation gives N

∑ i =1

if V =

∑

i

pVi =

⎛ N ⎞ ni RT or p ⎜ Vi ⎟ = nTRT ⎜ i =1 ⎟ i =1 ⎝ ⎠ N

∑

∑

or pV = nTRT

Vi .

This is known as the Amagat’s law of partial volumes. It is evident that the Amagat’s law of partial volumes is equivalent to the Dalton’s law of partial pressures since both of them imply that the component gases and the mixture behave ideally. This is the first characteristic of an ideal gas mixture.

21

Some authors, particularly in chemical engineering thermodynamics, denote this by yi and reserve the symbol x for mole fractions in liquid and solid mixtures.

Chapter 9:

Properties of Substances

231

The second characteristic of an ideal gas mixture is that the partial pressure exerted by each gas on both sides of a membrane permeable to it, is the same. This is called the characteristics of membrane equilibrium. Since all the component gases are inert, they do not interact among themselves. Hence, the changes of internal energy, enthalpy and the entropy due to mixing are zero. Thus in each case, the internal energy, enthalpy and entropy of the mixture are the sum of those of the component gases. This set is the third characteristic of ideal gas mixture. This characteristic (particularly, on the entropy of mixing) was first shown by Gibbs and, therefore, is sometimes called the Gibbs theorem. It is now shown that all these characteristics of an ideal gas mixture are incorporated in the chemical potential of the ‘ith gas, defined as

mi = mi0 + RT ln pi = mi0 + RT ln xi + RT ln p where as before, mi0 (i = 1, 2, … , N), are only functions of temperature. Note that the second equality is obtained since the partial pressure of the ith component is defined as pi = xip. Now, partially differentiating with respect to the total pressure p, keeping all other variables constant gives ⎛ ∂ mi ⎞ RT = ⎜⎝ ∂ p ⎟⎠ p T , n ,n i

The property relation dG = Vdp – SdT +

j

∑ midni shows that i

⎛ ∂V ⎞ ⎛ ∂ mi ⎞ =⎜ = Vi ⎟ ⎜ ⎟ ⎝ ∂ p ⎠T ,ni ,n j ⎝ ∂ ni ⎠T ,ni ,n j where Vi is the partial molar volume of the gas i. The above two relations give RT = Vi p

or pVi = RT

or pVi = niRT

where the last step arises due to the fact that the gases are inert and, hence, Vi = Vi/ni. This shows that each component gas behaves ideally. Now, summing this and using the Amagat’s law of partial volumes gives N

∑ i =1

⎛ N ⎞ pVi = p ⎜ Vi ⎟ = pV = ⎜ i =1 ⎟ ⎝ ⎠

∑

N

∑ n RT = n RT i

T

i =1

which shows that the mixture also behaves like an ideal gas. These form the first characteristic of an ideal gas mixture.

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Engineering Thermodynamics

In Chapter 8, during the discussions on equilibrium, it was shown that for each species which is in membrane equilibrium (i.e. in equilibrium across a membrane permeable to it), the chemical potential should be the same, i.e. mAi = miB, i = 1, 2, … , where the superscripts A and B denote both sides of the membrane. Then, equating the relations for chemical potentials gives

mi0,A + RTA ln piA = mi0,B + RTA ln pBi Since the system is in thermal equilibrium, TA = TB = T (say). Since the same gas is on both A B sides of the membrane, mi0,A = m0,B i . Thus the above relation reduces to pi = pi , which is the second characteristic of the ideal gas mixtures. Next, consider the variation of mi with temperature T. Writing the equation for mi as

mi/T = mi0/T + R ln xi + R ln p and partially differentiating it with respect to T gives

⎡ ∂ ⎛ mi0 ⎡ ∂ ⎛ mi ⎞ ⎤ = ⎢ ⎜⎜ ⎢ ⎜ ⎟⎥ ⎣ ∂ T ⎝ T ⎠ ⎦ p,ni , n j ⎢⎣ ∂ T ⎝ T

⎞⎤ d ⎛ mi0 = ⎟⎟ ⎥ ⎜⎜ ⎠ ⎥⎦ p, ni , n j dT ⎝ T

⎞ ⎟⎟ ⎠

where the first equality is obtained from differentiation and the second one is due to mi0 being only function of T. Now, by definition, mi = Gi = Hi – TSi, and as was shown in Chapter 8,

⎡ ∂ ⎛ mi ⎞ ⎤ Hi d ⎛ mi0 ⎞ = − = ⎢ ∂ T ⎜⎝ T ⎟⎠ ⎥ T2 dT ⎜⎝ T ⎟⎠ ⎣ ⎦ p, ni , n j where the last equality is obtained by substituting in the above equation. The last equality shows that since mi0 is only a function of temperature, Hi is also a function of T only. Then, if the mole fraction xi is increased from zero till it reaches unity, Hi will not change. Hence, H i = hi, where hi is the molar enthalpy of the species i. Applying this argument to all the species, it is clear that the total enthalpy of the mixture is, n h . Using similar arguments, it can be shown that, U = i niui. This shows that the H= i i i internal energy and enthalpy of mixing are zero. This is the third characteristic of an ideal gas mixture. Taken together, all of the above arguments show that the relation, mi = mi0 + RT ln pi represents the behaviour of a mixture of ideal gases completely. The molecular weights Mi are defined by the relation, miMi = ni. The mass fractions are

S

S

S

defined as wi = mi/ i mi = mi/m. Then, the following relations, which define the specific internal energy, u, and specific enthalpy, h, of the mixture can be obtained directly. N

U = nT u =

N

∑ n u = mu = ∑ m u i i

i =1

i i

i =1

N

and

H = nT h =

N

∑ n h = mh = ∑ m h i i

i =1

i i

i =1

Chapter 9:

Properties of Substances

233

where to avoid multiplicity, the same symbols have been used to denote molar quantities as well.22 Their units will be clear from the context. Similarly, the definitions of specific heats of the mixture give N

cv =

∑x c

N

N

i v ,i

=

i =1

∑w c

cp =

and

i v ,i

∑

N

x i c p, i =

i =1

i =1

∑w c

i p, i

i =1

The equations for changes of the entropy and free energy (the Gibbs function) of the mixing can be obtained as follows. Since G is a property, its value for the mixture is N

G=

N

∑n m = ∑n m i

i

i =1

i

0 i

N

+ RT

i =1

∑n

i

ln pi

i =1

Before mixing, let the ith gas be at a pressure Pi, where i = 1, 2, …, N, but at the same temperature. Then, the Gibbs function before mixing is N

G=

∑

ni mi =

i =1

N

∑

ni mi0 + RT

i =1

N

∑n

i

ln pi

i =1

Then, ⎛p ⎞ ( DG) mixing = RT ln ⎜ i ⎟ ⎝ Pi ⎠ Since the mixing is at constant temperature and the enthalpy of mixing is zero, the defining relation becomes DG = DH – TDS = – TDS. Applying this, the entropy change of the mixture is ( DG) mixing

⎛p ⎞ = − R ln ⎜ i ⎟ T ⎝ Pi ⎠ If all the gases are at the same pressure, i.e. P1 = P2 = … = PN = P (say), (as in the case of constant volume mixing) then the above relation reduces to ( DS )mixing = −

N

( DG )mixing = RT

∑n

i

i =1

N

ln xi

and

( DS) mixing = − R

∑n

i

ln xi

i =1

since by definition, xi = pi/P. As an illustration, suppose air (assumed to be a mixture with xO2 = 0.23 and xN2 = 0.77) were to be made in the laboratory by mixing these gases, then (DS)mixing = – (8.3143) . [(0.23) ln (0.23) + (0.77).ln (0.77) ] = 4.44837 [kJ/kmol air] The Dalton’s law implies that, in a mixture of ideal gases, each gas exists independently of others, i.e. as if others are absent. For example, assume that the mixing of the above 22

Similar to the use of R to denote the specific as well as the universal gas constants.

234

Engineering Thermodynamics

illustration takes place as follows. This is a vessel divided into two compartments, one containing 0.21 [mol O2] and the other containing 0.79 [mol N2], both gases being at the same pressure and temperature. Suppose the partition is broken, O2 will diffuse into the compartment containing N2. However, since both are ideal gases, as far as the O2 is concerned, only vacuum, and not N2 exists (i.e. O2 does not see N2). Hence the diffusion is identical to free expansion. The same arguments can also be applied to diffusion of N2 in O2. This argument is generalized as follows: the mixing (and, hence, diffusion) of one ideal gas with another is equivalent to the first gas undergoing a free expansion. Then, the entropy of mixing is the sum of these entropy changes. For example, let a mixture be formed from N ideal gases, with the ith one being initially at Pi and Ti. After mixing, let them be at pi and T. Then, ( DS) mixing =

⎡

N

∑ (n ) × ⎢⎣⎢c i

i =1

p

⎛T ln ⎜ ⎝ Ti

⎞ ⎛ pi ⎟ − R ln ⎜ ⎠ ⎝ Pi

⎞⎤ ⎟⎥ ⎠ ⎦⎥

where the expression in the brackets on the right-hand side is the molar entropy change of the ith gas when it changes from state (Pi, Ti) to state (pi, T). Sometimes this approach is easier for computations. Then, (DG)mixing = T(DS)mixing The importance of the results derived so far lies in the fact that, in the range of operations of common thermal systems, all gases and most vapours behave like ideal gases. In the following chapters these results are used in the analysis of combustion and of the common thermodynamic cycles. In the rest of this chapter, another important and interesting application of these results is discussed.

9.5.1

Psychrometry: Air-Water Vapour Mixtures

It is well known that atmospheric air is a mixture of gases such as O2, N2, Ar, CO2, H2O, …, out of which, the H2O content varies widely with time of the day as well as with the season. Hence, it is considered separately. This air (with moisture) is called the moist air and the air without moisture is called the dry air. The steam at the room temperatures and pressures is generally called moisture or water vapour. It was shown at the beginning of this chapter that, at these conditions, both air as well as the water vapour (i.e. steam) behave like ideal gases. Moreover, at these conditions, the solubility of air in the water vapour is negligible, so that the mixture also behaves ideally. This section deals with such mixtures. The molar composition of dry air is: N2 = 78.08%, O2 = 20.95%, Ar = 0.93%, CO2 = 0.03%, Ne = 0.0018%, He = 0.0005%, Kr = 0.0001%, Xe = 0.00001%, … ; but, neglecting the trace elements and adjusting the N2 content, its composition is taken as: N2 = 78.09%, O2 = 20.95%, Ar = 0.93% and CO2 = 0.03%. Its molecular weight is assumed to be 28.966 [kg DA/kmol DA], where DA indicates dry air. The molecular weight of water vapour is taken as 18.016 [kg WV/kmol WV]. ASHRAE, ([ASH], Ch. 6) gives these as 28.9645 [kg DA/kmol DA] and 18.01528 [kg WV/kmol WV]. But for most of the cases discussed in this book, this accuracy is not needed.

Chapter 9:

Properties of Substances

235

The calculations of the properties of such mixtures form part of psychrometry.23 Traditionally, psychrometry was of interest to the refrigeration and air-conditioning engineers for controlling humidity for food preservation and human comfort.24 In the rest of this section, the properties of such mixtures will be discussed. However, the emphasis will be on showing how they can be treated as an ideal gas mixtures. The symbols used in psychrometrics vary widely. Since psychrometry is being discussed here as an application of thermodynamics, as far as possible, the symbols used so far will be followed. In case of new (and/or differing) symbols, the usage of ASHRAE is adopted. Wherever possible, the differences are pointed out. In general, the quantities pertaining to moist air are unsubscripted because it is a mixture. The subscript a refers to dry air. Quantities pertaining to water vapour are indicated by the subscript v [for vapour but ASHRAE uses w (for water)]. The subscript s will denote the saturated state of water vapour. Equations of psychrometry For understanding the concepts of psychrometry, the system in Figure 9.3(a) will be helpful. It is shown as a flow system to represent the process of humidification. The system is assumed to be very large (i.e. technically infinite). Moist air enters it with some humidity and during its flow over the water (for infinite time) it picks up additional moisture. The temperature of the moist air, at the inlet to the system, as indicated by an ordinary thermometer (e.g. mercury-in-glass thermometer), is called the dry-bulb temperature. It is denoted as T (since it is the mixture temperature, but many authors use the symbol TDB or DBT for it). The pressure of the moist air at the inlet (also called the total pressure) is denoted by p. The point 1 in Figure 9.3(b) shows this state (of moist air at the inlet to the system). WÄ p

ps (T )

T T

2

T

1

T* Inlet i

Moist air

Exit e

Water (a) Figure 9.3

23 24

4

Tdp

3

0

pv = ps (Tdp)

s (b)

(a) The system and (b) the processes of psychrometry.

Derived from Greek: psy chros = cold and metron = a measure. However, similar methods have also been used in other industries (e.g. drying of solids, textiles, paper, power, …), where the processes require humidity control.

236

Engineering Thermodynamics

Note that the water vapour (or steam) at this state is superheated. The partial pressure of dry air and water vapour are denoted by pa and pv, respectively. Now, since the mixture behaves like an ideal gas mixture, the Dalton’s law gives p = pa + p v

or

pa = p – p v

The quantity of water vapour in moist air is indicated by the humidity ratio25 and is denoted as W. It is defined as the ratio of the mass of water vapour in the given volume to that of dry air, at the given pressure and temperature. Since the mixtures behave ideally, the mathematical form of this definition becomes

W=

mv pvV /( R / M v )T ⎛ M v ⎞⎛ pv ⎞ ⎛ 18.016 ⎞ ⎛ pv ⎞ = =⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ ma paV /( R / M a )T ⎝ M a ⎠⎝ pa ⎠ ⎝ 28.966 ⎠ ⎝ pa ⎠ ⎛p ⎞ ⎛ pv ⎞ = 0.622 ⎜ v ⎟ = 0.622 ⎜ ⎟ ⎝ pa ⎠ ⎝ p − pv ⎠

where Ma and Mv represent the molecular weights of dry air and water vapour in [kg DA/kmol DA] and [kg WV/kmol WV], with DA and WV referring to dry air and water vapour, respectively. Following is an alternate (perhaps simpler) method of derivation: W=

mv nv M v ⎛ M v ⎞⎛ x v = =⎜ ⎟⎜ ma na M a ⎝ M a ⎠⎝ xa

⎞ ⎛ M v ⎞⎛ pv ⎟=⎜ ⎟⎜ ⎠ ⎝ M a ⎠⎝ pa

⎞ ⎛ pv ⎞ ⎟ = 0.622 ⎜ ⎟ ⎠ ⎝ pa ⎠

where the first equality is the definition, the second one is the definition of mole numbers, the third is obtained by dividing by the total mole number, and the fourth one is the definition of partial pressures of an ideal gas. Note that W has the unit of [kg WV/kg DA], and hence, is not dimensionless. Note also that the number 0.622 has the units of [(kg WV/kmol WV)/(kg DA/kmol DA)] and, hence, is not a pure (i.e. dimensionless) number. Steam tables show that near room temperatures, pv 0. Since,

Chapter 9:

9.

10.

11.

12.

13.

Properties of Substances

247

dh = Tds + vdp, (∂h/∂p)s = v. Then, we should show that (∂v/∂s)p > 0. By the rule of partial differentiation, (∂v/∂s)p = (∂v/∂T)p × (∂T/∂s)p = (bv) × (cp/T). Since all these quantities are positive, the result follows.] Show that constant pressure lines in the wet region of a vapour are straight but not parallel. [Hints: Same procedure as for Exercise 8.] Calculate the humidity ratio and dew-point for air at 1 [bar], 30 [°C] with a relative humidity of 60%. [Hints: Steam tables give, ps = 0.042415 [bar], so that pv = 0.025449 [bar]. Then, W = 0.622(0.025449/0.974551) = 0.016243 = 16.24 [g WV/kg DA]. Since, the dewpoint is the saturation temperature for this pv, by interpolation, the steam tables give Tdp = 21.8 [°C].] Atmospheric air at 1 [bar] and 300 [K] with 50% relative humidity is pumped into a furnace at 300 [°C]. Assuming that for dry air cp = 1.005 [kJ/kg.K], calculate the quantity of preheat per [kg] of moist air. [Hints: Same as for Exercise 10, but since the mass basis is the moist air, mass fractions should be determined. The standard equation with the steam tables data gives W = 0.011284 [kg WV/kg DA] so that wv = 0.011158. Substituting values from the steam tables (by interpolating for 0.018 [bar] and extrapolating to 27 [°C]) and evaluating the enthalpy change gives, (DH)v = 53.0 [kJ/kg A]. The standard formula gives, (DH)DA = 271.9 [kJ/kg A], and the linear mixing rule gives total enthalpy as 274.4 [kJ/kg A].] How much heat should be added to change the state of moist air from TDB = 30 [°C], f = 30% to TDB = 45 [°C], f = 12%? [Hints: Calculate enthalpy of moist air at both the states and take the difference. Ans. –176.5 [kJ/kg A]. The (–) sign is due to dehumidification being predominant. Verify this by calculating DH for dehumidification from f = 30% to f = 12% and then heating at f = 12% from 30 [°C] to 45 [°C] and then adding them up.] How much heat should be rejected to change the state of moist air from TDB = 30 [°C], f = 70% to TDB = 20 [°C], f = 50%? [Hints: Same as for Exercise 12. Ans. 47.64 [kJ/kg A].]

Chapter 10

Thermodynamics of Combustion

10.1 INTRODUCTION It was emphasized in Chapters 2 and 3 that work is the energy flow defined in mechanics and, therefore, it is primitive to thermodynamics. In Chapter 5 on the first law of thermodynamics, the other energy flow, namely heat, was operationally defined. Thus, heat is the real thermodynamic energy flow. It was also emphasized that thermodynamic systems exchange heat only with reservoirs, so that heat exchanged between two (or more) systems is modelled as one with the appropriate number of reservoirs interposed between them. In other words, so far as heat is available for absorption from some appropriate reservoirs, the method of its generation is irrelevant to thermodynamics. However, in practice, combustion is the oldest and most common method of generating heat, and therefore as pointed out in Chapter 1, this has contributed significantly to the development of thermodynamics. Moreover, heat engines, which also contributed fundamentally to the development of thermodynamics, utilize heat generated by combustion. Hence, combustion calculations are used in the analysis of thermodynamic cycles presented in subsequent chapters. These are the reasons why this chapter is placed here.

10.2

MAJOR ENERGY SOURCES AND CONVERTERS

Sun is the main source (directly or indirectly) of all energy on earth. It is a middle-aged medium-sized star producing approximately 3.86 ¥ 1020 [MW] of power. This power is radiated in all directions and only 1353 [W/m2] reaches the outside of earth’s atmosphere. This value appears to be constant and, therefore, is called the solar constant. Only half of this power reaches the earth’s surface, since 5% is reflected by air molecules, dust, …, of the atmosphere, 20% is absorbed by clouds, and 25% is reflected by clouds. About 30% of the radiation reaching the surface is by direct (beam) radiation and the rest is by diffuse (scattered) radiation. 248

Chapter 10: Thermodynamics of Combustion

249

All the energy sources found on earth are the indirect manifestations of solar energy. Wind and rain are due the heating effects of solar radiation. Similarly, tidal energies are due to sun (and moon and other planets) through gravitational effect. All the fossil1 fuels as well as those derived from biomass are indirect effect of solar heating (through plants, …,). However, for historical reasons, these are considered separately. Figure 10.1 is a schematic representation of the major energy sources and conversion techniques. The top line (of boxes) of Figure 10.1 lists the primary sources. The second line indicates the preliminary conversion processes (where applicable).The third line contains the intermediate form of energy. The fourth line lists the systems used for conversion. The last line is the form of energy used by consumers. Wind and tide

Chemical (fuels)

Geothermal

Combustion

Mech. energy (KE and PE)

Hydroelectric & wind power

Solar

Collection

Nuclear

Fission and fusion

Heat

Thermal power

Mechanical energy (shaft work)

Thermoelectric

Thermionic and photovoltaic

Direct energy converters

Electrical energy Figure 10.1

Major energy sources and conversion techniques.

The vertical dashed line separates the two types of converters. The methods to the left of this line generate electrical energy through shaft work (of engines, turbines, …). These devices are the conventional thermal power generators. On the other hand, the methods to the right of the dashed line generate electricity without the intermediate shaft work. Hence, these devices are called the direct energy converters. 1

From fossile (Fr.) and fossiles (L) = to dig.

250

Engineering Thermodynamics

This diagram is not exhaustive since it does not include all aspects of energy conversion. For example, it excludes an important category of direct energy conversion devices such as fuel cells. Moreover, uses such as space heating are not mentioned.

10.3

FUELS

Substances that produce heat by combustion are called fuels and combustion is the process (of chemical reactions) by which the energy stored in the chemical bonds of the fuels is released as heat. Only the fossil fuels are considered in this chapter because of the following reasons: 1. The main objective of this chapter is to show how thermodynamics is used to analyse the process of combustion. Now, a combustion process is basically a chemical reaction between fuel and oxidant. But, chemical reactions depend (among other things) on the composition of reactants. Hence, it is convenient to base the thermodynamic analysis of a combustion process on the specifications of the fuel composition. Fossil fuels are good examples of these specifications. 2. Moreover, the majority of the energy conversion techniques (in terms of quantity of the energy produced) make use of the energy released due to combustion of fossil fuels. Now, the composition of reactants (i.e. fuel and oxidant) may be specified either as mass fractions or as mole2 fractions. It should be noted that specification of the chemical formulae of the reactants is equivalent to specification of mole fractions (because, by definition, a mole equals the Avogadro number of molecules). Hence, it is convenient to classify the method of combustion calculations on this basis. This is done in this chapter. It should be noted that the mole and mass fractions are not pure fractions, since the numerator and the denominator of these refer to different species. For example, it is commonly assumed that air is an ideal gas mixture with molar composition of xO2 = 0.21 and xN2 = 0.79, and these are assumed to be pure numbers and are written without units. However, it will be clear from the definition of mole fractions that these figures should really be written as xO2 = 0.21 [kmol O2/kmol A] and xN2 = 0.79 [kmol N2/kmol A], where A denotes dry air. Since combustion calculations involve different species, such a procedure (of using additional symbols to distinguish the species) helps in proper bookkeeping of the units (and species). This practice is followed in this chapter (and the book). Fuels can be in the form of solid, liquid or gas. On the basis of convenience of experimentation, conventionally, compositions of solid and liquid fuels are specified as mass fractions. This is called gravimetric composition. There are two ways in which the compositions of coals are determined.3 In the proximate (abbreviation of approximate) analysis, only moisture, volatile matter, fixed carbon and ash are measured. The ultimate analysis gives the composition in terms of carbon, hydrogen, nitrogen, oxygen, sulphur, moisture and ash. 2 3

It will be recalled that mole is really the scaled mass (the scale factor being the molecular weight), but here it is convenient to treat mole fractions as if they are different. Different parts of IS: 1350 contain the procedures for proximate and ultimate analyses as per the Indian Standards.

Chapter 10: Thermodynamics of Combustion

251

It is easier to measure volumes occupied by gases (at specified pressure and temperature). Hence it is convenient obtain the composition of gaseous fuels as volume fractions. Evidently, this is called volumetric composition. Since the fuel gases are assumed to behave ideally, the ideal gas equation shows that volume fractions are identical to mole fractions.4 Thus, the compositions of gaseous fuels are specified in mole fractions. Since the main aim of this chapter is to illustrate the method of thermodynamic analysis of the combustion process, coal is chosen as the example of solid fuels, and only petroleum fuels are chosen for liquid and gaseous fuels. Before going into detailed discussions on these aspects, the origin of these fuels is briefly described first. When plants die, they decay and most of the carbohydrates are oxidized. This is the wellknown process of decay. Under some conditions (e.g. when plants are completely buried under the earth) the process of decay becomes incomplete due to lack of oxygen. This results in some part of the energy being retained by the vegetation. Later, geological actions further convert vegetation. During combustion this energy is released as heat. This process is believed to have produced coal, petroleum oils and petroleum gases.

10.3.1

Coal

In the first stage, bacterial action is believed to have converted vegetation into peat and lignite by bacterial actions as follows. (a) Vegetation growing in wet areas (e.g. swamps) dies and falls down. (b) In the presence of water, the anaerobic (without air i.e. oxygen) bacterial actions decompose the cellulose into an intermediate product, loosely termed humidified residue. (c) This is further converted to compounds such as lignin, resins, waxes and dead bacteria—the principal constituents of peat and lignite. The conversion of this peat and lignite into different stages of coal is by geological action, thought to be taking place in the following stages over millions of years. (a) The peat and lignite deposits is covered with layers of sand and silt and the biological processes ends. (b) As the depths of these layers increase, the pressure increases and the high pressure expels water and volatiles. (c) Hydrogen detaches from oxygen and combines with carbon to form hydrocarbons. Coal (including peat and lignite) is divided into the following categories. (a) Peat which occupies the lowest rank in the classification, is characterized by well-preserved remains of vegetation and exceptionally high water content (85–90%). (b) Lignite which is the next ranking coal is not very different from peat. (c) Bituminous5 coal has the next rank and is divided into three types, namely (i) sub-bituminous coal, (ii) bituminous coal, and (iii) semibituminous coal. (d) Semi-anthracite coal, of the next rank, is considered to be a high ranking coal. (e) Anthracite coal is highest ranking coal with about 90% carbon content. Both types of anthracite coals have a high carbon content. Hence, they are mainly used for metallurgical industries, being too precious to be burnt. 4 5

Of course, at the same temperature and pressure, such as NTP (or STP), laboratory temperature and pressure [like 1 [bar] and 300 [K]]. When this coal is distilled to produce coal gas and coke, the coal tar which remains looks like bitumen (the tar derived from petroleum). Hence this name.

252

Engineering Thermodynamics

Table 10.1 summarizes typical data on the above types of coals found abroad. The quality of the Indian coals is poorer than these. For example, the ash content of the Indian bituminous (and lower) coals is about 30%. However, this data is not relevant here because the objective of this chapter is to indicate the method of thermodynamic analysis of combustion processes. Table 10.1 Type Peat Lignite Sub-bitu Bitumin Semi-bitu Semi-anth Anthracite

Typical gravimetric composition (in %) of coals

Proximate analysis

Ultimate analysis

Cal. val.

W

VM

FC

Ash

C

H

O

N

S

[MJ/kg]

— 40 17 8 3 4 5

— 28 32 34 20 9 4

— 24 46 44 72 72 80

— 9 5 11 6 14 10

57 38 60 66 81 74 77

6.0 6 6 6 5 4 3

38 46 27 16 6 6 7

1.8 0.6 1.5 1.5 1.5 1 1

— 0.7 1 5 1 1 1

3 15 26 29 32 30 30

Note: W = Water, VM = Volatile Matter, FC = Fixed Carbon. The chemical symbols C, H, O, N and S have usual meanings. Remember that these numbers have units associated with them.

10.3.2

Liquid and Gaseous Fuels

Since this chapter aims at illustrating the method of thermodynamic analysis of combustion processes, only those liquid and gaseous fuels that are petroleum derivatives are described. Petroleum crude oil is thought to have been formed, over millions of years, as follows. (a) Dead plankton (microorganisms found floating in water) and other organic matter deposited on the sea bed. (b) Slow decomposition by anaerobic bacteria converts plankton into sapropel (an amorphous organic matter obtained by decay of plankton and similar organic matter). (c) The sapropel is converted to gas and oil under high temperature and pressure. In nature, petroleum occurs as oil, called crude oil. Basically, it is a mixture of hydrocarbons of various molecular weights. Depending upon the molecular weight, at atmospheric conditions, the component hydrocarbons occur as gases or liquids; the lightest components being gases and the heavier ones being liquids6 with gradual variation between these states. When crude oil is heated, these fractions distill off from it depending on the temperature. Evidently, the heavier (larger molecular weight) gases7 and lighter liquids evolve first. Thus, the properties of fuels derived from petroleum are related to each other. Hence they are called hydrocarbon fuels. Following are the primary hydrocarbon families into which the petroleum fuels can be classified.

6 7

The heaviest, bitumen (i.e. petroleum tar) is a solid. Remember that the boiling point is proportional to the molecular weight. The lighter gases do not form part of the crude. They occur as gases over the crude oil in the same field or as gases in neighbouring fields.

Chapter 10: Thermodynamics of Combustion

253

Alkanes (or Paraffin (or Methane series)). These are the saturated hydrocarbons with the general formula CnH2n+2. The most famous member of this family is methane. The molecular structure of these hydrocarbons is in general open-chain. Alkenes (or Olefins (or Ethylene series)). These are unsaturated hydrocarbons with the general formula CnH2n. This means that they possess double bonds. This is their characteristic feature. The most famous member of this group is ethylene. Naphthenes. These have the same formula as olefins, but with ring structure. Hence they are isomers of olefins and are also called cyclo-paraffins. However, the ring structure eliminates the double bond which characterizes the olefins. Cyclo-propane and cyclo-pentane are the most common members of this family. Alkynes (or Diolefins (or Acetylene series)). These are also unsaturated hydrocarbons, but with the general formula CnH2n–2. This formula implies that it contains triple bonds. Acetylene is the most famous member of this family. Aromatics (or Benzene series). These are characterized by the ring structure that is typical of benzene. Even though they possess double bonds, the bond resonance makes them stable. A detailed treatment of petroleum fuels is beyond the scope of this book. However, it should be noted that commercially available fuels are a mixture of several components and, therefore, it is difficult to assign unique formulae to them. For example, petrol, considered to be octane (molecular formula C8H18) is a mixture of compounds from C5H12 to C10H22. Hence, it is usual to specify the gravimetric composition for oil fuels and volumetric composition for gaseous fuels. Table 10.2 gives the compositions of some fuel oils (used in boilers) derived from Bombay-High crude. These are the heaviest distillates of crude oil and what remains after this is the bitumen. Table 10.2

Typical gravimetric composition (in %) of fuel oils Components

Oil type HHS LSHS Bunker ‘C’

C

H

N

S

O

Mineral matter

Cal. Val [MJ/kg]

88.25 84.07 86.14

10.04 11.05 10.77

0.33 0.16 0.09

3.98 0.13 3.00

0.34 3.72 0.0

0.06 0.07 0.0

43.35 45.24 43.30

Note: HHS = High Stock High Sulphur; LSHS = Low Stock High Sulphur. Remember that these numbers have units associated with them.

The lightest of all petroleum fuels is methane. It always occurs as gas. Other components like ethane, propane, butane, …, also occur as gases. The natural gas, available near (or, in) oil fields is a mixture of these. The volumetric compositions (in %) of two of these gases commonly available in India are as follows:

254

Engineering Thermodynamics

Common gas. Methane (CH4) = 79.8; Ethane (C2H 6) = 9.23; Propane (C3H8) = 4.81; Isobutane (C4H10) = 1.06; n-butane (C4H10) = 1.57; Isopentane (C5H12) = 0.46; n-pentane (C5H12) = 0.52; n-hexane (C6H14) = 0.04; H2O = 0.43; N2 = 0.79; and CO2 = 1.29. Methane rich gas. Methane (CH4) = 88.59; Ethane (C2H6) = 6.94; Propane (C3H8) = 1.82; Isobutane (C4H10) = 0.20; n-butane (C4H10) = 0.22; Isopentane (C5H12) = 0.03; n-pentane (C5H12) = 0.02; n-hexane (C6H14) = 0.00; H2O = 0.00; N2 = 0.93; and CO2 = 1.25.

10.4

ATMOSPHERIC AIR

In this chapter (and the book), fuels are assumed to burn in atmospheric air by reacting with the oxygen available in it. In Section 9.5 (on psychrometry), it was mentioned that atmospheric air is a mixture of different gases including water vapour. The air without water vapour was termed dry air and its molar composition was assumed to be, N2 = 78.09%, O2 = 20.95%, Ar = 0.93%, and CO2 = 0.03%. Its molecular weight was assumed to be 28.966 [kg DA/kmol DA], where DA indicates dry air. However, for the purposes of combustion (in this chapter and in the book), the air is assumed to contain only oxygen and nitrogen in the proportions of or

xO2 = 0.21 [kmol O2/kmol A]

and

xN2 = 0.79 [kmol N2/kmol A]

wO2 = 0.23 [kg O2/kg A]

and

wN2 = 0.77 [kg N2/kg A]

where, x’s and w’s denote the mole and mass fractions, respectively. Since many calculations involve the molecular weight of air directly, for the convenience of calculations, its molecular weight is retained as 29 [kg A/kmol A]. Moreover, since oxygen is involved in all the reactions, its molecular weight is also retained as 32 [kg O2/kmol O2]. Then, the molecular weight of nitrogen is adjusted as follows. Since air is a mixture of ideal gases, MA = xO2MO2 + xN2MN2 or MN2 = (MA – xO2MO2)/xN2 = 28.2 [kg N2/kmol N2] This value is ª 0.7 % larger than the normally assumed value of 28 [kg N2/kmol N2]. This correction becomes significant (in absolute values) when dealing with large quantities of air (e.g. in the case of large central power stations). Note that the air referred to, in the above discussions, is really dry air. However, even though water vapour content is calculated separately, unlike the psychrometric calculations, the compositions and molecular weights are not adjusted for moist air.

10.4.1

Interconversion of Mole and Mass Fractions

Almost all the calculations in thermal engineering are done on mass basis, since mass is always conserved—this provides a check on the calculations. However, since combustion is essentially

Chapter 10: Thermodynamics of Combustion

255

a chemical reaction involving molecules of fuel and oxygen, the natural unit for such calculations is the mole—a scaled (by the Avogadro number) unit of the number of molecules. This means that interconversions between these units are essential. Table 10.3 illustrates the systematic procedure for converting mole fractions into mass fractions. It is presented in a manner which is self-explanatory. It will be noticed that the total of the fourth column is really the molecular weight of air. Table 10.3 Species

Conversion of mole fractions to mass fractions

(1)

Mole fraction [kmol ‘i’/kmol A] (2)

Mol. weight [kg ‘i’/kmol ‘i’] (3)

Prop. mass [kg ‘i’/kmol A] (4) = (2) ¥ (3)

Mass fraction [kg ‘i’/kg A] (5) = (4)/S (4)

O2 N2

0.21 0.79

32 28

6.72 22.12

0.233 ª 0.23 0.767 ª 0.77

Total

28.84

Table 10.4 illustrates the systematic procedure for converting mass fractions into mole fractions, which is also self-explanatory. Table 10.4 Species

Conversion of mass fractions to mole fractions

(1)

Mass fraction [kg ‘i’/kg A] (2)

Mol. weight [kg ‘i’/kmol ‘i’] (3)

Prop. mole [kmol ‘i’/kg A] (4) = (2) ¥ (3)

Mole fraction [kmol ‘i’/kmol A] (5) = (4)/S(4)

O2 N2

0.23 0.77

32 28

0.007187 0.0275

0.207 ª 0.21 0.793 ª 0.79

Total

0.0346875

It will be evident that the procedures given in Tables 10.3 and 10.4 are logical reverse of each other. Since these procedures are general, they can be (and are) used in other cases as well. The tables show the errors due to approximation as well as those due to computational procedures. In this chapter (and the book), in order to avoid the latter errors, wherever possible, exact values (by definition or by calculations) are used. It should be noted that these methods of interconversions between mass and mole fractions is not new. They are used in chemistry. The following typical example illustrates this. EXAMPLE 10.1 Iron (called pig iron) is obtained by reducing iron ore which is a mixture of several iron oxides such as magnetite (Fe3O4), haematite (Fe2O3), siderite (FeCO3), pyrite (FeS2), … . Calculate the mass of iron obtained by reducing 1 [kg] of magnetite. Assume that the molecular weight of iron is 55.85 [kg Fe/kmol Fe]. Solution The reader my recall the myriad of problems of this type learnt in chemistry and will know that the answer is simply

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(3) (55.85) [kg Fe] = 0.7236 (3) (55.85) + (4) (16) [kg Mag] However, the objective here is to analyse the logic of this method. Let NA denote the Avogadro number. Now, the molecular formula of magnetite, namely Fe3O4 gives its molecular (really, formula) weight as (3 × 55.85) + (4 × 16) = 231.55 [g Mag/ mol Mag]. The molecular formula also shows that it contains

⎡ atom Fe ⎤ ⎛ ⎡ atom Fe ⎤ ⎞ ⎧⎪⎛ N A ⎞ ⎡ atom Mag/mol Mag ⎤ ⎪⎫ 3⎢ ⎟⎢ ⎥ = ⎜3⎢ ⎥ ⎟ × ⎨⎜ ⎥⎬ ⎣ molecule Mag ⎦ ⎝ ⎣ molecule Mag ⎦ ⎠ ⎩⎪⎝ N A ⎠ ⎣ atom Fe/mol Fe ⎦ ⎭⎪ ⎡ mol Fe ⎤ ⎛ ⎡ mol Fe ⎤ ⎞ ⎛ ⎡ g Fe ⎤ ⎞ ⎛ 1 ⎡ mol Mag ⎤ ⎞ = 3⎢ ⎥ = ⎜3⎢ ⎥ ⎟ × ⎜ 55.85 ⎢ ⎥⎟×⎜ ⎢ ⎥⎟ ⎣ mol Mag ⎦ ⎝ ⎣ mol Mag ⎦ ⎠ ⎝ ⎣ mol Fe ⎦ ⎠ ⎝ 231.55 ⎣ g Mag ⎦ ⎠

=

⎡ kg Fe ⎤ 167.55 ⎡ g Fe ⎤ = 0.7236 ⎢ ⎢ ⎥ ⎥ 231.55 ⎣ g Mag ⎦ ⎣ kg Mag ⎦

which is the same value as the earlier one. Thus, essentially, this is converting mole fractions to mass fractions. This should not be surprising since chemical formulae of compounds basically give the proportion of different constituent atoms. In other words, the chemical formula of each compound gives the proportion of moles of various atoms that constitute it. EXAMPLE 10.2 The gravimetric composition of a fuel is 85% of carbon and 15% of hydrogen. What will be its simplest formula? Solution This is again converting mass fractions to mole fractions, since as mentioned above, a chemical formula essentially represents mole fractions. Hence, the earlier tabular method is used. However, to save space, units are not written down. Species

Mol. wt.

Mass frac.

Prop. mole

Mole frac.

Ratio

C

12

0.85

0.07083

0.32074

1.0

H

1

0.15

0.15000

0.67926

2.1

The last column shows that this fuel contains two moles (or atoms) of hydrogen for every mole (or atom) of carbon. Thus its general formula is (CH2)n, i.e. CnH2n (an alkene), and the simplest formula is clearly, CH2. In thermal power plants, the product gases (also called flue gases) are routinely monitored to check the state of combustion. It is also possible to estimate the composition of fuel from that of its products. The following is an interesting example.

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EXAMPLE 10.3 Mycomycin (similar to penicillin or streptomycin) is found to contain only carbon, hydrogen and oxygen. For determining its (simplest) molecular formula, 0.1141 [g Myc] was burnt in pure oxygen. The product gases were first passed over calcium chloride which absorbed only the water vapour and then, over a mixture of sodium hydroxide and calcium oxide which absorbed the carbon dioxide. At the end of the experiment, it was found that the increase in weight of calcium chloride and sodium hydroxide-calcium oxide mixture, was 0.0519 [g] and 0.3295 [g], respectively. Find the simplest formula consistent with these results. Solution The procedure is similar to that of Example 10.1. The increase in the weight of calcium chloride is due to the absorption of water vapour. Hence, the mass of water vapour in the product gases is 0.0519 [g H2O/g Myc]. Similarly, the increase in weight of the mixture of sodium hydroxide and calcium oxide shows that the product gases contain 0.3295 [g CO2/g Myc]. Then, the calculations proceed as follows: 1. Moles of hydrogen is

⎧⎪⎛ 0.0519 ⎞ ⎡ g H 2 O/g Myc ⎤ ⎫⎪ ⎛ ⎡ mol H2 ⎤ ⎞ ⎡ mol H2 ⎤ ⎨⎜ ⎥ ⎬ × ⎜⎜ 2 ⎢ ⎥ ⎟⎟ = 0.00576 ⎢ ⎟⎢ ⎥ ⎪⎩⎝ 18.02 ⎠ ⎣ g H 2 O/mol H2 O ⎦ ⎪⎭ ⎝ ⎣ mol H 2 O ⎦ ⎠ ⎣ g Myc ⎦ 2. Then, the mass of hydrogen becomes

⎛ ⎡ g H2 ⎤ ⎞ ⎡ mol H2 ⎤ ⎞ ⎛ ⎡ g H2 ⎤ ⎜ 0.00576 ⎢ ⎟ × ⎜⎜ 1.088 ⎢ ⎥ ⎟⎟ = 0.0058 ⎢ ⎥ ⎥ ⎣ g Myc ⎦ ⎠ ⎝ ⎣ g Myc ⎦ ⎣ mol H2 ⎦ ⎠ ⎝ 3. Similarly, moles of carbon is

⎛ ⎡ g CO2 ⎤ ⎞ ⎛ 1 ⎡ mol CO2 ⎤ ⎞ ⎡ mol CO2 ⎤ ⎜ 0.3295 ⎢ ⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ = 0.00749 ⎢ ⎥ ⎥ ⎣ g Myc ⎦ ⎠ ⎝ 44 ⎣ g CO2 ⎦ ⎠ ⎣ g Myc ⎦ ⎝ 4. Then, since 1 [mol C] produces 1 [mol CO2], the mass of CO2 is

⎛ ⎡ mol C ⎤ ⎞ ⎛ ⎡ g C2 ⎤ ⎞ ⎡ gC ⎤ ⎜ 0.00749 ⎢ ⎟ × ⎜ 12.08 ⎢ ⎟ = 0.0905 ⎢ ⎥ ⎥ ⎥ ⎣ g Myc ⎦ ⎠ ⎝ ⎣ mol C ⎦ ⎠ ⎣ g Myc ⎦ ⎝ 5. Then, by mass balance, the mass of oxygen is ⎡ gO ⎤ 0.1141 – 0.0058 – 0.0905 = 0.0178 ⎢ ⎥ ⎣ g Myc ⎦ 6. Then, moles of oxygen becomes

⎛ ⎡ g O ⎤ ⎞ ⎛ 1 ⎡ mol O ⎤ ⎞ ⎡ mol O ⎤ ⎜ 0.0178 ⎢ ⎟×⎜ ⎢ ⎟ = 0.00111 ⎢ ⎥ ⎥ ⎥ ⎣ g Myc ⎦ ⎠ ⎝ 16 ⎣ g O ⎦ ⎠ ⎣ gO ⎦ ⎝ 7. Thus, the mole fractions of C : H : O = 0.00749 : 0.00576 : 0.00111. Dividing by the lowest number, this proportion becomes C : H : O = 6.7 : 5.2 : 1.

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8. Since the molecular formula should contain only integer number of atoms, this proportion further becomes C : H : O = 13 : 10 : 1. So, the simplest formula is C13H10O. Later in this chapter, this procedure is used to determine the fuel structure from the composition of the flue gases. However, this example, adapted from Pauling8 with suitable modifications, is intended to show that this procedure can be used for determining the molecular structure of any compound. This should not be surprising since fuels are also chemical compounds.

10.5 COMBUSTION PROCESS Basically, combustion is the chemical reaction between fuel and oxygen (of the atmospheric air), which depends on the following processes and conditions. 1. Particles of fuel and oxygen are brought together. 2. Sufficient energy is supplied to initiate the combustion reactions. 3. Sufficient time is provided to complete the combustion (i.e. the reactions) process (inside the furnace, combustor, cylinder, …). 4. Products of combustion are removed so that fresh fuel meets fresh oxygen. Thus, for efficient combustion, these processes should also be efficient. The first and the last steps involve mass transfer (including flow), while the second step is essentially one of heat transfer. The third step includes sizing of the furnace. These aspects are beyond the scope of this book. The rest of this chapter deals with the other aspect of the third step, namely the combustion that is taking place. The basic immutable feature of combustion is that it is exothermic. All other characteristics have exceptions. For example, generally, combustion is accompanied by flame, but there are flameless combustions as well. The process of combustion can be schematically represented as Fuel + Oxidant Æ Products + Heat where the symbol ‘+’ is interpreted as ‘and’ (in the sense of Boolean logic or algebra) and the symbol ‘Æ’ is read as ‘produces.’9 The main purpose of combustion is to release the chemical energy of the reactants (fuel and oxidant) as heat. However, to pump these reactants through the combustion chamber (or combustor or furnace), power must be supplied. Similarly, power must also be expended on pumping the combustion products. Since these powers depend upon the mass flow rates of reactants and products, they must be estimated. Moreover, the products of combustion carry away energy with them that is not available for further use. This energy depends on the thermodynamic properties of the products. However, since the product gas is an ideal gas mixture, by the linear mixing rule, its properties depend on the composition10 of the product gas. 8 9 10

Pauling, L., General Chemistry, Indian ed., Vakils, Feffer and Simons Pvt. Ltd., 1965, pp. 148–149. Or as ‘gives’ or as ‘goes to’ or as ‘becomes’ etc. Of course, they depend also on the properties of the component gases. But this is assumed to be known.

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Thus, combustion calculations aim at estimating the heat released, the masses of reactants and products as well as the composition of the products. The first aspect is known thermochemistry and the second one forms part of stoichiometry,11 reaction equilibrium or reaction kinetics. In this chapter, all these aspects, except the reaction kinetics, are dealt with. Most of this is a review of college chemistry and, therefore, only briefly discussed here. Reactions may be homogeneous (uniform throughout a phase) or heterogeneous12 (reactions on the surface of solids such as catalysts, coal, …). In this chapter, the reactants and products of the combustion are assumed to behave like ideal gases. This accounts for most of the industrial furnaces and combustors. Moreover, the aim of this chapter is show how the thermodynamics of an ideal gas mixture is applied to them. Hence, only the homogeneous reactions are dealt with. Consider the combustion of carbon monoxide in oxygen. This reaction can be represented as dnO2 dnCO2 1 dnCO O2 → CO2 which implies = 2 −1 −(1/ 2) 1 where the first relation is the schematic representation of the reaction. It is called the reaction stoichiometry or stoichiometry of reaction. It is only a figurative representation indicating the species involved in the reaction and their proportions. The second of the above relations is a quantitative one. It is called the law of combining weights.13 The ratios, i.e. dnCO/(– 1), dnO2/(– 1/2) and dnCO2/1, are called the extend of reactions and are denoted by dxCO, dxO2 and dxCO2, respectively. The rate of reaction [also called the (archaically) reaction velocity], denoted as R with appropriate subscript is defined as CO +

RCO =

dcCO ; dt

RO 2 =

dcO 2 dt

;

RCO2 =

dcCO2 dt

;

etc.

where the c’s denote concentration of the species. In this chapter, these are assumed to be molar concentrations. Note that these definitions include the sign of each term. For example, if CO is being consumed the dcCO/dt will be negative. Note also that the stoichiometry of the given reaction implies RO2 RCO2 RCO = = −1 (1/ 2) 1 For clarity of further discussions it is convenient to generalize these concepts as follows. A single-step reaction scheme is written as

dnA dnB dnC = = −a −b c where the capital (upper case) letters like A, B, C, …, are the labels of the various chemical aA + bB Æ cC

11 12 13

which implies

From Greek: stoicheion = an element, and metron = measure. From Greek: homos = same, heteros = other, and genos = kind. Or the law of reciprocal proportions or the law of equivalents.

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species and the lower case letters like a, b, c, …, represent the stoichiometric coefficients of the respective species. For example, for the above reaction, A = CO, a = 1, B = O2, b = (1/2), C = CO2 and c = 1. Then, the rates of reactions are defined as

R dc R R dc A dc RB = B ; RC = C ; which implies A = B = C ; − a −b c dt dt dt Experiments show that at constant temperature, the rate of reaction is proportional to the product of the active concentrations of the species, where the active concentration is defined as each concentration raised to some small power. This is called the law of mass action. Then, for example, RA =

RA μ (cA)a · (cB)b · (cC)g

or

RA = k (cA)a · (cB)b · (cC)g

where a, b and g are constants to be determined from the experimental data. The constant of proportionality, k, called the rate constant, is only a function of temperature. Experiments (and kinetic theories) also indicate that k = Ae–(E/RT);

(A = constant and T is in [K])

where E is called the activation energy and A the Arrhenius constant This equation, called the Arrhenius law, shows that as temperature increases, the rate constant k and, therefore, the reaction rate R increases exponentially. kb Experiments also show that, C can dissociate as per the scheme cC ⎯⎯ → aA + bB, whose rate of dissociation may be written (based on the above arguments) as RC = kb (cA)a1 · (cB)b1 · (cC)g 1 where to distinguish this rate constant, it is written as kb, the subscript ‘b’ implying the backward reaction. Then, the earlier rate constant for the forward reaction is written as kf, thus, kf cC. Therefore, the complete scheme may be written as the scheme becoming aA + bB ⎯⎯→ ⎯⎯ → cC; so that, RA = kf (cA)a · (cB)b · (cC)g – kb(cA)a1 · (cB)b1 · (cC)g 1 (10.1) aA + bB ←⎯ ⎯ where kf and kb are the forward and backward rate constants, respectively. The main disadvantage of using the complete equations of reaction kinetics is that each overall reaction consists of many individual steps. For each of these steps, all the constants (i.e. the exponents of the concentrations as well A and E of the rate constant) should be determined. Thus, the complete model requires many equations and associated constants. Moreover, as shown above, each of these steps gives a differential equation, which should be integrated over the required time period (generally, by numerical methods). Thus, the computations of dynamics of combustion reactions of even the simplest fuel becomes very involved. Hence, several types of assumptions are made to bring down the complexity, out of which, the following are the most important ones. For elementary reactions, the constants a, b, g, … equal the stoichiometric coefficients.14 For example, if the above forward reaction (i.e. aA + bB Æ cC) is elementary, then a = a, 14

The number of molecules taking part in the reaction (sometimes called molecularity).

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261

b = b and g = c. Even though most reactions are not elementary, they are assumed to be so. Then, the exponents of the concentrations need not be determined separately. This assumption is extended to the overall reactions as well. If the forward reaction is fast (ideally instantaneous) and the dissociation is negligible (ideally zero), then, the scheme is written as aA + bB = cC. Note that this is only a symbolic representation of the law of combining weights, because, the rate equation cannot be written since kf Æ •. This is the well-known form used in elementary stoichiometry. It is first reviewed here. Reaction equilibrium is defined to be the state at which the net production of any species is zero. Then, for RA = 0, Eq. (10.1) shows that Keq =

kf kb

= (cA)(a – a1) . (cB)(b – b1) . (cC)(g – g1)

where Keq, called the equilibrium constant, is defined as kf /kb. This shows that, if the value of Keq is available, then, the concentrations can be obtained by solving algebraic equations of the above type rather than the set of differential equations of kinetics. This approximation, corresponding to the condition of reaction equilibrium, is also discussed later in this chapter.

10.6

COMBUSTION STOICHIOMETRY

The stoichiometric model of combustion reactions is important for the following reasons: 1. This is familiar due to its extensive use in chemistry. 2. It involves simple uncoupled algebraic equations enabling each equation to be solved independently. 3. This approximation gives adequate accuracy for most cases. 4. Hence, it is useful at the design and many other stages when most data will have to be assumed. It should be noted that the stoichiometric scheme, commonly written as aA + bB = cC, is not an equation, since ‘+’ is to be read as ‘AND’ of the Boolean logic and ‘=’ as ‘produces’.15 Combustion involves many different chemical species. Hence, for the convenience of accounting, the species is also indicated along with its mass or mole unit. For example, the airfuel ratio, defined as the mass of air to that of fuel, is written as [kg A/kg F], where ‘A’ indicates air and ‘F’ indicates fuel. Similarly, ‘G’ (meaning gas) and ‘P’ (meaning products) are used to denote combustion products (exhaust or flue gases). In gravimetric (mass-based) calculations, the chemical symbols, C, H, O, N, and S are used, generically16 to denote the species. In addition, they are also used to indicate the atoms of these elements in gas phase. This should not create confusion, for they are used to different contexts. 15

16

Or “gives” or “goes to” or “becomes” etc. Strictly, it should be written as aA + bB Æ cC, but since all the books (on combustion, chemistry and chemical kinetics) use the equation form this chapter also follows the same practice. COD: generic = characteristic of a genus or class.

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Since combustion is the chemical reaction between the fuel and the oxidant, its natural representation is in terms of the molecules that take part in it. Other representations follow from this. Thus, for example, the reaction between one molecule of hydrogen and one molecule of oxygen is written as 2 [molecule H2] + 1 [molecule O2] = 2 [molecule H2O] or, for the Avogadro number, NA, of molecules, this becomes 2 NA [molecule H2] + NA [molecule O2] = 2 NA [molecule H2O] or, since a mole is defined as the Avogadro number of molecules, this reaction also becomes 2 [mol H2] + 1 [mol O2] = 2 [mol H2O] or, using the definition of molecular weight,17 this reaction reduces to (2 × 2) [kg H] + (1 × 32) [kg O] = (2 × 18) [kg H2O] or 1[kg H] + [kg O] = 9 [kg H2O] where the symbols, H and O, indicate hydrogen and oxygen species (and do not mean hydrogen and oxygen atoms). The above relations simply state that “2 [mol H2] uses 1 [mol O2] and produces 2 [mol H2O]”; or, equivalently, “1 [kg H] uses 8 [kg O] and produces 9 [kg H2O]”. Note that the total mass is constant, i.e. the mass of products equals that of reactants. In other words, the net change of mass (i.e. mass of products minus the mass of reactants) is zero. This follows from the principle of conservation of mass. On the other hand, the total number of moles is not constant, since the number of moles of the product is 2, while that of the reactants is 3. This implies that mole is not a conserved property. The above arguments can be applied to combustion of other species as well. Table 10.5 summarizes these well-known results. Table 10.5 Species Mol. Wt.

Scheme 1

Oxygen required

C

12

C+

C

12

C + O2 = CO2

H2

2

H2 +

S

32

S + O2 = SO2

2

O 2 = CO

1 2

Common combustion reactions

O2 = H 2 O

0.5 1

[kmol C]

[kmol O2 ] [kmol C]

0.5 1

[kmol O2 ]

or

[kmol O2 ] [kmol H2 ]

[kmol O2 ] [kmol S]

or

Products produced

4 [kg O] 3 [kg C]

8 [kg O] 3 [kg C] or 8

or 1

[kg O] [kg H]

[kg O] [kg S]

1

[kmol CO] 7 [kg CO] or [kmol C] 3 [kg C]

1

[kmol CO 2 ] 11 [kg CO2 ] or [kmol C] 3 [kg C]

1

[kmol H 2 O]

1

[kmol SO2 ]

[kmol H 2 ] [kmol S]

or 9 or 1

[kg H 2O] [kg H 2 ]

[kg SO2 ] [kg S]

Note: Symbols H and O denote the species and not their atoms. To avoid round-off errors, numbers are expressed as proper fractions or whole numbers. 17

For simplicity only the round numbers are used in this chapter.

Chapter 10: Thermodynamics of Combustion

10.6.1

263

Air Required for Combustion

In this chapter, the fuel is assumed to burn in air, i.e. the oxygen required for combustion is assumed to be drawn from air. If the fuel contains oxygen, then it is assumed to be used up first and only the remainder of oxygen is drawn from air. A fuel may be burnt with air, just adequate for the chemical reaction to take place according to the stoichiometric equation. Quantities corresponding to this condition are referred to with the adjective stoichiometric.18 Thus, for example, the quantities of oxygen, air, products, etc. are called stoichiometric oxygen, stoichiometric air, stoichiometric products, etc. The stoichiometric air-fuel ratio is defined as the air required for the complete combustion of one unit of the fuel (of course, under the stoichiometric conditions). It is denoted as Zs, the subscript ‘s’ indicates the stoichiometric conditions. Generally, for solid and liquid fuels, this unit is the unit mass. Then, for them, Zs = (ma/mf)s, where ma and mf denote the mass of air and fuel, respectively, and it has the unit of [kg A/kg F]. For gaseous fuels the unit is one mole so that Zs = na/nf, where na and nf are the number of moles19 of air and fuel, respectively. Then, their air-fuel ratio has the unit of [kmol A/kmol F]. It was mentioned earlier that the stoichiometric reactions are instantaneous. Since the reactions in real cases take finite time, excess air is always supplied so that the combustion is complete. The air-fuel ratio under these conditions20 are called (actual) air-fuel ratio and is denoted as Z. The excess air is also specified as excess air factor, defined as (Z – Zs)/Zs, and usually expressed as percentage. Sometimes, the excess air is also specified as dilution ratio, defined as Z/Zs. In combustion literature, the excess air is specified as equivalence ratio, denoted as f and defined as f = (mf /ma)/(mf /ma)s. Note that, f = Zs/Z. Depending upon the quantity of air supplied, the carbon of the fuel can be oxidized to CO or CO2. However, it is conventional to assume, unless stated otherwise, that it burns to produce only CO2. In practic too, due to the finite reaction times, CO will be produced even if excess air is supplied. Since air contains 23% oxygen by mass, the mass of air supplied is ma = (1/0.23) × mO, where mO is the oxygen needed. Similar arguments show that number of moles of air required equals na = (1/0.21) × nO2, where nO2 is the mole number of oxygen required. Nitrogen contained in air is assumed to be inert and, therefore, does not take part in the reactions. In other words, nitrogen of the air is simply a mass that flows through the combustion chamber taking away energy and absorbing pumping power. The quantity of nitrogen drawn from air is mN = (0.77/0.23) × mO, or, nN2 = (0.79/0.21) × nO2, where mN and nN2 denote mass and moles of nitrogen, respectively.21 18 19 20 21

And also with the adjectives theoretical, chemically correct, etc. Remember, m and n denote mass and mole, respectively. In general, these quantities are unsubscripted. Many books give these ratios as (1/0.23) = 4.35, (1/0.21) = 4.76, (77/23) = 3.35 and (79/21) = 3.76. These are not used in this book since they do not show how these numbers arise. Moreover, as explained in Chapter 1, with electronic calculators, accuracies of desired degree can always be achieved easily.

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Combustion is basically the chemical reaction between fuel and oxygen. Hence, the basis of all computational strategies of combustion stoichiometry is the reaction scheme. However, in practice, on the basis of how the chemical composition of the fuel is specified, it is convenient to identify two broad methods. These are now discussed. Given the molecular formula of the fuel In this method, the equation of reaction stoichiometry is first written down. From this, the oxygen (and air) required for combustion and the mass as well as the composition of the products are computed. The following example illustrates this procedure. EXAMPLE 10.4 Ethane (C2H6) is burnt in air at stoichiometric proportion. Calculate (a) the air-fuel ratio, (b) the mass of products and, (c) the composition of the product gases on mass basis. Solution The molecular formula of the fuel is given here. Hence, it is convenient to start from the reaction equation. It should be pointed out that since ethane is a gas, the calculations should really be done on mole basis. However, here, mass basis is used to emphasize the fact that even the massbased calculations start with molar quantities. 1. The reaction scheme is C2H6 (2 × 12 + 6 × 1)

+

⎛ 7⎞ ⎜⎝ ⎟⎠ O2 2 ⎛7 ⎞ ⎜⎝ × 32⎟⎠ 2

=

2CO2

+

(2 × 44)

3H2O (3 × 18)

where the bottom line shows the mass of species taking part in the reaction. This implies that, “30 [kg C2H6] uses (7/2) × 32 [kg O] to produce (2 × 44) [kg CO2] and (3 × 18) [kg H2O]”. 2. Then, the stoichiometric oxygen required is

⎛ 7 ⎡ kmol O2 ⎤ ⎞ ⎛ 32 [kg O2 /kmol O2 ] ⎞ ⎡ kg O ⎤ mO = ⎜ ⎢ ⎟×⎜ ⎟ = 3.7333 ⎢ ⎥ ⎥ ⎣ kg F ⎦ ⎝ 2 ⎣ kmol F ⎦ ⎠ ⎝ 30 [kg F/kmol F] ⎠ 3. Since the fuel does not contain any oxygen, all the oxygen required for combustion should come from air. Now, air contains 23% oxygen by mass, so that the mass of stoichiometric air is ⎛ 7 ⎡ kmol O 2 ⎤ ⎞ ⎛ 32 [kg O 2 /kmol O2 ] ⎞ ⎛ 1 ⎡ kg A ⎤ ⎞ ma = ⎜ ⎢ ⎟×⎜ ⎥⎟ × ⎜ ⎢ ⎥⎟ ⎝ 2 ⎣ kmol F ⎦ ⎠ ⎝ 30 [kg F/kmol F] ⎠ ⎝ 0.23 ⎣ kg F ⎦ ⎠ ⎡ kg A ⎤ = 16.232 ⎢ ⎥ = Zs ⎣ kg F ⎦

[Ans. (a)]

4. The composition of the products is calculated as follows. Since nothing is specified about oxidation of carbon, by default assumption, only CO2 is produced, whose mass is

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265

⎛ ⎡ kmol CO2 ⎤ ⎞ ⎛ 44 [kg CO2 /kmol CO2 ] ⎞ ⎡ kg CO2 ⎤ mCO2 = ⎜ 2 ⎢ ⎟×⎜ ⎟ = 2.9333 ⎢ ⎥ ⎥ [kg F/kmol F] ⎠ ⎣ kg F ⎦ ⎝ ⎣ kmol F ⎦ ⎠ ⎝ 30 5. Similarly, the mass of H2O in the products is

⎛ ⎡ kmol H 2 O ⎤ ⎞ ⎛ 18 [kg H2 O/kmol H2 O] ⎞ ⎡ kg H2 O ⎤ mH2 O = ⎜ 3 ⎢ ⎟×⎜ ⎟ = 1.8000 ⎢ ⎥ ⎥ [kg F/kmol F] ⎠ ⎣ kg F ⎦ ⎝ ⎣ kmol F ⎦ ⎠ ⎝ 30 6. Air contains 77% nitrogen by mass. Since nitrogen is inert, it appears in the products. Thus, the mass of nitrogen in the products is

⎛ 7 ⎡ kmol O2 ⎤ ⎞ ⎛ 32 [kg O2 / kmol O2 ] ⎞ ⎛ 0.77 ⎡ kg N ⎤ ⎞ ⎡ kg N ⎤ mN = ⎜ ⎢ ⎥ ⎟ × ⎜ 30 [kg F/kmol F] ⎟ × ⎜ 0.23 ⎢ kg O ⎥ ⎟ = 12.498 ⎢ kg F ⎥ 2 kmol F ⎦⎠ ⎝ ⎠ ⎝ ⎣ ⎦⎠ ⎣ ⎦ ⎝ ⎣ 7. Then, the total mass of products mP (or, mg) = 17.231 [kg P/kg F]

[Ans. (b)].

8. The gravimetric composition (mass fractions) of (wet) products is wCO2 = 0.1702

wH2O = 0.1044

and

wN2 = 0.7253

[Ans. (c)]

The above method can be generalized as follows for any fuel specified by its molecular formula. 1. Let the formula of the fuel be CnHm, which means that each molecule of the fuel has n atoms of C and m atoms of H. 2. Assuming that only CO2 is produced (the default assumption), the reaction scheme for burning of carbon shows that each atom of C needs one molecule of O2 and produces one molecule of CO2. 3. Then, since one molecule of the fuel contains n atoms of C, each molecule of fuel will need, for oxidation of C alone, n molecules of O2 and will produce n molecules of CO2. 4. This can be generalized for NA (the Avogadro number) molecules (i.e. one mole) as follows: For burning of C alone, each mole of fuel requires n moles of O2 and produces n moles of CO2. 5. Similar arguments show that for burning of H alone, each mole of fuel requires (m/4) moles of O2 and produces (m/2) moles of H2O. 6. Thus, for the combustion of one mole of the fuel, (n + m/4) moles of O2 are needed. 7. Combustion of one mole of the fuel also produces n moles of CO2 and (m/2) moles of H2O. 8. Since O2 is drawn from air, (n + m/4) × (0.79/0.21) moles of N2 per mole of fuel appear in the combustion products. 9. Then, the molar composition of the products can be easily calculated.

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These expressions are not used in this book since (a) the reactions and equations are so simple that these generalized formulae are more cumbersome to use, and (b) the main aim of this book is to present the logic underlying the calculations and not a set of formulae. Given the composition (gravimetric or volumetric) of the fuel This procedure is similar to the one given above. The oxygen (and air) required as well as the exhaust produced are computed for each component. These are added together to get the overall results. The following examples illustrate this method. EXAMPLE 10.5 A coal has the following percentage gravimetric composition: C = 51.3, H = 3.5, N = 1.8, O = 7.3, S = 0.7, inherent moisture = 8 and the rest is ash. If it is burnt with 10% excess air, calculate (a) the stoichiometric air-fuel ratio, (b) the actual air-fuel ratio, and (c) the actual gravimetric composition of products. Solution It is convenient to solve this example in the following steps. The reactions are the same as those for Example 10.4. Hence, the calculations are direct. Note that since this is gravimetric analysis, the symbols H, O and N indicate the species (of hydrogen, oxygen and nitrogen, respectively) and not their atoms. 1. Theoretical (stoichiometric) oxygen required for the combustion of carbon in the fuel is

⎛ ⎡ kg C ⎤ ⎞ ⎛ 32 ⎡ kg O ⎤ ⎞ ⎡ kg O ⎤ ⎜ 0.513 ⎢ ⎟×⎜ ⎢ ⎟ = 1.368 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 12 ⎣ kg C ⎦ ⎠ ⎣ kg F ⎦ ⎝ 2. Also, the stoichiometric oxygen needed for the combustion of hydrogen of the fuel is

⎛ ⎡ kg H ⎤ ⎞ ⎛ 16 ⎡ kg O ⎤ ⎞ ⎡ kg O ⎤ ⎜ 0.035 ⎢ ⎟×⎜ ⎢ ⎟ = 0.280 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 2 ⎣ kg H ⎦ ⎠ ⎣ kg F ⎦ ⎝ 3. Similarly, the stoichiometric oxygen needed for the combustion of sulphur in the fuel is

⎛ ⎡ kg S ⎤ ⎞ ⎛ 32 ⎜ 0.007 ⎢ ⎥⎟ × ⎜ ⎣ kg F ⎦ ⎠ ⎝ 32 ⎝

⎡ kg O ⎤ ⎞ ⎡ kg O ⎤ ⎢ kg S ⎥ ⎟ = 0.007 ⎢ kg F ⎥ ⎣ ⎦⎠ ⎣ ⎦

4. Hence, the total oxygen needed for the combustion of fuel is, mO,t = 1.655 [kg O/kg F]. 5. Oxygen in the fuel (assumed to be used up first), is mO, f = 0.073 [kg O/kg F]. 6. Therefore, oxygen to be drawn from air is, mO,a = 1.582 [kg O/kg F]. 7. Then, the stoichiometric (theoretical) air-fuel ratio is

⎛ ⎡ kg O ⎤ ⎞ ⎛ 1 ⎡ kg A ⎤ ⎞ ⎡ kg A ⎤ Z s = ⎜1.582 ⎢ ⎟×⎜ ⎟ = 6.8783 ⎢ ⎥ ⎢ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 0.23 ⎣ kg O ⎦ ⎠ ⎣ kg F ⎦ ⎝

[Ans. (a)]

Chapter 10: Thermodynamics of Combustion

8. Thus, the actual air-fuel ratio is, Z = 1.1 Zs = 7.5661 [kg A/kg F]. 9. The gravimetric composition of the products is calculated as follows: 10. The CO2 produced is

267

[Ans. (b)]

⎛ ⎡ kg C ⎤ ⎞ ⎛ 44 ⎡ kg CO2 ⎤ ⎞ ⎡ kg CO2 ⎤ mCO2 = ⎜ 0.513 ⎢ ⎥ ⎟ × ⎜ 12 ⎢ kg C ⎥ ⎟ = 1.881 ⎢ kg F ⎥ kg F ⎣ ⎦⎠ ⎝ ⎣ ⎦⎠ ⎣ ⎦ ⎝ 11. Since the fuel contains 8% inherent moisture, the total H2O in products is

⎛ ⎡ kg H ⎤ ⎞ ⎛ 18 ⎡ kg H2 O ⎤ ⎞ ⎡ kg H2 O ⎤ ⎡ kg H2 O ⎤ mH2 O = ⎜ 0.035 ⎢ = 0.395 ⎢ ⎟×⎜ ⎢ ⎟ + 0.08 ⎢ ⎥ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 2 ⎣ kg H ⎦ ⎠ ⎣ kg F ⎦ ⎣ kg F ⎦ ⎝ 12. The oxygen in the products comes from the excess air, which is mO2 = 0.1 × 1.582 = 0.1582 [kg O/kg F]. 13. The mass of SO2 in the products is

⎛ ⎡ kg S ⎤ ⎞ ⎛ 64 ⎡ kg SO2 ⎤ ⎞ ⎡ kg SO2 ⎤ mSO2 = ⎜ 0.007 ⎢ ⎟×⎜ ⎢ ⎟ = 0.014 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 32 ⎣ kg S ⎦ ⎠ ⎣ kg F ⎦ ⎝ 14. The fuel also contains 1.8% of nitrogen, therefore, total nitrogen in the products is

⎛ ⎡ kg A ⎤ ⎞ ⎛ ⎡ kg N ⎤ ⎞ ⎡ kg N ⎤ ⎡ kg N ⎤ mN2 = ⎜ 7.5661 ⎢ = 5.8439 ⎢ ⎟ × ⎜ 0.77 ⎢ ⎟ + 0.018 ⎢ ⎥ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg A ⎦ ⎠ ⎣ kg F ⎦ ⎣ kg F ⎦ ⎝ 15. Thus, the total mass of the products, mg = 8.2921 [kg P/kg F]. 16. Then, the gravimetric composition (mass fractions) of the products is wCO2 : wH2O : wO2 : wSO2 : wN2 :: 0.2268 : 0.0476 : 0.0191 : 0.0017 : 0.7048 As the following example illustrates, the above procedure can also be used for gaseous fuels. EXAMPLE 10.6 The composition (in per cent) of a natural gas is found to be: methane (CH2) = 79.8, ethane (C2H6) = 9.23, propane (C3H8) = 4.81, isobutane (C4H10) = 1.06, n-butane (C4H10) = 1.57, isopentane (C5H12) = 0.46, n-pentane (C5H12) = 0.52, n-hexane (C6H14) = 0.04, H2O = 0.43, N2 = 0.79, and CO2 = 1.29. It is burnt with 5% of excess air. Calculate (a) the stoichiometric air-fuel ratio, (b) the actual air-fuel ratio, and (c) the composition of the exhaust gas. Solution At the outset, note that, since this is a gaseous fuel, the composition is volumetric (molar) composition. Thus, it is convenient to do all the calculations on mole basis. Secondly, since ‘iso-’ and ‘n-’ of a compound are only isomers, their values can be added.

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1. Since the scheme for oxidation of methane is CH4 + 2O2 = 2CO2 + 2H2O, the oxygen required for the combustion of methane of the fuel. ⎡ kmol O2 ⎤ = (0.798) × (2) = 1.596 ⎢ ⎥ ⎣ kmol F ⎦

2. The oxygen required for the combustion of ethane of the fuel ⎡ kmol O2 ⎤ ⎛ 7⎞ = (0.0923) × ⎜ ⎟ = 0.32305 ⎢ ⎥ ⎝ 2⎠ ⎣ kmol F ⎦ 3. The oxygen required for the combustion of propane of the fuel

⎡ kmol O2 ⎤ = (0.0481) × (5) = 0.2405 ⎢ ⎥ ⎣ kmol F ⎦ 4. The oxygen required for the combustion of butane of the fuel ⎡ kmol O2 ⎤ ⎛ 13 ⎞ ⎥ = (0.0263) × ⎜ ⎟ = 0.17095 ⎢ ⎝ 2⎠ ⎣ kmol F ⎦ 5. The oxygen required for the combustion of pentane of the fuel

⎡ kmol O2 ⎤ = (0.0098) × (8) = 0.0784 ⎢ kmol F ⎥ ⎣ ⎦ 6. And the oxygen required for the combustion of hexane of the fuel ⎡ kmol O2 ⎤ ⎛ 19 ⎞ = (0.0004) × ⎜ ⎟ = 0.0038 ⎢ ⎥ ⎣ kmol F ⎦ ⎝ 2⎠ 7. Hence, adding all the above values, the stoichiometric oxygen required for the combustion of the fuel, nO = 2.4127 [kmol O2/kmol F]. 2 8. Since the fuel contains no oxygen, all of the above requirement should come from air. Hence, the air-fuel ratio is

⎡ kmol O2 ⎤ 1 ⎡ kmol A ⎤ ⎛ kmol A ⎞ Zs = 2.4127 ⎢ × ⎢ ⎥ = 11.489 ⎜ ⎥ 0.21 ⎣ kmol O2 ⎦ ⎝ kmol F ⎟⎠ ⎣ kmol F ⎦

[Ans. (a)]

9. Since the excess air is 5%, the actual air-fuel ratio is

⎡ kmol A ⎤ Z = 1.05 × Zs = 12.063 ⎢ ⎥ ⎣ kmol F ⎦

[Ans. (b)]

10. The composition of the exhaust gas is also computed as above. Since the fuel contains 1.29% of CO2, the total CO2 in products is nCO2 = (0.798 × 1) + (0.0923 × 2) + (0.0481 × 3) + (0.0263 × 4) + (0.0098 × 5) ⎡ kmol CO 2 ⎤ + (0.0004 × 6) + (0.0129) = 1.2964 ⎢ ⎥ ⎣ kmol F ⎦

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269

11. Similarly, since the fuel contains 0.43% H2O, the total H2O in products is nH2O = (0.798 × 2) + (0.0923 × 3) + (0.0481 × 4) + (0.0263 × 5) + (0.0098 × 6) + ⎡ kmol H 2 O ⎤ (0.0004 × 7) + (0.0043) = 2.2627 ⎢ ⎥ ⎣ kmol F ⎦ 12. The oxygen contained in the products because of the excess air supplied is

⎛ ⎡ kmol O2 ⎤ ⎡ kmol A ⎤ ⎞ ⎡ kmol O2 ⎤ nO2 = (0.05) × ⎜ 11.489 ⎢ = 0.12063 ⎢ ⎥ ⎥ ⎟ × (0.21) ⎢ ⎥ ⎣ kmol F ⎦ ⎣ kmol F ⎦ ⎠ ⎣ kmol A ⎦ ⎝ 13. Finally, since the fuel contains 0.79% of N2, the total N2 in products is

⎡ kmol N2 ⎤ ⎡ kmol A ⎤ nN2 = 12.063 ⎢ × 0.79 ⎢ ⎥+ ⎥ ⎣ kmol A ⎦ ⎣ kmol F ⎦ ⎡ kmol N2 ⎤ ⎡ kmol N 2 ⎤ 9.53767 ⎢ ⎥ = 9.5380 ⎢ kmol F ⎥ ⎣ ⎦ ⎣ kmol F ⎦ 14. Hence the total products, ng,t = 13.218 [kmol P/kmol F]. 15. And the molar composition (mole fractions) of the products is xCO : xH 2

2O

: xO : xN :: 0.0981 : 0.1712 : 0.0091 : 0.7216 2

2

The above examples illustrate the basic strategy of computing the air required and the quantity and composition (on mass or mole basis) of the exhaust gas. The important points to be noted in these calculations are: 1. Conventionally (i.e. unless specifically required), the calculations for solid and liquid fuels are done on mass basis. The molar basis is used for gaseous fuels. 2. Unless stated otherwise, carbon in the fuel is assumed to burn to produce CO2. 3. For calculating the oxygen required from air (and hence, the air-fuel ratio), the oxygen in the fuel should be subtracted from the total oxygen required for combustion. 4. To get the final composition of the products, the mass (or moles) of CO2, H2O, N2,… contained in the fuel (if any) should be added to those obtained from the combustion calculations. 5. Conventionally, the composition of the exhaust gas is reported on volumetric basis (i.e. as mole fractions, since it is a gas), and for reasons explained in the next section it does not include H2O and SO2 (and, therefore, is called dry gas).

10.6.2

Exhaust Gas Analysis

The composition of exhaust gas from an engine, a furnace, a boiler, …, indicates how efficient the combustion is. The presence of CO in the exhaust shows incomplete combustion and consequent loss of heat. On the other hand, the presence of a large quantity of O2 (with no CO) indicates that too much of air is being supplied, leading to wastage as pumping power as well as the energy of these gases.

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The Orsat’s apparatus is a simple instrument,22 which is popular in thermal power engineering for measuring the exhaust gas composition. It consists of a eudiometer (a graduated cylinder), the one end of which is connected to a glass tube for collecting the gas sample. The other end of the eudiometer is connected to an aspirator bottle that can be raised or lowered. The eudiometer and the aspirator bottle are partly filled with water. The apparatus also has three other bottles that can be connected, as and when needed, to the eudiometer through stop-cocks. The first of these contains caustic soda (NaOH) (or caustic potash (KOH)) which absorbs only CO2. The second one contains caustic pyrogallol (pyrogallic acid (C6H3(OH)3) in NaOH or KOH) for absorption of CO. The last bottle contains cuprous chloride in HCl (or ammonia), which absorbs O2. The experiment consists first in aspirating a known volume of the gas. Since the gas is collected over water, the steam in the gas condenses and only CO2, CO, O2 and N2 appear23 in the sample. For this reason, this gas is called the dry products (or dry exhaust gas or dry flue gas). This sample is first passed through the caustic soda (NaOH) so that CO2 is absorbed. It is then passed through the pyrogallol for absorption of O2. Finally, it is passed through the cuprous chloride in HCl which absorbs O2. At each stage, the gas is repeatedly bubbled through the solutions till there is no change in the volume. It should be noted that, in the experiments, the order of absorption of the components is important. For example, if the gas is first passed through pyrogallol, it will absorb CO2 in addition to O2. At each stage, the reduction in volume is proportional to the volume (or moles) of the gas absorbed. The volume of the remaining gas is that of N2. The following example illustrates this procedure. EXAMPLE 10.7 In an experiment 70 [cc] of dry exhaust gas was passed through caustic soda, pyrogallol and cuprous chloride. The readings after each stage were 67.5 [cc], 64 [cc] and 60 [cc] respectively. What is the composition of the dry exhaust gas? Solution

As shown below, the solution is straight arithmetic.

1. The original volume of the sample is 70 [cc]. After passing through caustic soda, the volume reduces to 67.5 [cc]. The reduction in volume in the first stage (= 2.5 [cc] is the volume of the CO2 in the sample. 2. Similarly, the volume after the second stage is 64 [cc], therefore, the reduction in the volume = 3.5 [cc] is the volume of O2. 22

23

In the laboratory, depending upon the required accuracy, the composition of the product gas can be measured by a number of methods, such as the absorption method (Example 10.3 above illustrates this), gas chromatography, mass spectrometry, …. In industrial furnaces (including boilers), concentrations of CO2 and O2 in exhaust gases are continuously monitored by appropriate instruments; the most common type uses value of thermal conductivity of the gas, which, for the same temperature, depends on the mole fractions of the species. Since SO2 is soluble in water, the water vapour in the exhaust gas absorbs it anyway.

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3. Finally, the volume after the third stage is 60 [cc], and, therefore, the volume of CO is 4 [cc]. 4. Then, the final volume at the end of the third stage (= 60 [cc]) is the volume of N2. 5. Since the gases are assumed ideal, the volume fractions are also the mole fractions. Hence, the composition of the dry exhaust gas is xO2 = 0.0500 xCO = 0.0571 and xN2 = 0.8571 xCO2 = 0.0357 Determination of the actual air-fuel ratio As mentioned earlier, the most frequent use of the exhaust gas analysis is to determine the actual air-fuel ratio. In the following paragraphs, the logic of this procedure is developed.24 It basically uses the principle of conservation of chemical species. Given the composition of the dry exhaust gas. This is the input data for the problem. It is available from measurement. Let the mole fractions of the species be xCO2, xCO, xO2 and xN2, respectively. Case I. No other information is available. Then, we can proceed as follows: 1. By the linear mixing rule, the molecular weight of the dry products is MDP = (xCO2 × MCO2) + (xCO × MCO) + (xO2 × MO2) + (xN2 × MN2) 2. The mass of carbon due to CO2 of the dry products

⎛ ⎡ kmol CO2 ⎤ ⎞ ⎛ ⎡ kg CO2 ⎤ ⎞ ⎛ 12 ⎡ kg C ⎤ ⎞ = ⎜ xCO2 ⎢ ⎥ ⎟⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ ⎥ ⎟ × ⎜⎜ 44 ⎢ ⎣ kmol DP ⎦ ⎠ ⎝ ⎣ kmol CO2 ⎦ ⎠ ⎝ 44 ⎣ kg CO2 ⎦ ⎠ ⎝ ⎛ ⎡ kg C ⎤ ⎞ ⎛ 1 ⎡ kmol DP ⎤ ⎞ 12 ⎡ kg C ⎤ xCO2 ⎢ = ⎜ 12xCO2 ⎢ ⎟×⎜ ⎟= ⎥ ⎢ ⎥ ⎥ ⎣ kmol DP ⎦ ⎠ ⎝ MDP ⎣ kg DP ⎦ ⎠ M DP ⎣ kg DP ⎦ ⎝ Alternatively, since 1 [molecule CO2] contains 1 [atom C], or, NA [molecule CO2] contains NA [atom C], i.e. 1 [mol CO2] contains 1 [mol C], the mass of C due to CO2 of the dry products

⎛ ⎡ kmol CO2 ⎤ ⎞ ⎛ ⎡ kmol C ⎤ ⎞ ⎛ ⎡ kg C ⎤ ⎞ = ⎜ xCO2 ⎢ ⎥ ⎟⎟ × ⎜ 12 ⎢ ⎥ ⎟ × ⎜⎜ 1 ⎢ ⎥⎟ ⎣ kmol DP ⎦ ⎠ ⎝ ⎣ kmol CO2 ⎦ ⎠ ⎝ ⎣ kmol C ⎦ ⎠ ⎝ ⎡ kg C ⎤ ⎡ kg C ⎤ 12 = 12 x CO2 ⎢ = x CO2 ⎢ ⎥ ⎥ ⎣ kmol DP ⎦ M DP ⎣ kg DP ⎦ 3. Similarly, the mass of carbon due to CO of the dry products is

12 xCO M DP 24

⎡ kg C ⎤ ⎢ kg DP ⎥ ⎣ ⎦

This is essentially the dimensional analysis used in fluid mechanics in another form.

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4. Hence, the total carbon content of the flue gas is

mC,DP =

⎡ kg C ⎤ 12 ( xCO2 + xCO ) ⎢ ⎥ M DP ⎣ kg DP ⎦

5. Then, by dividing the above expression by the previous one, the ratio of C in CO to total C in dry flue gas =

12 x CO xCO = 12( xCO2 + xCO ) ( xCO2 + x CO )

6. Using similar arguments, the mass of N in the dry flue gas =

28x N ⎡ kg N ⎤ 2 2 M DP ⎢⎣ kg DP ⎥⎦

7. Finally, the oxygen balance (the principle of conservation of oxygen atoms) gives the ratio of carbon atoms to hydrogen atoms (but not their actual values). In this case, the general expressions are long and, therefore, Example 10.9 illustrates the procedure. Case II. Carbon content of the fuel is known. Here, the carbon balance (the principle of conservation of carbon atoms) is used as follows: 1. Let the carbon content of the fuel be mC,F [kg C/kg F]. 2. Then, equating the carbon atoms in the fuel to those in the dry flue gas gives the mass of the dry products as mDP =

mC,F mC,DP

=

mC,F × M DP 12( x CO2

⎡ kg DP ⎤ + xCO ) ⎢⎣ kg F ⎥⎦

3. Also, the mass of the carbon burnt to CO =

mC,F × xCO ⎡ kg C ⎤ (xCO2 + xCO ) ⎢⎣ kg F ⎥⎦

Case III. Nitrogen content in the fuel is known. Here, as explained below, the nitrogen balance is used to obtain the air-fuel ratio. 1. From the expression derived in Case I, the nitrogen in the dry flue gas

⎛ 28x N2 ⎡ kg N2 ⎤ ⎞ ⎛ ⎡ kg DP ⎤ ⎞ 28x N2 × mDP ⎡ kg N2 ⎤ =⎜ ⎟ × ⎜ mDP ⎢ ⎢ ⎥ ⎥⎟ = ⎢ kg F ⎥ MDP ⎣ kg F ⎦ ⎠ ⎣ ⎦ ⎝ MDP ⎣ kg DP ⎦ ⎠ ⎝ 2. Let the nitrogen content of the fuel be mN2,F [kg N/kg F]. 3. Then, the nitrogen supplied from air

⎛ 28x N2 × mDP ⎞ ⎡ kg N 2 ⎤ =⎜ − mN2 ,F ⎟ ⎢ ⎥ M DP ⎝ ⎠ ⎣ kg F ⎦

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273

4. Since the nitrogen content of air is 0.77 [kg N/kg A], the mass of the air supplied is

⎛ 28 x N2 × mDP ⎞ ⎛ 1 ⎞ ⎡ kg A ⎤ ma = ⎜ − mN2 ,F ⎟ × ⎜ ⎟⎢ ⎥ =Z M DP ⎝ ⎠ ⎝ 0.77 ⎠ ⎣ kg F ⎦ where Z is the air-fuel ratio. 5. Most of the fuels contain very little or no nitrogen, i.e. mN,F = 0. For them, substituting the expression for mDP and simplifying gives Z=

x N2 × mC,F 0.33( xCO2 + xCO )

Following example illustrates the above procedure. EXAMPLE 10.8 When hexane (C6H14) is burnt in air, the gas analysis by the Orsat’s apparatus showed: CO2 = 12%, CO = 4% and N2 = 84%. What is the excess air factor? Solution The molecular formula for hexane is C6H14. Thus, it does not contain oxygen, nitrogen or sulphur. Moreover, even though the fuel is a gas, since only mass is conserved (and mole is not), the calculations should be done on mass basis. 1. The molecular weight of the dry exhaust gas is

⎡ kg DG ⎤ MDG = (0.12 × 44) + (0.04 × 28) + (0.84 × 28) = 29.92 ⎢ ⎥ ⎣ kmol DG ⎦ 2. By nitrogen balance, the actual air supplied

⎛ ⎡ kmol N2 ⎤ ⎞ ⎛ 1 ⎡ kmol A ⎤ ⎞ ⎛ 1 ⎡ kmol DG ⎤ ⎞ ⎛ ⎡ kg A ⎤ ⎞ = ⎜ 0.84 ⎢ ⎢ ⎥ ⎟⎟ × ⎜ ⎥ ⎟ × ⎜⎜ ⎢ ⎥ ⎟ × ⎜ 29 ⎢ ⎥⎟ ⎣ kmol DG ⎦ ⎠ ⎝ 0.79 ⎣ kmol N 2 ⎦ ⎠ ⎝ 29.92 ⎣ kg DG ⎦ ⎠ ⎝ ⎣ kmol A ⎦ ⎠ ⎝ ⎡ kg A ⎤ = 1.0306 ⎢ ⎥ ⎣ kg DG ⎦ 3. As explained above, the carbon in the dry flue gas

⎛ ⎡ kg C ⎤ ⎞ ⎛ 1 ⎡ kmol DG ⎤ ⎞ 1.92 ⎡ kg C ⎤ = ⎜12(0.12 + 0.04) ⎢ ⎥⎟ × ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ 29.92 ⎣ kg DG ⎦ ⎠ 29.92 ⎣ kg DG ⎦ ⎝ 4. The carbon in the fuel is

6 × 12 72 ⎡ kg C ⎤ = (6 × 12) + (14 × 1) 86 ⎢⎣ kg F ⎥⎦

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5. Thus, by carbon balance, the mass of the dry flue gas

⎛ 72 ⎡ kg C ⎤ ⎞ ⎛ 29.92 ⎡ kg DG ⎤ ⎞ ⎡ kg DG ⎤ =⎜ ⎢ ⎟×⎜ ⎟ = 13.046 ⎢ ⎥ ⎢ ⎥ ⎥ ⎣ kg F ⎦ ⎝ 86 ⎣ kg F ⎦ ⎠ ⎝ 1.92 ⎣ kg C ⎦ ⎠ 6. Hence, the actual air-fuel ratio is ⎛ ⎡ kg DG ⎤ ⎞ ⎛ ⎡ kg A ⎤ ⎞ ⎡ kg A ⎤ Z s = ⎜ 13.046 ⎢ ⎟ × ⎜ 1.0306 ⎢ ⎥ ⎥ ⎟ = 13.445 ⎢ kg F ⎥ kg F kg DG ⎣ ⎦⎠ ⎝ ⎣ ⎦⎠ ⎣ ⎦ ⎝ 7. Since the stoichiometric reaction is C6H14 + (19/2)O2 Æ 6CO2 + 7H2O, the stoichiometric air-fuel ratio is ⎛ ⎡ kg O2 ⎤ ⎞ ⎛ 1 ⎡ kg A ⎤ ⎞ ⎡ kg A ⎤ 9.5 × 32 Zs = ⎜ ⎢ kg F ⎥ ⎟ × ⎜⎜ 0.23 ⎢ kg O ⎥ ⎟⎟ = 15.369 ⎢ kg F ⎥ × + × (6 12) (14 1) ⎣ ⎦⎠ ⎝ ⎣ ⎦ ⎣ 2 ⎦⎠ ⎝ 8. Hence, the excess air factor is Z 13.445 = = 0.8748 Z s 15.369 which means that there is 12.52% deficiency of air. One of the interesting applications of the exhaust gas analysis is in finding the ratio of carbon atoms to hydrogen atoms in the fuel. Since the gas composition is in terms of mole fractions, this gives the minimum general formula. For this analysis, it is assumed that (a) the fuel is a hydrocarbon and (b) all the nitrogen in the flue gases comes from air. The basic principle has already been explained in Examples 10.2 and 10.3. It is now illustrated by the following example. EXAMPLE 10.9 The analysis of flue gas by the Orsat’s apparatus showed: CO2 = 13.2%, CO = 1.8%, O2 = 3.2% and N2 = 81.8%. Assuming that the fuel contains only the carbon and hydrogen atoms, calculate (a) the ratio of the carbon to hydrogen atoms (i.e. the simplest formula) and (b) the molar composition of the wet flue gases. Solution

The principle as explained above is used here too.

1. Since the fuel contains only hydrogen and carbon, all the nitrogen in the flue gas comes from air. Then, the air supplied for combustion

⎛ ⎡ kmol N2 ⎤ ⎞ ⎛ 1 ⎡ kmol A ⎤ ⎞ ⎡ kmol A ⎤ = ⎜ 0.818 ⎢ ⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ = 1.03544 ⎢ ⎥ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ 0.79 ⎣ kmol N2 ⎦ ⎠ ⎣ kmol DG ⎦ ⎝ 2. Thus, the oxygen in the air supplied

⎛ ⎡ kmol N2 ⎤ ⎞ ⎛ 0.21 ⎡ kmol O2 /kmol A ⎤ ⎞ ⎡ kmol O2 ⎤ = ⎜ 0.818 ⎢ ⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ = 0.217443 ⎢ ⎥ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ 0.79 ⎣ kmol N2 /kmol A ⎦ ⎠ ⎣ kmol DG ⎦ ⎝

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275

3. Since the exhaust gas contains 3.2% O2, the oxygen used-up in combustion

⎛ ⎡ kmol O2 ⎤ ⎞ ⎛ ⎡ kg O2 ⎤ ⎞ ⎡ kg O 2 ⎤ = ⎜ 0.185443 ⎢ ⎥ ⎟⎟ = 5.9342 ⎢ ⎥ ⎟ × ⎜⎜ 32 ⎢ ⎥ kmol DG kmol O ⎣ ⎦⎠ ⎝ ⎣ ⎣ kmol DG ⎦ 2 ⎦⎠ ⎝ Since only mass is conserved in a chemical reaction, to calculate oxygen in combined forms, such as in CO2, CO, H2O, etc., the mass basis should be used. 4. Thus, the oxygen in CO2

⎛ ⎡ kmol CO2 ⎤ ⎞ ⎛ ⎡ kmol O2 ⎤ ⎞ ⎛ ⎡ kg O2 ⎤ ⎞ ⎡ kg O2 ⎤ = ⎜ 0.132 ⎢ ⎥ ⎟⎟ × ⎜⎜ 32 ⎢ ⎥ ⎟⎟ = 4.224 ⎢ ⎥ ⎟ × ⎜⎜ 1 ⎢ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ ⎣ kmol CO 2 ⎦ ⎠ ⎝ ⎣ kmol O2 ⎦ ⎠ ⎣ kmol DG ⎦ ⎝ 5. Similarly, the oxygen in CO

⎛ ⎡ kmol CO ⎤ ⎞ ⎛ ⎡ kmol O2 ⎤ ⎞ ⎛ ⎡ kmol O2 ⎤ ⎞ ⎡ kg O2 ⎤ = ⎜ 0.018 ⎢ ⎟ × ⎜1 ⎢ ⎟ × ⎜⎜ 32 ⎢ ⎥ ⎟⎟ = 0.228 ⎢ ⎥ ⎥ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ ⎣ kmol CO ⎦ ⎠ ⎝ ⎣ kmol O2 ⎦ ⎠ ⎣ kmol DG ⎦ ⎝ 6. Hence, the total oxygen in CO2 and CO = 4.512 [kg O/kmol DG]. 7. Then, the oxygen used-up in hydrogen combustion is

⎡ kg O2 ⎤ (5.9342 − 4.512) = 1.4222 ⎢ ⎥ ⎣ kmol DG ⎦ 8. Thus, the mass of H2 in the fuel

⎛ ⎡ kg O2 ⎤ ⎞ ⎛ 1 ⎡ kg H 2 ⎤ ⎞ ⎡ kg H 2 ⎤ = ⎜1.1342 ⎢ ⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ = 0.177775 ⎢ ⎥ ⎥ ⎣ kmol DG ⎦ ⎠ ⎝ 8 ⎣ kg O2 ⎦ ⎠ ⎣ kg DG ⎦ ⎝ 9. Since the exhaust gas composition is on mole basis, the mole number of each component should be calculated. Then, H2O in products

⎛ 1.1342 ⎡ kg H 2 ⎤ ⎞ ⎛ ⎡ kg H 2 O ⎤ ⎞ ⎛ 1 ⎡ kmol H2 O ⎤ ⎞ ⎡ kmol H2 O ⎤ =⎜ ⎟ × ⎜⎜ 9 ⎢ ⎥ ⎟⎟ × ⎜⎜ ⎢ ⎥ ⎟⎟ = 0.088888 ⎢ ⎢ ⎥ ⎥ ⎣ kg DG ⎦ ⎝ 8 ⎣ kg DG ⎦ ⎠ ⎝ ⎣ kg H2 ⎦ ⎠ ⎝ 18 ⎣ kg H2 O ⎦ ⎠ 10. Similarly, using the expression derived earlier, the total C in CO and CO2

⎛ ⎡ kmol CO2 ⎤ ⎞ ⎛ ⎡ kmol C ⎤ ⎞ ⎛ ⎡ kg C ⎤ ⎞ = ⎜ 0.132 ⎢ ⎥ ⎟⎟ × ⎜12 ⎢ ⎥ ⎟ × ⎜⎜1 ⎢ ⎥⎟ ⎣ kmol DG ⎦ ⎠ ⎝ ⎣ kmol CO2 ⎦ ⎠ ⎝ ⎣ kmol C ⎦ ⎠ ⎝ ⎛ ⎡ kmol CO ⎤ ⎞ ⎛ ⎡ kmol C ⎤ ⎞ ⎛ ⎡ kg C ⎤ ⎞ + ⎜ 0.018 ⎢ ⎥ ⎟ × ⎜1⎢ ⎥ ⎟ × ⎜ 12 ⎢ ⎥⎟ ⎣ kmol DG ⎦ ⎠ ⎝ ⎣ kmol CO ⎦ ⎠ ⎝ ⎣ kmol C ⎦ ⎠ ⎝ ⎡ kg C ⎤ = 1.692 ⎢ ⎥ ⎣ kmol DG ⎦

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11. The molecular formula represents mole numbers. Hence, the C : H ratio

⎛ 1.8 ⎡ kg C ⎤ ⎞ ⎛ 1 ⎡ kmol C ⎤ ⎞ ⎛ ⎡ kg H 2 ⎤ ⎞ ⎡ kmol C ⎤ =⎜ × × 2 = 1.58 ⎢ ⎥ ⎜ 0.141775 ⎢ kg H ⎥ ⎟⎟ ⎜ 12 ⎢ kg C ⎥ ⎟ ⎜⎜ ⎢ kmol H ⎥ ⎟⎟ ⎣ ⎦⎠ ⎝ ⎣ 2 ⎦⎠ ⎝ 2 ⎦⎠ ⎣ ⎣ kmol H2 ⎦ ⎝ 12. Hence, the simplest formula will be (C2H)n. 13. The total moles of wet products = 1.089 [kmol G/kmol DP]. 14. Therefore, the composition of the wet products is xCO2 = 0.1212

xCO = 0.0165

xO2 = 0.0294

xN2 = 0.7511 and xH2O = 0.0816

10.7 REACTION EQUILIBRIUM At the beginning of this chapter, it was mentioned that combustion (and also general chemical) reactions proceed forward as well as backwards. In other words, the reaction aA + bB = cC involves the forward reaction aA + bB Æ cC and the backward reaction cC Æ aA + bB. It was also mentioned that since the calculations of the complete reaction scheme are quite involved, simplifying assumptions are made. The simplest of these is the assumption of stoichiometric reactions, presented in Section 10.6, which assumes that the forward reactions are instantaneous and the reverse reactions are absent. It was seen that these give simple algebraic decoupled equations, each of which can be solved independently. A better assumption is that reaction equilibrium prevails, which implies that the forward rate (of consumption of reactants) equals the backward rate (of dissociation of the products). It was shown that for this condition, the ratio of the multiplication of the concentrations of the products to that of the reactants is a constant, in the sense that it is independent of time. This constant was called the equilibrium constant and was denoted as Keq. In this section, this aspect is further elaborated.

10.7.1

Equilibrium Constant

In this section, the conditions for reaction equilibrium are considered. In Chapter 8 on auxiliary functions, it was mentioned that reactions are assumed to take place under conditions of constant pressure and temperature. This gives N

( dG) p ,T =

∑ m dn i

i =1

i

=0

(10.2)

The system used for this purpose is the van’t Hoff box, shown schematically in Figure 10.2.

Chapter 10: Thermodynamics of Combustion

277

aA + bB = cC A C B Q

Semipermeable membranes Figure 10.2

van’t Hoff box.

Figure 10.2 indicates equilibrium of the reaction aA + bB = cC. The following points, mentioned in Section 10.5, should be noted about this equilibrium. ∑ This is a dynamic equilibrium between the forward reaction, aA + bB Æ cC, and the backward (dissociation) reaction cC Æ aA + bB. ∑ The reaction aA + bB Æ cC is assumed to take place in the following series of infinitesimal simultaneous steps. 1. The reactants A and B are introduced into the system. 2. They are converted into product C at the same temperature and pressure. 3. The reactants and products are exchanged between the system and their reservoirs (appropriately labelled) through semipermeable membranes in such quantities so as to keep the concentration at the required value. 4. The pistons in the species reservoirs maintain the proper pressure. 5. Appropriate heat, Q, is supplied (endothermic reaction) or extracted (exothermic reaction) from a suitable reservoir. ∑ The reverse reaction is also assumed to take place in such steps. To generalize this scheme further, it is first written as 0 = cC – aA – bB, because by convention, the stoichiometric coefficients of products are positive and those of the reactants are negative. Then, the stoichiometric coefficients are written as ni, (i = 1, …), and the species are denoted as Mi, (i = 1, …). In terms of these symbols, the general step involving N species is written as N

0=

∑n M i

i

i =1

Thus, in the new notation, the above reaction will have: N = 3, with n1 = c, M1 = C, n2 = – a, M2 = A, n3 = – b, and M3 = B.

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As explained in Section 10.5, for this scheme, the law of combining weights (or reciprocal proportions) implies the relation, dn1

n1

="=

dni

ni

=" =

dnN

dni =

or

nN

ni dn1 n1

Substituting this in Eq. (10.2) and simplifying gives N

∑mn

i i

=0

(10.3)

i =1

This equation is valid for all reactions (including combustion of solid, liquid or gaseous fuels), since the only assumption made so far is that the reaction takes place in a single step, represented by 0 = SiViMi. Most of the combustion reactions take place in gaseous phase. Even liquids fuels first vaporize before reacting. Moreover, these fuel gases and vapours as well as the product gases behave like ideal gases. Consequently, the reactants and products are considered to be ideal gas mixtures. In Chapter 9 on properties of substances, it was shown that for a mixture of ideal gases

mi = mi0 + RT ln pi

(10.4)

where pi is the partial pressure of the ith component and mi0 is a function of only temperatures. Substituting Eq. (10.4) in Eq. (10.3) gives N

∑ n (m i

0 i

+ RT ln pi ) = 0

i =1

which can be rewritten as follows: RT

∑n

i

ln pi = −

∑n m

1 RT

∑n m

i

0 i

or

∑n

or

ln

i

ln pi = −

1 RT

∑n m i

0 i

or

∑ ln ( p ) i

vi

=−

i

0 i

∏ (p )n i

i

=−

1 RT

∑n m i

0 i

where ’ is the product function defined as ’zi = z1 × z2 × … × zN. The last form of this equation can be written as ln Kp = –

1 RT

∑n m

where Kp is the equilibrium constant, defined as Kp = ’(pi)vi

i

0 i

(10.5)

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279

Thus, the condition of equilibrium implies the existence of equilibrium constant, Kp, with the following properties. ∑ Since the right-hand side of Eq. (10.5) is only a function of temperature, Kp is also only a function of temperature. ∑ The defining equation of Kp shows that, because of the sign convention implied in the reaction scheme (i.e. coefficients of products are positive and those of reactants negative), its numerator is the product of the partial pressures of the products of reaction, each raised to the power of its stoichiometric coefficient and its denominator is a similar expression, but in reactants. For calculations of compositions under reaction equilibrium, it is more convenient to convert the definition of Kp in terms of the mole fractions, as shown below. Since the reactants and products are ideal gas mixtures, the definition of partial pressure gives, pi = xi × p. Substituting this in the definition of Kp shows that ⎡∏( xi )ni ⎤ ⎣ ⎦ prod Kp = = ’(xi = × [(p) ] = C × ⎡∏( xi )ni ⎤ ⎣ ⎦ reactants Sni where C defined as C = (p) is a constant, since the total pressure p is constant. This implies that at reactions equilibrium, the concentrations are only a function of the temperature and not how they are reached. In other words, they are algebraic and not differential equations. This is the biggest advantage of using Kp. However, it should be remembered that the following assumptions are made in the above arguments. ∑ Since chemical reactions take place at constant pressure and temperature, (dG)p,T = 0.

’(pi)ni

∑ But, since (dG)p,T =

p)ni

∑

N i =1

[’(xi)ni]

Sni

midni = 0. This implies that

∑ For a single step reaction of the form 0 = ∑ For ideal gas reactions, ln Kp = –

1 RT

∑

N i =1

∑

N i =1

midni = 0.

niMi, this reduces to

∑

N i =1

mini = 0.

∑ ni mi0.

Evaluation of equilibrium constant In the following paragraphs, the procedure for evaluating Kp for all conditions is described. This is based on the following ideas developed above. N 1. By definition, the equilibrium constant is Kp = ∏i =1 (pi)vi, where the stoichiometric coefficients, ni, are positive for products and negative for reactants, and P is the product function. N n m 0, where mi0 are 2. The condition of reaction equilibrium implies, RT ln Kp = – i =1 i i functions of only temperature. 3. The chemical potential of an ideal gas may be written as m = m0 + RT ln p, and by its definition, m = g, the molar Gibbs function. 4. When the pressure is unity (in some units), m = m0. This unit of pressure is chosen as 1 [atm].

∑

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280

5. Then, for a mixture of ideal gases, as shown in Chapter 9, ni mi0 = nimi = nigi = (DG)p = 1[atm]. 6. Then, ln Kp = – (DG)p = 1[atm]/RT. 7. The reference condition at which the values of G are measured is 1 [atm] and 25 [°C] (i.e. 298.15 [K]). For reasons explained below (in Section 10.8.1 on heats of formation), these values are called the Gibbs free energy of formation. They are denoted as (DG)f,i0, where the subscript f implies formation. The superscript 0 (zero) indicates the reference condition. For all temperatures the pressure is always assumed (the default assumption) to be 1 [atm]. 8. Then, this will also be the reference state for the Kp values. Table 10.6 gives the values of Gibbs free energy of some common gases (assumed ideal) that form combustion products. These species are indicated in the first column. The last column of the table gives (DG)f0, the Gibbs free energy of formation. This is the change in the Gibbs function when a species is formed from its elements at the standard state (of 1 [atm] and 298.15 [K]); for example, CO is formed from C(graphite) and O2(g). By convention, the Gibbs free energy of formation of an element in its standard state at the reference condition is taken as zero. Thus, for example, at the reference condition of 1 [atm] and 298.25 [K], hydrogen occurs as H2 in gaseous state. Hence, (DG)f,H2(g) = 0. This table shows these. Columns 2–5 of Table 10.6 give the coefficients of a polynomial, in temperature, of the variation of the Gibbs free energy of formation as the deviation e(DG)f, defined as e(DG)f = (DG)fT – (DG)f0 The standard data part of the table are presented by all handbooks (like Perry’s, CRC, etc.). However, the data of JANAF tables25 is widely accepted in combustion engineering. Table 10.6 and Table 10.7 (given later) are generated from these data. Since Kp is only a function of temperature, in order to obtain its values at all states, it is necessary to derive an expression for its variation with T. This is done as follows. Table 10.6

e(DG) = (DG)fT – (DG)f0

Species

(DG)f0

f

a CO2 H2O CO SO2 OH C(g) H(g) O(g)

Gibbs function (free energy) of some gases

–393.352816 –242.446520 –109.354041 –284.403926 39.712793 716.478817 218.087902 249.697978

b –4.008157 43.487662 –93.765557 –77.943903 –16.978250 –155.809309 –47.723722 –58.464620

c 1.644179 7.627609 3.092295 92.069943 1.325032 –2.200027 –5.663545 –4.131481

d –0.204684 –1.317555 0.186471 –17.860617 –0.170718 0.626264 0.852871 0.673863

–394.4 –228.6 –137.2 –300.2 34.76 669.6 203.3 231.8

Notes: 1. e(DG)f = a + bq + cq2 + dq3, where q = T/1000, and T is in [K]. 2. (DG)f0, (DG)fT and e(DG)f are in [kJ/mol] (i.e. [MJ/kmol]). 3. (DG)f0 = 0 for H2(g), O2(g), N2(g) and C(graphite) since these are their standard states at the reference condition of 1 [atm] and 298.15 [K]. So they are not included in the table. 25

D.R. Stull and H. Prophet, JANAF Thermochemical Tables, 2nd ed., NSRDS-NBS 37, June 1971.

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281

The above equation for Kp can be rewritten as ( DG ) ln K p = − =− RT

∑

N ng i =1 i i

or

RT

N

⎛ 1⎞ ln K p = − ⎜ ⎟ × ⎝ R⎠

⎛ gi ⎞

∑ n ⎜⎝ T ⎟⎠ i

i =1

Then, d ⎛ 1⎞ ln K p = − ⎜ ⎟ × ⎝ dT R⎠

N

d ⎛ gi ⎞ vi dT ⎜⎝ T ⎟⎠ i =1

∑

d ln K p = dT

or

∑

N nh i =1 i i 2

RT

=

DH RT 2

where the last step has been shown in Chapter 8. The values of Kp for all temperatures are obtained by integrating this last equation, called the van’t Hoff’s equation. Since, by definition, hi = Úcp,i dT, once cp,i (T) are known, the above equation becomes d ln K p = dT

∑

N

n i =1 i

∫c

RT

p,i dT

2

=

1 R

N

∑n ∫

c p,i dT

i

T2

i =1

which, called the Kirchhoff’s equation, can be integrated and the Kp values can be obtained for all temperatures. The main disadvantage of this equation is the error it creates owing to large difference (three orders of magnitude) between the enthalpies and the specific heats. Typically, the former (as seen in the table) is in [kJ/mol] while the specific heats are in [J/mol]—a thousand times less. Hence, Table 10.6 gives variation of the Gibbs free energy of formation as a function of temperature. This can be used to calculate (DG)fT for any temperature T. The equilibrium constant Kp can then be evaluated from the standard formula. The following example illustrates this procedure. EXAMPLE 10.10 Solution

Calculate the values of Kp at 1500 [K] for the species O, H, OH and H2O.

For clarity, the solution is worked out in the following steps:

1. The basic relation of chemical equilibrium is

RT ln K Tp,i = − ( DG)Tf ,i

or

( DG)Tf ,i

ln K Tp,i = −

RT

2. Since the tabulated values of (DG)f,iT are in [kJ/mol], this relation becomes ln

K Tp,i

=−

( DG)Tf ,i × 1000 8.3143 × T

or

ln

K Tp,i

=−

( DG)Tf ,i 8.3143 × q

where q = T/1000. 3. Table 10.6 lists the coefficients a, b, c and d of (DG)f,iT = (DG)f,i0 + ai + biq + ciq 2 + diq 3, where q =

T 1000

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4. Substituting this, the above relation for Kp becomes

ln K Tp,i = −

ai + biq + ciq 2 + diq 3 8.3143 × q

5. Evaluating this relation for q = 1.5 for the different species gives the following results. Species

H

O

OH

H2O

(DG)f,i1500[K] [kJ/mol] 136.6378

154.9795

16.65057

– 164.49965

– 12.42674

1.3350947

–13.190098

ln Kp

1500[K]

–10.956046

6. It is seen that even at this high temperature, the only significant reaction is H2 + (1/2)O2 = H2O Some tables list the standard entropy of formation, (DS)f0, together with the standard enthalpy of formation, (DH)f0. Then, the standard Gibbs free energy of formation is calculated from the standard relation, namely, (DG)f0 = (DH)f0 – (298.15) × (DS)f0. Most of the older books and handbooks (including JANAF tables) also list log Kp, i.e. the natural logarithm (logarithm to the base 10) including the number 2.30258, which is the value of ln 10. This is the remnant of the days of slide rule when accurate calculations were done with tables of common logarithms. It is clear that, with electronic calculators, these are not necessary. Sometimes, the following equivalent forms of equilibrium constants are useful. 1. By definition, Kp = ’i(pi)vi. 2. Using the ideal gas equation, piV = niRT, the equation for Kp then becomes ∑ ⎛ RT ⎞ i i Kp = ⎜ × [∏i ( ni )ni ]. ⎟ ⎝ V ⎠ n

3. Analogous to Kp, define the equilibrium constant based on the mole numbers as Kn = ’i(ni)vi. ⎛ RT ⎞ 4. Then, the equation for Kp becomes, Kp = Kn × ⎜ ⎝ V ⎟⎠

∑i n i

.

5. Similarly, theni equilibrium constant based on molar concentrations cis defined as Kc = ’i(ci)Si can be shown to be Kp = Kc × (RT)Sini. 6. The equationni for the equilibrium constant based on the mole fractions, namely, ni Kx = ’i(xi)Si has been shown earlier to be Kp = Kx × (p)Si . All the preceding equilibrium constants show that they are functions of other properties in addition to the temperature. Note that they are all different forms of the same equilibrium constant, assumed here to be Kp. Note also the advantage of Kp — it is only a function of T.

Chapter 10: Thermodynamics of Combustion

10.7.2

283

Calculation of Equilibrium Composition

This is best illustrated by an example. EXAMPLE 10.11 Calculate the composition when 1 [kmol H2] reacts with 1 [kmol O2] and reaches equilibrium at 1 [atm] and 1500 [K]. Solution

The following steps give the details of the procedure.

Step I. Given data. Consider that the reaction is taking place in a van’t Hoff box. Then, the chemical species, initially put in the box, are 1 [kmol H2] (i.e. 2000 NA atoms of H, where NA is the Avogadro number, or 2 [kmol H]) and 1 [kmol O2] (i.e. 2000 NA atoms of O, or 2 [kmol O]). In other words, NH = 2 [kmol H] and NO = 2 [kmol O] are put in the box at T = 1500 [K] and p = 1 [atm]. Step II. Choice of significant species. To some extent, this choice is arbitrary. Example 10.10 shows that even at this temperature only H2O production is significant. Thus, it is assumed that the significant species are H2 and O2 (the reactants) as well as H2O. So the unknown variables (to be determined) are xH2, xO2 and xH2O. Mathematics requires that to solve for these variables uniquely, there must be three independent equations. Step III. Species conservation equations. These are the mathematical forms of the basic physical (of course in chemistry) law. In the present case, since there are two chemical species, namely, hydrogen and oxygen, there are two conservation equations, one for each species, which may be written as follows. A. Conservation of hydrogen atoms. Each molecule of H2 contains two H atoms; or nH2 [kmol H2] will contain 2nH [kmol H]. By similar arguments, nH O [kmol H2O] 2 2 will contain 2nH2O [kmol H]. 1. Thus, the equation of conservation of hydrogen becomes 2nH2 + 2nH2O = NH. 2. By the definition of mole fractions, xi = ni/nT; or, ni = nT xi, where nT = Sini, is the total number of moles in the box. 3. Substituting for ni in the equation of conservation of hydrogen atoms and dividing throughout by nT gives 2xH2 + 2xH2O = (NH/nT) = (2/nT)

(A)

B. Conservation of oxygen atoms. With similar arguments, the equation of conservation of oxygen atoms reduces to 2xO2 + xH2O = (NO/nT) = (2/nT)

(B)

Step IV. Auxiliary equations. These equations provide the additional information needed to solve the problem. They are chosen such that the total number of equations equals the total number of unknowns. In the present case, there are four unknowns, the three mole fractions, namely, xH2, xO2 and xH2O, as well as the total mole number nT. There are two species conservation equations, one each for H2 and O2. Thus, two extra equations are now needed. These are chosen as follows.

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A. Defining relations. These arise from the basic definitions of the basic physical laws. In this case, the basic law of conservation of species implies that nH2 + nO2 + nH2O = nT or xH2 + xO2 + xH2O = 1 (C) B. Kp relations. These relations are applicable since the reaction is under equilibrium. In this case, since only one additional equation is needed, and only the species H2, O2 and H2O are significant, then the scheme for the equilibrium reaction is chosen as H2 + (1/2)O2 = H2O. From the definition of Kp, for this reaction, it becomes

Kp =

pH2 O 1/ 2

( pH2 ) × ( pO2 )

=

x H 2O ( x H2 ) × ( x O2 )1/ 2 × ( p)1/ 2

where the second relation is, by the definition of mole fractions for an ideal gas, xi = pi/p. Since Kp is only a function of temperature, the above equation can be rewritten as xH2O = K (xH2) × (xO2)1/2, where K = e13.19

(D)

and the value of ln Kp = 13.19 for T = 1500 [K], from Example 10.10 is used. Step V. Simplification of the equations. In principle, the four equations (A–D) can be solved for the four unknowns. However, since Eq. D (involving Kp) is non-linear, the whole system is nonlinear. Thus, numerical methods should be used, for which some simplification is necessary. This is done in the following steps. 1. Subtracting Eq. (B) from Eq. (A) gives (E) 2xH2 + xH2O – 2xO2 = 0 2. To prevent accumulation of round-off errors, Eq. (C) should be used to solve for that species which has the smallest mole fraction. In the present case, since there is excess oxygen, most of the hydrogen will be used up, making xH2 to be the smallest. Then, Eq. (C) is rewritten as xH2 = 1 – xO2 – xH2O (F) 3. Eliminating xH2 from Eq. (E) with Eq. (F) and solving for xO2 gives (G) xO2 = (2 – xH2O)/4 4. Equation (D) is xH2O = K (xH2) × (xO2)1/2, where K = e13.19 Step VI. Solution procedure. Equations (F) and (G) and Eq. (D) are solved by iteration. It is possible to reduce the number of equations to one by eliminating all the mole fractions except one, but the resulting equation in the remaining unknown mole fraction (e.g. xO2) will be too long and cumbersome to handle. So this set of equations is solved as it is. It is convenient to use methods that do not involve use of derivatives, since getting a closed expression for the derivatives, may be difficult or the expression obtained may be too involved. Thus simple search methods are ideal for this type of problems.

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The method of bisection is a good procedure that does not involve gradients. Using this method, these equations are solved as follows: 1. 2. 3. 4.

Guess a value of xH2O. Evaluate xO2 from Eq. (G). Solve for xH2 from Eq. (F). Use Eq. (D) to check the value of xH2O.

The stoichiometric values of the mole fractions are good initial guesses. In this case, since the reaction is H2 + O2 = H2O + (1/2)O2, the initial value of xH2O = 2/3. After 20 iterations (including those guesses which gave negative values for some of the mole fractions) of calculations with the common pocket calculator, the final result was xH2O = 0.66666375, xO2 = 0.33333425, xH2 = 2 × 10–6 and nT = 1.000022 [kmol]. Combustion in air In most cases the fuels burn in air, i.e. take oxygen of the air. Then, N2 in air is assumed to be inert and its partial pressure is calculated as pN2 = (0.79/0.21)pO2. However, for pollution calculations involving oxides of nitrogen, reaction between N2 and O2 should be considered. For example, in simple cases, the reaction steps, (1/2)N2 = N and (1/2)N2 + (1/2)O2 = NO, may be adequate. This also gives rise to the equation of conservation of nitrogen atom as 2nN2 + nN + nNO = NN

or 2xN2 + xN + xNO = NN/nT

where the second equation is obtained using the definition of mole fraction. Moreover, the equation of conservation of oxygen should now contain the additional term, xNO, corresponding to NO. There will also be the equations for two Kp’s, corresponding to the two reaction steps, defined as K p,N =

pN ( pN2 )1/ 2

and

K p,NO =

( pN2

pNO × pO2 )1/ 2

Hydrocarbon combustion All the conventional fuels used for combustion contain hydrocarbons (e.g. volatiles of coals) or they are hydrocarbons themselves (e.g. petroleum fuels). When they burn, CO2 (and perhaps, CO) will be present in their products. Furthermore, carbon by itself may also be present in solid form. In these cases, carbon and its compounds should also be considered in the analysis. Consider the simplest case which corresponds to the significant species being carbon, CO and CO2. One difficulty encountered is due to the fact that the standard state of carbon at the reference condition is the solid phase. Out of the several forms of solid carbon, the graphite form is chosen as the standard state, so the (DG)0f, graphite = 0. However, since solids (theoretically) exert zero partial pressure, the reaction scheme for combustion is assumed to take place in two steps corresponding to sublimation of carbon

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followed by the combustion of carbon in gaseous phase. This assumption makes the definition of Kp still applicable. Thus, the combustion of solid carbon, Cgraphite + O2(g) = CO2(g) [or, CO(g)] is assumed to take place in two stages corresponding to the schema Cgraphite = C(g)

and

C(g) + O2(g) = CO2(g)

or C(g) + (1/2)O2(g) = CO(g)

Then, these give rise to the following Kp’s. Kp,C(g) = pC(g)

and

pCO2

Kp,CO2 =

pO2

and

Kp,CO =

pCO ( pO2 )1/ 2

The above equilibrium constants are given in Table 10.6. Now, the equation of conservation of carbon atoms gives xCgraphite + xC(g) + xCO + xCO2 = NC/nT However, in the presence of excess oxygen, solid carbon is generally absent in the industrial flames. In such cases, the above scheme is modified to CO(g) + (1/2)O2(g) = CO2(g), whose Kp can be expressed in terms of the above Kp’s as follows:

Kp =

pCO2 1/ 2

( pCO ) × ( pO2 )

=

pCO2 /pO2 1/ 2

pCO /( pO2 )

=

K p,CO2 K p,CO

and which have been tabulated. Then, the equation of conservation of carbon atoms reduces to xCO + xCO2 = NC/nT. These equations can be solved as shown in Example 10.11. Since no new ideas are involved, these are not elaborated further.

10.8

HEAT GENERATED BY COMBUSTION

It was mentioned that the main purpose of combustion is to generate heat which can be used elsewhere. This aspect is discussed in this section.

10.8.1

Heat of Formation

When a substance (or compound) is formed by a chemical reaction, heat is generated due to exothermic reactions. Heat absorption by endothermic reactions is simply negative heat generation. Since reactions are assumed to take place at constant pressure and temperature, and the reactants and products are simple compressible substances, the heat generated equals the change in enthalpy. The heat generated during a reaction not only depends upon the pressure and temperature at which the reaction takes place, but also on the enthalpy of the reactants. Hence, for reference, some condition is needed as the standard for temperature, pressure and enthalpy of reactants. When the reaction is one in which the substance is formed from elements (e.g. H2O formed by the reaction between H2 and O2), then this heat is known as the heat of formation of the substance.

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For the reactions in which substances are formed, the reference condition is chosen as p = 1 [atm], T = 298.15 [K] (25 [°C]). The reference for the enthalpy of reactants is defined by assigning zero enthalpies to elements that are in their standard states at the reference temperature and pressure. The standard state for an element is that state in which it occurs naturally at this temperature and pressure. For example, hydrogen, oxygen and nitrogen are elements and at 1 [atm] and 298.25 [K], they are in gaseous form as26 H2(g), O2(g) and N2(g). As explained in Section 10.7, the standard state for carbon is solid in graphite form. All their enthalpies are assumed to be zero. Thus, the standard heat of formation of a substance (compound) is defined as the change of enthalpy when the substance is formed at 1 [atm] and 298.25 [K] from its elements at 1 [atm] and 298.25 [K]. The standard heat of formation is denoted as (DH)f,i0, where the subscript f means formation, the subscript i denotes the species (substance, compound) and the superscript 0 (zero) indicates the reference condition (of p = 1 [atm] and T = 298.15 [K]). Since the reactants and products are at the same pressure and temperature, the heat generated during exothermic reactions should be extracted. Now, by the sign convention for heat, this heat is negative since it is transferred from the system. Thus, standard heats of formation of substances formed by exothermic reactions are negative. Table 10.7

eHg = HgT – Hg0

Species H2 O2 N2 CO2 H2O CO SO2 OH C(g) H(g) O(g) C(graph)

Enthalpies of some common gases

a – 8.30039 – 8.70795 – 7.98290 – 11.5910 – 8.71478 – 8.06397 – 12.7630 – 8.19895 – 6.24428 – 6.18182 – 6.37666 – 3.39294

b 27.7344 27.6353 25.6601 34.2036 27.2090 25.7726 36.8022 27.3719 20.9802 20.6876 21.6182 7.80612

c 1.06219 4.38141 4.40898 12.8474 8.27636 4.67387 10.0577 1.76354 – 0.199546 0.09197 – 0.464973 8.921902

(DH)f0 d 0.184288 – 0.605826 – 0.603409 – 2.079260 – 0.769569 – 0.677615 – 1.71170 0.014672 0.06492 – 0.022474 0.088306 – 1.55894

0 0 0 – 393.5 – 241.8 – 110.5 – 296.8 39.46 715.0 218.0 249.2 0

eHg = a + bq + cq2 + dq3, where q = T/1000, and T is in [K]. (DH)f0, HgT, Hg0 and eHg are in [kJ/mol] (i.e. [MJ/kmol]). Generated from JANAF tables cited above.

The last column of Table 10.7 gives (DH)f0, the standard heat of formation at the standard state (of 1 [atm] and 298.15 [K]) for some common gases (assumed ideal) which form combustion products.

26

In these symbols, ‘g’ in the parentheses indicates that the elements are in gaseous phase. However, this method of specification of the phase is used for clarity, only when confusion may arise.

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Bond energies and heat of formation The measured values of standard heat of formation are available for most of the substances in standard handbooks. However, in an emergency, as an approximation, they can be estimated from bond energies as shown below. The procedure of Pauling27 who pioneered the approach of assigning bond energies, is now briefly explained as an illustration. 1. Let it be required to assign energy to the O–H bond at the reference condition of 1 [atm] and 298.15 [K]. 2. It is known that the H2O molecule contains two O–H bonds. 3. Even though both the bonds are not identical and their energies are not the same, assume that they are. 4. It then follows that this bond energy will be half of the heat of reaction (DH)R0 of the step 2H(g) + O(g) = H2O + (DH)R0 since breaking of the H2O molecule will produce 2H(g) and O(g). 5. Now, the equation for the heat of formation of H2O is H2(g) + (1/2)O2 = H2O – 241.8 0 (DH)f, H2O is taken from Table 10.7

[kJ/mol H2O]

where 6. The equations of formation of H(g) and O(g) are by the reactions H2(g) = 2H(g) + 2 × 218.0 [kJ/mol H] and

(1/2)O2(g) = O(g) + 249.0 [kJ/mol O]

where the standard heats of formation have been taken from Table 10.7. 7. Then, subtracting the two equations in step 6 from that in step 5, the Hess law of heat addition, (the energy (first law) equation used in thermochemistry) gives 2H(g) + O(g) = H2O – 927.0 [kJ/mol] 8. Then, the energy of O–H bond = (1/2) × 927 = 463.5 [kJ/mol], i.e. 463.5 [MJ/kmol], which agrees well with the values measured by other methods. Energies of some typical bonds are given below: Energies of some typical bonds (in [MJ/kmol]) Type Energy

C–C 358.0

C=C 598.7

C∫C 813.5

C–H 410.7

C–O 359.2

O–O 138.6

O–H 463.1

C–N 339.1

N–H 368.4

Some substances possess several possible structures. At different times, they keep assuming different structures. This phenomenon is called resonance and the energy possessed due to this dynamics is called resonance energy. For example, benzene oscillates between ring structures in which the C=C and C–C are interchanged among themselves.28 The resonance energy for benzene is 204.7 [MJ/kmol]. 27 28

Pauling, L., The Nature of Chemical Bond, Ind. ed., Oxford-IBH, 1967, § 3–5. Perhaps, the double and single bonds go round the ring (like children’s play).

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The following example illustrates the procedure for estimating the heat of formation from bond energies. EXAMPLE 10.12 Aniline is a popular rocket propellant. It has the benzene structure with one of the H atoms replaced N–H2. The resonance energy for aniline is 291.4 [MJ/kmol]. What will be its standard heat of formation? Solution The reference structure for benzene is the hexagonal ring in which the single and double bond of carbon–carbon alternate. A hydrogen atom is attached to each carbon atom. Thus, the reference structure has three C–C, three C=C and six C–H bonds. Now, in the reference structure of aniline, one of the H atoms is replaced with NH2. 1. Then, this reference structure has three C–C, three C=C, five C–H, one C–N, and two N–H bonds. 2. Hence, the total energy of the reference structure is (3 × 358.0) + (3 × 598.7) + (5 × 410.7) + (1 × 339.1) + (2 × 368.4) = 5999.5 [MJ/kmol] where the bond energies tabulated before are used. 3. The resonance energy for aniline is 291.4 [MJ/kmol]. 4. Thus, the total bond energy of aniline 6290.9 [MJ/kmol]. 5. The molecular formula of aniline is C6H5NH2. 6. This implies the basic formation reactions of 6C(graphite) = 6C(g) + 6 × 750 [MJ/kmol] (for carbon) (7/2)H2(g) = 7H(g) + 7 × 218.0 [MJ/kmol]

(for hydrogen)

(1/2)N2(g) = N(g) + 471.7 [MJ/kmol] (for nitrogen) 7. Then, by the Hess’ law, adding these equations gives 6C(graphite) + (7/2)H2(g) + (1/2)N2 = C6H5NH2 – 3.2 [MJ/kmol aniline] which is the standard heat of formation of aniline. It agrees reasonably with the measured value. 8. However, similar calculations for methane gives (DH)f0 = – 206.8 [MJ/kmol aniline], which agrees reasonably well with experimental values. The above discussion only indicates the basic approach. With seventy years of research since then, more accurate and elaborate data are available now. There is also better knowledge about the differences between bonds existing between the same elements (e.g. C–C bonds) at different locations in the molecules. These differences increase as the number of atoms in a molecule increases. Thus, at best, these can only be very rough estimates.

10.8.2

Heat of Reaction

Consider the formation of CO2 by the combustion of CO in oxygen at the reference condition. The reaction scheme is CO + (1/2)O2 = CO2

(at T = 298.15 [K], p = 1 [atm])

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In this case, the term heat of formation cannot be applied since one of the reactants, namely CO, is not an element. In such cases, the change of enthalpy is called the heat of reaction and the reaction is written as CO + (1/2)O2 = CO2 + (DH)R0

(at T = 298.15 [K], p = 1 [atm])

(DH)R0

where denotes the heat of this reaction. It will be clear that heat of formation is the special case of heat of reaction when the reactants are elements in their standard states. However, since it is the heat of formation that is experimentally measured, this is taken as the basis. Then the expression for the heat of reaction for the above scheme can be derived as follows. The schema for the formation of CO2 at the reference condition is Cgraphite + O2 = CO2 + (DH)f,0CO2 where (DH)f,0CO2 = – 393.5 [MJ/kmol CO2]. Similarly, the schema for the formation of CO at the reference condition is Cgraphite + (1/2)O2 = CO + (DH)f,0CO where (DH)f,0CO = – 110.5 [MJ/kmol CO]. Now, subtracting the second equation from the first and transposing CO to the left gives CO + (1/2)O2 = CO2 + (DH)R0 where (DH)R0 = (DH)f,0CO2 – (DH)f,0CO = – 283 [MJ/kmol CO] (since 1 [kmol CO] produces 1 [kmol CO2]), which is the Hess’ law of heat addition. Thus, the heat of reaction for a general single-step reaction is defined as the difference between the enthalpy of the products and that of the reactants. For example, for the reaction scheme of aA + bB = cC, at the reference condition, the heat of reaction is (DH)R0 = c(DH)C0 – a(DH)A0 – b(DH)B0 which can be generalized for the reaction, 0 = ( DH )0R =

Â n M , as i

N

i

∑ n ( DH ) i

i =1

i

0 Mi

where the sign convention is: ni’s of products are positive and ni’s of the reactants are negative. The main difference between heats of formation and heat of reaction are: ∑ Heat of formation refers to a chemical species while heat of reaction is for a reaction (see the CO combustion discussed above). ∑ The heat of formation is generally applied for species at the reference conditions, while the heat of reaction is applicable to any temperature (and of pressure).

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For reactants that are ideal gases, the enthalpy is independent of pressure. Then, for the general reaction scheme discussed earlier, the heat of reaction at any temperature T can be calculated as (DH)RT = (DH)R0 +

⎛ N ⎞ ⎜ n i c p ,i ⎟ dT ⎟ 298.15 ⎜ ⎝ i =1 ⎠

∫

T

∑

This is called the Kirchhoff’s equation. If the data on the enthalpy of gases are available instead of their specific heats, then the integral in the Kirchhoff’s equation can be evaluated and this equation can be written as (DH)RT = (DH)R0 + (HgT – Hg0) where HgT and Hg0 are the enthalpies at T and 298.15 [K], respectively. The data on this difference in the enthalpies, denoted as eHg = HgT – Hg0, from JANAF tables has been expressed as a polynomial. That is,

eHg = a + bq + cq2 + dq3 where q = T/1000 with T in [K]. The values of the coefficients that fit this data best have been presented in columns 2–5 of Table 10.7.

10.8.3

The Calorific Value

In thermal engineering, the heat released by combustion per unit quantity of a fuel is called the calorific value of the fuel. It is denoted as Qcal. For solid and liquid fuels whose composition is specified on gravimetric basis (i.e as mass fractions), the unit quantity is the unit mass. On the other hand, for gaseous fuels whose composition is given on volumetric basis (i.e as mole fractions), this unit quantity is the unit kilomole (or m3). Historically, the concept of calorific value is at least a century older than the recent concept of heats of formations and heats of reactions. Thus, the latter data are more accurate than the former. Hence, it is good to know the relation between them. This is explained below. Consider the combustion of (solid) carbon burning to produce CO2 at the reference condition of 1 [atm] and 25 [°C]. 1. Table 10.7 shows that (DH)f,0CO2 = –393.5 [kJ/mol CO2], which is the heat released by combustion. 2. Since 1 [mol CO2] is produced by 1 [mol C], this heat released by combustion is also for 1 [mol C], i.e. Qcomb,C = (DH)f,0CO2 = –393.5 [kJ/mol C]. 3. Then, the calorific value of solid carbon is

⎛ ⎡ kJ ⎤ ⎞ ⎛ ⎡ mol C ⎤ ⎞ ⎛ 1 ⎡ kmol C ⎤ ⎞ ⎡ MJ ⎤ Qcal = ⎜ −393.5 ⎢ ⎟ × ⎜1000 ⎢ ⎟×⎜ ⎢ ⎟ = 32.79 ⎢ ⎥ ⎥ ⎥ ⎥ ⎣ mol C ⎦ ⎠ ⎝ ⎣ kmol C ⎦ ⎠ ⎝ 12 ⎣ kg C ⎦ ⎠ ⎣ kg C ⎦ ⎝ Thus, the following are the two aspects of the relation between the heat of formation (or reaction) and the calorific value that should be remembered.

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1. The calorific value of a fuel is the same as the heat of formation (or reaction), but expressed per unit quantity of the fuel (i.e. the reactant, C, here) rather than the product (CO2 here). 2. By convention, the calorific value of fuel (which react exothermically) is positive. This is the opposite of the sign convention for heats of formation (or reaction). In other words, for the single step reaction, 0 = ∑ i n i M i , occurring at the temperature T, the calorific value is N

Qcal = −

∑n H i

T g, i

or

Qcal = − ( DH )TR

i =1

Calculations similar to the above give the following calorific values. 1. 2. 3. 4. 5.

C burning to CO2 (from above illustration) = 32.79 [MJ/kg C]. C burning to CO (from Table 10.7), Qcal = 9.21 [MJ/kg C]. CO burning to CO2 (from Table 10.7), Qcal = 10.11 [MJ/kg C]. H2 burning to H2O(g) (from Table 10.7), Qcal = 120.9 [MJ/kg H2O]. S burning to SO2 (from Table 10.7), Qcal = 9.28 [MJ/kg S].

This calorific value for combustion of hydrogen, wherein H2O appears as steam, is called as the lower (or, net) calorific value (often denoted as LCV) since if the steam condenses, extra quantity of heat (corresponding to the enthalpy of vaporization) is also released. Clearly, the latter (in which H2O appears as water) is called the higher (or, gross) calorific value, popularly denoted as HCV. Generally, unless stated otherwise, lower calorific values are used in design and analysis. The calorific values of fuels are measured in the laboratory by burning a sample of the fuel and measuring the heat released. The experiments can be conducted at constant volume (Bomb calorimeters) or at constant pressure (Flow calorimeters). Now, by the energy (first law) equation, the heat transferred at constant volume for simple compressible substances equals the change of internal energy, and that at constant pressure equals the change of enthalpy. Thus we have Qcal,V = DU and Qcal,p = DH = DU + D(pV) = DU + RTDn = Qcal,V + RTDn where the last few steps of the second expression are due to the definition of enthalpy and to the ideal gas equation. Calculation of calorific values The calorific values of fuels can be calculated once their composition is known. The basic approach is to calculate the heat released by the combustible elements, namely, C, H2, CO (if any), and S of the fuel and add them up. The principle of conservation of energy guarantees that this is the calorific value of the fuel. Following examples illustrate this. To save space, the steps involved will be briefly indicated since the earlier examples were explained in detail.

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EXAMPLE 10.13

293

Calculate the calorific value of ethane.

Solution The molecular formula for ethane is C2H6. Then: 1. The carbon content of the fuel is ⎡ kg C ⎤ 2 × 12 24 = = 0.8 ⎢ ⎥ (2 × 12) + (6 × 1) 30 ⎣ kg F ⎦ 2. Similarly, the hydrogen content of the fuel is mC =

mH =

⎡ kg H ⎤ 6 ×1 6 = = 0.2 ⎢ ⎥ (2 × 12) + (6 × 1) 30 ⎣ kg F ⎦

3. Then, heat released by the combustion of carbon in the fuel

⎛ ⎡ kg C ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ ⎡ MJ ⎤ = ⎜ 0.8 ⎢ ⎟ × ⎜ 32.79 ⎢ ⎟ = 26.23 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg C ⎦ ⎠ ⎣ kg F ⎦ ⎝ 4. Similarly, heat released by the combustion of hydrogen in the fuel

⎛ ⎡ kg H ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ ⎡ MJ ⎤ = ⎜ 0.2 ⎢ ⎟ × ⎜ 120.9 ⎢ ⎟ = 24.18 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg H ⎦ ⎠ ⎣ kg F ⎦ ⎝ 5. Hence, the calorific value of the fuel is ⎡ MJ ⎤ Qcal = 26.23 + 24.18 = 50.41 ⎢ ⎥ ⎣ kg F ⎦

EXAMPLE 10.14 Calculate the calorific value of a coal with composition of C = 51.3%, H2 = 3.5%, N2 = 1.8%, O2 = 7.3%, S = 0.7%, and the rest being ash and moisture. Solution This is the same sample of coal as that in Example 10.5, wherein the air-fuel ratio as well as the composition of the exhaust gases were computed. Here, the calorific value is to be calculated. The procedure is the same as that of Example 10.13. The combustible elements are C, H2 and S. Hence, the calculation can be done in one step as follows.

⎛ ⎡ kg C ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ ⎛ ⎡ kg H ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ Qcal = ⎜ 0.513 ⎢ ⎟ × ⎜ 32.79 ⎢ ⎟ + ⎜ 0.035 ⎢ ⎟ × ⎜ 120.9 ⎢ ⎥ ⎥ ⎥ ⎥⎟ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg C ⎦ ⎠ ⎝ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg H ⎦ ⎠ ⎝ ⎛ ⎡ kg S ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ ⎡ MJ ⎤ + ⎜ 0.007 ⎢ ⎟ × ⎜ 9.28 ⎢ ⎟ = 20.76 ⎢ ⎥ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ ⎣ kg S ⎦ ⎠ ⎣ kg F ⎦ ⎝ The calorific values of gaseous fuels are calculated on mole basis, since their composition is on volume (mole) basis. Since the volumes of gases depend upon the pressure and temperature at which the volumes are measured, these must be specified along with the data. The default units are NTP.29 This is required for conversion to mass basis. 29

Normal Temperature and Pressure, also called STP, Standard Temperature and Pressure, i.e. 1 [atm] and 273.15 [K].

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EXAMPLE 10.15 A sample of gobar gas contains 55% methane and the rest is CO2. What will be its calorific value? Solution The calculations are done on mole basis. Since methane is the only combustible constituent, its calorific value should be determined first. 1. Since the molecular formula of methane is CH4, its calorific value is given by

⎛ ⎡ kmol C ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ ⎛ ⎡ kmol H ⎤ ⎞ ⎛ ⎡ MJ ⎤ ⎞ Qcal,CH4 = ⎜ 12 ⎢ ⎥ ⎟⎟ × ⎜ 32.79 ⎢ ⎥ ⎟⎟ × ⎜ 120.9 ⎢ ⎥ ⎟ + ⎜⎜ 4 ⎢ ⎥⎟ ⎜ ⎣ kg C ⎦ ⎠ ⎝ ⎣ kmol CH 4 ⎦ ⎠ ⎝ ⎣ kg H ⎦ ⎠ ⎝ ⎣ kmol CH 4 ⎦ ⎠ ⎝

2.

⎡ MJ ⎤ = 877.1 ⎢ ⎥ ⎣ kg CH 4 ⎦ Hence, the calorific value of fuel is given by

⎛ ⎡ ⎤⎞ ⎡ kmol CH 4 ⎤ ⎞ ⎛ ⎡ MJ ⎤ MJ Qcal,F = ⎜ 0.55 ⎢ ⎟ × ⎜⎜ 877.1 ⎢ ⎥ ⎟⎟ = 482.4 ⎢ ⎥ ⎥ ⎣ kmol F ⎦ ⎠ ⎝ ⎣ kmol F ⎦ ⎣ kmol CH4 ⎦ ⎠ ⎝

10.9

ADIABATIC FLAME TEMPERATURE

The adiabatic flame temperature is one of the important characteristics of combustion. This is explained in this section. Consider the combustion of a fuel with an oxidant occurring in an adiabatic closed system. Then, the products absorb all the heat released by combustion so that its temperature increases. At higher temperatures, dissociation reactions occur which absorb heat and thereby reduce the temperature of the system. However, when the temperature decreases, the forward reactions become predominant with the result that the net heat release increases the temperature. After all these processes have occurred and the system has reached equilibrium, the temperature attained is called the adiabatic flame30 temperature. It is denoted as Tad. Then, Tad = T0 +

m f Qcal Qcomb = T0 + mg c p, g mg c p, g

where mf is the mass of the fuel burnt, mg is the mass of the exhaust gases, and cp,g is the average specific heat of the products. Note that the conservation of mass implies that mg = mf + ma, where ma is the mass of the air supplied. This definition shows that: 1. In rich mixtures (of air and fuel), some part of the fuel will remain unburnt for want of oxygen. Hence, Qcomb, the heat released by combustion will be less. Consequently, Tad will be less. 2. On the other hand, in lean mixtures, there is extra air which appears in exhaust, thereby increasing mg. This increase in mg reduces Tad. 30

Since the flame is assumed to be a closed adiabatic system.

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The preceding arguments show that Tad is maximum when the air-fuel ratio is stoichiometric. Moreover, the second argument shows that Tad will be more if the combustion takes place in pure oxygen instead of air, which has more than three times (= 77/23) (inert) nitrogen to absorb the heat released due to combustion. Thus, Tad will be maximum when a fuel burns in pure oxidant in stoichiometric proportions. To understand the concept of adiabatic flame temperature better, we rewrite the preceding definition as (mgcp,g) × (Tad – T0) = Qcomb = mf × Qcal or Qabs = Qgen where T0 is the reference temperature. Note that, the left-hand side of the first equation is the heat absorbed by the exhaust gases and the right-hand side is the heat generated by combustion. The second equation shows this equality. Now, the principle of conservation of mass implies, mg = mf + ma. Moreover, by definition of the air-fuel ratio, Z = ma/mf. Then, the above equation can be solved for Tad as Tad = T0 +

Qcal 1+ Z

Before taking up the estimation of Tad, the method of calculating the enthalpies of exhaust gases is illustrated in Example 10.16 EXAMPLE 10.16 Compute the enthalpy of an exhaust gas at 1000 [K] with the composition of CO2 = 12.3%, CO = 1.74%, O2 = 3%, N2 = 76.4%, and H2O = 6.6%. Solution Since the default unit for specification of composition of gases (either fuels or products) is mole, the given composition is the molar composition. Table 10.7 lists the coefficients of polynomials that give the deviation of enthalpy, namely hg – hg0 in [MJ/kmol]. Taking hg0 as zero, at 298.15 [K] as reference, N

hg =

∑x

i

× ( ai + biq + ciq 2 + diq 3 )

i =1

or ⎡N ⎤ ⎡ N ⎤ ⎡ N ⎤ ⎡ N ⎤ hg = ⎢ ( ai × xi ) ⎥ + ⎢ (bi × xi ) ⎥ q + ⎢ (ci × xi )⎥ q 2 + ⎢ (di × xi )⎥ q 3 ⎢⎣ i = 1 ⎥⎦ ⎢⎣ i =1 ⎥⎦ ⎢⎣ i = 1 ⎥⎦ ⎢⎣ i =1 ⎥⎦

∑

∑

∑

∑

where q = T/1000. The terms in the brackets can be evaluated since the coefficients are known from Table 10.7 and the composition is given. The calculation is best done in a tabular form. Thus, the result is hg = (0.123) × (33.38074) + (0.0174) × (21.70488) + (0.03) × (22.70293) + ⎡ MJ ⎤ (0.764) × (21.48277) + (0.066) × (26.00101) = 23.2935 ⎢ ⎥ ⎣ kmol P ⎦

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Following example illustrates the procedure of computing the adiabatic flame temperature. Since the emphasis is on the procedure, compositions are computed assuming stoichiometry (i.e. neglecting the dissociation reactions). EXAMPLE 10.17 A sample of coal has the following composition: C = 51.3%, H = 3.5%, N2 = 1.8%, O2 = 7.3%, S = 0.7%, and the rest being inherent moisture and ash. It was burnt with 10% excess air. Calculate Tad. Solution

This fuel has already been discussed in Examples 10.5 and 10.14.

1. In Example 10.14, the calorific value of this fuel was calculated as Qcal = 21.12 [MJ/kg]. 2. Converting the gravimetric composition of the exhaust gases in Example 10.5 gives xCO2 = 0.1533, xH2O = 0.0786, xO2 = 0.0178, xSO2 = 0.0016, xN2 = 0.7487 making the molecular weight of the products equal to 24.8388 [kg G/kmol G]. 3. Calculations in Example 10.5 also gave the mass of the products equal to 8.2921 [kg G/kg F]. 4. The heat released by combustion and available for absorption is

⎛ ⎡ MJ ⎤ ⎞ ⎛ 1 ⎡ kg F ⎤ ⎞ ⎡ MJ ⎤ Qavail = ⎜ 21.12 ⎢ ⎟×⎜ ⎟ = 2.547002 ⎢ ⎥ ⎢ ⎥ ⎥ ⎣ kg F ⎦ ⎠ ⎝ 8.2921 ⎣ kg G ⎦ ⎠ ⎣ kg G ⎦ ⎝ 5. The expression for heat absorbed is shown in Example 10.16 as ⎡ N ⎤ ⎡ N ⎤ ⎡ N ⎤ ⎡ N ⎤ hg(q ) = ⎢ (ai × xi )⎥ + ⎢ (bi × xi ) ⎥ q + ⎢ (ci × xi ) ⎥ q 2 + ⎢ (di × xi )⎥ q 3 ⎣⎢ i =1 ⎦⎥ ⎢⎣ i =1 ⎦⎥ ⎣⎢ i =1 ⎦⎥ ⎣⎢ i =1 ⎦⎥

∑

∑

∑

∑

where q = T/1000. 6. The definition of the adiabatic flame temperature shows that it is the solution of the equation Qavail = hg(q) 7. In other words, the adiabatic flame temperature is calculated from the equation ⎡ N ⎤ ⎡ N ⎤ ⎡ N ⎤ 2 ⎡ N ⎤ 3 ⎢ ( ai × xi ) ⎥ + ⎢ (bi × xi ) ⎥ q + ⎢ (ci × xi ) ⎥ q + ⎢ ( di × xi )⎥ q = 2.547002 ⎢⎣ i =1 ⎥⎦ ⎢⎣ i =1 ⎥⎦ ⎢⎣ i =1 ⎥⎦ ⎢⎣ i =1 ⎥⎦

∑

∑

∑

∑

8. Since it is a cubic equation in q, it has to be solved by numerical methods. Using the search method initially and the method of bisection later, gives the following results: The available heat for absorption is Qavail = 2.547002 [MJ/kg G] T [K] Qabs

1000

2000

2500

3000

2900

2800

2850

0.9594

1.76

2.207

2.6

2.56

2.47

2.517 2.5385

Hence, Tad is a little less than 2885 [K].

2875

2885

2884

2.5472

2.5463

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9. Note that only 10 iterations are needed for getting a good answer. This is the advantage of the method of bisection (in n iterations, it decreases the interval by (1/2n)). Note also that in the initial stages of search, it is not necessary to know the exact value. Only towards the end, will accurate calculations be needed.

REVIEW QUESTIONS 1. Explain with the help of a diagram the different sources of energy and their conversion strategies. 2. What are fossil fuels? 3. What are gravimetric and volumetric compositions? Where are they used? 4. How is coal formed? What are its grades? 5. How is petroleum formed? 6. What are the primary hydrocarbon families of petroleum? 7. What is combustion? 8. What is the difference between homogeneous and heterogeneous reactions? 9. Define the terms: (a) reaction rate, (b) order of reaction. n i Mi . 10. Obtain an expression for equilibrium of the one-step gaseous reaction, 0 = 11. Define the equilibrium constant Kp. 12. Obtain the expression for the variation of Kp with T and p. 13. Define the terms—heat of formation and heat of reaction. How are they related? 14. Define calorific value. How is it related to the heat of reaction? 15. Define Tad and obtain an expression for it.

∑

EXERCISES 1. Calculate the stoichiometric (theoretical) air-fuel ratio and molar composition of dry products when ethane burns. [Hints: As in Example 10.4. Theoretical O2 = 3.5 [kmol CH4/kmol F], and hence Zs = 16.67 [kmol A/kmol F]. Molar composition of products is: nCO2 = 2 [kmol CO2/ kmol F], nH2O = 3 [kmol H2O/kmol F], nN2 = 13.167 [kmol N2/kmol F]. Dry products contain only CO2 and N2, so that the nDP = 15.167 [kmol DP/kmol F], and the mole fractions are xCO2 = 13.19% and xN2 = 86.81%.] 2. For the complete combustion of heptane with 10% excess air, calculate (i) the stoichiometric air-fuel ratio (Zs), (ii) the actual air-fuel ratio (Z), (iii) the composition of the products on mass basis (gravimetric) and mole basis (volumetric), (iv) the mass and moles of the product gas per unit mass of the fuel, and (v) the change in the number of moles between products and reactants.

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[Hints: (a) Since the molecular formula of heptane is C7H16, the equation of reaction scheme is used, i.e. C7H16 + 11O2 + (11 × 0.79/0.21) N2 = 7CO2 + 8H2O + (11 × 0.79/0.21) N2. Then, mO2 = (11 × 32/100) = 3.52 [kg O/kg F], where 100 is the molecular weight of heptane. Then, Zs = 15.3043 ª 15.30 [kg A/kg F], and Z ª 16.8343 ª 16.83 [kg A/kg F]. The compositions of the products are calculated as usual. Then, nCO2 = 0.07 [kmol O2/ kg F] and mCO2 = 3.08 [kg O2/kg F], nH2O = 0.08 [kmol H2O/kg F] and mH2O = 1.44 [kg H2O/kg F]; nO2 = 0.011 [kmol O2/kg F] and mO = 0.352 [kg O/kg F]; and nN2 = 0.45519 [kmol N2/kg F] and mN = 12.7453 [kg N/kg F]. Then, np = 0.61619 [kmol P/kg F] and mp = 17.6173 [kg P/kg F], so that Mp = 28.5907 [kg P/kmol P]. And the flue gas composition is wCO2 = 17.48% and xCO2 = 11.36%; wH2O = 8.17% and xH2O = 12.98%; wO2 = 2.00% and xO2 = 1.78%; and, wN2 = 72.35% and xN2 = 73.88%. The change in the mole numbers due to reaction is Dn = nP – nr = 15 – 12 = 3.] 3. A coal with composition of C = 74.4%, H = 12%, O = 2% and the rest as ash is burnt with 120% theoretical air. For the complete combustion, calculate (i) the stoichiometric air-fuel ratio (Zs), (ii) the actual air-fuel ratio (Z), (iii) the composition of the products on mass basis (gravimetric) and mole basis (volumetric), and (iv) the mass and moles of the product gas per unit mass of the fuel. [Hints: The same procedure as Exercise 2, but now the calculations are done on mass basis. Thus, mO2 = 0.744 × (32/12) + 0.12 × (16/2) – 0.02 = 2.924 [kg O/kg F], Zs = (2.924/0.23) = 12.713 ª 12.71 [kg A/kg F], and Z = 15.2556 ª 15.26 [kg A/kg F]. Then, mCO2 = 2.728 [kg CO2/kg F] and nCO2 = 0.062 [kmol CO2/kg F]; mH2O = 1.08 [kg H2O/kg F] and nH2O = 0.06 [kmol H2O/kg F]; mO = 0.701758 [kg O/kg F] and nO2 = 0.02193 [kmol O2/kg F], and, mN = 11.7468 [kg N/kg F] and nN2 = 0.419458 [kmol N2/kg F]; so that mp = 16.2566 [kg P/kg F], np = 0.563388 [kmol P/ kg F], and Mp = 28.8551 [kg P/kmol P]. Also, wCO2 = 16.78% and xCO2 = 11.00%; wH2O = 6.64% and xH2O = 10.65%; wO2 = 4.32% and xO2 = 3.89%; and, wN2 = 72.26% and xN2 = 74.46%.] 4. A coal with composition of C = 87.0%, H = 4%, and the rest as ash is burnt with 130% theoretical air. For the complete combustion, calculate (i) the stoichiometric air-fuel ratio (Zs), (ii) the actual air-fuel ratio (Z), (iii) the composition of the products on mass basis (gravimetric) and mole basis (volumetric), and (iv) the mass and moles of the product gas per unit mass of the fuel. [Hints: The same procedure as Exercise 3. Then, Z s = 11.4783 [kg A/kg F], Z = 14.9218 [kg A/kg F], mp = 16.0694 [kg P/kg F], and, np = 0.535025 [kmol P/kg F]. Also wCO2 = 19.85% and xCO2 = 13.55%; wH2O = 2.24% and xH2O = 3.74%; wO2 = 6.41% and xO2 = 6.01%; and wN2 = 71.5% and xN2 = 76.70%.] 5. When a coal of gravimetric composition C = 0.87, H = 0.04, and rest ash is burnt, the composition of the products as measured by Orsat’s apparatus was: CO2 = 12.6%, O2 = 6.2% and the rest N2. Determine (a) the mass of the dry products per kg of fuel, and (b) the excess air factor.

Chapter 10: Thermodynamics of Combustion

6.

7.

8.

9.

299

[Hints: Using the linear mixing rule, the molecular weight of the dry products is obtained as MDP = 30.264 [kg DP/kmol DP]. Now, the mass of carbon in dry products is (0.126 × 12/30.2645) × mDP, where mDP is [kg DP/kg F]. By the balance of carbon atoms, this equals the carbon content of the fuel = 0.87 [kg C/kg F], from which mDP = 17.4138 ª 17.41 [kg DP/kg F]. The N2 balance gives (0.812 × 28/30.264) × 17.4138 = 13.0822 [kg N/kg F], and Z = 13.0822/0.77 = 16.99 [kg A/kg F]. From Exercise 4, Zs = 11.4783 [kg A/kg F] so that the excess air factor = 0.480184 = 48.02%.] A gaseous fuel has the following composition: H2 = 48%, CO = 11%, CH4 = 8%, and the rest as N2. It is burnt with stoichiometric air. Determine (a) the stoichiometric Zs, and (b) the composition of the flue gases by the Orsat’s apparatus. [Hints: Here, the calculations are done on mole basis. Then, nO2 = (0.48 × (1/2)) + (0.11 × (1/2)) + (0.08 × 2) = 0.455 [kmol O2/kmol F], and Zs = 0.455/0.21 = 2.16667 ª 2.167 [kmol A/kmol F]. The Orsat’s apparatus shows the composition of dry products, which is obtained as, nCO2 = 0.11 + 0.08 = 0.19 [kmol CO2/kmol F], and nN2 = 1.66836 [kmol N2/kmol F], so that xCO2 = 10.22% and xN2 = 89.78%.] Determine the ratio of carbon to hydrogen of the fuel whose Orsat’s analysis is: CO2 = 12.5%; CO = 2.5%; and, N2 = 85%. [Hints: It is assumed that the fuel contains only the hydrogen and carbon atoms. Now, since each mole of CO2 and CO is produced by 1 [kmol C], nC,DP = 0.125 + 0.025 = 0.15 [kmol C/kmol DP]. Then, since all the nitrogen and oxygen come from air, nO2 = 0.85 × (0.21/0.79), and since 2 [kmol H2] combines with 1 [kmol O2], nH2 = 1.7 × (0.21/0.79) = 0.451899 [kmol H2/kmol F]. Then, C : H :: 1 : 3, or the formula will be (CH3)n.] Propane (C3H8) reacts with air in such a ratio that an analysis of the products by the Orsat’s apparatus gives the following values: CO2 = 11.5%, O2 = 2.7%, and CO = 0.7%. What is the excess air factor? [Hints: By the usual formulae the molecular weights of fuel and dry products are obtained as Mf = 44 [kg F/kmol F], and MDP = 29.948 [kg DP/kmol DP]. Then, the carbon balance is (12 × 0.122/29.948) × mDP = (3 × 12)/(3 × 12 + 8 × 1), which gives mDP = 16.737 [kg DP/kg F]. And, mN = (0.851 × 28 × 16.737)/29.948, and Z = mN/0.77 = 17.2944 [kg A/kg F], or na = 17.2944 × (44/29) = 26.2398 ª 26.24 [kmol A/kmol F]. From the reaction scheme, C3H8 + 5O2 = 3CO2 + 4H2O, na,s = 5/0.21 = 23.8095, so that the excess air is 27.36%.] A fuel gas of molar composition C3H8 = 70% and C4H10 = 30% is burnt with 110% theoretical air. Compute the product composition as measured by the Orsat’s apparatus. [Hints: Now the reaction schemes are: C3H8 + 5O2 = 3CO2 + 4H2O and C4H10 + (13/2)O2 = 4CO2 + 5H2O. Since the fuel is a gas, the calculations are done on mole basis. Then, nO2,s = (0.7 × 5) + (0.3 × (13/2)) = 5.45 [kmol O2/kmol F], and na,s = 5.45/0.21 = 25.9524 [kmol A/kmol F].

300

10.

11.

12.

13. 14.

Engineering Thermodynamics

Since the Orsat’s apparatus shows composition of dry products, nCO2 = (0.7 × 3) + (0.3 × 4) = 3.3 [kmol CO2/kmol F]; and with 110% air, nO2 = 0.1 × 5.45 = 0.545 [kmol O2/kmol F]; and nN2 = 1.1 × 5.45 × (0.79/0.21) = 22.5526 [kmol N2/kmol F]. Then, xCO2 = 12.50%; xO2 = 2.06%; and, xN2 = 85.44%.] An unknown hydrocarbon fuel CxHy was allowed to react with air. An Orsat analysis was made of a representative sample of the product gases with the following results: CO2 = 12.1%, O2 = 3.8%, and CO = 0.9%. Determine the simplest formula of the fuel. [Hints: Same procedure as Example 10.9. The molecular weight of the dry products is MDP = 30.06 [kg DP/kmol DP]. Then, mC,DP = 12 × (0.121 + 0.009)/30.06 = 0.0518496 [kg C/kg DP], and nC,DP = 0.13 [kmol C/kmol DP]. Since oxygen comes from air, mO = nN2 × (0.21/0.79) × 32 = 7.06876 [kg O/kmol F]. Now, mO,CO2 = 0.121 × 44 × (32/44) = 3.872 [kg O/kmol DP]; and, mO,CO = 0.009 × 28 × (16/28) = 0.144 [kg O/kmol DP]; and free oxygen in dry products is, mO,DP = 0.038 × 32 = 1.216 [kg O/kmol DP]; all of which add up to 5.232 [kg O/kmol DP]. Then, oxygen in steam is mO,H2O = 7.06876 – 5.232 = 1.83676 [kg O/kmol DP]; = 1.83676 × (2/18) = 0.20408 [kg H/kmol DP]; or nH = and m H,H2O 0.20504 H/kmol DP]. Then, nH/nC = 0.20504/0.13 = 1.5772 ª 1.5 = 3/2, so that the simplest formula is C2H3.] Calculate the calorific value of the following fuels: (a) the coal with gravimetric (mass basis) composition of C = 74.4%, H = 12%, O = 2% and, the rest as ash; (b) coal with gravimetric composition of C = 0.87, H = 0.04, and the rest as ash; (c) producer gas of (volumetric) composition: H2 = 14%, CH4 = 3%, CO2 = 6%, O2 = 2%, and the rest as N2. [Hints: These calorific values are calculated straightaway as follows. (a) Qcal = (0.744 × 32.79) + (0.12 × 120.9) = 38.90376 = 38.904 [MJ/kg F]. (b) Qcal = (0.87 × 32.79) + (0.04 × 120.9) = 33.363 [MJ/kg F]. (c) Qcal = (0.14 × 241.8) + 0.03 × [(12/16) × 32.79] + (4/16) × 120.9 = 35.51 [MJ/kmol F].] Given that for benzene (DH)0C6H6 = 82.963 [MJ/kmol C6H6], calculate DHR and DER at standard conditions, if H2O appears in the products as steam. [Hints: The scheme for this reaction is C6H6(g) + 7½O2 = 3H2O(g) + 6CO2(g). Then, by definition, DHR0 = [6 × (– 393.5)] + [3 × (– 241.8)] – 82.963 = – 3169.363 = – 3169.0 [MJ/kmol C6H6], where the negative sign implies that the reaction is exothermic. Since these are ideal gas reactions taking place at constant temperature, the defining relation gives DE = D(H – pV) = DH – D(pV) = DH – D(nRT) = DH – RTDn = – 4408.0 [MJ/kmol C6H6].] The same as Exercise 12 except that, now H2O appears as water. [Hints: Since steam condenses, the enthalpy of vaporization is released. So add Hfg.] Neglecting dissociation, calculate the adiabatic flame temperature when methane is burnt with stoichiometric air. [Hints: The reaction scheme is CH4 + 2O2 + 2 × (0.79/0.21)N2 = CO2 + 2H2O + 2 × (0.79/0.21)N2, so that the energy equation gives 1 × HCO2 (Tad) + 2 × HH2O (Tad) + 2 × (0.79/0.21) × HN2 (Tad) = Qcal, where for methane, Qcal = (12 × 32.79) + (4 × 120.9) = 877.08 [MJ/kmol CH4].

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301

Solving this equation gives Tad = 2375 [K]. This high value is the result of neglecting the dissociations.] 15. Same as Exercise 14, but now 5% excess air is preheated to 300 [°C]. [Hints: The same procedure as of Exercise 14, but now, there is excess air to absorb the heat and excess enthalpy input as preheat. Now, the reaction scheme is CH4 + 2.1O2 + 7.9 N2 = CO2 + 2H2O + 0.1O2 + 7.9N2. Then, solving the energy equation with the appropriate terms gives, Tad = 2481 [K].] 16. Calculate the enthalpy carried away by an exhaust gas at 500 [K] if its composition is: CO2 = 12.32%, CO = 1.68%, O2 = 2.99%, N2 = 76.38%, and H2O = 6.63%. Assume that the reference temperature is 25 [°C]. [Hints: Taking the values of enthalpies from Table 10.7 of the text, by the linear mixing rule, DHp = 34.41 [MJ/kmol P].]

Chapter 11

Ideal Cycles

11.1 INTRODUCTION In thermodynamics, cycles are defined as a series of processes executed by a closed system1 such that the end state of the last process is the same as the initial state of the first one. In the conventional presentation of thermodynamics, the concept that cycles exist is used for the following purposes: Definition of properties. A property is one whose net change is zero over a cycle. However, in this book, the definition used is as follows: A property is one whose change between two states is independent of the process.

v∫

v∫

The Planck–Poincaré form of first law. It is stated as d W = dQ . However, the Born– Carathéodory formulation, used in this book, employs adiabatic processes. Kelvin–Planck and Clausius forms of the second law. These are stated for engines (which continuously convert heat into work) and for refrigerators or heat pumps (which continuously pump heat from a cold reservoir). It is the continuousness of the processes that requires the concept of cycles. In this book, these ideas have been retained since they give a better physical feel of the second law. However, the Born–Carathéodory formulation of the second law has also been presented to show that cycles are not really needed for this purpose. Identity of thermodynamic (Kelvin) and ideal gas temperatures. The Carnot cycle is used for this purpose. However, in this book, the equivalence of the thermodynamic (Kelvin) scale of temperature to the ideal gas temperature has been derived as a consequence of the Joule’s law (for ideal gases). All these arguments show that the concept of existence of cycles is not necessary for the development of thermodynamics. However, heat engines and heat pumps are the most important 1

Periodic (cyclic) variations of state of a fluid (e.g. sinusoidal heating and cooling) in a flow system do not form cycles since the same ‘particle’ does not stay in the system at all times. 302

Chapter 11:

Ideal Cycles

303

applications of thermodynamics. Moreover, as mentioned in Chapter 1, thermal engineering is one of the ancestors of thermodynamics. Hence, cycles are discussed in this and the following two chapters.

11.2 TYPES OF CYCLES For thermodynamic analysis, it is convenient to classify cycles on the basis of some thermodynamic criteria. The most obvious of these is the type of processes that constitute a cycle. For using this criterion, a definition similar to that of an nT engine employed in the second law analysis, namely the nP cycle —a cycle consisting of n processes—is needed. The common processes employed in cycles of interest in thermal engineering are: Isochoric. This is the constant volume process which is the idealization of processes taking place instantaneously (e.g. combustion in S.I. engines). Isobaric. This is the constant pressure process which is the idealization of flow processes or combustion in C.I. engines. Isothermal.

This is the constant temperature process which implies perfect heat transfer.

Isentropic.

This idealizes the perfectly insulated process.

Unless specifically stated otherwise, all the above processes are assumed quasi-static and reversible. The most important working fluid of thermal systems is the ideal gas. Figure 11.1 shows the above processes for an ideal gas. For ease of comprehension, they are shown as the special cases of the polytropic process, pV n = C. p

p1

2

1

Legends 1–2: Isobaric process (n = 0) 1–3: Isothermal process (n = 1) 1–4: Polytropic process (n = n) 1–5: Isentropic process (n = g)

p2

1–6: Isochoric process (n = •)

6

5

4

3

V Figure 11.1

Common processes of ideal gas.

304

Engineering Thermodynamics

By logarithmic differentiation, it is easy to see that the slope of the polytropic process in the p–V plane is, dp/dV = – n(p/V), which implies that the curves get steeper as n increases. The above figure shows this. It is also easy to see that the slope of isobars in the h–s plane diverge, since dh = Tds + vdp implies that (∂h/∂s)p = T. Similarly, in the T–s plane, the isochors are steeper than the isobars. This is evident from the following arguments. The definition, cp = T × (∂s/∂T)p gives (∂T/∂s)p = (T/cp). Similarly, the definition, cv = T × (∂s/∂T)v gives (∂T/∂s)v = (T/cv). Since cp > cv, these definitions show that (∂T/∂s)p < (∂T/∂s)v, which is the required result. It is convenient to discuss different cycles using these processes as the criteria. This is done in the following paragraphs. 1P cycles. It will be noticed that the above processes are monotonic and single valued, which means that they cannot intersect themselves. This implies that in spite of their extensive theoretical use, the 1P cycles are impossible to construct. 2P cycles. The above processes are single valued and monotonic. Then, geometry asserts that they can intersect only at one point. In other words, the 2P cycles cannot be constructed. It is common to draw a 2P cycle as shown in Figure 11.2. It is easy to show that both the processes are shown identical since, m = n and CA = CB (see Exercise 1 at the end of this chapter). p 1

Legends

B

n

1–A–2: Polytropic process pV = CA m

1–B–2: Polytropic process pV = CB A

2

V

Figure 11.2

A typical two process cycle.

3P cycles. The number of distinct 3P cycles can be calculated as follows. The aim is to choose three processes from the four. Labelling of any process as first is arbitrary because together all the processes form a closed curve. Then, this number = 4C3 = 4, where 4C3 is the number of combinations of three things taken at a time out of four. (see Exercise 2 at the end of this chapter for another approach). Since the number of distinct 3P cycles are four, they can easily be enumerated. This will help to clarify the logic of the approach.

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305

Consider the following symbolism. Let the alphabets be written sequentially to indicate the cycle. For example, the sequence, pVS, indicates a 3P cycle with one isobar (constant pressure, p = C), one isochor (constant volume, V = C) and one isentrope (constant entropy, S = C), with the isobar as the first process. Note that the cycle VSp only means that the first process is now the isochor instead of the isobar. Similarly, the cycle VpS is the pVS cycle run in the reverse direction. Hence, as far as thermodynamics is concerned, they are all the same. In other words, combinations and not permutations are important. The other distinct 3P cycles are pVT, pTS and VTS. It is difficult to execute 3P cycles in closed systems (i.e. as I.C. engines, in cylinder–piston assemblies), since, in the conventional design, it requires an odd number of strokes for completion of the cycle. Thus, the completion of the thermodynamic cycle and the cycle of operation of the mechanical parts of the engine become asynchronous. Moreover, there have been very few attempts to execute 3P cycles in flow (open) systems. 4P cycles. These are the most common cycles used in thermal engineering. The possible distinct 4P cycles of different types can be calculated as shown above (see Exercise 4 at the end of this chapter), but the following are the only ones used in practice. Cycles using 2 distinct processes. There are five of these since with a choice of the first process, there are only three processes remaining; and then, for another chosen as the last process, only two distinct processes are possible. The five cycles are: STST (the Carnot cycle); SVSV (the Otto cycle); SpSp (the Brayton, Joule, Rankine and vapour compression refrigeration cycles); TVTV (the Stirling cycle); and TpTp (the Ericsson cycle). Cycles using 3 distinct processes. SpSv (the Diesel cycle) is used.

Out of all the possible cycles of this type, only the

Cycles using 4 distinct processes. None in this class is known to be in use. 5P cycles.

The only one in use is the SVpSV (the dual-combustion) cycle.

Section 11.3 deals with the basic methods of cycle analysis. Sections 11.4 and 11.5 present the reversible cycles that are the ideal cycles. The chapter closes with a discussion of the ideal and real cycles in Section 11.6. Chapter 12 deals with cycles that are generally executed in closed systems,2 i.e. in the cylinder–piston assembly. Since these cycles were developed for running internal combustion engines, they are also called the I.C. engine cycles. They are the Otto, the Diesel and the dualcombustion cycles. Since their basic work mode is the expansion (i.e. pdV) work, the p–V coordinates are the natural ones for representing these cycles. However, since the net work done in a cycle is the difference between the heat absorbed and the heat rejected, and since heat absorbed and the rejected can be represented as areas of curves in the T–S plane,3 these cycles are also represented on the T–S plane. Chapter 13 presents cycles in which each process is executed in one flow subsystem. Then, the overall system is obtained by interconnecting these subsystems. This strategy was 2 3

It will be recalled that, theoretically, only the closed systems can execute cycles. Of course, all the processes are assumed to be reversible.

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developed for steam and gas turbine power plants used for electrical power generation and later adopted for many other systems including gas turbine plants. Basically, they are the Brayton cycle (i.e. the SpSp cycle) in different forms. Brayton cycle for gas turbines and the Rankine cycle (the Brayton cycle for a vapour) for steam turbines are first discussed there. Since the work done and heat exchanged by flow systems are expressed in terms of changes of enthalpy, the h–s coordinates are the natural choice for representing these cycles. Chapter 13 also includes the cycles that are executed in the reverse direction. They are used for refrigeration and heat pumping. The Joule cycle (reversed Brayton cycle) and the vapour-compression cycle (reversed Rankine cycle) are explained there. Since these applications deal mainly with heat exchanged, the T–s coordinate system is a natural choice. Moreover, since the component are flow systems, the p–h coordinate system is also a convenient choice.

11.3 CYCLE CALCULATIONS In this section different methods for calculating the cycle parameters are discussed. Since the emphasis is on explaining the methodology, a simple 3P cycle shown in Figure 11.3 is used. p

p2

2 Legends 1–2: Isochoric process 2–3: Isentropic process 3–1: Isobaric process

p1

3

1

V Figure 11.3

A three process (3P) cycle.

The default assumptions in cycle analysis are that: (a) the working fluid is a simple compressible substance, (b) the only work mode is Wx, the expansion work (shaft work for flow systems), and (c) all the processes are quasi-static and reversible. The basic method All the procedures are based on the following facts. 1. For each process i of the cycle, the energy (first law) equation becomes Qi = (DU)i + Wi.

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2. Since a cycle consists of a series of processes such that the end state of the last process is the same as the initial state of the first process, the energy equation for the whole cycle becomes N

N

∑

∑

Qi =

N

( DU )i +

i =1

i =1

∑W

i

i =1

where N is the total number of processes in the cycle. N

3. Since U is a property,

∑ ( DU )

i

= 0. This provides a check that the equations are

i =1

formulated properly. Since, the above methodology is based on the energy conservation principle, it is also valid for flow systems. The only difference is that their energy equation4 of the ith process becomes Qi = He – Hi, while the (shaft) work done is given by Wi = Hi – He. It will be recalled from the discussions on flow systems that, generally, a subsystem is either heat exchanging (combustors, boilers, condensers, heat exchangers, …) or work exchanging (turbines, pumps, …), so that the above form of the energy equation is adequate. However, if some subsystems exchange both heat and work, then, the energy equation becomes Qi – Wi = He – Hi. This procedure is now illustrated for the cycle in Figure 11.3. It is convenient to do the calculations in tabular form. Type

DU

W

Q

1–2

V=C

mcv(T2 – T1)

0

mcv(T2 – T1)

2–3

S=C

mcv(T3 – T2)

(p2V2 – p3V3)/(g – 1)

0

3–1

p=C

mcv(T1 – T3)

p1(V1 – V3)

mcp(T1 – T3)

Process

Remember:

The operator Dx = xfinal – xinitial (includes the sign).

Verification of the formulation The equations formulated in the table above are verified as follows: 1. Since U is a property, (DU) for the cycle should be zero. This means that the sum of the third column of this table should be zero. This is seen to be true. 2. The energy (first law) equation for each process is Q = DU + W. This implies that the sum of the third and fourth columns of each row should equal the fifth column. This is clear for the first row. For the second row, this implies that mcv (T3 – T2) + or mcv (T3 – T2) + 4

p2V2 − p3V3 =0 g −1 mRT2 − mRT3 =0 g −1

Of course, it is assumed that the system is operating at steady state and steady flow with Ek = 0 = Ep.

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which reduces to cv = R/(g – 1). Since, this is the relation for ideal gases which is the working fluid, this shows that the second row is properly formulated. Similarly, the third row can be shown to give cp = cv + R, which shows that the third row is also properly formulated. 3. The energy equation for the cycle dQ = d W , which implies that sum of the fourth column equals that of the fifth. From the above table, it is seen that this gives an identity.

v∫

v∫

These results show that the above equations are correctly formulated. Now, all the cycle parameters can be easily calculated. This is done in the next section. The main advantage of this method is that it can be applied to any cycle. Consequently, this method should be learnt thoroughly and memorized. However, in many cases, such detailed analysis is not required. For such cases this method can be simplified as given below. A simplified approach In many cases, only the energy fluxes, that is, Q1, the heat absorbed, W, the work done and Q2, the heat rejected are required. In such cases, the above procedure is cumbersome. For these cases, the following approach is useful. Step I. Identify the processes of heat absorption and heat rejection. Remember that, the heat is absorbed during higher temperature processes. In this case, since 2–3 is an isentropic process, heat should be absorbed and rejected only during processes 1–2 and 3–1. Out of these, since process 1–2 takes places at higher temperatures, heat is absorbed in it and the value of heat absorbed is Q1 = mcv (T2 – T1). Similar argument shows that the heat rejected during the process 3–1 is Q2 = mcp (T3 – T1), where the sign of heat transfer is excluded. Step II. Calculate the work done. By the energy (first law) equation, this is W = Q1 – Q2. In the present case, writing this expression is not worth it, since it does not give any further insight. Step III. Calculate the cycle efficiency. Using the definition of cycle efficiency and those of the heat fluxes and the energy (first law) equation gives h = W/Q1 = 1 – (Q2/Q1). Note that the sign of heat rejected is already included in the second relation. For the present case, this reduces to

h=1–

c p × (T3 − T1 ) Q2 g × (T3 − T1 ) =1– =1– Q1 T2 − T1 cv × (T2 − T1 )

Step IV. Eliminate the unknown temperatures. Now, the unknown temperatures are eliminated in terms of the known (given) one using the process relations. In the case of vapour (e.g. steam, refrigerants, …), appropriate tables are used for this purpose. Generally, the temperature ratios can be expressed in terms of ratios of some parameters of the cycle, which determine its geometry in the state space. In this case, the cycle can be completely drawn, if the initial point ‘1’ together with the ratios of V3/V1 and p2/p1 is given. These ratios are called the compression ratio and the pressure

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ratio, respectively. They are denoted, respectively, as rv and rp. Then, T2 and T3 are expressed in terms of T1, rv and rp as follows. Since the working fluid is ideal gas for the isobar 1–2, the equation of the process becomes ⎛p ⎞ T2 = T1 × ⎜ 2 ⎟ = T1 × rp ⎝ p1 ⎠

and

⎛V ⎞ T3 = T1 × ⎜ 3 ⎟ = T1 × rv ⎝ V1 ⎠

Step V. Express the cycle efficiency in terms of cycle parameters. For this cycle, the parameters are the compression ratio, defined as rv = V3/V1 and the pressure ratio, defined as rp = p2/p1. In terms of these, the equation for cycle efficiency given in step III reduces to

h=1–

g × (T3 − T1 ) T2 − T1

=1–

g × (rv − 1) rp − 1

Step VI. Verify the calculation of h. Using the basic definition of efficiency, W = h × Q1, We verify, independently, that the derivation of the efficiency equation is correct. From the geometry of the processes in the state space, it is clear that the compression and pressure ratios are not independent for this cycle, because, starting with the point ‘3’, knowing rv, point ‘1’ can be reached. Now, point ‘2’ is the intersection of the vertical line (isochor) from ‘1’ and the isentropic from ‘3’. Mathematically, this can be shown as follows. Since 2–3 is an isentrope, p2V2g = p3V3g; or, since p3 = p1 and V2 = V1, g

⎛V ⎞ p2 = ⎜ 3⎟ p3 ⎝ V2 ⎠

g

or

⎛V ⎞ p2 = ⎜ 3⎟ p1 ⎝ V1 ⎠

or

rp = (rv)g

11.4 CYCLE AND PERFORMANCE PARAMETERS Engines are compared on the basis of their performance. The performance of an engine depends on the cycle on which it works as well as other parameters. In this section, the broad class of these parameters is discussed. Those parameters that are applicable to particular cycles (and engines) are explained in later chapters during discussions on them. The general aim of cycle calculations is to estimate these parameters.

11.4.1

Cycle Parameters

These determine the geometry of the cycle in the state space. The compression ratio rv and the pressure ratio rp are examples of these. However, as shown in the above section, all of these ratios may not be independent. In addition, the following are also important cycle parameters. Indicated work (IW). It was mentioned in earlier chapters that, the work done in a cycle equals the area of the p–V diagram and that this used to be measured by a device called the indicator.

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Thus, the work obtained by measuring the area of the p–V diagram by an indicator is called the indicated work. Thus, IW =

v∫ pdV .

Indicated mean effective pressure (IMEP, pi,m). As the name indicates, this is the constant average pressure that, by acting throughout the cycle, produces the same work. In the cycle shown in Figure 11.3, V1 is the clearance volume, Vcl, and V3 is the cylinder volume, Vcyl. Then, Vcyl – Vcl is the volume swept by the piston in one stroke. Hence, it is called the stroke or swept volume and is denoted as Vs. If IW is the work done in one cycle, then the indicated mean effective pressure, denoted as pi,m, is defined by the equation pi,m = Since the work done is the area enclosed by the cycle, the pi,m will be the value of the ordinate projected on the stroke volume which will enclose the same area as the cycle. The term, brake mean effective pressure is explained later. This pressure is an important indicator, since it is the average pressure to which the cylinder is subjected and, therefore, the mechanical design is based on it. If A is the area of the cylinder (and of course the piston) and L is the length of stroke, then the stroke volume is Vs = A × L. Combining this with the above quantities, it is easy to see that IW = pi,m × Vs = pi,m × A × L IW IW = Air standard efficiency (has). One of the mostV important fluids of engines (which Vcyl − Vworking s cl includes turbines) is air. At states encountered in most engines, air behaves like an ideal gas. Thus, W, Q1, Q2, … can be calculated assuming that air is the only working fluid in a cycle and that it behaves like an ideal gas with constant specific heats. The objective air standard refers to these quantities. For example, in the cycle analysis of Section 11.3, if the ideal gas can be taken as air, then, the above cycle is called an air standard cycle; and the efficiency calculated above is called the air standard efficiency (denoted as has). Generally (i.e. unless specified), all gas cycles are assumed to be air standard cycles.

11.4.2

Performance Parameters

These correspond to the parameters that are determined by the operating conditions. The same engine may run at different conditions. For example, a petrol engine of a car runs at different speeds depending upon the road conditions. Thus the values of the performance parameters depend on the operating conditions, while the cycle parameters are fixed once the cycle is designed. Plant efficiency (hp). This term is applied to the actual plant (engine and turbine). It is obtained from the relations

hp =

W (by definition) Q1

and

hp = 1 –

Q2 (by energy (first law) equation) Q1

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where, W, Q1 and Q2 are the values obtained by measurements on the plant. Alternatively, the plant efficiency can also be calculated taking into account the actual state of the working fluid at different points in the cycle (if appropriate data is available). Indicated power (IP). This is calculated as IP = IW × Nc, where Nc is the number of cycles executed per unit time (generally one second). Friction power (FP). This is the power spent in overcoming the friction in the cylinder– piston contact area, bearings, etc. Generally, this cannot be measured directly.5 So, it is obtained from the measurements of indicated and brake powers. Brake power (BP). This is the power available at the shaft of an engine to drive a load. It is used to be measured by braking the engine. Hence this name. It can be measured by dynamometers, rope brakes, etc. It can also be calculated from the measured values of IP and FP as BP = IP – FP; or BW = IW – FW, where BW and FW denote the brake work and friction work, respectively. Brake mean effective pressure (BMEP, pb,m). The definition of this term is similar to that of the indicated mean effective pressure, pi,m. Denoting it as pb,m gives BW BW = Vs Vcyl − Vcl Q1 volumes, respectively. where Vcyl and Vcl are the cylinder and clearance Employing the same arguments as used for relating the indicated mean effective pressure to the indicated power gives the relation

Pb,m =

BW = pb,m × Vs = pb,m × A × L where A is the area of the cylinder and L is the length of the stroke. Mechanical efficiency (hm). This is defined as, hm = BP/IP = BW/IW. Indicated thermal efficiency (hi,th). This is defined as, hi,th = IW/Q1 = IP/ Q1 . Brake thermal efficiency (hb,th). This is defined as, hb,th = BW/Q1 = BP/

.

Note that the relation between these efficiencies is hb,th = hi,th × hm. Relative efficiency (hr). This is defined in two different ways: The first definition is hr = hth/hR, where hR is the efficiency of a reversible (Carnot) engine. This is similar to one used in refrigeration, but is not widely used in thermal power engineering. An alternate definition, which is much more useful is, hr = hth/has, where has is the air standard efficiency. Specific fuel consumption (sfc). As the name indicates, this is defined as the mass of fuel required to produce unit power output. Its unit is [kg F/kWh]. Depending upon the choice of 5

The motoring test of an I.C. engine is excluded.

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the power used in calculations, there are two ways of defining this parameter as indicated below: isfc = mf /IW =

/IP

and bsfc = mf /BW =

/BP

where isfc and bsfc denote the indicated and brake specific fuel consumptions, respectively, and mf is the fuel consumed. These parameters are generally used for internal combustion (I.C.) engines. For turbines, the following are the appropriate parameters. Specific output. This is defined as the power output per unit mass of the working fluid flowing through the system. It is mainly used in turbine plants. Its unit is [kJ/kg of working fluid]. Steam rate (SR). As the name indicates, this is used in steam turbine plants. It is the rate of mass of steam needed to produce unit power. Its unit is [kg S/kWeh], where ‘kWe’ indicates kilowatt of electricity. Notice that this is the inverse (in a different unit) of the specific output. Heat rate (HR). This is defined as the rate of heat input required to produce unit power, so that HR = Q1/SP, where SP denotes ‘shaft power’. Its unit is [kJ/kWeh]. Note that this parameter is similar to the specific fuel consumptions of I.C. engines. It is also related to the above two parameters. When the cycles are used in the reverse direction for operating heat pumps or refrigerators, the following performance parameters are applicable. 2000 × 144removed6 from the cold reservoir (or the f heat The refrigerating effect. This is defined asm room, or refrigerated space, …). The unit of this is 24 called the ton of refrigeration. Originally, it was defined as the heat removed in 24 [h] to produce 1 [ton] [the short (or American) ton] of ice at atmospheric pressure and 32 [°F] from water at the same conditions. An American ton is defined as equal to 2000 [lb]. At the time when this unit was defined, the latent heat of ice was taken to be 144 [Btu/lb]. Thus, 1 [TR] =

= 12,000 [Btu/h] = 200 [Btu/min]

where [TR] denotes ton of refrigeration effect. Even though this is the primary definition, in metric system this becomes, 1 [TR] = 200 [Btu/min] × 0.252 [kcal/Btu] = 50.4 [kcal/min] Thus, in metric units, 1 [TR] has been defined as equal to 50 [kcal/min]. To eight digit accuracy, this gives 1 [TR] = 3.5168667 [kW]. However, even if this is taken as 1 [TR] = 3.52 [kW], it gives a maximum error of 0.1 per cent, which is acceptable in most cases. For more accurate calculations, the original definition of 1 ton (= 200 [Btu/min]) should be used. Coefficient of performance of refrigerators [(COP)ref]. The main function of a refrigerator 6

Some authors use the term “cold produced”: but for obvious reasons, this term is not used since it is unscientific.

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313

is to extract heat from a cold reservoir (room, inside of a domestic refrigerator, …). Hence, for refrigerators, the COP is defined as COP ref = =

(by definition) Q2 [by energy (first law) equation] Q1 − Q2

where Q2 is the heat pumped out of cold reservoir, W is the work absorbed by the machine, and Q1 is the heat rejected to the hot reservoir. Coefficient of performance of heat pumps [(COP)hp]. Heat pumps are used to pump heat to the hot reservoir (e.g. cold room during winter in cold climates). Hence, for heat pumps, the COP is defined as

Q1 (by definition) W Q1 = [by energy (first law) equation] Q1 − Q2

COP hp =

where Q2 is the heat pumped out of the cold reservoir, W is the work absorbed by the machine and Q1 is the heat rejected to the hot reservoir. Note that this parameter is the inverse of the efficiency when an engine operates on the cycle. 2 Relative COP [COPr]. This is defined as Q COP r = COP/COPR, where COPR is the COP W of a reversible (Carnot) heat pump or refrigerator.

11.5 THE CARNOT CYCLE In this section, the ideal cycles will be discussed with emphasis on the Carnot cycle, the most well-known of them. It logically follows from the Carnot (2T) engine as follows: 1. Since it is a reversible engine, all its processes are reversible. 2. It absorbs heat from a hot reservoir (at T1) and rejects heat to a cold reservoir (at T2). 3. Hence, the cycle should have two reversible isothermals, one at T1 for heat absorption and other at T2 for heat rejection. 4. Since two different isothermals cannot intersect, some other reversible processes should connect them. 5. Since there can be no heat exchange during these two processes, they should be isentropics (i.e. reversible adiabatics). 6. Then, only heat is exchanged in two of the four processes (i.e. in the isothermals) and only work is exchanged in the other two (i.e. the isentropics). So, it is an STST cycle. 7. Thus, the Carnot cycle should consist of two isothermals and two isentropics. 8. Hence, in the T–S plane, the Carnot cycle is a rectangle.

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This result depends only upon the geometry of the state space and the processes and is independent of the working substance of the Carnot engine. Figure 11.4 shows the Carnot cycle with ideal gas as well as a vapour (say, steam) as the working fluid. Notice that they are on different state spaces. The legends in this figure mention the nature of the processes. Note that the initial point ‘1’ is the beginning of the isentropic compression. This convention, generally, is followed for all cycles. With vapour as the working substance, heat absorption is evaporation and heat rejection is condensation of the vapour. T

p 2

T1

2

3

3

1

T2

1

4

4

S1 S2 (a) All working substances

V

S (b) Ideal gas cycle

T Liquid

T1

Superheated vapour

Wet vapour

2

3

Legends 1–2: Isentropic compression 2–3: Isothermal heat absorption

T2

1

3–4: Isentropic expansion

4

S1

4–1: Isothermal heat rejection

S2

S

(c) Vapour cycle Figure 11.4

The Carnot cycle.

The tabular form below shows the Carnot cycle with some other working substances. Working substance

Coordinates

S1 Isentropic

Ideal gas Vapour Elec. cell Paramag.

p–V T–S E–Q M–H

Compression Compression Charging Magnetizing

T1 Isothermal Heat Absorption Evaporation Discharging Demagnetizing

S2 Isentropic Expansion Expansion Discharging Demagnetizing

E = EMF; Q = Charge; M = Magnetization; and H = Magnetic Field Strength.

T2 Isothermal Heat Rejection Condensation Charging Magnetizing

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It should be noted that these processes indicate that the Carnot cycle runs clockwise for gases and vapours and counterclockwise for electrical and magnetic systems. These characteristics are due to the nature of the working substances. Since the Carnot cycle is reversible, all its processes are also reversible. Then, by the definitions of the thermodynamic temperature, T, and of entropy, S, the heat absorbed, Q1 and the heat rejected, Q2 may be written as Q1 = T1 × (S1 – S2)

and

Q2 = T2 × (S1 – S2).

Then, the energy (first law) equation for a cycle gives, W = Q1 – Q2. The above equations show that the examples on the Carnot cycle is exactly the same as those on Carnot engines, except that, for the cycle calculations the nature of the working substance should also be specified. The following examples illustrate this aspect. EXAMPLE 11.1 A Carnot cycle operates between reservoirs at the normal boiling point of water and its triple point. If it produces 100 [kJ] of work, calculate (a) h (b) Q1 and (c) DS. Solution 1. In this case, T1 = 100 [°C] = 373.15 [K] and T2 = 273.16 [K]. 2. Hence, the thermal efficiency is, h = 1 – (273.26/373.15) = 0.268. 3. By the definition of h, Q1 = W/h = 100/0.268 = 373.13 [kJ]. 4. From Figure 11.4(a), DS = S2 – S1 = Q1/T1 = Q2/T2 = 373.13/373.15 = 1.0 [kJ/kg]. EXAMPLE 11.2 The definition of the efficiency of a Carnot engine, h = 1 – (T2/T1), shows that it can be increased either by increasing the high temperature, T1, or by decreasing the low temperature, T2. Which is the better way? Solution This example will be first solved analytically, which corresponds to the engine. The result is then explained with the aid of the T–S diagram of the Carnot cycle. 1. Partial differentiation of the efficiency equation with respect to T1 gives,

⎛ T2 ⎞ ⎛ ∂h ⎞ ⎜ ⎟ = ⎜⎜ 2 ⎟⎟ ⎝ ∂T1 ⎠T2 ⎝ T1 ⎠ 2. Similarly, partial differentiation with respect to T2 gives, ⎛ ∂h ⎞ ⎛1⎞ ⎜ ⎟ =– ⎜ ⎟ ⎝ ∂T2 ⎠T1 ⎝ T1 ⎠ where the negative sign shows that h increases when T2 decreases. 3. Thus,

( ∂h / ∂T1 )T2 ( ∂h / ∂T2 )T1

=

T2 T12

× (–T1) = –

T2 T1

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4. This expression shows that, since T1 > T2, h increases more slowly with increase in T1. 5. To understand the physical significance of this result, note that an increasing T1 shifts this isotherm upwards in the T–S diagram, thereby increasing the area of the rectangle (corresponding to the work done). In other words, any increase in T1 increases W, but more Q1 is also absorbed, so that their ratio changes slowly. 6. On the other hand, a decrease in T2 shifts the lower isotherm in the T–S diagram downwards, indicating that more work is obtained by rejecting less heat to the cold reservoir. Thus, this is the better alternative. The following two illustrative examples have fundamental values. EXAMPLE 11.3 Obtain a relation for the efficiency of the Carnot cycle with an ideal gas as the working substance [see Figure 11.4(b)]. Solution 1. For clarity, let q denote the ideal gas temperature, i.e. q = pv/R; or, pv = Rq. 2. To avoid confusion of subscripts, let qH and qC be the temperatures of the hot and cold reservoirs, respectively. 3. Then, the cycle efficiency is h = 1 – (Q2/Q1). 4. Since the processes of heat absorption and heat rejection are isothermals, for an ideal gas the relation for cycle efficiency reduces to

h=1–

mRqC ln(V4 / V1 ) mRq H ln(V3 / V2 )

5. Since 1–2 is an isentropic process and the working substance is an ideal gas 1/(g −1)

⎛q ⎞ V1 = ⎜ 2⎟ V2 ⎝ q1 ⎠ 6. Similarly, since 3–4 is isentrope of an ideal gas

q1V1g –1 = q2V2g –1 or

q3V3g –1 = q4V4g –1 or

1/(g −1)

⎛q ⎞ V4 = ⎜ 3⎟ V3 ⎝ q4 ⎠

1/(g −1)

⎛q ⎞ = ⎜ H⎟ ⎝ qC ⎠

1/(g −1)

⎛q ⎞ = ⎜ H⎟ ⎝ qC ⎠

7. These two relations show that V V1 = 4 V3 V2

or

V4 V = 3 V1 V2

8. Substituting this relation in the equation for cycle efficiency (step 4) and cancelling the appropriate terms reduces it to, h = 1 – (qC/qH). 9. During the discussions on the second law of thermodynamics, it was shown that the efficiency of a Carnot engine working between the reservoirs TH and TC (TH > TC) is, h = 1 – (TC/TH).

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317

10. Comparing these two relations for efficiency, it is seen that the ideal gas temperature is identical to the thermodynamic (Kelvin) temperature. During the discussion on the thermodynamic scale of temperature, this was shown to be the consequence of the Joule’s law. However, most books on thermodynamics follow the above arguments to establish identity of these two scales. EXAMPLE 11.4 As an interesting example of the generality of the Carnot cycle (and, Carnot engine), consider a perfectly evacuated closed system with perfectly reflecting walls containing thermal radiation with energy density (radiant energy per unit volume), u. Then, if V is the volume of the system, the total radiant energy is U = u × V. 1. Consider the thermal radiation as the ‘working substance’. 2. The classical (Maxwell’s) electromagnetic theory shows that this radiation field exerts a pressure, p = u/3. 3. Consider an infinitesimal Carnot cycle executed by the radiation field. 4. The heat absorbed during the isothermal process is

4u dV d−Q = dU + d−W = udV + pdV = (u + p)dV = 3 5. During the infinitesimal isentropic expansion, let the temperature fall by dT and let the radiation pressure reduce by dp. 6. Let the cycle be completed by another set of infinitesimal isotherm and isentrope. 7. Then, the net work done in the infinitesimal cycle is

du × dV d−W = dp × dV = 3 8. Then, the cycle efficiency is

h =

dW du (du × dV/3) dT = = = dQ 4×u T (4u × dV/3)

where the last step follows from the definition of the Kelvin scale of temperature. 9. Integrating the last equation gives, u = s T 4, where the integration constant is written as s. 10. This law is called the Stefan-Boltzmann law of thermal radiation and was first derived as shown above by Boltzmann.

11.5.1

The Stirling and Ericsson Cycles

Substituting the reversible isentropic of the Carnot cycle by other reversible processes, some other reversible cycles are obtained. Conceptually, since the Carnot cycle is an STST cycle, other cycles that can be generated from this by replacing the isentropics are VTVT and pTpT

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cycles. These are called the Stirling cycle and the Ericsson cycle, respectively. They are shown in Figure 11.5. p p 3

3

2

TH = constant

2

TH = constant

TC = constant

4 TC = constant

1

4

1 V (a) Stirling cycle

V (b) Ericsson cycle

Figure 11.5

The Stirling and Ericsson cycles.

It should be noted that the isothermal processes are reversible. Now, if the isochorics of the Stirling cycle and the isobarics of the Ericsson cycle are made reversible, then these cycles also become reversible. Stirling made the isochorics reversible by regeneration. In this, during the isochoric expansion, 4–1, the hot gases pass through a regenerator and give up their heat to the matrix of the regenerator. During the isochoric compression, 2–3, the cold gas passes through the regenerator and picks up this heat from the regenerator matrix. Since the regenerator is part of the engine, these heat exchanges remain within the system. Thus, with ideal regeneration, the only heat exchanged with external sources takes place during the two isothermals. Then, it can easily be shown that the cycle efficiency becomes 1 – (TC/TH). Similar arguments show that with ideal regeneration, the Ericsson cycle also becomes reversible with efficiency equal to 1 – (TC/TH). The following example illustrates the Stirling cycle calculations. EXAMPLE 11.5 The inlet air to a Stirling cycle is at 1 [bar], 300 [K]. It is compressed isothermally and then heated at constant volume till the pressure and temperature reach 10 [bar] and 600 [K]. Calculate (a) has, (b) W, and (c) Q1. Solution

The Stirling cycle is shown in Figure 11.5(a).

1. Since the isochoric processes of regeneration are internal to the system, heat is exchanged only during the isothermals. 2. And the heat exchanged during isothermals of an ideal gas are ⎛V ⎞ Q1 = mRTH ln ⎜ 4 ⎟ ⎝ V3 ⎠

and

⎛V ⎞ Q2 = mRTC ln ⎜ 1 ⎟ ⎝ V2 ⎠

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319

3. Defining the compression ratio as in Otto cycle, rv = V1/V2 = V4/V3, the above relations become Q1 = mRTH ln (rv) and Q2 = mRTC ln (rv) 4. The compression ratio is determined from the equation of state of ideal gas as follows.

V1 ⎛ p ⎞ ⎛T ⎞ ⎛ 300 ⎞ 10 = ⎜ 3 ⎟ × ⎜ 1 ⎟ = ⎛⎜ ⎞⎟ × ⎜ ⎟ = 5.0 T V3 ⎝ 1 ⎠ ⎝ 600 ⎠ ⎝ p1 ⎠ ⎝ 3 ⎠ 5. Then, the air standard efficiency is p1V1 pV = 3 3 T1 T3

or

has = 1 –

T 300 Q2 =1– C =1– = 0.5 600 TH Q1

6. The work done is ⎛ 8.3143 ⎞ W = Q1 – Q2 = mR(TH – TC) ln rv = 1 × ⎜ × 300 × ln (5) ⎝ 29 ⎟⎠ ⎡ kJ ⎤ = 138.428 = 138.4 ⎢ ⎥ ⎣ kg A ⎦

7. Then, the heat input is Q1 =

11.5.2

W

has

=

⎡ kJ ⎤ 138.428 = 276.856 = 276.8 ⎢ ⎥ 0.5 ⎣ kg A ⎦

The Reversed Carnot Cycle

This is the cycle on which the Carnot refrigerator or heat pump operates. This Carnot cycle run in the reverse direction and Figure 11.6 shows this cycle. T

T1

3

2 Legends

1–2: Isentropic compression 2–3: Isothermal heat rejection 3–4: Isentropic expansion 4–1: Isothemal heat absorption T2

4

1 S2

S1 Figure 11.6

S

The reversed Carnot cycle.

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Engineering Thermodynamics

The legends in the figure explain the nature of the processes. Hot refrigerant at low pressure enters the compressor at the state ‘1’. It is then isentropically compressed to high pressure and high temperature. External fluids like atmospheric air or water then cool it. The cold refrigerant is expanded isentropically to low pressures. Evidently, this expansion also cools it. This cold refrigerant is then passed through the space to be cooled. Then, the refrigerant picks up heat from the space and gets warmer. The cycle is then repeated. From the above figure it is clear that: 1. The heat absorbed from the refrigerated space is, Q2 = T2 × (S2 – S1). 2. The heat rejected to atmosphere is, Q1 = T1 × (S2 – S1). 3. Then, the coefficient of performance is COPref =

Q2 Q2 T2 = = W Q1 − Q2 T1 − T2

This relation was obtained earlier as the consequence of the definition of thermodynamic (Kelvin) scale of temperature. The following examples illustrate some typical cases. EXAMPLE 11.6 What will be the maximum COP of a refrigerator that freezes water when the ambient temperature is 27 [°C]? Solution The solution is direct. The COP will be maximum when the refrigerator is reversible. Then, its COPref = T2/(T1 – T2) = 273.15/(300.15 – 273.15) = 10.12. EXAMPLE 11.7 A refrigerator removes 10 [kW] of heat from a body to maintain its temperature at – 10 [°C] and rejects the heat to atmosphere at 30 [°C]. If its relative COP is 0.75, calculate (a) the refrigerating effect in [TR], (b) the power input, and (c) the heat rejected. Solution One ton of refrigerating effect is defined as, 1 [TR] = 3.52 [kW]. Then: 1. The refrigerating effect is

⎛ 1 ⎡ TR ⎤⎞ Q2 = (10 [kW]) × ⎜ = 2.841 [TR] ⎝ 3.52 ⎢⎣ kW ⎥⎦⎟⎠

[Ans. (a)]

2. The definition of relative COP gives ⎛ T2 ⎞ 263.15 = 6.579 COPref = COPr × COPR = COPr × ⎜ ⎟ = 0.75 × 40 ⎝ T1 − T2 ⎠

3. Hence, by the definition of COP, the power input, is W=

10 Q2 = = 1.52 [kW] 4.934 COPref

[Ans. (b)]

Chapter 11:

Ideal Cycles

4. Then, by the energy (first law) equation, the heat rejected is Q1 = Q2 + W = 10 + 1.52 = 11.52 [kW]

321

[Ans. (c)]

As the following examples show, the procedures illustrated by the above examples are also applicable to heat pumps. The only difference is in the definition of the COP of a heat pump. EXAMPLE 11.8 A Carnot heat pump supplies 25 [kW] of heat at 25 [°C] when the environment temperature is 0 [°C]. Calculate the COP and the power input. Solution 1. By the definition of the COP of a heat pump, for a Carnot heat pump, COPhp =

298.15 T1 = = 11.926 ª 11.93 25 T1 − T2

2. Hence, the power input is

W =

=

25 = 2.096 ª 2.10 [kW] 11.926

EXAMPLE 11.9 In winter, when the outside temperature is 0 [°C], a room is to be maintained at 25 [°C]. The heat loss from the room is estimated to be 5 [kW]. Two alternative strategies to keep the room temperature constant are: (a) direct electrical heating and (b) by 1 reversing the room air-conditioner (of course, theQquality of air has to be maintained). Which COP hp is the better proposition? Solution Since the two alternatives are only to be compared, it is convenient to assume that all devices are ideal. 1. Then, the direct heating requires We = 5 [kW]. 2. Assuming that the window air-conditioner is a Carnot refrigerator, its COP, when working as a heat pump is COPhp =

288.15 T1 = = 11.93 25 T1 − T2

3. Then, by the definition of the COP of heat pump, its power input is

W =

=

5 = 0.42 [kW] 11.93

4. Since the second option requires only 0.42 [kW] of electricity (for running the motor of the air-conditioner), it is the better option. However, this theoretical result has some difficulties in its implementation.

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11.6

IDEAL AND REAL CYCLES

The cycles discussed in this chapter are all ideal cycles. For ease of discussion, it is convenient to classify the reasons for the differences between the ideal and real cycles into those owing to the processes and those owing to the working substance. It has already been mentioned that quasi-static processes cannot be executed in practice since they require infinite time. Thus, in all real processes one has to deal with the finite rates of processes. Since only finite time is available for the execution of these processes, the combination of these two factors results in significant deviation from ideal. Secondly, the isentropic process is a popular process incorporated in all cycles. Even conceptually, it is impossible to prevent heat transfer whenever there is a temperature difference. The perfect insulators do not exist. Then, this may be used as an approximation to any process which takes place so fast that ‘there is no time for heat to be transferred’. These two points indicate why, for example, the Carnot engine cannot be built. The isothermal processes will require infinite times, i.e. they would have be executed very slowly, while the isentropics would be executed with infinite speed. Such speed fluctuations in an engine are not acceptable. The finiteness of the processes also implies that sharp corners of the cycle diagrams are rounded off. In other words, it means that no process is completed abruptly, i.e. the processes overlap. To understand the effect of the working fluids, consider the S.I. engine. It has been shown that the working of the Carnot engine is independent of the working substance. It means that the performance of the Carnot cycle is so general that it can only give the values of energy exchanged as Q1, Q2 and W. Hence, a better approximation (to the real cycle) is to assume that the working fluid is air which behaves like an ideal gas with constant specific heats. The heats absorbed (and rejected) are assumed to be from (and to) reservoirs external to the system. Such cycles are called airstandard cycles. The next possible approximation is assuming that the specific heats of air change with temperature, but maintaining all other assumptions. The next step is to consider the mixture of air and fuel as well as the product gas as mixtures of ideal gases with variable specific heats. The actual nature of the gases and the processes if taken into account will lead to such a complex model that it cannot be solved with any reasonable effort. In other words, such analysis is useless. The following two chapters present different cycles used in practice.

REVIEW QUESTIONS 1. How are cycles used in thermodynamics? 2. What is an NP cycle? 3. What are the common processes used in thermal cycles?

Chapter 11:

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Ideal Cycles

323

Show that, in the p–V plane, the isentropics are steeper than the isothermals. Show that, in the T–S plane, the isochorics are steeper than the isobars. How many 3P cycles can be constructed? Enumerate them. What are the main I.C. engine cycles? What is a Brayton cycle? How is it used? Using a 3P cycle, explain the basic method of cycle analysis. Using the same cycle, explain the simplified approach. What are (a) cycle parameters and (b) performance parameters of any cycle? What are the performance parameters of a refrigerator/heat pump? Define the unit 1 [TR] (one ton of refrigeration). Draw the Carnot cycle on the p–V and T–S planes for (a) an ideal gas and (b) for a vapour. Explain its operation. Draw the Stirling and Ericsson cycles on the p–V and T–S planes and explain their operation. What are the differences between the ideal and real cycles?

EXERCISES 1. Show that 2P cycles cannot be constructed. [Hints: Consider the ideal gas cycle shown in Figure 11.2. Let ‘i’ and ‘f’ denote the initial and final states. Then, since, A and B form a cycle, pf,A = pi,B; Vf,B = Vi,A; pf,B = pi,A; and, Vf,A = Vi,B. Since the processes (i.e m and n as well as the initial and final pressures and volumes) are arbitrary, and since the equation for process A is pVn = CA, and that for process B is pVm = CB, take the ratio pf,A/pi,A and show that the above condition gives CA = CB and m = n, i.e. both processes are identical.] 2. Determine the number of 3P distinct cycles which can be constructed from the four processes. [Hints: In this case all of the processes should be distinct. Then, the first process can be chosen in four ways. The second cannot be the first one since, then, it is simply the extension of the first and only the end state is changed. Thus, the second process can be chosen in three ways. By the same argument, the last process cannot be the first or the second one. Hence, it can be chosen in two ways. Since the choices of the first process as well as its direction are arbitrary, only the combinations should be taken. This is equal to 4C3.] 3. Obtain equations for work done, heat absorbed and heat rejected for the following 3P cycles: (a) pVT, (b) pST, and (c) VST. [Hints: Proceed as explained in Section 11.3] 4. Determine the number of distinct 4P cycles which can be constructed using the four common processes.

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[Hints: The number of 4P cycles using two distinct processes has already been shown to be five. Using the same method of enumeration as of Exercise 2, the number using three distinct processes is 108 (if the third and first processes are the same), and is 72 (if they are different). The number using four distinct processes is 1. Thus, the total number is 111 or 75.] 5. In an air-standard cycle, heat is supplied at constant volume raising the air temperature from T1 to T2. It is then expanded isentropically till the temperature falls to T1. It is finally returned to its original state by an isothermal compression. Show that

h=1–

T1 T ln 2 T2 − T1 T1

[Hints: Note that since heat is absorbed during the constant volume process it must be rejected during the isothermal process because the third process is isentropic. Hence, it is better to calculate the heat absorbed and heat rejected separately, take their ratio and simplify the resulting expression.] 6. A heat engine drives a heat pump (COP = 5) by supplying 75 per cent of its work output. A process is heated solely by the heat rejected by the engine and by the heat pump. The total heat supplied for the process equals twice that supplied to the engine. Draw the system diagram and calculate the efficiency of the engine? [Hints: Let the subscripts E and P denote the engine and the heat pump respectively. Then, by the first law Q1,E = WE + Q2,E. Then, the above statements give the following WE equations. Q2,P = 5WP, WP = 0.75WE; Q2,E + Q2,P = 2Q2,E; and h = . Q1, E 1 .] Solving all these simultaneously, we get h = 2.75 7. A thermodynamic system executes a cycle consisting of three processes, during which the amounts of heat transferred are 100 [kJ] 300 [kJ], and – 200 [kJ]. Work done in the first process is 50 [kJ]. The change in internal energy in the second process is 100 [kJ]. Determine the work done and the change of internal energy in the third process. [Hints: Applying the first law for the first two processes gives, DE1 = 50 [kJ] and W2 = 200 [kJ]. All the three processes form a cycle for which the net change of internal energy is zero. Thus, DE3 = – 150 [kJ]. From the first law, W3 = – 50 [kJ].] 8. A closed system containing 1 [kg] of air at 1 [bar] and 300 [K] executes the following cyclic processes: (i) air is compressed at constant pressure till the volume reduces to one-tenth of the original, (ii) it is then heated at constant volume, and (iii) finally it is expanded isentropically to the initial state. (a) Draw the processes on the p–V state space. (b) Calculate the work done and the heat transferred in each process. (c) Calculate the net work done in the cycle. Assume that air behaves like an ideal gas with molecular weight = 29 [kg/kmol] and g = 1.4 and that only expansion/compression work is done.

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325

[Hints: First calculate p, V and T at all corner points. From the given data of p1 = 1 [bar] and T1 = 300 [K], using the ideal gas law, we get V1 = 0.8601 [m3]. Then, V2 = V1/10 = 0.08061 [m3]. Since 1–2 is an isobaric process, p2 = 1 [bar]. Then, using the equation of state, T2 = 30 [K]. Now, the 3–1 is an isentropic process. Hence, p3V3g = p1V1g and T3V3g –1 = T1V1g –1. Substituting the values, we get p3 = 25.12 [bar] and T3 = 753.6 [K]. Next, we evaluate the specific heats. Since air behaves like an ideal gas, cv = R/(g –1) and cp = cv × g, where R is the specific gas constant. Substituting the values, we get cv = 0.717 [kJ/kg.K] and cp = 1.004 [kJ/kg.K]. Finally, we evaluate Q and W for all the processes. Since 1–2 is an isobaric process, W12 = p1(V2 – V1) = – 77.41 [kJ], and Q12 = mcp(T2 – T1) = – 271.0 [kJ]. Since the 2–3 is an isochoric process, W23 = 0 and Q23 = mcv(T3 – T2) = 518.8 [kJ]. Since, the 3–1 is an isentropic process, W31 = (p3V3 – p1V1)/(g – 1) = 325.1 [kJ] and Q31 = 0. Thus, Wnet = 247.7 [kJ].] 9. An engine using 0.1 [kg] of an ideal gas (M = 30 [kg/kmol]) as its working substance works in a Carnot cycle between 300 [K] and 1000 [K]. The pressure at the end of isentropic expansion is 1 [bar] and that at the beginning of the isentropic compression is 5 [bar]. Calculate (a) the heat rejected, (b) the heat absorbed, and (c) the work done. [Hints: Heat rejected in an isothermal process of an ideal gas = mRT ln r, where r is the inverse of ratio of pressures. Since r = 5 (data), Q2 = 13.38 [kJ] [Ans. (a)]. For a Carnot cycle, Q1 = Q2(T1/T2) = 44.6 [kJ] [Ans. (b)]. So, the work done = 31.22 [kJ], [Ans. (c)].]

Chapter 12

Internal Combustion (I.C.) Engine Cycles

12.1 INTRODUCTION In this chapter, the cycles on which internal combustion (I.C.) engines work are described. In particular, the Otto cycle, the Diesel cycle, and the dual combustion cycles are explained. The reciprocating steam engines were used extensively at the time when the I.C. engines were invented. The steam for running these engines was generated by boilers, which were external to the cylinder of the engine in which the power was produced. Hence, it was only natural that the new engines that produced power by the combustion of fuel inside their cylinders were termed internal combustion engines. In engines which use gaseous fuels or liquid fuels that evaporate easily (e.g. petrol), the fuel and air are mixed before they enter the engine. The combustion is then initiated by an external source such as electric sparks. Naturally, these engines are called spark ignition (S.I.) engines. Since the combustion of premixed fuel and air is very fast (theoretically, instantaneous), it was assumed to take place at constant volume, and the Otto cycle was used to represent the working of such engines. Hence, they are also called Otto engines.1 With less volatile fuels (like diesel oil, kerosene, …), only air is taken in the engine. It is then compressed to very high pressures so that the resulting temperatures are high. Then, the fuel is sprayed into the cylinder so that the droplets evaporate and burn. Hence, these engines are called compression ignition (C.I.) engines. Since combustion of these fuels is slow, the expansion stroke (see below) would have already started so that the combustion takes place essentially at constant pressure. This is idealized by the Diesel cycle. Hence, these engines are called diesel engines.2

1 2

More correctly, since Nicholas A. Otto successfully built such an engine in 1876. Of course, Rudolf Diesel successfully built such an engine in 1897. 326

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327

12.2 OPERATION OF AN I.C. ENGINE 12.2.1 Four-Stroke Engines Figure 12.1, which presents the schematic diagram of an I.C. engine together with that of the Otto cycle, will help in understanding the working of an I.C. engine. For reasons mentioned at the end of this section, this is known as a four-stroke engine. Inlet valve Cylinder head

Spark plug Exhaust valve

p

TDC

Gudgeon pin

3

TDC

Piston BDC

2

4

BDC

Cylinder

Connecting rod Axis

pi

0

1

Crank Crank shaft (a) Schematic diagram

Vcly

Vcl

V

(b) Process diagram

Figure 12.1 The Otto cycle.

It is assumed to be an S.I. engine and that all of the processes are ideal.3 Then, it operates as follows. 1. At the beginning, let the piston be at the position marked TDC (called the Top Dead Centre) as shown in Figure 12.1(a). The inlet valve and the exhaust valve are closed. This state is shown as ‘0’ (zero) in the cycle diagram in Figure 12.1(b). 2. The piston starts moving downwards. The inlet valve opens. A mixture of air and petrol is drawn (or sucked) into the cylinder due to partial vacuum created. 3. This suction stroke ends when the piston reaches the position marked as BDC (called the Bottom Dead Centre). This suction stroke is shown as 0–1 in the cycle. 4. At BDC, the inlet valve closes. This state, shown as ‘1’ in the cycle, corresponds to the complete cylinder being filled with the combustible mixture at the inlet pressure, pi. 5. Next, the piston starts moving upwards. Both the valves are closed so that the mixture is compressed. This compression stroke is shown as 1–2 in the cycle diagram. 3

That is, the valves open instantaneously, the flow takes place without pressure drop and of course, the processes are all quasi-static and reversible.

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6. This compression stroke ends when the piston reaches TDC. This state is shown as ‘2’ in the cycle. 7. At TDC, an electrical spark across the electrodes of the spark plug ignites the mixture. Since the combustion is ideal it is instantaneous and since the volume remains constant, the pressure rises. The state at the end of this constant volume combustion is shown as ‘3’. 8. Exhaust gases at high pressure and temperature produced owing to combustion of the mixture act on the piston and drive it downwards. This expansion stroke is shown as 3–4 in the cycle. Since the main power is produced in this stroke, it is also called the power stroke. 9. The state at the end of the expansion is shown as ‘4’. 10. At BDC, labelled as ‘4’ in the cycle, the exhaust valve opens. 11. Since the processes are ideal, the pressure in the cylinder drops from that at ‘4’, to the atmospheric pressure immediately. This process is shown as 4–1 in the cycle. 12. When the piston starts moving upwards again, the product (exhaust) gases are expelled (exhausted) from the cylinder. This exhaust stroke, shown as 1–0, ends at TDC when the exhaust valve closes. 13. The piston has now reached its starting position and is ready to begin the next cycle of operations. The power produced in the cylinder is transmitted to the crankshaft through the connecting rod and the crankshaft drives the load that is connected to this shaft. Assuming that one stroke is needed to complete one process, this engine requires four strokes of the piston to complete the cycle. Hence, such cycles are called four-stroke cycles. Since each stroke requires half a revolution, this means that one four-stroke cycle is completed in two revolutions. Thus, the number of four-stroke cycles per minute is Nc = N/2, where N is the rotational speed in [RPM].

12.2.2

Two-Stroke Engines

In two-stroke cycles, all the operations of the cycle are completed in two strokes, i.e. in one revolution. This is achieved by combining the different strokes. The most significant feature of the two-stroke engines is that they have ports instead of valves. As it moves the piston covers/uncovers these ports. Moreover, instead of being connected directly to the inlet pipe (manifold), the inlet port of the cylinder is connected to the crankcase; which is connected to the inlet pipe. However, the exhaust port opens directly to the exhaust pipe (manifold). The operation of the cycle is as follows. 1. Let the piston be at BDC. Then, the crankcase is full of combustible mixture and port to the inlet manifold is covered, so the combustible mixture does not enter cylinder. 2. When the piston moves upwards, the inlet and exhaust ports are open so that pressurized combustible mixture in the crankcase enters the cylinder. This is equivalent of the suction stroke.

the the the the

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Internal Combustion (I.C.) Engine Cycles

329

3. During the last part of the upward movement of the piston, it covers the inlet and exhaust ports so that the mixture is compressed. 4. At the same time, the piston uncovers the port between the inlet manifold and the crankcase and so the fresh mixture enters the crankcase. 5. Thus, this stroke combines the compression and the initial part of the suction strokes. 6. At TDC, ignition takes place and the high-pressure, high-temperature gases push down the piston. This is the power (or expansion) stroke. 7. At the same time, the specially designed shape of the crank pressurizes the combustible mixture. 8. A little before the BDC, the exhaust port is uncovered and the hot gases are discharged into the exhaust manifold. 9. A little after this, the inlet port is uncovered so that the fresh mixture enters the cylinder from the crankcase. 10. From BDC, the cycle repeats itself. Note that in two-stroke engines, one cycle is completed in two strokes, i.e. in one revolution, i.e. in the usual symbols, Nc = N. Two-stroke engines are compact and produce more uniform torques (since there is one power stroke during every revolution). However, its fuel efficiency cannot become high, since the valve overlap (the duration when both the inlet and exhaust valves are simultaneously open) cannot be reduced and this results in loss of combustible mixture.

12.3

AIR-STANDARD CYCLES

The working substance of the air-standard cycles is air, assumed to be an ideal gas with constant specific heats. Heat released owing to combustion of the fuel inside the cylinder is assumed to be absorbed from suitable high temperature reservoirs. Similarly, during the exhaust stroke, heat is assumed to be transferred to suitable number of cold reservoirs. Thus, engines working on air-standard cycles are truly closed systems. Consequently, the suction and exhaust strokes of the earlier engines, (i.e. processes 0–1 and 1–0 of Figure 12.1) do not exist. In this section, different types of air-standard cycles are described.

12.3.1

Air-Standard Otto Cycle

Since, air-standard cycles truly correspond to closed systems, the Otto cycle shown in Figure 12.1 gets modified to that shown in Figure 12.2, i.e. it becomes an SVSV cycle. As illustration of the nature of the processes, the cycle is also drawn on the T–S plane in this figure. However, only the p–V diagram is used here. The air-standard Otto cycle is analyzed as follows. 1. Since two of the processes are isentropic, i.e. no heat is transferred, it is easier to calculate the efficiency and obtain other results. 2. Since heat is absorbed in the constant volume process 2–3, Q1 = mcv(T3 – T2)

Engineering Thermodynamics

330

T p

TDC

BDC 3

3

2

4

2

1

1

Vcyl

Vcl

Figure 12.2

4

TDC

BDC

V

S1

S3

S

The air-standard Otto cycle.

3. Similarly, the heat rejected in the constant volume process 4–1, Q2 = mcv (T4 – T1) where the sign convention (that heat rejected is negative) is already included in this expression. 4. Then, the air-standard efficiency is

Q2 T4 − T1 has = 1 – Q = 1 – T − T 1 3 2 5. The temperatures are eliminated as follows. 6. Since 1–2 is an isentropic process of an ideal gas g −1

T1V1

g –1

=

T2V2g –1

⎛ V1 ⎞ or T2 = T1 ⎜ ⎟ ⎝ V2 ⎠

or

T2 = T1(rv)g –1

where rv is the compression ratio, defined as V1/V2. 7. Similarly, since 3–4 is an isentrope of an ideal gas T3V3g –1 = T4V4g –1

or

T3 = T4(rv)g –1

8. Using the dividendo rule and substituting in the equation for the air-standard efficiency and simplifying it gives

has = 1 –

1 (rv )g −1

9. Then, the work done may be evaluated as W = has × Q1.

Internal Combustion (I.C.) Engine Cycles

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331

The above analysis implies that the expansion and compression are perfect adiabatic processes. In order to bring in some realism into the analysis, these are assumed to be polytropic processes, pVn = C. The above relation shows that the air-standard efficiency increases as the compression ratio, rv, increases. However, it decreases as the polytropic index n increases. Many factors determine the value of the compression ratio, rv. For most petrol engines it is about 7, and the value of the polytropic index is about 1.3.

12.3.2

Air-Standard Diesel Cycle

Figure 12.3 shows this (an SpSV) cycle in both p–V and T–S planes. T

p

3 2

3

TDC 4 BDC 2

4

BDC

TDC 1

1

Vcl

Vcyl

V

Figure 12.3

S1

S3

S

The Diesel cycle.

The following arguments show that the geometry of the Diesel cycle requires two parameters for its complete specification. For simplicity, the p–V plane is used in the arguments. Since the point ‘2’ is the intersection of the isentropic through the point ‘1’ and the vertical line at ‘2’, V2 should be known to fix it. This is obtained from the compression ratio, rv. Similarly, since the point ‘3’ is the intersection of the isentropic through the point ‘4’ and the isobar through the point ‘2’, it can be fixed only if V3 is known. This is specified in many ways. The most convenient method for calculations is to specify, rc = V3/V2, where rc is called the cut-off ratio. However, the most convenient (and hence common) method is to state that “the cut-off is x% of the stroke”. Then, the cut-off ratio can be calculated as rc =

V + ( x × Vs ) V + x (V1 − V2 ) V3 = 2 = 2 = 1 + x(rv – 1) V2 V2 V2

where Vs is the stroke volume and, by definition, is equal to (V1 – V2). The air-standard efficiency of the Diesel cycle can be derived using the same method as that employed for the Otto cycle. This is done below.

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1. By definition, the air-standard efficiency is

has = 1 –

Q2 c × (T4 − T1 ) T4 − T1 =1– v =1– Q1 c p × (T3 − T2 ) g × (T3 − T2 )

2. As before, since 1–2 is an isentrope of an ideal gas, T1V1g –1 = T2V2g –1

or T2 = T1(rv)g –1

3. For the isobar, 2–3, ⎛V ⎞ T3 = T2 × ⎜ 3 ⎟ = T2 × rc = T1 × rc × (rv)g –1 ⎝ V2 ⎠ 4. Since, 3–4 is an isentrope of an ideal gas g −1

g −1

g −1

g −1

g −1

⎛V ⎞ ⎛r ⎞ ⎛V ⎞ ⎛V ⎞ ⎛V ⎞ T4 = T3 × ⎜ 3 ⎟ = T3 × ⎜ 3 ⎟ = T3 × ⎜ 3 ⎟ × ⎜ 2 ⎟ = T3 × ⎜ c ⎟ ⎝ V1 ⎠ ⎝ rv ⎠ ⎝ V4 ⎠ ⎝ V2 ⎠ ⎝ V1 ⎠ 5. Substituting for T3 from above and simplifying gives T4 = T1 × (rv)g –1 6. Substituting in the equation for air-standard efficiency and simplifying gives ⎡ 1 ⎤ has = 1 – ⎢ g −1 ⎥ × ⎣ (rv ) ⎦

⎡ (rc )g − 1 ⎤ ⎢ ⎥ ⎣ g × (rc − 1) ⎦ The above relation for the air-standard efficiency of a Diesel cycle is more complicated than that of the Otto cycle. The value of the term in the second brackets on the RHS of the efficiency relation is greater than unity. Thus, for the same compression ratio, the Otto cycle is more efficient than the Diesel cycle. However, this comparison is spurious, since the compression ratios of Otto engines are limited by spontaneous ignition and are generally between 7 and 10. On the other hand, the compression ratios of Diesel engines are chosen so that high temperatures are obtained at the end of compression to make the fuel droplets burn easily. Hence, the compression ratio for diesel engines generally ranges from 12 to 22. The effect of the cut-out ratio, rc, on the air-standard efficiency can be understood more easily from physical arguments. The diesel cycle shows that as the cut-off ratio is increased, more heat is released, but the expansion ratio, i.e. V4/V3, decreases. Thus, increasing the cutoff ratio increases the power output, but decreases the cycle efficiency. The equation for the airstandard efficiency also shows this. As a thumb rule, the cut-off is never more than 10% of the stroke.

12.4 REAL ENGINES Fortunately, the ideal analysis is applicable for most of the real engines with minor modifications in the form of suitable efficiencies. The following modifications are two of the more important ones.

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Volumetric efficiency (hV). This is defined as the ratio of the actual volume of air drawn at NTP to the cylinder volume. In other words, hV = (VNTP/Vs), where VNTP is the actual volume of air inducted at NTP and Vs is the stroke volume. It is usual to use inlet conditions instead of NTP, so that the definition becomes, hV = Vi/Vs. This definition of volumetric efficiency is generally used for C.I. engines. For S.I. engines, the equivalent definition is as given below. Charge efficiency (hch). This is defined as the ratio of the mass of air-fuel mixture actually inducted to that which can be theoretically inducted. The latter equals the mass of the mixture whose volume is the stroke volume [of course at inlet conditions (really, at NTP)]. In other words, hch = (m/ms)i, where m is the mass of the actual air inducted and ms is mass of the mixture filling the stroke volume.

12.4.1

Spark Ignition (S.I.) Engines

The following example illustrates the method of calculating the performance of S.I. engines. EXAMPLE 12.1 A 4-cylinder, 4-stroke, 10 [cm] × 12 [cm], petrol engine runs at 1500 [RPM]. It has a clearance of 14% of cylinder volume. Its relative, volumetric and mechanical efficiencies are 60%, 85% and 80%, respectively. Its air-fuel ratio is 18 to 1. The calorific value of the fuel is 45 [MJ/kg F]. The inlet conditions are 1 [bar] and 300 [K]. Calculate the engine performance. Solution In the following calculations, six significant digits are retained for the intermediate results and the final results have four significant digits. This improves the accuracy of the calculations. 1. The compression ratio is V1 1 = = 7.14286 ª 7.143 0.14 V2

rv =

2. Since his is an S.I. engine, it works on the Otto cycle, and hence its air-standard efficiency is g −1

⎛1⎞ has = 1 – ⎜ ⎟ ⎝ rv ⎠

= 1 – (0.14)0.4 = 0.544539 ª 0.5445

3. Next, the masses of air and fuel inducted are calculated. Since the fuel-air ratio is 18 to 1, the mass of fuel is small compared to that of air. Hence, the properties of the mixture can be assumed to be the same as those of air. 4. Then, the density of the mixture at the inlet is

r=

p 100 ⎡ kg ⎤ = = 1.16266 ª 1.163 ⎢ 3 ⎥ RT (8.3143 / 29) × 300 ⎣m ⎦

5. The calculations are done for 1 [s], but it can also be done for one cycle.

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6. Since this is a 4-stroke cycle, one cycle is completed in two revolutions, so that the number of cycles per second is

⎛ 1 ⎡ cycle ⎤⎞ ⎛ ⎛ 1 ⎡ min ⎤⎞ ⎡ cycle ⎤ ⎡ rev ⎤ ⎞ Ncyc = ⎜ 1500 ⎢ × ⎜ ⎢ × ⎜ ⎢ = 12.5 ⎢ ⎟ ⎥ ⎟ ⎥ ⎟ ⎥ ⎥ ⎝ 2 rev ⎠ ⎝ ⎠ ⎝ 60 ⎣ s ⎦⎠ ⎣ s ⎦ ⎣ ⎦ ⎣ min ⎦ 7. When the engine dimensions are written as 10 [cm] × 12 [cm], the first number represents the cylinder bore (i.e. diameter) and the second number is the stroke. Then, the stroke (or swept) volume is

⎡ m3 ⎤ × (0.1)2 × (0.12) × (12.5) = 0.0122718 ª 0.01227 ⎢ ⎥ ⎣ s ⎦

Vs =

8. Assuming that the mixture properties are the same as those of air implies that charge efficiency is related to the volumetric efficiency. Then, the mass of air actually inducted is ⎡ kg A ⎤ m a = 0.85 × 4 × 1.16266 × 0.0122718 = 0.048511 ª 0.04851 ⎢ ⎥ ⎣ s ⎦

9. The specification that the air-fuel ratio is 18 to 1 means that 1 [kg F] is supplied with 18 [kg A] so that in the mixture, Z = 18 [kg A/kg F]. 10. Then, the mass of the fuel inducted is 0.0465708 paf ⎞ ⎛m ⎡ kg F ⎤ ⎜⎝ ⎟⎠19 Z4 m f = = = 0.002695 ª 0.002695 ⎢ ⎥ ⎣ s ⎦ 11. Then, the rate of heat release is

Q1 =

× Qcal = 0.0022695 × 45000 = 121.275 ª 121.3 [kW]

12. By the definition of relative efficiency, hr, the indicated thermal efficiency is

hi,th = hr × has = 0.6 × 0.544539 = 0.326723 ª 0.3267 13. And, the indicated power is IP = hi,th × Q1 = 0.326723 × 121.275 = 39.6233 ª 39.62 [kW] 14. Using the definition of mechanical efficiency, the brake thermal efficiency is

hb,th = hi,th × hm = 0.8 × 0.326723 = 0.261378 ª 0.2614 15. And the brake power is BP = hm × IP = 0.8 × 39.6233 = 0.261378 ª 0.2614 [kW] 16. By the definition of the mean effective pressures,

Chapter 12:

pi,m =

=

Internal Combustion (I.C.) Engine Cycles

335

36.0372 ⎡ kN ⎤ = 807.202 ⎢ 2 ⎥ = 8.072 [bar] 4 × 0.011781 ⎣m ⎦

17. And, pb,m = hm × pi,m = 0.8 × 8.07202 = 6.45762 ª 6.458 [bar] 18. The specific fuel consumptions are calculated from their definitions. Then, isfc =

m f IP

⎡ kg F ⎤ = 0.244856 ª 0.2448 ⎢ ⎥ ⎣ kWh IW ⎦

=

and, bsfc =

12.4.2

isfc

hm

=

0.244843 ⎡ kg F ⎤ = 0.30607 ª 0.3061 ⎢ ⎥ 0.8 ⎣ kWh BW ⎦

Compression Ignition (C.I.) Engines

The procedure of Subsection 12.4.1 is useful for calculating the performance of the C.I. engines too. The following example illustrates this. EXAMPLE 12.2 A diesel engine receives 0.7 [kg A/s] at 1 [bar] and 60 [°C]. The temperature at the end of compression is 900 [K]. The cut-off occurs at 9% of the stroke. The relative efficiency with respect to the brake power is 60%. The calorific value of the fuel is IP(c) h . × 3600 43.03 [MJ/kg F]. Calculate (a) has, (b) BP, and0.0245096 b,th Vs 36.0372 Solution Here also the intermediate results are given to six significant digits and the final answers four. Even though only some final answers are asked, all the above calculations will have to be done except those related to the indicated power. 1. First, the compression ratio should be calculated since the relative efficiency is given. This is obtained from the temperature at the end of isentropic compression. Thus, 1/(g −1)

T2 = T1 × (rv)

g –1

⎛T ⎞ or rv = ⎜ 2 ⎟ ⎝ T1 ⎠

900 ⎞ = ⎛⎜ ⎝ 333.15 ⎟⎠

1/ 0.4

= 11.9952 ª 12

2. As shown in the preceding section, the cut-off ratio is rc = 1 + 0.09 × (11.9952 – 1) = 1.98957 3. Then, the air-standard efficiency is

⎡ 1 ⎤ ⎡ (r )g − 1 ⎤ has = 1 – ⎢ g −1 ⎥ × ⎢ c ⎥ = 0.567212 ª 56.72% ⎣ (rv ) ⎦ ⎣ g × (rc − 1) ⎦ 4. Then, the definition of the relative efficiency gives

hb,th = hr × has = 0.6 × 0.576212 = 0.340327 ª 34.03%

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336

5. The air-fuel ratio should be first determined, before the heat input and the BP can be found. This is done as follows. 6. The heat released by combustion of fuel is given by Q1 = m f × Qcal 7. The heat absorbed by the combustion gases is given by Q1 = × (T3 – T2), × where m g is the mass of exhaust gases which, by the conservation of mass principle, + is equal to , and p is the average specific heat of the gases. 8. It was shown earlier that the cut-off ratio implies that T3 = T2 × rc. 9. Equating the two expressions in steps 6 and 7 and substituting the expression for T3 above, dividing throughout by m f and rearranging gives Z=

=

ma Qcal = –1 mf c p × T2 × (rc − 1)

10. Since the air-fuel ratio in diesel engines is very large, the properties of the gases may assumed to be those of air. Then, taking c p = cp,a = 1.00 [kJ/kg.K], the above relation gives Z = 47.3150 ª 47.32 [kg A/kg F]. 11. Then, by the definition of Z m f =

=

⎡ kg F ⎤ = 0.0147945 ª 0.01480 ⎢ ⎥ ⎣ s ⎦

12. Then the heat input is

c0.014801 pagaff×0.7 m ×1000 3600×=3600 Q1 = m f × Qcal = 0.0147945 × 43.03 636.607 ª 636.6 [kW] 216.767 47.2940 mZf BP 13. Hence, the brake power is given by BP = hb,th × Q1 = 0.340327 × 636.607 = 216.655 ª 216.6 [kW] 14. Finally, the brake specific fuel consumption is bsfc =

12.4.3

=

⎡ kg F ⎤ = 0.24583 ª 0.2458 ⎢ ⎥ ⎣ kWh BW ⎦

The Air-Standard Dual Combustion Cycle

This section deals with the dual combustion cycle and not the dual combustion engines, since, unlike the Otto (S.I.) and Diesel (C.I.) engines, the dual combustion engines do not exist for the following reasons. The actual combustion in engines always takes finite time. With increase in the operating speeds of these engines, the time available for each stroke reduces. This means that some part of the fuel burns near TDC (under constant volume conditions) and the rest burns after the expansion stroke has begun (under constant pressure conditions). The dual combustion cycle was invented (‘cooked-up’ to be more precise) to account for this fact. Thus, one has a SVpSV cycle.

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The general formula for the air-standard efficiency of a dual combustion cycle is complicated and, therefore, is not worth remembering. As the following example illustrates, the calculations for this cycle can be easily done from first principles. EXAMPLE 12.3 This is the same diesel engine as in Example 12.2, but with the additional parameter, i.e. the pressure ratio, rp = p3/p2 = 1.25. For air, assume, cp = 1.0 [kJ/kg.K] and g = 1.4, so that cv = 1/g = 1/1.4. Solution The cycle is shown in both p–V and T–S planes in Figure 12.4. Comparing this with Figure 12.3, it is seen that owing to the isochoric heat addition, the isobaric heat absorption has shifted upwards and, therefore, the isentropic expansion is suitably modified. p

T 4

3

4 3 TDC

BDC

2

5

5

2 BDC

TDC 1

1

Vcl

Vcyl Figure 12.4

V

S1

S3

The dual combustion cycle.

1. First the temperatures at the corner states are calculated. 2. Since 1–2 is an isentrope of an ideal gas, T1V1g –1 = T2V2g –1, or g −1

⎛ V1 ⎞ T2 = T1 × ⎜ ⎟ ⎝ V2 ⎠

= T1 × (rv)g –1 = (333.15) × (12)0.4 = 900.145 [K]

3. Since 2–3 is an isochor of an ideal gas, the equation of state gives ⎛p ⎞ T3 = T2 × ⎜ 3 ⎟ = T2 × rp = 900.145 × 1.25 = 1125.18 [K] ⎝ p2 ⎠

4. Similarly, for isobar 3–4 of an ideal gas, the equation of state shows ⎛V ⎞ T4 = T3 × ⎜ 4 ⎟ = T3 × rc = 1125.18 × 1.99 = 2239.11 [K] ⎝ V3 ⎠

S

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Engineering Thermodynamics

5. Again, since 4–5 is an isentrope of an ideal gas, T5V5g –1 = T4V4g –1, or g −1

⎛V ⎞ T5 = T4 × ⎜ 4 ⎟ ⎝ V5 ⎠

g −1

⎛V ⎞ = T4 × ⎜ 4 ⎟ ⎝ V1 ⎠

⎡⎛ V = T4 × ⎢⎜ 4 ⎣⎢⎝ V3

⎞ ⎛ V3 ⎞ ⎤ ⎟ × ⎜ ⎟⎥ ⎠ ⎝ V1 ⎠ ⎦⎥

g −1

g −1

⎛r ⎞ = T4 × ⎜ c ⎟ ⎝ rv ⎠

= T1 × (rp) × (rv)g = (333.15) × (1.25) × (1.99)1.4 = 1091.30 [K] where the last step is obtained by recursively substituting for T4. 6. The quantities of heat exchanged in the cycle are Q23 = m × cv × (T3 – T2); Q34 = m × cp × (T4 – T3);

and

Q51 = m × cv × (T5 – T1)

7. Then, the air-standard efficiency is

has = 1 –

Q51 T5 − T1 =1– = 0.575156 ª 57.52% Q23 + Q34 (T3 − T2 ) + g × (T4 − T3 )

8. Then, the brake thermal efficiency is

hb,th = hr × has = 0.6 × 0.575156 = 0.345094 = 34.51% 9. As in Example 12.2, the air-fuel ratio is evaluated by equating the heat released by the processes and by combustion. Then,

Q1 =

× Qcal = (

+

) [cv × (T3 – T2) + cp × (T4 – T3)] 0.021369 1aaff0.7 m ×as3600 Q which can be solved for the air-fuel ratio × 3600 32.7578 Zf BP m317.317 ⎡ kg A ⎤ Qcal Z= = – 1 = 32.7582 ª 32.76 ⎢ ⎥ cv × (T3 − T2 ) + c p × (T4 − T3 ) ⎣ kg F ⎦ 10. Then, the rate of fuel flow is m f =

=

⎡ kg F ⎤ = 0.0213690 ª 0.02137 ⎢ ⎥ ⎣ s ⎦

11. And the heat release rate is

Q1 =

× Qcal = 0.0213690 × 43.03 × 1000 = 919.508 [kW] ª 919.5 [kW]

12. Then, the brake power is BP = hb,th ×

= 0.345094 × 919.508 = 317.317 ª 317.3 [kW]

13. The brake specific fuel consumption is bsfc =

=

⎡ kg F ⎤ = 0.242434 ª 0.2424 ⎢ ⎥ ⎣ kWh BP ⎦

14. The brake mean effective pressure is calculated as follows.

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339

15. The equation of state gives the density of air as

ra =

⎡ kg A ⎤ p 100 = = 1.04696 ⎢ 3 ⎥ RT ⎣ m ⎦ (8.3143 / 29) × 333.15

16. Then, the volume of air inducted is

V =

=

⎡ m 3A ⎤ 0.7 = 0.668602 ⎢ ⎥ 1.04696 ⎣ s ⎦

17. Assuming that the volumetric efficiency is 100%, this equals the stroke volume, so that the brake mean effective pressure is pb,m =

BP 317.317 ⎡ kN ⎤ = = 474.598 ⎢ 2 ⎥ = 4.746 [bar] 0.668602 Vs ⎣m ⎦

REVIEW QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What is the main difference in the fuels used in S.I. and C.I. engines? Draw a single cylinder I.C. engine, name its parts and explain its working. What is a 4-stroke engine? m What are the major differences betweena a 4-stroke engine and a 2-stroke engine? ra Explain the operation of a 2-stroke engine. What are air-standard cycles? Draw the Otto and Diesel cycles on the p–V and T–S planes and explain their working. Derive expressions for the air-standard efficiencies of the Otto and Diesel cycles. Define volumetric and charge efficiencies. Draw the dual combustion cycle on the p–V and T–S planes and explain its working.

EXERCISES Notes: Make the following default (in the absence of any information to contrary) assumptions: 1. Air to behave like an ideal gas with molecular weight of 29 [kg/kmol] and g of 1.4. 2. The value of the universal gas constant as 8.3143 [kJ/kmol.K]. 3. The state of air at the inlet is 1 [bar] and 300 [K]. 1. Determine the change in the efficiency in the air-standard Diesel cycle with a compression of 15 when its cut-off changes from 5% to 15%.

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Engineering Thermodynamics

[Hints: Let subscript ‘1’ denote the case with x = 5%, and let the other case be indicated by the subscript ‘2’. Then, the standard formula gives rc,1 = 1.7 and rc,2 = 3.1. Substituting in the expression for the air-standard efficiency of the Diesel cycle gives, has,1 = 0.619363 and has,2 = 0.553931; so that Dhas = (has,2 – has,1)/has,1 = –0.118123 = – 11.81%; i.e. has decreases as rc increases (as expected).] 2. In an air-standard Otto cycle the clearance and stroke volumes are 7 [cc] and 63 [cc], respectively. The cycle MEP (mean effective pressure) is 5 [bar]. Calculate the heat absorbed during one cycle. [Hints: By definition, rv = V1/V2 = Vcyl/Vcl = (Vs + Vcl)/Vcl = 70/7 = 10. Then, has = 1 – 1/rvg –1 = 0.601893. By the definition of the mean effective pressure, W = pm × Vs = 500 × 63 × 10–6 = 0.0315 [kJ/cycle]. The definition of efficiency gives Q1 = W/has = 0.052335 ª 0.05234 [kJ/cycle].] 3. The compression ratio of an Otto cycle is 6. The temperature at the end of expansion is 800 [K]. Calculate (a) the air-standard efficiency, (b) the heat absorbed, (c) the work done, and (d) the mean effective pressure. [Hints: (a) h = 1 –

1 rv( g −1)

= 51.2%; (b) Now, T2 = T1(rv)g –1 = 614.3 [K], and

T3 = T4(rv)g –1 = 1638.1 [K]. Then, Q1 = cv(T3 – T2) = 733.8 [kJ/kg]; (c) W = hQ1 = 375.7 [kJ/kg]; and, W (d) pm = V − V = 1 2

W rvW (W )(rv )( p1 )( M ) = = = 5.242 [bar].] − − V ( r 1) ( T )( r 1) R ⎛ ⎞ 1 1 v W 1 v mf V1 ⎜1 − ⎟ rv ⎠ ⎝

4. A 6-cylinder 4-stroke petrol engine with 80 [mm] bore and 90 [mm] stroke developing 50 [kW] at 3000 [RPM] uses 18 [kg/h] of fuel with calorific value of 44 [MJ/kg]. Calculate (a) the indicated thermal efficiency, and (b) the BMEP if the mechanical efficiency is 90%. × Qcal) = 50 × 3600/(18 × 44000) = 22.7%; (b) pi,m = / [Hints: (a) hi,th = W /( (Vs × Ncyc × Ncyl) = 7.368 [bar]; and, (c) BMEP = pm,b = hm × pm,i = 6.631 [bar].] 5. A 6-cylinder 4-stroke petrol engine of 10 [cm] bore and 12.5 [cm] stroke was run on full throttle at a constant speed of 1500 [RPM] from which the following data were obtained. Determine the engine performance for each case. pm,l [bar] pm,b [bar] sfc [g/kWh BP]

7.82 6.27 547

8.32 6.62 486

8.14 6.76 432

8.02 6.72 337

8.32 6.69 346

8.06 6.34 334

6.91 5.79 346

6.06 5.14 401

Other data: Compression ratio = 5; Fuel calorific value = 44.5 [MJ/kg]. pm,i = IMEP (Indicated Mean Effective Pressure); pm,b = BMEP (Brake Mean Effective Pressure); IP = Indicated Power and BP = Brake Power.

[Hints: The computations are explained for the first case and the results are tabulated for others. Let Ncyl and Ncyc denote the number of cylinders and the number of cycles per second. Then, BP = (BP/cycle/cylinder) × Ncyc × Ncyl, and by the definition of

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341

BMEP (i.e. pm,b), BP = (pm,b · Acyl · L) × [(RPM/2)/60] × Ncyl, (where, the term in brackets arises because the engine is 4-stroke) = [(6.27 · (p ) · (0.1)2/4 × (0.125)] × (1500/120) × (6) = 46.1667 [kW]. Similarly, using the definition of IMEP (i.e. pm,i), the indicated power is IP = 57.5795 [kW]. Now, = (0.547) × (46.1667) = 25.2532 [kg/h]; and × Qcal = (25.2532/3600) = (44,500) = 312.157 [kW]. Then, the definition of the brake thermal efficiency gives, hb,th = BP/ = 14.79%. Similarly, the definition of indicated thermal efficiency gives, hi,th IP/ = 18.44%. The mechanical efficiency defined as hm = BP/IP gives hm = 80.12%. Results for other cases are tabulated below.] IP [kW] BP [kW] [kW] hi,th hb,th hm

61.26 78.74 292.83 0.2092 0.1664 0.7956

59.94 49.77 265.80 0.2222 0.1872 0.8303

50.05 49.48 206.12 0.2428 0.2400 0.9886

61.26 49.26 210.68 0.2908 0.2338 0.8041

59.35 46.68 192.73 0.3079 0.2422 0.7865

50.88 42.63 182.34 0.2790 0.2338 0.8378

44.62 37.85 187.60 0.2378 0.2018 0.8483

Note: The hb,th and hm values of the 3rd column are too high and so data are erroneous.

6. During a test on a 6-cylinder 4-stroke petrol engine (10 [cm] bore and 12 [cm] stroke) the following readings were obtained. N = 1500 [RPM]; BMEP = 6.72 [bar]; bsfc = 350 [g/kWh]; Z = 15 [kgA/kg F]; compression ratio = 7; Qcal = 42 [MJ/kg]; and, Zs = 14.5 [kgA/kg F]. Calculate (a) BP, (b) , (c) hth,b, (d) , and (e) has. Obtain BP m T1loss the energy balance if Ti = 300 [K], Q eaf = 600 [K], cp,g = 1.10 [kg.K] and cp,a = 1.0 [kJ/kg.K]. Q1 [Hints: Using the well-known formula, BP = pb,m ¥ A ¥ L ¥ Ncyc ¥ Ncyl, where pb,m is the BMEP; A is the area of the piston, L is the stroke, Ncyc is the number of cycles per second, and Ncyl is the number of cylinders, we get BP = 47.5 [kW]. Then, = (bfsc) (BP) = 4.618 × 10–3 [kg/s]. Also, Q1 = ( ) Qcal = 193.96 [kW]. Hence, hb,th = = 0.245. Also m a = (

)(Z) = 0.06252 [kg/s]. Since this is an Otto cycle, using

the usual formula we get has = 0.5408. The energy balance gives = 124.4 [kW].] 7. In a Diesel engine with a compression ratio of 16, the maximum cycle temperature is 2500 [K]. Calculate (a) the cut-off ratio, (b) the heat absorbed, (c) the work done, and (d) the cycle efficiency. V3 T = 3 = 2.75; (b) Q1 = V2 T2

[Hints: (a) T2 = T 1(rv) g –1 = 909.4 [K], and rc =

g −1

⎛ V3 ⎞ T cp (T3 – T2) = 1590.6 [kJ/kg]; (c) Now, 4 = ⎜ ⎟ T3 ⎝ V4 ⎠

g −1

⎡⎛ V ⎞ ⎛ V ⎞ ⎤ = ⎢⎜ 3 ⎟ ⎜ 2 ⎟ ⎥ ⎣⎝ V2 ⎠ ⎝ V4 ⎠ ⎦

g −1

⎛r ⎞ =⎜ c ⎟ ⎝ rv ⎠

so that T4 = 1236.0 [K]. Then, Q2 = cv(T4 – T1) = 670.9 [kJ/kg], and W = Q1 – Q2 = 919.7 [kJ/kg]; (d) h =

W = 57.82%.] Q1

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8. The compression ratio in a dual combustion cycle is 18. The heat absorbed in the cycle is 1500 [kJ/kg A]. Assume that equal quantity of heat is absorbed in the constant volume and the constant pressure processes. If the inlet state of the air is 1 [bar] and 300 [K], calculate (a) Tmax, (b) pmax, and (c) has. [Hints: Here, the pressure ratio, rp and the cut-off ratio, rc are given (indirectly), through the heat released. Using the usual formula for the isentropic process of an ideal gas, T2 = T1 × rvg –1 = 953.301 [K]. Since the heat released in the isochoric combustion is 750 [kJ/kg A], T3 = T2 + Q1,v/(m × cv) = 1999.69 [K].Similarly, T4 = T3 + Q1,p/(m × cp) = 2749.69 [K] and T5 = 982.889 [K]. Thus, Tmax = 2750 [K]. Since, 1–2 is the isentrope of an ideal gas, p2 = p1 × (V1/V2)g = (1) × (15)1.4 = 44.3127 [bar], and for the isochor 2–3, p3 = p2 × (T3/T2) = 46.4103 = 46.41 [bar]. This is the maximum cycle pressure, pmax. Finally, the air-standard efficiency is obtained as has = 1 – Q2/Q1 = 0.674255 = 67.42%.] 9. The compression ratio of a 6-cylinder 4-stroke petrol engine is 5. The clearance volume of each cylinder is 110 [cc]. Its indicated relative efficiency is 66%. While running at 2400 [RPM], it is found to consume 2.75 [g/s] of fuel with the calorific value of 43.75 [MJ/kg]. Determine the indicated mean effective pressure. [Hints: Using the standard formulae, has = 1 – 1/50.4 = 0.474694; so that hi,th = 0.313298. Then, indicated work = has × Q1 = (0.313298) × (2.75 × 10–3) × (43.75 × 103) = 37.6937 [kW]. Hence, work done per cycle = 37.6937/[(6) × (2400/(2 × 60))] = 0.314114 [kJ/cycle]. The stroke volume per cylinder is Vs = Vcyl – Vcl = rvVcl – Vcl = 440 [cc] = 440 × 10–6 [m3]. Then, by definition, the indicated mean effective pressure = W/Vs = 713.895 [kPa] ª 7.139 [bar].] 10. Calculate the heat supplied and the net work done in an air-standard Otto cycle with a compression ratio of 8, if the initial and maximum temperatures are 335 [K] and 1670 [K], respectively. [Hints: From the definition, h as = 1 – 1/80.4 = 0.564725. For the isentropic compression, T2 = 335 × 80.4 = 769.628 [K], so that, the heat absorbed is Q1 = 0.71675 × (1670 – 769.628) = 645.342 ª 645.3 [kJ/kg]. Then, W = has × Q1 = 364.441 ª 364.4 [kJ/kg].] 11. Same exercise as above, but now the indices of compression and expansion are both 4/3. [Hints: Same procedure as above. has = 0.464113, T2 = 625.132 [K], so that Q1 = 748.909 [kJ/kg] and W = 347.578 [kJ/kg].] 12. How does the air-standard efficiency of an Otto cycle vary with the compression ratio and the polytropic index? [Hints: Calculate has for rv = 5, 6, 7, 8, 9 and 10 and plot the values on the graph with rv on the X-axis and has on the Y-axis. Repeat the calculations for n = 1.0, 1.1, 1.2, 1.3 and 1.4 and plot the graph with n on the X-axis and has on the Y-axis.] 13. In an Otto cycle with a compression ratio of 7, the pressure and temperature are 0.9 [bar] and 27 [°C], respectively. If the polytropic index for compression and expansion

Chapter 12:

Internal Combustion (I.C.) Engine Cycles

343

is 1.3, calculate (a) the pressure and temperature at the end of compression, and (b) the air-standard efficiency. [Hints: Using the standard formulae, T2 = 300 × (7)0.3 = 537.837 ª 537.8 [K], and p2 = 0.9 × (7)1.3 = 11.2946 ª 11.29 [bar], and has = 1 – 1/70.3 = 0.44221 = 44.22%.] 14. In S.I. engines, the compression ratio is limited by the condition that the air-fuel mixture should not spontaneously start burning due to the high temperature reached after compression. Assume that this will correspond to the ignition (also called autoignition or spontaneous ignition). Experiments show that the auto-ignition temperatures for 70-octane (i.e. petrol with octane number of 70) and 100-octane petrol are, respectively, 300 [°C], and 470 [°C]. Determine the maximum compression ratios in these cases. [Hints: Use the usual formula, rv = (T2/T1)1/(g –1). Then, (a) for 70-octane: rv = (573/300)1/0.4 = 5.04177 ª 5, and (b) for 100-octane: rv = (743/300)1/0.4 = 9.65315 ª 10. Commercial automobile engines use compression ratios between 7 and 10.] 15. In an ideal engine, the clearance volume is 17% of the stroke and the inlet pressure is 0.95 [bar]. If the pressure at the end of constant volume heating is 35 [bar], calculate the IMEP. [Hints: By the usual formula, rv = Vcyl/Vcl = (Vs + Vcl)/Vcl = (Vs + cVs)/cVs = (1 + c)/c = 1.17/0.17 = 6.88235. Then, p2 = 0.95 × (6.88235)0.4 = 14.1434 [bar]; T2 = T1 (6.88235)0.4 = 2.16319T1 [K]; and T3 = T2 × (35/0.95) = 2.47365T2 = 5.35314T1 [K]. Now, the definition of IMEP gives pm,i =

Wrv hQ1rv W W = = = V1 − V2 V1 × (1 − 1/ rv ) V1 (rv − 1) V1 (rv − 1)

But, Q1 = mcv(T3 – T2) = 1.47465(mcvT2) = 3.18995(mcvT1). Substituting this in the expression for IMEP and recalling that, for an ideal gas, cv = R/(g – 1) and using the ideal gas equation reduces it to pm,i = [1 – 1/rvg –1)] × [rv/(rv – 1)] × [3.18995/(g – 1)] p1= 501.725 = 5.107 [bar].] 16. The inlet conditions to an Otto cycle are 1 [bar] and 37 [oC]. The pressures at the end of compression and combustion are 25 [bar] and 75 [bar], respectively. Determine (a) rv, (b) the clearance, and (c) has. [Hints: Since 1–2 is an isentropic process, rv = V1/V2 = (p2/p1)g = 9.96618. If c is the clearance expressed as fraction of the stroke, i.e. c = Vcl/Vs, then, as shown in the text, rv = (1 + c)/c, or c = 0.11153 = 11.53%. The usual formula gives, has = 1 – 1/rv(g –1) = 0.601353 = 60.14%.] 17. A 6-cylinder, 10 [cm] × 9 [cm], 4-stroke S.I. engine with a compression ratio of 7, develops 75 [kW] at 3200 [RPM] and consumes 22 [kg/h] of fuel with the air-fuel ratio of 14. Assume that the charge behaves like air and that the inlet conditions are 1 [bar] and 30 [°C]. Determine the charge efficiency. [Hints: The mass of actual charge inducted = 22 × 15 = 330 [kg/h]. Since air behaves like an ideal gas its density is 100/[(8.3143/29) × 303] = 1.15114 [kg/m3]. The ideal mass inducted = (p/4) × 0.12 × 0.09 × 6 × 1600 × 60 × 1.15114 = 468.687 [kg/h]; and, charge efficiency = 330/468.687 = 0.704095 = 70.41%.]

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18. A 8-cylinder, 9.5 [cm] × 8 [cm], 4-stroke, automotive petrol engine running at 3000 [RPM] has a charge efficiency of 75% and air-fuel ratio of 16. If the inlet state is 1 [bar] and 30 [°C], what will be the fuel consumption? [Hints: The density of air = 100 × 29/(8.3143 × 303) = 1.15114 [kg/m3]. The mass of charge inducted is m a = (p/4) × 0.0952 × 0.08 × (3000/2) × 60 × 8 × 0.75 × 1.15114 = /Z = 352.492/16 = 22.0308 = 352.492 [kg/h]. Hence, the fuel consumption is ª 22.31% [kg/h].] 19. A Diesel engine uses 0.25 [kg] of air and its inlet state is 1 [bar], and 27 [°C]. The pressure at the end of compression is 40 [bar]. If it absorbs 300 [kJ] of heat in each cycle, determine (a) rv and (b) has. [Hints: From the usual formula, rv = (p2/p1)1/g = 13.9421 ª 13.94. Then, T2 = 300 × 13.94210.4 = 860.701 [K], and T3 = T2 + Q1/mcp = 2060.7 [K]; so that, rc = V3/ V2 = T3/T2 = 2.39421. Then, the standard formula gives, has = 0.572339 = 57.23%.] 20. A Diesel engine uses 0.5 [kg] of air with inlet state of 0.95 [bar] and 45 [°C]. The pressure at the end of compression is 33 [bar] and the cut-off is at 6% of stroke. Determine (a) rv, (b) Q1, (c) has, and (d) W. [Hints: Same procedure as above. rv = (33/0.95)1/1.4 = 12.6056 ª 12.6, so that, T2 = 318 × (12.6056)0.4 = 876.299 [K]. Now, rc = V3/V2 = [V2 + 0.06(V1 – V2)]/V2 = 1 + 0.06(rv – 1) = 1.69634. Then, T3 = T2(V3/V2) = 1539.08 [K], so that, Q1 = (mcp) (T3 – T2) ª 305.101 ª 305.1 [kJ]. The standard formula gives, has = 0.592154 = 59.22%; so that W = has × Q1 = 180.667 ª 180.7 [kJ].] 21. An air-standard dual combustion cycle works with 0.5 [kg] of air with inlet state at m 0.97 [bar] and 27 [°C]. Its compressionaf ratio is 9 and at the end of the isochoric combustion, the pressure is 32.5 [bar]. If 100 [kJ] of heat is added during the isobaric heating, determine (a) Q1, (b) Q2, and (c) has. Assume that cp = 1.0 [kJ/kg.K] and g = 1.4. [Hints: Calculate the corner temperature first. T2 = 300 × 90.4 = 722.467 [K]; p2 = 0.97 × 91.4 = 21.0238 [bar], so that, T3 = T2 × (p3/p2) = 1116.84 [K]; T4 = T3 + Q1,p/mcp = 1316.84 [K]. Now, rc = V4/V3 = T4/T3 = 1.17908, so that, T5 = T4 (rc/rv)(g – 1) = 584.054 [K]. Then, proceeding as above, Q1 = Q1,v + Q1,p = 240.846 [kJ], and Q2 = 101.448 [kJ], so that, has = 0.421.212 = 42.12%.] 22. A dual combustion cycle with rv = 12, inducts air at 1 [bar] and 60 [°C]. The pressure at the end of isochoric combustion is 63 [bar] and the cut-off ratio, rc = 1.5. Determine, per [kg] of air, (a) Q1, (b) Q2, and (c) has. [Hints: Same method as above. T2 = 333 × 120.4 = 899.739 [K]; p2 = 1 × 121.4 = 32.423 [bar]; T3 = T2 × (p3/p2) = 1748.25 [K]; T4 = T3 × rc = 2622.38 [K]; T5 = T4 × (rc/rv)(g–1) = 1141.46 [K]; Q1 = 1485,32 [kJ]; Q2 = 579.464 [kJ]; and, has = 0.390127 = 39.01.]

Appendix A

Additional Problems

A.1 INTRODUCTION All the textbooks and examinations (university and others) generally follow a structured approach to problems. Conventionally, textbooks are arranged in the form of chapters and examinations have one part that contains some theory questions. These give ample hints to the approach for solving the problems. However, real life is not so well arranged. One has to guess the approach and sometimes try out more than one approach to determine the proper one. It will be difficult to produce the same ambience in a textbook. However, to give some taste of reality, the problems in this chapter are not arranged in any manner. These problems are taken from the examinations of Kerala University (KU) and Calicut University (CU). Vague and confusingly worded problems, those with wrong and/or insufficient and routine equation substitution problems [e.g. property determination (of steam, gas pressure by ideal gas equation and van der Waal equation, etc.)] have been discarded. Some problems have been reworded to make them clearer. Since books/notes/guides containing adequate solved problems from national examinations like, Central Civil Services, Central Engineering Services, GATE, etc. are available in the market, these are not included. Since the hints to the exercises in the main text have been sufficiently elaborate to conserve space, the hints in this chapter are brief. Default Assumptions For air cp = 1.0 [kJ/kg·K] and g = 1.4, so that cv = cp/g = 1/1.4 = 0.714 [kJ/kg·K] and R = cp – cv = 0.286 [kJ/kg·K]. 1. A vessel of volume 0.04 [m3] contains a mixture of saturated water and saturated steam at a temperature of 250 [°C]. The mass of liquid present is 9 [kg]. Find the pressure, the mass of the mixture and the specific volume. 373

374

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2.

3.

4.

5.

6.

Additional Problems

[Hints: Since the tables in this book do not have an entry for 250 [°C], interpolate. Then, comparing the data with those in steam tables shows that the steam is wet. Then, with the usual formulae give psat = 39.78 [bar] [Ans. (a)], m = 0.496587 [Ans. (b)] and v = 0.0042119 [m3/kg] [Ans. (c)].] A gas of mass 1.5 [kg] undergoes a quasi-static expansion that follows the relation p = a + bV, where a and b are constants. The initial and final pressures are 1000 [kPa] and 200 [kPa] respectively and the corresponding volumes are 0.2 [m3] and 1.2 [m3]. The specific energy of the gas is given by u = 1.5 pV – 85 [kJ/kg], where p is in [kPa] and v is in [m3]. Calculate the work done and the heat transfer. [Hints: The data gives a = 1160 and b = – 800. Then W = 600 [kJ] and DU = 60 [kJ], so that Q = 660 [kJ].] Air flows steadily at the rate of 0.5 [kg/s] through an air compressor entering at 7 [m/s], 100 [kPa], and 0.95 [m3/kg] and leaving at 5 [m/s], 700 [kPa] and 0.19 [m3/kg]. The internal energy of air at exit is 90 [kJ/kg] greater than at entry. Cooling water in the jacket absorbs 58 [kW] of heat. Calculate the shaft power. [Hints: Substituting in SFEE gives W = – 6589.5 [kJ].] A heat engine is used to drive a heat pump. The heat transfers from the heat engine to the heat pump which is used to heat the water circulating through the radiator of the building. The efficiency of the heat engine is 27% and COP of the heat pump is 4. Evaluate the ratio of heat transfer to the circulating water to that to the heat engine. [Hints: Let the heat absorbed and rejected by engine be Q1 and Q2. Similarly, let Q3 and Q4 be the heat absorbed and rejected by the heat pump. Then, W = 0.27Q1, Q2 = 0.73Q1, Q4 = 4 × 0.27Q1 and the heat transferred to water is Q2 + Q4, so that (Q2 + Q4)/Q1 = 1.81 [Ans.].] A closed system consists of 1 [kg] of air that is initially at 1.5 [bar] and 67 [°C]. The volume doubles as the system undergoes a process according to the law pV1.2 = constant. Determine the work done, the heat transfer, and the change in entropy. [Hints: Usual procedure. W = 63.1 [kJ], DU = –31.5 [kJ], so that Q = 31.6 [kJ] and DS = 0.9939 [kJ/K].] 0.5 [kg] helium and 0.5 [kg] nitrogen are mixed at 20 [°C] at a total pressure of 100. Find (i) the volume of the mixture, (ii) the partial pressures of the components, (iii) the specific heats of the mixture, and (iv) the gas constant of the mixture. Assume MWHe = 4 [kg/kmol], MWN2 = 28 [kg/kmol] and gHe = 1.667 and gN2 = 1.4. [Hints: Usual approach. Since gases behave ideally, partial volumes are VHe = 3.0451 [m3] and VN2 = 0.4350 [m3], so that by Amagat’s law, VT = 3.48 [m3] [Ans. (i)]. By Dalton’s law, pHe = 87.5 [kPa] and pN2 = 12.5 [kPa] [Ans. (ii)]. Then, linear mixing rule gives cv = 1.9293 [kJ/kg·K], cp = 3.1171 [kJ/kg·K] [Ans. (iii)] and R = 1.1878 [kJ/kg G-K] [Ans. (iv)].]

Appendix A: Additional Problems

375

7. Saturated water vapour at 200 [°C] is contained in a cylinder fitted with a piston. The initial volume of steam is 0.01 [m3]. The steam then expands in a quasi-equilibrium isothermal process until the final pressure is 200 [kPa] and in doing so does work against the piston. (a) Determine how much work is done against the piston, (b) how much error would be made by assuming the steam to behave as an ideal gas. [Hints: Steam tables show that pi = 15.5 [bar] and m = 0.0784 [kg]. Using numerical integration (trapezoidal rule will do), W = 31.85 [kJ].] 8. A rigid vessel having a volume of 5 [m3] contains 0.05 [m3] of saturated liquid later and the rest of the volume as saturated vapour at 0.1 [MPa]. Heat is transferred until the vessel is filled with saturated vapour. Determine the heat transfer and the work done during the process. [Hints: The data implies wet steam at 100 [kPa]. Then, since the vessel is rigid, W = 0. The data also shows that initially, mw = 47.92 [kg] and ms = 2.9226 [kg], so that v2 = 0.098343. Then, by interpolation, the steam tables give p2 = 20.26 [bar]. Then, by the usual procedure, Q = DU = 104778 [kJ].] 9. Air of mass 0.5 [kg] is compressed reversible and adiabatically from 80 [kPa], 60 [°C] to 0.4 [MPa], and is then expanded at constant pressure to the original volume. Sketch the process on p~v plane. Determine the heat transfer and the work transfer. Assume R = 0.287 [kJ/kg.K] and cv = 0.713 [kJ/kg·K]. [Hints: There are two common processes. So, the usual formulae give W = 100.6 [kJ] and Q = – 178.3 [kJ].] 10. In a steam generator, compressed liquid water at 10 [MPa], 30 [°C] enters a 30 [mm] diameter tube at the rate of 5 [litres/s]. Steam at 9 [MPa], 400 [°C] exits the tube. Find the rate of heat transfer to the water. [Hints: SFEE gives Q = m (he – hi) = 6918.3 [kJ].] 11. Two reversible heat engines A and B are arranged in series with A rejecting heat directly to B through an intermediate reservoir. Engine A receives 200 [kJ] of heat from a reservoir at 421 [°C] and engine B is in thermal communication with a sink at 4.4 [°C]. If the work output of A is twice that of B, find (a) the intermediate temperature between A and B, (b) the efficiency of each engine, and (c) the heat rejected to the cold sink. [Hints: Let Q1 and Q2 be the heat absorption and heat rejection of engine A. Since engine A rejects heat directly to the engine B, the intermediate reservoir can be eliminated. However, for calculations let its temperature be T2. Since the engines are reversible, Q2/Q1 = T2/T1, Q3/Q2 = T3/T2 and Q3/Q1 = T3/T1. We also have WA = Q1 – Q2 = 2 WB = Q2 – Q3. Solving these equations with the given data gives T2 = 416.4 [K] [Ans. (a)], hA = 0.4 and hB = 0.33 [Ans. (b)] and Q3 = 79.94 [kJ] [Ans. (c)].] 12. A cylinder contains 0.5 [m3] of a gas at 0.1 [MPa] and 90 [°C]. The gas is compressed to a volume of 0.125 [m3]. The final pressure is 600 [kPa]. Determine the work done and the change in entropy of the gas during the process. Assume R = 0.287 [kJ/kg·K] and cv = 0.713 [kJ/kg·K].

376

13.

14.

15.

16.

17.

18.

Appendix A:

Additional Problems

[Hints: Usual procedure. The data gives, the index of the process, pV n = C, where n = 1.2925, so that W = – 85.47 [kJ] and DS = – 0.07205 [kJ/K], implying thereby that the process cannot be executed.] A mixture of 2 [kg] oxygen and 2 [kg] argon is in an insulated piston cylinder arrangement at 100 [kPa], 300 [K]. The piston now compresses the mixture to half its initial volume. Find the final pressure, the temperature, and the piston work. Molecular weight of oxygen is 32 and of argon it is 40. Ratio of specific heats for oxygen is 1.39 and for argon it is 1.667. [Hints: Usual procedure. xO2 = 0.5556, so that xAr = 0.4444. Also, since gases behave ideally, VO2 = 1.55893 [m3] and VAr = 1.24715 [m3]. Then. by Amagat’s law V1 = 2.80608 [m3]. Since the compression is isentropic, p2 = 142.7 [kPa] and T2 = 214.] A vessel contains 0.12 [m3] of air at 1 [bar] and 90 [°C]. The air is compressed reversibly to a volume of 0.03 [m3], the final pressure being 6 [bar]. Determine the mass of air, the change in internal energy of air, and the heat transfer. [Hints: Usual procedure. m = 0.1152 [kg], T2 = 544.4 [K], DU = 14.92, process index, n = 1.29, W = – 20.55 [kJ] and Q = – 5.63 [kJ].] Air is supplied to the combustion chamber of a gas turbine at the rate of 12 [kg/s]. The temperature of the air is 120 [°C] and its velocity is 86 [m/s] with its enthalpy being 176 [kJ/kg]. Liquid fuel, at 15 [°C], flows into the combustion chamber at the rate of 700 [kg/h]. Products of combustion leave the chamber at 760 [°C], a velocity of 200 [m/s] and enthalpy 775 [kJ/kg]. Determine the specific enthalpy of entering fuel. Neglect fuel velocity and heat transfer to the atmosphere. [Hints: Substituting in SFEE gives h = 38, 767.9 [kJ/kg].] A certain gas has cv = 0.3 [kJ/kg.K]. Find the change in entropy per kg of gas in each of the following processes. (i) When it is expanded reversible and adiabatically from a specific volume of 1 [m3/kg]. And temperature 555 [°C] to a specific volume of 3 [m3/kg] its temperature falls by 166 [°C]. (ii) When it is expanded adiabatically from the same initial state to the same final specific volume while the temperature falls by 25 [°C] only. [Hints: For (i) DS = 0 (isentropic process) and data to this part gives g = 1.2, R = 0.0612. (ii) Then, for the second preprocess, DS = 0.058 [kJ/kg·K].] Steam initially at 1.5 [MPa], 300 [°C] expands reversible and adiabatically in a steam turbine to 40 [°C]. Determine the ideal work output of the turbine per kg of steam. Also show the process on T–s and h–s diagrams. [Hints: Usual method, s1 = s2 and W = h1 – h2 = 883.3 [kJ/kg].] A system receives 200 [kJ] of energy as heat, at constant volume. Then it is cooled at constant pressure when 50 [kJ] of work was done on the system while it rejects 70 [kJ] of heat. Supposing the system is restored to the initial state by an adiabatic process, how much work will be done by the system? [Hints: Usual method, W = 320 [kJ].]

Appendix A: Additional Problems

377

19. Steam at 10 [bar], 260 [°C] expands hyperbolically to a pressure of 2 [bar]. Determine the properties of the steam after expansion. [Hints: Since the expansion is hyperbolic, p1v1 = p2v2, so that x2 = 0.2677, h2 = 1094.1 [kJ/kg] and s2 = 3.0283 [kJ/kg·K].] 20. A cylinder contains 0.5 [m3] of a gas at 0.1 [MPa] and 90 [°C]. The gas is compressed to a volume of 0.125 [m3]. The final pressure is 600 [kPa]. Determine the work done and the heat transfer from or to the gas during the process. Assume R = 0.287 [kJ/kg·K] and cv = 0.714 [kJ/kg·K]. [Hints: Usual procedure. The process index, n = 1.292, so that, W = – 81.62 [kJ] and Q = – 24.62 [kJ].] 21. Air flows steadily at the rate of 0.5 [kg/s] through an air compressor entering at 7 [m/s] velocity, 100 [kPa] pressure and 0.95 [m3] specific volume, and leaving at 5 [m/s], 700 [kPa] and 0.19 [m3/kg]. The internal energy of air leaving is 90 [kJ/kg] greater than that of the air entering. Cooling water in the compressor jackets absorb heat at the rate of 58 [kW]. Calculate the rate of shaft work input to the compressor. [Hints: SFEE gives W = –122 [kW].] 22. A reversible heat engine operates between two systems at 600 [°C] and 40 [°C]. The engine drives a reversible refrigerator that operates between 40 [°C] and – 20 [°C]. The heat transfer to the engine is 2000 [kJ] and the network output of the combined engine-refrigerator plant is 350 × 103 [N.m]. Evaluate (a) the heat transfer to the refrigerant and the net heat transfer to the system at 40 [°C]. Also (b) recalculate the same, if the efficiency of the heat engine and the COP of the refrigerator are each 40% of their maximum possible values. [Hints: In the usual symbols, for engine, WE = Q1 – Q2, Q2/Q1 = T2/T1 and for refrigerator, WR = Q4 – Q 3, Q 4/Q3 = T4/T3. Given data are Q1 = 2000 [kJ] and WE – WR = 350 [kJ]. Solving these equations gives Q3 = 3933.7 [kJ] and Q4 + Q2 = 3802.2 [kJ].] 23. A 2 [kg] piece of iron is heated from room temperature of 25 [°C] to 400 [°C] by a heat source at 600 [°C]. What is the irreversibility in the process? Assume for iron cp = 0.450 [kJ/kg·K]. [Hints: I = Wu,max – Wu = Wu,max (since Wu = 0) = 149.8 [kJ].] 24. A valve connects two heavily insulated tanks. Tank A has a volume of 5 [m3] and contains initially oxygen at 15 [°C], 400 [kPa]. Tank B has a volume of 35 [m3] and initially contains nitrogen at 35 [°C], 150 [kPa]. The valve is now opened and remains open until the mixture comes to uniform state. Determine (i) the final pressure and temperature, and (ii) the entropy change during the process. Assume gN2 = 1.4 and gO2 = 1.39. [Hints: Since the gases behave ideally, mA = 26.7277 [kg], mB = 57.4038 [kg], so that xA = 0.2897 and xB = 0.7105. The linear mixing rule gives T2 = 302.2 [K], so that p2 = 180.4 [kPa].]

378

Appendix A:

Additional Problems

25. Air at 100 [kPa], 20 [°C] is taken into a gas turbine power plant at a velocity of 150 [m/s] through an opening of 0.15 [m2] cross-sectional area. The air is compressed, heated, expanded through a turbine, and exhausted at 0.18 [MPa], 10 [°C] through an opening of 0.10 [m2] cross-sectional area. The power output is 375 [kW]. Calculate the net amount of heat added to the air in [kJ/kg]. Assume that air obeys the law pv = 0.287 (t + 273), where p is the pressure in [kPa], v is the specific volume in [m3/kg], and t is the temperature in [°C]. Take cp = 1.005 [kJ/kg·K]. [Hints: Substituting in SFEE gives Q = 510.7 [kW].] 26. A heat pump is used to heat a house in winter and then reversed to cool the house in summer. The interior temperature is to be maintained at 20 [°C] during winter and summer. The heat transfer through walls and roof is estimated to be 2400 [kJ] per hour per degree temperature difference between the inside and outside. Determine (a) the minimum power to drive the heat pump during winter, if the outside temperature is 4 [°C], and (b) the minimum power required to run the machine during summer, if the outside temperature is 42 [°C]. [Hints: Usual method. For heat pump, Whp = 0.5825 [kW] and Wref = 1.101 [kW].] 27. An aluminium block (cp = 400 [J/kg.K]) with a mass of 10 [kg] is initially at 50 [°C] in room air at 27 [°C]. The aluminium block is allowed to cool by natural convection to room air until block reaches 27 [°C]. The room temperature is unchanged during the process. Compute (a) the change in entropy for the block, (b) the change in entropy for the room air, and (c) the net change of entropy for the universe. [Hints: The usual formulae give (DS)block = – 0.2955 [kJ/K], (DS)air = 0.3067 [kJ/K] and hence, (DS)U = 0.0112 [kJ/K].] 28. A diaphragm divides a closed rigid cylinder into two equal compartments each of volume 1 [m3]. Each compartment contains air at a temperature of 20 [°C]. The pressure in one compartment is 2 [MPa] and in the other compartment is 1 [MPa]. The diaphragm is ruptured, the air in both the compartments mixes to bring the pressure to a uniform value throughout the cylinder that is insulated. Find the net change in entropy for the mixing process. [Hints: Since air behaves ideally, each process is a free expansion, so that (DS)U = (DS)A + (DS)B = 7.096 [kJ/K].] 29. The temperature t on a thermometric scale is defined in terms of a property K by the relation t = a ln k + b, where a and b are constants. The values of K are found to be 1·83 and 6·78 at the ice point and steam point which are assigned values 0 and 100 respectively. Determine the temperature corresponding to a reading of K equal to 2·42 on the thermometer. [Hints: Standard procedure. t = 21.3° (Ans.)] 30. A gas occupies 0·024 [m3] at 700 [kPa] and 95 [°C]. It is expanded according to the law pV1.2 = constant to a pressure of 70 [kPa] after which it is heated at constant pressure back to its original temperature. Calculate for the whole process the heat transfer and the work done. Take cp = 1·05 and cv = 0·775 [kJ/kg.K] for the gas. [Hints: There are two standard processes. Then, ee gives W = 26.78 [kJ] and Q = 11.69 [kJ].]

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379

31. A turbo-compressor delivers 2·33 [m3/s] at 0·276 [MPa], 43 [°C] which is heated at this constant pressure to 430 [°C]. and finally expanded in a turbine which delivers 1860 [kW]. During the expansion there is a heat transfer of 0·09 [MJ/s] to the surrounding. Calculate the turbine exhaust temperature, if changes in potential and kinetic energy are negligible. [Hints: Solving the SFEE with the values gives Te = 303.9 [°C].] 32. Two reversible heat engines A and B are arranged in series. A rejecting heat directly to B. Engine A receives 200 [kJ] at a temperature of 421 [°C] from a hot source. While engine B is in communication with a cold sink at a temperature of 4.4 [°C]. If the work output of A is twice that of B, find (i) the intermediate temperature between A and B, (ii) the efficiency of each engine, and (iii) the heat rejected to the cold sink. [Hints: Let Q1 and Q2 be the heats absorbed and rejected by A from reservoirs at T1 and T2 respectively. Then, B will absorb Q2 at T2 and reject Q3 at T3. The given data is WA = Q1 – Q2 = 2WB = 2(Q2 – Q3) with Q1 = 200 [kJ]. Since the engines are reversible, Q3/Q1 = T1/T2. Solving all these equations give T2 = 417.6 [K], hA = 0.40, hB = 0.33 and Q3 = 79.48 [kJ].] 33. A tube of negligible volume connects two vessels A and B each of volume 3 [m3]. Vessel A contains air at 0·7 [MPa], 95 [°C]. While vessel B contains air at 0·35 [MPa], 205 [°C]. Find the change in entropy, when air in A mixes with B adiabatically and completely. For air cp = 1 [kJ/kg.K] and R = 0·287 [kJ/kg.K]. [Hints: Linear mixing rule gives, T2 = 398.6 [K]. Then, air behaves ideally, the standard formula gives (DS)U = (DS)A + (DS)B = 5.6183 [kJ/K].] 34. Steam at 0·8 [MPa], 250 [°C] flowing at the rate of 1 [kg/s] passes into a pipe carrying wet steam at 0·8 [MPa], 0·95 dry. After adiabatic mixing the flow rate is 2·3 [kg/s]. Determine the condition of steam after mixing. [Hints: ce gives m = 1.3 [kg/s] and the SFEE shows that exit steam is superheated. Interpolation gives its temperature as 180.4 [°C].] 35. A gaseous mixture of 1 [kg] oxygen and 2 [kg] nitrogen is at a pressure of 150 [kPa] and temperature of 20 [°C]. Determine the changes in internal energy, enthalpy and entropy of the mixture, when the mixture is heated to a temperature of 100 [°C] in a constant volume process. MWN2 = 28; MWO2 = 32 [kg/kmol], cpN2 = 1·04 [kJ/kg·K] and cpO2 = 0·92 [kJ/kg.K]. [Hints: Standard problem on ideal gas mixtures. xO2 = 0.3043 and xN2 = 0.6957. Then, linear mixing rule gives Mmix = 29.22 [kgmix/kmolmix], Rmix = 0.284, cp.mix = 1.0 [kJ/kg.K], cv = 0.716 [kJ/kg.K], so that (DU) = 171.8 [kJ], (DH) = 240 [kJ], and (DS) = 0.5185 [kJ/K].] 36. Two boilers one with superheater and other without superheater are delivering equal quantities of steam into a common main. The pressure in the boilers and main is 20 [bar]. The temperature of steam from a boiler with a superheater is 350 [°C] and temperature of the steam in the main is 250 [°C]. Determine the quality of the steam. [Hints: Linear mixing rule gives h2 = 2666.2 [kJ/kg], so that x2 = 0.953.]

380

Appendix A:

Additional Problems

37. An insulated cylinder having an initial volume of 25 litres contains oxygen at 150 [kPa], 227 [°C]. The gas is compressed to 1.5 [MPa] in a reversible adiabatic process. Calculate the final temperature and work, assuming oxygen behaves as an ideal gas with cp = 0.922 [kJ/kg.K]. [Hints: Since O2 behaves ideally, RO2 = 0.260 [kJ/kg.K], so that cv = cp – R = 0.662 [kJ/kg.K] and g = 1.39. Since the process is isentropic, the process equation gives V2 and T2. Then, the usual formula gives W = – 8.731 [kJ] and T2 = 286.2 [K].] 38. 10 [kg/s] of chilled water for air-conditioning enters a tall building with a velocity of 50 [m/s] at an elevation of 30 [m] from the ground. The water leaves the system with a velocity of 10 [m/s] at an elevation of 60 [m]. The enthalpies of water entering and leaving are 21 [kJ/kg] and 43 [kJ/kg] respectively. The ratio of work done by a pump in the line is 35 [kW]. Calculate the rate at which heat is removed from the building. [Hints: SFEE gives Q = 174.5 [kW].] 39. One [kg] mass of air is compressed reversibly from 6.5 [bar], 0.0135 [m3] to a final volume of 0.01 [m3]. Find the final pressure, the final temperature, the work done, and the change in internal energy and heat interaction; if the compression is (a) adiabatic and (b) polytropic with n = 1.3. [Hints: Usual problem on gas expansion/compression. p2 = 9.89 [bar], T2 = 34.47 [K], W = – 2.798 [kJ], DU = + 2.798 [kJ] and Q = 0 [Ans. (a)] and p2 = 9.602 [bar], T2 = 33.45 [K], W = – 2.757 [kJ], DU = + 2.056 [kJ] and Q = – 0.701 [kJ] [Ans. (b)].] 40. Air at 20 [°C] and 1.05 [bar] occupies 0.025 [m3]. The air is heated at constant volume until the pressure is 4.5 [bar] and then cooled at constant pressure back to original temperature. Calculate (a) the net heat flow from air, and (b) the net change in entropy. [Hints: Q = – 8.62 [kJ] [Ans. (a)] and, since the end state of the second process is the same as the initial state of the first process DS = 0 [Ans. (b)].] 41. A mixture containing 1 [kg] of H2 and 2.5 [kg] of N2 at 300 [K] and 100 [kPa] is compressed in reversible adiabatic process to 700 [kPa]. Determine (a) the partial pressures of the constituents at the end of compression, (b) the final temperature, and (c) the change in internal energy of the mixture. [Hints: The usual problem on ideal gas mixtures, so that xH2 = 0.8485, xN2 = 0.1515, pH2 = 84.85 [kPa], pN2 = 15.15 [kPa], cp, cv (by linear mixing rule) and so, g, which is process index. Then, T2 = 523.1 [K] and DU = 2732.5 [kJ].] 42. Estimate the density of nitrogen at 170 [K] and 4 [MPa], if compressibility factor is 0·85. [Hints: By standard formula, 92.2233 [kg/m3].] 43. An insulated 2 [kg] box falls from a balloon 3·5 [km] above the earth. What is the change in K.E. of the box after it hits the earth’s surface? [Hints: Dek = – Dep = [mgh] = 68.67 [kJ].] 44. An inventor claims an efficiency of 52% for his heat engine working between temperature limits of 500 [K] and 1000 [K]. Justify his claim. [Hints: hmax = hCarnot = 0.5, so that the claim is false.]

Appendix A: Additional Problems

381

45. Air enters a nozzle at a pressure of 2700 [kPa] at a velocity of 30 [m/s] and with an enthalpy of 900 [kJ/kg] and leaves with a pressure of 700 [kPa] and enthalpy of 660 [kJ/kg]. Find the exit velocity (i) for adiabatic condition, (ii) for a heat loss of 1 [kJ/kg]; if mass flow rate is 0·2 [kg/s]. [Hints: By SFEE, (i) Ve = 693.5 [m/s], and (ii) Ve = 693.2 [m/s].] 46. 3 [kg] of air is heated from 300 [K] to 800 [K] while the pressure changes from 100 [kPa] to 500 [kPa]. Calculate the change in entropy. [Hints: Standard formula gives DS = 1.5568 [kJ/K].] 47. An isolated expansion of air occurs in a piston-cylinder from 298 [K] and 800 [kPa] to 225 [kPa]. Determine the change in entropy per kg. [Hints: Isolated expansion means free expansion, so that the standard formula gives, Ds = 0.3641 [kJ/kg.K].] 48. 500 [kJ] of heat is removed from a constant temperature reservoir at 835 [K]. The heat is received by a system at constant temperature of 700 [K]. The temperature of surroundings is 280 [K]. Find the net loss of available energy due to this irreversible heat transfer. [Hints: As in Example 7.5, I = T0 (DS)U = 32.34 [kJ].] 49. Determine the specific gas constant and the molecular mass of a mixture containing 20% oxygen, 20% nitrogen and 60% carbon dioxide by volume. [Hints: Since the gases are ideal, volume fractions are mole fractions, so that linear mixing rule gives Mmix = 47.2 [kg/kmol]. Then, R = 8.3143/Mmix = 0.2165 [kJ/kgmixK].] 50. A vessel of spherical shape of capacity 0.8 [m3] contains steam at 10 [bar], 0.95 dryness. Steam is blown off until the pressure drops to 5 [bar]. The valve is then closed and the steam is allowed to cool until the pressure falls to 4 [bar]. Assuming that the enthalpy of steam in the vessel remains constant during blowing of periods. Determine (1) the mass of steam blown off, (2) the dryness fraction of stream in the vessel after cooling, and (3) the heat lost by steam per kg during cooling. [Hints: Usual procedure. m1 = 4.333 [kg]. Since h remains constant during blowing, h1 = h2, so that x2 = 0.9658. Since the vessel is rigid, m2 = 2.2105 [kg], so that mblown off = 2.1225 [kg]. Then, x3 = 0.7808, so that h3 = 2270.1 [kJ/kg]. Then, DU = DH – D(pV) = –1014.2 [kJ].] 51. 5 [kg] of nitrogen is cooled in a rigid tank from 250 [°C] to 27 [°C]. The pressure is 25 [bar]. Calculate the changes in entropy, internal energy and enthalpy. Assume nitrogen to be an ideal gas with cp = 1.042 [kJ/kg.K] and cv = 0.745 [kJ/kg.K]. [Hints: Standard problem. DU = – 830.1 [kJ], DH = – 1161.8 [kJ], and DS = – 2.8958 [kJ/K].] 52. 10 [kg] of fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are: pressure p1 = 1.5 [bar], density r1 = 26 [kg/m3], velocity = 110 [m/s] and internal energy = 910 [kJ/kg] and at exit are: pressure p2 = 5.5 [bar], density r2 = 5.5 [kg/m3], velocity = 190 [m/s] and internal energy = 710 [kJ/kg]. During the passage, the fluid rejects 55 [kJ/s] and rises 55 metres. Determine (1) the change in enthalpy and (2) the work done during the process. [Hints: Using the definition h, Dh = –105.77 [kJ/kg], so that SFEE gives W = – 39.33 [kW].]

382

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Additional Problems

53. A fluid contained in a horizontal cylinder fitted with a frictionless leakproof piston is continuously agitated by means of a stirrer passing through the cylinder. The cylinder, diameter is 400 [mm]. During a stirring process lasting 10 [min], the piston slowly moves outwards a distance of 500 [mm] against the atmosphere at 1 [bar]. The net work done by the fluid during the process is 2000 [N.m]. The speed of the electric motor driving the stirrer is 800 [RPM]. Compute the torque acting on the driving shaft and the shaft power output of the motor. The electric motor is supplied with current from 24 [V] accumulator. If the drawn is 1.2 [A], compute the network done by the accumulator. [Hints: There are Wx and Ws. The data gives Wx = 6.2832 [kJ], so that Ws = – 4.2832 [kJ], which means that t = – 85.21 [N.m]. Usual formula gives We = 1.728 [kW].] 54. A cyclic heat engine operates between a source temperature of 1000 [°C] and a sink temperature of 40 [°C]. Find the least rate of heat rejection per [kW] net output of the engine. [Hints: Minimum Q2 implies maximum W, which corresponds to reversible (Carnot) engine. Then, standard formula is manipulated to give Q2/W = 0.326.] 55. (1) One kg of water at 300 [K] is heated to 500 [K] by bringing it in contact with a heat reservoir at 500 [K]. Determine the entropy change of the universe. (2) If instead the water is first heated to 400 [K] by bringing it in contact with an intermediate heat reservoir at 400 [K] and then to 500 [K] as before, what will be entropy change of the universe in this case? [Hints: Since (DS)U = (DS) + (DS)R, the standard formula gives, (1) (DS)U = 0.464 [kJ/K] and (2) (DS)U = 0.255 [kJ/K].] 56. A vessel of 0.2 [m3] capacity contains 2 [kg] of CO2 and 1.5 [kg]. of N2 at 300 [K]. Determine (1) the pressure in the vessel, (2) the mole fraction of each constituent, and (3) the R and M of the mixture. [Hints: Standard problem in ideal gas mixtures. p = pCO2 + pN2, where each partial pressure is obtained using ideal gas equation for the constituent gas. So, p = 1235.2 [kPa], xCO2 = 0.459, xN2 = 0.541 and linear mixing rule gives Mmix = 35.34 [kJ/kg.K], so that Rmix = 8.3143/Mmix = 0.235 [kJ/kg.K].] 57. The internal energy of certain substance is given by the following equation; u = 3.56 pv + 84, where u is in [kJ/kg], p is in [kPa] and v is in [m3/kg]. A system composed of 5 [kg] of this substance J expands from an initial pressure of 500 [kPa] and volume of 0.22 [m3] to a final pressure of 100 [kPa] in a process in which pressure and volume are related by pv1.25 = constant. If the expansion is quasi-static, find Q, DU and W for the process, where Q is the heat transfer, DU is the change in internal energy and W is the work transfer. [Hints: The process law gives v2 = 0.7972 [m3/kg], so that W = 12.12 [kJ]. The given equation gives DU = – 33.308 [kJ], so that ee gives Q = 87.812 [kJ].] 58. A pump is used to inflate a perfectly elastic balloon. The air pressure inside the pump and the balloon are the same and equal to the atmospheric pressure. Estimate the work done on the piston if it is moved very slowly. Assume adiabatic boundary. The volume

Appendix A: Additional Problems

59.

60.

61.

62.

63.

383

of inflated balloon is 0.001 [m3], the filling is done in one stroke. Assume suitable value for atmospheric pressure. [Hints: W = pDV (since p is constant) = 0.1 [kJ].] A closed system consists of water in a vessel. The water is stirred by a paddle wheel. The work done for stirring the water is 34 [kJ/hr]. The initial internal energy is 120 [kJ] and the final internal energy is 144 [kJ] after half an hour. The rate of heat transfer is uniform with respect to time. The variation of internal energy with time is also uniform. Find the heat transfer in two hours. [Hints: W = 34 [kJ/h], D U = (144 – 120)/0.5 = 48 [kJ/h] and ee gives Q = 82 [kJ/h], so that Q = 164 [kJ].] A reversible heat engine operates between two reservoirs at temperatures of 1000 [°C] and 30 [°C]. The engine drives a reversible refrigerator that operates between reservoirs at temperatures of 30 [°C] and –10 [°C]. The heat transfer to the heat engine is 2000 [kJ] and the corresponding net work output of the combined engine refrigerator plant is 400 [kJ]. Determine (a) the heat transfer to the refrigerator and (b) the net heat transfer to the reservoir at 30 [°C]. [Hints: Since engine and refrigerators are reversible, the heats exchanged are proportionate to temperatures. Thus, Q2 = 2000(303/1273) = 476 [kJ], but WE = Q1 – Q2 = 1524 [kJ]. So, Wref = 1524 – 400 = 1124 [kJ]. For the combined system, Q3 = 2000(263/1273) = 413.2 [kJ] [Ans. (a)]. Then, Q4 = Q3 + Wref = 1537.2 [kJ] so that Q2 + Q4 = 2013.2 [kJ] [Ans. (b)].] A cylinder containing 1 [mole] of CO at 172 [kPa] and 30 [°C] is connected through a valve to a closed cylinder containing 1 [mole] of [N2] at 103.4 [kPa] and 15 [°C]. The valve is opened and adiabatic mixing takes place. Find the change in entropy. [Hints: Since the gases are ideal, these are free expansions so that the standard formula gives (DS)U = (DS)CO2 + (DS)N2 = 0.0119 [kJ/K].] A rigid tank contains N2 at a pressure of 0.8 [MPa] and 50 [°C]. A leak occurs in the tank that not detected until the pressure falls to 0.4 [MPa] at which time the temperature is 30 [°C]. Find the mass of nitrogen that has leaked out, if the original mass was 40 [kg]. [Hints: Since N2 behaves ideally, V = 4.7966 [m3] so that m2 = 21.32 [kg] and mleak = 18.68 [kg].] A piston and cylinder machine contains a fluid system which passes through a complete cycle. During a cycle, the sum of all heat transfers is equal to 150 [kJ]. The system completes 100 [cycles/min]. Compute the power of the machine.

v∫

[Hints: ee for a cycle is W = dQ = 150 [kJ/cycle] so that P = 150 × (100/60) = 250 [kW].] 64. At the inlet of a nozzle, the enthalpy of the fluid passing is 3000 [kJ/kg] and the velocity is 50 [m/s]. The nozzle is horizontal and there is negligible heat loss from it. At the discharge end the enthalpy is 2762 [kJ/kg]. (a) Find the velocity at exit from

384

Appendix A:

Additional Problems

the nozzle; (b) If the inlet area is 15 m2 and the specific volume at inlet is 0.187 [m3/kg], find mass flow rate and (c) if the specific volume at the nozzle exit is 0.498 [m3/kg], find the exit area of the nozzle.

65.

66.

67.

68.

69.

70.

[Hints: Standard formulae for nozzle give Ve = 691.7 [m/s] [Ans.(a)], m = 40.11 [kg/s] [Ans. (b)] and Ae = 0.0289 [m2].] A heat engine operates between a source temperature 900 [°C] and a sink temperature of 27 [°C]. What is the least rate of heat rejection per kW of net output of the engine? [Hints: Minimum heat rejection implies maximum work done implies reversible engine. Then, the usual formulae can be manipulated to get Q2/W = 0.3436.] A reversible engine interacts with three heat reservoirs. During one cycle of operation it draws 5000 [kJ] from a 400 [K] reservoir and does 840 [kJ] of work. Find the amount and direction of heat interaction with the other two heat reservoirs maintained at 300 [K] and 200 [K]. [Hints: Let Q1, Q2 and Q3 be the heats exchanged with reservoirs at 400 [K], 300 [K] and 200 [K]. Then, 5000 + Q2 – Q3 = 840 [kJ], Q2 = 5000(300/400) = 3750 [kJ] so that Q3 = 5000 – 840 – 3750 = 410 [kJ].] A mixture of ideal gases consists of 6 [kg] of Oxygen and 4 [kg] of Nitrogen at a pressure of 100 [kPa] and temperature 27 [°C]. Find (a) the equivalent molecular weight of the mixture (b) the equivalent gas constant of the mixture; (c) partial pressure; and (d) volume of the mixture. [Hints: Linear mixing rule gives Mmix = 30.27 [kg/kmolmixK] [Ans. (a)], Rmix = 8.3145/30.27 = 0.275 [kJ/kgmixK] [Ans. (b)], pO2 = 56.76 [kPa] [Ans. (c)], pN2 = 43.23 [kPa] [Ans. (c)] and by ideal gas law, Vmix = 2.75 [m3].] An automobile tube of volume 0.014 [m3] has air at 20 [°C] and 130 [kPa] (gage). If the pressure inside the tube has to be raised to 220 [kPa] (gage), how much air needs to be pumped in? Assume the temperature and volume to remain constant and take atmospheric pressure to be 101 [kPa]. [Hints: Since V and T remain constant, ideal gas law gives m1 = 0.038458 [kg] and m2 = 0.053442 [kg] so that Dm = 0.01498 [kg].] A 0.5 [m3] tank contains air at pressure 7 [MPa] and 250 [°C] and is perfectly insulated from the surroundings. A valve on the tank is opened and air discharged until the pressure drops to 400 [kPa]. Calculate the mass of air discharged from the tank. [Hints: Ideal gas law gives m1 = 23.3176 [kg]. Perfectly insulated process implies isentropic process whose process index is 3. Then, the process law gives T2 = 230.9 [K] so that ideal gas law gives m2 = 3.01803 [kg] and Dm = 22.41 [kg].] Air at 20 [°C] and 1.05 [bar] occupies 0.025 [m3] that is heated at constant volume until the pressure is < 1.5 [bar], and then at constant pressure back to original temperature. Calculate (i) the net heat flow from the air and (ii) the net entropy change. [Hints: Standard problem on gas expansion, but there are two successive processes involved, so that Q = – 8.60 [kJ] [Ans. (i)] and DS = 0.01304 [kJ/K].]

Appendix A: Additional Problems

385

71. A pressure cooker contains 1.5 [kg] of steam at 5 [bar] and 0.9 dryness when the gas was switched off. Determine the quantity of heat rejected by the pressure cooker, when the pressure in the cooker falls to 1 [bar]. [Hints: Since pressure cooker is a rigid vessel, the process is an isochoric cooling, v1 = v2, so that x2 = 0.1987, H2 = 1299.2 [kJ]. Then Q = DU = DH – D(pv) = DH – pD(v) = – 2708.3 [kJ].] 72. A vessel of 0.2 [m3] capacity contains 2 [kg] of CO2 and 1.5 [kg] of N2 at 300 [K]. Determine (i) the pressure in the vessel; (ii) the mole fraction of each constituent; and (iii) the R and M of the mixture. [Hints: Standard method. xCO2 = 0.4590 and xN2 = 0.5410 [Ans. (ii)]. Since component gases behave ideally, pCO2 = 567 [kPa] and pN2 = 891 [kPa], so that p = 1458 [kPa] [Ans. (i)], Mmix = 35.34 [kgmix/kmolmix] and Rmix = 0.235 [kJ/kgmixK].] 73. Calculate the internal energy of 0.3 [m3] of steam at 4 [bar] and 0.95 dryness. If this steam is superheated at constant pressure through 30 [°C], determine the heat added and the change in internal energy. [Hints: Usual steam problem, Q = DH. Then, m = 0.6816 kg, h1 = 2631.0 [kJ/kg] and u1 = 2455.0 [kJ/kg]. Superheated by 30 [°C] implies T2 = 174 [°C]. Then, by interpolation, h2 = 2804.0 [kJ/kg] and u2 = 2603.5 [kJ/kg], so that Q = 117.92 [kJ] and DU = 101.22 [kJ].] 74. A mass of 0.25 [kg] of an ideal gas has a pressure of 300 [kPa] at 80 [°C] and a volume of 0.07 [m3]. The gas undergoes an irreversible adiabatic process to a final pressure of 300 [kPa] and final volume of 0.1 [m3] during which the work done on the gas is 25 [kJ]. Evaluate the cp, cv and increase in entropy of the gas. [Hints: Initial data gives the specific gas constant as R = 0.238 [kJ/kg.K]. Ideal gas equation also gives T2 = 504.3 [K], so that Du = cv(T2 – T1) = – W (since Q = 0 for adiabatic process). This gives cv = 0.165 [kJ/kg.K] and cp = cv + R = 0.403 [kJ/kg.K], so that the standard formula on gas expansion gives DS = 0.03594 [kJ/K].] 75. Air flows steadily at the rate of 1 [kg/s] through an air compressor. The properties of air at entry are: Velocity = 7 [m/s], pressure 100 [kPa], specific volume 0.95 [m3/kg]. The properties at exit are velocity = 5 [m/s], pressure = 700 [kPa], specific volume = 0.19 [m3/kg]. The internal energy of the air increases by 90 [kJ/kg], as it flows through the compressor: Cooling water in the compressor jaket removes heat from the air at the rate of 60 [kW]. (i) Compute the rate of shaft work input to the air in [kW], (ii) Find the ratio of inlet pipe diameter to out let pipe diameter. [Hints: With the given data, SFEE gives W = – 188 [kW] and the continuity equation (ce) is AiVi/vi = AeVe/ve, which can be solved di /de = 1.891.] 76. The properties of a certain substance are related as follows. u = 400 + 0.32t, pv = 60(t + 273), where u is the specific internal energy, t is temperature in [°C], p is pressure and v is the specific volume. Determine cp and cv. [Hints: Using the definition, cv = 0.32 and since h = u + pv, cp = 60.32.]

386

Appendix A:

Additional Problems

77. A reversible heat engine operates between two reservoirs at temperatures 700 [°C] and 50 [°C]. The engine drives a reversible refrigerator, which operates between reservoirs at temperatures of 50 [°C] and – 25 [°C]. The heat transfer to the engine is 2500 [kJ] and the net work output of the combined engine refrigerator plant is 400 [kJ]. (1) Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir at 50 [°C]. (2) Reconsider (1) given that the efficiency of the heat engine and coefficient of performance of the refrigerator are each 45 per cent of their maximum possible values. [Hints: (1) For the given data, Q2 = 2500(323/973) = 829.9 [kJ], WE = Q1 – Q2 = 1670.1 [kJ], WR = WE – 400 = 1270.1 [kJ], Q3 = 2500(248/973) = 637.2 [kJ] [Ans.], Q4 = Q3 + WR = 1907.3 [kJ] and Q2 + Q4 = 2737.2 [kJ] [Ans.]. Using similar procedure for (2) gives Q3 = 523.0 [kJ] and Q2 + Q4 = 2623 [kJ].] 78. One kg of water at 0 [°C] is brought in contact with a heat reservoir at 90 [°C]. When the water has reached 90 [°C], find (1) entropy change of water, (2) entropy change of the heat reservoir and (3) entropy change of the universe. [Hints: Usual method. (DS)water = 1.19295 [kJ, (DS)R = – 1.03805 [kJ] and (DS)U = 0.1549 [kJ].] 79. A vessel contains 8 [kg] of O2, 7 [kg] of N2 and 22 [kg] of CO2, The total pressure in the vessel is 400 [kPa] and the temperature is 100 [°C]. Calculate (1) the partial pressure or each gas in the vessel, (2) the volume of the vessel and (3) the total pressure in the vessel when the temperature is raised to 200 [°C]. [Hints: Usual method. pO2 = 100 [kPa], pN2 = 100 [kPa], pCO2 = 200 [kPa]. [Ans. (1)]. By Dalton’s law the volume of the vessel equals the volume of any component gas at its partial pressure, so that V = VO2 = 7.753 [m3] [Ans. (2)]. Then, by ideal gas law for the mixture, p = 507.2 [kPa].] 80. Determine the pressure in a steel vessel having a volume of 0.015 [m3] and containing 3.4 [kg] of N2 at 400 [°C]. [Hints: By ideal gas law, p = 45306.4 [kPa].] 81. A gas obeys p(v – b) = RT, where b is positive constant. Find the expression for the Joule–Thomson coefficient of this gas. Could this gas be cooled effectively by throttling? [Hints: The Joule–Thomson (Kelvin) coefficient is defined as mJ–T = (∂T/∂p)h, which can be shown to be = (1/cp)[T(∂v/∂T) – v]. For the present gas this = – (b/cp). Since b > 0 (data) and cp > 0, mJ–T < 0, which implies during throttling when p decreases T can only increase.] 82. Steam enters a steam turbine at a pressure of 15 [bar] and 350 [°C] with a velocity of 60 [m/s]. The steam leaves the turbine at 1.2 [bar] with a velocity of 180 [m/s]. Assuming the process to be reversible adiabatic, determine the work done per kg of steam flow through the turbine. Neglect the change in potential energy. [Hints: SFEE gives W = 524.3 [kJ/kg].]

Appendix A: Additional Problems

387

83. 2.5 [kg] of air at 6 [bar], 90 [°C] expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surrounding which is at 1 [bar], 5 [°C]. For this process, determine (i) the maximum work, (ii) the change in availability, and (iii) the irreversibility. [Hints: Since this is a change of state, Wu,max = –DF = – 347.4 [kJ] [Ans. (i) and (ii)] and I = T0 DS = 5.1329 [kJ].] 84. In an air turbine the air expands from 7 [bar] and 460 [°C] to 1.012 [bar] and 160 [°C]. The heat loss from the turbine can be assumed to be negligible. (i) Show that the process is irreversible, and (ii) calculate the change of entropy of per kg of air. [Hints: DS = 0.1771 [kJ/kg.K] [Ans. (ii)] > 0 and so the process is irreversible. [Ans. (i)].] 85. 0.2 [m3] of air at 4 [bar] and 130 [°C] is contained in a system. A reversible adiabatic expansion takes place till the pressure falls to 1.02 [bar]. The gas is then heated at constant pressure till enthalpy increases by 72.5 [kJ]. (1) Ca1culate the work done and (2) the index of expansion; if a single reversible polytropic process giving the same work between the same initial and final states replaces the above processes. Take cp = 1.005 [kJ/kg.K] and cv = 0.714 [kJ/kg.K]. [Hints: Standard method. W = 85 [kJ] [Ans. (1)] and n = 1.011 [Ans. (2)].] 86. A mixture of hydrogen (H2) and (O2) is to be made, so that the ratio of H2 to O2 is 2:1 by volume. If the pressure and temperature are 1 [bar] and 25 [°C] respectively, calculate (i) the mass of O2 required and (ii) the volume of the container. [Hints: Standard method. mO2 = 8/9 [kgO2/kgmix] [Ans. (i)] and V = 2.0647 [m3/kgmix] [Ans. (ii)].] 87. A fluid is contained in a cylinder by a spring loaded, frictionless piston so that the pressure in the fluid is a linear function of the volume (p = a + bV). The internal energy of the fluid is given by the following equation u = 42 + 3.6 pv, where U is in [kJ], p in [kPa], and V in [m3]. With no work other than that done on the piston, find the direction and magnitude of the work and heat transfer. [Hints: By integrating the expression for p gives W = a (V2 – V1) + (b/2) (V22 – V12). Then, DU = (3.6a)(V2 – V1) + (3.6b)(V22 – V12), so that Q = (4.6a)(V2 – V1) + (4.1b)(V22 –V12).] 88. 3 [kg] of air at a pressure of 1 [bar] and a temperature of 27 [°C] is compressed to a pressure of 4 [bar] and a temperature of 93 [°C]. During the process heat loss from the air is 100 [kJ] and the work done on the air is 64 [kJ]. (i) Calculate the index of compression. Following the above process the air undergoes a second process at constant pressure to a temperature of 27 [°C]. (ii) and (iii) Determine the magnitude and direction of heat transfer during the second process. (iv) Also calculate the increase in internal energy of air after it has undergone both processes in sequence. [Hints: The given end states of the first process p1V1n = p2V2n gives n = 1.16 [Ans. (i)]. Then W = (p1V1n = p2V2n)(n – 1) = –378.8 [kJ]. However, the given W = 64 [kJ] gives n = 0.11. Thus, for the first process, the specified work done is inconsistent with the given end states. Hence, we discard the given work done so that DU = –278.8. For the isobaric second process DU = –141.4 [kJ] [Ans. (ii)], W = 56.84 [kJ] so that Q = –198.2 [kJ] [Ans. (iii)] and DU = –420.2 [kJ] [An2s. (iv)].]

388

Appendix A:

Additional Problems

89. A stream of gases at 7.5 [bar], 800 [°C] and 150 [m/s] is passed through a turbine of a jet engine. The stream comes out of the turbine at 2.0 [bar], 600 [°C] and 200 [m/s]. The process may be assumed adiabatic. The enthalpies of gas at the entry and exit of the turbine are 960 [kJ/kggas] and 700 [kJ/kggas] respectively. Determine the capacity of the turbine, if the gas flow is 4 [kg/s]. [Hints: SFEE gives W = 1005 [kW].] 90. Two Carnot engines A and B are connected in series between two thermal reservoirs maintained at 1000 [K] and 100 [K] respectively. Engine A receives 500 [kJ] of heat from the high temperature reservoir and rejects heat to the Carnot engine B. Engine B takes in heat rejected by engine A and rejects heat to the low temperature reservoir. If engines A and B have equal thermal efficiencies, determine (i) the heat rejected by engine B, (ii) the temperature at which heat is rejected by engine A, and (iii) the work done by engines A and B. [Hints: Standard procedure for reversible engines. Let the intermediate temperature be T. Then, Q3 = 500(100/1000) = 50 [kJ] [Ans. (i)], T = 316 [K] [Ans. (ii)], WA = 842 [kJ] [Ans. (iii)], and WB = 108 [kJ].] 91. The analysis by weight or a perfect gas mixture at 20 [°C] and 1.3 [bar] is 10% O2, 70% N2, 15% CO2 and 5% CO. For a reference state of 0 [°C] and 1 [bar], determine (i) the partial pressures of the constituents and (ii) the gas constant of the mixture. [Hints: Standard problem on ideal gas mixture. pO2 = 0.0911 [bar], pN2 = 0.7286 [bar], pCO2 = 0.994 [bar], pCO = 0.081 [bar] [Ans. (i)], Rmix = 0.285 [kJ/kgmixK] [Ans. (ii)].] 92. A system receives 10000 [kJ] of heat at 500 [K] from a source of 1000 [K]. The temperature of the surroundings is 300 [K]. Assuming that the temperature of the system and source remains constant during heat transfer, find (i) the entropy production due to above-mentioned heat transfer and (ii) the decrease in available energy. [Hints: Sp = DS = 10 [kJ/K] [Ans. (i)] and I = T0DS = 3000 [kJ]. [Ans. (ii)].] 93. A quantity of steam at 10 [bar] and 0.85 dryness occupies 0.15 [m3]. Determine the heat supplied to raise the temperature of the steam to 300 [°C] at constant pressure and percentage of this heat which appears as external work. Take specific heat of superheated steam as 2.2 [kJ/kg.K]. [Hints: Since this is an isobaric process, Q = DH = 268.8 [kJ] and W = pDV = 84.04 [kJ].] 94. Air flows steadily through an air compressor at the rate of 20 [kg/min], entering at 6 [m/s] velocity, 10 [N/cm2] pressure and 0.6 [m3/kg] volume. It leaves at 8 [m/s] velocity, 70 [N/cm2] pressure and 0.16 [m3/kg] volume. The internal energy of air leaving is 160 [kJ/kg] greater than that of air entering. Heat absorbed by cooling water in compressor jacket is 6800 [kJ/min]. Find the rate of work input and the ratio of inlet and outlet pipe diameters. [Hints: SFEE gives W = 164.0 [kW] and continuity equation (ce) gives di/de = 2.236.] 95. One [m3] of N2 at 5 [bar] and 200 [°C] expands into an evacuated insulated container so that volume is doubled. Calculate the change in entropy per kg of gas. Assume the gas to be a perfect gas. [Hints: This is a free expansion and the standard formula gives DS = 0.7327 [kJ/K].]

Appendix A: Additional Problems

389

96. The mass analysis of a gas mixture shows that it consists of 60% N2, 30% CO2 and 10% CO. If the temperature and pressure of the mixture are 35 [°C] and 3 [bar], compute (i) the partial pressure of the components; (ii) molecular weight of the mixture; and (iii) the gas constant of the mixture. [Hints: Standard method gives pN2 = 199.17 [kPa], pCO2 = 60.51 [kPa] and pCO = 49.32 [kPa] [Ans. (i)]. Linear mixing rule gives Mmix = 29.58 [kgmix/kmolmix] and Rmix = 8.3143/Mmix = 0.281 [kJ/kgmixK].] 97. 0.44 [kg] of air at 180 [°C] expands adiabatically to three times its original volume and during the process, there is a fall in temperature to 15 [°C]. The work done during the process is 52.5 [kJ]. Calculate cu and cp. [Hints: Adiabatic process means W = DU = mcv(T2 – T1), which gives cv = 0.723 [kJ/kg.K] and cp = cv + R = 1.01 [kJ/kg.K].] 98. The working fluid, in a steady flow process flows at a rate of 220 [kg/min]. The fluid rejects 100 [kJ/s] passing through the system. The condition of the fluid at inlet and outlet arc given as: V1 = 320 [m/s], p1 = 6.0 [bar], u1 = 2000 [kJ/kg], v1 = 0.36 [m3/kg] and V2 = 140 [m/s], p2 = 1.2 [bar], u2 = 1400 [kJ/kg], v2 = 1.3 [m3/kg]. The suffix 1 indicates conditions at inlet and 2 indicates at outlet of the system. Determine the power capacity of the system in MW. The change in PE can be neglected. [Hints: SFEE gives W = 2471.8 [kW].] 99. A cyclic heat engine operates between a source temperature of 1000 [°C] and a sink temperature of 40 [°C]. Find the least rate of heat rejection per kW net output of the engine. [Hints: Least heat rejection means maximum work done, i.e., reversible engine. Then, the usual equations may be manipulated to give Q2/W = 0.326.] 100. Calculate the decrease in available energy, when 20 [kg] of water at 90 [°C] mixes with 30 [kg] water at 30 [°C], the pressure being taken as constant and the temperature of the surroundings being 10 [°C]. Take C of water as 4.18 [kJ/kg.K]. [Hints: Since no data is given on heat loss from the system, assume adiabatic mixing. Then, linear mixing rule gives T2 = 327 [K] and usual formula gives DF = (U2 – U1) + p0(V2 – V1) – T0(S2 – S1). Now, ee shows that the first term is zero and since water is incompressible, V2 = V1. Then, the usual formula gives DF = – T0DS = – 234.2 [kJ].] 101. Steam at a pressure of 5 [bar] passes into a tank containing water where it condenses. The mass and temperature in the tank before the admission of steam are 50 [kg] and 20 [°C] respectively. Calculate the dryness fraction of steam as it enters the tank, if 3 [kg] of steam condenses and resulting temperature of the mixture becomes 40 [°C]. Take water equivalent of tank as 1.5 [kg]. [Hints: Usual energy balance problem. Then, ee gives h2 = 2743.4 [kJ/kg], so that x2 = 0.998.] 102. Given that air consists of 21% oxygen and 79% nitrogen by volume. Determine (i) the moles of nitrogen per mole of oxygen, (ii) the partial pressure of oxygen and nitrogen; if the total pressure is 1 [atm], and (iii) the kilograms of nitrogen per kilogram of mixture.

390

103.

104.

105.

106.

107.

Appendix A:

Additional Problems

[Hints: Usual problem on ideal gas mixtures. nN2/nO2 = 0.79/0.21 = 3.762 [Ans. (i)], pO2 = 0.21 atm and pN2 = 0.79 [atm] [Ans. (ii)], and mN2/mT = 0.767 [Ans. (iii)].] Helium contained in a cylinder fitted with a piston expands reversibly according to the law pV1.5 = constant. The initial pressure, temperature and volume are 5 [bar], 222 [K] and 0.055 [m3]. After expansion, the pressure is 2 [bar]. Calculate the work done during the process. [Hints: Standard problem on gas expansion. W = 14.48 [kJ].] A fluid system undergoes a non-flow frictionless process following the pressure– volume relation as p = (5/V) + 1.5, where p is in bar and V is in [m3]. During the process the volume changes from 0.15 [m3] to 0.05 [m3] and the system rejects 45 [kJ] of heat. Determine (i) the change in internal energy and (ii) the change in enthalpy. [Hints: ee is DU = Q – W = – 39.357 [kJ] [Ans. (i)] and by definition, DH = DU + D(pV) = – 39.51 [kJ].] Two Carnot engines work in series between the source and sink temperatures of 550 [K] and 350 [K]. If both the engines develop equal power, determine the intermediate temperature. [Hints: Standard formulae can be manipulated to get T = 450 [K].] An insulated cylinder of volume capacity 4 [m3] contains 20 [kg] of nitrogen. Paddle done on the gas by stirring it till the pressure in the vessel gets increased from 4 [bar] to 8 [bar]. Determine (i) the change in internal energy, (ii) the work done, (iii) the heat transferred, and (iv) the change in entropy. [Hints: Usual formulae give DU = 3787.8 [kJ] [Ans. (i)], Q = 0 (insulated) [Ans. (ii)], (by ee) W = – DU = – 3787.8 [kJ] [Ans. (iii)] and DS = 5.6284 [kJ/K].] A vessel of 0.35 [m3] capacity contains 0.4 [kg] of CO and 1 [kg] of air at 20 [°C]. Calculate (i) the partial pressure of each constituent and (ii) the total pressure in the vessel. The gravimetric analysis of air is to be taken as 23.3% oxygen and 76.7% nitrogen. [Hints: This is actually a mixture of CO, O2 and N2, all of which are assumed to be have ideally. Then, since the mixture behaves ideally, its equation of state gives p = 554.3 [kPa] [Ans. (i)], so that pCO = 216.5 [kPa], pO2 = 70.9 [kPa] and pN2 = 266.9 [kPa] [Ans. (ii)].]

Appendix B

Explanatory Notes

This appendix contains notes elaborating the concepts, ideas, etc. presented in the main text. These should be selectively read or omitted altogether on first reading. 1. Primitives. The best and most authoritative statement about primitives is: “Following Peano, we shall call the undefined ideas and the undemonstrated propositions primitive ideas and primitive propositions, respectively. The primitive ideas are explained by means of description intended to point out to the reader what is meant, but the explanations do not constitute definitions because they really involve ideas they explain.”1 In sciences, the primitives are assigned their intuitive2 meanings by the exemplar3 method. In general, the disciplines of sciences developed later borrow concepts, terms and methods from those developed earlier. They are also primitives, but their meanings will remain unchanged. 2. The principle of operational definition (pioneered by Bridgman (see [BRI1]), Ch. 2) is based on the fact that the basic test of truth in science is experimental verification.4 Experimental verification consists of conducting the actual experiments and processing the data through appropriate calculations (Bridgman, P.W. [BRI2] calls this the paperand-pencil operation). Then, it seems logical to demand that the definition of a quantity should be in terms of the operations for its experimental measurement. A consequence of this is that only questions operationally asked5 are meaningful and, therefore, only answers operationally given are acceptable. The following are 1 2 3 4 5

Whitehead, A.N. and B. Russel, Principia Mathematica, Cambridge University Press, 2nd abridged ed., p. 91 (1964). Intuition: immediate apprehension by mind without reasoning or by senses (The Concise Oxford Dictionary) (COD). From the French word exemplaire or the Latin word exemplarium, both meaning ‘example’. In fact, Feynman uses (see [FEY], Vol. I, pp. 1–1 and 2–7) this as the definition of science. Feynman calls this “asking the question experimentally” ([FEY], Vol. III, pp. 2–9). 391

392

Appendix B: Explanatory Notes

some interesting examples of questions which are indeterminate (Bridgman calls them meaningless) because they cannot be confirmed or denied experimentally. (a) “Is it possible that as time goes on the dimensions of the universe may be continually changing, but in such a way that we can never detect it because all our measuring sticks are also shrinking in the same way as everything else?” (Bridgman ([BR11], pp. 11–12)). (b) Attempts by H.A. Lorentz to integrate gravitational and electric forces by assuming a slight difference (not experimentally detectable) in the electrical force between equal positive and negative charges, by adjusting the force constant ([BR11], pp. 73–74). (c) The question about the exact position of an elementary particle (forbidden by uncertainty principle). Feynman ([FEY], Vol. I, pp. 38–9) cites the following case: “Suppose we have two theories one of which contains an idea which is not directly testable (i.e. it does not appear directly in the result; it is only an aid in its construction) while the other does not. Then the question that arises is: “if they disagree, can we say that the first one is wrong?” The answer is ‘no’, since we cannot test the idea directly.” In fact, the question itself is indeterminate since it cannot be asked experimentally. In this appendix and the book, unless specifically stated, definition means operational definition. 3. Axiomatic method. The steps involved are: (a) specification of primitives, (b) definition of the terminology to be used in the discipline in terms of the primitives, (c) statement of the axioms6, (d) statement of the theorems7, and (e) proofs8 of theorems based on the rules of (mathematical) logic. These form the body of the discipline, i.e. its theory. 4. The axiomatic method is useful for identifying (a) the primitives, (b) the irrelevant or superfluous terms and concepts, (c) the minimum number of axioms (laws) needed, and (d) the theorems which logically follow. Thus, it is a powerful method for streamlining a theory and is used in later stages of development when sufficient information (empirical theory, data, results, etc.) has been obtained by the heuristic (or intuitive) method. 5. With the advancement of science, better experimental results were demanded and experimentation required more resources (manpower, materials money and time). Hence, alternatives to experimentation, at least in part, were thought of. Modelling and simulation is one such (and, in fact, the most fruitful) alternative. Models are used when direct experimentation on a system becomes impossible due to one or more of the following reasons: (a) Too small a system (e.g. viruses) or too large a system (e.g. the system of galaxies) 6 7 8

These are statements accepted as true (in science these are called ‘laws’) which cannot generally be proved directly. These are the main results developed and stated in the terminology developed in step (b) above and based on the axioms. The two kinds of proof are: (a) the constructive proof which not only shows that a quantity exists but also shows how to measure it, and (b) the non-constructive proof which only shows that some quantity exists. “Note on Mathematical Logic” in The Mathematical Encyclopedia, Kluwer Acd. Publishers (1990).

Appendix B:

Explanatory Notes

393

(b) Too fast a process (e.g. interactions between fundamental particles) or too slow a process (e.g. astronomical processes) (c) Complex interactions between the system and the environment (e.g. ecological processes) (d) Direct experimentation is not allowed (e.g. national economies) (e) Direct experimentation is impossible due to resource constraints. Under the above circumstances, the models of the systems are built. A model is a simplified system obtained by neglecting irrelevant variables. This process is called idealization. Point mass, rigid body, harmonic oscillator, frictionless fluid, market economy, middle-class citizen, etc. are some examples. Similarly, the processes can also be idealized to give (process) models. The models can be classified as: Scale models. A smaller-sized or scaled unit of the same system (or phenomenon) is used (e.g. a scaled model of a ship for towing in the laboratory). Analogue models. A physical model built from an analogous phenomenon (e.g. heat conduction being studied with the help of electrical field). Mathematical models. A set of mathematical equations (also called the system equations) set up to describe the behaviour of the system. Verbal models. A set of verbal descriptions of the system behaviour (e.g. theory of personality, theory of social interactions, etc.). The process of building models (modelling) was pioneered by hydraulic engineers who wanted to understand the flow of water around abutments, piers, etc. They built smaller-sized (scaled down) structures (of abutments, piers, jetties, etc.) and tested them in the laboratory. Naval architects used models to determine the drag on the ship hull. Later, aeronautical engineers also used models to obtain data needed for designing aeroplanes. Once a model is built, it is studied (experimented with). Such a study of a model is called simulation. The results of simulation are then translated to the system. Whether a model is good or not is determined from the results of simulation it gives. For a specified accuracy in simulation result, we generally build the simplest possible model. Currently, mathematical models are extensively used in all disciplines of science (natural and social). Then, simulation is the process of solving these equations for different values of the system parameters. For very large or complex (or both), computers are used for solving the system equations. That is why, these days, simulation automatically means computer simulation. 6. The following interesting explanation regarding the difference between the microscopic and macroscopic approaches is due to Callen (see [CAL], § 1.1). The macroscopic approach is a coarse-grained description of physical phenomena, while the microscopic approach is their fine-grained description. Moreover, the macroscopic properties are obtained from microscopic ones by statistical averaging over a number of atoms as well as the durations of the phenomena (e.g. typical times for atomic

394

Appendix B: Explanatory Notes

phenomena are 10–15 [s], while the macroscopic phenomena take place in [ms]). This is the domain of the statistical mechanics (or thermodynamics). The point to be noted is that during the process of averaging, some microscopic properties are lost, i.e. they do not possess macroscopic equivalents. As an example, consider the motion of a system of atoms shown in Figure B.1, where the central atom is fixed and the two atoms shown by filled circles have a charge on them. In Figure (a), the charged atoms are assumed stationary. This mode of vibrations is lost in the process of averaging. Since the motion in Figure (b) changes the interatomic distances, it appears as change in volume. In Figure (c), since the charged atoms also vibrate, this motion appears as an electric dipole moment. Filled circles are atoms with unit charge. Arrows indicate directions of motion

(a) Random vibrations

(b) Dilation Figure B.1

(c) Electric dipole

Some modes of molecular motions.

In short, it is seen that some modes of atomic motion survive the process of statistical averaging, while others do not. The study of the consequences of the modes which survive are dealt with in mechanics and electricity (e.g. Figure (b) falls in mechanics while Figure (c) falls in electricity). Thermodynamics is the study of the consequences of the atomic motions lost during (i.e. which do not survive) the process of statistical averaging.9 7. The following concepts are basic to all disciplines of science: (a) relevant quantity is an undefined concept and possesses its intuitive meaning; (b) universe is the totality of all things relevant to a study10; (c) body or system is that part of the universe which can be separated from rest of its universe (in laboratory, in imagination, etc.) for controlled investigation11 and suitably idealized; (d) system boundaries are the surfaces (real or imaginary) which separate the system from the rest of the universe and which are specified either verbally (e.g. a star) or through drawings (e.g. a freebody diagram), or generally, both (for safety); (e) interactions are those by which a 9 10 11

The energy transfer associated with this mode has the characteristics of heat, while those associated with surviving modes of motion are work. In philosophy of science, this is called the universe of discourse; in set theory it is known as the universal set etc. This means that a body (or system) and its environments will form a bigger system. See [BER], for interesting examples from biological sciences where such separation is not possible.

Appendix B:

Explanatory Notes

395

system and its universe influence each other and which are determined from experiments conducted on the system boundaries; (f) environment is that part of the universe which directly interacts with a system (i.e. immediate part of the universe); (g) properties are quantities whose values depend only on the state of a system; (h) state is that which determines the condition of a system and is defined by fixed values of a set of its properties; (i) behaviour is the variation of state as a function of interactions; (j) steady state is when the state of a system does not change with time and interactions are non-zero; and (k) equilibrium is when the state does not change with time and when the interactions are zero.12 8. An illustration of the basic concepts common to all disciplines of science mentioned above (and in the main text), consider the example of describing the motion of a body13 of mass m, initially at rest at the origin, subjected to a constant force F and moving along a straight line.14 (a) The body is isolated from among the collection of other bodies, which form the environment. (b) The body is characterized by its parameters (here its mass, m). (c) The environment interacts with the body by exerting the (specified constant) force F. (d) The equations of motion show that the state of the body may be specified in terms of t, the time instant and position s(t) of the body at that instant.15 (e) The body responds to the applied force by moving along a straight line, i.e. changing x(t), (changing its state). (f) The motion (behaviour) of the body may be drawn as a graph in a space with the state variables as coordinates, called state space.16 The graph of the motion is also called the trajectory of the body in state space. (g) The equation of motion, written as d2x/dt2 = F/m, symbolically becomes change of state = f (interactions) This is called the equation of behaviour. 9. The correspondence to thermodynamics is straightforward and is shown in Table B.1 Table B.1

Correspondence to thermodynamics

Mechanics

Thermodynamics

Mechanics

Thermodynamics

Body Parameters State State space

System Properties State State space

Environment Interaction State variables Trajectory

Environment Interaction Properties Process

Also Process = Equation of Behaviour/Trajectory/... 12

13 14 15

16

The only difference between equilibrium and steady state is that in the former the interactions are also zero. Consequently, both require measurements made inside the system (to determine change of state) and on the boundaries (to determine the fluxes) simultaneously. This should not be surprising because scientific analysis started from mechanics, i.e. a modification of this problem. The answer is well-known, viz. a = F/m; V(t) = at; s(t) = (1/2) at2. Alternatively, the pair (V(t), t) can also be used instead of (x(t), t). In advanced mechanics (Lagrangian and Hamiltonian mechanics), the pairs (q(t), q (t)) and (q(t), p(t)) (where q. q and p are position, velocity and momentum respectively of the body) are found to be more convenient. In kinetic theory of gases and statistical mechanics, the space with q(t) and p(t) as axes is called phase space.

396

Appendix B: Explanatory Notes

10. The definition of property mentioned earlier implies that its change between two states is independent of the process, i.e. F is a property if and only if

∫

2

1, A

dF =

∫

2

1, B

d F . This

v∫ dF = 0 for a closed path. This is its operational definition. Properties

is equivalent to

can be classified as (a) primitive, (b) defined within the discipline, and (c) mathematically defined (as combinations of (a) and (b)) above. Then, it is evident that to avoid circularity in definition only the primitive properties should be used in the definition of state. 11. The following paragraphs briefly review the concepts from mechanics. More detailed explanations are available in college physics textbooks. Geometrical concepts essential to mechanics (and hence, appear in thermodynamics) are summarized in Table B.2. The concepts of a point, a (straight) line, its length, a coordinate system, its origin, etc. are not defined in geometry. Hence, their intuitive meanings17 are assumed. Table B.2

(Euclidean) Geometrical concepts

Name & symbol

Definition

Position (x) Area (A) Volume (V)

Length of the line joining the point to the origin. Calculated from geometry of the shape. Calculated from geometry of the shape.

12. Length is not defined in mechanics as well. Hence, it is defined in terms of some method accepted as standard by all.18 Then other quantities are defined in terms of this standard length. For example, the position of a point (denoted as x in Table B.2) is (operationally) defined as: (a) choose an arbitrary point as the origin for a coordinate system, (b) construct a coordinate system with this point as the origin,19 (c) join the required point to the origin by a straight line, and (d) measure the length of this line. Similarly, the area of a circle is defined by: (a) draw a circle, (c) measure its diameter (say, D), and (c) calculate (a paper-and-pencil operation) the area as (pD2/4). The main idea is: “a mathematical formula for a quantity is its operational definition.” The verbal statements of the operational definitions are only explanations of the formula. For example, the operational definition of the position x of a point given above is only the verbal statement of the formula x = L. Similarly, the operational definition of the area of a circle is the verbal explanation of the formula A = (pD2/4). This is general nature of all formulae. Hence, from here onwards only the mathematical equations defining a quantity will be stated as its operational definition. 17 18 19

In fact, the word geometry arises from geo = earth and mētron = measure, i.e. geometry was the science of land survey used for tax purposes. Euclidean (plane) geometry is considered here. For example, metre, the SI unit of length. This method of definition is called exemplar (meaning model, pattern (COD)) method. By definition, the length of this point from the origin is zero.

Appendix B:

Explanatory Notes

397

13. Time is also an undefined concept in mechanics.20 Hence, it is defined in terms of some standard. Then, the concepts of velocity and acceleration are operationally defined as: V = dx/dt and a = dV/dt. 14. Mass is a concept developed basically by chemists—through weighing of chemicals for reactions. Thus, the gravitational measure of mass becomes its accepted intuitive meaning.21 Newton’s first law may be stated as follows: In the absence of any external influence a body continues to be in the state of rest or of uniform motion along a straight line. The external influences are called forces. This law implies the existence of a property which ensures that, in the absence of forces, a body continues to be in its state (of rest or uniform motion along a straight line). This property is called (inertial) mass. This law only states that mass exists but does not show how to measure it.22 Hence, mechanics assumes that the inertial mass of a body is identical with its gravitational mass. 15. In terms of the above concepts, the other concepts of dynamics (dealing with forces) are defined as shown in Table B.3. Table B.3

Concepts in dynamics

Name & symbol

Definition

Density (r) Kinetic energy (Ek) Momentum (p) Force (F) Pressure (p)

Defined as (dm/dV) Defined as (1/2)(mV2) Define as mV Defined as (dp/dt)(second law) Defined as (dF/dA)

16. Newton’s second law (i.e. the rate of change of momentum of a body is proportional to the applied force) is the (operational) definition of force since it involves the following operations: (a) take the standard body, (b) let the unknown force act on it, (c) measure its velocity, (d) calculate its momentum, and (e) calculate the rate of change of momentum. This is the value of the applied force.23 Note that this is the verbal description of the formula F = dp/dt given above as the defining relation for force. 17. The principles of dynamics (mechanics) are so general that they cannot specify the details of any motion.24 In the example cited above, further calculations are possible only if the force F is specified. In other words, force becomes a primitive concept in dynamics if the details of motion of a body are to be calculated, i.e. Newton’s second 20 21 22 23 24

In fact, Einstein arrived at the special theory of relativity by trying to define time operationally. For example, 1 [kg] is the mass of the international prototype kept in the Bureau International des Poids of Mesures at Sévres near Paris. Hence, this is a non-constructive definition. Unlike the first law, the second law not only defines the existence of the quantity called force, but also shows how to measure it. Hence this is a constructive definition. The basic principle is: “The more general (or, broader, i.e. it covers a larger number of disciplines) a theory, the less will be its content (or, depth, i.e. details of any discipline)”.

398

Appendix B: Explanatory Notes

law can be used to calculate the motion (dynamics) of a body only when the force is specified. It is generally specified by some other rules (laws), e.g. Newton’s law for gravitational forces, Coloumb’s law for electrical forces and Lennard Jones for molecular forces. This gives rise to expressions for the above concepts in different disciplines. For example, the basic laws of force in electricity, in magnetism and in electromagnetism give rise to appropriate equations for motion, work done (voltage) and energy, but these are only applications of the above concepts. The only new concept is the principle of conservation of electric charge. 18. Consider the following interesting example. In note 14 presented above, it was stated that gravitational mass is assumed to have intuitive meaning. Now, consider the process of ordering inertial masses by weighing by a spring balance. A spring balance really shows the force acting on a mass due to the action of gravity. Assuming the spring to be linear, the expression for force, F, exerted by it is given by F = kx, where k is called the spring constant and x is the displacement of the end of the spring (where the hook is attached). For this case, the Newton’s second law can be written as kx = mg, where m is the mass of the body and g is the acceleration due to gravity which is assumed to be constant. This relation can be rewritten as m = (k/g)x. It shows that m μ x which implies that mass m can be ordered according to the displacement x it produces on weighing. That is why this m is called the gravitational mass.

∫

2

Fds, where F is the force whose point 19. Work (W) is defined in mechanics as W = 1 of application moves a distance ds in the direction of force. The value of the integral

∫

2

F ⋅ ds (and hence, the work) depends on the values of F(s, t) at all points (s, t). Such functions are called path functions. Moreover, mathematics25 requires that the path, F(s, t), should be the continuous function26 for the integral to exist. In thermodynamics, the processes which give rise to such continuous functions are called the quasi-static processess. Then, this implies that in thermodynamics this expression can be used to calculate work only for quasi-static processes. This is a mathematical requirement and not thermodynamic. Note that this definition shows that work done depends upon the mass of the system. 20. The kinetic energy, Ek, of a body of mass m moving with velocity V is defined as Ek = (mV2/2). This definition shows that Ek depends only on the value of V. Then, since V is a state variable, the value of Ek depends only on the state of the body and not on how that was reached. Hence, in the state space with V and t as axes, the value of Ek depends only on the point representing that state. Such quantities are called point functions. In thermodynamics, these quantities are called properties. Note also that Ek depends on mass of the body. In thermodynamics, such properties are called extensive27 properties. 1

25 26 27

Riemann’s definition of an integral. A function with limited total fluctuations and a function with a finite number of ordinary discontinuities are also Riemann integrable ([WHI], p. 63). Because they depend upon the extent (size, in this case the mass) of the system.

Appendix B:

Explanatory Notes

399

21. The Work–Energy theorem of mechanics states that “work done on a body by a force is stored as the increase in its kinetic energy.” This is shown as follows. Consider the motion of a body through a small distance ds under the influence of force F. Then, 2 ⎛ dV ⎞⎛ ds ⎞ ⎛ dV ⎞ m⎜ ds = m ⎜ ⎟ ⎟⎜ ⎟ ds = 1 1 1 ⎝ dt ⎠ ⎝ ds ⎠⎝ dt ⎠ 2 2 ⎛ mV 2 ⎞ ⎛ V2 ⎞ = md ⎜⎜ ⎟⎟ = d ⎜⎜ ⎟⎟ = E k ,2 − Ek ,1 1 ⎝ 2 ⎠ 1 ⎝ 2 ⎠

W=

∫

2

F ⋅ ds =

∫

2

∫

∫

∫

2

1

⎛ dV ⎞ m⎜ ⎟ (V) ds ⎝ ds ⎠

∫

(B.1)

Using the intuitive idea that anything stored by a body must have flown into it, Eq. (B.1) implies that work is an energy flow. Consequently, work exists only as long as the energy flows. Hence, work stored is a meaningless concept. It also shows that energy is neither destroyed nor created. This is the basic form of the energy conservation principle. 22. A force is defined to be conservative if the work done by it around a closed loop is zero.28 This implies that the work done by a conservative force equals the change in a property. Mathematically, this is shown as follows. By definition, a force F is conservative if and only if

∫

v∫

C

F ⋅ ds = 0 . Then, by the Stokes’ theorem of vector

× F) ⋅ dA = 0, which implies that F = –—Ep, with Ep as the potential function. Now, the above definition implies that Ep is a property (i.e. it is a point function since its net change around a closed path is zero) and it depends only on the state. 23. For one dimension the relation in note 22 becomes, F = –(dEp/dx). Separating the variables and integrating, this becomes. analysis this becomes

∫

2

1

A (∇

F ⋅ dx = −

∫

2

1

dE p = − ∇E p = E p1 − E p 2

Following the interpretation from the work–energy theorem that work is an energy flow, the preceding equation shows that Ep is an energy. It is called the potential energy because it is the potential of a conservative force. Now, since work is an energy flow, the principle of conservation of energy demands that the potential energy should decrease when the force does work; i.e. work is done at the expense of the potential energy. Hence the negative sign is essential in the definition. This equation is called the potential energy theorem, which states that: “There exists a property called potential energy whose change between two states equals the work done by a conservative force.” 24. It should be noted that the adjectives point and path to functions are based on whether the value of the defined quantity depends only on the value of the variable or its integral (sum over all past values). Moreover, whether or not an integral exists depends 28

Since dissipative effects (like friction, hysteresis, etc.) always oppose the motion, work done by them cannot be zero around a closed loop. Hence, in a conservative force field these must be absent.

400

Appendix B: Explanatory Notes

only on whether the function is continuous or not. Both of these are purely mathematical characteristics and are independent of the discipline (including thermodynamics) which uses them. 25. The energy conservation theorem states that “the total energy of a body during motion in a conservative force field is constant.” This is shown below for a onedimensional motion. Since the force F is conservative, F = –(dEp/ds). Substituting in the differential form of Eq. (B.1), namely F . ds = d(mV2/2) gives ⎛ mV 2 ⎛ dE p ⎞ −⎜ ⎟ ds = d ⎜⎜ ⎝ ds ⎠ ⎝ 2

⎡ ⎞ ⎛ mV 2 ⎟⎟ or d ⎢ ( E p ) + ⎜⎜ ⎢⎣ ⎠ ⎝ 2

⎞⎤ ⎟⎟ ⎥ = 0 or ( E p + Ek ) = C ⎠ ⎥⎦

where the constant C is the total energy of the body. 26. The principle of conservation of momentum may be illustrated as follows. Newton’s third law states that “the action and reaction are equal and opposite.” For a pair of bodies this can be written as F12 = – F21. By second law this becomes

dp12 dp21 + =0 ds dt

or

d ( p12 + p21 ) =0 dt

or

p12 + p21 = C

where C is the total momentum of the body. Thermodynamics does not use this principle. 27. The properties of working substances are specified in properties of matter—a discipline of the chemistry (and, also common to physics): Based on its chemical properties, a working substance can be classified as: (a) element if it consists entirely of atoms of one type; (b) compound if it is chemically formed from one or more elements combined in definite proportion by moles; or (c) mixture (i) if the constituents can be separated by suitable physical or mechanical means, (ii) if the constituents can occur in all proportions, (iii) if heat of formation is zero, and (iv) if the overall properties are aggregate (e.g. weighted average) properties of the constituents. A mixture is homogeneous if its composition is same throughout the mixture. Otherwise, it is called heterogeneous. A homogeneous molecular mixture of two or more substances of dissimilar molecular structures (generally, solids in liquids) is called a solution. Although a solution has all the properties of a mixture, its solubility (which determines the proportion of combining) has a restricted range. 28. The quantitative specifications of working substances are in terms of (a) the p–v–T data represented by the general relation f (p, v, T) = 0, known as the equation of state. For liquids and vapours, the form of f (.) is complex. The most well-known of these a⎞ ⎛ are the ideal gas equation, pv = RT, the van der Waal equation, ⎜ p + 2 ⎟ ( v − b) = RT , ⎝ v ⎠ and (b) the specific heat data in the form of a polynomial in temperature, i.e. cp (or cv) = a + bT + cT2 + . . . . 29. For thermodynamics, these are primitives. The nature of the function f (.) in the equation of state and the values of the constants a, b, . . ., in the specific heat equation can only be determined from experiments (either macroscopic or microscopic)

Appendix B:

Explanatory Notes

401

conducted on the substance. However, thermodynamics imposes restrictions on the relations between the properties. For example, once these are known, thermodynamics shows how to calculate the variation of specific heats with p, V etc. Thermodynamics also demands that for all substances 2

⎛ ∂v ⎞ ⎛ ∂ p⎞ c p − cv = − T ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂ v ⎠ T 30. The approximate years of development of concepts and devices in heat, thermodynamics and temperature are listed below. ∑ Hero of Alexandria more than 2000 years ago developed his ‘engine’ which was really a turbine. ∑ In 1592, Galileo used a glass bulb containing air extended into a vessel of water. It was called a barothermoscope, since it measured atmospheric pressure as well as temperature. ∑ As early as 1620, Francis Bacon concluded that “heat itself, its essence and quiddity, is motion and nothing else”. ∑ The same idea was expressed in the seventeenth century by Boyle and by Hooke. ∑ In 1631, Jean Rey had used the same set-up as that used by Galileo but the stem was filled with water and inverted into the vessel of water. This was the first liquid expansion thermometer. ∑ In 1641, Grand Duke Ferdinand II of Tuscany developed the first thermoscope with its stem sealed and filled with alcohol. ∑ In 1665, Robert Boyle, Robert Hooke and Christian Huygens independently proposed the one fixed point method of calibration. ∑ In 1680, Huygens experimented with a gunpowder engine, which may be considered as the forerunner of the I.C. engine. ∑ In 1694, Carlo Renaldini proposed the two fixed point method of calibration using ice and steam points. ∑ Savery patented, in 1698, the first commercially successful steam ‘engine’ for pumping water from mines. However, it was not an engine in the present sense of the term. ∑ Newcomen, in 1705, devised the first steam engine which used a cylinder and piston assembly. ∑ In 1702, Ole Roemer (Romer) constructed a thermometer with 0° and 60° as fixed points. ∑ In 1717, Daniel Fahrenheit used cylindrical rather than spherical bulbs and constructed his thermometer. ∑ The centigrade system is believed to have been suggested by Elvius, in 1710. It was later proposed independently by Linnaeus (the eminent Swedish Botanist) in 1740 and Christian of Lyons in 1743. ∑ Anders Celsius, a Swedish astronomer used a centesimal system as early as 1742. However, it was inverted with respect to the centigrade in that he assigned the values 100° and 0° to the ice and steam points.

402

Appendix B: Explanatory Notes

∑ James Watt, in 1763–64, introduced the idea of a separate condenser. Thus, the early (circa 1769) Watt engines were really Newcomen engines. ∑ Joseph Black was the pioneer in heat measurement. Until Black (Joseph Black) made his discoveries, there was no clear distinction in people’s minds between the concepts of ‘quantity of heat’ and ‘degree of hotness’ or ‘temperature’. Between the years 1766–1799, Black measured specific heats, latent heat of fusion and vaporization, etc. ∑ In 1779, William Cleghorn proposed an extension to the material theory to take into account the Black’s discoveries of specific and latent heats. ∑ In 1782, Watt introduced the double acting engine and he also invented several mechanisms for its successful operation. ∑ In 1781, Hornblower patented the first compound engine. ∑ This ‘matter of heat’ was named ‘caloric’ by Lavoisier in 1787. Thus, this theory became known as the caloric theory.29 ∑ In 1794, Street patented an engine which ran on air and turpentine. ∑ Between 1794 and 1838, Lebon, Cecil, Brown, Wright, and Barnett made attempts to use the then newly discovered coal gas and hydrogen for combustion. ∑ In 1799, Benjamin Thompson (or Count Rumford) found that caloric has no weight. ∑ The results of the now famous cannon boring experiments of Benjamin Thompson (Count Rumford) were announced in 1798. ∑ Experiments of Humphry Davy in 1799 on production of heat by friction were also along the same lines. ∑ In 1802, the first marine and locomotive engines were patented. ∑ In 1821, Thomas Seebeck discovered the thermocouple effect. ∑ In 1824, the basics of the second law of thermodynamics were first published by Sadi Carnot although in an imprecise form. ∑ The experimental works of Mayer (but published in 1842) and of Joule (in 1843) in determining the mechanical equivalent of heat are well-known and need not be repeated. Mayer’s statement is the first statement of the first law of thermodynamics. ∑ In 1847, Helmholtz recognized the importance of the Joule’s experiments and formulated the equivalence of the present form of the first law of thermodynamics. ∑ In 1848, William Thomson (Lord Kelvin) developed the thermodynamic (Kelvin) scale of temperature. ∑ In 1850, Clausius stated his form of the second law of thermodynamics. ∑ In 1851, Kelvin stated his form of the second law of thermodynamics. ∑ In 1860, Lenoir developed a really practicable I.C. engine. ∑ In 1860, William Siemens discovered resistance thermometry. ∑ In 1867, Otto and Langlen exhibited their free piston engines. ∑ Around 1890s, the second law was made more precise by Clausius, Kelvin, Planck, Caratheodory and others. 29

In 1775, Lavoisier also refuted a similar theory called the ‘phlogiston theory’ which was developed (in 1660–1734) by Becher and Stahl to explain chemical reactions (oxidation reactions were thought to be combustion (from phlogizeinin Greek meaning ‘to burn’)).

Appendix B:

Explanatory Notes

403

∑ In 1893, Diesel invented his engine with solid injection of fuel. The development of I.C. engines (both Otto and Diesel) was more rapid after this. ∑ In 1897, Maxwell stated the zeroth law (but it was given this name by Fowler in 1929). ∑ The third law was first enunciated in the unattainability form (as the unattainability of 0 [K]) by Nernst in 1907. The main points to be noted are: (a) The concepts that are supposed to be developed in thermodynamics were in existence long before thermodynamics itself came into existence. (b) The strong contribution of engineering in the development of thermodynamics. (c) Temperature was measured using non-thermal (i.e. mechanical, electrical or thermoelectrical) effects, but mainly thermal expansion of liquids (and, gases). This is the reason why thermodynamic concepts (all except entropy) were taken as primitives. Thermodynamics only defines them precisely through operational definitions and thereby shows how they should be used. 31. The Born diagram30 is a diagrammatic representation of the Legendre transformation.31 The method of its construction is explained below. (a) Draw a square (or a rectangle) and label its sides with the energy functions A, G, H and U alphabetically, starting from top and proceeding clockwise [see Figure B.2(a)]. (b) Label the corners of Figure B.2(a) with the variables p, V, T and S such that each energy function is flanked by its natural coordinates (i.e. A with V and T, G with p and T, H with p and S, and U with S and V) as shown in Figure B.2(b). (c) Using any of the energy relations, draw the arrows inside the square such that they point away from the natural coordinates if the term is positive (and towards it if negative). Figure B.2(c), has been drawn for the relation dU = TdS – pdV. A

U

V

G

H (a)

A

V

T

U

G

S

H (b)

p

A

T

U

G

S

H (c)

p

Figure B.2 The Born diagram (first method).

The Born diagram need not be constructed in the above manner only. Another method of construction is now described. (a) Assume that the given relation is dU = TdS – pdV. (b) Draw a square and write U on its bottom side. (c) Write S and V (the natural coordinates of U) at the two bottom corners. (d) Label the corner opposite to S as T 30 31

Apparently, it was first used by Max Born. Hence Tisza (see [TIS], p. 64) has given it by this name (see also [CAL], p. 29). See Appendix C for a brief discussion.

404

Appendix B: Explanatory Notes

and that opposite to V as p (i.e. by the coefficients of the equation). (e) Since the first term in this expression is positive, draw the arrow from S to T. (f) Similarly, draw the arrow corresponding to the second term from p since it is negative. (g) Write the other energy functions on the sides between their natural variables. (h) This gives Figure B.3(a). Note that in this figure the energy functions occur alphabetically anticlockwise starting from the right side of the square. This method can be used for constructing the diagram for any other pair of variables. As an example, Figure B.3(b) is drawn on the basis of the relation dU = TdS + Smidni. Note that the direction lebels and arrows are arbitrary and may be reversed as well. But be consistent. p

G

ni

T

A

H

S

U

T

U

V

(a) Figure B.3

A

i

S (b)

The Born diagram (second method).

The above diagrams can be used to write down: The basic relations. The property relations can be directly written down immediately, once care is taken to assign appropriate signs depending on the directions of the arrows. For example, we can immediately write dG = Vdp – SdT. Maxwell’s relations. These use only the corners. The sign of a term is positive if the arrows are symmetric. The full arrows outside the square in Figure B.4 illustrate this for the relation ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜⎝ ⎟⎠ = ⎜ ⎟ ∂S p ⎝ ∂ p⎠ s where the LHS and RHS are written down beside the arrows. Similarly, the relation ⎛∂S ⎞ ⎛∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂ T ⎠p ⎝ ∂ p ⎠T can also be easily written down. However, this case has a negative sign because the arrows are asymmetrically placed. V

LHS

A

T

U

G S

H

RHS

P

LHS RHS

Figure B.4 The Born diagram (usage).

Appendix B:

Explanatory Notes

405

32. This part illustrates the method of using the Jacobians to systematically transform the partial derivatives. In the next part below, the Bridgman’s tables have been developed using this. A Jacobian is the determinant (of partial derivatives) which acts as the factor that scales (up or down) the areas during coordinate transformations (see Appendix C for a brief review). In thermodynamics, the Jacobians are useful mainly due to the fact that

∂ (T , S) = 1. To ‘prove’ this, note that: ∂( p, V ) ∑ For a cycle, the net work done is equal to the net heat transferred. ∑ The net work done equals the area of the cycle in the p–V plane. ∑ If the cycle is reversible, the net heat absorbed equals the area of the cycle in the T–S plane. ∑ Hence, using the area property of the Jacobian while transforming the processes of a cycle from the p–V coordinates to T–S coordinates, the above result follows directly. EXAMPLE B.1 Derivation of the Maxwell’s relations. The characteristic of Jacobian used in deriving the relations among properties is

∂ (T , S) ∂ ( p, V ) = ∂ ( x , y) ∂ (x, y) where x and y are the natural coordinates of a property. From the equation of U. The natural variables for U are S and V. Hence, taking x = S and y = V, the above relation becomes

∂ (T , S) ∂ ( p, V ) = , which, using the ∂ ( S, V ) ∂ (S, V )

⎛ ∂T ⎞ ⎛ ∂ p⎞ properties of the Jacobian, can be reduced to – ⎜ ⎟ = ⎜ ⎟ . ⎝ ∂V ⎠ S ⎝ ∂ S ⎠ V From the equation of H. The natural variables for H are S and p. Hence, taking x = S and y = p, the above relation becomes

∂ (T , S) ∂ ( p, V ) , which, using the = ∂ (S, p) ∂ ( S, p)

⎛ ∂T ⎞ ⎛ ∂V ⎞ properties of the Jacobian, can be reduced to – ⎜ ⎟ = – ⎜ ⎟ . ⎝ ∂S ⎠ p ⎝ ∂ p⎠ S

From the equation of A. The natural variables for A are T and V. Hence, taking x = V and y = T, the above relation becomes

∂ (T , S ) ∂ ( p, V ) = , which, using the ∂ (V , T ) ∂ (V , T )

⎛ ∂S ⎞ ⎛ ∂ p⎞ properties of the Jacobian, can be reduced to – ⎜ ⎟ = − ⎜ ⎟ . ⎝ ∂V ⎠ T ⎝ ∂T ⎠ V

406

Appendix B: Explanatory Notes

From the equation of G. The natural variables for G are p and T. Hence, taking x = p and y = T, the above relation becomes

∂ (T , S ) ∂ ( p, V ) , which, using the = ∂ ( p, T ) ∂ ( p, T )

⎛ ∂S ⎞ ⎛ ∂V ⎞ properties of the Jacobian, can be reduced to – ⎜ ⎟ = ⎜ ⎟ . ⎝ ∂ p ⎠ T ⎝ ∂T ⎠ p

⎛ ∂T ⎞ EXAMPLE B.2 Here an expression for ⎜ ⎟ is obtained. ⎝ ∂ p⎠ S Now,

⎛ ∂T ⎞ ∂(Τ , S) ∂(Τ , S) ⎜⎝ ∂ p ⎟⎠ = ∂ ( p, S) = ∂ ( p, T ) S

⎛ ∂S ⎞ ∂( p, S) = −⎜ ⎟ ⎝ ∂ p⎠ T ∂ ( p, T )

⎛ ∂S ⎞ ⎜⎝ ⎟⎠ ∂T p

Using a Maxwell’s relation this becomes ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜⎝ ∂ p ⎟⎠ = − ⎜⎝ ∂ T ⎟⎠ p S

TV b ⎛ ∂S ⎞ ⎜⎝ ⎟⎠ = cp ∂T p

EXAMPLE B.3 It has been stated in the main text that all property variations can be expressed in terms of b (the volume expansivity), kT (the isothermal compressibility), and cp. In this example, the cv is expressed in terms of b, kT and cp.

⎛ ∂S ⎞ By definition, cv = T ⎜ ⎟ . Hence, the derivative is transformed. Now ⎝ ∂T ⎠ v

∂ ( S, V ) ∂ ( S , V ) Ê ∂S ˆ = ÁË ˜¯ = ∂ T v ∂ (T , V ) ∂ (T , p)

∂ (T , V ) ∂ ( S, V ) Ê ∂ V ˆ = ∂ (T , p) ∂ (T , p) ÁË ∂ p ˜¯ T

In terms of the definition of kT, the last derivative may be written as – (V kT). Hence, consider only the Jacobian. Since it cannot be reduced, it is expanded as

Ê∂Sˆ ÁË ˜¯ ∂T p

∂ ( S, V ) = ∂ (T , p) Ê ∂ S ˆ ÁË ∂ p ˜¯ T

Ê ∂V ˆ ÁË ∂ T ˜¯ p Ê ∂V ˆ ÁË ∂ p ˜¯ T

ÈÊ ∂ S ˆ Ê ∂ V ˆ Ê ∂ S ˆ Ê∂V ˆ ˘ = ÍÁ ˜ Á ˜ ˙ ˜ Á ˜ Á ÍÎ Ë ∂ T ¯ p Ë ∂ p ¯ T Ë ∂ p ¯ T Ë ∂ T ¯ p ˙˚

The quantity in the brackets can be written in terms of cp, kT b and V as

⎡⎛ c p ⎞ ⎤ 2 ⎢⎜ ⎟ (−k T V ) + ( b V ) ⎥ ⎢⎣⎝ T ⎠ ⎥⎦ Substituting all these results in the original equation, gives cv = cp –

T b 2V

kT

.

Appendix B:

Explanatory Notes

407

33. The Bridgman’s tables are one of the most used tables of relations among the thermodynamic properties. These are mentioned here as an example of the use of Jacobians. Consider the following transformation

∂ (z, y) ⎛ ∂ (z, y) ⎞ ⎛ ∂ (q, r ) ⎞ ∂ (z, y)/∂ (q, r ) ⎛ ∂z ⎞ = = ⎜⎝ ⎟⎠ = ∂ x y ∂ ( x , y) ⎜⎝ ∂ (q, r ) ⎟⎠ ⎜⎝ ∂ ( x, y) ⎟⎠ ∂ ( x, y)/∂ (q, r ) ∂ (z ) y ⎛ ∂z⎞ Suppose we formally write ⎜ ⎟ = , then it is seen that ⎝ ∂ x ⎠ y ∂ (x ) y

(∂ z ) y =

∂ (z, y) ∂ (q, r )

and

∂ (x, y) ∂ (q, r )

(∂ x ) y =

In his tables, Bridgman chooses q and r to be p and T, respectively. Consequently,

(∂ z ) y =

∂ ( z , y) ∂ ( p, T )

and

(∂ x ) y =

∂ (x, y) ∂ (q, T )

Note that by choosing q = T and r = p, though the preceding relations become

(∂ z) y =

∂ (z, y) ∂ (T , p)

and

(∂ x ) y =

∂ (x, y) ∂ (T , p)

it does not change their ratio. Bridgman also used the Maxwell’s relations to convert all derivatives of entropy since he wanted to express the first partial derivatives of all properties in terms of (∂V/∂ T) p,(∂V/∂ T) T, and (∂H/∂T)p. These quantities are volumetric expansivity, isothermal compressibility and the specific heat at constant pressure, all of which can be measured directly. Table B.4 shows the 28 derivatives so developed. Before the use of these tables is illustrated, let us verify the expression listed against (∂S)V as an example. (a) In this case take z to be S and y to be V. (b) Then, the above relation for (∂z)y becomes, (∂S)V = (c) Expanding the Jacobian gives ⎛ ∂S ⎞ ⎜⎝ ⎟⎠ ∂T p

⎛ ∂V ⎞ ⎜⎝ ⎟⎠ ∂T p

⎛ ∂S ⎞ ⎜⎝ ∂ p ⎟⎠ T

⎛ ∂V ⎞ ⎜⎝ ∂ p ⎟⎠ T

(d) Evaluating this determinant gives ⎛ ∂ S ⎞ ⎛ ∂V ⎞ ⎛ ∂ S ⎞ ⎛ ∂V ⎞ ⎜⎝ ⎟⎠ ⎜ ⎟ − ⎜ ⎟ ⎜⎝ ⎟ ∂T p ⎝ ∂ p ⎠ T ⎝ ∂ p ⎠ T ∂ T ⎠ p

∂ (S , V ) ∂ (T , p)

408

Appendix B: Explanatory Notes

(e) The definition of cp is used to convert the first derivative and a Maxwell’s relation is used to convert the third one. The result is ⎛ cp ⎞ ⎛ ∂ V ⎞ ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎜ ⎟ ⎜ ⎟ −⎜ ⎟ ⎜ ⎟ ⎝ T ⎠ p ⎝ ∂ p ⎠T ⎝ ∂ T ⎠ p ⎝ ∂ T ⎠ p (f) This can be rewritten as 2 ⎡ 1 ⎤ ⎡ ⎛ ∂V ⎞ ⎛ ∂V ⎞ ⎤ c + T ⎜⎝ ⎟⎠ ⎥ ⎢ T ⎥ ⎢ p ⎜⎝ ∂ p ⎟⎠ ∂T p ⎥⎦ ⎣ ⎦ ⎢⎣ T (g) From Table B.4, we see that it is correct. As an illustration of the use of Table B.4, consider the expression for the Joule–Kelvin coefficient, viz. (∂T/∂p)H.

Table B.4

Bridgman’s table of partial derivatives

Deriv.

Deriv.

Expression

(∂T)p (∂V)p (∂S)p (∂U)p (∂H)p (∂G)p (∂A)p

= = = = = = =

– (∂p)T = – (∂p)V = – (∂p)S = – (∂p)U = – (∂p)H = – (∂p)G = – (∂p)A =

1 (∂V/∂T)p (cp/T) cp – p(∂V/∂T)p cp –S – S – p(∂V/∂T)p

(∂V)T (∂S)T (∂U)T (∂H)T (∂G)T (∂A)T

= = = = = =

– (∂T)V = – (∂T)S = – (∂T)U = – (∂T)H = – (∂T)G = – (∂T)A =

– (∂V/∂p)T (∂V/∂T)p T(∂V/∂T)p + p(∂V/∂p)T –V + T(∂V/∂T)p –V p(∂V/∂p)T

(∂S)V

= – (∂V)S =

(∂U)V (∂H)V (∂G)V (∂A)V

= = = =

– (∂V)U = – (∂V)H = – (∂V)G = – (∂V)A =

cp(∂V/∂T)p + T(∂V/∂p)2T cp(∂V/∂T)p + T(∂V/∂p)2T – V(∂V/∂T)p –V(∂V/∂T)p – S(∂V/∂p)T –S(∂V/∂p)T

(∂U)S (∂H)S (∂G)S (∂A)S

= = = =

– (∂S)U = – (∂S)H = – (∂S)G = – (∂S)A =

pcp(∂V/∂p)T /T + p(∂V/∂T)2p – Vcp/T – Vcp/T + S(∂V/∂T)p pcp(∂V/∂p)T /T + p(∂V/∂T)2p + S(∂V/∂T)p

(∂H)U (∂G)U (∂A)U

= – (∂U)H = = – (∂U)G = = – (∂U)A =

(∂G)H (∂A)H

= – (∂H)G = = – (∂H)A =

– V(cp + S) + TS(∂V/∂T)p – [S + p(∂V/∂T)p][V – T(∂V/∂T)p] + p(∂V/∂p)T

(∂A)G

= – (∂G)A =

– S[V + p(∂V/∂p)T] – pV(∂V/∂T)p

cp(∂V/∂T)p/T + (∂V/∂T)2p

–V[cp – p(∂V/∂T)p] – p[cp(∂V/∂p)T + T(∂V/∂T)2p] –V[cp – p(∂V/∂T)p] + S[T(∂V/∂T)p + p(∂V/∂p)T] + p[cp(∂V/∂p)T + T(∂V/∂T)2p] + S[T(∂V/∂T)p + p(∂V/∂p)T]

Appendix B:

Explanatory Notes

⎛ ∂T ⎞ (∂Τ ) H ∑ First write formally, ⎜ ⎟ = ⎝ ∂ p ⎠ H (∂ p) H

∑ Now, from the table, (∂p)H = –cp

and

⎛ ∂V ⎞ (∂ T)H = V – T ⎜ ⎟ . ⎝ ∂T ⎠ p

∑ Dividing one by the other ⎛ 1 ⎛∂T ⎞ ⎜ ⎟ = ⎜⎜ ⎝ ∂ p ⎠H ⎝ c p

⎤ ⎞⎡ ⎛ ∂V ⎞ ⎟⎟ ⎢T ⎜ ⎟ −V⎥ ⎥⎦ ⎠ ⎣⎢ ⎝ ∂ T ⎠ p

409

Appendix C

Problem-Solving

This appendix, in which some aspects of problem-solving are discussed, begins with units. It is followed by discussions on accuracy and a methodology for solving problems in thermodynamics.

C.1

UNITS

Numerical calculations are fundamental to thermodynamics.1 To get correct answers, the quantities involved have to be sufficiently accurate and be in proper units. The accuracy of data and calculations are dealt with in a later section. Here units, specifically SI units2 are briefly discussed. No attempt is made to be exhaustive since there is adequate literature on SI units.

C.1.1

Units and Dimensions

Each physical quantity has a dimension and is measured in some unit3. The dimension of a physical quantity is unique4. However, the unit in which it is measured varies. For example, height (of a man, child, table, wall, etc.) has the dimension of length, but it can be measured in inches, feet, metres or in any other units of length. Adding or subtracting quantities of different dimensions is physically meaningless since this implies adding or subtracting different physical quantities.5 This is achieved by ensuring 1 2 3 4 5

In fact, quantitative calculations are the basic hallmark of all sciences and, hence, engineering. This is a brief review of what is contained in college physics courses. The CRC Handbook of Chemistry & Physics states: “A physical quantity is equivalent to the product of the numerical value, i.e. a pure number and a unit”, 68th ed., 1987–88, p. F-256. A dimension has the same name as the physical quantity it represents (e.g. mass, force, energy, etc.), except, of course, because of historical reasons, length, breadth, height, thickness, etc. have the same dimensions. Mathematics is only concerned with pure numbers. Therefore, in it, the nature (and consequently, its dimensions) of the terms of an equation is irrelevant. However, the terms in our equations are physical quantities and therefore should have the same dimensions. 410

Appendix C: Problem-Solving

411

that all terms in an equation have the same dimension.6 For example, all the terms of the energy equation should have the dimensions of energy. Because of its importance, the units of a quantity will always be written (at least in the early stages) in brackets following the number representing its magnitude (e.g. a mass of 5 [kg]). The fundamental constants (the best values known as of now) used extensively in physics and engineering are given in Table C.1. Some of these are also used in defining derived units. For example, second, the unit of time, is defined in terms of the time taken by light to travel 1 [m] in vacuum. Table C.1 Fundamental constants

C.1.2

Name

Value

Avogadro number Electronic charge Electron rest mass Molar (universal) gas constant Planck’s constant Speed of light in vacuum

6.022045 ¥ 1023 [mole–1] 1.60199 ¥ 10–19 [C] 0.9109534 ¥ 10–30 [kg] 8.3144 [J/mol.K] 6.626176 ¥ 10–34 [J.s] 2.99792458 ¥ 108 [m/s]

The SI Units

The SI units are used in this book. The standard symbol of each unit is written within brackets. In this section, only a brief mention will be made since this is only a review. Basic and derived units In some sense the definitions of fundamental and derived units are arbitrary, because using fundamental physical constants a basic unit can always be redefined as a derived one. For example, originally the unit of length, metre, was defined as the distance between two marks on a standard platinum rod kept in Paris. However, in 1984 it was redefined as the distance travelled by light in vacuum during the time of (1/c) [s], where c is the velocity of light (a fundamental constant shown in Table C.1). There have been some who have advocated a definition of metre in terms of the Bohr radius.7 Then, the unit of time, second, can be defined as the ratio of metre to velocity of light. However, the relative merits of these definitions are not discussed here. Table C.2 contains the basic (and supplementary) units (without definitions). Table C.3 contains the more important derived units. The explanatory notes contained in Table C.4 are meant as a reminder. 6 7

Dimensional analysis is based on this idea. The well-known physical constant—the radius of electron orbit in hydrogen. This has the advantage that it is constant throughout our physical universe.

Appendix C: Problem-Solving

412

Table C.2 Quantity

Symbol

Basic and supplementary SI units Unit

Quantity

Symbol

Unit

m T n

[kg] (kilogram) [K] (kelvin) [mol] (mole)

—

—

—

[sr] (steradian)

Basic units Length Time Electric current

L t I

[m] (metre) [s] (second) [A] (ampere)

Luminuous intensity

Iv

[cd] (candella)

Plane angle

—

[rad] (radian)

Mass Temp. Amount of substance —

Supplementary units

Table C.3

Solid angle

Derived SI units

Quantity

Name

Symbol

Definition

Force Pressure Energy Power Charge (elec.) Potential difference (elec.) Resistance (elec.) Capacitance (elec.) Conductance (elec.) Inductance (elec.) Magnetic flux Magnetic flux density

newton pascal joule watt coulomb volt ohm farad siemens henry weber tesla

N Pa J W C V W F S H Wb T

[kg.m/s2] [N/m2] [N.m] [J/s] [A.s] [J/C] [V/A] [C/V] [A/V] [V.s/A] [V.s] [Wb/m2]

Table C.4

Explanatory notes

1. Mole is a measure of the quantity of any substance. By definition, 1 [mol] is equal to the Avogadro number of particles. These particles may be atoms, molecules, ions, electrons, photons, etc. Hence, the type of particles must be specified with [mol]. 2. By definition, Avogadro number is the number of atoms in 0.012 kg of C12. As of now, its value is 6.022045 ¥ 1023 [number/mol]. 3. Compound units are formed by writing the base units side-by-side. These may be separated by a period (e.g.[kg m/s2] or [kg.m/s2].) 4. Hence, to avoid confusion, all units are written only in singular. For example, writing ten kilograms as 10 [kgs] might be interpreted as 10 [kg.s], i.e. 10 [kg] ¥ 1 [s]. Hence it should be written only as 10 [kg]. 5. Temperature is written as [K] and not as [°K]. 6. When writing units in full, the first letter in the names of scientists is not capitalized (e.g. newton and not Newton; joule and not Joule, etc.). However, when they are single letters they are written as usual (capital or small) (e.g. [m] for metre; [N] for newton).

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413

The prefixes commonly used in SI units are presented in Table C.5. Prefixes used in SI units

Table C.5 Name

Symbol

exa peta tera giga mega kilo hecto deca

C.1.3

Factor 18

E P T G M k h da

10 1015 1012 109 106 103 102 101

Name

Symbol

Factor

atto femto pico nano micro milli centi deci

a f p n m m c d

10–18 10–15 10–12 10–9 10–6 10–3 10–2 10–1

Other Units and Conversion Factors

Definitions of some additional units used in engineering and some conversion factors are given in Table C.6. They are definitions and hence their values are exact. Table C.6

Additional units and conversion factors for engineering

All these units are definitions and hence their values are exact. SI and MKS units 1 1 1 1 1

[tonne] [Pa] [atm] [ton] [litre]

= = = = =

1000 [kg] 1 [N/m2] 101325 [Pa] (1/760) [atm] 0.001 [m3]

1 1 1 1 1

[kgf] [bar] [ata] [mm Hg] [Ps]

= = = = =

9.80665 [N] 105 [Pa] l [kgf/cm2] 1 [torr] 75 [kgf.m/s]

[tonne] is metric ton; [Pa] is pascal; [atm] is standard atmosphere. [ata] is technical atmosphere; [torr] is torricelli. [Ps] is “Pferde Starke” or “Cheval-vapeur”—the “metric horse power”. FPS units 1 1 1 1

[ft] [mile] [US ton] [lbf]

= = = =

12 [inch] 5280 [ft] 2000 [lb] 32.174 [lbl]

1 1 1 1

[yard] [UK ton] [lbl] [hp]

= = = =

3 [ft] 2240 [lb] 1 [lb.ft/s2] 550 [ft.lbf/s]

[UK ton] is ‘long’ ton; [US ton] is ‘short’ ton; [lbl] is poundal. Conversion factors 1 [abs.cal] 1 [inch] 1 [btu/lb.°F]

= 4.1840 [J] = 2.54 [cm] = 1 [kcal/kg.K]

1 [lT cal] 1 [lb] 1 [°F]

= 4.1868 [J] = 0.45359237 [kg] = 1.8 [°C] + 32.0

[abs.cal] (‘absolute’ calorie) is Thermochemical Calorie. [IT cal] is International Steam Tables Calorie and = (1/860) [W.h].

414

C.1.4

Appendix C: Problem-Solving

Systematic Derivation of Conversion Factors

This is not a handbook and therefore the tables of conversion factors are not given. Instead, examples are presented to illustrate the basic principle involved so that the method for systematically converting units is learnt8. Another equally important reason is that with the widespread availability of electronic calculators, these factors can always be generated to any desired degree of accuracy. All conversions needed for exercises in this book and in most of others can be obtained beginning with the factors in Table C.6. By definition, they are exact. Most of them will be familiar. Those which are not should be memorized now. The following examples illustrate a systematic way of converting units. The main point to be remembered is “when quantities are multiplied (or divided), their units also get multiplied (or divided).’’ EXAMPLE C.1

Obtain the value of the standard acceleration due to gravity in British units.

Solution 1 [in] = 2.54 [cm] and 1 [ft] = 12 [in]. Hence, 1 [ft] = 0.3048 [m]. This can also be written as 1 = 0.3048 [m/ft]. Note that the number on the LHS has no units. It is a pure number. Now, the standard acceleration due to gravity is defined as g0 = 9.80665 [m/s2]. Therefore, 9.80665 [m/s2 ] g0 = = 32.174048 [ft/s2 ] 0.3048 [m/ft] Note that, in the final answer, when the numbers were divided, their units were also divided. The standard value is defined as 32.174 [ft/s2]. This example is just to show the procedure of conversion. It will not be used later. EXAMPLE C.2 Solution

In FPS units, what is the value of a standard atmosphere?

By definition, 1 [atm] = 101325 [Pa]. This is converted into FPS units as follows: ⎡ kg.m ⎤ 101325 [Pa] = 101325 [N/m2] = 101325 ⎢ 2 ⎥ ⎣ m .s ⎦ ⎛ ⎡ 1b ⎤ ⎞ ⎛ 1 ⎡ kg ⎤ ⎞ ⎛ ⎡m ⎤⎞ = ⎜ 101325 ⎢ ⎟ × ⎜ 0.3048 ⎢ ⎥ ⎟ ⎟×⎜ ⎢ ⎥ ⎥ ⎣ m.s ⎦ ⎠ ⎝ 0.45359237 ⎣ kg ⎦ ⎠ ⎝ ⎣ ft ⎦ ⎠ ⎝

On cancelling the units, this will reduce to the units of [lbl/ft2], which can be converted to [lbf/in2] (i.e. [psi]) by dividing first by gc, which, in FPS units, is written as (9.80665/0.3048) [lbf/lbf] and then by 144 [in2/ft2]. This gives ⎛ ⎡ lb ⎤ ⎞ ⎛ 1 ⎡ kg ⎤ ⎞ ⎛ ⎡m ⎤⎞ 1[atm] = ⎜ 101325 ⎢ ⎟ × ⎜ 0.3048 ⎢ ⎥ ⎟ × ⎟×⎜ ⎢ ⎥ ⎥ ⎣ m.s ⎦ ⎠ ⎝ 0.45359237 ⎣ kg ⎦ ⎠ ⎝ ⎣ ft ⎦ ⎠ ⎝

⎡ in 2 ⎤ ⎞ ⎛ 0.3048 ⎡ lbf ⎤ ⎞ ⎛ ⎜ ⎟ × ⎜⎜ 144 ⎢ 2 ⎥ ⎟⎟ ⎢ ⎥ ⎝ 9.80665 ⎣ lbl ⎦ ⎠ ⎝ ⎣ ft ⎦ ⎠ 8

However, for the purposes of examinations, it will be good to prepare tables for oneself and learn them by heart. Such an exercise will be a confidence builder.

Appendix C: Problem-Solving

415

On carrying out the indicated calculations this shows to 7 significant digits that 1 [atm] = 14.69595 [psi]. EXAMPLE C.3 Pressure units defined in terms of the length of a liquid column result in significant uncertainty. As an illustration, consider the pressure 1 [mm Hg]. As per an earlier definition, this unit, named as [torr], was defined as 1 [mm Hg] = (1/760) [atm]. By fluid statics, the expression for the pressure exerted by a fluid column of height h is p = rgh, where r is its density and g is the acceleration due to gravity. This means that both r and g must be specified before the pressure can be calculated. Moreover, since r depends upon temperature it should also be specified. For example, consider the pressure due to 760 [mm Hg] at NTP at a place where the acceleration due to gravity is its standard value, i.e. 9.80665 [m/s2]. From standard tables (The CRC Handbook of Chemistry & Physics, 1988, p. F-6), for mercury at [0°C], r = 13.5595 [g/ml]. Then, the pressure exerted will be ⎛ ⎡ kg ⎤ ⎞ ⎛ ⎡ m ⎤⎞ p = ⎜ 13559.5 ⎢ 3 ⎥ ⎟ × ⎜ 9.80665 ⎢ 2 ⎥ ⎟ ¥ 0.760 [m] = 101059.6857 [Pa] π 101325 [Pa] ⎝ ⎣ m ⎦⎠ ⎝ ⎣ s ⎦⎠

Hence, in SI unit, 1 [torr] is defined as (1/760) of 1 [atm], but the use of the unit 1 [mm Hg] is discouraged because 1 [mm Hg] is only approximately equal to 1 [torr]. EXAMPLE C.4 The FPS unit of dynamic (absolute) viscosity is [lbf.s/ft2], while its SI unit is [kg/m.s]. Convert the FPS unit into SI unit. Solution

The calculations are done as before. ⎛ ⎡ 1 ⎤⎞ ⎛ ⎡ lbf ⋅ s ⎤ ⎡ lb ⋅ ft ⎤ ⎞ ⎛ ⎡ 1 ⎤ ⎞ 1 ⎢ 2 ⎥ = (1[lbf]) × ⎜1 ⎢ 2 ⎥ ⎟ × 1[s] = ⎜ 32.174 ⎢ 2 ⎥ ⎟ × ⎜ 1 ⎢ 2 ⎥ ⎟ × 1[s] ⎝ ⎣ ft ⎦ ⎠ ⎝ ⎣ ft ⎦ ⎣ s ⎦ ⎠ ⎝ ⎣ ft ⎦ ⎠ ⎛ ⎡ lb ⎤⎞ ⎛ ⎡ kg ⎤⎞ ⎛ ⎛ 1 ⎞ ⎡ ft ⎤⎞ = ⎜ 32.174 ⎢ × ⎜ 0.45359237 ⎢ ⎥⎟ × ⎜ ⎜ ⎟ ⎢ ⎥⎟ ⎟ ⎥ ⎝ ⎣ ft ⋅ s ⎦⎠ ⎝ ⎣ lb ⎦⎠ ⎝ ⎝ 0.3048 ⎠ ⎣ m ⎦⎠ ⎡ kg ⎤ = 47.8802 ⎢ ⎥ = 478.802 [P] ⎣m ⋅s⎦

where [P], called poise, is the cgs unit of dynamic viscosity defined as 1 [P] = 1 [g/cm.s]. Verify the following conversion factors. 1 [Btu/h.ft.°F] = 1.73073 [W/m.K] 1 [Btu/h.ft2.°F] = 3.15459 [W/m2] 1 [Btu/h] 1 [Btu/h.ft2]

= 5.67826 [W/m2.K]

1 [Btu]

= 6894.76 [Pa]

Exercise

1 [lbf]

= 1055.06 [J] = 4.44822 [N]

1 [psi] 3

1 [lb/ft ]

= 0.203071 [W] = 16.0185 [kg/m3]

416

C.2

Appendix C: Problem-Solving

ACCURACY OF CALCULATIONS

In order to obtain proper answers, numerical calculations9 have to be sufficiently accurate and also be in proper units. Units have already been discussed in the previous section. This section contains a brief review of the concept of accuracies in data and how they affect the accuracy of calculations.10 However, the same principles are also used in the analysis of experimental errors. The discussion will begin with the intuitive meanings of the words accuracy and error and the relation between them, namely that the greater accuracy implies the lesser extent of error. This will be followed by a discussion of how they are specified and computed.

C.2.1

Values of Physical Quantities

No physical quantity can be measured exactly (i.e. with infinite accuracy or zero error) because, as sensitivities of measuring instruments increase, molecular effects11 also increase. Moreover, as the demanded accuracy increases, the resources required (time, money, materials and manpower) increase almost exponentially. Consequently, the accuracy demanded should just be adequate for the purpose on hand. The value of any physical quantity can be written as x ± dx, where x is called its nominal value and dx is known as its error.12 Then, the quantity x is said to be known to an accuracy of ± dx. Alternatively, ± dx is called the error in the value of the quantity x. Either of these expressions may (and, will) be used as per convenience.

C.2.2

Approximate Numbers and Physical Quantities

Following the usage of Scarborough ([SCA], p. 2), the term approximate number will mean the approximate value of a number. For example, by definition, the value of an irrational number can never be expressed exactly. Consequently, they are always represented by approximate numbers. From the above discussion, it will be clear that the value of a physical quantity can never be known exactly. Hence, the number indicating it is also an approximate number. Thus, all the numbers in thermodynamics (indeed in all sciences) are approximate numbers. For example, let the length of a rod be 1 [m] ± 1 [mm]. This means that its length lies between 1.001 [m] and 0.999 [m] and the number representing the length is accurate only to the third decimal place.

C.2.3

Representation of Numbers

A significant figure is any one of the digits 1, 2, 3, ..., 9. The digit 0 (zero) is significant only when it is not used as a place holder. For example, the zeros in the number 0.00123 9 10 11 12

They are the fundamental to all sciences and therefore to thermodynamics. This has already been done in college physics courses. Experimentally, Brownian movement effects come into play. Even theoretically, at the microscopic level the uncertainty principle of quantum mechanics specifies upper limits on possible accuracy. Strictly speaking, this is called the absolute error and dx/x is called the relative error.

Appendix C: Problem-Solving

417

indicate the position of the non-zero digits, by holding the first two decimal places. Hence the zeros in these places are not significant. Similarly, the zeros is the number 12300 are place holders, merely indicating that the digit 3 is in the hundredth place. The ambiguity about zero, i.e. whether it is significant or not, arises only in the cases of leading zeros of a decimal number and of trailing zeros of a whole number. The accuracy of a number is represented either by the number of decimal places or by the number of significant digits.13 To avoid confusion, the number of decimal places or the number of significant digits should be specified in each case. This ambiguity can also be eliminated by using the normalized exponential notation. For example, the number 987600 is written to 4 significant digits as 9.876 ¥ 105 and, to 6 significant digits as 9.87600 ¥ 105. The point to be noted is: using the normalized exponential notation automatically eliminates the ambiguity about the significance of the digit zero. In this book, as far as possible, this method will be followed.

C.2.4

Accuracy of Computed Results

The following are some important results regarding the accuracy of calculations (see [SCA], Chapter 1. for a detailed discussion). The usual rule for rounding off a number is well known. If the current (one under consideration) digit is greater than 5, increase the previous (next higher) digit by one and discard the current digit. If the current digit is less than 5 simply discard it. When the current digit is 5, the next higher digit is increased only if it is odd. This ensures that the rounded-off number is always even. For example, to two decimal places, the numbers 1.235 and 1.245 will both be rounded off as 1.24. The logic of this is as follows. Odd and even numbers occur with roughly the same frequency and, therefore, in the long run, the round-off errors due to this rule will cancel out. However, the resultant (an even number) is divisible, in general, by more numbers than are the odd ones. Consequently, the errors in subsequent calculations will be less. Note that when 1.24 is obtained by rounding off the numbers 1.235 and 1.245 to 1.24, the resulting error is one unit in the second decimal place. This result can be generalized as follows: The error in a number containing n significant digits is one unit in the nth place. The following result, that is, the relative error in a number with n significant digits is less than l/(k ¥ 10n–1), where k is its first digit and its converse, that is, if the relative error in a number is l/[(k + 1) ¥ 10n–1], where k is its first digit, then the number is accurate to n significant digits follow directly from the round-off rule presented above. The basic principle of all calculations is: The results are only as accurate as the least accurate data. The following steps are recommended for getting maximum accuracies in calculations. 13

The number of decimal places is related to the absolute error and the number of significant digits is related to the relative error.

418

Appendix C: Problem-Solving

1. Arrange the data into groups of comparable accuracies and treat these groups first by carrying out appropriate mathematical operations on them. 2. In all calculations, keep the number of digits in the more accurate data to be one more than that in the least accurate one. After calculations have been completed, round the result off to the number of significant digits of the least accurate data. For example, 987.6 + 5.432 = 987.6 + 5.43 = 993.03 = 993.0. 3. During subtraction, especially of two nearly equal numbers, keep the additional significant digits to compensate for their loss. For example, 987.65342 – 987.60211 = 0.05131. Thus, only 4 significant digits could be obtained by subtraction of two 8-digit numbers. 4. Use mathematical relations to transform equations such that the resulting operations are more accurate. For example, for small values of x, 1 – cos(x) will result in subtraction of two nearly equal numbers. Hence, convert this to 2 sin2(x/2). 5. Do all calculations at one stretch and finally round off the results. This avoids accumulation of round-off errors. For example, 9.876 + (4.567 ¥ 1.234 – (–2.453)/ 6.1 = 10.39775 = 10 because the least accurate data (i.e. 6.1) has only 2 significant digits. 6. Use the error propagation formulae for estimating error in the result of a set of complex calculations. 7. Use the ‘principle of equal effects’ to obtain a preliminary estimate of the permissible error. 8. Do the calculations in steps 6 and 7 only in the case of a doubt. Generally, in all calculations in this book, 5 significant digits are kept and the final result is rounded off to 4 digits.

C.3 A PROBLEM-SOLVING METHODOLOGY Solving a physical problem systematically involves the following steps. Formulation. From its verbal description, form the relevant equations. They are called system equations or governing equations. This is the responsibility of a scientist (engineer/...). Solution. Using the appropriate mathematical methods, solve the system equations. This is purely a mathematical exercise. When needed, the help of mathematicians is sought. Verification. Check whether the results are reasonably accurate. If not, check whether the error is due to poor formulation of equations or due to bad mathematical tools used. In this appendix the first two steps are illustrated with respect to thermodynamics. However, they can also be used in solving problems in other fields. A systematic procedure has the advantage that it helps (a) in identifying where different problems are similar and where they are different,14 and (b) in ensuring that the problem specification is complete (i.e. there is enough information to solve it uniquely). 14

Thus, one may find that there are only a few types of problems which occur in different disguises.

Appendix C: Problem-Solving

419

Its main disadvantage is that for simple problems such a detailed procedure is a wastage of effort. For a unique solution, mathematics requires that the number of independent equations should be equal to the number of unknowns. Generally, difficulty arises only in counting the number of independent equations, particularly, when the same equation is used in different (disguised) forms in different places.15 All good problem-solving procedures have a built-in systematic approach to make this accounting easy. The present procedure is based on the fact that, in thermodynamics, it is possible to divide specifications (or models) of any problem into the following components: System specification.

This can be isolated, closed or open (flow).

State specification. Generally, the initial state (denoted by subscript ‘1’) is given and the final state (denoted by ‘2’) is to be determined. Process specification. This can be constant pressure (isobaric), constant volume (isochoric), constant temperature (isothermal) and adiabatic (no beat transfer). The flow processes are not classified. This may also include specification of the energy flows in the form of work and heat. Substance specification. This is done by giving the equation of state and other thermodynamic properties in the form of equations, tables or charts. Each of these specifications gives rise to one or more equations. Keeping an account of them ensures that an equation arising out of a specification is used in the same form at all places. It also ensures that all the equations are independent. This is called mathematical consistency. For the purposes of counting equations, the given data should also be put in the form of equations. For example, a specification such as: ‘a constant pressure process at 1 [bar]’ contains two equations, namely p1 = 1 [bar], p2 = const., which also implies p1 = p2; and therefore p2 = 1 [bar]. Generally, this is compactly written as p1 = p2 = 1 [bar]. It is clear that only two of these three equations are independent since any one can be derived from the other two. The major steps of this methodology are presented in Table C.7. The following example illustrates the above steps. Problem statement. Two-and-a-half kilograms of saturated water at 10 [bar] is adiabatically mixed with 1 [kg] of superheated steam at 10 [bar] and 250 [C°]. For the system, determine the work done. Solution. is used.

This problem is solved first by a closed system model and then a flow system model

Closed system model. The number of the following items generally corresponds to that of the steps shown in Table C.7. 1. The system diagram is shown in Figure C.1

15

They are then not independent equations.

Appendix C: Problem-Solving

420

Table C.7

Steps of the problem-solving methodology

No.

Description of the step

Reason

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Draw the system diagram. Describe the system in words. State the nature of the system. Name the working substance. Draw the process diagram. State the initial conditions. State the process. Obtain the governing equations. No. of eqs. = no. of unknowns? Use auxiliary equations if needed Solve the equations Evaluate the unknowns

Only then W and Q will have a meaning. A precautionary measure. The form of equations depends on this. Its properties can be obtained. Helps in visualizing the system. Given data. Forms part of the basic set of equations. For obtaining the quantitative relations. No unique solution otherwise. To satisfy step 9. To determine the unknowns. To get numerical answers.

W

W

Piston

Piston

System A

System B

System C

System C A + B

Figure C.1

The system diagram: closed system model.

2. The system has two compartments, one of which (A) contains 2.5 [kg] of saturated water. The other [B] contains 1 [kg] of superheated steam. Hence, their properties should be obtained from steam tables. 3. This system is closed. Hence its mass is constant. 4. In this case the process diagram is not very helpful. 5. The water in compartment A is saturated at 10 [bar]. The steam in compartment B is superheated to 250 [°C] at 10 [bar]. 6. This is a mixing process at adiabatic and isobaric conditions. During this process, the water absorbs energy from the steam and gets heated (and perhaps, evaporates) while the steam condenses. Hence it is a non-quasi-static process. However, the isobaric expansion is quasi-static. Hence, only the initial and final equilibrium states are considered. 7. The governing equations are put in a concise form (see Table C.8).

Appendix C: Problem-Solving

Table C.8

Equations of the closed system model

No.

Governing equation

Reason

1. 2. 3. 4. 5.

m1 = mA + mB = m2 Q = DE + W Q=0 E=U W = Wx

Closed system and so m = constant. First law for the closed system. Insulated system. Pure substance. All other work modes are absent.

6.

Wx = ∫12 pdV Wx = pDV = D(pV) 0 = DU + D(pV) DH = 0 H1 = H2 H1 = mAhA + mBhB H2 = (mA + mB)h2

Quasi-static process.

7. 8. 9. 10. 11. 12.

421

p = constant during the process. By substituting in the first law. By the definition of H. From step (i). H is a property and so is additive. By the definition of specific.

8. Solving for h2, from steps (k) and (1) of the above table, gives h2 = (mAhA + mBhB)/(mA + mB). 9. To get numerical results, the values of masses, enthalpies, etc. are substituted in the above relation. This gives, h2 = 1385.6 [kJ/kg]. 10. Thus, the final state (‘2’) is p2 = 10 [bar] and h2 = 1385.6 [kJ/kg]. Then, the steam tables show that the steam at this state is wet. 11. Using the definition of quality (dryness fraction) gives, x2 = 0.309. 12. Using the definition of volume of wet vapour gives, v2 = 0.06082 [m3/kg]. 13. Then, DV = –0.0227 [m3]; W = –22.7 [kJ]; and DU = + 22.7 [kJ]. Flow system model. Now the flow system model is used to formulate the governing equations. It is solved in the same way as the closed system model. Hence, the above steps of calculations are applicable and they are not repeated. The system is shown in Figure C.2. W

Piston

System (CV)

Inlet i

Figure C.2

The system diagram: flow system model.

422

Appendix C: Problem-Solving

Table C.9 indicates only the major steps of formulation. It is evident that the same governing equations are obtained.16 However, for this class of problems the open system model is simpler. Table C.9 Equations of the flow system model No.

Governing equation

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

m2 – m1 = mi DECV = Q – W + mihi Q=0 E= U W = Wx Wx = ∫12 pdV Wx = (pDV)CV = D(pV)CV DUCV = D(pV)CV + mihi DHCV = mihi m 2h 2 – m 1h 1 = m i h i

Reason Mass conservation equation. First law for open systems. Insulated system. Pure substance. All other work modes are absent. Quasi-static process. p = constant during the process. By substituting in the first law. By the definition of H. From the first law.

To compare with the closed system model, use CV = 1 = A, and i = B.

With minor modifications to these steps, the following types of problems (which are really mixing) can be solved: (a) mixing of gases; (b) mixing of liquids; (c) contact type fluid heaters (e.g. feed water heater, heat exchanger, etc.); (d) desuperheaters (water injected into superheated steam); and (e) quenching or chilling or other types of heat treatment (e.g. solids immersed into liquids). In these cases, appropriate equations (or tables or charts) for thermodynamic properties should be used. The energy (first law) equation should also be modified to account for heat transfer and additional modes of work done (e.g. stirrer) as applicable.

16

Otherwise there is something wrong with the formulation (or theory), because the same physical process is being described.

Appendix D

Some Mathematical Principles

Thermodynamics deals with only scalar quantities (e.g. mass, volume, work, heat, internal energy, etc). Hence, all its basic equations are algebraic. Simple differential equations are encountered when rate processes are analyzed; most of which may be solved by elementary methods of integration. Only in the case of evaluation of properties of substances at different states (as in the case of developing the steam tables) numerical methods are needed. Thus, thermodynamics is mathematically the simplest. In this appendix, the mathematical formulae most frequently employed in thermodynamics are reviewed. In the interest of clarity, only the simplest forms are dealt with. Extensions to more complicated cases (e.g. higher dimensions) will be obvious. These are given in any textbook on college mathematics.

D.1

DIFFERENTIALS AND DERIVATIVES

Let f ¢(x) be the derivative of a variable y with respect to the variable x. Then, the differential dy is defined by the relation dy = f ¢(x) dx whenever dx π 0. Based on this trait, differentials may be treated as algebraic quantities. For example, given the differentials dy = f ¢(u)du and du = g¢(x)dx and substituting the second differential in the first gives dy = f ¢(u) g¢(x)dx, which can be rewritten as dy/dx = f ¢(u) g¢(x). This is the well-known chain rule of differentiation.1 In engineering, derivatives are always assumed to be the same as the ratio of differentials in the limit. However, the finer distinction is worth remembering.

D.2

INTEGRATION

The process of integration is generally defined in the following two ways. 1

The method of separation of variables used in solving differential equations is another example of this. 423

Appendix D: Some Mathematical Principles

424

D.2.1

Anti-derivation

Let the derivative of the function F(x) with respect to x be denoted as f(x), i.e. f(x) = (dF(x)/dx). This process, called differentiation, is well known. The inverse process, called integration, defined as F(x) = Ú f(x)dx, is also well known. Substituting the first equation in the second gives F(x) = Ú(dF(x)/dx)dx. This shows that the process of integration is the inverse of that of differentiation. In other words, integration is anti-derivation. Since the relations are symmetric, derivation (differentiation) may be called antiintegration. Since the derivative of a constant is zero, it is usual to write the integral as F(x) = Ú f (x)dx + C, where the integration constant is denoted as C. This interpretation can be used to integrate only elementary functions. Hence, a more general procedure (definition) is needed.

D.2.2

Area of a Curve

In this definition,

∫

2

1

f ( x ) dx is interpreted as the area enclosed by the x-axis and the curve

f (x) between the abscissa xl and the abscissa x2, as shown in Figure D.1. y

1

y = f (x)

2

x1

xi xi+1 Figure D.1

x2

x

Integral as area.

Now dividing the interval into a large number of small subintervals, as shown in Figure D.1, the curve can be approximated by a straight line. Then, the area of the infinitesimal trapezoidal strip between xi and xi+1 may be written as dA = (1/2)[ f (xi) + f (xi+1)]dx = (yi + yi+1)(dx/2) The total area A can be obtained by summing these small areas. That is, N

A=

∑ (1/ 2)[ f ( x ) + f ( x i

i =1

i +1 )] d x

Appendix D:

Some Mathematical Principles

425

This is called the trapezoidal rule for numerical integration. As the subintervals become smaller, better approximation to the integral is obtained, so that the limit gives

∫

2

1

N

f ( x )dx = lim

N →∞

∑ f (x )d x i

i =1

where dx = (x2 – x1)/N. If the limit of the sum on the RHS exists and is unique, the function f (x) is called Riemann integrable and the value of the integral equals this sum. This is the Riemann’s definition of an integral (see [WHI], Chapter 4 for a rigorous discussion). A continuous function with a finite number of ordinary discontinuities is Riemann integrable. A function with limited fluctuation and a finite number of ordinary discontinuities is also Riemann integrable (ibid., p. 63). However, if f (x) takes on random values in the interval, the sum on the RHS will not converge to any unique value and therefore is not Riemann integrable.

D.3 PFAFFIANS All the equations in thermodynamics possess the general form dF ∫ Mdx + Ndy + Pdz + º where, M, N, P, … are functions of the coordinates x, y, z. … For example, the first law for a non-flow quasi-static infinitesimal expansion process is dQ = du + pdv. Similarly, the differential equation of the variation of a property (say s) of a simple compressible substance is ds = (cv dT/T)dT + (∂p/∂T)v dv, where T and v are the state variables (see the discussion on properties in the main text). In mathematics, such an expression as above (on the RHS) is called a Pfaffian2 differential expression. The corresponding equation, namely dF ∫ Mdx + Ndy + Pdz + º = 0 is called a Pfaffian differential equation. Both of them possess identical mathematical characteristics.3 In this section, the important properties of Pfaffians are briefly reviewed. For clarity, only two and three dimensions are considered. Extensions to higher dimensions are straightforward. All the Pfaffians in thermodynamics (except the first and second law expressions) are exact. Hence, essentially, exact Pfaffian differential expressions and equations are dealt with. For this purpose, the process of deriving an exact Pfaffian differential equation is first investigated. 2 3

In honour of J.F. Pfaff (pronounced as ‘phuf’) who first investigated this form and correctly interpreted its significance ([FOR], pp. 80–83 or [INC], § 2.83). Except that an expression gives a value, while an equation deter