Basic Thermodynamics [1 ed.] 9788120341128

Citation preview

BASIC THERMODYNAMICS

B.K. Venkanna Swati B.V.

BASIC THERMODYNAMICS

Basic Thermodynamics B.K. VENKANNA Professor Department of Mechanical Engineering Basaveshwar Engineering College Bagalkot

SWATI B.V. Assistant Professor Department of Mechanical Engineering Basaveshwar Engineering College Bagalkot

New Delhi-110001 2010

BASIC THERMODYNAMICS B.K. Venkanna and Swati B.V. © 2010 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-978-81-203-4112-8 The export rights of this book are vested solely with the publisher. Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Mohan Makhijani at Rekha Printers Private Limited, New Delhi-110020.

Contents Preface Acknowledgements

1.

xi xiii

Fundamental Concepts and Definitions 1.1 Thermodynamics: Definition and Scope 1 1.1.1 Practical Application of Engineering Thermodynamics 1 1.2 Microscopic Versus Macroscopic Approaches 2 1.2.1 Microscopic Point of View 2 1.2.2 Macroscopic Point of View 2 1.3 Thermodynamic Systems 3 1.3.1 Closed System 3 1.3.2 Open System, Control Volume and Control Surface 3 1.3.3 Isolated System 4 1.4 Thermodynamic Property 5 1.4.1 Intensive Property 5 1.4.2 Extensive Property 6 1.5 Thermodynamic State 6 1.6 Thermodynamic Process or Change of State, Path 6 1.6.1 Point and Path Function 7 1.6.2 Equilibrium 8 1.6.3 Quasi-static Process 8 1.7 Thermodynamic Cycle or Cyclic Process 9 1.8 Reversible Process 9 1.8.1 Internally Reversible Process 9 1.8.2 Totally Reversible Process 9 v

1–34

vi

Contents

1.9

Thermodynamic Equilibrium 9 1.9.1 Thermal Equilibrium 9 1.9.2 Mechanical Equilibrium 10 1.9.3 Chemical Equilibrium 11 1.10 Diathermic Wall 11 1.11 Temperature Concept 11 1.11.1 Equality of Temperature 11 1.11.2 Zeroth Law of Thermodynamics 12 1.12 The Thermometer and the Thermometric Property 1.13 Temperature Scale 12 1.13.1 Standard Scale of Temperature 13 1.14 Temperature Measurements 14 1.14.1 Method in Use before 1954 14 1.14.2 Method in Use after 1954 15 1.15 International Practical Temperature Scale 15 1.16 Comparison of Thermometers 16 1.17 Comparison of Temperature Scale 16 Solved Examples 17 Exercises 32

12

2.

Work and Heat 2.1 Work 35 2.1.1 Work—According to Mechanics 35 2.1.2 Limitation 36 2.1.3 Work—According to Thermodynamics 36 2.2 Sign Conventions for Work and Heat 37 2.3 Work Done in a Frictionless Quasi-equilibrium Process or PdV Work or Displacement Work 37 2.4 PdV Work in Various Quasi-static Processes 38 2.5 Other Types of Work Transfer 40 2.6 Heat 43 2.7 Comparison of Heat and Work 44 2.7.1 Common between Heat and Work 44 2.7.2 Difference between Heat and Work 44 Solved Examples 45 Exercises 84

35–87

3.

First Law of Thermodynamics 3.1 Joule’s Experiment: Equivalence of Heat and Work 88 3.2 First Law of Thermodynamics for a System Undergoing a Thermodynamic Cycle 89 3.3 First Law of Thermodynamics for a Process in a Closed System [First Law of Thermodynamics for a Non-cyclic Process] 90 3.3.1 Components of Stored Energy of a System: Modes of Energy 3.4 Internal Energy—A Property of the System 91 3.5 Enthalpy (H) 92

88–170

91

Contents

Specific Heat (C) 93 3.6.1 Specific Heat at Constant Volume (Cv) 93 3.6.2 Specific Heat at Constant Pressure (Cp) 93 3.6.3 Specific Heat of Solids and Liquids 94 3.7 First Law to a Closed System for Different Processes 94 3.7.1 The Constant Volume (Isochoric) Process 94 3.7.2 The Constant Pressure (Isobaric) Process 95 3.7.3 The Constant Temperature Process (Isothermal) 95 3.7.4 Adiabatic Process or Isentropic Process 95 96 3.7.5 Polytropic Process (PV n = C) 3.8 Application of First Law—Steady Flow Energy Equation 97 3.9 Steady-State, Steady Flow Process 100 3.9.1 SFEE for One Stream 100 3.10 Applications of Steady Flow Energy Equation 101 3.10.1 Work Absorbing Systems 101 3.10.2 Non-work Developing and Non-work Absorbing Systems 3.10.3 Work Developing System 104 3.10.4 Some Other Systems 106 3.11 Transient Analysis or Unsteady Flow Process Analysis 108 3.11.1 Tank Filling Process 108 3.11.2 Tank Discharging Process 110 3.12 Analysis of the Open System for Different Processes 111 Solved Examples 111 Exercises 167

vii

3.6

4.

Second Law of Thermodynamics 4.1 Limitations of First Law of Thermodynamics 171 4.2 Heat Engine 175 4.2.1 Thermal Efficiency 177 4.3 Reversed Heat Engines 177 4.3.1 Refrigerator 177 4.4 Heat Reservoirs (HR) 179 4.4.1 Heat Source 179 4.4.2 Heat Sink 179 4.5 Statements of Second Law of Thermodynamics 179 4.5.1 Kelvin-Planck Statement 179 4.5.2 Clausius Statement 181 4.6 Equivalence of Kelvin-Planck and Clausius Statements 181 4.7 Reversible and Irreversible Processes 183 4.7.1 Conditions for Reversibility 183 4.7.2 Internally and Externally Reversible Process 183 4.7.3 Factors That Support Reversible Processes 185 4.7.4 Examples of Reversible Processes 185 4.7.5 Irreversible Process 187 4.7.6 Factors That Support Irreversible Processes 188

104

171–231

viii

Contents

4.7.7 Classification of Irreversible Process 188 4.7.8 Examples of Irreversible Processes (External Irreversibility) 4.8 Carnot Cycle 191 4.8.1 Working Principle of Carnot Cycle 191 4.9 Carnot’s Theorem 194 4.9.1 Corollary-I 194 4.9.2 Corollary-II 195 4.10 Thermodynamic Temperature Scale 196 4.10.1 Corollary-III 196 4.11 Zero Temperature on Thermodynamic Temperature Scale 198 4.12 Corollary-I 199 4.13 Corollary-II 200 Solved Examples 201 Exercises 228

5.

6.

Entropy 5.1 Introduction 232 5.2 Entropy and Heat 232 5.3 Carnot Theorem 233 5.4 Clausius Theorem 234 5.5 Entropy—Property of a System 235 5.6 Clausius Inequality 235 5.7 Entropy Change of an Irreversible Process of a Closed System 5.8 Principle of Increase of Entropy 239 5.9 The Combined First and Second Law 240 5.10 Entropy Change for an Ideal Gas 241 5.11 Change of Entropy for Different Processes 242 5.12 Isentropic Process for Solid or Incompressible Fluid 244 5.13 Isentropic Work in a Steady Flow System 245 5.14 Available and Unavailable Energy 246 Solved Examples 247 Exercises 277

189

232–280

238

Availability and Irreversibility 281–321 6.1 Introduction 281 6.2 Concept of Availability 281 6.3 Maximum Work and Maximum Shaft Work 283 6.4 Availability of a System 284 6.4.1 Reversible Work 284 6.4.2 Reversible Work for a Steady Flow Process 285 6.4.3 Availability for a Steady Flow Process 286 6.4.4 Reversible Work for a Non-flow Process or Closed System 287 6.4.5 Availability for a Non-flow Process or Closed System 287 6.5 Irreversibility 289 6.6 Availability for Ideal Gas Systems 290 6.7 Availability for Substances in the Solid and Liquid Phases 292

Contents

6.8 Availability Associated with Heat Flow from Source 6.9 Second Law Efficiency or Effectiveness 294 Solved Examples 294 Exercises 319

ix

292

7.

Pure Substances 322–367 7.1 Pure Substances—Definition and Explanation 322 7.2 Two Property Rule 324 7.3 Formation of Steam at Constant Pressure or Phases of Pure Substances 324 7.4 Specific Volume 328 7.4.1 Specific Volume of Saturated Water 328 7.4.2 Specific Volume of Dry Steam or Saturated Vapour 328 7.4.3 Specific Volume of Wet Steam 328 7.4.4 Specific Volume of Superheated Steam 329 7.5 External Work of Evaporation (W) 329 329 7.5.1 External Work of Evaporation of Dry Saturated Steam (Wg) 329 7.5.2 External Work of Evaporation of Wet Steam (Ww) 7.5.3 External Work of Evaporation of Superheated Steam (Wsup) 329 7.6 True or Internal Latent Heat 329 329 7.6.1 Internal Latent Heat of Dry Saturated Steam = (hfg – PVg) 7.7 Internal Energy of Steam (U) 329 330 7.7.1 Internal Energy of Dry Saturated Steam (ug) 7.7.2 Internal Energy of Wet Steam (uw) 330 330 7.7.3 Internal Energy of Superheated Steam (usup) 7.8 PT, PV, TV, TP, PH Diagrams 330 7.9 State Changes of a System Involving Pure Substances 331 7.9.1 Constant Volume Process or Isochoric Process 332 7.9.2 Constant Pressure Process or Isobaric Process 332 7.9.3 Constant Temperature Process or Isothermal Process 333 7.9.4 Isentropic or Reversible Adiabatic Process 334 7.9.5 Throttling Process 335 7.10 Steam Tables 336 7.11 Measurement of Dryness Fraction or Quality of Steam 337 7.11.1 Throttling Calorimeter 337 7.11.2 Combined Separating and Throttling Calorimeter 338 Solved Examples 340 Exercises 364

8.

Ideal Gases and Mixtures of Ideal Gases 368–433 8.1 Ideal Gas 368 8.2 Ideal Gas Laws 368 8.2.1 Boyle’s Law 369 8.2.2 Charles’ Law 369 8.3 Ideal Gas Equation or Characteristic Equation for the Gases 369 8.4 Avogadro’s Hypothesis 370 8.5 Relation between Properties of an Ideal Gas 371 8.5.1 Internal Energy and Enthalpy Change of an Ideal Gas 371

x

Contents

8.5.2 Relation between Specific Heats of Gases and Gas Constant 8.5.3 Entropy Change of an Ideal Gas 373 8.6 Analysis of Non-flow Processes for an Ideal Gas 374 8.6.1 Constant Volume Process (Isochoric Process) 374 8.6.2 Constant Pressure Process (Isobaric Process) 375 8.6.3 Constant Temperature Process (Isothermal Process) 376 8.6.4 Reversible Adiabatic Process (Isentropic Process) 377 8.6.5 Polytropic Process 378 8.6.6 Specific Internal Energy 381 8.6.7 Specific Enthalpy 381 8.7 Reversible Adiabatic Process 382 8.8 Other Relations for an Isentropic Process 383 8.9 Analysis of the Steady Flow Process (Open System) 384 8.10 Ideal Gas Mixture 389 8.11 PVT Behaviour of an Ideal Gas Mixture 389 8.11.1 Mass Fraction 389 8.11.2 Mole Fraction 390 8.11.3 Molecular Weight of the Mixture of Ideal Gases 391 8.11.4 Dalton’s Law of Partial Pressure or Dalton’s Law of Additive Pressure 391 8.11.5 Amagat’s Law of Partial Volume or Amagat’s Law of Additive Volume 392 8.11.6 Characteristic Gas Constant of the Mixture of an Ideal Gas 8.11.7 Internal Energy and Constant Volume Specific Heat of an Ideal Gas Mixture 395 8.11.8 Enthalpy and Constant Pressure Specific Heat of an Ideal Gas Mixture 395 8.11.9 Entropy of an Ideal Gas Mixture 396 8.11.10 Change in Entropy 397 Solved Examples 397 Exercises 430

9.

Real Gases 9.1 Introduction 434 9.2 van der Waals Equation of State 434 9.3 van der Waals Constants, a and b 435 9.3.1 Non-dimensional Properties 436 9.4 Compressibility Factor (Z) 438 Solved Examples 441 Exercises 451

Bibliography

372

394

434–452

453

Model Question Papers (With Answers)

455–509

Index

511–514

Preface Thermodynamics is a fundamental subject, which deals with energy, heat and work. The thermodynamic laws have wide applications in day-to-day life. It is a fundamental subject in thermal stream. A perfect knowledge of this can make one a competent engineer in thermal applications. The book provides the foundation for subsequent subjects like Fluid Mechanics, Applied Thermodynamics, Turbo Machines, Heat and Mass Transfer, Refrigeration and Air Conditioning. Generally, the books are of three types. First, written exclusively for practising engineers, second, written exclusively for research purpose where rigorous mathematical treatment is dealt with little importance on the art of problem solving and dealing basic principles and third, presenting a huge number of problems without much emphasis on basic fundamental principles. This book has been written keeping in mind to stress the fundamentals, physics of phenomena and art of solving the problems. The book covers fundamentals of all the topics of Basic Thermodynamics. The objective in this book is to present a comprehensive and rigorous treatment of Basic Thermodynamics to engineering applications. The subject is dealt in a most elegant and simple manner, so that the students can grasp easily and quickly. A large number of problems have been illustrated, graded in the order of increasing complexity, so that the concepts and principles will be understood without confusion. The solved examples also help the students to develop confidence in this subject. Despite the stringent efforts, the book may not be free from few errors, technical or otherwise. There is always a scope for the improvement. The authors, therefore, will be grateful to all of those who bring errors to their notice. Suggestions made for the improvement are most welcome. B.K. Venkanna Swati B.V. xi

Acknowledgements We express our gratitude to the publisher, PHI Learning, New Delhi, who encouraged us to take up the task. We would like to thank the members of B.V.V. Sangha and Principal of Basaveshwar Engineering college, Bagalkot, Karnataka, for excellent and congenial academic environment of the college that inspired us to accept the challenge. We wish to take this opportunity of thanking our friends, colleagues and H.O.D. of Mechanical Enginering of the college, for their moral support and encouragement to take up this work. We acknowledge the students whose active interactions, comments and thought-provoking questions prompted us to present the topics in a more elegant manner. We recall ourselves to the advice of my father, Late B.V. Krishnamurthy, who used to say, “Money comes and goes, but morality stays and grows”. The book demanded our overall attention even at the cost of small pleasures of our beloved children, B.V. Arpita and B.V. Vinayak. We thank them profusely for their understanding, beyond their age, of the seriousness of their parents’ involvement in bringing out this book. B.K. Venkanna Swati B.V.

xiii

1 CHAPTER

Fundamental Concepts and Definitions

1.1

THERMODYNAMICS: DEFINITION AND SCOPE

The term thermodynamics is derived from the Greek word therme (heat) and dynamics (power or motion); thus thermodynamics means heat power or heat in motion. Thermodynamics is defined as the science of engineering that deals with the storage of energy, entropy, heat and work, mutual conversation of heat and work, exchange of energy and those properties of substances that govern the relation between heat and work. Earlier, thermodynamics was considered as a subject of science by Joule (1818–1889), Kelvin (1824–1907), Clausius (1822–1888) and Carnot (1792–1832). Later, Gibbs (1839–1903) extended its scope and now it is extensively used in branches of science and engineering to analyze physical and chemical changes. Experimental observation is the basis for thermodynamics. Based on these observations and results, certain basic laws are developed which are known as the Zeroth, First, Second and Third laws of thermodynamics. A large part of the subject of thermodynamics deals with a study of energy. Energy can be defined as a capacity to produce change. The energy output of an I.C. engine supplies the capacity to move from one location to another. The energy output of a power plant supplies the capacity to produce a wide variety of changes to operate motors, television sets and lights, etc.

1.1.1 Practical Application of Engineering Thermodynamics 1. Internal combustion engines 2. Steam power plants 1

2

Basic Thermodynamics

3. Gas turbines 4. Heat exchangers 5. Refrigerators

1.2

MICROSCOPIC VERSUS MACROSCOPIC APPROACHES

The system is made up of substances (material or matter). The determination of properties of the substances is very important. The properties can be determined on certain quantity of matter or on a molecular level.

1.2.1 Microscopic Point of View Consider a piston cylinder arrangement with air in it. The air in the cylinder is assumed to be consisting of a very large number of identical particles. The volume of air contains approximately 1025 molecules. Each molecule is moving with an independent velocity. To describe the state of each molecule, three co-ordinates and three velocity components are required. To describe the state of each molecule, at least six variables are required. Hence, 6 × 1025 equations are required to calculate any properties of the given substance (in this example, air). Then, mathematical calculations are to be done to determine the properties from statistical averaging of the behaviour of identical molecules. In the above example, air in a piston cylinder, if pressure is to be measured, the change in the momentum of all the molecules colliding the wall per unit area should be taken into account. Then from statistical averaging of individual molecules, the pressure can be determined. The analysis based on the ‘average’ value of all the particles in the system is known as microscopic point of view. This is called as statistical thermodynamics, because here individual molecule is studied and the analysis is done to collective molecular action by statistical methods.

1.2.2 Macroscopic Point of View In the macroscopic point of view, the properties are measured on a large scale. In the above example of piston and cylinder, air occupies a certain volume for each position of the piston. For any given position, the volume can be measured. Similarly, consider the pressure, a gas exerts on the wall of its container. Pressure is the average rate of change of momentum due to all the molecular collision made on a unit area. From a macroscopic point of view, we are not interested in the action of individual molecules but we are interested with the time averaged force on a given area, which can be measured by a pressure gauge. The effect of pressure can be felt. Similarly, the temperature is measured with the help of a thermometrer, and this can also be felt. We have described the material (air) in a piston and cylinder by specifying three quantities— volume, pressure and temperature. These quantities refer to the gross characteristic or large-scale properties of the system and provide a macroscopic description. It is also called as classical thermodynamics. Classical thermodynamics involves observations and measurement of properties on a large scale. However, a clear understanding of the macroscopic behaviour of the system is possible only if we study microscopic behaviour.

Fundamental Concepts and Definitions

1.3

3

THERMODYNAMIC SYSTEMS

A thermodynamic system or simply a system is a region in space, or a finite collection of matter, enclosed by a boundary, on which study is focussed. The system boundary is an envelope or surface, which surrounds the system and separates the system and surroundings. The boundary can be real or imaginary. The boundary can be rigid or flexible. Hence, the system can be either fixed or moving in space. The imaginary boundary often coincides with a physical boundary. Figure 1.1 shows representation of a system. Thermodynamic systems are interacting with their surroundings, which involve the transfer or exchange of goods (mass or energy) across the system boundary. Hence, the surroundings are defined as everything external to the system, i.e. everything away from the boundary. The universe is a combination of system and surroundings. Thermodynamic systems can be classified into three classes, namely, the closed system, open system and isolated system. Figure 1.1 gives clear idea of each type of system.

Figure 1.1 Representation of (a) a closed system, (b) an open system and (c) an isolated system.

1.3.1 Closed System A closed system is one through which no mass crosses the boundary of the system (fixed quantity of matter). Only energy transfer can take place. Hence, in a closed system, mass does not cross the system boundary, even though energy (heat or work) may cross. Figure 1.2 shows an example of closed system. The imaginary system boundary indicated by dashed line separates the system (gas) and surroundings. System may be heated with the help of external source or work may be done on the system; in both the cases, the volume of system may change due to the movement of piston. Since the piston is assumed to fit tightly within the cylinder, no mass is permitted to cross the boundary. In this system, mass is controlled.

1.3.2 Open System, Control Volume and Control Surface An open system is one in which mass as well as energy cross the boundary of the system. In the closed system, the analysis is focussed on a fixed mass of matter, whereas in the open system,

4

Basic Thermodynamics

analysis is focussed on the region in space through which matter flows. An open system is more frequently referred to as control volume. A control volume is a volume in space through which matter, mass, momentum and energy may flow. Boundary of a control volume is called the control surface. The control surface can be real or imaginary [Figure 1.3(c)], and it can be fixed [Figure 1.3(b)] or moving boundary [Figure 1.3(a)].

W Piston

Cylinder Gas

Figure 1.2 Mass in

Piston movement

Boundary

Closed system.

Mass out

Fixed boundary Control volume (CV)

Control surface Moving boundary

Piston

(a) Hot H2O Water heater (Control volume) Cold H2O

Compressor

Imaginary boundary

(Control volume)

Heat (b)

Figure 1.3

Control surface

Real boundary

(c)

Open system: (a) A CV with fixed and moving boundary, (b) an open system with fixed control surface, (c) a CV with real and imaginary boundary.

Consider Figure 1.3(b) in which cold water enters the control volume, hot water leaves the control volume and at the same time heat transfer also takes place. Hence, such a system is known as open system.

1.3.3 Isolated System An isolated system is one in which neither mass nor energy crosses its boundary. This is purely a theoretical one. Very few real life applications qualify as an isolated system. But this concept is useful in the study and analysis of thermodynamic principles and laws. This is shown in Figure 1.1(c).

Fundamental Concepts and Definitions

1.4

5

THERMODYNAMIC PROPERTY

A property is any measurable characteristics of a system. Examples of properties are mass, pressure, temperature and volume. Some other properties are viscosity, modulus of elasticity, coefficient of thermal expansion, coefficient of friction, etc. Some properties are defined in terms of other properties. For example, the density r is defined as mass per unit volume; here both mass and volume are properties. Property, either measured or calculated, should have a single P value for each state, i.e. same unique value for a given state. This is possible only when a quantity (P, T, V, etc.) depends on A B the state of system and is independent of the path by which the C system has reached the given state, i.e. volume and pressure at 1 is same irrespective of the path the system follows in coming 1 to the point 1 as shown in Figure 1.4. The change in value of a property can be calculated by end states of the system. We know that both P and V are properties of a system. In V order to calculate the change in pressure and volume of a system, Figure 1.4 Concept of property. we should know the values of P and V at states 1 and 2, but we are not interested how the system has reached from state 1 to state 2, i.e. change in pressure and volume does not depend on the path. Thermodynamic properties can be divided into two classes, namely 1. Intensive and 2. Extensive

1.4.1 Intensive Property Intensive properties are those properties that are independent of the mass (size) of the system. Examples are pressure, temprature, density, velocity, etc. This is shown in Figure 1.5. As shown in Figure 1.5(a), divide the given system into two equal parts with partition wall. Measure the pressure and temperature of the system, and it is same, for both, before and after the partition. Similarly as in Figure 1.5(b), density is also same, for both, before and after the partition. Mass-based intensive properties are represented by lower case letters, for example, specific volume (n), kinetic energy (k.e.), potential energy (p.e.), density (r), etc. Mole-based intensive properties are represented by lower case letters with a bar on head, for example, specific volume ( ), kinetic energy (k.e.) , potential energy ( p.e.) , etc. (a)

(b)

Pressure = P Temperature = T

Volume = 2 m Mass = 5 kg

=

3

Pressure = P1 Temperature = T1

3

= 3

\ r = 2.5 kg/m

P1 = P 2 = P T1 = T2 = T

3

Volume = 1 m Mass = 2.5 kg

Volume = 1 m Mass = 2.5 kg 3

\ r = 2.5 kg/m

Figure 1.5

Pressure = P2 Temperature = T2

r1 = r2 = r 3

\ r = 2.5 kg/m

Intensive property.

6

Basic Thermodynamics

1.4.2 Extensive Property Extensive properties are those properties whose values depend on mass of the system. Examples are mass, volume and total energy, etc. This is shown in Figure 1.6. Volume = V

=

Volume = V1

Volume = V2

Mass = m

=

Mass = m1

Mass = m2

Figure 1.6

1.5

V1 π V V2 π V m1 π m m2 π m

Extensive property.

THERMODYNAMIC STATE

A state of a system is the condition of the existence of the system and is specified by the value of its properties. Consider a gas in a piston and cylinder arrangement as shown in Figure 1.7. This is represented by state 1, and it is having a fixed value of pressure (P1), volume (V1), temperature (T1), etc. Gas can be cooled or heated either at constant pressure or constant temperature or constant volume, etc. Then the gas is changed to another condition. The new condition is represented by state 2. At state 2, it is having another fixed value of pressure (P2), volume (V2), temperature (T2), etc. P1 T

P

P1 Piston

P1 = P2

Piston

1

2

2

Gas Gas

1

Heat

V1

V2 V

S

Figure 1.7 Thermodynamic state and process.

Therefore, the state may be represented, identified or described by certain observable microscopic properties like P, V, T, C, U, H, S, etc.

1.6

THERMODYNAMIC PROCESS OR CHANGE OF STATE, PATH

The transformation of a system from one equilibrium state to another is called a process. Consider a gas as a system in a piston and cylinder arrangement as shown in Figure 1.7. The initial condition of the system is represented by state 1 and its macroscopic properties are P1, V1, T1, etc. Gas is heated slowly from an external heat source, outside the system boundary. Piston is loaded in such way that constant pressure is acting on the system during the heating process. Due to the heat addition to the system, its temperature and volume increase while the pressure remains constant. At the end of the process at state 2, its pressure, temperature and volume are P2, T2 and V2, respectively. The macroscopic property values can be plotted on process diagrams, which are graphical

Fundamental Concepts and Definitions

7

representation of the changes in properties that occur between the initial and final states. Here, P1 and P2 are same. Therefore, process has to take place as shown in Figure1.7. If the condition specified is constant temperature, then the path is represented as shown in Figure 1.8(a) (T–S diagram). Similarly, in order to reach state 2 from state 1, the path followed may be infinite. These paths are called as processes. This is shown in Figure 1.8(b). The series of states through which a system passes during a process is called the path of the process. Hence, thermodynamic process is also called as a path. Thermodynamic process can also be defined as the series of states through which a system passes from the initial state 1 to a final state 2.

Figure 1.8 Thermodynamic state and process.

1.6.1 Point and Path Function 1.6.1.1 Point function A system can be taken from state 1 to state 2 along many quasi-static paths (processes), such as A or B as shown in Figure 2.4. Value of pressure and volume at states 1 and 2 are fixed and hence, these properties are called point functions. Hence, thermodynamic properties are point functions, i.e. for a given point on a diagram, there is a definite value for each property. The differentials of point functions are exact or perfect differentials, and the integration is simply,

Ú

V2

V1

dV = V2 - V1 ;

Ú

P2

P1

dP = P2 - P1

(1.1)

i.e. the change in volume or pressure thus depends only on the end states of the system irrespective of the path the system follows. 1.6.1.2 Path function Work, on the other hand, is a path function. A system can be transferred from state 1 to state 2 along many quasi-static paths (processes), such as A or B (Figure 2.4). The area under the curve is called as work. Quantity of work involved in each path is not same, and is not depending on the end points. Hence, work is a path function, i.e. it depends on the path the system follows in changing from state 1 to state 2. The differentials of path functions are inexact differentials and represented by d. Therefore, for work, we write dW.

8

Basic Thermodynamics

On integration

Ú

2

1

dW = 1W2 or W1- 2 ;

Ú

2

1

dW = W2 - W1

(1.2)

The cyclic integral of any point function is zero i.e.,

Ú dV = 0, Ú dP = 0, Ú dT = 0, Ú dW = 0, etc. 1.6.2

(1.3)

Equilibrium

A system in equilibrium experiences no changes in microscopic properties of a state when it is isolated from its surroundings, i.e. when a system is in equilibrium state, there are no unbalanced potentials within the systems which may lead to change of state. Consider a gas as a system as shown in Figure 1.9. The initial condition of the gas is represented by state 1. The moment the weight is removed from the piston, the equilibrium condition is disturbed, and as a result the piston is moved upwards until equilibrium condition is attained. The system at this condition is represented by state 2 [Figure 1.9(b)]. P Piston

Cylinder

1 Piston

Piston

1a 1b

System boundary

2 Gas

Gas

Gas 3

(b) State 2

(a) Initial position State 1

Figure 1.9

(c) State 3

V

Equilibrium and quasi-static process.

1.6.3 Quasi-static Process We know that the properties describe the state of a system only when it is in equilibrium. In the above example (Figure 1.9) states 1, 2 and 3 are in equilibrium. Hence, properties can be described at these states. But what about the properties between the states 1 and 2 and 2 and 3? Between 1 and 2 and 2 and 3, states are not in equilibrium. Hence, properties cannot be described. The answer for this depends on the definition of the quasi-equilibrium process. When a process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times, it is called a quasi-static or quasi-equilibrium process, i.e. the process goes on so slowly that the state of equilibrium exists at every moment. Apply this definition to the above example (Figure 1.9). The system changes from state 1 to state 2 so slowly that the state of equilibrium exists at every moment, i.e. equilibrium exists between states 1 and 1a, 1a and 1b, 1b and 2, etc.

Fundamental Concepts and Definitions

9

The thermodynamic deviation between states 1 and 2 is infinitesimal and the process 1–2 is known as quasi-state process. Hence, all the states (1, 2, 3, etc.) the system passes through during a quasi-static process can be considered as equilibrium state. Quasi-equilibrium process is also known as internally reversible process.

1.7 THERMODYNAMIC CYCLE OR CYCLIC PROCESS When a system in a given state goes through a number of different processes and finally returns to its initial state, the system is said to have undergone a cycle or cyclic process, i.e. for a thermodynamic cycle, the initial and final states remain same [Figure 1.8(b)], for example, steam power plant (Rankine cycle), closed cycle gas turbine, vapour compression refrigeration cycle, etc. Internal combustion engines go through a mechanical cycle once in every two revolutions, but the working fluid does not go through a thermodynamic cycle, since the products of combustion are exhausted to the atmosphere. 1.8 REVERSIBLE PROCESS Reversible process can be divided into two classes: 1. Internally reversible 2. Totally reversible 1.8.1 Internally Reversible Process An internally reversible process is one in which the system goes through a series of equilibrium states, and if the direction is reversed, the system returns to its initial equilibrium state without leaving any permanent change in the system. Therefore, quasi-equilibrium process is called as internally reversible process. 1.8.2 Totally Reversible Process A system changes from state A to B through a series of equilibrium states 1, 2, 3, etc. During the change of state from A to B, energy transfer may be there between the system and the surroundings. At any point in the process if the direction is reversed, the system returns to its initial equilibrium state and all the energies transferred during the process can be completely restored in both the system and the surroundings. This is called as totally reversible process. Thus, a totally reversible process does not leave any net permanent change in the system or in the surroundings. 1.9 THERMODYNAMIC EQUILIBRIUM There are three types of equilibrium: 1. Thermal equilibrium means equality of temperature 2. Mechanical equilibrium means equality of pressure 3. Chemical equilibrium means equality of chemical potentials 1.9.1 Thermal Equilibrium If the temperature is same throughout the entire system and the system is isolated from its surroundings, the system is said to be in thermal equilibrium. This is shown in Figure 1.10(a).

10

Basic Thermodynamics

1.9.2 Mechanical Equilibrium If there is no change in pressure at any point of the system with time and the system is isolated from its surroundings, the system is said to be in mechanical equilibrium. With elevation, the pressure may vary within the system due to gravitational effects. A bottom layer is having the higher pressure and it is balanced by the extra weight acting on it. Pressure variation due to gravity is very small and is neglected in many cases. This is shown in Figure 1.10(b). With time gap 60°C

70°C

90°C

80°C

50°C

10°C

(i) No thermal equilibrium

st

5 bar

5.5 bar

1 layer

6 bar

6.5 bar

2

nd

layer

rd

7 bar

7.5 bar

3 layer

8 bar

8.5 bar

4 layer

9 bar

9.5 bar

5 layer

th

th

(i) No mechanical equilibrium With time gap

40°C

40°C

40°C

40°C

40°C

40°C

40°C

40°C

40°C

5 bar

5 bar

6 bar

6 bar

7 bar

7 bar

8 bar

8 bar

9 bar

9 bar

(ii) Thermal equilibrium

(ii) Mechanical equilibrium

(a)

(b)

CO

CO2 H2O

CO2 H2O

CO2 H 2O

N2

N2

N2

CO2

O2

N2 H2O

With time gap

With time gap (i) No chemical equilibrium

(ii) Chemical equilibrium (c)

Figure 1.10 (a) Thermal equilibrium, (b) mechanical equilibrium and (c) chemical equilibrium.

Fundamental Concepts and Definitions

11

1.9.3 Chemical Equilibrium If there is no change in chemical composition of the system with time, i.e. no chemical reaction or diffusion occurs, the system is said to be in chemical equilibrium. This is shown in Figure 1.10(c). If the system is isolated from its surroundings, the properties of the system do not change with respect to time and the conditions of all types of equilibrium are satisfied, then that system is said to be in thermodynamic equilibrium.

1.10 DIATHERMIC WALL If two systems are separated by a diathermic wall (Figure 1.11), the temperature will change spontaneously until an equilibrium state of the combined system is attained. Two systems are then said to be in thermal equilibrium with each other.

System 1

System 2

System 1

System 2

60°C

70°C

65°C

65°C

Diathermic wall (a)

(b)

Figure 1.11 (a) Before thermal equilibrium and (b) after thermal equilibrium.

1.11 TEMPERATURE CONCEPT We know that the temperature is the property with which we all are familiar. Still it is very difficult to define. We are aware of temperature as measure of ‘hotness’ or ‘coldness’ of a substance. The level of temperature can be expressed qualitatively like freezing, cold, warm, hot, etc. Numerical values cannot be assigned to temperature based on our sensation effect. We know that when hot and cold bodies are brought into contact, cold body becomes warmer and hot body becomes cooler. These bodies attain the same temperature, if they remain in contact for some time.

1.11.1

Equality of Temperature

Let the two blocks of copper, one hot and the other cold, be brought into thermal communication with each other. Each block is in contact with mercury in glass thermometer. After some time, height of the mercury in both the thermometers is same. Therefore, it is concluded that two bodies have equality of temperature, if no change in any observable property occurs when they are in thermal contact.

12

Basic Thermodynamics

1.11.2

Zeroth Law of Thermodynamics

The Zeroth law of thermodynamics states ‘if two bodies are in thermal equilibrium with the third body, they are also in thermal equilibrium with each other’. Consider the third body as mercury in thermometer. Let the first block of copper be brought into contact with the thermometer. Remove the thermometer after attaining equality of temperature and note down the height of mercury in the thermometer. Now the same thermometer is brought into contact with the second block of copper. If there is no change in the mercury level, then we may conclude that both the blocks are in thermal equilibrium with the given thermometer.

1.12 THE THERMOMETER AND THE THERMOMETRIC PROPERTY Zeroth law made us to compare the temperature of the two bodies ‘A and B’ by bringing a particular body into contact with each of them in turn, say the particular body is a thermometer. Without the communication between ‘A and B’, this law says that the temperature of A is same as that of ‘B’. Let the body ‘A’ be a standard body. (If certain states of a body are easily reproducible, then that body is called as standard body. Examples for easily reproducible states are ice point, steam point, etc.) Thermometer is brought into thermal equilibrium with body ‘A’. Mercury in the thermometer shows a particular height and that is represented by ‘X’. Now thermometer is brought into thermal equilibrium with body ‘B’. If the height of the mercury column is also ‘X’, then we can say that the body ‘B’ has attained equality of temperature with the thermometer, and hence with the body ‘A’, i.e. the height of mercury column is same in both the cases. Therefore, we can conclude that the body ‘A’ has a temperature of body ‘B’. Now we can measure the temperature. To measure the temperature with the help of a thermometer, the instrument should have an easily observable and measurable characteristic called thermometric property. The height of a mercury column in the above example is the thermometric property, i.e. X (similarly, the resistance of a wire, the pressure of a gas in a closed vessel and the emf generated in a junction of two dissimilar metal wires are commonly used thermometric properties). Now we can assign the value to the height ‘X’, because this height ‘X’ is corresponding to the easily reproducible state of the standard body. Therefore, the temperature scale is established by assigning numerical value to certain easily reproducible states. It is customary to use the following two fixed points.

1.13

TEMPERATURE SCALE

Two temperature scales are in use. I. Celsius scale [(formerly called the Centigrade scale, 1948 onwards, it was renamed after the Swedish astronomer Anders Celsius (1701–1744), who derived it)]. II. Fahrenheit scale [named after the German instrument maker G. Fahrenheit (1686–1736)]. On Celsius scale, the temperatures of ice and steam points are fixed and are assigned a value of 0° and 100°C, respectively. There are 100 degrees between the ice point and the steam point. On Fahrenheit scale, the temperatures of ice and steam points are 32°F and 212°F, respectively. There are 180 degrees between the ice point and the steam point.

Fundamental Concepts and Definitions

13

Ice point—The equilibrium temperature of ice with air-saturated water at standard atmospheric pressure and is assigned a value of 0°C. Steam point—The equilibrium temperature of pure water with its own vapour at standard atmospheric pressure and is assigned a value of 100°C. The two absolute scales, Kelvin and Rankine, are most commonly used in engineering applications. Kelvin scale—A value of 273.15 K, 373.15 K and 273.16 K are assigned to ice point, steam point and triple point, respectively. Rankine scale—A value of 491.67 R, 671.67 R and 491.69 R are assigned to ice point, steam point and triple point of water, respectively.

1.13.1

Standard Scale of Temperature

Let initially the bulb (B) be kept at ice point and note down pressure of gas. Say it is 1000 mm of Hg (Pi). The bulb is kept in steam point and pressure is represented by Ps. The ratio (Ps/Pi) be determined and it is represented by (Ps/Pi)1000. Then remove some gas and repeat the above procedure. Let Pi be 900 mm of Hg, 800, ..., 600, etc., then this is represented by (Ps/Pi)900, (Ps/Pi)800, ..., (Ps/Pi)600, etc.

B—Gas bulb (with ideal gas), U—‘U’ tube and

Figure 1.12

A—Tube connecting gas bulb to U tube, F—Flexible hose pipe.

Standard temperature scale.

Then plot the ratio of (Ps/Pi) versus Pi as shown in Figure 1.12. It is important to note that a straight line is obtained. Now repeat the whole procedure for different gases. For all the gases, a straight line is obtained. When these lines are extrapolated, meet at a fixed point which is (Ps/Pi) = 1.366006 Ts P (1.4) i.e. = lim s = 1.366086 Æ 0 P Ti Pi 1 also

Ts – Ti = 100

(1.5)

These equations define Kelvin or absolute Celsius scale and are independent of the property of individual substance.

14

Basic Thermodynamics

Solve Eqs. (1.4) and (1.5) Ts = 373.16 K, Ti = 273.16 K \ \

1.14

T K = t°C + 273.16 = t°C + 273

(1.6)

T P = lim Ti Pi Æ 0 Pi

(1.7)

For other temperatures (T),

TEMPERATURE MEASUREMENTS

There are two methods for temperature measurement.

1.14.1

Method in Use before 1954

This is based on two standard fixed-point methods. This is an old method, which is based on two easily reproducible fixed points of a standard system, i.e. ice and steam points. Let

q (X) = The temperature of the system to be measured. q (X1), q (X2) = The temperature of a standard system with easily reproducible states, i.e. steam point and ice point, respectively. X, X1, X2 = Thermometric property of the system to be measured, steam point and ice point, respectively.

It is assumed that the variation of a temperature is a linear function of X, i.e. thermometric property. First, the thermometer is placed in contact with a system whose temperature q (X) is to be measured. The height recorded by the thermometer is X. Then, the thermometer is placed in contact with an arbitrarily chosen standard system with easily reproducible state, i.e. steam point. The temperature corresponding to this point is 100°C [i.e. q (X1) = 100°C, assigned value]. The height recorded by the thermometer is X1. Thus, ( X1 ) X1 = (X) X

(1.8)

( X2 ) X2 = (X) X

(1.9)

Similarly with the ice point,

From Eqs. (1.8) and (1.9), we have [ ( X1 ) - ( X 2 )] X1 - X 2 = (X) X

Therefore,

(X) =

[ ( X1 ) - ( X 2 )] X ( X1 - X 2 )

(1.10)

It is very difficult to identify the condition of equilibrium between pure ice and air-saturated water in obtaining ice point accurately. Hence, this method was abandoned in 1954.

Fundamental Concepts and Definitions

1.14.2

15

Method in Use after 1954

This is based on single standard fixed-point method. This is based on single easily reproducible fixed point of a standard system, i.e. triple point of water, the state at which ice, liquid water and water vapour coexist in equilibrium. A value of 0.01°C (273.16 K) is arbitrarily assigned to this state (i.e. triple point of water). It is assumed that the variation of temperature is directly proportional to thermometric property, i.e. X. Let

qtp = The temperature of a triple point of water q = The temperature of a body to be measured X, Xtp = Thermometric property of a system whose temperature to be determined and triple point, respectively. tp

= aXtp ;

= aX ;

=

a=

=

tp

Xtp

tp

Xtp

¥X

273.16 ¥X Xtp

(1.11)

The triple point of water is an easily reproducible state and can be realized and identifiable; hence now it is in use. The function ‘f’ can be expressed as follows: Directly proportional: Linear variation: Quadratic relation: Polynomial relation:

q (X) = aX q (X) = a + bX q (X) = aX2 + bX + g q (X) = A¢ + B¢X + C¢X2 + D¢X3

1.15 INTERNATIONAL PRACTICAL TEMPERATURE SCALE In a general conference on weights and measures, in 1968, the international committee has adopted the international practical temperature scale. This is given in Table 1.1. Table 1.1

Equilibrium states at 1 atm pressure NBP of oxygen MP of ice

International practical temperature scales

Assigned value, temperature °C –182.97

K 90.19

0.0

273.16

NBP of water

100.0

373.16

NBP of sulphur

444.60

717.76

MP of silver

960.80

1233.96

MP of gold

1063.00

1336.16

16

Basic Thermodynamics

1.16 COMPARISON OF THERMOMETERS The information of thermometers, thermometric property and temperature in use after 1954 to all the thermometers are listed in Table 1.2. Table 1.2 Comparison of thermometers with thermometric property

Sl. No.

Thermometer

Thermometric property

Temperature

1

Constant volume gas thermometer

Pressure—P

( P ) = 273.6

P Ptp

2

Constant pressure gas thermometer

Volume—V

( P ) = 273.6

V Vtp

3

Electric resistance thermometer

Resistance—R

( P) = 273.6

R Rtp

4

Thermocouple

Thermal emf—e

( P) = 273.6

e etp

5

Mercury in glass thermometer

Length—L(X)

( P) = 273.6

X Xtp

1.17 COMPARISON OF TEMPERATURE SCALE Let the temperature q vary linearly with the thermometric property, height (X), of the mercury column. Case 1: Celsius scale q(X) = a + bX

(1.12)

where a and b are constants. Apply the above equation for steam and ice points. Therefore,

100 = a + bXs

(1.13)

0 = a + bXi

(1.14)

Solving Eqs. (1.13) and (1.14), a = – bXi b=

100 ( X s - Xi )

where Xs = height of mercury column corresponding to steam point Xi = height of mercury column corresponding to ice point

(1.15) (1.16)

Fundamental Concepts and Definitions

17

Figure 1.13 Comparison of temperature scale.

Substituting Eqs. (1.15) and (1.16) in Eq. (1.12), (X) =

( X - Xi ) ¥ 100 ( X s - Xi )

Similarly for Fahrenheit, Rankine and Kelvin scales, È ( X - Xi ) ˘ ( X ) ∞F = Í ¥ 180 ˙ + 32 Î ( X s - Xi ) ˚ È ( X - Xi ) ˘ (X) R = Í ¥ 180 ˙ + 491.67 Î ( X s - Xi ) ˚ È ( X - Xi ) ˘ (X) K = Í ¥ 100 ˙ + 273.15 Î ( X s - Xi ) ˚

Comparison of temperature scale is shown in Figure 1.13.

SOLVED EXAMPLES EXAMPLE 1.1 The limiting value of the ratio of the pressure of gas at the steam point and at the triple point of water when the gas is kept at constant volume is found to be 1.36605. What is the ideal gas temperature of the steam? Solution:

We have the following relation for constant volume bulb thermometer: T = 273.16 lim

Pt Æ 0

= 373.15 K

Ps = 273.16 ¥ 1.36605 Pt

È ˘ Ps = 1.36605 given ˙ Í∵ Plim t Æ 0 Pt Î ˚

18

Basic Thermodynamics

EXAMPLE 1.2 The resistance of a platinum wire is found to be 11.00 (W) at the ice point, 15.247 (W) at the steam point and 28.887 (W) at the sulphur point. Find (a) the constants A and B in the equation R = R0 [1 + At + Bt2] and (b) plot R against t in the range of 0° to 500°C, where R0 is resistance at zero temperature. Solution: (a) The constants A and B Apply the above equation to ice point, 11.00 = R0 [1 + 0 + 0] = R0 Apply to steam point, 15.247 = R0[1 + A × 100 + B × 1002] 15.247 = 11[1 + 100A + 104B]

(1)

Apply to sulphur point; (sulphur point = 444.6) 28.887 = R0[1 + 444.6A + (444.6)2B] Solving Eqs. (1) and (2), 1 A = 3.92 ¥ 10 -3 , ∞C 1 B = - 5.905 ¥ 10 -7 2 ∞C Substitute A and B in Eq. (1),

(2)

R = 11.0[1 + 3.92 × 10–3t – 5.905 × 10–7 t2]

(3)

(b) Plot R against t in the range 0° to 500°C in step of 100°C Substitute t = 0, 100, 200, 300, 400 and 500 in Eq. (3) and calculate R. t °C

0

100

200

300

400

500

R(W)

11.00

15.247

19.364

23.351

27.208

30.879

RW 40 35 30 25 20 15 10 5 0 0

100

200

300

400

Figure E1.2

500

600

°C

Fundamental Concepts and Definitions

19

EXAMPLE 1.3 One junction of a thermocouple is kept at the ice point and the other junction is kept at the Celcius temperature t. The emf ‘e’ of the thermocouple is given by the equation,

e = At + Bt2, where A = 0.25 Calculate:

mV , B = - 5.0 ¥ 10 -4 °C

(a) The emf when t = –100, 0, + 100, 200 and 400 and draw a graph of e against t in this range. (b) If emf e is considered as a thermometric property, and temperature scale, t*, be defined by the relation t* = ae + b, such that t* = 0 (ice point), t* = 100 (steam point). Find the constants a and b and draw a graph of e against t*. (c) Find the value of t* at t = –100, 0, + 100, 200, 300 and 400°C and draw a graph of t* against t. Solution: (a) Given relation e = At + Bt2 t in °C, e in mV,

A = 0.25

mV , °C

(1) B = –5.0 × 10–4

mV (°C)2

Substitute A and B in Eq. (1). e = [0.25t – 5.0 × 10–4 t2] mV t °C

–100

0

100

200

300

400

e mV

–30

0

20

30

30

20

Figure E1.3(a)

(b)

t* = ae + b At ice point, t* = 0, e = 0 (e from the above table) Substitute the above boundary conditions in Eq. (2), 0 = 0 + b \ b=0

(2)

20

Basic Thermodynamics

At steam point, t* = 100°C, e = 20 (e from the above table) Substitute the above boundary conditions in Eq. (2), 100 = a × 20 + b a=5

\

100 = 20a + 0

Substitute a and b in Eq. (2), t* = 5e + 0 t* °C

–100

0

100

200

300

400

e mV

–20

0

20

40

60

80

100 e mV 80 60 40 20

t°C –200

0

–100

100

200

300

400

500

–20 –40

Figure E1.3(b)

(c) We have t* = 5e t °C e mV t* °C

–100 –30 –150

0 0 0

100 20 100

Figure E1.3(c)

200 30 150

300 30 150

400 20 100

Fundamental Concepts and Definitions

21

EXAMPLE 1.4 For the same number of moles of gas at the same temperature, we have, lim ( Pv) = universal constant for all gases,

pÆ0

Construct a temperature scale (a) by using a triple point of water and (b) by using both steam point and ice point as fixed point. Solution:

We know that, lim ( Pv) = f ( ) only

pÆ0

Define the temperature scale such that f(q) = R(q), where R = constant. (a) By using triple point of water lim Pv R( ) = = lim ( Pv)t R( t ) \

= t

273.16

lim Pv lim ( Pv)t

= 273.16

(b) By using both steam point and ice point as fixed point lim (Pv) = Rq, lim (Pv)ice = Rqice, lim (Pv)steam = Rqsteam lim (Pv)steam – lim (Pv)ice = Rqsteam – Rqice = R[qsteam – qice] \

R ( steam - ice ) R = lim ( Pv) [lim ( Pv)steam - lim ( Pv)ice

=

But Let

(

steam

-

ice )

È lim ( Pv)steam ˘ - 1˙ Í Î lim ( Pv) ice ˚ qsteam – qice = 100 lim ( Pv)steam = Rs lim ( Pv) ice

\

=

=

lim ( Pv) lim ( Pv)ice

100 lim ( Pv) ( Rs - 1) lim ( Pv) ice

where Rs can be determined experimentally. EXAMPLE 1.5 A Celsius scale and a Fahrenheit scale are immersed in a fluid. Calculate the temperature of a fluid when, (a) the numerical reading is identical in both the thermometers and (b) the Fahrenheit reading is numerically twice that of the Celcius reading. Calculate the values in R and K. Solution: We have the relation between T °F and T °C as, T ∞C =

[T ∞F - 32] 1.8

(1)

22

Basic Thermodynamics

where T °C = temperature in the Celsius scale T °F = temperature in the Fahrenheit scale (a) Temperature, when numerical reading is identical in both thermometers T °C = T °F = T Substitute the above condition in Eq. (1), T=

\

(T - 32) or T = - 40 1.8

T = – 40°C or –40°F

(2)

To express in terms of R and K, we have the following relations, T(R) = T °F + 460, Substitute Eq. 2 in Eq. 3,

T(K) = T °C + 273,

T(R) = 1.8 T(K)

(3)

T(R) = – 40°F + 460 = 420 T(K) = – 40°C + 273 =233

(b) Fahrenheit reading is numerically twice that of the Celsius reading Equation 1 can be written as, T °C =

[2T °C - 32] 1.8

\ T °C = 160°C

Given that T °F = 2T °C = 2 × 160 = 320°F (c) To express in R and K T(R) = T °F + 460 = 320 + 460 = 780 T(K) = T °C + 273 = 160 + 273 = 433 EXAMPLE 1.6 Define a new temperature scale, say °M in which the boiling and freezing points of water are 500°M and 100°M, respectively. (a) Corelate this scale (°M) with Centigrade scale. (b) If °M reading on the scale is the same number on the corresponding absolute temperature scale, what is the absolute temperature on °M? (c) What is this absolute temperature at 0°M? Solution: (a) Correlate this scale °M with Centigrade scale Let us denote Ls and Li for steam and ice point on the new scale, °M, and assume a linear relation as follows: Define a new scale as, T °M = aL + b For steam point,

500 = aLs + b

For ice point,

100 = aLi + b

(1)

Fundamental Concepts and Definitions

a=

\

Substitute a and b in Eq. (1),

400 , Ls - Li

23

b = 100 - aLi

È 400 ˘ T ∞M = Í ˙ L + [100 - aLi ] Î Ls - Li ˚ È 400 ˘ È 400 Li ˘ =Í ˙ L + Í100 ˙ ( Ls - Li ) ˚ Î Ls - Li ˚ Î ( L - Li ) T °M = 400 + 100 ( Ls - Li ) Similarly on the corresponding Celsius scale

\

T °C =

( L - Li ) ¥ 100 ( Ls - Li )

(2)

(3)

Compare Eqs. (2) and (3), 400 T ∞C + 100 = 4T ∞C + 100 100 (b) Corresponding absolute temperature scale T °M =

Ts P = lim s = 1.366086 Ti Pt Æ 0 Pi

We have,

Ts – Ti = 500 – 100 = 400

(4) (5)

Solving Eqs. (4) and (5), Ti = 1092.64°Mabs Ts = 1492.64°Mabs \

T °Mabs = t °M + 992.64

(6)

(c) Absolute temperature at 0°M Substitute t = 0°M in Eq. (6) T °Mabs at 0°M = 0°M + 992.64 = 992.640 Mabs EXAMPLE 1.7 Define a new temperature scale, say °F in which the steam point and ice point are 212°F and 32°F, respectively. If °F reading on this scale is same number on the corresponding absolute temperature scale (also called Rankine scale), what is this absolute temperature at 0°F and 100°F? Solution:

We have Ts P = lim s = 1.366086 Ti Pt Æ 0 Pi

Ts – Ti = 212 – 32 = 180°F

24

\

Basic Thermodynamics

1.36608Ti – Ti = 180 Ti = 491.7°Fabs or 491.7 R Ts = 491.7 + 180 = 671.1°Fabs or 671.69 R T(R) = (t °F – 32) + 491.7 = t °F + 459.7

(1)

Equation (1) is the absolute temperature scale corresponding to °F scale, where 212°F and 32°F for are steam and ice points respectively. Temperature at 0°F and 100°F Substitute t °F = 0° and 100 in Eq. (1). T(R) = 0 + 459.7 = 459.7 R T(R) = 100 + 459.7 = 559.7 R EXAMPLE 1.8 A constant volume gas thermometer is used to determine the temperature of boiling water, which at t1 = 100°C reads 20 cm Hg. (a) Determine, if this same thermometer is immersed in other fluid at a temperature of 120°C, the height of the mercury column in cm. (b) What is the temperature if the thermometer is immersed in another fluid and reads 50 cm Hg? (c) What is the temperature if the height of the mercury column is 0 cm? (d) What is the temperature °C of the fluid if the height of mercury column is –10 cm? (e) Derive the relation between height of the Hg column and temperature °C for the gas thermometer. Barameter records 76 cm Hg. Solution: The given thermometer is a constant volume gas thermometer. The gas used in the bulb of this thermometer must be ideal gas, then it satisfies,

ÈP˘ ÍT ˙ = C Î ˚v The atmospheric pressure = 76 cm Hg = h P1 P2 = T1 T2 (h0 + h1 ) (h0 + h2 ) = T1 T2

(1)

1 Æ Fluid 1 (Boiling water, t1 = 100°C, h1 = 20 cm Hg) 2 Æ Fluid 2 (a) h0 = 76 cm Hg,

h1 = 20 cm Hg, h2 = ?

Substitute these boundary conditions in Eq. (1), (76 + h2 ) (76 + 20) = (100 + 273) (120 + 273)

h2 = 25.15 cm Hg

t1 = 100°C t2 = 120°C

Fundamental Concepts and Definitions

(b) h0 = 76 cm Hg,

h1 = 20 cm Hg,

t1 = 100°C

h2 = 50 cm Hg

t2 = ?

25

Substitute the boundary conditions in Eq. (1), (76 + 20) (76 + 50) = (100 + 273) T2 T2 = 489.56 K = 216.56°C (c) h0 = 76 cm Hg,

h1 = 20 cm Hg,

t1 = 100°C

h2 = 0 cm Hg

t2 = ?

Substitute these boundary conditions in Eq. (1), (76 + 20) (76 + 0) = (100 + 273) T2 T2 = 295.3 K = 22.3°C (d) h0 = 76 cm Hg,

h1 = 20 cm Hg,

t1 = 100°C

h2 = –10 cm Hg

t2 = ?

Substitute these boundary conditions in Eq. (1), (76 + 20) [76 + ( -40)] = (100 + 273) T2 T2 = 256.44 K = –16.56°C (e) Substitute these boundary conditions in Eq. (1), [76 + h2 ] (76 + 20) = (100 + 273) T2

h2 = 0.2574t2 – 5.7372 EXAMPLE 1.9 The temperature t on a thermometric scale is defined in terms of a property P by the relation t = a ln P + b, where a and b are constants. Experiments give values of P as 1.86 and 6.81 at the ice point and steam point, respectively. Evaluate the temperature t on the Celsius scale corresponding to a reading of P = 2.5 on the thermometer. Solution:

The given relation is t = a ln P + b

At ice point,

T = 0°C, P = 1.86

At steam point,

T = 100°C, P = 6.81

(1)

Substitute the above conditions in Eq. (1), 0 = a ln 1.86 + b

(2)

100 = a ln 6.81 + b

(3)

Solving Eqs. (2) and (3), we get, a = 77.0525,

b = –47.8169

26

Basic Thermodynamics

Substitute a and b in equation 1, t = 77.0525 ln P – 47.8169

(4)

If P = 2.5, then t = 77.0525 ln 2.5 – 47.8169 = 22.785°C EXAMPLE 1.10 A constant volume gas thermometer containing H2 records a gas pressure p as 1200 and 1400 mm of Hg at ice and steam points respectively. Assume a linear relationship of t = ap + b. Derive a relation for gas thermometer Celsius temperature t in terms of p. If pressure recorded by the gas thermometer is 1050 mm Hg, what is the temperature in t°C? Solution: A relation for gas thermometer Celsius temperature t in terms of p The given linear relation is t = ap + b At ice point,

t = 0°C, p = 1200 mm Hg

At steam point,

t = 100°C, p = 1400 mm Hg

(1)

Substitute these boundary conditions in Eq. (1), 0 = a1200 + b

(2)

100 = a1400 + b

(3)

Solving Eqs. (2) and (3), we get, a = 0.5,

b = – 600

Substitute a and b in Eq. (1), t°C = 0.5p – 600

(4)

If pressure recorded by the gas thermometer is 1050 mm Hg, what is the temperature in t°C? t°C = 0.5 × 1050 – 600 = –75°C EXAMPLE 1.11 A thermocouple with test junction at t°C on a gas thermometer scale and reference junction at ice point is given by the relation e = (0.25t – 5.1 × 10–4 t2) mV The millivoltmeter is calibrated at ice and steam points. If the gas thermometer reads 50°C and 80°C, what will be the reading on this thermometer? Solution:

The given relation is e = (0.25t – 5.1 × 10–4 t2) mV \ ei = 0

At ice point,

ti = 0

At steam point,

ts = 100°C es = [0.25 × 100 – 5.1 × 10–4 × 1002] = 19.9 mV

For tb = 80°C,

eb = 0.25 × 80 – 5.1 × 10–4 × 802 = 16.736 mV

For ta = 50°C, ea = 0.25 × 50 – 5.1 × 10–4 × 502 = 11.225 mV Therefore, when the gas thermometer reads 50°C and 80°C, the thermocouple reading t is given by,

Fundamental Concepts and Definitions

27

ts t = es e

ts = steam point = 100°C , es = 19.9 mV ea = 11.225 mV, eb = 16.735 mV, then ta = ? ; tb = ?

If

ta =

ts 100 ¥ ea = ¥ 11.225 = 56.4071∞C 19.9 es

tb =

ts 100 ¥ eb = ¥ 16.735 = 84.1∞C 19.9 es

EXAMPLE 1.12 The length h of the mercury column of a mercury in glass thermometer records 50 mm (at ice point) and 200 mm Hg (at steam point). The temperature scale (t°C) and property (length of the mercury column h) are related by, t = ah2 + b where t = 0 at ice point, t = 100°C at steam point. Compare this scale with Celsius scale where t is linearly defined in terms of h with the same values of t at the ice and steam point. Solution:

We have the relation, t = ah2 + b

At ice point, At steam point,

0= 100 =

a52

\

+b

a202

(1)

b = –25a

+ b = 400a – 25a = 375a

a = 0.267,

b = – 6.675

Substitute a and b in Eq. (1), t = 0.267h2 – 6.675

(2)

Let the Celsius scale is given by the relation, t* = Ph + Q At ice point, At steam point,

0=P×5+Q 100 = P × 20 – 5P

\ \

(3)

Q = –5P Q = –33.33

Substitute P and Q in Eq. (3), t* = 6.667h – 33.33

(4)

Dividing Eq. (2) by Eq. (4), we get, (0.267 h2 - 6.675) t = t * (6.667 h - 33.33) EXAMPLE 1.13 The resistance of a platinum wire is found to be 11000 ohms at the ice point, 15.247 ohms at the steam point and 28.887 ohms at the sulphur point. Find the constant A and B in the equation, R = R0[1 + At + Bt2] and plot R against t in the range of 0 to 660°C. (KUD – IInd year II-91)

28

Basic Thermodynamics

Solution:

The constants A and B

At ice point,

R = R0[1 + At + Bt2]

(1)

t = 0, R = 11000

(2)

Substitute condition Eq. (2) in Eq. (1), 11000 = R0[1 + 0 + 0]

\

R0 = 11000 W

t = 100°C, R = 15.247 W, R0 = 11000 W

At steam point,

(3)

Substitute condition Eq. (3) in Eq. (1), At Sulphur point,

15.247 = 11000 [1 + A × 100 + B × 1002]

(4)

t = 444.6°C, R = 28.887 W, R0 = 11000 W

(5)

Substitute condition Eq. (5) in Eq. (1), 28.887 = 11000 [1 + A × 444.6 + B × 444.62]

(6)

Solving Eqs. (4) and (6), we get, A = – 0.0122, B = 2.3 × 10–5 Substitute R0, A and B in Eq. (1), R = 11000 × [1 – 0.0122t + 2.3 × 10–5 t2] t°C R

0

100

200

300

400

500

600

660

11000

110

–5720

–6490

–2200

7150

21560

32634.8

35000 R 30000 25000 20000 15000 10000 5000 0 100

200

300

400

500

600

t 700

–5000 –10000

Figure E1.13

EXAMPLE 1.14 A constant volume gas thermometer containing helium gives reading of gas pressure P of 1000 mm Hg and 1366 mm Hg at ice point and steam point, respectively. Assuming a linear relationship of the form t = a + bP, express the gas thermometer Celsius temperature t in terms of gas pressure P. What is the temperature recorded by the thermometer when it registers a pressure of 1074 mm Hg? (KUD I-94)

Fundamental Concepts and Definitions

Solution:

29

The gas thermometer Celsius temperature t in terms of gas pressure P t = a + bP

At ice point,

(1)

t = 0°C, P = 1000 mm Hg

(2)

0 = a + b × 1000

(3)

Substitute condition 2 in Eq. (1), At steam point,

t = 100°C, P = 1366 mm Hg

(4)

Substitute condition 4 in Eq. (1), 100 = a + b × 1366

(5)

Solving Eqs. (3) and (5), we get, a = –273.22, b = 0.273224 Substitute a and b in Eq. (1), t = –273.22 + 0.273224P

(6)

The temperature recorded by the thermometer when it registers a pressure of 1074 mm Hg. t = –273.22 + 0.273224 × 1074 = 20.22°C EXAMPLE 1.15 What type of temperature measuring instruments would you expect in the following applications? (a) (b) (c) (d)

A heat treatment bath or furnace (400 to 1000°C) A boiler furnace (1660°C) A boiler flue gas (370°C) A fuel calorimeter water system (20°C)

Solution: (a) 0–660°C Æ Platinum resistance thermometer 660–1063°C Æ Thermocouple (i) First wire, platinum (ii) Second wire, alloy of 90% platinum and 10% rhodium EXAMPLE 1.16 The recordings tA and tB of two Celsius thermometers A and B agree at the ice and steam points, but elsewhere are related by the equation tA = L + MtB + NtB2, where L, M and N are constants. When both thermometers are immersed in a steam of fluid, A registers 10°C while B registers 11°C. Determine the reading on B when A registers 40°C. (KUD I-96) Solution: Ice point Steam point Elsewhere

Thermometer A

Thermometer B

0°C 100°C 10°C

0°C 100°C 11°C

30

Basic Thermodynamics

Given that tA = L + MtB + NtB2

(1)

At ice point,

tA = tB = 0

(2)

\

0 = L + M× 0 + N× 0

At steam point,

L=0

(3)

tA = tB = 100°C

(4)

Substitute condition 4 in Eq. (1), (5) 100 = 0 + M × 100 + N × 1002 When these two thermometers are immersed in a fluid, thermometer A registers 10°C, and B registers 11°C, i.e. tA = 10°C, tB = 11°C Substitute the above conditions in Eq. (1), 10 = 0 + M × 11 + N × 112

(6)

M = 0.8978, N = 1.02 × 10–3, L = 0

(7)

Solving Eqs. (5) and (6), we get, Substitute condition 7 in Eq. (1), tA = 0 + 0.8978tB + 1.02 × 10–3 tB2

(8)

If thermometer A records 40°C, temperature recorded by B? Substitute tA = 40°C, in Eq. (8), \

40 = 0 + 0.8978tB – 1.02 × 10–3 tB2

\

tB = 42.5°C

EXAMPLE 1.17 There are two scales. One is Celsius scale and another one is excellent scale.

Steam point

100°C

1000°E

0°C

0°E

Celsius scale: ice point 0°C, steam point = 100°C Excellent scale: ice point 0°C, steam point = 1000°C. Find the relation between these two scales. (KUD-II-83, I-83)

Ice point

Solution: Let T be the linear function of property L (length) of Hg column. Then we have, t = AL + B where A and B are constants. First consider Celsius scale, ice point, \ Steam point, t = 100°C

(1) t = 0°C 0 = ALi + B

Figure E1.17

(2)

Fundamental Concepts and Definitions

\

100 = ALs + B

31

(3)

Solving Eqs. (2) and (3), we have 100 100 and B = ( Ls - Li ) ( Ls - Li )

(4)

100 Li 100 ( L - Li ) 100 L = ( Ls - Li ) ( Ls - Li ) ( Ls - Li )

(5)

A= Substitute A and B in Eq. (1), t=

Now consider the excellent scale, ice point, t = 0°E \ Steam point, t = 1000°E, \ Solving Eqs. (6) and (7), we have A=

0 = ALi + B

(6)

1000 = ALs + B

(7)

1000 Li 1000 ; B= ( Ls - Li ) ( Ls - Li )

(8)

Substitute Eq. (8) in Eq. (1), t=

1000 ( L - Li ) ( Ls - Li )

(9)

È 100 ( L - Li ) ˘ t ∞C Í ( Ls - Li ) ˙ 1 ˙= =Í t ∞E Í 1000 ( L - Li ) ˙ 10 Í (L - L ˙ s i Î ˚ t°E = 10t°C

\

EXAMPLE 1.18 Two temperature scales with the following relations are given. Calculate temperature in Rankine. t °C = T °K – 273.16, t °F = T °R – 459.7, t°F = 2t°C Solution: Given that t °F = 2t °C We know that T °R = 1.8T °K \ \

(KUD II-82, I-81)

T °R – 459.7 = 2 × [T °K – 273.16] = 2T °K – 546.32 1.8T °K – 459.7 = 2T °K – 546.32 T °K = 433.1 T °R = 1.8T °K = 1.8 × 433.1 = 779.58

EXAMPLE 1.19 The readings tA and tB of two Celsius thermometers A and B agree at the ice point (0°C) and the steam point (100°C), but elsewhere are related by the equation tA = l + mtB +

32

Basic Thermodynamics

nt2B; where l, m and n are constonats. When both the thermometers are immersed in a well-stirred oil bath, A registers 51°C, while B registers 50°C. Determine the reading on B when A reads 25°C. Which thermometer is correct? (New scheme July/Aug–2004) Solution: tA and tB agree at ice point, 0°C and at steam point, \ tA = tB = 0 and tA = tB = 100°C At 0°C, 0=l+0+0 \ l=0 At 100°C, 100 = l + m × 100 + n × 1002 When tA = 51°C, tB = 50°C ; 51 = 0 + m × 50 + n × 502 1.02 = m + 50n From Eqs. (1) and (2), we get, m = 1.04; n = − 0.0004 Now the equation becomes tA = 0 + 1.04tB + (– 0.004)tB2 The reading on B, when A reads 25°C 25 = 0 + 1.04tB + (– 0.004)tB2 \ tB = 24.96°C But A has higher accuracy, so A is correct.

(1) (2)

EXERCISES 1.1 Distinguish between 1. Closed system and open system 2. Intensive property and extensive property 3. Process and cyclic process

(KUD II-98)

1.2 Define thermodynamic system and point out the difference between closed system, open system and isolated system. (KUD March 2001) 1.3 Distinguish between intensive and extensive properties. State to which category the following belong. 1. 2. 3. 4. 5. 6. 7. 8.

Pressure Volume Enthalpy Mass density Temperature Entropy Elevation Specific volume, strain, PE, KE

1.4 Define the following terms: 1. Process 2. Cycle 3. Thermodynamic equilibrium

(KUD March 2001, II-93)

Fundamental Concepts and Definitions

4. Quasi-equilibrium process 5. Temperature

33

(KUD March 2001)

1.5 Distinguish between the following: 1. Microscopic and macroscopic points of view 2. Path function and point function 3. Flow process and non-flow process

(KUD March 2001)

1.6 Describe briefly the working of the following: 1. Thermocouple 2. Electric resistance thermometer 3. Constant volume gas thermometer

(KUD Aug 2001)

1.7 Explain the following terms: 1. 2. 3. 4. 5.

System, boundary and surroundings Isolated and adiabatic system Homogeneous and heterogeneous system Point and path function Zeroth law of thermodynamics

(KUD July 2002)

1.8 Explain the terms: 1. 2. 3. 4.

Thermal equilibrium Mechanical equilibrium Chemical equilibrium Thermodynamic equilibrium

1.9 Distinguish between Celsius and Fahrenheit temperature scale. 1.10 Explain equality of temperature.

(KUD II-97) (KUD I-96) (KUD I-97)

1.11 Distinguish between classical thermodynamics and statistical thermodynamics. (KUD I-97) 1.12 The resistance of a platinum wire is found to be 11000 ohms at the ice point, 15.247 ohms at the steam point and 28.887 ohms at the sulphur point. Find the constants A and B in the equation R = R0[1 + At + Bt2] and plot R against t in the range of 0 to 660°C. 1.13 Over a limited temperature range, the relation between electrical resistance R and temperature T for a resistance temperature detector is R = R0[1 + a(T – T0)] where R0 is the resistance in ohms (W) measured at reference temperature T0 (in °C) and a is a material constant with units of (°C – 1). The following data are obtained for a particular resistance thermometer. T °C R(W) Test 1 0 51.39 Test 2 91 51.72 what resistance reading would correspond to a temperature of 50°C on this thermometer?

34

Basic Thermodynamics

1.14 Two liquids in glass thermometers have Celsius and Fahrenheit scales, respectively. Both the thermometers are used to measure the temperature of an object. If each thermometer indicates the same numerical reading, what is the temperature in Kelvin? 1.15 A new absolute temperature scale is proposed. On this scale, the ice point of water is 150°S and the steam point is 300°S. Determine the temperatures in 0°C that corresponds to 100°S and 400°S, respectively. What is the ratio of the size of the °S to the Kelvin? 1.16 The relation between resistance R and temperature T for a thermistor closely follows, È Ê1 1 ˆ˘ R = R0 exp Í Á - ˜ ˙ ÎÍ Ë T T0 ¯ ˚˙ where R0 is the resistance in ohms (W) measured at temperature T0 (K) and b is a material constant with units of K. For a particular thermistor, R0 = 2.2 W at T0 = 310 K. From a calibration test it is found that R = 0.31 W at T = 422 K. Determine the value of b for the thermistor and make a plot of resistance versus temperature.

2 CHAPTER

Work and Heat 2.1 WORK Energy can interact between the surroundings and the system. There are two modes of energy transfer between the surroundings and the system. They are 1. Work energy 2. Heat energy The energy crossing the boundary of a closed system, if it is not heat, it must be work and if it is not work, it must be heat. We know that heat transfer between the system and the surroundings of a closed system is purely due to temperature difference. If energy interaction between the system and the surroundings is not due to temperature difference, we can simply say that the energy is work energy only.

2.1.1 Work—According to Mechanics In mechanics, the result of a force on a moving body is known as work. Work is defined as the product of the force (F) and the distance moved (ds) in the direction of the force or simply force acting through a distance. This is shown in Figure 2.1. Figure 2.1 35

Work done by a force acting through a distance (concept of mechanics work).

36

Basic Thermodynamics

Let

F = Force applied on a body at an angle q 1 = Initial position 2 = Final position ds = Body moves through a differential distance or small distance dW = Differential work q = Angle between force (F) and displacement (ds)

\

dW = Fs ds = F cos q ds

(2.1)

On integrating between initial to final position, i.e. over a finite path

Ú 2.1.2

Ú

dW = F cos ds

W1- 2 =

Ú

S2

S1

F cos ds = F cos (S2 - S 1 )

(2.2)

Limitation

According to the definition of work in mechanics, a force (load) and action of this force through a distance are necessary. Work cannot be calculated without the actual movement of the load.

2.1.3 Work—According to Thermodynamics According to Obert, work is defined as “the energy transferred across the boundary of a closed system because of an intensive property difference excluding temperature that exists between the system and surroundings.” The intensive property difference, i.e. pressure differences between the system and the surroundings at the boundary of the system, gives rise to a force (F = PA) and the effect of this over a distance is called mechanical work. If energy transfer is due to the electrical potential difference between the surroundings and the system, the energy is known as electrical work. With respect to thermodynamics, work should be tied with the concepts of systems, properties and processes. According to J. H. Keenan, “work is said to be done by a system if the sole effect of the system on things external to the system, i.e. surroundings, can be reduced to the raising of a weight.” According to this definition, the weight may not be actually raised but the net effect external to the system should be the raising of a weight. Therefore, the thermodynamic definition of work is more general than that used in mechanics. Consider a system, battery, as shown in Figure 2.2(a). When the switch is closed, resistor gets warmer, i.e. transfer of electrical energy takes place between the system and the surroundings because of potential difference. This clearly shows that the system has interaction with the surroundings. There is no displacement of the system in this example. According to the definition of work in mechanics, this interaction cannot be called work. Now, modify this example according to Figure 2.2(b). Replace the resistor by a motor, which draws the same amount of current as the resistor. A pulley and weight arrangement is coupled to the motor. When the switch is closed, the motor runs and the weights are raised. Thus, the sole effect external to the system has been reduced to the raising of weight. Hence, according to the definition of work in thermodynamics, this interaction can be called work, though actually no weight has been raised through any distance.

Work and Heat

37

Figure 2.2 Concept of thermodynamic work.

A force acting on the boundary and movement of the boundary are the minimum requirements for a work interaction between a system and the surroundings to exist. If the forces on either side of the system boundary are not equal in magnitude, then the movement of the boundary is said to be partially resisted. If the magnitude of the opposing force is same as the actual force in the direction of movement of the boundary, then the boundary movement is called fully resisted. If the boundary movement is unresisted on either system side or the surroundings side, the work done is zero. The following conditions must be satisfied for work interaction: • The process must be quasi-static. • The boundary of the system should move. • There must be resisting force at the boundary.

2.2

SIGN CONVENTIONS FOR WORK AND HEAT

The sign conventions for work and heat are shown in Figure 2.3. The figure is self-explanatory.

Figure 2.3 Sign conventions for work and heat.

2.3 WORK DONE IN A FRICTIONLESS QUASI-EQUILIBRIUM PROCESS OR PdV WORK OR DISPLACEMENT WORK Consider a system (gas) in thermal equilibrium in a frictionless piston and cylinder arrangement as shown in Figure 2.4. At initial state 1, gas is represented by P1, V1, T1, etc. Now a small weight is removed from the piston. There will be small drop in pressure on the gas. The gas undergoes a

38

Basic Thermodynamics

Figure 2.4 Work in quasi-static process: W1- 2 =

Ú

2

1

PdV .

slight expansion. The piston will be raised by small distance, ds; hence, the process is a quasistatic process. The work done by the system is given by where

dW = Fds = PAd P = Pressure on inside of the piston ds = Small distance movement A = Cross-sectional area of the piston d = Diameter of the piston F = PA Ads = dV = Volume displaced = Change in volume dW = PAds = PdV

On integrating between states 1 and 2,

Ú

2

1

2.4

W=

Ú

2

1

PdV ; W1- 2 =

Ú

2

1

PdV

PdV WORK IN VARIOUS QUASI-STATIC PROCESSES 1. Constant Volume Process (Isochoric Process) (Figure 2.5)

Figure 2.5 Constant volume process.

(2.3)

39

Work and Heat

W.D. = Area under the curve 1–2 W1–2 =

Ú

2

1

PdV = P (V2 - V1 ) = 0

2. Constant Temperature Process (Isothermal Process) (Figure 2.6)

[∵ P P1

V1 = V2 ]

(2.4)

1

W.D. = Area under the curve 1–2 PV = P1V1 = C ; P =

Ú

W1- 2 = PdV =

Ú

2

1

PV = C

P1V1 V2 P1 ; = V V1 P2

P1V1 dV V

W1–2

P2

2 V2

V1

V P = P1V1 ln 2 = P1V1 ln 1 V1 P2

(2.5)

V

Figure 2.6 Constant temperature process.

3. Constant Pressure Process (Isobaric or Isopiestic Process) (Figure 2.7) P P1 = P 2

1

2

Piston

P1

Piston

P Gas (System)

W1–2

1 PV = C 1

V1

2

V2

V

Figure 2.7 Constant pressure process.

W.D = Area under the curve 1–2 W1- 2 =

Ú

2

1

PdV = P(V2 - V1 )

4. Polytropic Process (Figure 2.8)

(2.6) P

W.D. = Area under the curve 1–2

P1

1 n

PV = C

1/ n

n

ÈP ˘ V ÈV ˘ PV n = P1V1n P = P1 Í 1 ˙ ; 2 = Í 1 ˙ V1 Î p2 ˚ ÎV ˚ =

Ú

2

1

PdV =

= P1V1n

Ú

2

1

Ú

2

1

n

ÈV ˘ P1 Í 1 ˙ dV ÎV ˚ V

È V - n +1 ˘ 2 = Í ˙ V n ÎÍ - n + 1 ˚˙V dV

1

W1–2 P2

2

V1

V2

V

Figure 2.8 Polytropic process.

40

Basic Thermodynamics

=

P1V1n [V21- n - V11- n ] ( P2V2nV21- n - P1V1nV11- n ) = (1 - n) (1 - n)

=

( P2V2 - P1V1 ) ( P1V1 - P2V2 ) = (1 - n) ( n - 1)

(2.7)

n -1 ˘ È P1V1 Í Ê P2 ˆ n ˙ 1(n - 1) Í ÁË P1 ˜¯ ˙ Í ˙ Î ˚

or

(2.8)

All the processes can be shown on a PV diagram (Figure 2.9). P

n = ±• (V = C)

T

n = –2.5 n = –1.5

n=g

n = –0.5

n = ±• (V = C) n = –1 n=0 (P = C)

n = 0 (P = C) n = 1 (T = C) n = 1 (T = C) n = g (S = C) > 1 V

(a)

Figure 2.9

2.5

(b)

S

All the processes on common PV and TS diagrams (for quasi-static process): (a) PV diagram and (b) TS diagram.

OTHER TYPES OF WORK TRANSFER

1. Electrical Work (Figure 2.10) When a current flows through a resistor, there is work transfer in the system. Here current can run a motor, the motor can drive a pulley and the pulley can raise a weight. System boundary 1

1

Figure 2.10 Electrical work.

I = Current =

where

C = Charge in Coulombs, t = Time in second, E = Voltage potential

dC d

Work and Heat

W = E dC = EI d

Ú

2

1

W = EI

Ú

2

1

d

41

(2.9)

W1–2 = EI 2. Shaft Work (Figure 2.11)

T Motor

Shaft

N

Figure 2.11 Shaft work.

When a shaft (system) is rotated by a motor, there is work transfer into the system. If T is the torque applied to the shaft and dq is the angular displacement of the shaft, then shaft work is given by,

W1- 2 =

Ú

2

1

Td = T (

2

-

1)

kN m =

Ú

2

1

T

d =T d

kW

(2.10)

where w = angular velocity 3. Paddle Wheel Work or Stirring Work (Figure 2.12) Stirrer (or paddle wheel) Pulley System

Weight

Figure 2.12 Paddle wheel work.

As the weight is lowered, the paddle wheel rotates, thus work is transferred into the system. Here volume is constant.

Ú PdV = 0

Hence,

If m is the mass of the weight lowered through a distance ds and T is the torque applied by the shaft to the system in rotating the shaft through an angle q W1- 2 =

Ú

2

1

mg ds =

Ú

2

1

Td

(2.11)

42

Basic Thermodynamics

4. Work in Straining a Bar If the length of a bar in tension or compression is changed from L to (L + dL), the work is given by, dW = F dL F = Total axial load on the bar = sA s = Stress A = Cross-sectional area dL L V = Volume = A L

d = Strain =

E = Young’s modulus \

\ dW = sA dL = s A L de = s V de

dL = de × L

On integrating, we get a finite change in strain, W1- 2 =

Ú

2

Vd

or

1

W1- 2 = V

Ú

2

d

1

Stress is given by s=E \

W1- 2 = V

Ú

2

E d

or

1

W1- 2 =( VE

2

-

1)

(2.12)

5. Work in Changing the Area of a Surface W1- 2 =

Ú

2

1

dA

(2.13)

HdI

(2.14)

where s = Surface tension, N/m 6. Magnetization of a Paramagnetic Solid W1- 2 =

Ú

2

1

where H = Field strength I = Component of the magnetization field in the direction of the field 7. Free Expansion Process—No Work Transfer We know that the work is a boundary phenomenon and is identified at the boundary of a system and surroundings. Consider a gas as shown in Figure 2.13, in which a membrane separates a gas from vacuum. Now the partition is removed and the gas rushes to fill the entire volume. This expansion is called as Free Expansion. Here gas and vacuum together act as a system. If we neglect work required to remove the membrane, then there is no work transfer involved here, since no work crosses the system boundary.

Work and Heat

Figure 2.13

43

Free expansion process.

In Figure 2.14, consider only the fluid as the system. Here system boundary moves and volume of the system does change from V1 to V2. However, it is not a quasi-equilibrium process, (PV diagram represented by dotted line), even though the initial and final states are in equilibrium. Therefore, work cannot be calculated by using the relation W1–2 = Ú PdV. P

Fluid

1

Vaccum 2 (a)

(b)

Figure 2.14

V

Free expansion process.

This can be made a quasi-static process by dividing the vacuum into the number of small compartments and then expand the fluid by removing each compartments one by one. Then every state passing through the system is an equilibrium state and the work done can be calculated by using the relation

Ú

2

1

PdV .

But in free expansion of fluid, there is no resistance to the fluid at the system boundary even though volume increases to fill the vacuum space. Therefore, there is no work transfer involved in free expansion. V1 = Volume of fluid V2 = Volume of fluid + volume of vacuum W1- 2 =

Ú

2

1

PdV = 0

(2.15)

2.6 HEAT Energy transfer across a boundary as a result of temperature difference between a system and its surroundings is called heat. Heat is transferred from the system at the higher temperature to the system at the lower temperature, and the heat transfer occurs solely because of the temperature

44

Basic Thermodynamics

difference between the two systems. Thus, heat is a transient phenomenon. Temperature difference is the ‘cause’ or the ‘potential’ and flow a heat is the flux.

2.7

COMPARISON OF HEAT AND WORK

2.7.1 Common between Heat and Work 1. Heat and work are both transient phenomena (exist during interactions only). It is wrong to use ‘Heat content of a body’ or ‘Work content of a body’. Heat or work crosses the boundary when a system undergoes a change of state. 2. Heat and work are both boundary phenomena, i.e. always referred with reference to the system boundary. 3. Heat and work are both path functions and inexact differentials.

2.7.2 Difference between Heat and Work 1. Consider an example as shown in Figure 2.15 to show the difference between heat and work transfer. System boundary

System (gas)

Battery

–

–

Battery

Electric resistance coil

System (gas) +

+

Electric resistance coil

System boundary (a) Heat transfer

(b) Work transfer

Figure 2.15

Difference between heat and work.

In Figure 2.15(a), the electric resistance coil is placed outside the system boundary. As soon as circuit is closed, resistance coil gets heated and its temperature increases. There is a temperature difference between the system (gas) and the surroundings (resistance coil) and heat transfer takes place. In Figure 2.15(b), the system includes both gas and the resistance coil. As soon as circuit is closed, electrical energy crosses the boundary of the system. We know that electrical energy is work energy. Hence work crosses the boundary of the system. For both the cases, the amount of energy transfer to the gas is the same. Depending on the selection of the system, one can say whether heat crosses the boundary or work crosses the boundary. Q1- 2

Ú dQ = Ú TdS

(2.16)

Work and Heat

45

2. The energy interaction due to temperature difference is called heat; on the other hand, the energy interaction due to other than the temperature difference is called work. 3. The percentage conversion of heat to work is less; on the other hand the percentage of conversion of work to heat is more; hence, work is a high grade energy, whereas heat is a low grade energy. Example: (a) 1000 kJ of heat % of conversion =

(b) 1000 kJ of work

300 kJ of work

H.E. 300 ¥ 100 = 30% 1000

Resistor

900 kJ of heat

(1000 kJ electrical work) 900 % of conversion = ¥ 100 = 90% 1000

SOLVED EXAMPLES EXAMPLE 2.1 If 70 kg of water is evaporated at atmospheric pressure until a volume of 10 m3 is occupied, how much work is done? Solution: V1 = 70 kg = 70 × 10–3 m3, V2 = 10 m3 70 ˘ kN È W.D. = PdV = P[V2 – V1] = 101.325 ¥ Í10 ¥ m 3 = 1006.16 kN m ˙ 2 1000 ˚ m Î

EXAMPLE 2.2 Steam is admitted to the cylinder of a steam engine at a constant pressure of 20 bar. The bore of the cylinder is 20 cm and the stroke of the piston is 30 cm. How much work is done per stroke? Solution:

Ú

W.D. = P dV = 20 ¥ 100 ¥

4

¥ (0.2)2 ¥ 0.3 = 18.85 kN m

È ˘ = F Ds = [ P ¥ A] ¥ Ds = Í 20 ¥ 100 ¥ ¥ (0.2)2 ˙ ¥ 0.3 = 18.85 kN m 4 Î ˚

EXAMPLE 2.3 A thin walled metal bomb of volume VB contains n (mole) of gas at high pressure. A capillary tube and a stopcock are connected to the bomb. (a) When the stopcock is opened slightly, the gas leaks slowly into a cylinder equipped with a non-leaking frictionless piston where the pressure remains constant at the atmospheric value P0. After as much gas as possible is leaked out, calculate work done. (b) How much work would be done if the gas is leaked directly into the atmosphere?

46

Basic Thermodynamics

Solution: (a) W.D. = P0 DV = P0[VB – n V0] (b) W.D. = Same as that of (a) EXAMPLE 2.4 In a closed system, air is compressed from 0.5 m3 to 0.1 m3 by a piston. The pressure and volume are related by P = 5 – 3V, where P is in bar and V in m3. Calculate the work done on the system. Solution: W1–2 = Work done =

Ú

V2

V1

PdV =

Ú

V2

V1

0.1

È 3V 2 ˘ 5 (5 - 3) dV ¥ 10 = Í 5V ˙ ¥ 10 2 ˙˚ ÍÎ 0.5 5

= [5(0.1 – 0.5) – 3(0.12 – 0.52)]105 = – 105 × 1.64 J EXAMPLE 2.5 A shaft as shown in the figure rotates at a rate of 120 revolution per minute (rev/min) against a constant torque of 1000 Nm. Calculate the power required to rotate the shaft. Also, calculate the work required to rotate the shaft through 60 rev. Solution: Power = Tw =1000 Nm × 120 rev ×

2 rad min

Nm 753952.2 Nm = = 753982.2 min 60 s

T System

= 12566.37 W

M

Figure E2.5

W.D. = T(q2 – q1) = 1000 × 60 × 2 × p = 376991.2 Nm or J EXAMPLE 2.6 A well-built horse can work at a rate of 250 W, and it works 6 hours per day. It takes food equivalent to 42000 kJ of food energy per day. What is the ratio of work output to food energy input for the horse. Solution:

Work output 0.25 kJ × 6 × 2600 s = = 0.1286 Food energy input s × 42000 kJ

EXAMPLE 2.7 Calculate the energy received by the earth across the diametral plane of the earth for a period of one year. kW kJ Solution: Assume: Solar constant = 1.39 2 = 1.39 m s m2 Mean earth radius = 6371 km Diametral plane area of the earth = p × [6371 × 1000]2 = 1.275 × 1014 m2

Work and Heat

\

47

The energy received by the earth over a period of one year = 1.39

kJ s m2

× 1.275 × 1014 m2 × 365 × 24 × 3600 s = 5.59 × 1021 kJ

EXAMPLE 2.8 The combustion gases of an I.C. engine expand within an enclosed piston and cylinder arrangement and follow the path PV1.6 = C. The pressure at the beginning of the power stroke is 5 MPa, and volume 50 cm3. At the end of the stroke the volume is 1500 cm3. Calculate (a) The work developed during the power stroke. (b) Average power developed by the gas if there are 20 power strokes per second. Solution: (a) The work developed during the power stroke (W.D.) W.D. =

Ú

V2

V1

PdV =

Ú

C

V2

V1

V

1- n

dV =

P1V1 - P2V2 (n - 1)

This is a polytropic process where index n = 1.6. n

1.6

ÈV ˘ È 50 ˘ P2 = P1 Í 1 ˙ = 5 MPa ¥ Í ˙ Î 1500 ˚ Î V2 ˚

= 0.0217 MPa

È 5 ¥ 50 - 0.0217 ¥ 1500 ˘ 3 -6 W .D. = Í ˙ 10 ¥ 10 = 0.3624 kJ (1.6 - 1) Î ˚ (b) Power developed (P) P = W.D. ¥ 20

stroke 20 = 0.3624 kJ ¥ = 7.248 kW s s

EXAMPLE 2.9 A gas at a pressure of 0.2 MPa is contained in a closed system as shown in the figure. The gas in the cylinder is held at the pressure by a pair of studs riveted to the cylinder. The studs are removed instantaneously, and the piston moves up a distance 120 mm before coming to rest. Assume that the piston has a negligible mass. Ignoring the work required to remove the studs, determine the work done by the gas held in the cylinder. Assume c/s area of the piston 0.01 m2, atmospheric pressure is 0.1 MPa. Solution: This is a case of partially resisted expansion against the ambient pressure. At the interface between the system and lower face of the piston, the resisting force is f = Atmospheric pressure × Area = 100 kPa × 0.01 m2 = 1 kN Wg = Work done by the gas = f × Ds = 1 kN × 0.12 m = 0.12 kN m = 0.12 kJ [∵ Ds = 120 mm, given]

48

Basic Thermodynamics

Piston 120 mm Piston System (gas)

Studs System (gas) 0.2 MPa System boundary

Figure E2.9

Note: It is important to note that opposing force is used to calculate work done, instead of the force exerted by the gas on the moving boundary. This is the correct measure of the effect equivalent to the rise of a weight produced in the surrounding. EXAMPLE 2.10 Modify Example 2.9 in such a way that, there are no studs, weight of the piston is so that the gas (system) pressure is maintained at 0.2 MPa and heat is added so that the piston moves through a distance of 120 mm. The remaining conditions are same. Determine the work done. Solution: Work is done by the system across the boundary with the lower face of the piston, which moved through a distance 120 mm. Thus, assuming that the gas expansion is fully resisted, Wg = Work done by the gas =

Ú

2

1

PdV = 200

kN m2

¥ 0.01 m 2 ¥ 0.12 m = 0.24 kN m W

Atm. 100 kPa Piston 120 mm Piston

System (gas) 0.2 MPa

System (gas) 0.2 MPa

Q

Figure E2.10

Work and Heat

EXAMPLE 2.11 Modify Example 2.10 in such a way that the system includes gas and piston. Determine work done by the system.

49

Atm. pr. 0.1 MPa

Solution: The boundary, when the system includes the gas and the piston, is shown in the figure. The force opposing the movement of the system boundary is atmospheric pressure. Thus,

Piston

System boundary

Gas 0.2 MPa

WS = Work done by the system [gas + piston]

= 100

kN m

2

Figure E2.11

¥ 0.01 m 2 ¥ 0.12 m = 0.12 kN m

EXAMPLE 2.12 A fixed amount of air is contained in frictionless leak proof pistons A and B as shown in the figure. Cross-sectional area of piston A is 0.01 m2 and that of B is 0.005 m2. Piston ‘A’ is exposed to the atmosphere at 100 kPa. The weight of piston ‘A’ is sufficient to maintain the pressure of the gas at 130 kPa. 100 kPa

A

B 170 mm (XA)

A 130 kPa

XA B

Value

130 kPa

Figure E2.12(a)

Piston ‘A’ moves up slowly through a distance of 170 mm when the valve is opened and a horizontal force is applied to the piston ‘B’ to maintain a pressure of 150 kPa in a cylinder ‘B’. Calculate the work done. (a) (b) (c) (d)

By piston ‘A’ at the interface of the upper surface of the piston and surrounding. By the atmosphere at the interface with the piston A. By piston A. By the system comprising the air trapped in both the cylinders A and B.

Solution: (a) By piston ‘A’ at the interface of the upper surface of the piston and surrounding (Wua)

50

Basic Thermodynamics

Consider piston A as the system. The force opposing the movement of the upper surface of piston A is, ftop = 100

kN 2

¥ 0.01 m 2 = 1 kN

m Wua = 1 kN × 0.17 m = 0.17 kN m (b) By the atmosphere at the interface with the piston A (Wsur) Wsur = –Wsystem = –Wua = – 0.17 kN m (c) By piston A (Wa) The net work done by the piston is to be calculated. Consider piston A as the system, Work done on the piston at the lower surface of piston A (Wla)

Wla = 130

kN m

2

100 kPa

¥ 0.01 m 2 ¥ 0.17 m = 0.221 kN m = 0.221 kN m

[Note: Force acting on the lower face of the piston is assumed to be fully resisted, i.e. processes occur slowly, i.e. quasi-static.] \

130 kPa

Wa = Net work on piston = Work done by the piston (system) on the surrounding – Work done on the piston (system)

Figure E2.12(b)

= Wua – Wla = 0.17 – 0.221 = –0.051 kN m i.e. 0.051 kN m of work done on the piston (system). (d) By the system comprising the air trapped in both the cylinders A and B (Ws) The system does positive work at the interface with piston A, while work is done on the system by piston B. Volume decrease in cylinder B = Volume increase in cylinder A DVB = DVA

A

170 mm (XA)

B XB

System

System

Figure E2.12(c)

Work and Heat

51

XB × 0.005 = 0.17 × 0.01 \ \

XB = 0.34

WS = 130

kN m

2

¥ 0.01 m 2 ¥ 0.17 m - 150

kN

¥ 0.005 ¥ 0.34 m = – 0.034 kN m

m2

EXAMPLE 2.13 A lump of a substance (ball) is falling in a gravitational field through an evacuated space. The substance takes 1 second to reach the ground. Can work interaction be considered during the fall? The lump of substance is defined as the system. Solution: We know that whatever forces are acting on the system side of the boundary, there is zero force on the surroundings side since the environment is an evacuated space. Thus, there is no equivalent effect of raising or lowering a weight in all environment that does not offer any resistance to the movement of the system through it. \ The work done by the system = 0 [Note: Whenever the boundary movement is unresisted on either system side or the surroundings side, the work done is ZERO.]

Figure E2.13

EXAMPLE 2.14 A liquid is compressed isothermally. Calculate the work done if the process follows V ln = - A ( P - P0 ) where A, V0 and P0 are constants. V0 Process is quasi-static. Solution: W=

We know that

ln

\ Substitute Eq. (2) in Eq. (1),

Ú

2

1

P dV

(1)

V = - A ( P - P0 ) V0 dV = –AVdP

Ú

(2)

Ú

W = - AVPdP = - A VPdP

\

Generally volume of a liquid remains constant, if it is not sensitive to pressure change. \

We get W = - AV

Ú

2

1

PdP = - AV

( P22 – P12 ) 2

EXAMPLE 2.15 1 kg mole of a gas is contained in a piston and cylinder arrangement. The gas is expanded from 20 bar to 10 bar at a constant temperature of 40°C. Calculate the work done for the following cases:

52

Basic Thermodynamics

(a) The expansion rate is large against a pressure of 101.325 kPa. (b) The expansion rate is negligible. (c) The expansion rate is so large that it approaches infinity. Assume no losses. Solution: (a) The expansion rate is large against a pressure of 101.325 kPa (W) Here, expansion rate is large means, partially resisted expansion against a pressure of 1 bar (P = resisting pressure) W=

\

Ú

2

1

PdV = P (V2 - V1 )

(1)

P2V2 = RT2 \

V2 =

RT2 RT ; V1 = 1 P2 P1

(2)

This is a constant temperature process \

T1 = T2 = 40° + 273 = 313 K

Substitute Eq. (2) in Eq. (1) È1 1˘ W = PRT Í - ˙ Î P2 P1 ˚

\ = 101.325

kN m

2

¥ 8.314

kJ 313 K È 1 1 ˘ 1 kJ ¥ ¥ = 131.84 Í ˙ 2 kg mole K 100 Î 10 20 ˚ kNm kg mole

(b) The expansion rate is negligible (W) The expansion rate is negligible means, fully resisted expansion, i.e. resisting pressure is equal to that of the system pressure. \

W=

Ú

V2

V1

PdV =

= - 8.314

Ú

V2

V1

V P RT dV = RT ln 0 = - RT ln 2 V V1 P1

kJ 10 kJ ¥ 313 K ¥ ln = 1803.76 kg mole K 20 kg mole

(c) The expansion rate is so large that it approaches infinity (W) Here, resisting pressure is zero \ Work done = 0 EXAMPLE 2.16 A gas has an initial volume of 0.05 m3 and expands to a final volume of 0.1 m3 by heating the gas. The initial pressure of the gas is 200 kPa and is maintained with the help of weights. Considering the gas in the chamber as a system, calculate

Work and Heat

53

(a) The work done by the system, while heating is going on. Weights are removed from the piston, as the gas expands according to the following relation. Calculate work done if initial conditions are same and final volume is 0.1 m3. (b) PV = C (c) PV1.4 = C (d) The piston is fixed, for the same initial conditions calculate the work done if heat is removed from the system and the pressure drops to 100 kPa. Solution:

V1 = 0.05 m3 V2 = 0.1 m3 P1 = 200 kPa P2 = 100 kPa

Initial volume Final volume Initial pressure Ambient pressure

(a) The work done by the system (W) Since pressure remains constant (this is a fully resisted system) W=

Ú

V2

V1

= 200

Gas (system) P1, V1

PdV = P1 (V2 - V1 )

Q

kN

3

¥ (0.1 - 0.05) m = 10 kN m m2 (b) Work done, if PV = C(W) This is an isothermal process. \

W = P1V1 ln

Figure E2.16

V2 0.1 = 200 ¥ ln = 6.93 kJ V1 0.05

(c) Work done, if PV1.4 = C(W) This is a polytropic process. \

W=

( P1V1 - P2V2 ) ( n - 1)

n

4

ÈV ˘ È 0.05 ˘ P2 = P1 Í 1 ˙ = 200 ¥ Í ˙ = 75.79 kPa Î 0.1 ˚ Î V2 ˚

\

È 200 ¥ 0.05 - 75.79 ¥ 0.1 ˘ W =Í ˙ = 6.052 kJ (work done by the gas) 1.4 - 1 Î ˚

(d) Work done (W) if the piston is fixed, no change in its volume This is a constant volume process. W=

Ú

V2

V1

dW = 0

54

Basic Thermodynamics

EXAMPLE 2.17 1.5 m3 of a gas at 400 kPa is contained in a frictionless piston. The piston is held in position by applying an external force. [The weight of the piston is so that the presure of the gas (400 kPa) remains constant]. Calculate the amount of work done if (a) This external force is released gradually until the volume is 3 m3 in such a way that PV is always a constant. (b) The external force is released suddenly to half of its initial value. Solution: (a) Work done (W) (this is a fully resisted expansion), if external force is released gradually according to the law of PV = C W = P1V1 ln

V2 kN 3 = 400 2 ¥ 1.5 m 3 ¥ ln = 415.9 kN m 1.5 V1 m

Figure E2.17

a to b Æ Gradually load is removed (reversible) a to c Æ Suddenly load is removed (irreversible) (b) Work done (W) if external force is released suddenly to half of its initial value As soon as external force is reduced to half of its initial value, the gas will undergo a sudden expansion. Eventually the system will return to an equilibrium condition similar with the final state attained in the reversible process. \ Change in volume is same in both the cases. This is a case of partial resistance expansion. P2 =

\

P1V1 400 ¥ 1.5 = = 200 kPa V2 3

W = 200

kN m

2

¥ (3 - 1.5) m 3 = 300 kJ

Work and Heat

\

55

This process is clearly irreversible.

The efficiency of the second process =

300 ¥ 100 = 72.13% 415.9

EXAMPLE 2.18 A gas with a volume of 0.05 m3 and at a pressure of 70 kPa is contained in a container filled with a frictionless leak proof piston. The spring holding the piston is in tension. The force on the spring varies linearly with length. The cylinder is heated until the volume increases to 0.25 m3 and pressure rises to 200 kPa. Assume no losses and pressure of the atmosphere = 100 kPa. Calculate (a) The work done by the gas. (b) The work done if the system consists of the gas, piston and spring. Solution:

System

Q (a)

Figure E2.18

We know that

(b)

(a) Before heat addition, (b) after heat addition.

Fs = Spring force = Kx (linearly given)

\ \

P=

FS Kx = A A

K Ê ˆ = constant ˜ Pressure also varies linearly with displacement. Á\ Ë ¯ A

(a) The work done by the gas (W1–2)

W1-2 =

Ú

2

1

Ê P + P1 ˆ PdV = Á 2 (V - V1 ) Ë 2 ˜¯ 2

[If pressure varies linearly, then for work done calculation, the pressure can be taken as the average of the initial and final pressure.] \

W1–2 =

(70 + 200) [0.25 – 0.05] = 27 kJ 2

56

Basic Thermodynamics

(b) The work done if the system consists of the gas, piston and spring W1- 2 =

Ú

V2

V1

PdV = 100 ¥ (0.25 - 0.05) = 30 kJ

Piston

Figure E2.18(c)

EXAMPLE 2.19 Air expands inside a piston and cylinder arrangement. Let the initial volume be 0.2 m3 and the initial pressure be 20 bar. An external force is applied to keep the 750 kg piston in position, which is open to the atmosphere of 100 kPa. When the external force is removed, the piston occupies 0.5 m3. The gas in the cylinder is kept at the same temperature. Assume area of the piston is 0.1 m2. Calculate, (a) The total amount of the work done by the gas. (b) The amount of work done by the gas against the piston. (c) The velocity of the piston as it moves. (d) The total vertical displacement of the piston. Solution: (a) The total work done by the gas (Wg) [constant temperature process] Wg = P1V1 ln

V2 kN 5 = 20 ¥ 100 2 ¥ 0.2 m 2 ¥ ln = 366.51 kJ 2 V1 m

(b) The amount of work done by the gas against the piston (W) W = Wg – Ws Ws = Work due to effective surrounding pressure Weight of the piston = Effective surrounding pressure Area of the piston

Ps = Patm + = 100

kN m

2

+

750 ¥ 10 kN kN = 175 2 2 1000 ¥ 0.1 m m

Certain amount of work must be used to move the effective surrounding pressure.

Ú

Ws = Ps DVsys = 175

kN m

2

¥ (0.5 - 0.2) m 3 = 52.5 kN m

Work and Heat

57

The energy used to accelerate the piston OR work done against the piston is, W = Wg – Ws = 366.51 – 52.5 = 314.01 kJ (c) Velocity of the piston (V) The change in K.E. of the piston is 314.01 kJ. 1 mV2 = 314.01 ¥ 103 Nm 2 gc V2 =

\

È kg m ˘ Í gc = 1 ˙ N s2 ˚˙ ÎÍ

314.01 × 10 3 N m × 1 kg m × 2 Ns2

750 kg

= 837.36

m2 s2

V = 28.94 m/s at the exit of the cylinder.

(d) Vertical displacement of the piston (Z) At the heighest point, P.E. = K.E. imparted to the piston P.E. =

mgz = K.E. = 314.01 ¥ 103 Nm gc

WpZ = 314.01 × 103 Z=

[Wp = Weight of the piston]

314.01 ¥ 103 Nm = 418.68 m 750 N

EXAMPLE 2.20 An elastic balloon has a diameter of 0.5 m and is filled with a gas at a pressure of 200 kPa. The gas is heated, so that its diameter increases to 0.6 m and a pressure to 250 kPa. During the process, the pressure is proportional to the balloon’s diameter. Calculate, (a) The work done by the gas during the process. (b) The work done by the baloon on the atmosphere. Assume atmospheric pressure = 100 kPa. Solution: P1 = 200 kPa

d1 = 0.5 m

P2 = 250 kPa

d2 = 0.6 m

4 3 r = d3 3 6 V1 = Initial volume of the balloon

V = Volume of the balloon =

= V2 =

4 3 r1 = d13 = ¥ 0.52 = 0.065 m 3 3 6 6 6

¥ 0.6 2 = 0.1131 m 3

58

Basic Thermodynamics

Final

P1

Tank

P2

W.D.

Valve Initial

V1

V2

V

Figure E2.20

(a) Work done by the gas (W) We know that Ê 6V ˆ P = Cd = C Á ˜¯ Ë

Ê 6ˆ =CÁ ˜ Ë ¯

1/ 3

V 1 / 3 = CV 1 / 3

3 4/3 [V - V 4/3] 4

W1- 2 =

Ú

V2

=

3 4

È P2V14 / 3 P1V14 / 3 ˘ 3 Í 1 / 3 - 1 / 3 ˙ = [ P2V2 - P1V1 ] V1 ÍÎ V2 ˙˚ 4

=

3 [250 ¥ 0.1131 - 200 ¥ 0.065] = 11.46 kJ 4

V1

PdV =

Ú

V2

1/3

V1

CV 1 / 3 dV = C ¥

(b) The work done by the balloon on the atmosphere (W) (This is a case of partially resisted) W=

Ú

V2

V1

Pa dV = 100 (V2 - V1 ) = 100

kN m2

¥ (0.1131 - 0.065) m 3 = 4.81 kJ

EXAMPLE 2.21 0.05 m3 gas at a pressure of 250 kPa is contained in a piston and cylinder arrangement. A spring with a spring constant of 150 kN/m is touching the piston but it does not exert any force on the piston. Gas is heated till the volume becomes double and compress the spring. The cross-sectional area of the piston is 0.3 m2. Calculate, (a) The final pressure of the gas. (b) The total work done by the gas. (c) The fraction of this work done against the spring to compress it. Solution: (a) The final pressure of the gas (P2g) Force exerted by the spring (F) F=K×S

(1)

Work and Heat

59

Figure E2.21

Pressure exerted by the gas is Pg = Pp + Ps

where

(2)

Pg = Gas pressure Pp = Piston pressure Ps = Spring pressure =

F A

S = Displacement F K¥S = Pp + A A Displacement of the spring in terms of initial and final volume. Pg = Pp +

S= \

Substitute Eq. (4) in Eq. (1),

(V - V1 ) A

(3)

(4)

(V - V1 ) A For initial condition, force exerted by the spring on the piston is ZERO, F=K

\

F =0=K

\

V = V1

\

(V - V1 ) A (5)

Substitute Eq. (4) in Eq. (3), Pg = Pp + K

(V - V1 ) A2

(6)

60

Basic Thermodynamics

\

For state 1, V = V1, P1g = Pp + K

\

(V1 - V1 ) A2

= Pp + 0 = 250 kPa

(7)

For state 2, V = V2, P2 g = Pp + K

(V2 - V1 ) A2

From Eq. (7), P1g = Pp = 250 kPa P2 g = 250 + 150

kN (0.1 - 0.05) m 3 kN ¥ = 333.33 2 3 m m (0.3) m

S = (x2 – x1) = Displacement of the piston =

(V2 - V1 ) (0.1 - 0.05) = = 0.1667 m A 0.3

We know that F = Kx Let x1 = 0 (Assumed)

x2 = 0.1667

F2 = Force applied by the spring at the final position = Kx2 – Kx1 = K(x2 – x1) = KS = 150

kN ¥ 0.1667 m = 25.005 kN m

Additional pressure applied on the piston at the final position (Pad)

Pad = \

F2 25.005 kN kN = = 83.35 2 2 A 0.3 m m

P2g = Final pressure of the gas = Pad + P1g = 83.35 + 250 = 333.35 kPa

(b) Total work done by the gas (Wg) Wg = W1- 2 =

Ú

V2

V1

PdV

(8)

Substitute Eq. (6) in Eq. (8), Wg =

Ú

V2

V1

È (V2 - V1 )2 Ê V - V1 ˆ ˘ dV = P ( V V ) + K Í Pg + K ÁË ˙ 2 1 p ˜ A2 ¯ ˚ 2 A2 Î

From Eq. (7), P1g = Pp = 250 kPa Wg = 250 ¥ (0.1 - 0.05) +

150 ¥ (0.1 - 0.05)2 2(0.3)2

= 14.58 kJ

Work and Heat

61

Assume atmospheric pressure is acting on the spring (Pa). Force balance at any position, Gas pressure = Atmospheric pressure + Pressure exerted by the spring PA = PaA + Kx Volume V = Area A PA = PaA + KV

Displacement = \

KV = PA2 – PaA2

P = Pa + Wg =

Ú

V2

V1

PdV =

Ú

V2

V1

= Pa (V2 - V1 ) +

KV

(9)

A2

KV ˘ K È (V22 - V12 ) Í Pa + 2 ˙ dV = Pa (V2 - V1 ) + 2 A ˚ 2A Î

K È ˘ (V2 + V1 )(V2 - V1 ) = (V2 - V1 ) Í Pa + (V2 + V1 ) ˙ 2A 2 A2 Î ˚ K

2

(10)

Substitute Eq. (9) in Eq. (10), È ( P2 g A2 - Pa A2 + P1g A2 - Pa A2 ) ˘ ˙ Wg = (V2 - V1 ) Í Pa + 2 A2 ÍÎ ˙˚ È 2 Pa A2 + P2 g A2 - 2 Pa A2 + 2 P1g A2 ˘ (V2 - V1 )( P2 g + P1g ) ˙= = (V2 - V1 ) Í 2 2 A2 ÍÎ ˙˚

(11)

Substitute P2g = 333.33 kPa, P1g = 250 kPa, V2 = 0.1 m3, V1 = 0.05 m3 in Eq. (11), (0.1 - 0.05)(333.33 + 250) = 14.58 kJ 2 (c) Work required to compress the spring (Ws) Wg =

1 1 K ( x22 - x12 ) = ¥ 150 ¥ [0.1672 - 0 2 ] = 2.091675 kJ 2 2 Or from P–V diagram, Ws =

Ws =

Ws =

1 (333.33 - 250)(0.1 - 0.05) = 2.0825 kJ 2

Ú

V2

V1

P dV =

Ú

V2

V1

K

(V - V1 ) A2

dV =

K (V2 - V1 )2 2 A2 F K s K (V - V1 ) ˘ È Í∵ Pa = A = A = ˙ A2 Î ˚

62

Basic Thermodynamics

\

Ws =

150 ¥ (0.1 - 0.05)2 2(0.3)2

= 2.0833 kJ

(d) The work required to raise the piston without spring, i.e. due to piston (Wp) Wp =

Ú

V2

V1

Pp dV = Pp (V2 - V1 ) = 250

kN m2

(0.1 - 0.05) = 12.5 kJ

[From Eq. (7), P1g = Pp = 250 kPa] (e) The total work done by the gas (Wg) (third method) Wg = Wp + Ws = 12.5 + 2.0833 = 14.5833 kJ EXAMPLE 2.22

Show that for a Van der Waals gas equation of state which is described by a˘ È Í P + ˙ ( - b) = RT Î ˚

the isothermal work per unit mass is given by È1 - b) 1 ˘ - aÍ ˙ 1 - b) 2˚ Î 1 where u1 and u2 are the initial and final specific volumes. RT ln

Solution:

( (

2

We have a ˘ È Í P + 2 ˙ ( - b) = RT Î ˚

RT a = 2 ( - b)

P=

\ Work done is given by W1- 2 =

Ú

V2

V1

PdV =

= RT ln

( (

2 1

Ú

V2

V1

È RT a ˘ - 2 ˙ dV Í Î ( - b) ˚

È1 - b) 1 ˘ - aÍ ˙ - b) 2˚ Î 1

EXAMPLE 2.23 Consider a piston and cylinder as shown in the figure. Initial conditions of the gas are V1 = 0.1 m3 and P1 = 4 bar. The spring force exerted through the piston is proportional to the volume. Atmospheric pressure of 1 bar acts on the spring side of the piston. The gas is heated until the volume becomes 0.5 m3. Calculate (a) The work done against the atmosphere. (b) The work done against the spring. (c) The work done by the gas.

Work and Heat

63

Solution: P1 = 4 bar, Pa = 1 bar 3 V1 = 0.1 m , V2 = 0.5 m3 Pressure balance at any position Gas pressure (P1) = Atmospheric pressure (Pa) + Pressure exerted due to the spring (Ps) 4 = 1 + Ps \

Ps = 3 bar

Let

A = Area of the piston

\

x1 =

(1)

Figure E2.23

Volume 0.1 = A A Fs = Ps × A = Kx1

(2)

3 × 10 5 N × Am 2 N = 3 ¥ 10 6 A2 m È 0.1 ˘ m2 Í ˙ m A Î ˚ (a) Work done against the atmosphere (Wa) K = Spring constant =

(3)

The gas when expands has to do work against atmosphere and also against spring.

Wa = 100

kN

m2 (b) Work done against the spring (Ws) Ws =

Ú

x2 x1

Kx dx =

Ú

x2 x1

(0.5 - 0.1) = 40 kJ

3 ¥ 10 6 A2 x dx È x2 ˘ = Í3 ¥ 103 A2 ˙ 1000 2 ˙˚ ÍÎ

1 2 1 È (0.5)2 (0.1)2 ˘ 3 ( x2 - x12 ) = 3 ¥ 10 3 ¥ A2 ¥ Í 2 ˙ = 0.36 × 10 kJ 2 2 2 ÍÎ A A ˙˚ (c) The total work done by the gas (Wg) = 3 ¥ 103 ¥ A2 ¥

Wg = Wa + Ws = 40 kJ + 360 kJ = 400 kJ Or force balance at any position, PA = Pa A + Kx

(4)

A = Cross-sectional area of piston x = Displacement = \

Volume V = Area A

(5)

Substitute Eq. (5) in Eq. (4),

PA = Pa A +

KV A

P = Pa +

KV A2

(6)

64

Basic Thermodynamics

Wg =

Ú

V2

V1

PdV =

Ú

V2

V1

= Pa (V2 - V1 ) +

KV ˘ K È (V22 - V12 ) Í Pa + 2 ˙ dV = Pa (V2 - V1 ) + 2 A A 2 Î ˚ K È ˘ (V2 + V1 )(V2 - V1 ) = (V2 - V1 ) Í Pa + (V2 + V1 ) ˙ (7) 2 2A 2A Î ˚ K

2

From Eq. (6), KV1 2

KV2

= P1 - Pa ;

A2

A Substitute Eq. (8) in Eq. (7),

= P2 - Pa

(8)

( P - Pa ) ( P1 - Pa ) ˘ È È ( P + P1 ) ˘ + = (V2 - V1 ) Í 2 Wg = (V2 - V1 ) Í Pa + 2 ˙ ˙ 2 2 2 Î ˚ Î ˚ P2 = Pa +

KV2 A

= 1 ¥ 10 5

2

N m

2

+

3 ¥ 10 6 A2 ¥ 0.5 A

2

= 16 ¥ 10 5

(9)

N m2

Given,

P1 = 4 ¥ 10 5 \

N m

2

, V1 = 0.1 m 3 , V2 = 0.5 m 3

È 16 ¥ 10 5 + 4 ¥ 10 5 ˘ 5 Wg = (0.5 - 0.1) Í ˙ = 4 ¥ 10 J = 400 kJ 2 ÍÎ ˙˚

EXAMPLE 2.24 Gas is used to inflate an inelastic flexible substance (balloon) [initially folded] to volume 0.5 m3. The barometer reads 760 mm Hg. What is the work done by the balloon on the atmosphere? Solution:

Ú

Ú

Wd = Work done = PdV (cylinder) + PdV (balloon)

= 0 + P(V2 - V1 ) = 101.325 = 50.6625 kN m

kN m2

[0.5 - 0] m 3

or kJ

Wall of the cylinder is rigid, hence no PdV work is involved. Cylinder Patm Tank

Final volume 3

= 0.5 m

Valve

Initial volume 3

=0m

Figure E2.24

Work and Heat

65

[Note: Assumption—Balloon material is light, inelastic and unstressed. Hence, pressure in the balloon is always atmospheric.] Work done by the atmosphere is—50.6625 kJ EXAMPLE 2.25 2.0 m3 of air enters into an evacuated cylinder from atmosphere when the value is opened. The atmospheric pressure is 1 bar. Calculate the work done by the air. Solution:

Ú

Ú

W = Work done = PdV (cylinder) + PdV (atmosphere)

= 0 + 1 ¥ 10 5

N

[V2 - V1 ] m 3 m2 = 105 [0 – 2] = –2 × 105 J = – 200 kJ [–ve sign indicates work done on the system] Initial volume System

3

=2m Valve

Final volume 3

=0m

Figure E2.25

EXAMPLE 2.26 The figure shows the piston and cylinder arrangement with a stirring device. Piston is held against the fluid due to atmospheric pressure equal to 100 kPa. The stirrer rotates 8000 revolutions with an average torque of 1.5 Nm, then the piston moves out 0.5 m. The diameter of the piston is 0.6 m. Calculate the net work done. Solution: Ws = Work done by the motor (stirrer device) = 2pNT = 2p × 8000 × 1.5 Nm = 75398.2 Nm = 75.4 kJ [This work is done on the system] Wp = Work done by the system or done by the piston movement on the surroundings

Figure E2.26

66

Basic Thermodynamics

kN

¥ ¥ (0.6)2 ¥ 0.5 m2 4 [This work is done by the system]

= PdV = Pa AL = 100 = 14.14 kJ

Net work done on the system (Wn) Wn = Ws + Wp = – 75.4 + 14.14 = – 61.26 kJ EXAMPLE 2.27 200 kJ of work is supplied to a closed system. The pressure and volume relation is P = 8 – 5V, P is in bar and V in m3. The initial volume is 0.5 m3. Calculate final volume and pressure. Solution: Final volume (V2) and pressure (P2) W.D. =

Ú

V2

V1

V

2 È 5V 2 ˘ PdV = (8 - 5V ) dV ¥ 10 = 10 Í8V ˙ 2 ˙˚ ÎÍ 0.5

Ú

5

5

= 105 [8(V2 – 0.5) – 2.5(V22 – 0.52 )] – 200 × 103 J = 105 [8V2 – 4 – 2.5V22 – 0.625] J 2.5V22 – 8V2 + 2.625 = 0 V2 =

\ \

8 ± 64 - 4 ¥ 2.5 ¥ 2.625 = 0.3712 m 3 2 ¥ 2.5

P2 = 8 – 5V2 = 8 – 5 × 0.3712 = 6.144 bar

EXAMPLE 2.28 The temperature of oxygen changes from 200 K to 300 K during an adiabatic frictionless process. The specific heat at constant pressure is given by the expression Cp = (0.2 + 0.00006T) kJ/kg K. Calculate the work done by the gas. T is in absolute temperature. Solution: R=

kJ R 8.314 = = 0.26 32 kg K M

We have DH = DU + PdV Q = DU + PdV Given, adiabatic process, \

Q = 0 = DU + PdV

\

PdV = – DU = – CV dT

\

Work done =

Ú

T2

T1

PdV =

Ú

T2

T1

-Cv dT =

Ú

T2

T1

-(C p - R) dT =

Ú

T2

T1

( R - C p ) dT

Work and Heat

=

Ú

T2

T1

(0.26 - 0.2 - 0.00006T ) dT = 0.06(T2 - T1 ) -

= 0.06(300 - 200) - 0.00003(300 2 - 200 2 ) = 4.5

67

0.00006(T22 - T12 ) 2

kJ kg

EXAMPLE 2.29 Modify Example 2.28. Heat is added to the system so that temperature changes from 200°C to 300°C. Rest is same. Calculate the heat added to the system. Solution: Q = DH = heat =

Ú

T2

T1

Cv dT =

= 0.2 (T2 - T1 ) + 0.00006

Ú

T2

T1

(0.2 + 0.00006T ) dT

(T22 - T12 ) 2

kJ kg EXAMPLE 2.30 In a closed system during a process, a volume changes from 4 to 8 m3. The process takes place according to the relation P = V3 + 10/V, where P is in bar and V is in m3. Calculate the work done. = 0.2 (300 - 200) - 0.00003(300 2 - 200 2 ) = 21.5

Solution: W = Work done =

Ú

V2

V1

ÈV 4 ˘ 10 ˆ Ê + 10 ln V ˙ PdV = 10 5 Á V 3 + ˜ dV = 10 5 Í Ë V¯ ÍÎ 4 ˙˚

Ú

8

È (V 4 - V14 ) V ˘ = 10 Í 2 + 10 ln 2 ˙ = 10 5 4 V1 ˙˚ ÎÍ 4 5

È (84 - 4 4 ) 8˘ + 10 ln ˙ = 96693 kJ Í 4 4 ˙˚ ÎÍ

EXAMPLE 2.31 A pressure and volume relation is given by P = 500/u + 0.05u3, where P is pressure and u specific volume. In the process, specific volume changes from 5 m3/kg to 3 m3/kg. Calculate (a) Work done if losses are taken into account. (b) Work done if frictionless piston is considered. Solution: (a) Work done if losses are taken into account

Ú

W + Losses = PdV =

Ú

3

5

Ê 500 + 0.05 ÁË

= 500 (ln 3 - ln 5) + W = – 262.2

(Losses)

3ˆ

˜¯ dV

0.05 4 kJ (3 - 54 ) = - 262.2 4 kg

68

Basic Thermodynamics

(b) Work done if losses are neglected

Ú

W = PdV = - 262.2

kJ kg

EXAMPLE 2.32 Air is compressed in an air compressor from 1 bar to 6 bar according to the relation PV1.3 = C. Calculate (a) The work done during the compression processes if the entrance and exit velocities are assumed to be the same, (b) The shaft work of the compressor. Take specific volume of air = 0.9 m3/kg (V1). Solution: (a) Work done during the compression process (W.D.) W.D. =

Ú

V2

V1

PdV ; PV 1.3 = C

This is a polytropic process with index n = 1.3. ( P V - P2V2 ) \ W.D. = 1 1 (n - 1) 1 / 1.3

V2 È P1 ˘ =Í ˙ V1 Î P2 ˚

1 / 1.3

\

È1˘ V2 = 0.9 Í ˙ Î6˚ W.D. =

= 0.2268

m3 kg

Figure E2.32

10 5 [1 ¥ 0.9 - 6 ¥ 2268] = –1.536 × 105 J 0.3

Work is done on the system

(b) Work done by the shaft (Ws), i.e. power required to drive the shaft Ws = -

Ú

P2

P1

VdP +

(V12 - V22 ) r + ( Z1 - Z 2 ) = 2g gc

PV1.3 = C 1 / 1.3

\

ÈC ˘ V =Í ˙ ÎP˚

Ws = -

Ú

P2

P1

1 / 1.3

ÈC ˘ ÍP˙ Î ˚

dP =

( - P2V2 + P1V1 ) 0.3 1.3

Ú

P2

P1

VdP + 0 + 0

Work and Heat

69

1.3 ¥ 10 5 [ -6 ¥ 0.4953 + 1 ¥ 0.9] = - 8.98 ¥ 10 5 J 0.3 Due to losses, the work done in each case is not the same. =

EXAMPLE 2.33

The equation of state for an ideal elastic substance is =

A0 1 ˘ È KT Í - 2 ˙ A Î ˚

where K is a constant, l is the extension ratio L/L0, and the ratio of no load area to area of the stressed specimen, A0/A converts the stress s from the nominal value (based on the no load area) to the actual value. L0 is the no load length and is a function of temperature only. (a) Using the formula for work in the form dW = Vs de Show that,

W 1 ˘ È = KT Í - 2 ˙ d V0 Î ˚

for isothermal, quasi-static extension or compression. (b) For a sample of rubber, L0 = 0.3 m, A0 = 0.65 cm2, and for which

K = 1.86 ¥ 103

N

at 300 K m 2K Calculate W for elongation of the sample from L = L0 to L = 2L0. Solution: s dee = –ALs s dee (a) dW = Vs But \

=

A0 1 ˘ È KT Í - 2 ˙ A Î ˚

1 ˘ È dW = A0 KT Í - 2 ˙ L d Î ˚

=

d =

L L0

or L = L0

and dL = L0

(at constant T )

dL L0 d = d = L L

\

È 1 ˘ dW = A0 KT Í ˙ L0 0˚ Î

Hence,

W 1 ˘ È = KT Í - 2 ˙ d V0 Î ˚

d

1 ˘ È = V0 KT Í - 2 ˙ d Î ˚

70

Basic Thermodynamics

(b) Integrating at constant temperature È( W = V0 KT Í ÍÎ

2 2

2 1)

2

Ê 1 1 ˆ˘ +Á - ˜˙ Ë 2 1 ¯˙ ˚

For l2 = 2 and l1 = 1 È (4 - 1) Ê 1 1 ˆ ˘ + Á - ˜ ˙ = V0 KT = A0 L0 KT W = V0 KT Í Ë 2 1¯ ˚ Î 2 \

W = 0.3 × 0.65 × 300 × 1.86 × 103 = 108.81 × 103 J = 108.81 kJ

EXAMPLE 2.34 Calculate the work done of a gas during a quasi-static isothermal expansion from an initial volume V1 to final volume V2, when the equation of state is P(V – b) = RT, where ‘b’ is +ve constant. Solution:

Ú

W.D. = PdV =

(V2 - b) 1 - b)

RT

Ú (V - b) dV = RT ln (V

EXAMPLE 2.35 Compute the quasi-static work done in compressing 0.04 m3 of mercury at a constant temperature of 0°C from a pressure of 1 bar to a pressure of 3000 bar. The isothermal compressibility of mercury at 0°C is given as a function of pressure by, K = 3.9 × 10–6 – 0.1 × 10–9 P, where P is in bar and K is in bar–1. Solution: W =-

Ú

P2

P1

PKV dV = -

Ú

P2

P1

(3.9 ¥ 10 -6 - 0.1 ¥ 10 -9 P) PV dP

Assume V = V0 as the compressibility is small.

P1 >> P2 then,

È 3.9 ¥ 10 -6 ( P22 - P12 ) 0.1 ¥ 10 -9 ( P23 - P13 ) ˘ W = - V0 Í ˙ 2 3 ÍÎ ˙˚ È 3.9 ¥ 10 -6 ¥ 3000 2 0.1 ¥ 10 -9 ¥ 30003 ˘ W = - 0.04 Í ˙ 2 3 ÎÍ ˚˙ N = - 0.666 ¥ 10 5 2 ¥ m 2 = - 66.6 kJ m

EXAMPLE 2.36 (a) The tensile stress in a metal bar is increased quasi-statistically and isothermally from s1 to s2. If the volume of the bar (V) and Young’s modulus (E) remain constant, show that the work is given by, W=

V ( 2E

2 2

-

2 1)

Work and Heat

71

(b) The tensile stress in a steel bar of 300 cm long and 0.2 cm2 in area is increased quasistatistically and isothermally from 18 bar to 90 bar. What is the work done ?

E = 52537 ¥ 10 5

N m2

Solution: (a) We have d W = Vs s dee Ê∂ ˆ E=Á at constant T Ë ∂ ˜¯ d E Considering V and E to be constant, we have

Ú

\

W=

(b)

W=

V E

Ú

Ú

W= V

d =

V(

2 2

2E

2 1)

3 ¥ 0.2 ¥ 10 -4 m 2 ¥ m 2 5

52537 ¥ 10 N ¥ 2

[90 2 - 182 ] ¥ 10 5

N m

2

¥

N m2

= 0.444 J

EXAMPLE 2.37 5 litres of water is heated from 20°C to 60°C. Fuel is used at the rate of 0.0001 kg/s for this purpose. Calculate the time required. Take calorific value of the fuel as 40000 kJ/kg. Assume no heat transfer to the surrounding. Solution: Q = heat supplied to water = m Cp dT = 5 kg × 4.187

kJ (60 – 20)K = 837.4 kJ kgK

Time required to heat this water (t) Q = mf × C.V. × t (t = time in seconds) \ \

837.4 kJ = 0.0001

kg kJ ¥ 40000 ¥ts s kg

t = 209.35 s

EXAMPLE 2.38 A gas is taken in a piston and cylinder arrangement at an initial pressure of 25 bar. It undergoes a cyclic process as follows. (a) The gas is expanded reversibly according to the relation PV 2.5 = C until the volume is doubled. (b) Then, the gas is cooled reversibly at a constant pressure until the piston reaches the initial position.

72

Basic Thermodynamics

(c) Now piston is kept fixed and heat is added until the pressure rises to the original value of 25 bar. Calculate the net work done by the fluid. Take initial volume 0.05 m3 and mass 1 kg. Solution:

P

V1 = 0.05 m3, V2 = 2V1

P1 = 25 × 100 kPa, V3 = V1 ,

m = 1 kg, n = 2.5,

1

T

1

(i) (i)

PV

V=C PV

2.5

=C

(iii)

P=C

(ii)

=C

2

V=C 3

2.5

(iii)

2

3

P=C

(ii)

V

S

Figure E2.38

Considering the process 1–2 n

ÈV ˘ È 0.05 ˘ P2 = P1 Í 1 ˙ = 25 Í ˙ Î 0.1 ˚ Î V2 ˚

2.5

= 4 bar

Process 1–2 is an isentropic process with index, n = 2.5 \

W1- 2 =

( P1V1 - P2V2 ) (25 ¥ 0.05 - 4.42 ¥ 0.1)10 5 = = 53.87 kJ (n - 1) (2.5 - 1)

Work done on gas from 2 to 3, P = C, W2–3 = P2[V2 – V3] = 4.42 × 100

kN m2

× [0.1 – 0.05] = 22.1 kJ

Work done during constant volume process = ZERO Net work done by the fluid (Wn) \

Wn = W1–2 + W2–3 + W3–1 = 53.87 – 22.1 + 0 = 31.77 kJ

EXAMPLE 2.39 A gas undergoes a cyclic process as shown in the figure. Calculate (a) the law of process 3–1 and (b) net work interaction. Solution: Process 1–2 = P = C Process 2–3 = T = C Process 3–1 = ?

Work and Heat

(a) The law of process 3–1, i.e. index n

73

P bar

Consider the process 2–3, i.e. T = C P2V2 P3V3 = T2 T3

8 bar

P2V2 P3

5 bar

V3 =

\

3

1

0.05

n

\

ln

= P1V1

0.1

3

Vm

Figure E2.39

\ Let us assume that process 3–1 is a polytropic with index n. P3 È V1 ˘ =Í ˙ ; P1 Î V3 ˚

n

2 P=C

È5˘ V3 = Í ˙ ¥ 0.1 = 0.0625 m 3 Î8˚

P3 V3n

T=C

?

ln

P3 V = n ln 1 P1 V3

8 0.05 = n ln 5 0.0025

\

n = – 2.10764

\

Process 3–1 follows the law, PV–2.1076 = C

(b) Net work done (Wn) Work done during 1–2 (W1–2), P = C W1–2 = P(V2 – V1) = 5 × 100 × [0.1 – 0.05] = 25 kJ Work done during 2–3(W2–3), T = C W2–3 = P2V2 ln

V2 0.1 = 5 ¥ 100 ¥ 0.1 ¥ ln = - 23.5 kJ V3 0.0625

Work done during 3–1 (W3–1), PVn = C \ \

W1- 2 =

( P3V3 - P1V1 ) (8 ¥ 0.0625 - 5 ¥ 0.05) ¥ 10 2 = kJ = - 8.045 kJ (n - 1) ( -2.10764 - 1)

Wn = W1–2 + W2–3 + W3–1 = 25 – 23.5 – 8.045 = – 6.55 kJ

EXAMPLE 2.40 A battery charger is used to charge a cell. The charger operates for 2 hours, at 12 V and 15 ampere. Calculate the work done by the charger on the cell. Solution: Wb = Work done by the charger = EI = 12 V × 15 I × 2 × 3600 s = 12 ¥

kg m 2 Cs

2

¥ 15

C ¥ 2 ¥ 3600 s = 1296000 J = 1296 kJ s

74

Basic Thermodynamics

EXAMPLE 2.41 Modify Example 2.40 so that during the charging process 0.1 kmole of gas was liberated at constant pressure of 1 bar and temperature of 300 K. Assume that the gas behaves ideally, what is the net work interaction? Other conditions are same, i.e. 12 V, 15 I, 2 hours. Solution: During the charging, gas is released at constant pressure, P = C; hence work is done by the gas, i.e. PdV work (Wg). Wg = PdV = P(V2 – V1) = PV2 – PV1 = n2 RT2 - n1 RT1

= RT (n2 - n1 )

[Assuming temperature constant]

Dn = Amount of H2 gas released = 0.1 kg mole Wg = RT ( D h ) = 8.314

kJ ¥ 300 K ¥ 0.1 kg mole = 249.42 kJ kg mole K

Wn = Net work done = Wb – Wg = 1296 – 249.42 = 1046.58 kJ 1046.58 kJ of work done on the CELL by the charger. EXAMPLE 2.42

An experiment was conducted as shown in the figure.

Duration of the experiment Mass of ideal gas (m) Initial pressure of the gas (P1) Initial temperature (T1) Torque applied (T) Speed (N)

= 2 hours = 1 kg of air = 1 bar = 300 K = 25 N m = 200 rpm

Calculate work done. Solution: Ws = Shaft work =

Ú

2

1

T d = -T(

2

-

1)

Gas

M

= –T × 2pN × duration of experiment =

-25 kN m 200 ¥2 ¥ ¥ 2 ¥ 60 min = - 3769.9 kJ 1000 min

Figure E2.42

EXAMPLE 2.43 Modify Example 2.42 so that, during the work interaction, volume will be 20 times the initial volume at constant pressure. Remaining conditions are same. Calculate the net work interaction. Assume molecular weight of the gas is 29 kg/kg mole. Solution: V2 = V1 × 20 We = Work during the expansion process (P = C)

Ú

= PdV = P1 (V2 - V1 ) = P1 [20V1 - V1 ] = 19 P1V1

Work and Heat

V1 =

\ \

mRT1 mR T1 kN m = = 1 kg ¥ 8.3143 ¥ kg mole K P1 M P1

75

300 K = 0.86 m 3 kg kN 29 × 100 2 kg mole m

kN

¥ 19 ¥ 0.86 m 3 = 1634 kJ m2 Wn = Net work interaction = Ws + We = –3769.9 + 1634 = – 2136 kJ We = 100

This is the work done by the stirrer on the gas (–ve). EXAMPLE 2.44 A steel wire of 5 mm diameter is gradually subjected to an axial force of 10 kN. The length of the bar is 5 m, Young’s modulus (E) = 2.067 × 108 kPa. Calculate the work done during the stretching of the bar. Solution: W =-

Ú

2

1

F = 10 kN, L = 5 m

F dL ;

dL = change in length of the wire = eL W = - FdL = - F L =

W=

F¥F¥L E¥A

È F ˘ Í = strain = ˙ E ¥ A˚ Î

10 2 kN ¥ 5 m ¥ 4 kN 2.067 ¥ 10 ¥ m2 8

W = - 0.1232 kJ

2

Ê 5 ˆ ¥Á m2 Ë 1000 ˜¯

EXAMPLE 2.45 Modify Example 2.44 so that during the stretching of the wire, work is done by the wire at constant pressure of 100 kPa. Calculate the net work interaction. Other conditions are same. Assume wire diameter as 5 mm. Solution: E = 2.067 × 108 kPa, L = 5 m, F = 10 kN, d = 5 mm We have,

=

\

dL F = = L EA

10 kN ¥ 4 kN 2.067 ¥ 10 ¥ m2 8

2

Ê 5 ˆ m2 ¥Á Ë 1000 ˜¯

= 0.002464

dL = eL = 0.002464 × 5 m = 0.01232 m 2

Ê 5 ˆ ¥ 0.01232 = 0.242 ¥ 10 -6 DV = Volume change = A ¥ dL = Á ˜ Ë ¯ 4 1000

Ws = Work done by the wire during the stretching

Ú

= PdV = 100

kN m

2

¥ 0.242 ¥ 10 -6 m 2 = 24.2 ¥ 10 -6 kJ

76

Basic Thermodynamics

\

Wn = net work done = W + Ws = [–0.1232 + 24.2 × 10–6] = – 0.1232 kJ

Ws is too small, hence can be neglected. EXAMPLE 2.46 Two fans (100 watts each), one pump (735 watts) and one heater of 1 kW capacity equipment are housed in a house. Fans run for 6 hours per day. Pump runs for 2 hours per day and heater is used for 1 hour per day. Electricity is charged at the rate of 3 Rs. per unit. What is the monthly bill of that house (Take 1 month = 30 days). Solution: Energy consumed per day = 2F + 1P + 1H = 2 × 100 × 6 + 1 × 735 × 2 + 1 × 1000 × 1 =

3670 watts × hr 3670 watts × hr × 30 days = days days

Energy consumed for one month, i.e. 30 days =

3670 watts ¥ hr ¥ 30 days kW hr = 110.100 days month

We know that 1 kW hr = 1 unit = Rs. 3.00 \

Monthly bill = 110.1 units × 3 = Rs. 330.3

EXAMPLE 2.47 A gas is contained in two cylinders A and B connected by a piston of two different diameters as shown in the figure. The mass of the piston is 10 kg and the gas pressure inside the cylinder ‘A’ is 300 kPa. Calculate the pressure in cylinder B. Solution: Fb = Force acting on the gas in cylinders = Pb × Ab Fa = Force acting on the piston in the cylinder A = Force due to gas pressure on piston in cylinder A + Force due to weight of the piston – Force due to air pressure on piston of cylinder A (excluding stem area).

Figure E2.47

= (P × A)A + Piston Force – (P × A)B = 200 ¥

2 È Ê 100 ˆ 2 Ê 50 ˆ 2 ˘ 100 È 100 ˘ Í ˙ + 10 ¥ 9.81 ¥ -Á Ë 1000 ˜¯ ˙ 4 ÍÎ 1000 ˙˚ 1000 4 Í ÁË 1000 ˜¯ Î ˚

= 0.9967 kN We know that

Fa = Fb

Work and Heat

1.0798 kW = Pb ¥

\

Ê 50 ˆ 4 ÁË 1000 ˜¯

77

2

Pb = 507.678 kPa

EXAMPLE 2.48 The acceleration due to gravity as function of elevation above sea level is given by g = 9.807 – 4.0 × 10–6z, where g is in m/s2 and z is in m. A satellite with a man of 300 kg is boosted 400 km above the earth surface. Calculate the work required to overcome gravity. Solution: g = f [9.807 – 4.0 × 10–6z] For small elevation change dz, dgg = [9.807 – 4.0 × 10–6z] dz Integrating between 0 to 400 km gg =

Ú

400

0

9.807 dz =

= 9.807z ]0

400

Ú

400

0

4 ¥ 10 -6 z dz

- 4 ¥ 10 -6

= 3602800 m

z2 ˘ ˙ 2 ˙˚

400

= 9.807 ¥ 400000 0

4 ¥ 10 -6 ¥ (400000)2 2

m

s2 \ W.D. = Work done by the satellite with a man = m × gg = 300 kg × 3602800 m

m = 10.808 × 108 J or 1080.8 MJ s2

EXAMPLE 2.49 An elastic sphere of one meter diameter contains a gas at 150 kPa. Heating of the sphere increases the pressure to 450 kPa and during this process the pressure of the gas is proportional to the diameter cubed of the sphere. Determine (a) the work done by the gas. If the process is assumed to be of the form PVn = C, find (b) the value of n. (KUD I-96, II-98) Solution: (a) Index of polytropic process (n) D1 = 1 m, Sphere volume \

P1 = 150 kPa, V=

dV =

P1 D13

=

4 R3 = 3

6

¥ 3D 2 dD

P2 D23

P2 = 450 kPa, D3 6

P = D3

78

Basic Thermodynamics

\

D23 =

We have,

P2 3 450 D1 = ¥ (1)3 = 3 m 3 P1 150

P1 È V2 ˘ =Í ˙ P2 Î V1 ˚

n

P1 150 ln P2 = 450 = - 1.0009 n= V2 3 ln ln 1 V1 ln

\ (b) Work done by the gas

W.D. =

Ú

V2

V1

PdV =

Ú

V2

V1

2

KD dV =

È D2 D2 ˘ P1 KÍ 2 - 1 ˙= 2 ÎÍ 6 6 ˙˚ 2 D12 = 314.16 kN m or kJ

=

Ú

D2

D1

D6 ˘ KD D dD = K ˙ 2 2 6 ˙˚ 2

D2

2

kN È9 1˘ Í 6 - 6 ˙ = 150 2 ¥ m Î ˚

D1

(1.5 - 0.167) m 2 2 ¥ 1 m2

EXAMPLE 2.50 A gas confined in a cylinder by a piston at a pressure of 3 × 105 N/m2 and a volume of 12000 cm3 changes its states in such a way that the pressure is inversely proportional to the volume. In this process, if the pressure falls to 1/3 of its initial value, find the magnitude and direction of work flow. (KUD II-93) Solution: P 1 N N ; P1 = 3 ¥ 10 5 2 ; P2 = 1 = 10 5 2 ; PV = C V 3 m m P1V1 = P2V2 P=

V2 P1 = V1 P2

\

This is a constant temperature process V1 = 12000 cm3 Work done (W) W=

Ú

V2

V1

PdV = P1V1 ln

= 3955 Nm

Ê 10 5 ¥ 3 ˆ V2 N = 3 ¥ 10 5 2 ¥ 12000 ¥ 10 -6 m 3 ¥ ln Á ˜ V1 m Ë 10 5 ¯

Work is done by the system

EXAMPLE 2.51 In a piston and cylinder arrangement, the non-flow reversible process is given by V = 200/P m 3, where P is in N/m2. Find (a) the work done when the pressure

Work and Heat

79

increases from 1 × 105 N/m2 to 10 × 10 5 N/m2. Indicate whether the process is compression or expansion. (KUD I-93, II-97) Solution:

V=

200 3 N m , P 2 P m

Work done (W) W=

Ú

V2

V1

PdV =

Ú

V2

V1

V 200 dV = 200 ln 2 V V1

Ê 200 /10 ¥ 10 5 ˆ = 200 ln Á ˜ = - 460.5 J Ë 200 /1 ¥ 10 5 ¯ –ve sign indicates the process is compression. EXAMPLE 2.52 1. Q = +W,

Comment on the following statement. 2. du = –W,

3. du = Q,

4. du = 2W,

5. W = 0

Solution: 1. Q = +W Q = du + W First law of thermodynamics. If du = 0, then Q = +W. This is possible if T2 = T1; it is an isothermal process. 2. du = –W This is possible if Q = 0, i.e. adiabatic process. 3. du = Q This is possible if work done is zero, i.e. gas should expand in the evacuated space where there is no resistance at all or free expansion. 4. du = 2W This is possible if Q = 3W. 5. W = 0 This is possible if dV = 0, i.e. constant volume process. EXAMPLE 2.53 A liquid is irregularly stirred in a container. A rise in temperature of the liquid is observed after some time. If the liquid is considered as the system, answer the following by drawing the boundary of the system. 1. 2. 3. 4.

Has heat been transferred? Has work been transferred? What is the direction of work? Is the process reversible?

Solution: 1. No

2. Yes

3. Surrounding to system

4. Irreversible.

80

Basic Thermodynamics

EXAMPLE 2.54 A piston and cylinder arrangement is containing fluid at 10 × 105 N/m2 and a volume of 0.05 m3. Find the work done by the fluid when it expands reversibly for the following cases (a) (b) (c) (d) (e)

At constant pressure to a final volume of 0.2 m3 According to a law PV = C to a final volume of 0.2 m3 According to a law PV1.3 = C to a final volume of 0.2 m3 At constant volume Sketch all the processes on a PV diagram (KUD I-94)

Solution: (a) W.D. (W) at constant pressure to a final volume at 0.2 m3, P = C W=

Ú

V2

PdV = P(V2 - V1 ) = 10 ¥ 10 5

N

(0.2 - 0.05) m 3 = 150 kJ

2

m (b) W.D. (W) according to a law PV = C, V2 = 0.2 m3 V1

V2 kN 0.2 = 10 ¥ 10 2 2 ¥ 0.05 m 3 ln = 69.3 kJ V1 0.05 V1 m (c) W.D. (W) according to a law PV1.3 = C, V2 = 0.2 m3 We know that W=

Ú

V2

PdV = P1V1 ln

n

1.3

ÈV ˘ kN È 0.05 ˘ P2 = P1 Í 1 ˙ = 1000 2 ¥ Í ˙ m Î 0.2 ˚ Î V2 ˚

= 164.94

kN m2

( P1V1 - P2V2 ) [(1000 ¥ 0.05) - (164.94 ¥ 0.2)] kN 3 = m = 56.71 kJ V1 (n - 1) 0.3 m2 (d) W.D. (W) at constant volume, V = C W=

Ú

V2

PdV =

W=

Ú

V2

V1

PdV = 0

(e) Sketch of all the processes on a PV diagram P n = 0, P=C

n = 1, PV = C n=• V=C

n = 1.3, PV

1.3

=C V

Figure E2.54

Work and Heat

81

EXAMPLE 2.55 State whether the heat, Q, and the work, W, are positive, negative or zero in each of the following cases. The system to be considered is bold. 1. One kilogram of air flows adiabatically from the atmosphere into an evacuated bottle through a valve. 2. Hydrogen and oxygen in a combustible mixture within a non conducting and rigid vessel are ignited by a spark, the gases combine to experience pressure and temperature rise. Solution: 1. Work transfer from atmosphere into the system, \ W = –ve, Adiabatically work enters, \ Q = 0 2. Combustion takes place within the non-conducting vessel, \ Q = 0 Combustion takes place within the rigid vessel, \ W = 0 (V = C) EXAMPLE 2.56 A closed system containing a gas expands slowly in a piston-cylinder from 600 kPa and 0.1 m3 to a final volume of 0.5 m3. Determine the work done if the pressure distribution is determined to be (a) P = C, (b) PV = C, (c) PV1.4 = C, (d) P = –300 V + 630, where P is in kPa and V is in m3. (KUD I-97) Solution: (a) W.D. (W), if P = C

W = P(V2 - V1 ) = 600

kN m2

(0.5 - 0.1) m 3 = 240 kJ

(b) W.D. (W), if PV = C W = P1V1 ln

V2 kN 0.5 = 600 2 ¥ 0.1 m 3 ln = 96.57 kJ V1 0.1 m

(c) W.D. (W), if PV1.4 = C 1.4

ÈV ˘ We know that, P2 = P1 Í 1 ˙ Î V2 ˚ W=

1.4

È 0.1 ˘ = 600 Í ˙ Î 0.5 ˚

= 63.037 kPa

( P1V1 - P2V2 ) [(600 ¥ 0.1) - (630.04 ¥ 0.5)] kN 3 = m = 71.204 kJ (n - 1) (1.4 - 1) m2

(d) W.D. (W), if P = – 300V + 630 W=

Ú

V2

V1

PdV =

Ú

V2

V1

V2

[ -300V + 630] dV = - 150V 2 ˘˚

V1

+ 630V ]V2 V

1

= - 150[V22 - V12 ] +630[V2 – V1] = –150[0.52 – 0.12] + 630[0.5 – 0.1] = 216 kJ EXAMPLE 2.57 The piston of an oil engine of area 0.0045 m2 moves downwards 75 mm drawing in 0.00028 m3 of fresh air from the atmosphere. The pressure in the cylinder is uniform during the process at 80 kPa while the atmosphere pressure is 101.325 kPa. The difference being

82

Basic Thermodynamics

due to the flow resistance in the induction pipe and the inlet valve. Estimate the displacement work done by the air finally in the cylinder. (KUD II-97) Solution:

Figure E2.57

A = Piston cross-sectional area = 0.0045 m2 L = Distance moved = 75 mm V = Volume of fresh air drawn = 0.00028 m3 Pa = Atmospheric pressure = 101.325 kPa Pi = Pressure of air inside the cylinder = 80 kPa Work done by the free air = –ve, free air boundary contracts Work done by the piston = +ve, piston boundary increases Wn = Net work done =

kN

= 0.8 ¥ 100

m

2

Ú PdV (piston) + Ú PdV (free air)

¥ 0.0045 m 2 ¥

75 kN m - 101.325 2 ¥ 0.00028 m 3 = - 0.001371 1000 m

EXAMPLE 2.58 A fluid expands frictionlessly in a closed system from a volume 0.1 m3 to 0.16 m3 in such a manner that the pressure is given by P = C/V2, where C is a constant. The initial pressure is 300 Pa. Calculate the amount of work done. (KUD, March 2001) Solution: P1 = 300 C = 300 W=

Ú

V2

V1

kN m

2

kN m2

; V1 = 0.1 m3; V2 = 0.16 m3; C = P1V12

P 1 2

PV = C

× (0.1)2 m6 = 3 kN m4

PdV =

Ú

V2

V1

V

È1 1˘ È 1˘ 2 dV = C Í- V ˙ = C ÍV - V ˙ 2 V Î ˚V1 2˚ Î 1 C

1 ˘ È 1 = 3 kN m 4 Í ˙ = 11.25 kN m Î 0.1 0.16 ˚

2

Figure E2.58

V

Work and Heat

83

EXAMPLE 2.59 A spherical balloon has a diameter of 25 cm and contains air at a perssure of 1.5 bar. The diameter of the balloon increases to 30 cm due to heating and during this process, the pressure is directly proportional to the diameter. Calculate the work done by air. (KUD, March 2001) Solution: P1 = 1.5 bar,

D1 = 0.25 m, D2 = 0.3 m,

C = 1.5 ¥ 10 5

N m

2

¥

P2 = CD2 = 6 × 105 V2

W=

Ú

W=

C 2

V1

Ú

PdV =

Ú

V2

V1

D3 dD =

P=D

\

P = CD

P1= CD1

1 N = 6 ¥ 10 5 3 0.25 m m

N m

× 0.3 m = 1.8 × 105

3

CD

2

D 2 dD

N m2

È ˘ 3 2 ÍV = 6 D ; dV = 2 D dD ˙ Î ˚

C 6 ¥ 10 5 ¥ ( D24 - D14 ) = 8 8

[0.34 - 0.54 ] = 988.13 J

EXAMPLE 2.60 The surface tension of a spherical weather balloon can be described by the relationship s (N/m) = 10000 + 500A2, where A is the surface area in m2. Determine the total work done by the gas inside the balloon to increase its diameter from 1 to 1.5 m. Solution: Work done by the matter against the surface tension of the material of the balloon [Work done in stretching the balloon, +ve] (Ws) A1 = 4p (0.5)2 = 3.142 m2 A2 = 4p (0.75)2 = 7.1 m2 V1 =

4 (0.5)3 = 0.52 m 3 3

V2 =

4 (0.75)3 = 1.77 m 3 3

ws =

Ú

A2 A1

dA =

Ws = 10000 A]

Ú

7.1 3.14

7.1

3.14

P = 1.01325 bar

10000 dA +

A2 ˘ + 500 ˙ 2 ˙˚

Ú

7.1

3.14

500 A2 dA

7.1

= 92.701 kJ 3.14

Work done against the atmospheric pressure (Wa) Ws =

Ú

V2

V1

PdV = P [V ]0.52 = 101.33[1.77 - 0.52]

Wa = 126.662 kJ

1.77

Figure E2.60

84

Basic Thermodynamics

Total work done by the gas inside the balloon to increase its diameter from 1 to 1.5 m (Wn) Wn = Ws + Wa = 92.701 + 126.662 = 220.363 kJ EXAMPLE 2.61 A battery indicates 12 V at no load. But while being chardged at a rate of 24 A, it requires a charging voltage of 12.3 V. Find the rate at which work is being done on the battery. Solution:

24 A

24 A Battery

Battery

12 A

12 A

No load

With load

Figure E2.61

V = Potential difference I = Current dz = Electrical energy (a) Rate at which work is being done on the battery d w/dt)with load (d d w/dt)no load, (d dw = –V dz I=

\

dz (t = time) dt

dw = –V I dt w = - V I watt dt

(d d w/dt)no load = –12 × 24 = –288 watt (d d w/dt)with load = –12.3 × 24 = –295.2 watt

EXERCISES 2.1 Substantiate the statement “work and heat are path functions”.

(KUD II-93)

2.2 State whether the Q and the work W are +ve, –ve or zero in each of the following cases. The system to be considered is bold. (a) One kilogram of air flows adiabatically from the atmosphere into an evacuated bottle through a valve.

Work and Heat

85

(b) Hydrogen and oxygen in a combustible mixture within a non-conducting and rigid vessel are ignited by a spark. The gases combine to experience pressure and temperature rise. (KUD II-93) 2.3 Explain the following processes. 1. 2. 3. 4.

Isothermal Isobaric Isochoric Isenthalpic

(KUD Dec 94)

2.4 Give thermodynamic definition of work and heat. Establish the relation between the two. (KUD Dec 94) 2.5 Derive an expression for the work done during the following compression processes. 1. 2. 3. 4. 5.

Constant volume Constant pressure Isothermal Isentropic Polytropic

(KUD I-95)

2.6 Explain an example to show that the thermodynamic definition of work is superior to the work as defined in mechanics. (KUD I-96) 2.7 State the similarities between heat and work. Explain an example to illustrate the difference between heat and work. (KUD I-96) 2.8 Starting from a common state point draw the following expansion processes on the PV plane and write down the expression for the work done in each case if the expansion ratio is ‘r’. 1. 2. 3. 4. 5.

Isobaric process Isentropic process Isothermal process Polytropic process of index n > g, where g is the ratio of specific heats Isochoric process

(KUD I-90)

2.9 Give thermodynamic definition of work and heat and mention their common features. (KUD I-93) 2.10 A liquid is irregularly stirred in a container. A rise in temperature of the liquid is observed after some time. If the liquid is considered as the system, answer the following by drawing the boundary of the system. (KUD I-93) 1. 2. 3. 4.

Has heat been transferred? Has work been transferred? What is the direction of work? Is the process reversible?

86

Basic Thermodynamics

2.11 A gas system confined by a piston and cylinder undergoes a change of state such that a product of pressure and volume remains constant. If the process begins at a pressure 3 bar and volume 15000 cm3 and proceeds until the pressure falls to half its initial value, determine the magnitude and direction of the work flow. (KUD I-89) 2.12 A fluid system undergoes a non-flow frictionless process from V1 = 6 m3 to V2 = 2 m3. The relation between the pressure and volume is P = [15/V + 2] N/m2, where V is in m3. Determine the magnitude and direction of work. (KUD I-95) 2.13 Calculate the work done for the following processes if P1 = 5 bar, V1 = 5 m3, 1. 2. 3. 4. 5. 6.

V2 = 25 m3

P=C V=C PV = C PVn = C (n = 1.3) PVg = C (g = 1.4) PV3 = C [Ans: 10000 kJ, 0, 4220 kJ, ____, 2965.6 kJ, 1200 kJ, 7200 kJ]

2.14 A gas contained in a piston and cylinder arrangement expands from 1.5 m3 to 2 m3 and receives 200 kJ of heat from a paddle wheel. The pressure on the gas remains constant at 600 kPa. Calculate the net work done by the system. [Ans: 100 kJ] 2.15 A gas contained in a piston and cylinder arrangement as shown in the figure is at 3 bar. The spring force exerted through the piston is proportional to the volume of gas. An additional atmospheric pressure of 1 bar acts on the spring side piston. Determine the work done by gas in expansion from 0.1 m3 to 0.5 m3. [Ans: 280 kJ]

Gas

Figure Ex. 2.15

2.16 Calculate the work done by a perfect gas during an adiabatic frictionless process when its temperature changes from 200°C to 300°C. The molecular weight of the gas is 36, and the specific heat at constant pressure is given by the expression Cp = 0.3 + 0.000059T. 2.17 Prove that the work required to stretch a wire of length L within the elastic region is given by the equation W = – 0.5 AEL(e)2. A is the cross-sectional area of the wire. E is the Young’s modulus and e, the unit strain. 2.18 A vertical cylinder fitted with a weighted piston contains air. The pressure required to support the piston is found to be 1.5 bar. Determine the work by the system in kJ and kW h. When the cylinder is heated in such a way that the volume changes from 0.17 m3 to 0.51 m3. [Ans: 51 kJ, 14.166 × 10–3 kW h]

Work and Heat

87

2.19 In a piston and cylinder arrangement, 1 kg of air at a pressure of 1 bar and volume 0.2 m3 expands under constant pressure to a volume of 0.8 m3. Then it undergoes a constant volume process in such a manner that under isothermal compression, which follows after constant volume process, the air returns to its initial state. Represent the cycle on a P-V coordinating and determine the net work done by the system. [Ans: 32.274 kJ] 2.20 A shaft rotates at a rate of 100 rpm against a constant torque of 103 N m. Calculate the power required to rotate the shaft. Also calculate the work required to rotate the shaft through 50 revolutions. [Ans: 10.5 kW, 314 kJ] 2.21 14.5 litres of gas at 172 MN/m2 is expanded at constant pressure until its volume becomes 130.5 litres. Determine the work done by the gas. (KUD July 2005) [Ans: 19952 Nm] 2.22 A mass of gas is compressed from 80 kPa and 100 litres to 0.4 MPa and 30 litres. If the pressure and volume of the gas are related by the equation PVn = C, find the work done. (KUD Aug 2001) [Ans: W.D. = –12.422 kJ, n = 1.336]

3 CHAPTER

First Law of Thermodynamics

3.1

JOULE’S EXPERIMENT: EQUIVALENCE OF HEAT AND WORK

J.P. Joule conducted a series of experiments between 1843 and 1848 that led to the formulation of the first law of thermodynamics. Of the several experiments conducted by Joule, the paddle wheel experiment is considered as the classic. Figure 3.1 shows the experimental set-up. A known quantity of fluid has taken in a rigid and insulated container [Figure 3.1(a)]. The initial state of the system is represented by 1 [Figure 3.1(c)]. Two-process cycles are carried out on the system. During the first process 1–2, work is done on the system by means of a paddle wheel. The state of the system was changed to 2. The work done (W1–2) on the system computed in terms of the change in the potential energy, as the weight, mg, fell through a distance h (Z1 – Z2). Due to work done on the system, the temperature of the fluid found to increase. The system is then placed in contact with a water bath [Figure 3.1(b)] and the system is allowed to come to the initial state 1 (process 2–1). Heat is, thus, transferred from the fluid to the water bath during the process 2–1. The amount of heat transfer, Q2–1, from the fluid to the water is estimated in terms of the temperature rise of bath. The original state 1 of the system was reestablished. Here, the system has undergone a two series of processes and completed one thermodynamic cycle. Joule found that the amount of heat rejected (QR) by the fluid was equal to the increase in energy of the water bath and was estimated in terms of rise of temperature. Joule carried out many experiments for a wide variety of work transfer and for different amounts of work. Joule found that the net work done on the system (W1–2) was always proportional to the net heat rejected from the system (Q2–1) irrespective of the type of work and rate of work done. Thus, Joule concluded that during a cyclic process, the net work done on the system was proportional 88

First Law of Thermodynamics

89

Figure 3.1 Process diagram to show Joule’s experiment.

to the net energy rejected as heat from the system. From these experiments, Joule established that heat is a form of energy (prior to Joule, heat was considered to be an invisible fluid flowing from a higher calorie to a lower calorie body).

3.2

FIRST LAW OF THERMODYNAMICS FOR A SYSTEM UNDERGOING A THERMODYNAMIC CYCLE

The results of the Joule experiments lead to the foundation of the first law of thermodynamics. The first law of thermodynamics states, “when a closed system undergoes a cyclic process, the cyclic integral of heat transfer is equal to the cyclic integral of work transfer”. Mathematically, the first law can be written as

Ú

w=

Ú

Q

(3.1)

d represents both work and heat are inexact differentials. The above said law has the following constraints. 1. Applicable to closed systems only. 2. Applicable to a thermodynamic cycle. In Eq. (3.1), units of work and heat are same.

90

Basic Thermodynamics

For different units of heat and work, a proportionality constant called the mechanical equivalent of heat (J) is used. Then the first law of thermodynamics of Eq. (3.1) becomes J where

Ú

Q=

Ú

w

J = 778.17 ft lbs/Btu = 427 kgf m/kCal

3.3

(3.2) (3.3)

FIRST LAW OF THERMODYNAMICS FOR A PROCESS IN A CLOSED SYSTEM [FIRST LAW OF THERMODYNAMICS FOR A NON-CYCLIC PROCESS]

The first law of thermodynamics stated in Eq. (3.1) is applicable to cyclic processes only. There are many situations where the system undergoes a change of state during which both heat and work transfer are involved. Therefore, the first law of thermodynamics for a process is also important. If Q1–2 is the amount of heat transferred to the system and W1–2 is the amount of work transferred from the system during a process 1–2 [Figure 3.1(d)], the net energy transferred (Q1–2 – W1–2) will be stored in the system. E denotes this energy and is called as the stored energy. E is neither heat nor work. Q1–2 – W1–2 = DE

\

(3.4)

where DE is the increase in the energy of the system. Q1–2 = W1–2 + DE = W1–2 + (E2 – E1)

\

(3.5)

This can be shown in a different method. The change in the total energy of the system (DEsys) be compensated by an equal but opposite change in the total energy of the surrondings (DEsur) so that there is no net change in the energy of the universe in any process. The change in the total energy of the surroundings, DEsur, must be equal to the net energy transferred from the surroundings as heat and work. Since Q is the heat transferred to the system from the surroundings and W is the work transferred from the system to the surroundings during the process, then we have (DE)sur = Net energy change of the surrounding = – Q + W For a closed system the total energy change of the system, (DE)sys, is given by (DE)sys = DK.E. + DP.E. + DU We know that

DEsys = –DEsur DK.E. + DP.E. + DU = + Q – W

For a steady non-flow process, DK.E. and DP.E. are neglected. \

DU = Q – W

In differential form du = dQ – dW

First Law of Thermodynamics

91

3.3.1 Components of Stored Energy of a System: Modes of Energy The stored energy, E, is in the form of macroscopic and microscopic. Macroscopic forms of stored energies are kinetic (K.E.) and potential (P.E.). Microscopic forms of energies are associated with the position and motion of molecules (the energy stored within the molecular and atomic strucure of the system, internal energy, I.E.), the structure and arrangement of atoms (chemical energy, C.E.), electrical and magnetic charges, etc. \

E = K.E. + P.E. + I.E. + C.E. + etc.

(3.6)

In the absence of chemical energy and others, Eq. (3.6) can be written as, E = K.E. + P.E. + I.E. dE = d(K.E.) + d(P.E.) + d(I.E)

(3.7)

U represents the internal energy. \

dE = d(K.E.) + d(P.E.) + d(U)

È mV 2 ˘ È mgZ ˘ dE = d Í ˙ +dÍ ˙ + dU ÍÎ 2 gc ˙˚ Î gc ˚ Equation (3.5) can be written as dQ = dW + dE

(3.8)

(3.9)

Substituting Eq. (3.8) in Eq. (3.9) È mV 2 ˘ È mgZ ˘ Q =dÍ ˙ + dÍ ˙ + dU Î gc ˚ ÎÍ 2 gc ˚˙

(3.10)

Ê V 2 - V12 ˆ Ê Z - Z1 ˆ + mg Á 2 + (U 2 - U1 ) Q1- 2 = W1- 2 + m Á 2 ˜ Ë gc ˜¯ Ë 2 gc ¯

(3.11)

\ On integration,

If we neglect K.E. and P.E. changes, then dQ = dW + dU Q = W + DU Q1–2 = W1–2 + (U2 – U1)

(3.12)

For specific internal energy, q1–2 = w1–2 + u2 – u1 In the absence of K.E., P.E.; chemical energy, electrical and magnetic charges, etc., the stored energy is called as internal energy (U). It can be shown that internal energy has a definite value at every state of a system and is, therefore, a property of the system.

3.4

INTERNAL ENERGY—A PROPERTY OF THE SYSTEM

Consider a system that undergoes a cyclic process from state 1 to state 2 by the path A and returns to the state 1 by two paths B and C as shown in Figure 3.2.

92

Basic Thermodynamics

First, consider cyclic processes, 1–A–2–B–1 Apply the first law of thermodynamics Eq. (3.1) to this cyclic process.

Ú

2A

Q+

1A

Ú

1B

Q=

2B

Ú

2A

W+

1A

Ú

1B

1

(3.13) A

Now consider another cyclic process 1–A–2–C–1, similarly

Ú

2A

Q+

1A

Ú

1C

Q=

2C

Ú

2A

W+

1A

Ú

2

V

1C

W

2C

(3.14) Figure 3.2

Subtracting Eq. (3.14) from Eq. (3.13)

Ú

1B

2B

Ú

1B

2B

QQ-

Ú

1C

2C

Ú

1B

2B

Q=

Ú

W=

Ú

1B

2B

1C

2C

WQ-

C B

W

2B

P

Ú

1C

W

2C

Ú

1C

2C

Internal energy—a property of the system.

W or

Ú

1B

2B

( Q - W) =

Ú

1C

2C

( Q - W)

(3.15)

The result of Eq. (3.15) is of great significance. Since return processes B and C are arbitrarily selected between states 2 and 1, the quantity (dQ – dW) remains the same for all the processes between the two states. Thus, (dQ – dW) is always the same irrespective of the path followed for a given change of state and depends only on the initial and final states and not on the path followed between the two states. Therefore, (dQ – dW) is a point function and is a property of the system. This property is known as stored energy. Particularly for a closed system, stored energy is called as internal energy (because K.E., P.E., etc. are neglected for a closed system). Substitute Eq. (3.12) in Eq. (3.15),

Ú

1B

2B

3.5

dU =

Ú

1C

2C

dU

ENTHALPY (H)

The total enthalpy of a substance is given by, H = U + PV

(3.16)

It is an extensive property. Substitute W= PdV in Eq. (3.12), \

dQ = dU + PdV

At constant pressure, dQ = dU + d(PV)

[PdV = d(PV)]

= d[U + PV] dQ = dH Q = DH

(3.17)

First Law of Thermodynamics

93

For an ideal gas, the enthalpy is given by H = U + RT

(3.18)

Since U for an ideal gas is a function of temperature, then Eq. (3.18) becomes H = f(T)

3.6

(3.19)

SPECIFIC HEAT (C)

It is the amount of heat required to raise the temperature of a unit mass of the substance by one degree. Q˘ È q ˘ È1 C=Í ¥ ˙=Í ˙ Î m dT ˚ Î dT ˚

\

(3.20)

dq is a path function, hence C is also a path function. However, specific heat of constant volume and constant pressure processes are point functions and hence thermodynamic property.

3.6.1 Specific Heat at Constant Volume (Cv) It is defined as the heat required to raise the temperature of 1 kg of the substance by one degree, keeping the volume constant. \

PdV = 0

[Constant volume, dV = 0]

Substitute PdV = 0 in Eq. (3.12), dQ = dU

(3.21)

Substitute Eq. (3.21) in Eq. (3.20),

1 m

Cv =

Ú

1

1

du =

Ú

2

1

Du = Cv

È dU ˘ È du ˘ Í dT ˙ = Í dT ˙ Î ˚V Î ˚V Cv dT

Ú

2

1

dT

Du = CvDT

(3.22)

u2 – u1= Cv(T2 – T1)

3.6.2 Specific Heat at Constant Pressure (Cp) It is defined as the heat required to raise the temperature of 1 kg of the substance by one degree, keeping the pressure constant. Substitute Eq. (3.17) in Eq. (3.20), \

Cp =

1 È dH ˘ È dh ˘ = m ÍÎ dT ˙˚ p ÍÎ dT ˙˚ p

94

Basic Thermodynamics

dh = CpdT

Ú

1

1

dh =

Ú

2

1

(3.23)

C p dT

Dh = C p

Ú

2

1

dT

Dh = CpDT

(3.24)

(h2 – h1) = Cp(T2 – T1) Equations (3.22) and (3.24) are essentially used for gases.

3.6.3 Specific Heat of Solids and Liquids In case of solids and liquids, specific volume is very small, so its change is neglected. Then, d(PV) is negligible, i.e. d(PV) ª 0. dH = dU + d(PV) ∫ dU From Eqs. (3.22) and (3.24), Cp[T2 – T1] = Cv[T2 – T1] \

Cp = Cv = C

Specific heat (C) for solids and liquids in any process, dq = dh = du = C dT q = dh = Du = C DT

3.7

(3.25)

FIRST LAW TO A CLOSED SYSTEM FOR DIFFERENT PROCESSES

3.7.1 The Constant Volume (Isochoric) Process Work done (W1–2) We know that for constant volume process, W1- 2 =

Ú

2

1

PdV

Heat transfer (Q1–2), dQ = dW + dU

P System boundary

2

Gas (System)

Q = W + DU

1

Q = 0 + (U2 – U1) Q = [U2 – U1] q = (u2 – u1)

Heat or cool

V1 = V2

Figure 3.3 Constant volume process.

q = Cv [T2 – T1] [Du = Cv DT]

(3.26)

95

First Law of Thermodynamics

3.7.2 The Constant Pressure (Isobaric) Process Work done (W1–2) We know that for constant pressure process, W1–2 = P(V2 – V1)

P W P1 = P2

Piston

Heat transfer (Q1–2), dQ = dW + dU = W1–2 + (U2 – U1)

V1

Heat or cool

Q1–2 = H2 – H1

2

Gas (System)

= P(V2 – V1) + (U2 – U1) = (U2 + P2V2) – (U1 + P1V1)

1

V2

V

Figure 3.4 Constant pressure process.

For perfect gas, Q1–2 = mCp(T2 – T1)

(3.27)

3.7.3 The Constant Temperature Process (Isothermal) Work done (W1–2) We know that work done for constant temperature process, W1- 2 = P1V1 ln

V2 V = mRT1 ln 2 V1 V1

Heat transfer (Q1–2), dQ = dW + dU Q1–2 = W1–2 + DU = P1V1 ln

V2 + mCc (T2 - T1 ) V1

Figure 3.5 Constant temperature process.

Constant temperature process, T2 = T1, \

Q1–2 = P1V1 ln V2 = W1–2

(3.28)

3.7.4 Adiabatic Process or Isentropic Process Heat transfer (Q1–2), Q1–2 = 0 dQ = dW + dU 0 = PdV + mCv dT Ideal gas equation PV = mRT

P 1

PV = C

(3.29) 2

On differentiation, PdV + VdP = mRdT

V

Figure 3.6 Isentropic process.

96

Basic Thermodynamics

dT =

( PdV + VdP ) mR

(3.30)

Substitute Eq. (3.30) in Eq. (3.29),

Cv ( PdV + VdP) R R PdV + Cv (PdV + VdP) = 0

\

PdV +

(Cp – Cv) PdV + CvPdV + CvVdP = 0 Cp PdV – CvPdV + Cv PdV + CvVdP = 0 Cp PdV + CvVdP = 0 Divide by Cv,

Cp Cv

PdV +

Cv VdP Cv

i.e. g PdV + VdP = 0

Divide by PV, PdV VdP + =0 PV PV

i.e.

dV dP + =0 V P

On integration, g ln V + ln P = ln (constant) = constant PVg = econstant = constant For adiabatic process, P1V1 = P2V2 Work done for isentropic process, W1- 2 =

( P1V1 - P2V2 ) -1

Heat transfer (Q1–2), dQ = dW + dU 0 = dW + dU W1–2 = – (U2 – U1) = (U1 – U2)

3.7.5 Polytropic Process (PV n = C) Work done for polytropic process, W1- 2 =

mR (T1 - T2 ) n -1

Change in internal energy (DU) (U2 – U1) = mCv (T2 – T1)

(3.31)

First Law of Thermodynamics

97

Heat transfer (Q1–2), dQ = dW + dU Q1- 2 = W1- 2 + (U 2 - U1 ) =

mR (T1 - T2 ) mR (T2 - T1 ) + n -1 -1

È 1 = mR (T1 - T2 ) Í În -1 =

È Í∵ Cv = Î

R ˘ ˙ - 1˚

1 ˘ ( - n) mR (T1 - T2 ) ˙= - 1˚ ( - 1 (n - 1)

(3.32)

( - n) ( - n) ¥ W1- 2 = ¥ Polytropic work done ( - 1) ( - 1)

(3.33)

PVT Relations P1V1n = P2V2n

(3.34) n

P 1

1/ n

ÈP ˘ P2 È V1 ˘ V =Í ˙ ; 1 =Í 2˙ P1 Î V2 ˚ V2 Î P1 ˚ Ideal gas equation,

n

PV = C

(3.35) 2

P1V1 P2V2 = T1 T2

(3.36) V

Figure 3.7 Polytropic process.

P2 V1T2 = P1 V2 T1

(3.37)

V1 P2 T1 = V2 P1T2

(3.38)

Equating Eqs. (3.35) and (3.37), n

È V1 ˘ V1T2 Í ˙ = V2T1 Î V2 ˚ Equating Eqs. (3.35) and (3.38), 1/ n

È P2 ˘ Í ˙ Î P1 ˚

3.8

PT = 2 1 P1T2

ÈT ˘ È V ˘ \ Í 2˙=Í 1˙ Î T1 ˚ Î V2 ˚

n -1

T ÈP ˘ \ 1 =Í 1˙ T2 Î P2 ˚

n -1 n

(3.39)

APPLICATION OF FIRST LAW—STEADY FLOW ENERGY EQUATION

Control volume concept is used to open systems. Control volume is any volume, fixed in space, fixed shape and of fixed volume relative to the observer but mass crosses its boundary in addition to heat and work interactions. dQ = dW + dE

(3.40)

98

Basic Thermodynamics

Internal energy: Each unit mass of the fluid brings with it a certain amount of internal energy, Ui. The incoming internal energy is mui. Similarly, for outgoing, it is mue. Potential energy: The fluid has a potential energy above the reference plane equal to the work done against gravity in raising it to a height, Zi. The potential energy of the fluid at section ‘i’ is mgzi. Similarly, for outgoing, it is mgze. Kinetic energy: Each unit mass of the fluid brings with it a certain amount of kinetic energy due to the velocity at which it enters into the system, (V )2 . Its value is

m(Vi )2 . Similarly, for outgoing, 2

m(Ve )2 . 2 Entrance work or flow work or flow energy: Each unit mass of the fluid at the entry is being pushed by the fluid behind it by a certain force. This force multiplied by the distance through which the force acts is equal to the flow energy.

it is

Force on the fluid = P × A Entrance work = Pi Ai × distance = Pi ui The flow energy at the inlet is mpiui. Similarly, at the outlet, it is mpeue. Let Dmi, Dme = Small masses entering and leaving the control volume, respectively during time Dt for the process dW = Work done by the system = dWd + dWs dWd = Displacement work dWs = Shaft work V = Velocity dE = Total energy for this closed system = Energy of the contents of the CV + Flow energy gc =

1 kg m N s2

Figure 3.8 The first law of thermodynamics applied to open system (control volume).

99

First Law of Thermodynamics

The heat transfer dQ to the system in time period Dt is given by, dQ = QDt

(3.41)

The shaft work dWs to the system in time period Dt is given by, dWs = Ws Dt

(3.42)

The displacement work dWd is given by, dWd = (Dme)Pene – (Dmi)Pini dW = dWs + dWd = Ws Dt + (Dme) Pene – (Dmi) Pini

(3.43)

È (Vi )2 gZ ˘ + (ECV)2 = (ECV)t+Dt = Energy of the contents of the CV at 2 = m2 Íu + ˙ 2 gc gc ˙˚ ÎÍ 2 È (Vi )2 gZ ˘ + (ECV)1 = (ECV)t = Energy of the contents of the CV at 1 = m1 Í u + ˙ 2 gc gc ˙˚ ÎÍ 1

(ECV)2 – (ECV)1 = Stored energy in the control volume È (V )2 gZ ˘ DEi = Energy entering with the mass (Dmi) = Dmi Íui + i + i ˙ 2 gc gc ˙˚ ÍÎ È (V )2 gZ ˘ DEe = Energy leaving with the mass (Dme) = Dme Íue + e + e ˙ 2 gc gc ˙˚ ÍÎ ÏÔ È È (V )2 gZ ˘ ¸Ô (V )2 gZ ˘ dE = Ì( ECV )2 + Dme Íue + e + e ˙ ˝ ( ECV )1 + Dmi Íui + i + i ˙ 2 gc gc ˙˚ Ô 2 gc gc ˙˚ ÔÓ ÎÍ ÎÍ 1˛

(3.44)

Substitute Eqs. (3.41), (3.42), (3.43) and (3.44) in Eq. (3.40) QDt – (dWs + dWd) = dE È È (V )2 gZ ˘ (V )2 gZ ˘ QDt – (dWs + dWd) = (ECV)2 – (ECV)1 + Dme Íue + e + e ˙ - Dmi Íui + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ QDt – Ws Dt – [(Dme) Pene – (Dmi)Pini] È È (V )2 gZ ˘ (V )2 gZ ˘ = ( ECV )2 - ( ECV )1 + Dme Íue + e + e ˙ - Dmi Íui + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ QDt – WsDt = (Dme)Pene – (Dmi)Pini + (ECV)2 – (ECV)1 È È (V )2 gZ ˘ (V )2 gZ ˘ + Dme Íue + e + e ˙ - Dmi Íui + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ È (V )2 gZ = ( ECV )2 - ( ECV )1 + Dme Íue + e + e + Pe 2 gc gc ÍÎ

˘

È (V )2 gZ - Dmi Íui + i + i + Pi 2 gc gc ˙˚ ÍÎ

e˙

˘

i˙

˙˚

100

Basic Thermodynamics

Q - Ws =

È È dECV (V )2 gZ ˘ (V )2 gZ ˘ + me Í he + e + e ˙ - mi Í hi + i + i ˙ dt 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ

(3.45)

Equation (3.45) is used for unsteady flow and for single (inlet and outlet) stream purpose. For multi-streams, Q - Ws =

dECV + dt

È È (V )2 gZ ˘ (V )2 gZ ˘ me Í he + e + e ˙ mi Í hi + i + i ˙ e i 2 gc 2 gc gc ˙˚ gc ˙˚ ÎÍ ÎÍ all streams all streams

Â

Â

(3.46) Multiplying the term (dECV/dt) in Eq. (3.45) by dt and integrating between states 1 and 2, we obtain, È Ê Ê (V )2 gZ ˆ (V )2 gZ ˆ ˘ ˙ Q - Ws = Í m2 Á u + + - m1 Á u + + ˜ 2 gc 2 gc gc ¯ gc ˜¯ ˙ ÍÎ Ë Ë 2 ˚

È Ê Ê (V )2 gZ ˆ (V )2 gZ ˆ ˘ + Í me Á he + e + e ˜ - mi Á hi + i + i ˜ ˙ 2 gc 2 gc gc ¯ gc ¯ ˙˚ Ë ÍÎ Ë

3.9

(3.47)

STEADY-STATE, STEADY FLOW PROCESS

The term steady state means no variation with time occurs for any property at any point in the control volume. As per the definition of steady state, the following assumptions are considered. 1. No accumulation of matter in the control volume. mi = me = m

Â

e all streams

=

Â

i all streams

(3.48)

=

Â

m all streams

2. The state of the fluid at any point in the control volume remains constant.

È dECV 3. The energy accumulation rate for the control volume must be zero, i.e. Í Î dt Substitute conditions in Eqs. (3.45) and (3.46).

˘ ˙ = 0. ˚ (3.49)

3.9.1 SFEE for One Stream È È (V )2 gZ ˘ (V )2 gZ ˘ Q - Ws = m Í he + e + e ˙ - m Í hi + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ

(3.50)

First Law of Thermodynamics

101

SFEE for multi-stream Q - Ws =

È È (V )2 gZ ˘ (V )2 gZ ˘ me Í he + e + e ˙ mi Í hi + i + i ˙ e i 2 gc 2 gc gc ˙˚ gc ˙˚ ÍÎ ÍÎ all streams all streams

Â

Â

3.10 APPLICATIONS OF STEADY FLOW ENERGY EQUATION 3.10.1

Work Absorbing Systems

3.10.1.1 Rotary compressor Applying SFEE for single stream, Eq. 3.50 and 1 for inlet and 2 for exit È È (V )2 gZ ˘ (V )2 gZ ˘ Q = m Í he + e + e ˙ - m Í hi + i + i ˙ + Ws 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ È È (V )2 ˘ (V )2 ˘ Q = m Í he + e ˙ - m Í hi + i ˙ + Ws 2 gc ˙˚ 2 gc ˙˚ ÍÎ ÍÎ Control surface Q=0

Ws

Work is done on the system Adiabatically covered, Q = 0

Adiabatic shield

2 Ws

Potential energy neglected

1

m

m

Figure 3.9 Adiabatically covered rotary compressor.

Here Ws is the work done on the system, hence –ve. È È (V )2 ˘ (V )2 ˘ 0 = m Í h2 + 2 ˙ - m Í h1 + 1 ˙ + Ws 2 gc ˙˚ 2 gc ˙˚ ÍÎ ÍÎ \

È (V )2 (V )2 ˘ Ws = W1- 2 = m Í h2 - h1 + 2 - 1 ˙ 2 gc 2 gc ˚˙ ÎÍ

If K.E. is neglected, Ws = W1- 2 = m [ h2 - h1 ]

(3.51)

102

Basic Thermodynamics

If rotary compressor is not adiabatically covered, then applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit,

Q

Q = –ve, heat lost to cooling water, i e. surroundings W = –ve, work done on the system, K.E. and P.E. changes are neglected

2 Ws

1

-Q1- 2 = m [ h2 - h1 ] - W1- 2

m

W1- 2 = Ws = m [ h2 - h1 ] + Q1- 2

Figure 3.10 Non-adiabatically covered rotary compressor.

3.10.1.2 Blowers Here, DZ = 0, PV = C, Q = 0 (isentropic process), Ws = –ve Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, \

È È (V )2 gZ ˘ (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ gc ˙˚ 2 gc 2 gc ÍÎ ÍÎ È È (V )2 gZ ˘ (V )2 gZ ˘ 0 + m Íu1 + P1V1 + 1 + 1 ˙ = m Íu2 + P2V2 + 2 + 2 ˙ + Ws 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ È È ˘ (V )2 ˘ (V )2 m Íu1 + 1 ˙ + m Íu2 + 2 - Ws ˙ 2 gc ˙˚ 2 gc ÍÎ ÍÎ ˙˚

È (V )2 (V )2 ˘ Ws = m Íu2 - u1 + 2 - 1 ˙ 2 gc 2 gc ˙˚ ÍÎ For fans and blowers, the rise in temperature is small, \

\

(u2 – u1) = Cv(T2 – T1) = 0

\

Ws =

(V2 )2 - (V1 )2 2 gc

Inlet velocity is usually neglected, \

Ws = W1- 2 =

m

(V2 )2 2 gc

3.10.1.3 Reciprocating compressor In this case, K.E. and P.E. changes are neglected, heat, Q, is –ve, work is –ve. Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, È (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÎÍ

First Law of Thermodynamics 1

2

m m

Receiver

1

2 Compressor Control surface Q

Ws

Figure 3.11 Reciprocating compressor.

È (V )2 gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÍÎ – Q + m[h1] = m[h2] – Ws Ws = m[h2 – h1] + Q1–2 3.10.1.4 Centrifugal water pump In this case,

Q = 0, Du = 0, n1 = n 2

Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, È È (V )2 gZ ˘ (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc gc ˙˚ 2 gc ÍÎ ÍÎ

Water out

Z1

Centrifugal pump

Ws

0

Z1

Water in

Control volume surface

Figure 3.12 Centrifugal water pump.

103

104

Basic Thermodynamics

È 0 + m Íu1 + P1 ÍÎ È m Í P1 ÍÎ \

3.10.2

1

1

+

È (V1 )2 gZ1 ˘ + ˙ = m Íu2 + P2 2 gc gc ˙˚ ÍÎ

+

È (V1 )2 gZ1 ˘ + ˙ = m Í P2 gc ˙˚ 2 gc ÍÎ

È Ws = m Í( P2 ÎÍ

2

- P1 1 ) +

2

+

2

+

(V2 )2 gZ 2 ˘ + ˙ + Ws gc ˙˚ 2 gc

(V2 )2 gZ 2 ˘ + ˙ + Ws gc ˙˚ 2 gc

((V2 )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + ˙ 2 gc gc ˚˙

Non-work Developing and Non-work Absorbing Systems

3.10.2.1 Boilers In this case, DZ = 0, DK.E. = 0, Ws = 0 Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit,

2

Control Boiler

Q + m[h1] = m[h2] \

2 Steam out

Water in

Q = m[h2 – h1] Q

Figure 3.13 Boiler.

3.10.2.2 Condenser Here, the system is insulated, Q = 0, Ws = 0, this is a type of heat exchanger.

Steam in

Water

Heat lost by steam = Heat gained by water ms(hs1 – hs2) = mw(hw2 – hw1)

Q=0

m s and m w are flow rate of steam and water, respectively. Suffix s and w represent steam and water, respectively.

Control

Water in

Condensate out

Figure 3.14 Condenser.

3.10.3

Work Developing System

3.10.3.1 Turbine (Steam and Gas) In this case, shaft work done by the system, Ws = W1–2 = +ve, adiabatically covered, Q = 0, changes in P.E. and K.E. are neglected, i.e. DK.E. = 0, DP.E. = 0

105

First Law of Thermodynamics

Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit,

Insulated (adiabatic)

Control surface

Ws

È (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÍÎ

Q=0

Turbine

È (V )2 gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÍÎ 0 + m[h1] = m[h2] + Ws \

1

Ws = m[h1 – h2]

2

Figure 3.15

3.10.3.2 I.C. Engine Here, shaft work done by the system, Ws = +ve, adiabatically covered, Q = 0. Changes in P.E. and K.E. are neglected, i.e. DK.E. = 0, DP.E. = 0 Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit,

1

2 Out

In 2

1

Cylinder

Q=0

È (V ) gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÎÍ 2

Adiabatically covered

Ws Piston

Control volume surface

È (V ) gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÎÍ 0 + m[h1] = m[h2] + Ws 2

\

Figure 3.16 I.C. Engine.

Ws = m[h1 – h2]

3.10.3.3 Hydraulic Turbine In this case, shaft work done by the system, Ws = +ve, heat transfer, Q = 0. Stored water (Dam)

Z1

Water in

Ws

] Z2

Turbine.

Water out

Tail race

Figure 3.17

Hydraulic turbine.

106

Basic Thermodynamics

Changes in temperature of water is neglected, hence Du = 0. Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, È (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÍÎ È (V )2 gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÎÍ È Q + m Í P1 ÍÎ

1

+

(V1 )2 gZ1 ˘ + ˙ gc ˙˚ 2 gc

È = m Í P2 ÎÍ

2

+

\

È Ws = m Í( P1 ÍÎ

1

- P2

\

ÈÊ P P ˆ ((V )2 - (V2 )2 ) g( Z1 - Z 2 ) ˘ Ws = m Í Á 1 - 2 ˜ + 1 + ˙ gc 2 gc 2¯ ÎÍ Ë 1 ˚˙

3.10.4

(V2 )2 gZ 2 ˘ + ˙ + Ws gc ˙˚ 2 gc 2)

+

((V1 )2 + (V2 )2 ) g( Z1 - Z 2 ) ˘ + ˙ 2 gc gc ˙˚

Some Other Systems

3.10.4.1 Steam Nozzle In this case, no work done, W s = 0, adiabatically covered, Q = 0. Changes in P.E. i.e. DP.E. = 0 Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, È (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÍÎ

Control surface 2 Adiabatic covered

Nozzle

in

È (V )2 gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÎÍ \

È (V )2 - (V2 )2 ˘ m (h1 - h2 ) + m Í 1 ˙=0 2 gc ÎÍ ˚˙

\

(V2 )2 + (V1 )2 + 2(h1 - h2 ) ¥ gc ¥ 1000

Q=0

out

Ws = 0 2

Figure 3.18 Steam nozzle.

V2 = (V1 )2 + 2(h1 - h2 ) ¥ gc ¥ 1000 If inlet velocity of the steam is neglected, V2 = 2( h1 - h2 ) ¥ gc ¥ 1000

First Law of Thermodynamics

107

3.10.4.2 Throttling process In this case, shaft work done, Ws = 0, adiabatically covered, Q = 0. Changes in P.E. and K.E. are neglected, i.e. DK.E. = 0 and DP.E. = 0 Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, Thermometer

È (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ gc ˙˚ 2 gc ÎÍ È (V )2 gZ ˘ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ 2 gc ÍÎ m[h1] = m[h2] \

1

1 Fluid out

Fluid in

h1 = h 2

Control volume surface

T1 = T2 for gases only

Insulation

Figure 3.19 Throttling process.

3.10.4.3 Flow through pipe In this case, shaft work done, Ws = 0, adiabatically covered, Q = 0. Changes in temperature are neglected, i.e. DT = 0, \

Du = 0

Applying SFEE for single stream, Eq. (3.50) and 1 for inlet and 2 for exit, È È (V )2 gZ ˘ (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ = m Í h2 + 2 + 2 ˙ + Ws gc ˙˚ gc ˙˚ 2 gc 2 gc ÎÍ ÎÍ È 0 + m Í P1 ÍÎ

1

+

È (V1 )2 gZ1 ˘ + ˙ = m Í P2 2 gc gc ˙˚ ÍÎ

2

+

(V2 )2 gZ 2 ˘ + ˙+0 2 gc gc ˙˚ Figure 3.20 Flow throughout pipe.

È P1 (V1 )2 gZ1 ˘ È P2 (V2 )2 gZ 2 ˘ + + + Í + ˙=Í ˙ 2 gc gc ˙˚ ÎÍ 2 2 gc gc ˚˙ ÎÍ 1 This is the same as Bernouli’s equation for fluid flow. If the fluid is incompressible like water, r1 = r2 = r

È P1 (V1 )2 gZ1 ˘ È P2 (V2 )2 gZ 2 ˘ + + Í + ˙=Í + ˙ 2 gc gc ˙˚ ÍÎ 2 gc gc ˙˚ ÍÎ During the flow of fluid through pipe, there is always loss due to friction (hf), if this friction is taken into account, then the above equation, È P1 (V1 )2 gZ1 ˘ È P2 (V2 )2 gZ 2 ˘ + + Í + ˙ - hf = Í + ˙ gc ˙˚ gc ˙˚ 2 gc 2 gc ÍÎ ÍÎ

108

Basic Thermodynamics

3.11 TRANSIENT ANALYSIS OR UNSTEADY FLOW PROCESS ANALYSIS Many engineering applications involve a transient, or time varying, analysis. Whenever flow conditions are transient, the mass of the system can change with respect to time. For example, a filling or emptying a tank is an unsteady (transient) flow process in which both internal energy and mass change with respect to time. Only two cases of the following will be discussed. (a) Tank filling process (b) Tank emptying process

3.11.1

Tank Filling Process

[Flow of fluid from a large reservoir (container) into a rigid container.] Let subscript 1 represent initial and subscript 2 represent final condition of the tank. Subscript e represents condition of the fluid in the main pipe line or large reservoir. m1, P1, n1, T1, u1—Initial mass, pressure, specific volume, temperature and specific internal energy of the fluid in the tank. m2, P2, n2, T2, u2—Final mass, pressure, specific volume, temperature and specific internal energy of the fluid in the tank. Pi, ni, Ti, Ci, ui, hi—Pressure, specific volume, temperature, velocity, specific internal energy and specific enthalpy of the entering fluid (entering fluid into tank). m2 – m1 = Mass of fluid entering into the tank. Ei = Total specific energy of the fluid entering (fluid inlet). [Properties and energies essentially remain constant because fluid entering into the tank (small) from a large reservoir.] Ê Ei = Á ui + Pi Ë

i

+

Ci2 gZ ˆ + i˜ 2 gc gc ¯

Q1–2 and W1–2 = Heat and work interactions. Applying the conservation of energy, (m2 – m1)Ei + Q1–2 = W1–2 + m2u2 – m1u1 Main pipe line (Large reservoir)

i

Q

i

Tank m1, P1, T1, V1

Figure 3.21 Tank filling process.

Control surface

(3.52)

First Law of Thermodynamics

109

If tank is insulated and rigid, then Q1–2 and W1–2 = 0 (m2 – m1)Ei = m2u2 – m1u1 Ê (m2 - m1 ) Á ui + Pi Ë

i

+

Ci2 gZ ˆ + i ˜ = m2 u2 - m1u1 2 gc gc ¯

If K.E. and P.E. of entering fluid are neglected, then (m2 – m1) (ui + Piui) = m2u2 – m1u1 If the tank is initially empty, i.e. m1 = 0, then m2(ui + Piui) = m2 u2 \

hi = u2

(3.53)

If the fluid is an ideal gas, the temperature of the fluid after filling the tank is, C pi Ti = Cv2 T2 \

T2 =

C pi Ti Cv2

(3.54)

Or Eq. (3.45) can be used for unsteady flow process, i.e. È È (V )2 gZ ˘ (V )2 gZ ˘ Q - Ws = m2 Í u + + + ˙ - m1 Í u + ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ Í ÎÍ Î 2 1

È È (V )2 gZ ˘ (V )2 gZ ˘ + me Í he + e + e ˙ - mi Í hi + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ If tank is insulated and rigid, i.e. Q = 0, Ws = 0. If mass (me) is not leaving the system, i.e. me = 0 then the above equation reduces to, È È È (Vi )2 gZ i ˘ (V )2 gZ ˘ (V )2 gZ ˘ 0 = m2 Íu + + + + ˙ ˙ - m1 Íu + ˙ - mi Í hi + 2 gc gc ˙˚ 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ ÎÍ 2 1 1

(3.55a)

If K.E. and P.E. of entering fluid are neglected, then we have, È È (V )2 gZ ˘ (V )2 gZ ˘ 0 = m2 Í u + + m u + + ˙ ˙ - mi [ hi ] 1Í 2 gc gc ˙˚ 2 gc gc ˙˚ Í ÎÍ Î 2 1

(3.55b)

If K.E. and P.E. of contents of the tank are neglected at states 1 and 2, then the above equation reduces to 0 = m2[u2] – m1[u1] – mi[hi]

(3.55c)

If the tank is initially empty, i.e. m1 ª 0, 0 = m2[u2] – mi[hi]

(3.55d)

110

Basic Thermodynamics

Mass of entering fluid (mi) is equal to the mass of the fluid (m2) retained in the tank. \

m2 = mi

Hence the above equation reduces to u2 = hi

(3.55e)

Therefore, Eqs. (3.53) and (3.55e) are same.

3.11.2

Tank Discharging Process

Here, fluid flows from tank to the main pipe line or surroundings. The main pipe line or surroundings are considered as sufficiently large compared to the tank being emptied. Hence, specific energy of the fluid at the exit of the control volume (tank) remains constant. \

(m1 – m2) Ee = Total specific energy of the fluid leaving the control volume or entering into the surroundings [Properties, energies essentially remain constant] Ê = Á ue + Pe Ë

e

+

Ce2 gZ ˆ + e˜ gc ¯ 2 gc

Applying the conservation of energy, \

(m1 – m2) = mass of fluid leaving the tank

\

(m1 – m2)Ee + Q1–2 = W1–2 + m1u1 – m2u2

If tank is insulated and rigid, then Q1–2 and W1–2 = 0 \

(m1 – m2)Ee = m1u1 – m2u2

Ê (m1 - m2 ) Á ue + Pe Ë If K.E. and P.E. are neglected, then

e

+

Ce2 gZ ˆ + e ˜ = m1u1 - m2 u2 2 gc gc ¯

(m1 – m2) [ue + Peue] = (m1u1 – m2u2) (m1 – m2)he = m1u1 – m2u2

Figure 3.22 Tank emptying process.

(3.56)

111

First Law of Thermodynamics

If tank is completely emptied, then m2 = 0 m1he = m1u1 \

he = u1

(3.57)

Equation (3.45) can be used for unsteady flow process, i.e. È È (V )2 gZ ˘ (V )2 gZ ˘ Q - Ws = m2 Í u + + + ˙ - m1 Í u + ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ 2 1

È È (V )2 gZ ˘ (V )2 gZ ˘ + me Í he + e + e ˙ - mi Í hi + i + i ˙ 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ If tank is insulated and rigid, i.e. Q = 0, Ws = 0, If mass (mi) is not entering into the system, then mi = 0. The above equation reduces to, È È È (Ve )2 gZ e ˘ (V )2 gZ ˘ (V )2 gZ ˘ 0 = m2 Íu + + + + ˙ ˙ - m1 Íu + ˙ + me Í he + 2 gc gc ˙˚ 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ ÍÎ 2 1

(3.58a)

If K.E. and P.E. of leaving fluid are neglected, then we have, È È (V )2 gZ ˘ (V )2 gZ ˘ 0 = m2 Í u + + + ˙ - m1 Í u + ˙ + me [ he ] 2 gc gc ˙˚ 2 gc gc ˙˚ ÍÎ ÍÎ 2 1

(3.58b)

If K.E. and P.E. of contents of the tank are neglected at states 1 and 2, then the above equation reduces to, (3.58c) 0 = m2[u2] – m1[u1] + me[he] If the tank is empty at state point 2, i.e. m2 ª 0 0 = – m1[u1] – me[he]

(3.58d)

Mass of leaving fluid (me) is equal to the mass of the fluid (m1) in the tank at state 1. \

m1 = m e

Hence, the above equation reduces to, u1 = h e

(3.58e)

Therefore, Eqs. (3.57) and (3.58e) are same.

3.12 ANALYSIS OF THE OPEN SYSTEM FOR DIFFERENT PROCESSES Refer to Section 8.9 (page 384).

SOLVED EXAMPLES EXAMPLE 3.1 The indicator diagram as shown in the figure is a quasi-static process carried out on a fluid. During each process, heat and work is transferred to or from the fluid. Determine W and DU for each process of the cycle and for the complete cycle. Take Qab = 700 J, Qbc = – 900 J and Qca = 300 J.

112

Basic Thermodynamics

Solution: Work and I.E. from each process (Wab, Wbc, Wca, DUab, DUbc, DUca)

P bar

Work can be calculated directly from the diagram, i.e. area under each process.

2

a

Wab = Area under a-b, i.e. Area abdeca = Area abca + Area cbdec

=

2

d

1 N (3 - 1) ¥ 10 -3 m 3 ¥ (2 - 1) ¥ 10 5 2 2 m + 1 ¥ 10 5

N

b

c 1

3

V (L)

Figure E3.1

¥ (3 - 1) ¥ 10 -3 m 3

m2 Wab = 102 N m + 2 × 102 Nm = 300 J (W.D. by the system) ( Pa + Pb ) (2 + 1) N ¥V = ¥ 10 5 2 ¥ (3 - 1) ¥ 10 -3 m 3 = 300 J 2 2 m From the first law of thermodynamics, Or Wab =

DUab = Qab – Wab = 700 – 300 = 400 J Wbc = Area under cb = 1 × 105 × (3 – 1) × 10–3 = 2 × 102 J (W.D. on the system) DUbc = Qbc – Wbc= – 900 – (– 200) = – 700 J Wca = 0

[∵ V = C]

DUca = Qca + Wca = 300 + 0 = 300 J The values of W, Q and DU for each process are tabulated. Process

DU (J)

Q (J)

W (J)

ab

400

700

300

bc

700

–900

–200

ca

300

300

0

0

100

100

Cycle abca

EXAMPLE 3.2 System A is in thermal contact with system B. The composite system undergoes an adiabatic process during which no work is exchanged between systems A and B. Determine the consequences of the first law. Solution: \ Similarly for B,

Apply the first law of thermodynamics to each system A and B, QA = DUA + WA DUA = QA – WA DUB = QB – WB

(1) (2)

First Law of Thermodynamics

Adding Eqs. (1) and (2),

DUA + DUB = QA + QB – (WA + WB)

113

(3)

Apply the first law of thermodynamics to the combined system \ Compare Eqs. (3) and (4), \

DU = W

(∵ Q = 0, adiabatic)

(4)

DU = DUA + DUB W = – (WA + WB) QA + QB = 0

and

i.e.

QA = – QB

Thus, under adiabatic conditions, the heat lost (or gained) by system A is equal to the heat gained (or lost) by system B. This result finds frequent application to problems in calorimeter and heat exchanger. EXAMPLE 3.3 One kilogram of fluid is decelerated from a velocity of 5 m/s to 0 m/s and undergoes a simultaneous decrease in elevation of 3 m. Determine change in K.E. and P.E. Solution:

V = Velocity m/s DK.E. =

m( DV )2 1 kg ¥ (0 2 - 52 ) m 2 N s2 = ¥ = - 12.5 J 2 gc 1 kg m 2s2

DP.E. = Change in potential energy =

mg ( DZ ) m ( -3) m N s2 = 1 kg ¥ 9.81 2 ¥ = - 29.43 J gc 1 kg m s

EXAMPLE 3.4 The figure shows an idealized throttling process. Initially, the gas is held on left-hand side of the porous plug. Initial conditions (Figure a) Pressure P1 (higher), Volume = V1 Now both the pistons moved simultaneously and slowly in such a way that the constant pressure P1 (higher) is maintained on the left side of the plug and constant pressure P2 (lower) is maintained on the right side of the piston. As the piston moves, gas passes through the plug until all the gas occupies on the right-hand side of the plug as shown in Figure b. Now show that H1 = H2. Solution:

But

We have H = U + PV DH = Q – W + D(PV) H2 – H1 = Q – W + (P2V2 – P1V1) Q = 0 (∵ adiabatically covered) W1 = W2 =

Ú

0

Ú

V2

V1

0

P1dV = P1 (0 - V1 ) = - P1V1 P2 dV = P2 (V2 - 0) = P2V2

(1) (2)

114

Basic Thermodynamics

Piston

a

P2 V2

Piston

P1 V1

Piston

Piston

Porous plug

b

Figure E3.4

\

W = (W2 + W1) = (P2V2 – P1V1)

(3)

Substitute Eqs. (2) and (3) in Eq. (1) \

H2 – H1 = 0 – (P2V2 – P1V1) + P2V2 – P1V1 H2 – H 1 = 0

\

H2 = H1

i.e. change in enthalpy is zero when a gas undergoes expansion through a porous plug. EXAMPLE 3.5 When a system is taken from state ‘a’ to state ‘b’ along the path acb, 80 kJ of heat flows into the system, and the system does 30 kJ of work. (a) How much heat is flowed into the system along the path adb, if the work done is 10 kJ? (b) When the system is returned from ‘b’ to ‘a’ along the curved path, the work done on the system is 20 kJ. Does the system absorb or liberate heat and how much? (c) If Ua = 0 and Ud = 40 kJ, find the heat absorbed in the processes ad and db. Solution:

We know that Q = DU + W P

Along the path acb, \

DUacb = Qacb – Wacb (Ub – Ua) = 80 – 30 = 50 kJ

c

b

a

d

(1)

(a) Heat flow into the system along the path adb, if the work done is 10 kJ (Qadb) Along the path adb, DUadb = Qadb – Wadb Ub – Ua = Qadb – 10 \

50 = Qadb – 10 Qadb = 50 + 10 = 60 kJ

(2)

V

Figure E3.5

First Law of Thermodynamics

115

(b) Does the system absorb or liberate heat and how much? (Qba) DUba = Qba – Wba Ua – Ub = Qba – (– 20) \

–50 = Qba + 20

\

Qba = –70 kJ

(c) If Ua = 0, Ud = 40 kJ, find Qad, Qdb Consider process db, V = C, \

\ Wdb = 0 DUdb = Qdb – Wdb = Qdb

Ub – Ud = Qdb From Eq. (1), we have Ub = 50 kJ \

50 – 40 = Qdb

\

Qdb = 10 kJ

Now consider process ad,

DUad = Qad – Wad Ud – Ua = Qad – Wad Qad = (Ud – Ua) + Wad = 40 – 0 + 10 = 50 kJ [Wadb = Wad + Wdb 10 = Wad + 0]

EXAMPLE 3.6 An exhausted chamber with non-conducting walls is connected through a valve to the atmosphere, where the pressure is Pa. The valve is opened and air flows into the chamber until the pressure within the chamber is Pa. Prove that Ha = Uf where Ha = molal enthalpy of the air at atmospheric pressure and temperature Uf = molal internal energy of the air in the chamber Solution: Imagine, exhausted chamber is connected to a frictionless, leak-proof piston and cylinder arrangement, where the cylinder to contain exactly the amount of air that will enter the chamber when the valve is opened. As soon as the first small quantity of air enters the chamber, the pressure in the cylinder is reduced a small amount below the atmospheric pressure, and the outside air pressurizes the piston in. Adiabatically covered, \ Q=0 From the first law of thermodynamics, DU = Q - W = - W = \ \

Ú

0

Va

Pa dV = - Pa (0 - Va )

Uf – Ua = PaVa Uf = PaVa + Ua = Ha Ha = Uf

116

Basic Thermodynamics

Figure E3.6

EXAMPLE 3.7 A bomb of volume Vb contains n mole of gas at high pressure. A capillary tube is connected to the bomb, through which the gas may slowly leak out into the atmosphere, where the pressure is Pa. Bomb and capillary are immersed in the water bath in which heat is added through the electric resistor. Now the gas is allowed to leak slowly through the capillary tube into the atmosphere and at the same time electric energy is dissipated in the resistor in such a rate that temperature of the gas, the bomb, the capillary and water is kept equal to that of the outside air. Show that after as much gas as possible has leaked out during time ‘t’, the change of internal energy is, DU = VIt – Pa(nVa – Vb) Solution:

W = Pa DV = Pa [nVa – Vb]

From the first law of thermodynamics DU = Q – W Q = Heat added is obviously equal to the electrical energy expended, which is VIt D DU = VItt – Pa[nVa – Vb]

\

Valve (closed) Bomb Vb n moles

Initial position Valve (opened)

Pa, Va

Figure E3.7

Bomb capillary tube and water are at the same temperature equal to atmosphere

First Law of Thermodynamics

117

EXAMPLE 3.8 A thick walled insulated metal chamber contains n1 moles of gas at high pressure P1. It is connected through a valve with a large almost empty gas holder in which the pressure is maintained at a constant value, Pa, very nearly atmospheric. The valve is opened slightly, and the gas flows slowly and adiabatically into the gas holder until the pressure on the two sides of the valve is equalized. n2 U1 - H a Prove that, = n1 U 2 - H a where, n1 = Number of moles of gas initially in the chamber n2 = Number of moles of gas left in the chamber U1 = Initial molar internal energy of the gas in the chamber U2 = Final molar internal energy of the gas in the chamber Ha = Molar enthalpy of helium in the gas holder Solution: Chamber n1 u1

n2 u2

Pa, Va Ua

Figure E3.8

From the first law of thermodynamics, Q = DU + W \

[Q = 0, \

adiabatic]

DU = –W = –Pa [(n1 – n2)Va]

(1)

also we have, DU = Final internal energy of the gas in the cylinder + Final internal energy of the gas (left) in the chamber – Initial internal energy of the gas in the chamber = (n1 – n2)Ua + U2n2 – n1U1

(2)

Equating Eqs. (1) and (2), n1Ua – n2Ua + U2n2 – n1U1 = –Pan1Va + Pan2Va n1Ua – n1U1 + Pan1Va = Pan2Va – U2n2 + n2Ua n1[Ua – U1 + PaVa] = n2[–U2 + Ua + PaVa] \ \

n1[–U1 + Ha] = n2[–U2 + Ha] n2 -(U1 - H a ) U1 - H a = = n1 -(U 2 - H a ) U 2 - H a

118

Basic Thermodynamics

EXAMPLE 3.9 Calculate the heat capacities, Cv, and Cp, for one mole of a gas which obey Van der a a ˘ È Waals equation of state Í P + 2 ˙ (V - b) = RT and its internal energy is given by, U = CT – , V V ˚ Î where a, b, C and R are constants.

Solution:

\

Cv =

dQ ˘ , but for constant V, dU = dQ dT ˙˚ v

Cv =

dU ˘ dT ˙˚ v

U = CT -

a V

dU ˘ =C dT ˙˚ v

\ \

(1)

(2)

Substituting Eq. (2) in Eq. (1) Cv = C CP =

We have,

dU ˘ dT ˙˚ p

(3)

and for constant P, dU = dQ – PdV dU ˘ dQ ˘ dV ˘ = -P dT ˙˚ p dT ˙˚ p dT ˙˚ p

\

U = CT -

a V

dU ˘ a È dV ˘ =C + 2 Í ˙ ˙ dT ˚ p V Î dT ˚ p

\ Substitute

(4)

(5)

dU ˘ from Eq. (5) in Eq. (4), dT ˙˚ p

dQ ˘ a È dV ˘ dV ˘ RT È dV ˘ Ê a ˆ È dV ˘ =C + 2 Í +P = C + Á 2 + P˜ Í =C + ˙ ˙ ˙ ˙ Ë ¯ dT ˚ p dT ˚ p (V - b) ÍÎ dT ˙˚ p V Î dT ˚ p V Î dT ˚ p dV ˘ R /(V - b) = dT ˙˚ È RT 2a ˘ Í ˙ 2 V2 ˚ Î (V - b)

From Eqs. (3), (6) and (7), Cp = C +

R 2 T/(C – V ) 2 È RT 2a ˘ – 2˙ Í 2 V ˚ Î (V – b)

(6)

First Law of Thermodynamics

119

EXAMPLE 3.10 A system undergoes various processes in which the initial state is identical to the final state. The work and heat interactions during the processes are tabulated. Complete the table and calculate the net work done by the system. (KUD I-82)

Solution:

Process

Q (kJ/s)

W (kJ/s)

DU (kJ/s)

1–2 2–3 3–4 4–1

— 50 –50 —

35.13 25.0 — 17.6

64.85 — — –17.57

From the first law of thermodynamics, consider process 1–2, Q = DU + W Q1- 2 = 64.85 + 35.13 = 99.98

kJ s

Process 2–3,

D2U3 = Q2–3 – W2–3 = 50 – 25 = 25

kJ s

Process 4–1,

Q4–1 = 4W1 + D4U1 = 17.6 + (–17.6) = 0

We know that, total change in internal energy = 0 DUn = 1U2 + 2U3 + 3U4 + 4U1 0 = 64.85 + 25 + 3U4 – 17.6 \

3U4

Now consider process 3–4, \ \

= -72.25

kJ s

Q3–4 = W3–4 – D3U4

W3–4 = Q3–4 + D3U4 = –50 – (–72.25) = 22.25

kJ s

Wn = net work done = 1W2 + 2W3 + 3W4 + 4W1 = 35.13 + 25.0 + 22.25 + 17.6 = 99.97

kJ s

Qn = Net change in heat transfer = 1Q2 + 2Q3 + 3Q4 + 4Q1 = 99.98 + 50 – 50 + 0 = 99.98 \

kJ s

ÚW = ÚQ

EXAMPLE 3.11 An ideal gas is heated at constant volume until its temperature is doubled, then it is expanded isothermally till it reaches the original pressure. Finally the gas is cooled at constant

120

Basic Thermodynamics

pressure till it is restored to the original state. Draw PV and TS diagram and determine the net work done per kg of gas when the initial temperature is 300 K. R = 0.287 kJ/kg K. (KUD II-98) Solution: T

2

P

3

1

2

1

3

S

V

Figure E3.11

T1 = 300 K T2 = 2 × T1 = 600 K

Net work done (Wn)

Wn = W1–2 + W2–3 + W3–1 = 0 + RT2 ln

= RT2 ln = 0.287

P2 + P1 [V1 - V3 ] P3

È ˘ P2 P + R (T1 - T2 ) = R ÍT2 ln 2 + (T1 - T2 ) ˙ P3 P3 Î ˚

˘ T2 kJ È + (T1 - T2 ) ˙ K Í 600 ln kg K Î T1 ˚

600 È ˘ = 0.287 Í 600 ln + (300 - 600) ˙ = 33.26 kJ/kg 300 Î ˚

È P2 P RT2V2 T ˘ = RT2 ln 2 = RT2 ln = RT2 ln 2 ˙ Í RT2 ln P3 P1 V2 RT1 T1 ˚ Î

EXAMPLE 3.12

A gas undergoes a thermodynamic cycle as follows:

(i) Process 1–2, constant pressure, P = 3 bar, V1 = 0.5 m3, 1W2 = 20 kJ (ii) Process 2–3, isothermal T = C, U3 = U2 (iii) Process 3–1, constant volume, U1 – U3 = –50 k Neglecting the changes in K.E. and P.E. Determine, (a) Sketch the cycle on the PV diagram (b) Net work done (c) Net heat transfer (d) Net internal energy

121

First Law of Thermodynamics

Solution: (a) The cycle on the PV diagram (Figure E. 3.12) (b) Net work done (Wn) (i) Consider process 1–2, P = C; P1 = P2 1W2 = P1[V2 – V1] = 20 kJ kN \ 20 kJ = P1[V2 – 0.5] m3 × 100 2 m 20 = 3[V2 – 0.5]100 kJ \

P 3

2

1

V2 = 0.567 m3

Figure E3.12

V

From the first law of thermodynamics, 1Q 2

= 1W2 + D1U2 = 20 + (U2 – U1) = 20 – (–50) = 70 kJ

[Given that, U3 = U2 U1 – U3 = –50 kJ

U1 – U2 = –50 kJ]

(ii) Consider process 2–3, PV = C 2Q 3

= 2W3 + D2U3 = 2W3 + U3 – U2

2 Q3

= 2W3 = P2V2 ln

= 300

kN m

2

V3 V = P2V2 ln 1 V2 V2

¥ 0.567 m 3 ln

0.5 = - 21.39 kJ 0.567

È ˘ Ê V2 ˆ Ê V2 ˆ kN 0.567 = 340.2 kPa ˙ ÍV3 = V1 , P2 = P1 , P3 = P2 Á ˜ = P1 Á ˜ = 3 ¥ 100 2 ¥ 0.5 Ë V1 ¯ Ë V3 ¯ m ÍÎ ˙˚ (iii) Consider the process 3–1, V = C, 3W1 = 0 3Q1

\

= 3W1 + D3U1 = 3W1 + U1 – U3 = 3W1 – 50 = 0 – 50 kJ

Wn = 1W2 + 2W3 + 3W1 = 20 – 21.39 + 0 = –1.39 kJ

(c) Net heat transfer (Qn) Qn = 1Q2 + 2Q3 + 3Q1 = 70 – 21.39 – 50 = –1.39 kJ Hence

ÚW = ÚQ

Proved.

(d) Net internal energy (Un) Un = D1U2 + D2U3 + D3U1 = 50 + 0 – 50 = 0 EXAMPLE 3.13 In an I.C. engine, during the compression stroke the heat rejected to the cooling water is 50 kJ/kg and the net work input is 100 kJ/kg. Calculate the change in internal energy of the working fluid stating whether it is a gain or loss. (KUD July 2002)

122

Basic Thermodynamics

Solution: From the first law of thermodynamics, 1Q2

= D1U2 + 1W2

–50 kJ = (U2 – U1) – 100 kJ \

(U2 – U1) = –50 + 100 = 50

kJ kg

i.e. gain in internal energy EXAMPLE 3.14 A volume of 115 litres of air at 105 N/m2 and 90°C is compressed adiabatically until the volume is reduced to 11.5 litres. Find the change in internal energy and change in enthalpy. If 0.091 kg of air from a source kept at 180°C is allowed to flow into this space of 11.5 litres, find the temperature and pressure at the end of the operation. CV = 0.72

kJ kJ and R = 0.287 kg K kg K

Solution:

(KUD II-97) 3

P

P1 = 105 N/m2 V1 = 115 litres V2 = 11.5 litres m2 = 0.091 kg T1 = 90°C T = 180 + 273 = 453 K

(m1 + m2) 2 g

PV = C

(a) Change in internal energy (D DU) and enthalpy D H) (D 1

Consider the process 1–2, P1V12 = P2V22

V

1.4

ÈV ˘ N È 115 ˘ P2 = P1 Í 1 ˙ = 10 5 2 ¥ Í ˙ m Î 11.5 ˚ Î V2 ˚

P2 = 25.12 ¥ 10 5 ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚

W=

-1

Figure E3.14

N m2 È 25.12 ¥ 10 5 ˘ = 363 Í ˙ 10 5 ÍÎ ˙˚

1.4 - 1 1.4

= 911.83 K

( P1V1 - P2V2 ) (10 5 ¥ 0.115 - 25.12 ¥ 10 5 ¥ 0.0115) = = - 43.465 kJ ( - 1) (1.4 - 1)

DU = Q – W = 0 – [– 43.465] = 43.465 kJ DH = m1Cp dT = 0.1104 × 1.005 × (911.83 – 363) = 60.894 kJ È ˘ P1V1 N kg K 1 = 10 5 2 ¥ 115 ¥ 10 -3 m 3 ¥ ¥ = 0.1104 kg ˙ Í m1 = RT1 287 Nm 363 K m Î ˚

First Law of Thermodynamics

123

(b) Temperature (T3) and pressure at the end of the operation (P3) Internal energy at 2 + Enthalpy of new air = IE at 3. Taking absolute zero as datum for IE, m1Cv T2 + m2Cp T = (m1 + m2) Cv T3 0.1104 × 0.72 × 912 + 0.091 × 1.005 × 453 = (0.1104 + 0.091] × 0.72 × T3 \

T3 = 786.2 K or P3V3 = m3 RT3

We have \

P3 =

513.2°C

(m3 = m1 + m2)

m3 RT3 0.2014 ¥ 0.287 ¥ 786.2 kN = = 3951.6325 2 V3 0.0115 m

EXAMPLE 3.15 A system undergoes a process in which the heat transfer to the system is 40 kJ and the work done by the system is 45000 Nm. Evaluate (a) the change in the energy of the system. In a second process between the same end states, the same system does 35000 Nm of work with heat transfer. Determine (b) the magnitude and sign of the heat transfer. (KUD I-96) Solution:

P 1 W2

= + 45000 Nm

1Q 2

= 40 kJ

1

D1U2) (a) The change in the energy of the system (D From the first law of thermodynamics D1U2 = Q1–2 – W1–2 = 40 – 45 = –5 kJ 2

(b) The magnitude and sign of the heat transfer (dQ) Q = 1W2 + D1U2

1Q2

= 35 + (–5) = 30 kJ

V

Figure E3.15

EXAMPLE 3.16 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 10 kJ and the work done on the air is 346.06 Nm. Calculate (a) index of compression and (b) change in internal energy. Following the above process, the air undergoes a second process at constant pressure to a temperature of 27°C. Determine the (c) magnitude and direction of heat transfer during the second process. (KUD I-92) Solution:

P

m = 3 kg

P1 = 1 bar

T1 = 27 + 273 = 300 K

P2 = 4 bar

2

3

T2 = 93 + 273 = 366 K T3 = 27 + 273 = 300 K (a) Index of compression (n) We have

1

P1V1 = mRT1 Figure E3.16

V

124

Basic Thermodynamics

\

mRT1 kN m 300 K m 2 = 3 kg ¥ 0.287 ¥ kg K 1 × 100 kN P1 = 2.6343 m3 P2V2 = mRT2

V1 =

mRT2 kN m 366 K m 2 = 3 kg ¥ 0.287 ¥ = 0.8034 m 3 P2 kg K 1 ¥ 100 kN Process 1–2 is a polytropic process \

V2 =

ln P1 / P2 ln V2 / V1

n=

= \

W.D. =

or

ln 1 / 4 ln 0.8034 / 2.6343

-346.06 = \

n = 1.1674

( P1V1 - P2V2 ) ( n - 1)

(100 ¥ 2.6343 - 400 ¥ 0.8034) (n - 1)

n = 1.1674

DU) (b) Change in internal energy (D 1Q2

= 1W2 + D1U2

–10 kJ = –346.06 kJ + DU

\ DU = 336.06 kJ

(c) The magnitude and direction of heat transfer during the second process Consider the process 2–3, P = C V3 = V2 ¥

\

2 Q3

T3 300 = 0.8034 ¥ = 0.6585 m 3 T2 366

= mC p (T3 - T2 ) = 3 kg ¥ 1

kN m ¥ (300 - 366) K = - 198 kJ kg K

EXAMPLE 3.17 A closed system contains 0.5 m3 of air at 5 bar and 130°C. A reversible adiabatic expansion takes place till the pressure falls to 1 bar. The gas is then heated by a constant pressure till enthalpy increases by 75 kJ. Calculate the following: (a) The work done (b) The index of expansion, if the above processes are replaced by a single reversible polytropic process giving the same work between the same initial and final state. Take Cp = 1 kJ/kg K, Cv = 0.714 kJ/kg K.

P

1 Adiabatic expansion

Solution: V1 = 0.5 m3

P1 = 5 bar,

P2 = 1 bar

t1 = 130°C

P2 = P3

H3 = H2 + 75 kJ

P=C

(a) The work done (Wn) Consider process 1–2, P1V1 = P2V2

2

Figure E3.17(a)

3 V

125

First Law of Thermodynamics 1

\

1

ÈP ˘ È 5 ˘ 1.4 V2 = V1 Í 1 ˙ = 0.5 ¥ Í ˙ = 1.578 m 3 Î1˚ Î P2 ˚ Cp È 1 kJ ˘ = = 1.4 R = (C p - Cv ) = 1 - 0.714 = 0.286 Í = ˙ kg K ˚ Cv 0.714 Î

ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚ m=

-1

È1˘ = 403 Í ˙ Î5˚

1.4 - 1 1.4

= 254.6 K

P1V1 5 ¥ 100 kN ¥ 0.5 m 3 kg K = = 2.169 kg RT1 m 2 ¥ 0.286 kN m ¥ 403 K

Consider process 2–3, P = C, Q2–3 = m Cp (T3 – T2) = 2.169 × 1 × [T3 – 254.5] H3 – H2 = 75 = 2.169 × 1 × [T3 – 254.5] \

T3 = 289.1 K V3 = V2 ¥

W1- 2 =

T3 1.578 ¥ 289.1 = = 1.793 m 3 T2 254.5

( P1V1 - P2V2 ) 10 2 ¥ 15 ¥ 0.5 - 1 ¥ 1.578 = = 230 kJ ( - 1) 1.4 - 1

W2–3 = P2 × [V3 – V2] = 1 × 100 × [1.793 – 1.578] = 21.5 kJ Wn = W1–2 + W2–3 = 230.5 + 21.5 = 252 kJ (b) Index of expansion (n) W1–3 = Wn = 252 kJ Condition corresponding to state 1 and state 3 are same.

P

1 n

PV = C

Consider the process 1–3, PVn = C

\

W1-3 =

( P1V1 - P3V1 ) (n - 1)

252 =

100 ¥ (5 ¥ 0.5 - 1 ¥ 1.793) n -1

3 V

Figure E3.17(b)

n = 1.285

EXAMPLE 3.18 Consider an adiabatically covered cylinder with frictionless piston as shown in the figure. Piston moves slowly as the paddle work is done on the system. Prove that the paddle work is equal to change in enthalpy.

126

Basic Thermodynamics

Solution:

From the first law of thermodynamics, Q = DU + W Wpaddle = DU + PdV = DU + D(PV) = D(U + PV) = DH

Piston

[Paddle work is fully converted into heat. High grade energy to low grade energy]

Moves outward (Work done by the system)

Paddle work on the system

Figure E3.18

EXAMPLE 3.19 Calculate (a) Work done, (b) Q, (c) DH and Cp for the following data. V1 = 1.5 m3, T2 = 120°C,

m = 5 kg, T1 = 20°C, Solution:

Here P1 = P2,

V2 = 0.3 m3, DU = 2500 kJ.

P1 = P2 = 10 bar

hence P = C

(a) Work done (W)

P

W1–2 = P × [V2 – V1] = 10 × 100 × [0.3 – 1.5] = –1200 kJ 1

(b) Heat interaction (Q)

2

Q1–2 = DU + W1–2 = 2500 – 1200 = 1300 kJ DH) (c) Change in enthalpy (D DH = Q1–2 = 1300 kJ

[∵ P = C]

(d) Specific heat at constant pressure (Cp) Cp =

V

Figure E3.19

dh DH 1300 kJ = = = 2.6 dT m ¥ (120 - 20) 5 ¥ 100 kg K

EXAMPLE 3.20 Twenty members are sitting in a conference hall of dimension 5 × 7 × 3 m3. Each member occupies 100 cm3 and rejects heat of 350 kJ per hour. Calculate the rise in temperature of air within 30 min of the start of the conference. The room is completely insulated and sealed. Cv = 0.715 kJ/kg K.

First Law of Thermodynamics

127

Solution: VR = Volume of the room = 5 × 7 × 3 = 105 m3 Vm = Volume of the 20 members = 20 × 100 × 10–6 m3 = 2 × 10–3 m3 Va = Volume of air = VR – Vm = 105 – 10–3 × 2 = 104.997 ª 105 m3 Assume constant volume heat addition. Initial room temperature = 25°C, pressure = 1 bar. \ \

P1Va 100 ¥ 105 kN kg K 1 = ¥ m3 ¥ ¥ = 122.16 kg 2 kN m K RT1 0.287 ¥ 298 m

m = mass of air =

For constant volume heat addition, we have DU = m Cv dT

From the first law of thermodynamics, Q = W + DU Here,

W=0 350 ¥ 20

[∵ V = C]

kJ = 122.16 ¥ 0.715 ¥ (T2 - T1 ) hr

Total time of conference = 30 min. 350 ¥ 30 ¥ 20 min kJ = 122.16 ¥ 0.715 ¥ (T2 - 298) 60 min

\ \

T2 = 337.87 K

EXAMPLE 3.21 Consider a system as shown in the figure where 2 kg of saturated liquid water at 2 MPa is mixed with 2 kg of vapour at 2 MPa and 700°C. During the process the pressure in the cylinder kept constant and no energy is lost between the cylinder and water. Calculate the final quality of the steam. T 700

Vapour

212.4

1

3

2

1

1

2M

Pa

Piston

3

2

2 MPa

1 Liquid (a)

(b)

S

(c)

Figure E3.21

Solution: \

No energy lost between the cylinder and water. Q=0 -

P = 2 MPa = 20 bar

Ú

3

1

PdV = U3 - U1

W = DU

U

128

Basic Thermodynamics

\

PV1 – PV3 = U3 – U1

i.e. PV1 + U1 = PV3 + U3

H1 = H3 \

(1)

H1 = U1 + PV1 = ml1Ul1 + mv1uv1 + P[ml1Vl1 + mv1Vv1] = ml1hl1 + mv1hv1

From steam tables, at 2 MPa = 20 bars

hf hg h at 700∞C ˘ È Í ˙ 3916.5 ˙ Í 20 bars 908.6 2797.2 Í ˙ hl1 h2 hv1 Î ˚ H1 = 2 kg ¥ 908.6

kJ kJ + 2 kg ¥ 3916.5 = 9650.2 kJ kg kg

(2)

The final quality of the steam (x3) The final state is represented by H3. H3 = U3 + PV3 = m [u + Pu]3 = mh3 \

(3)

Substitute Eqs. (2) and (3) in Eq. (1) mh3 = 9650.2 h3 =

\

9650.2 9650.2 kJ = = 2412.55 4 kg m

Enthalpy corresponding to state 3 can be written as, h3 = h1 + x3 (h2 – h1) 2412.55 = 908.6 + x3 [2797.2 – 908.6] \

x3 = 0.796

EXAMPLE 3.22 A system with a volume of 0.1 m3 is fixed with a piston enclosing 0.7 kg of steam at 0.4 MPa. Calculate (a) the amount of work done and (b) heat transferred when the steam is heated to 300°C at constant pressure. Solution: (a) Work done (W) 2

W=

Ú

u1 =

V1 0.1 m3 = = 0.1429 m 0.7 kg

1

PdV = mP (u2 - u1 ) = 0.7 × 400 + (0.6548 – 0.1429) = 143.332 kJ

From steam tables at 0.4 MPa and 300°C u2 = 0.6548

m3 kJ , h2 = 3066.8 kg kg

First Law of Thermodynamics

129

(b) Heat transfer (Q) Q = DU + Wm(u2 – u1) + m(P2u2 – P1u1) Q = m(h2 – h1) = 0.7[3066.8–1260.67] = 1264.3 kJ u1 = 0.1429 = uf + x1ufg = 0.001084 + x1(0.4614) \

x1 = 0.3074 h1 = hf + x1hfg = 604.74 + 0.3074(2133.8) = 1260.67 kJ/kg

EXAMPLE 3.23 A fluid is continuously overflowing over a tank at a height of 120 m from ground level. Calculate, (a) the potential energy of the liquid at the top of the tank with respect to its base. (b) the kinetic energy of the liquid just before it strikes the floor. After the liquid enters the flow below, what change has occurred to its state? Assume 1 kg of mass of liquid, no exchange of energy with the surrounding. Solution: (a) Potential energy of the liquid at the top of the tank (P.E.1)

P.E.1 =

mZg 1 kg × 120 m × 9.81 m = = 1177.2 J gc 1 kg m s2 N s2

(b) Kinetic energy of the liquid just before it strikes the floor (K.E.2) During fall, no change in internal energy. \

DU = 0

\

DK.E. + DP.E. = 0

We know that

K.E.1 = 0,

i.e.

K.E.2 – K.E.1 + P.E.2 – P.E.1 = 0

P.E.2 = 0

[State 1 represents liquid at a height of 120 m, state 2 represents liquid at ground level, state 3 represents, after mixing with the main stream.] \

K.E.2 = P.E.1 = 1177.2 J

Du) (c) After the liquid enters into the main stream of flow, changes in its state (D As soon as the liquid enters into the main stream of flowing liquid, turbulence takes place; hence kinetic energy will be converted into internal energy. \ Here

DU + DK.E. = 0

U3 – U2 + K.E.3 – K.E.2 = 0

K.E.3 = 0

[Velocity of liquid is very small (\ V3 = 0) after entering into the main stream.] \

DU = K.E.2 = 1177.2 J

EXAMPLE 3.24 A system is having 5 kg water as shown in the figure. A paddle work of 5000 J is done on the system. Calculate

130

Basic Thermodynamics

(a) the change in specific and total internal energy of the system (assume adiabatic process). (b) change in internal energy if 500 J of heat lost to the surrounding (assume non-adiabatic). Solution: (a) The change in specific and total internal energy of the system. (Assume adiabatic Du, DU) process) (D From the first law of thermodynamics Q = W + DU

O = W + DU

DU = –5000 J

System

DU -5000 J J = = - 1000 m 5 kg kg [Adiabatic, \ Q = 0] [work done on the system]

Du =

Insulation

Figure E3.24

(b) Change in internal energy if 500 J of heat lost to the surrounding (assume nonDU) adiabatic) (D –500 J = –5000 J + DU EXAMPLE 3.25 Consider a piston and cylinder arrangement as shown in the figure. The system is having 2 kg of air at 15 bar. 100 kJ of work is done on the system (from paddle wheel) and there is a rise in temperature of 500 K to 800 K. Assume the system is adiabatic and at constant pressure process. Calculate (a) the work done (b) the change in internal energy (c) the enthalpy change for the system.

m

System

Piston

DU = 4500 J

Piston

\

1

2

Figure E3.25

Solution: (a) Work done (W) W = Work done by the system in moving the piston from state 1 to state 2 at constant pressure. W=

\

Ú

2

1

PdV = P [V2 - V1 ]

V1 =

mRT1 2 ¥ 0.287 ¥ 500 = = 0.1913 m 3 P1 15 ¥ 100

V2 =

mRT2 2 ¥ 0.287 ¥ 800 = = 0.306 m 3 P2 15 ¥ 100

W = 15 × 100 × [0.306 – 0.1913] = 172.05 kJ

DU) (b) The change in internal energy (D DU = m Cv (T2 – T1) = 2 × 0.714 × [800 – 500] = 428.4 kJ

First Law of Thermodynamics

131

DH) (c) The change in enthalpy (D DH = m Cp(T2 – T1) = 2 × 1.005 × (800–500) = 603.0 kJ (d) Net work done (Wn) Wn = Work done on the system – Work done by the system = –100 kJ + 172.05 kJ = 72.05 kJ EXAMPLE 3.26 Consider a (piston and cylinder arrangement) closed system with constant pressure process. The following data are considered. Suffix 1 represents initial state. Suffix 2 represents final state. P1 = 2 bar, u1 = 0.5 m3/kg, u2 = 0.75 m3/kg, t1 = 25°C, t2 = 300°C. Specific heat of the fluid is given by, È 20 ˘ kJ C p = Í 0.5 + ˙ (T + 30) ˚ kg K Î where T is in °C. Calculate (a) Heat added, (b) Work done, (c) Du , (d) Dh. Solution: (a) Heat added (Q) Q=

Ú

T2

T1

C p dT =

Ú

300 È

25

20 ˘ 30 Í 0.5 + ˙ dT = [0.5T + 20 ln (T + 30)]25 (T + 30) ˚ Î

= 0.5[300 - 25] + 20 ln

(300 + 30) kJ = 173.74 (25 + 30) kg

Piston

System

Piston

W

Q

Figure E3.26

(b) The work done by the system (W) W=

Ú

V2

V1

PdV = P (V2 - V1 ) = 2 ¥ 100 ¥ [0.75 - 0.5] = 50

Du) (c) Change in internal energy (D Du = Q – W = 173.34 – 50 = 123.34

kJ kg

kJ kg

132

Basic Thermodynamics

Dh) [P = C] (d) Change in enthalpy (D Dh = Q = 173.34

kJ kg

EXAMPLE 3.27 Consider a piston and cylinder arrangement as shown in the figure. During the process the volume changes from 0.2 m3 to 0.04 m3 and the system rejects 10 kJ of heat. Process follows pressure volume relation of, P = [10/V + 2], where ‘P’ is in kPa and V is in m3. Calculate (a) The work done (b) Change in internal energy (c) Change in enthalpy. Solution: (a) The work done (W) W=

Ú

V2

V1

PdV =

Ú

V2

V1

È 10 ˘ Í V + 2 ˙ dV Î ˚

P2 = \

10 10 +2= + 2 = 252 kPa V2 0.04

DH = 6.41 + (252 × 0.04 – 52 × 0.2)

2

1

Piston

10 10 P1 = +2= + 2 = 52 kPa V1 0.2

Piston

0.04

È ˘ V = Í10 ln 2 + 2(V2 - V1 ) ˙ = - 16.41 kJ V1 Î ˚0.2 DU) (b) Change in internal energy (D DU = Q – W = –10 kJ – (–16.41) = 6.41 kJ DH) (c) The change in enthalpy (D DH = DU + D(PV) = DU + (P2V2 – P1V1)

Q

Figure E3.27

= 6.41 – 0.32 = 6.09 kJ. EXAMPLE 3.28 Initially a football is in flate position. Then air is filled in the foot ball at a pressure of 100 kPa (gauge). After filling the air, the volume is 2500 cm3 and at a temperature of 30°C. Now the ball is taken onto the field and started playing. After sometime, the temperature of air inside the football dropped to 20°C, but there is no significant change in its volume. Calculate, (a) (b) (c) (d)

The mass of air in the ball. The pressure of the air in the ball as soon as the temperature dropped to 20°C. The amount of heat transfer from the ball to the surroundings. The pressure required in order to maintain a pressure of 100 kPa (gauge) at 20°C.

Solution: (a) The mass of air filled (m) m=

P1V1 (100 + 101.325) ¥ 2500 ¥ 10 -6 = = 5.788 ¥ 10 -3 kg RT1 0.287 ¥ 303

First Law of Thermodynamics

V1 = 2500 cm T1 = 30°C P1 = 100 kPa

3

V2 = 2500 cm T2 = 20°C P2 = ?

133

3

Q (a) Initial position

(b) After filling air

(c) Final position

Figure E3.28

(b) The pressure (P2) as soon as the temperature dropped to 20°C (V1 = V2) P2 =

mRT 5.788 ¥ 10 -3 ¥ 0.287 ¥ 293 = V2 2500 ¥ 10 -6

= 194.68 kPa (ab) or 93.4 kPa (gauge) (c) The amount of heat transfer (Q1–2) Q1–2 = D1U2 + W1–2

[Here W1–2 = 0 no PdV work,

∵ V = C]

From the air tables, at 30°C, u1 = 214.32, at 20°C, u2 = 195.0 kJ/kg \

Q1–2 = (u2 – u1) m = 5.788 × 10–3 × [195.0–214.32] = – 0.112 kJ

(d) The initial pressure required (P1) Assumption : V1 = V2 We have, P1 = P2

P1V1 P2V2 = T1 T2

T1 (100 + 101.325) ¥ 303 = 208.2 kPa (ab) or 106.87 kPa (gauge) = T2 293

EXAMPLE 3.29 Consider a piston and cylinder arrangement containing R-12 exposed to the sun. The volume of the tank is 25 m3 and mass of the R-12 is 1300 kg at a pressure of 6 bar. Heat transfer takes place from the sun to the R-12. The R-12 changes to saturated vapour. Calculate (a) The initial temperature and state of the R-12. (b) The final temperature and heat transfer if heating is a constant pressure process. (c) The final temperature, pressure and amount of heat transfer if heating is a constant volume process Solution: P1 = 6 bar

P2 = 6 bar

State 1 = liq + vap.

m = 1300 kg

V1 = 25 m3

V2 = ?

State 2 = dry vapour

(a) The initial temperature (t1) and state of the R-12 u1 = Specific volume at inlet condition =

V1 25 = 0.01923 m3/kg = m 1300

134

Basic Thermodynamics

Figure E3.29

From R-12 tables, at 6 bar,

tsat = t1 = 22°C

u4 = 0.00075663 m3/kg

u2 = 0.029373 m3/kg.

u1 is in between u2 and u4, hence initial condition is a mixture of liq + vap. t1 = initial temperature = tsat = 22°C We know that, u1 = u4 + x1(u2 – u4) 0.01923 = 0.00075663 + x1[0.029373 – 0.00075663] x1 = 0.6456 (b) The final temperature (t2) and heat transfer (q) if heating is a constant pressure process i.e. P1 = P2 \

h1 = h4 + x1(h2 – h4) = 57.278 + 0.6456[197.93 – 57.278] = 148.08 kJ/kg q = h2 – h1 = 197.93 – 148.08 = 49.85 kJ/kg

\

Q = 49.85 × 1300 = 64805 kJ t2 = t1 = 22°C

(c) The final temperature (t3), pressure (P3) and amount of heat transfer (Q1–3) if heating is a constant volume process Now consider process 1–3 (V = C), \

i.e. V1 = V3

V1 = V3 = 0.01923 m3/kg

From R-12 tables, corresponding to V1 = V3 = 0.01923 m3/kg, pressure P3 is 0.91 MPa = 9.1 bar and saturation temperature, t3 is 37.8°C. We know that, Q1–3 = W1–3 + D1U3 But

W1–3 = 0

\

Q1–3 = D1U3

[∵ V1 = V3] u1 = u4 + x1(u3 – u1)

= 56.824 + 0.4076 × [186.8 – 56.824] = 140.73 kJ/kg \

q1–3 = u3 – u1 = 186.8 – 56.83 = 129.97 kJ/kg

\

Q1–3 = 129.97 × 1300 = 168961.00 kJ

Heat transfer to the system.

First Law of Thermodynamics

135

EXAMPLE 3.30 2 kg of a substance at 100 kPa is contained in a piston and cylinder arrangement. The initial volume is 0.5 m3. Heat transferred to the substance causes a slow expansion at constant temperature. The process ceases when the final volume is double the initial volume. Calculate the amount of heat transfer if the substance is (a) O2 and (b) water. Solution: P, kPa

P, kPa

T=C System T1 = T2

Piston

100

V1 = 0.5 m3

1

T=C 100

W

3 1

2 4

2

V2 = 2V1 0.5

Q (a)

1.0

3

V, m

(b)

0.5

1.0

3

V, m

(c)

Figure E3.30

(a) Heat transfer (Q1–2) if the substance is O2 The given pressure of O2, i.e. 100 kPa is well below its critical pressure, hence it behaves like an ideal gas. We have D1u2 = u2 – u1 = 0 \

W1- 2 = mRT1 ln

[isothermal process, T = C]

V2 V 2V kN = P1V1 ln 2 = 100 2 ¥ 0.5 m 3 ¥ ln 1 = 34.66 kJ V1 V1 V1 m

Q1–2 = W1–2 + D1U2 = 34.66 + 0 = 34.66 kJ (b) Heat transfer (Q1–2) if the substance is water (Figure c) u1 = specific volume =

0.5 m3 = 0.25 2 kg

From steam tables, u3 = 0.0010431 m3/kg u4 = 1.694 m3/kg \

tsat = 99.63°C.

State 1 is a wet vapour, lies in between states 3 and 4.

We have q1–2 = W1–2 + u2 – u1 = P(V2 – V1) + u2 – u1 = PV2 + u2 – (PV1 – u1) = h2 – h1 u1 = u3 + x1 [u4 – u3] 0.25 = 0.0010431 + x1[1.694 – 0.0010431] x1 = 0.1471

136

Basic Thermodynamics

u2 = u3 + x2 [u4 – u3]

V2 1 È ˘ Íu2 = m = 2 = 0.5˙ Î ˚

0.5 = 0.0010431 + x2[1.694 – 0.0010431] x2 = 0.2947 h1 = h3 + x1 [h4 – h3] = 417.51 + 0.1471[2675.1–417.51] = 749.6 kJ/kg h2 = h3 + x2 [h4 – h3] = 417.51 + 0.2947 [2675.1 – 417.51] = 1082.82 kJ/kg \

Q1–2 = (h2 – h1)m = (1082.82 – 749.6) × 2 = 666.44 kJ

EXAMPLE 3.31 An adiabatic frictionless piston and cylinder arrangement as shown in the figure contains 0.75 kg of air at 100°C and 300 kPa. Electrical work is added, and rises the temperature of air to 200°C at constant pressure. Calculate (a) the net work and (b) electrical work. Solution:

Figure E3.31

(a) The net work (W1–2) From the first law of thermodynamics or energy conservation law, Q1–2 – W1–2 = D1u2 = m[u2 – u1] Adiabatically covered,

∵ Q1–2 = 0

W1–2 = –m[u2 – u1] = –m Cv(T2 – T1) = - 0.75 kg ¥ 0.714

kJ (200 - 100) K = - 53.55 kJ kg K

(b) Electrical work (We) W1–2 = We + PdV PdV = P[V2 – V1] = mR(T2 – T1) = 0.75 × 0.287 × (200 – 100) = 21.53 kJ \ or

We = W1–2 – PdV = –53.55 – 21.53 = –75.075 kJ W1–2 = We + W1–2 PdV = –m(u2 – u1) We = –W1–2 PdV – m(u2 – u1) = P(V1 – V2) – m(u2 – u1) We = P(V1 – V2) – u2 + u1 = – (H2 – H1)

First Law of Thermodynamics

137

EXAMPLE 3.32 A mass of 5 kg of ice at –10°C is added to 200 kg of hot tea at 90°C, both are at the ambient pressure of 1 bar. The container is well insulated. The thermodynamic properties of ice and hot tea are almost same. Calculate the final temperature. Solution:

Ice Mix of (ice + hot tea) Hot tea

(a) Initial state 1

(b) Final state 2

Figure E3.32

(a) Final temperature (t2) Q1–2 = W1–2 + u2 – u1 = P[V2 – V1] + u2 – u1 = H2 – H \

H1 = mice hice + mtea htea = 5 × (–354.1) + 200 × 398 = 77829.5 kJ Q1–2 ª 0

\

H2 = H1 = m1h2 = 77829.5 kJ

\

h2 =

77829.5 kJ = 379.66 205 kg

h2 = C p t2 ∞C = 4.2 379.66 \

kJ ¥ t2 ∞C kg

kJ = 4.2 ¥ t2 kg t2 = 90.39°C

EXAMPLE 3.33 A closed system as shown in the figure undergoes a process from state 1 to 2. During the process, heat at the rate of 7 kJ/s is transferred and work is done by the system at the rate of 3 kJ/s. System takes 1 hour to complete the process. Calculate (a) The rate of energy accumulation in the system (b) Total increase in the internal energy of the system.

138

Basic Thermodynamics

Solution: (a) The rate of energy accumulation in the system Du kJ =7-3=4 or kW dt s (b) Total increase in the internal energy of the system Q -W =

DU = U 2 - U1 =

Ú

time 2

time1

W

Q

dU kJ = 4 ¥ 3600 s dt s

Figure E3.33

= 14400 kJ EXAMPLE 3.34 A bag contains gas at 25 bar, 50°C and 20 cm3. The bag is kept in a container of volume 2500 cm3. The container is perfectly insulated and evacuated. Now the bag is bursted so that gas occupies the whole container. Calculate the final pressure after the equilibrium. Solution:

Apply the first law of thermodynamics to container

(a) Final pressure (P2) after the equilibrium Q + W = DU

[W = 0,

V2 = 2500 cm

T2 = T1 = constant temperature P1V1 P2V2 = T1 T2

P2 = P1

Q=0

∵ V = C, Q = 0]

DU = m(u2 – u1) = 0 \

W=0

3 3

Gas

V1 20 = 25 ¥ 100 ¥ = 20 bar V2 2500

V1 = 20 cm P1 = 25 bar T1 = 50°C

Figure E3.34

EXAMPLE 3.35 A system is shown in the figure. Data are given in the figure. The membrane is punctured and the gases are allowed to mix. Determine the temperature and pressure after equilibrium has been established. Take CvN2 = gas.

5R 3R and CvHe = . Assume N2 and He as perfect 3 2

Solution: (a) Final temperature (Tf) nN2 = Number of moles of nitrogen =

P2V2 500 ¥ 1 = RT2 8.314 ¥ 500

= 0.1203 kg mole After the equilibrium, [ nCv (T f - 400)]He = [ nCv (500 - T f )]N2

Figure E3.35

First Law of Thermodynamics

139

where Tf = Final temperature after equilibrium P1V1 200 ¥ 1 = 0.06 kg mole = RT1 8.314 ¥ 400

nHe = Number of moles of helium =

0.06 ¥ 3 ¥ 8.314 (T f - 400) 2

=

0.1203 ¥ 5 ¥ 8.314 (500 - T f ) 3

Tf = 469.01 K

(b) Final pressure (Pf) Pf V f = (nHe + nN2 ) RT f

Pf = (0.06 + 0.1203) × 8.314 × 469 = 703.05 kPa EXAMPLE 3.36 A rigid, insulated vessel is shown in the figure with data. The partition is removed and the water in both the compartment fills up the entire vessel. Determine (a) the final pressure and (b) heat interaction. Final temperature is 200°C. Solution: Compartment A Pgiven > Psat

\

liquid state, from tables

vA = 0.001153 m3/kg uA = 848.1 kJ/kg Compartment B Tgiven > Tsat

\ Super heated state, from tables

vB = 0.4249 m3/kg,

A 3 kg 200°C 5 MPa

A 2 kg 200°C 0.5 MPa

Figure E3.36

uB = 2642.5 kJ/kg

V = The volume of the container = VA + VB = mU Vf = Final volume = mAuA + mBuB = 3 × 0.001153 + 2 × 0.4249 = 0.8533 m3 mf = Total mass = mA + mB = 2 + 3 = 5 kg \

vf = Final specific volume =

Vf mf

=

0.8533 m3 = 0.171 5 kg

(a) Final pressure (Pf) Now we are having \

Tf = 200°C

From steam tables, Pf ª1.2 MPa,

Vf = 0.171 m3/kg uf ª 2612.3 kJ/kg

(b) Heat interaction (Q) Now consider the compartments A and B as a system. The container is rigid. \

W=0

Q = W + DU = U2 – U1 = Uf – Ui = mf uf – [mA uA + mB uB] = 5 × 2612.3 – [3 × 848.13 + 2 × 2642.5] = 5232.2 kJ

140

Basic Thermodynamics

EXAMPLE 3.37 A piston and cylinder arrangement with an initial volume of 0.1 m3 holds water at 1.4 MPa and 250°C. The piston is held in position by a pin while the pressure is raised to 2.0 MPa. At this constant pressure, the pin is removed and additional heat is transferred until the temperature reaches 400°C. Calculate the total heat transferred during the process to the system. Solution: P

2

3

1 Q State 2

State 1

Q State 3

V

Figure E3.37

State 1

V1 = 0.1 m3

P1 = 1.4 MPa T1 = 250°C

From steam tables, state 1 is a superheated vapour. \

v1 = 0.16347 m3/kg,

State 2

V1 = 0.1 m3 P1 = 2.0 MPa

m1 = m2 = m3 = m =

h1 = 2926.4 kJ/kg v1 = v2 h2 ª 3357.5 kJ/kg

V 0.1 = = 0.612 kg v f 0.16347

From the 1st law of thermodynamics, Q1–2 = W1–2 + U2 – U1

[W1–2 = 0,

∵

V1 = V2]

Q1–2 = m(u2 – u1) = m[(h2 – h1) – v1(p2 – p1)] = 0.612 × [3357.5 – 2962.4) – 0.16347(2.0 – 1.4)103] = 181.78 kJ State 2 to 3 is a constant pressure process \

Q2–3 = mCp(T3 – T2) = m[(h3 – h2) – P(v3 – v2)] + W2–3 W2–3 = mP(v3 – v2)

From steam tables at 2.0 MPa and 400°C, v3 = 0.15119 m3/kg

h3 = 3247.5 kJ/kg

\ W2–3 = 0.612 × 2 × 1000 × (0.15119 – 0.1634) = –14.95 kJ Q2–3 = 0.612 [(3247.5 – 3357.5) – 2 × 1000 × (0.15119 – 0.1634)] + (– 46.45) = – 89.35 \

Q1–2–3 = Total heat transfer = Q1–2 + Q2–3 = 181.78 + (–89.35) = 92.43 kJ

Heat is transferred to the system.

First Law of Thermodynamics

141

EXAMPLE 3.38 A vessel contains an inert gas at 60°C and 10 bar. The vessel is then heated until the pressure becomes 15 bar. For kg mole of inert gas, calculate (a) The heat transferred per kg. (b) The change in total internal energy (c) The change in total enthalpy Take

Cv = 1.00 + 3.02 ¥ 10 -3 T

kJ , where T is in Kelvin. kg mole K

Solution:

DU =

Ú

T2

T1

Cv dT =

Ú

T2

T1

(1 + 3 ¥ 10 -3T ) dT = (T2 - T1 ) + 3 ¥ 10 -3

= (499.5 - 333) + 3 ¥ 10 -3

(499.52 - 3332 ) 2

È P1 P2 ˘ P 15 \ T2 = T1 2 = 333 ¥ = 499.5 K ˙ Í = P1 10 Î T1 T2 ˚

Du = 582.33 kJ/kg mole W1–2 = 0 (Rigid vessel

(T22 - T12 ) 2

\ V = C)

(a) The heat transferred (q1–2) q1–2 = W1–2 + DU

q1- 2 = 582.33

kJ kg mole K

DU) (b) The change in total internal energy (D DU = n Du = 1 kg mole × 582.3

kJ = 582.33 kJ kg mole

DH) (c) The change in total enthalpy (D DH = DU + D(PV) = DU + (P2V2 – P1V1) = DU + V1[P2 – P1] = 582.33 kJ + 2.77 (15 – 10) × 100 kJ = 1967.33 kJ È ˘ Í ˙ nRT1 1 kg mole ¥ 8.314 kJ ¥ 333 K = = 2.77 m 3 ˙ Í P1V1 = nRT1 \ V1 = kN P1 Í ˙ 10 ¥ 100 2 kg mole K ÍÎ ˙˚ m

FIRST LAW—STEADY FLOW PROCESS EXAMPLE 3.39 In a heat exchanger, hot water enters pipe with 59 kJ/kg and leaves with 47.5 kJ/kg. Exit of the pipe is 10 m above the entry. The flow rate of heat is 400 kJ/s. Estimate the rate of water circulation through the pipe. (KUD II-83)

142

Basic Thermodynamics

Solution: Apply SFEE for single stream, i.e. Eq. (3.50) and 1 for inlet condition and 2 for exit condition. È È (V )2 gZ ˘ (V )2 gZ ˘ Q + m Í h1 + 1 + 1 ˙ = m Í h2 + 2 + 2 ˙ + W gc ˙˚ gc ˙˚ 2 gc 2 gc ÎÍ ÎÍ W=0

DK.E. = 0

h1 = 59.0 kJ/kg h2 = 47.5 kJ/kg

Z1 = 0

Z2 = 10 m

(a) Flow rate of water (m) È ((V1 )2 - (V2 )2 ) g( Z1 - Z 2 ) ˘ + Í(h1 - h2 ) + ˙ +Q =0 2 gc gc ÎÍ ˚˙ (10 - 0) ¥ 9.81 ˘ kJ È [–ve heat lost to the surroundings] Q = Í (47.5 - 59) + = - 11.402 ˙ 1000 kg Î ˚

\

m = mass flow rate of water =

Heat flow rate kJ kg kg = 400 ¥ = 35.08 Q s 11.402 kJ s

EXAMPLE 3.40 The steam nozzle shown in the following figure is having data as below. P1 = 15 bar u1 = 0.15 m3/kg U1 = 620 kJ/kg V1 = 1800 m/s 3 P2 = 1 bar u2 =1.7 m /kg U2 = 600 kJ/kg Z1 = Z2 Calculate exit velocity. Assume adiabatic process, i.e. Q = 0. (KUD II-82) Solution: (a) Exit velocity (V2) Apply SFEE to the nozzle, i.e. Eq. (3.50) and 1 for inlet condition and 2 for exit condition.

Control surface 1

2

Figure E3.40

È È (V )2 gZ ˘ (V )2 gZ ˘ m Í q + h1 + 1 + 1 ˙ = m Í w + h2 + 2 + 2 ˙ gc ˙˚ gc ˙˚ 2 gc 2 gc ÍÎ ÍÎ W=0 q=0 DZ = 0 [h1 = u1 + p1u1 = 620 + (15 × 100 × 0.15) = 845.00 kJ/kg h2 = u2 + p2u2 = 600 + (1 × 100 × 1.7) = 770.00 kJ/kg ] EXAMPLE 3.41 Modify Example 3.39 in such a way that there is heat transfer of 10 kJ/kg and calculate exit velocity V2 . Solution:

Here W = 0

q = 10 kJ/kg

First Law of Thermodynamics

143

(V2 )2 (V )2 1800 2 = (h1 - h2 ) + 1 + q = (845 - 770) + = 10 2 ¥ 1000 2 gc 2 ¥ 1000

\

V2 = 1835.76 m/s EXAMPLE 3.42 A steam turbine receives steam with a flow rate of 54000 kg/hr and experiences a heat loss of 50.4 MJ per hour. The exit pipe is 3 m below the level of the inlet pipe. Find the power developed by the turbine if the pressure decreases from 6.20 MPa to 9.86 kPa. Velocity increases from 30.5 m/s to 274.3 m/s, internal energy decreases by 938.5 kJ/kg and specific volume increases from 0.058 m3/kg to 13.36 m3/kg. (KUD II-98) Solution:

Figure E3.42

u1 = 0.058 m3/kg (u2 – u1) = –938.5 kJ/kg Q=

u2 = 13.36 m3/k (Z2 – Z1) = –3 m

50.4 ¥ 1000 kJ = 14 3600 s

Power developed (W ) We have SFEE, Eq. (3.50), È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q - W = m Í( P2 v2 - P1v1 ) + (u2 - u1 ) + 2 ˙ 2 gc gc ÍÎ ˙˚ È (274.32 - 30.52 ) 9.81( -3) ˘ -14 - W = 15 Í(9.86 ¥ 13.36 - 6200 ¥ 0.058) + ( -938.5) + + ˙ 2 ¥ 1000 1000 ˙˚ ÍÎ W = 16924.67

kJ or 16.92 MW s

EXAMPLE 3.43 A steam turbine operates under steady flow conditions and receives steam at the following state.

144

Basic Thermodynamics

P1 = 1.2 MPa h1 = 2785 kJ/kg

P2 = 20 kPa h2 = 2512 kJ/kg

T1 = 188°C T2 = ? V1 = 33.3 m/s V2 = 100 m/s

Z1 = 4 m

Z2 = 1 m Q = - 0.29 kJ/s

ms = 30 m/min = 0.5 kg/s

Calculate power output.

(KUD Aug 2001)

Solution: Refer to figure of Example 3.42. Power output (W ) Apply SFEE, Eq. (3.50), È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q - W = m Í(h2 - h1 ) + 2 ˙ 2 gc gc ÍÎ ˙˚ È (100 2 - 33.32 ) 9.81(1 - 4) ˘ -0.29 - W = 0.5 Í(2512 - 2785) + + ˙ 2 ¥ 1000 1000 ˙˚ ÍÎ W = 134.58 kJ/s or kW EXAMPLE 3.44 A steam turbine developes 600 kW with a heat loss of 525 kJ/min to the surroundings. Using the followings data and assuming steady flow condition, find the rate of steam flow through the turbine in kg/hr. (KUD II-93) Properties Pressure Velocity Elevation Enthalpy

Inlet 2 × 106 N/m2 180 m/s 5m 3200 kJ/kg

Outlet 1.1 × 106 N/m2 50 m/s 3m 2200 kJ/kg

Solution: The rate of steam flow through the turbine (m) Apply SFEE, Eq. (3.50), È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q - W = m Í(h2 - h1 ) + 2 ˙ 2 gc gc ÎÍ ˚˙ È (50 2 - 180 2 ) 9.81(3 - 5) ˘ -8.75 - 600 = m Í(2200 - 3200) + + ˙ 2 ¥ 1000 1000 ˚˙ ÎÍ \

m = 0.5988

kg kg or 2159.2 s hr

EXAMPLE 3.45 Air enters an air turbine at a pressure of 3 bar and a temperature of 58°C at a steady velocity of 45 m/s. The air leaves the turbine at a pressure of 1.1 bar and a temperature of 2°C and a velocity of 150 m/s. Neglecting changes in elevation determine the magnitude and direction of heat transfer per kg of air flow if the shaft output is 54 kJ/kg of air. (KUD II-92)

First Law of Thermodynamics

145

Solution: Heat transfer (Q) Apply SFEE, DZ = 0, Eq. (3.50) È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q - W = Í(h2 - h1 ) + 2 ˙ 2 gc gc ÎÍ ˚˙ È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q = W + ÍC p (T2 - T1 ) + 2 ˙ 2 gc gc ÎÍ ˚˙ Q = 54

kJ È (150 2 - 452 ) ˘ + Í1.005(2 - 58) + ˙ kg ÎÍ 2 ¥ 1000 ˙˚

Q = 7.9575 kJ/kg

Heat transferred to the system.

EXAMPLE 3.46 Water flows through a pump with a mass flow rate of 3 kg/s. The entering state of the water is 6.5 bar and 20°C. While the exiting state is 32 bar and 20°C. Negligible changes in P.E. and K.E. energies occur during the process, the flow is steady state and process is adiabatic. Calculate the power input required by the pump. (KUD) Solution: Q = 0 Adiabatic DP.E. = 0

DK.E. = 0

2

At state 1, the water is in compressed state. From steam tables at 20°C uf = 0.0010018 m3/kg

Psat = 2.3388 kPa

hf = 83.84 kJ/kg

Pump

P1 = 650 kPa

\ h1 = hf at 20°C + (P1 – Psat)uf Q=0

= 83.84 + (650 – 2.3388) × 0.0010018 = 84.49 kJ/kg h2 = 83.84 + (3200 – 2.3388) × 0.0010018 = 87.04 kJ/kg (a) The power input (W ) Apply SFEE, Eq. (3.50)

1

Figure E3.46

È (V )2 gZ ˘ È (V )2 gZ ˘ Q + Í h1 + 1 + 1 ˙ = Í h2 + 2 + 2 ˙ + W gc ˙˚ ÎÍ gc ˙˚ 2 gc 2 gc ÎÍ \

h1 = W + h2

\

W = (h1 + h2 ) = (84.49 - 87.04) = - 2.55 kJ/kg W = - 2.55

kJ kg ¥3 = - 7.66 kW kg s

W

146

Basic Thermodynamics

EXAMPLE 3.47 Water enters at 300 kPa and 30°C is mixed in an adiabatic mixing chamber with steam at 300 kPa and 250°C which is entering at the rate of 80 kg/s. The mixture leaves as a saturated vapour at 300 kPa. Determine (a) the mass flow rate of the water entering the chamber and (b) the mass flow rate of saturated vapour leaving the chamber. (KUD) Solution:

Water

1 3

Steam

Inlet water 300 kPa 30°C m2 = 80 kg/s

2

Steam 300 kPa 250°C

Exit vapour 300 kPa

Figure E3.47

(a) The mass flow rate of the water entering (m1) The conservation of energy equation, m1h1 + m2h2 = m3h3 = (m1 + m2)h3 = m1h3 + m2h3 È h - h2 ˘ m1 = m2 Í 3 ˙ Î h1 - h3 ˚ From steam tables at 30°C,

\

hf = 125.67 kJ/kg uf = 0.0010044 \

m3/kg

Psat = 4.2455 kPa P1 = 300 kPa

h1 = hf 20°C + (P1 – Psat)uf = 125.67 + (300 – 4.2455) × 0.0010044 = 125.967 kJ/kg

From steam tables at 300 kPa and 250°C, h2 = 2967.1 kJ/kg From steam tables at 300 kPa and saturated vapour, h3 = 2725.3 kJ/kg È (2725.3 - 2967.1) ˘ m1 = 80 Í ˙ = 7.442 kg/s Î (125.97 - 2725.3) ˚ (b) The mass flow rate of saturated water leaving (m3) \

m3 = m1 + m2 = 7.442 + 80 = 87.442 kg/s EXAMPLE 3.48 An air compressor is used to compress the air from 100 kPa and 20°C to 1 MPa. The heat transfer to the surrounding is 15% of the power input to the compressor. The air enters at 50 m/s and inlet area is 10 × 10–3 m2 and leaves at 110 m/s through an area 5 × 10–3 m2. Calculate (a) the exit air temperature, (b) power input to the compressor and (c) ratio of inlet pipe diameter to outlet pipe diameter.

First Law of Thermodynamics

Solution: Q = 15% W Inlet W

100 kPa 20°C 50 m/s –3 2 10 × 10 m

Air

Outlet 1000 kPa 110 m/s –3 2 0.5 × 10 m

2 1

Figure E3.48

(a) Exit temperature (T2) m1 =

A1V1 10 ¥ 10 -3 ¥ 50 ¥ P1 10 ¥ 10 -3 ¥ 50 ¥ 100 kg = = = 0.5946 v1 RT1 0.287 ¥ 293 s

È ˘ RT1 , treating air as ideal gas ˙ Í v1 = P1 Î ˚ We know that

m1 = m2 = T2 =

A1V1 P1 A2V2 P2 = RT1 RT2

A2V2 P2 0.5 ¥ 10 -3 ¥ 110 ¥ 1000 = = 322.3 K or 49.3°C Rm1 0.287 ¥ 0.5946

(b) Power input to the compressor (W) Apply SFEE, Eq. (3.50), È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ -Q - W = m Í(h2 - h1 ) + 2 + ˙ 2 gc gc ÎÍ ˚˙ From air tables,

h2 at 49.3°C = 321 kJ/kg

h1 at 20°C = 293.3 kJ/kg

È (110 2 - 50 2 ) ˘ - 0.15W - W = 0.5946 Í(321 - 293.3) + ˙ 2 ¥ 1000 ˙˚ ÎÍ -0.5946 È (110 2 - 50 2 ) ˘ Í(321 - 293.3) + ˙ = - 16.8 kW 1.15 ÎÍ 2 ¥ 1000 ˙˚ (c) Ratio of inlet pipe diameter to outlet pipe diameter (di/do) W=

We have,

A1V1 A2V2 = v1 v2

147

148

Basic Thermodynamics

A1 V2 n1 110 P2 RT1 110 ¥ 293 ¥ 1000 d2 = = = = 20 = 12 A2 V1n2 50 P1 RT2 50 ¥ 322.3 ¥ 100 d2

or

A1 d12 10 ¥ 10 -2 = = = 20 A2 d22 0.5 ¥ 10 -2

\

d1 = 4.47 d2

\

d1 = 4.47 d2

EXAMPLE 3.49 A centrifugal compressor draws 15 m3/min of air whose density is 1.3 kg/m3 and discharges at a density of 4.5 kg/m3. The inlet and outlet pressures are 1 bar and 6 bar, respectively. The work input to the compressor is 40 kW. Calculate the change in internal energy per kg of air. The heat lost to the surrounding is 15 kJ/kg. Assume changes in P.E. and K.E. are neglected. Solution: Apply SFEE, Eq. (3.50), È È (V )2 gZ ˘ (V )2 gZ ˘ Q + Í h1 + 1 + 1 ˙ = W + Í h2 + 2 + 2 ˙ gc ˙˚ gc ˙˚ 2 gc 2 gc ÍÎ ÍÎ DK.E. = 0 DP.E. = 0 \

Q + h1 = W + h2 Q + [u1+ P1u1] = W+ [u2 + P2u2] (u2 – u1) = Q – W + [P1u1 – P2u2]

È 1 1 m3 1 1 m3 ˘ = = = = = = 0.7692 0.2222 v v Í 1 ˙ 2 4.5 kg ˙ 1.3 mg 1 2 Í Í ˙ 3 Í m = volume ¥ density = 15 m ¥ 1.3 kg = 0.325 kg ˙ ˙˚ 60 s s m3 ÎÍ \

(u2 – u1) = –15 +

\

(U2 – U1) = 51.68

40 kJ + [(1 × 0.7692 – 6 × 0.2222) × 100] = 51.68 0.325 kg kJ kJ kJ kJ × m = 51.68 × 0.325 = 16.79 s s kg kg

EXAMPLE 3.50 A diffuser is shown in the figure. Following are the properties at inlet and outlet. Calculate the temperature of the air leaving the diffuser. Diffuser is well insulated. The air leaving the diffuser with a velocity that is very low compared to the inlet velocity. Solution: At inlet

P1 = 300 kPa

t1 = 120°C

m2

m1 = 5 kg/s

A1 = 0.01 At outlet

V2 ª 0

First Law of Thermodynamics

149

(a) Exit temperature (T2) 1

\

=

P1 300 kg = = 0.877 2 RT1 0.87 ¥ 393 m

1

Air

V1 = inlet velocity =

m1 5 m = = 57.01 0.877 ¥ 0.01 S 1 A1

Figure E3.50

For a diffuser, W = 0 DZ = 0 Q = 0 (adiabatically covered) \

2

SFEE reduces to (h1 - h2 ) +

((V1 )2 - (V2 )2 ) =0 2 gc

Since the air is an ideal gas and the temperature changes of the air is small, hence specific heats are nearly constant. C p (T1 - T2 ) +

\

((V1 )2 - (V2 )2 ) 2 gc

Given data: Exit velocity is very low and hence neglected, i.e. V2 ª 0 \

C p (T1 - T2 ) +

\

T2 =

(V1 )2 =0 2 gc

(V1 )2 570.12 + T1 = + 393 = 554.69 K 2 gc ¥ C p 2 ¥ 1000 ¥ 1.005

EXAMPLE 3.51 Refrigerant-12 enters a throttling valve in a refrigeration unit as a saturated liquid with a pressure of 2 MPa. It is discharged at 250 kPa. Calculate (a) inlet, (b) exit temperatures and (c) exit quality of the refrigerant. Assume DK.E., DP.E., Q, W = 0 Solution: (a) Inlet temperature (T1) Apply conservation of mass m1 = m2 = m

Apply conservation of energy, i.e. SFEE, Eq. (3.50)

Q - W = m [ h2 + K.E.2 + P.E.2 ] - m [ h1 + K.E.1 + P.E.1 ] DK.E., DP.E., Q,

W = 0 m1 = m2

From R-12 tables at 2 MPa, T1 = 73.07°C

\

h2 = h1

h1 = 111.54 kJ/kg

150

Basic Thermodynamics P 1

Throttle valve 1

2

Saturated liquid

4

2

3 V

Figure E3.51

(b) Exit temperature (T2) At exit, h1 = h2 = 111.54 kJ/kg and 250 kPa From R-12 tables at 250 kPa, (h4)hf = 30.293 kJ/kg, (h3)hg = 185.78 kJ/kg \ h1 = h2 = 115.54 kJ/kg lies in between hf and hg t2 = tsat = –6.27°C \ exit condition is mixture of liq and vap. (c) Exit quality (x2) h2 = h4 + x2[h3 – h4] 111.54 = 30.293 + x1[185.78–30.293] \ x2 = 0.5225 EXAMPLE 3.52 A heat exchanger with properties are shown in the figure. The flow rate of air is 5 kg/s. Assume outer shell is well insulated. Calculate the temperature of the air as it leaves the heat exchanger. Solution: Conservation of energy equation for air side, Qa = [ma2ha2 – ma1ha1] = ma [ha2 – ha1]

(1)

Conservation of energy equation for steam side, Qs = (ms1hs1 – ms2hs2) = ms[hs1 – hs2] Air out Pa2 = 100 kPa, ta2

Saturated steam out P = 200 kPa

Steam ms1 = 1.5 kg/s T = 200°C P = 100 kPa

Air in

Figure E3.52

ms1 = 3 kg/s T = 30°C, Pa1 = 100 kPa

(2)

First Law of Thermodynamics

151

We know that heat lost by steam is gained by air. \ Equating Eqs. (1) and (2), ma(ha2 – ha1) = ms(hs1 – hs2)

(3)

From air tables at 30°C, ha1 = 303.32 kJ/kg From steam tables at 200 kPa and 200°C, hs1 = 2870 kJ/kg

(4)

and at 200 kPa and saturated steam, hs2 = 2707 kJ/kg Substitute Eq. (4) in Eq. (3) \

3

kg kJ kg kJ [ ha 2 - 303.32] [2870 - 2707] = 1.5 s kg s kg

\ ha 2 = 384.82 kJ/kg

The temperature of the air at exit of the heat exchanger (Ta2) From air tables corresponding to 384.82 kJ/kg, temperature, Ta2 is 384 K \

Ta2 = 384 K or

111°C

EXAMPLE 3.53 A centrifugal pump is shown in the figure with properties at inlet and outlet. Volume flow rate of water is 50 litre/s. The pump is well insulated and heating of the water due to frictional effect is so small that it may be neglected. Calculate the temperature of the water at the pipe exit. Solution:

As per the continuity equation V1A1 = V2A1 = 50 litre/s V1 = Velocity at inlet P2 = P1 = 1 bar d1 = 190 mm T1 = ?

H 2O

100 m

W = 75 kW 1

Pa = 1 bar = P1

River

Figure E3.53

V1A1 = 50 Lit/s d1 = 125 mm T1 = 25°C

152

Basic Thermodynamics

=

50 litre 1 m3 ¥ 4 m ¥ = 4.07 2 2 s s 1000 litre ¥ ¥ (0.125) m

V2 = Exit velocity =

\

V1 A1 A2

2

m È 125 ˘ = 4.07 Í = 1.7635 ˙ s Î 190 ˚

m = mass flow rate =

1 A1V1 = 1000

kg m3

¥

50 litre 1 m3 kg ¥ = 50 s 1000 litre s

(a) The temperature of the water at the exit (T2) Apply SFEE, È È (V )2 gZ ˘ (V )2 gZ ˘ Q - W = m Í h2 + 2 + 2 ˙ - m Í h1 + 1 + 1 ˙ gc ˚˙ gc ˚˙ 2 gc 2 gc ÎÍ ÎÍ Q = 0 [Adiabatic] Dh = Cp(T2 – T1) [Water is incompressible fluid] ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ -( -75) È = ÍC p (T2 - T1 ) + 2 + ˙ m 2 gc gc ÍÎ ˙˚ ((1.764)2 - (4.07)2 ) 9.81( -100) ˘ -( -75) È = ÍC p (T2 - T1 ) + + ˙ 50 2 ¥ 1000 1000 ˙˚ ÍÎ È kJ ˘ Cp(T2 – T1) = 0.5257 Í∵ C p = 4.18 ˙ kg K˚ Î \ T2 = 298.125 K or 25.125°C EXAMPLE 3.54 A centrifugal pump delivers 60 kg of water per second. The inlet and outlet pressures are 10 kPa and 400 kPa, respectively. The suction is 2 m below and delivery is 8 m above the centre line of the pump. The suction and delivery pipe diameters are 20 cm and 10 cm, respectively. Determine the capacity of the electric motor to run the pump. (VTU model question paper) Solution: m 1 V1 = Velocity at inlet = A1 =

60 kg ¥ 1 m 3 ¥ 4 s ¥ 1000 kg ¥

2

¥ (0.2) m

V2 = Velocity at exit = =

60 ¥ 1 ¥ 4 1000 ¥

¥ (0.1)

2

2

m 2 A2

= 7.64

m s

= 1.91

m s

153

First Law of Thermodynamics

P2 = 400 kPa d2 = 10 cm 2 8m Water 2m 1 P1 = 10 kPa d1 = 20 cm

Figure E3.54

Capacity of the electrical motor to run the pump (W) Apply SFEE, Eq. (3.50), È Q - W = Íu2 + P2 ÎÍ È -W = Í( P2 ÎÍ

2

2

Q=0 +

u2 = u1 u1 = u2 (in compressible)

(V2 )2 gZ 2 ˘ È + ˙ - Íu1 + P1 gc ˙˚ ÎÍ 2 gc

- P1 1 ) +

1

+

(V1 )2 gZ1 ˘ + ˙ gc ˙˚ 2 gc

((V2 )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + ˙ 2 gc gc ˚˙

È 400 ¥ 1 10 ¥ 1 ˘ 7.642 - 1.912 9.81(10 - 0) =Í + + 1000 ˙˚ 2 ¥ 1000 1000 Î 1000 W = - 0.5155

kJ kg

W = - 0.5155

kJ kg kg ¥ 60 = - 30.93 or kW kg s s

[–ve sign represents work done on the system] EXAMPLE 3.55 A steam turbine is shown in figure with properties at inlet and outlet. The mass flow rate of steam is 40 kg/s. Calculate power developed by the turbine. Neglect heat interaction. 2 Solution:

From conservation of mass m1 = m2 = m A1V1 A2V2 = =m v1 v2

A1 = 0.05 m P1 = 15 MPa T1 = 600°C m = 30 kg/s

1

W

Steam

From steam tables at 15 MPa and 600°C, h1 = 3581.5 kJ/kg

u1 = 0.02490 m3/kg 2

at 100 kPa and x = 1, h2 = 2675.1 kJ/kg

u2 = 1.694 m3/kg

A2 = 0.30 m P2 = 100 kPa x=1

Figure E3.55

2

154

Basic Thermodynamics

V1 =

mV1 kg m3 1 m = 40 ¥ 0.02490 ¥ = 19.92 2 A1 s kg 0.05 m s

V2 =

mv2 kg m3 1 m = 40 ¥ 1.694 ¥ = 225.87 2 A2 s kg 0.3 m s

Power developed by the turbine (W) Apply SFEE, Eq. (3.50),

Q = 0 DZ = 0

È ((V )2 - (V2 )2 ) g( Z1 - Z 2 ) ˘ + W = Í(h1 - h2 ) + 1 ˙ 2 gc gc ÎÍ ˚˙ = (35815 - 2675.1) + W = 881.1

(19.92)2 - (225.87)2 kJ 881.1 2 ¥ 1000 kg

kJ kg ¥ 40 = 35243.62 kW kg s

EXAMPLE 3.56 A steam power plant is shown in the figure. Properties at various locations are tabulated below. State

Temperature (°C)

Pressure (MPa)

Velocity (m/s)

1

600°C

20

–

2

x2 = 0.95

0.075

150 m/s

3

50°C

0.075

–

4

80°C

20

–

The water flow rate is 10 kg/s. Calculate the following. (a) (b) (c) (d) (e)

Amount of heat transfer to the water in the boiler (Qs) Work developed by the turbine (Wt) Amount of heat transfer from the water (QR) Work done on the pump (WP) The thermal efficiency (nth)

Solution: From steam tables at 20 MPa and 600°C,

h1 = 3536.7 kJ/kg

At 75 kPa and x2 = 0.95,

hf = 384.43,

\

hg = 2662.5 kJ/kg

h2 = hf + x2(hg – hf) = 384.43 + 0.95[2662.5 – 384.43] = 2548.6 kJ/kg

From steam tables at 50°C and 75 kPa,

h3 = 209.33 kJ/kg

From compressed water table at 80°C and 20 MPa (It is a compressed liquid) h4 = 350.82 kJ/kg

First Law of Thermodynamics

155

1

Q=0

Boiler Qs

Wt

Turbine

4 2

Condenser 3 Pump

Wp

QR

Figure E3.56

(a) Amount of heat transfer to the water in the boiler (Qs) Qs = m(h1 – h4) = 10 × (3536.7 – 350.82) = 31858.8 kW (b) Work developed by the turbine (Wt) È È V2 ˘ 150 2 ˘ Wt = m Í(h1 - h2 ) - 2 ˙ = 10 Í(3536.7 - 2548.6) ˙ = 9768.5 kW 2 gc ˙˚ 2 ¥ 1000 ˚˙ ÎÍ ÎÍ (c) Amount of heat transfer from the water in the condenser (QR) È È V2 ˘ 150 2 ˘ QR = m Í(h3 - h2 ) - 2 ˙ = 10 Í(209.33 - 2548.6) ˙ = - 23505.2 kW 2 gc ˙˚ 2 ¥ 1000 ˙˚ ÍÎ ÍÎ (d) Work done on the pump (Wp) Wp = m[h3 – h4] = 10 × [209.33 – 350.82] = –1414.9 kW h th) (e) The thermal efficiency (h th

=

(Wt + Wp ) Qs

¥ 100 =

(9768.5 - 1414.9) ¥ 100 = 26.22% 31858.5

EXAMPLE 3.57 Modify Example 3.54 in such a way that the steam enters the turbine at 19 MPa and 550°C. Remaining properties are same. Calculate the following. (a) Heat transfer between boiler and turbine (b) Turbine work (c) Heat transfer in condenser (d) Heat transfer in the boiler Neglect DK.E., DP.E. = 0

156

Basic Thermodynamics

1

2

Q=0

Boiler Qs

Wt

Turbine

5

3 4

Condenser QR

(Pump Wp)

Figure E3.57

State

Temperature (°C)

Pressure (MPa)

1

600

20

2

550°

17.5

3

x2 = 0.95

0.075

4

50

0.075

5

80

20

Velocity (m/s)

150

Solution: From steam tables at 17.5 MPa, 550°C, h2 = 3421.7 kJ/kg

h1 = 3536.7 kJ/kg (taken from Example 3.55)

h3 = 2548.6 kJ/kg

h4 = 209.33 kJ/kg

h5 = 350.82 kJ/kg (taken from Example 3.55)

(a) Heat transfer between boiler and turbine (Q1-2) Q1–2 = (h2 – h1)m = (3421.7 – 3536.7) × 10 = –1150 kW (b) Turbine work (Wt) È È V2 ˘ 150 2 ˘ Wt = m Í(h2 - h3 ) - 3 ˙ = 10 Í(3421.7 - 2548.6) ˙ = 8618.5 kW 2 gc ˙˚ 2 ¥ 1000 ˚˙ ÎÍ ÎÍ (c) Heat transfer in condenser (QR) È È V2 ˘ 150 2 ˘ QR = m Í(h4 - h3 ) - 3 ˙ = 10 Í(209.33 - 2548.6) ˙ = - 23505.2 kW 2 gc ˙˚ 2 ¥ 1000 ˚˙ ÎÍ ÎÍ (d) Heat transfer in the boiler (Qs) Qs = m[h1 – h5] = 10 × [3536.7 – 350.82] = 31858.8 kW EXAMPLE 3.58 A fluid is flowing through a pipe as shown in the figure with properties at inlet and exit. Calculate the area of the inlet of the pipe. Specific volume of the fluid is 0.015 m3/kg.

First Law of Thermodynamics

Solution: V1 = 10 m/s h1 = 1000 kJ/s

1

Q = 50 kJ/s 10 m 2 m = 5 kg/s V2 = 5 m/s h2 = 1020 kJ/s

Figure E3.58

The area of the inlet of the pipe (A1) v = Specific volume = 0.015 m3/kg È ((V )2 - (V1 )2 ) g( Z 2 - Z1 ) ˘ + Q - W = Í(h2 - h1 ) + 2 ˙ 2 gc gc ÍÎ ˙˚ 50 (52 - 10 2 ) 9.81(0 - 10) - W = (1020 - 1000) + + 5 2 ¥ 1000 1000 -W = 9.8644

kJ kg

W = - 9.8644

kJ kg ¥5 = - 49.32 kW kg s

A1 = Area at inlet =

EXAMPLE 3.59

m 1 5 ¥ 0.015 = = 0.0075 m 2 V1 10

Calculate the work done for the following process.

2

2

P2 = 100 kPa Z2 = 2 m V2 = 3 m/s

Oil (r = 160 kg/m) P1 = 200 kPa Z1 = 0 V1 = 0 m/s

Wp

1

Figure E3.59

Length of the pipe = 2200 cm. Assume a loss of 50 kJ/kg per 100 cm.

157

158

Basic Thermodynamics

Solution: Work done (Wp) We have DP ( P2 - P1 )

+ DK.E. + DP.E. + W + losses = 0

((V2 )2 - (V1 )2 ) g( Z 2 - Z1 ) + + W + losses = 0 2 gc gc

+

(100 - 200) (32 - 0 2 ) 9.81(2 - 0) 0.5 ¥ 2200 + + + Wp + 160 2 ¥ 1000 1000 100

Wp = - 10.399 kJ/kg

UNSTEADY FLOW PROCESS EXAMPLE 3.60 An air compressor is used to supply air to a rigid tank that has a volume of 4 m3. Initially, the pressure and temperature of the air in the tank are 200 kPa and 40°C respectively. The supply pipe to the tank is 8 cm in diameter, and the velocity of the air in the inlet pipe remains constant at 12 m/s. The pressure and temperature of the air in the inlet pipe are constant at 600 kPa and 40°C, respectively. Calculate the following. (a) The mass flow rate of the change. (b) The mass of air added to the tank if the compressor stops operating when the tank reaches 400 kPa and 60°C. (c) The time that the compressor must be operated to produce a tank pressure of 400 kPa and at temperature of 60°C. Solution: (a) The mass flow rate of charge (mi) 1

=

P1 = RT1

600 kPa kPa m 3 0.287 ¥ 313 K kg K

= 6.679 kg/m3 mi = riViAi = 6.679

kg m3

2

¥ 12

= 0.513 kg/s

m Ê 8 ˆ ¥ ¥Á m2 s 4 Ë 100 ˜¯

Compressor

Pipe line V1 = 12 m/s P1 = 600 kPa T1 = 40°C d1 = 8 cm

Valve i

i

Tank air 3 P1 = 200 kPa, V = 4 m T1 = 400 C, P2 = 400 kPa T2 = 60°C

Control surface

Figure E3.60

(b) The mass of air added to the tank if the compressor stops operating when the tank reaches 400 kPa and 60°C (m) m2 = mass of air in the tank if air reaches 400 kPa and 60°C

First Law of Thermodynamics

=

P2V2 = RT2

159

400 kPa ¥ 4 m 3 = 16.74 kg kN m 0.287 ¥ 333 K kg K

m1 = initial mass in the tank before filling stated =

P1V1 200 ¥ 4 = = 8.906 kg RT1 0.287 ¥ 313

m = m2 – m1 = mass of air added = 16.74 – 8.906 = 7.834 kg D t) required to fill the tank at 400 kPa and 60°C (c) The time (D Dt =

(m - m1 ) m mass added 7.834 kg = 2 = = = 15.272 seconds mass flow rate m1 m1 0.513 kg/s

EXAMPLE 3.61 A tank initially has 10 kg of air at 200 kPa and 300 K. The tank is filled by using the air compressor. The supply main is at 600 kPa and 350 K. Calculate the following: (a) The quantity of air added to the tank. (b) The final temperature. Solution: Air compressor P1 = 600 kPa T1 = 350 K

Pipe line Valve i P2 = 600 kPa V2 = V1

i

Tank air m1 = 10 kg P1 = 200 kPa T1 = 300 K

Control surface

Figure E3.61

(a) The quantity of air added to the tank (m) We have from the Eq. (3.52), È (V )2 gZ ˘ (m2 - m1 ) Í h1 + 1 + 1 ˙ + Q1- 2 2 gc gc ˙˚ ÍÎ = m2u2 – m1u1 + W1–2 Assume tank is insulated and rigid Q1–2 = 0 W1–2 = 0 \

K.E. and P.E. = 0

(m2 – m1)hi = m2u2 – m1u1

160

Basic Thermodynamics

Assuming air as an ideal gas, (m2 – m1)Cp Ti = Cv(m2T2 – m1T1) \

Ti (m2 - m1 ) =

We have

(m2T2 - m1T2 )

m2 P2V2 RT1 P2T1 = ¥ = m1 RT2 P1V1 T2 P1

(1)

(V2 = V1; Rigid tank)

m2 600 ¥ 300 900 = = m1 200 ¥ T2 T2 m2 =

\

900 m1 900 ¥ 10 9000 = = T2 T2 T2

(2)

Substitute all the data into Eq. (1)

\

Ê ˆ 9000 ÁË m2 ¥ m - 10 ¥ 300 ˜¯ 2

(mT2 - 10 ¥ 300) = = 22.245 kg 1.4 1.4 m = m2 – m1 = mass of air added = 22.245 – 10 = 12.245 kg

350 (m2 - 10) =

(3)

(b) Final temperature (T2) From Eqs. (2) and (3) \

T2 =

9000 9000 = = 404.59 K m2 22.245

EXAMPLE 3.62 A tank containing air at 20°C, 400 kPa is connected to a piston and cylinder arrangement. Initially the cylinder contains 0.1 m3 of air at 50°C and the mass of the piston is such that a pressure of 200 kPa is required to raise it. Now the valve is opened and air is allowed to flow into the cylinder until its volume doubles, at which time the valve closes. Calculate (a) the final temperature and (b) the mass of air added to the cylinder. Assume no heat transfer during the process and air as ideal gas. Solution: (a) The mass of air added (m) m1 = Initial mass of air in the cylinder =

P1V1 200 ¥ 0.1 = = 0.2157 kg RT1 0.287 ¥ 323

V2 = Final volume of the cylinder after air addition = 2V1 = 2 × 0.1 = 0.2 m3 m2 = Final mass of air in the cylinder =

P2V2 RT2

(1)

First Law of Thermodynamics

Apply the first law of thermodynamics, i.e. unsteady flow process Subscript i and e = Condition of the air Tank entering and leaving the cylinder 1 and 2 = Initial and final condition of the air in the cylinder at stages 1 and 2 K.E. and P.E. of entering fluid are neglected and K.E. and P.E. of contents of the tank are neglected at states 1 and 2, then Eq. (3.47) Figure E3.62 reduces to Q1–2 + m1u1 + mihi = W1–2 + m2u2 + mehe Here, Q1–2 = 0 no heat transfer during the process (given condition) me = 0 no mass leaves from the cylinder W1- 2 =

Ú

2

1

PdV = P (V2 - V1 )

161

Weight Piston Cylinder

(2)

[given condition, P = C]

= 200 × [0.2 – 0.1] = 20 kJ We know that du = Cv dT dh = Cp dT mi = m2 – m1 Equation (2) reduces to miCpTi = m2CvT2 – m1CvT1 + W1–2 È (mi C pTi + m1CvT1 - W1- 2 ) ˘ T2 = Í ˙ m2Cv Î ˚

(3) (4)

Substitute T2 in Eq. (1), m2 =

P2V2 ¥ m2Cv R (mi C pTi + m1Cv T1 - W1- 2 )

=

200 ¥ 0.2 ¥ m2 ¥ 0.714 0.287[(m2 - 0.2157) ¥ 1.005 ¥ 293 + (0.2157 ¥ 0.714 ¥ 323) - 20]

=

28.56 m 2 86.5 m 2 - 18.229 + 29.75

\ m2 = 0.2016 kg (b) The final temperature (T2) Substitute data in equation 4,

or

T2 =

(0.2016 - 0.2157) ¥ 1.005 ¥ 293 + (0.2157 ¥ 0.714 ¥ 323) - 20 = 418.33°C 2016 ¥ 0.714

T2 =

P2V2 200 ¥ 0.2 = = 418.33°C m2 R 0.2016 ¥ 0.287

162

Basic Thermodynamics

EXAMPLE 3.63 In a pneumatic lift, tank and cylinder are connected with a valve in between as shown in the figure. Tank properties: 0.4 m3, 40°C, 7 bar. Lifting occurs when the valve is opened and air is allowed into the cylinder. Cylinder is lifted for 1 m. Cylinder properties : 1 bar, 20°C, 0.02 m3, 0.1 m2 (c/s) A cylinder pressure of 5 bar is maintained while the load is lifted up to 1 m. Air continues to flow into the cylinder after the load has reached the maximum lifting height and finally the pressure in the tank and cylinder will be equal and the air will attain a uniform temperature of 20°C. Calculate the following: (a) The final pressure of air (b) Amount of energy transferred as heat from surroundings to the air within the tank. Solution: W Piston

Cylinder

Air Vt = 0.4 m t1 = 40°C Pt = 7 bar

3

Tank Air Valve Q

Figure E3.63

W = Work done by the piston against a constant pressure of 3 bar =

Ú

2

1

PdV = 500 ¥ (V2 - V1 )

V2 = Final volume of the cylinder = V1 + AL = 0.02 m3 + 0.1 × 1 = 0.12 m3 A = Cross-sectional area of the piston m2 L = Height of piston, lifted m \

W = 500 × (0.12 – 0.02) = 50 kJ

Tank air rt = Tank air density =

P1 7 ¥ 100 kg = = 7.79 3 RT1 0.287 ¥ 313 m

mt = Tank air mass = volume of tank × density of tank air = 0.4 × 7.79 = 3.117 kg ut = Tank internal energy = Cvtt = 0.714 × 40 = 28.56

kJ kg

ut = Total internal energy of the tank air = 28.56 × mt = 28.56 × 3.117 = 89.022 kJ

First Law of Thermodynamics

163

Cylinder air rc = Cylinder air density in the begining =

P 100 kg = = 1.1892 3 RTc 0.287 ¥ 293 m

mc = Initial air mass in the cylinder = initial volume of the cylinder × rc = 0.02 × 1.892 = 0.0378 kg uc = Initial specific internal energy of the air in the cylinder = CvTc = 0.714 × 20°C = 14.28 kJ/kg Uc = initial total internal energy of the air in the cylinder = ucmc = 14.280 × 0378 = 0.5398 kJ (a) Final pressure in the tank and cylinder (Pf) Initial state Total mass and total energy of combined tank and cylinder at initial condition \

mi = mf = mt + mc = 3.117 + 0.0238 = 3.154 kg Ui = Ut + Uc = 89.022 + 0.3398 = 89.56 kJ

Final state Final temperature of air in cylinder and tank (tt) = 20°C Vf = Final total volume of the air (cylinder and tank) = Vt + V2 = 0.4 + 0.03 = 0.43 m3 rf = Final density of the air =

mf Vf

=

3.184 kg = 7.334 3 0.43 m

kN

or 6.17 bar m2 (b) The amount of energy transferred from the surroundings to the tank (Q) Pf = rf R Tf = 7.334 × 0.287 × 293 = 616.7

uf = Final specific internal energy = cvtf = 0.714 × 20 = 14.28 Uf = Final total internal energy = mf uf = 3.154 × 14.28 = 45 kJ \

DU = Increase in internal energy = Uf – Ui = 45 – 89.56 = – 44.56 kJ

(–ve, internal energy decreases) From the 1st law of thermodynamics, Q = W + DU = 50 kJ + (– 44.56) = 5.44 kJ EXAMPLE 3.64 Steam from boiler enters into an evacuated tank through the steam main pipe line. Steam pipe line conditions: 30 bar, 350°C, 60 m/s. Tank conditions: 0.1 m3, 350°C. (a) Calculate the quantity of heat transfer when the pressure at the final state in the tank becomes 20 bar.

164

Basic Thermodynamics

(b) If the tank initially contains steam at 2 bar and 250°C, calculate the amount of steam that will enter into the cylinder and the magnitude of heat transfer at the final state from the cylinder. Solution: Pi = 30 bar, Ti = 350°C, Vi = 60 m/s

Boiler

Valve

Tank 3

Vi = 0.1 m

Figure E3.64

(a) Quantity of heat transfer when the pressure at the final state in the tank becomes 20 bars (Q1–2) Properties Pressure, bar Temperature, °C Specific volume, m3/kg Enthalpy, kJ/kg IP internal energy, kJ/kg

Steam (i)

Tank

pipe line

Initial state 1

Final state 2

30 350

— —

20 350

3114.8 (hi)

3136.6 2859.4

Substitute W1–2 = 0 (tank is rigid) and P.E. of the entering fluid is neglected, then Eq. (3.52) can be reduced to, È (V )2 ˘ Q1- 2 + (m2 - m1 ) Í h1 + 1 ˙ = m2 u2 - m1u1 2 gc ˙˚ ÍÎ Initially the cylinder is evacuated. \

m1 = 0 È (V )2 ˘ Q1- 2 = m2 u2 - m2 Í h1 + 1 ˙ 2 gc ˚˙ ÎÍ m2 = mass of steam added to the tank at the final condition

=

V Total volume 0.1 = 1 = = 0.722 kg Specific volume u 0.13856

First Law of Thermodynamics

165

È 60 2 ˘ Q1- 2 = 0.722 ¥ 2859.4 - 0.722 Í3114.8 + ˙ = - 185.6984 kJ 2000 ˚˙ ÎÍ (b) Amount of steam entered into the tank (m) and magnitude of heat transfer (Q1–2) Condition of the tank and steam main pipe for the second part \

Properties

Steam (i)

Pressure P, bar Temperature, °C Specific volume, m3/kg Enthalpy, kJ/kg Specific internal energy, kJ/kg

Tank

pipe line

Initial state 1

Final state 2

30 350 — 3114.8

2 250 1.1988 2970.5

20 350 0.13856 3136.6

—

2730.8

2859.4

m1 = Initial mass of air in the tank for case II = m2 = Final mass of air in the tank for case II =

V 0.1 = = 0.08342 kg u1 f 1.1988

V 0.1 = = 0.722 kg u2 f 0.13856

m = m2 – m1 = 0.722 – 0.08342 = 0.6386 kg = mass of air added to the tank in case II We have energy balance, for case II È (V )2 ˘ Q1- 2 + (m2 - m1 ) Í h1 + 1 ˙ = m2 u2 - m1u1 2 gc ˙˚ ÍÎ È (V )2 ˘ Q1- 2 = (m2 u2 - m1u1 ) - (m2 - m1 ) Í h1 + 1 ˙ 2 gc ˙˚ ÍÎ Ê 60 2 ˆ = (0.722 ¥ 2859.4 - 0.08342 ¥ 2730.8) - 0.6386 Á 3114.8 + = - 153.56 kJ 2 ¥ 1000 ˜¯ Ë EXAMPLE 3.65 A tank contains air at a pressure of 5 bar, 100°C and capacity of 2 m3. The air is leaked to the atmosphere through a valve. Calculate the following. (a) The velocity of air at which it is coming from the tank. (b) The work obtainable from the leaked air to run a frictionless turbine. Solution: (a) The velocity of air at which it is coming from the tank (Ve) m1 = Initial mass of the air in the tank =

P1V1 5 ¥ 100 ¥ 2 = = 9.34 kg RT1 0.287 ¥ 373

166

Basic Thermodynamics

Tank P1 = 5 bar T1 = 100°C

Velocity = ? 100 kPa

3

Valve

V1 = 2 m

Figure E3.65

Assume that, the expansion is reversible adiabatic process. Final temperature of the air in the tank

ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚ \

-1

È 100 ˘ = 373 Í ˙ Î 500 ˚

1.4 - 1 1.4

= 235.4 K

m2 = Final mass of the air in the tank =

P2V2 100 ¥ 2 = = 2.96 kg RT2 0.287 ¥ 235.4

Substitute Q1–2 = 0, W1–2 = 0, P.E. of the leaving fluid is neglected, then Eq. (3.56) reduces to, È (V )2 ˘ (m1 - m2 ) Í he + e ˙ = m1u1 - m2 u2 2 gc ˙˚ ÍÎ (Ve )2 (m1u1 - m2 u2 ) (m C t - m2Cv t2 ) = - he = 1 v 1 - C p t2 2 gc (m1 - m2 ) (m1 - m2 ) =

(9.34 ¥ 0.714 ¥ 373 - 2.96 ¥ 0.714 ¥ 235.4) kJ - 1.005 ¥ 235.4 = 75.32 (9.34 - 2.96) kg

Cv = 75.32 ¥ 2 ¥ 1000 = 388.14

m s

(b) The work obtainable from the leaked air to run a turbine (W) W = K .E. = (m1 - m2 )

(Ve )2 388.142 = (9.34 - 2.96) = 480.58 kJ 2 gc 2 ¥ 1000

EXAMPLE 3.66 A tank contains air at 1 bar, 30°C and a capacity of 1 m3. A compressor is used to empty the tank at atmospheric pressure, i.e. 1 bar and operates isothermally at 30°C. Calculate the total work done by the compressor which is assumed to operate reversibly. Assume air as ideal gas. Solution: Quantity of air removed by the compressor in one cycle is small compared to total quantity of air to be removed. \ Work required to remove small quantity of air is given by

First Law of Thermodynamics

Tank P1 = 1 bar

167

1 bar e

V1 = 1 m3 T1 = 30°C

Compressor

e

Figure E3.66

w = - dm

Ú

Pa

Pt

VdP

dm = small amount removed Pa = discharge pressure (1 bar) Pt = compressor pressure during the cycle under consideration v=

We have

RT P

Total work done by the compressor (W) \

Ú

w = - dm

Pa

Ú

Pt

RT dP P

dw = - dm RT ln

We know that

Pa P = d m RT ln t Pt Pa

dm =

Vt dP1 RT

dw =

Vt dPt P P RT ln t = Vt dPt ln t RT Pa Pa

Integrating between Pt1 = 1 bar to Pt2 = 0 \

W = Vt [ Pt ln Pt - Pt ]1 bar = - Vt ¥ 1 = - Vt m3 ¥ 100 0 bar

= - 1 m 3 ¥ 100

kN m2

kN kJ m2

= - 100 kJ

EXERCISES 3.1 Derive an expression for the first law of thermodynamics applied to a closed system. Show how does it give a conception of internal energy. (KUD I-89) 3.2 Define enthalpy and give the thermodynamic definition of two specific heats. (KUD I-89)

168

Basic Thermodynamics

3.3 The steady flow energy equation for unit mass of any substance during a process 1-2 is È (V )2 gZ ˘ È (V )2 gZ ˘ Q1- 2 + Í h1 + 1 + 1 ˙ = Í h2 + 2 + 2 ˙ + W1- 2 gc ˙˚ ÎÍ gc ˙˚ 2 gc 2 gc ÎÍ Rewrite this in differential form.

(KUD II-93)

3.4 Define internal energy of the system. Show that it is a property of the system. (KUD Dec. 94) 3.5 Prove the following for an adiabatic process. ( P v - P2 v2 ) W.D. = 1 1 for closed system ( - 1) =

( P1 v1 - P2 v2 ) for open system ( - 1)

3.6 State the first law of thermodynamics for a closed system executing a cyclic process. (KUD I-96) 3.7 Write down the steady state steady flow energy equation and state the meaning of all the terms associated with it. Simplify the above equation for a throttle flow of the fluid in a pipeline stating the assumptions. (KUD I-96) 3.8 From the first law of thermodynamics write the general energy equation for a flow system. Modify this equation for the following cases: 1. Steady flow through nozzle 2. Throttling flow (KUD II-90) 3.9 Differentiate between internal energy and enthalpy. 3.10 State the first law of thermodynamics. Write down its formulations for: 1. Non-flow process 2. Thermodynamics cycle 3. Steady flow through isentropic horizontal nozzle 4. Steady flow through water cooled compressor

(KUD II-90)

(KUD II-93)

3.11 Deduce an expression for the work done when a system undergoes a quasi-static hyperbolic process. (KUD I-93) 3.12 State and explain the first law of thermodynamics. Give its equation with reference to a cyclic process and non-cyclic process. What is the importance of this law in thermodynamics. (KUD I-88) 3.13 Derive W.D. and heat transfer equations for the following processes: 1. Reversible constant volume 2. Reversible constant pressure 3. Reversible constant temperature

First Law of Thermodynamics

169

4. Reversible constant entropy 5. Polytropic 3.14 Derive the relation between R, Cp, Cv. Clearly explain why Cp is greater than Cv. (KUD) 3.15 Air passes through a gas turbine at the rate of 5 kg/s. It enters with a velocity of 200 m/s and an enthalpy of 6700 kJ/kg. At exit the velocity is 150 m/s and the enthalpy 5234 kJ/kg. The air has a loss of heat to the surroundings of 63 kJ/kg as it passes through the turbine. Determine the power developed by the turbine. (KUD Dec. 94) 3.16 A tank containing a fluid is stirred by a paddle wheel. Heat is transferred from the tank at the rate of 6000 kJ/hr. Write the work required is 3 kW. Determine the change in internal energy of the system in one hour. (KUD I-95) 3.17 During a steady flow process 5000 kg m/hr of fluid passes through a system in which the exit pipe 2 m below the level of the inlet pipe. Find the power developed of the system, if pressure decreases from 7 to 1.2 N/m2, velocity increases from 60 to 400 m/s, internal energy decreases by 50 kJ/kg. Specific volume increases from 0.03 to 0.2 m3/kg and heat loss by the system is 10000 kJ/hr. (KUD I-95) 3.18 In a steady flow apparatus 135 kJ of work is done by each kg of fluid. The specific volume of the fluid, pressure and velocity at the inlet are 0.37 m3/kg 600 kPa and 16 m/s. The inlet is 32 m above the floor and the discharge pipe is at floor level. The discharge conditions are 0.62 m3/kg, 100 kPa and 270 m/s. The total heat loss between the inlet and discharge is 9 kJ/kg of fluid. In flowing through this apparatus, does the specific internal energy increase or decrease and by how much? (KUD I-91) 3.19 A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa, the pressure remaining constant during the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal energy of the gas change? (KUD II-91) 3.20 A video theatre hall has a volume of 150 m3. Twenty persons are watching the video film. Each person occupies a volume of 0.07 m3 and gives out a heat of 132 watts. If the hall is completely sealed off and insulated, calculate the rise in temperature in air in 10 mins. For air assume Cp = 1.005 kJ/kg K, Cv = 0.743 kJ/kg K. 3.21 The following amount of heat transfer occurs during a cycle comprising four processes. 428.8 kJ, –117.9 kJ, –186.8 kJ, 88.9 kJ, respectively. Calculate network output of the cycle and type of the work. (KUD I-95) 3.22 A tank containing water is stirred by a paddle wheel. The paddle wheel requires 3.5 kW of power and the heat transfer from the tank takes place at the rate of 100 kJ/min. Calculate the change in energy of the tank and its contents per hour. (KUD II-93) 3.23 At the inlet to a certain nozzle, the enthalpy of fluid entering is 3000 kJ/kg and the velocity is 60 m/s. At the exit of the nozzle, the enthalpy of the fluid is 2000 kJ/min. Assuming that the nozzle is horizontal and adiabatic, find the velocity of the fluid at the exit. (KUD I-93)

170

Basic Thermodynamics

3.24 Air flows steadily at the rate of 0.5 kg/s through an air compressor entering with velocity 7 m/s, 100 kPa pressure and 0.95 m3/kg specific volume and leaving at 5 m/s, 700 kPa and 0.19 m3/kg. The increase in internal energy of the air leaving is 90 kJ/kg. The cooling water in the compressor jacket absorbs heat from the air at the rate of 58 kW. Calculate the shaft work input to the compressor. (KUD, IInd year I-94) 3.25 A gas is compressed at constant pressure of 0.105 MPa without friction from 0.4 m3 to 0.2 m3. 42.5 kJ of heat is transferred from the gas during the process. Calculate the change in internal energy. [Ans: 21.5 kJ] 3.26 A cyclic process is having the following process: 1. 82000 J of work is done by the piston on the air during the compression stroke and 45 kJ of heat are rejected to the surroundings. 2. During the expansion stroke 100000 J of work is done by the air on the piston. Calculate the quantity of heat added to the system. [Ans: 63 kJ] 3.27 A system comprises a stone of mass of 30 kg and a drum containing 1500 kg water. Initially the stone is 50 m above the water and the stone and water are at the same temperature. The stone is then made to fall into the water. Determine DU, DK.E., DP.E., Q and W for the changes of state given below. The valve of gravitational acceleration may be assumed 9.81 m/sec2. 1. The stone is to just enter water 2. The stone just comes to rest in drum 3. The heat transferred to surroundings is such that the water and stone remain at the same temp. 3.28 Air at 30°C and 1 bar is compressed reversibly and polytropically from 6 m3 to 1 m3. Calculate the final temperature, pressure, the work done and heat transferred if the index of compression is 1, 1.14 and 0.9, respectively. Cp = 1.05 kJ/kg K and

Cv = 0.714 kJ/kg K.

3.29 A cylinder of 8 cm diameter has a spring loaded piston. The stiffness of the spring is 15 N/ mm of compression. The initial pressure, volume and temperature of air in the cylinder are 3 bar, 0.00005 m3 and 25°C respectively. Determine the amount of heat added to the system so that the piston moves by 50 mm. W = 0.714 kJ/kg K

R = 0.287 kJ/kg K.

(KUD)

4 CHAPTER

Second Law of Thermodynamics

4.1

LIMITATIONS OF FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics states that (a) When a closed system undergoes a thermodynamic cycle, the cyclic integral of the heat is equal to the cyclic integral of the work. (b) The energy is neither created nor destroyed, but one form of energy can be converted into another form, i.e. heat energy and mechanical work are mutually convertible, i.e.

vÚ

Q=

vÚ

W

Consider the following examples. EXAMPLE 4.1 Figure 4.1 shows a closed system and energy exchange takes place between the system and surroundings. Work (W1–2) is done on the closed system with the help of paddle wheel by lowering the load W [Figure 4.1(a), (c), (e)]. This work is completely converted into heat and stored in the form of internal energy. Hence, the internal energy of the system increases. The cycle is completed by transferring heat (Q2–1) from the system to the surroundings [Figure 4.1(b), (c), (e)], thus bringing the system back to the initial state. The cycle is thus completed leading to complete conversion of work into heat. In the cycle, both W1–2 and Q2–1 are negative. Such a cycle is possible. \ When a work is converted into heat, we have always, –W1–2 = –Q2–1 or W1–2 = Q2–1 171

172

Basic Thermodynamics

Gas system

Gas system W2–1

W1-2

W

Q1–2

Q2–1

W

(a)

(b)

1–2–1

W1–2

Q2–1

1–2–1

Q1–2

(c) Possible

W2–1

(d) Impossible

P

P 1

Q2–1

Q2–1

1

W1–2

W2–1

2

2

V

V

(e)

(f)

Figure 4.1 Closed system in a cycle involving heat and work.

Now, if the same quantity of heat (Q1–2 = Q2–1) is added to the system, as shown by the dotted line [Figure 4.1(b), (d), (f)], the temperature of the gas will increase, but the paddle wheel will not turn and raise the weight (W), i.e. reverse process, raising the mass by transferring heat to the system does not occur in nature. We know that during the reverse process, the work (W2–1) is not recovered but the heat (Q1–2) transferred into the system is stored as an internal energy; therefore, it does not violate the first law of thermodynamics. \

Q1–2 = DE

Here both the heat and work are positive, which is impossible. [Energy is neither created nor destroyed, but one form of energy (Q1–2) is converted into another form (DE) of energy.]

Second Law of Thermodynamics

173

EXAMPLE 4.2 Consider the heating of water by passage of current through an electric resistor. The temperature of the water increases. Electrical energy is converted into heat and stored in the form of internal energy. As per the first of law of thermodynamics, the amount of electrical energy supplied to the resistance wire is equal to the amount of energy transferred to the water as heat (Figure 4.2), i.e. Electrical energy = DE Heat

Heat I>0

I=0

Figure 4.2 Process showing heat to a resistance wire will not generate electricity.

Now do an attempt to reverse this process, i.e. transferring same quantity of heat to the wire will not cause an equivalent amount of electrical energy to be generated in the wire. EXAMPLE 4.3 A running automobile can be stopped by using the brake, K.E. of automobile converted into heat, which increases the internal energy of the brake drum. But it is not possible to accelerate the automobile by the cooling of the hot brake drum. At the same time reversal process will not violate the first law of thermodynamics. It is clear from the above examples that processes take place in a certain direction and not in the reverse direction (Figure 4.3). Work = Heat One way

Electrical energy = Heat

K.E. of automobile = Heat

Figure 4.3 Processes occur in one direction and not in the reverse direction.

The first law does not put any restriction on the direction of a process, but satisfying the first law does not ensure that the process will actually occur (Figure 4.4).

Figure 4.4

Processes which are impossible.

From the above examples, it may be concluded that 1. The first law of thermodynamics does not help to predict whether the system will or will not undergo a particular change.

174

Basic Thermodynamics

2. No restriction has been imposed by the first law on the possibility of conversion of energy from one form into another. 3. The first law does not address what portion of heat may be converted into useful form of energy and work. This deficiency of the first law is remedied by introducing another general principle, the second law of thermodynamics. EXAMPLE 4.4 Consider two systems, one at a higher temperature and the other at a lower temperature undergoing a process in which heat is transferred from the high temperature system to the low temperature system. This process is possible. The reverse process is impossible, i.e. transferring heat from the low temperature system to the high temperature system by just heat transfer only (Figure 4.5).

High temperature

Possible

Q

Q

Impossible

Opportunities for developing work and Low temperature qualitative difference between heat and work Consider an example of paddle wheel and heat Figure 4.5 Process showing the transfer of engine as shown in Figure 4.1(b). Instead of heat from high temperature to low temperature and reverse is transferring the heat (Q1–2) to the system, heat could impossible. be transferred to another special system undergoing a cycle that would develop a net amount of work. The special systems (devices) are called heat engines (power cycles). In the same way, net amount of work can be developed in case of resistance coil also (Figure 4.2). In the above example, complete conversion of heat into work in a cycle is not possible. So heat and work are not completely interchangeable forms of energy. Consider a system as shown in Figure 4.6. Heat (Q1–2) is transferred to a system. Heat cannot be converted completely into work in a cycle. Some heat has to be rejected. W2–1 is the net work done and Q2–1 is the heat rejected to complete the cycle. When heat is converted into work in a complete closed cycle process, Q1–2 > W2–1. Hence, work is said to be a high-grade energy. The complete conversion of a low-grade energy into a high-grade energy in a cycle is impossible. This distinction is a dissymmetry in nature. The first law does not address it, but the distinction definitely exists. Q2–1

System

W2–1

Q1–2

Figure 4.6 Quantitative difference between heat and work.

Second Law of Thermodynamics

175

Difference between high-grade and low-grade energy

High-grade energy

4.2

Low-grade energy

Molecular actions are ordered.

Molecular actions are disordered.

Molecular actions are coherent.

Molecular actions are incoherent.

Energy is undispersed.

Energy is dispersed.

Atoms of high-grade energy are highly structured.

Atoms of low-grade energy are not structured.

HEAT ENGINE

A heat engine is a system (devices, special systems, power cycles) that can operate continuously in a cycle and produce work by absorbing heat (QH) from a heat source [high temperature (TH) thermal reservoir] and rejecting heat (QL) to a sink [low temperature (TL) thermal reservoir]. Figure 4.7 is a schematic representation of a heat engine.

Figure 4.7 Schematic diagram of a heat engine.

Many heat engine devices are available. These heat engines differ considerably from one another, but there is something common to all the heat engines. 1. Heat is absorbed from the heat source (solar energy, oil furnace, nuclear reactor, combustion chamber, etc.). 2. A part of this heat is converted into work (in the form of rotating shaft). 3. Rejects the remaining quantity of heat to a heat sink (river, atmosphere, etc.). 4. They operate on a cycle. The best example of a heat engine is the steam power plant. Figure 4.8 illustrates the simple steam power plant.

176

Basic Thermodynamics

Figure 4.8 Schematic representation of a steam power plant.

Qin quantity of heat is added to water in a boiler to produce steam (process 1–2). The steam is then expanded through a turbine that produces a positive work output (Wt) (process 2–3). The steam, which is leaving from the turbine, goes to a condenser where it is condensed by rejecting heat (Qout) to a heat sink (process 3–4). The condensate from the condenser is pumped to the boiler pressure and returned to the boiler to complete the thermodynamic cycle. Some of the work output of the turbine is used to drive the pump. The net output from the turbine (Wt – Wp) is available as the net work output (Wn) of the plant. i.e.

Wn = Wt – Wp

The four components of the steam power plant should be considered as open systems because mass flows in and out of each component. Let us consider the combination of components and pipelines as one system. Then, this is treated as a closed system, i.e. no mass enters or leaves the combination system, which is represented by dotted line (Figure 4.8). We know that for a closed system undergoing a cycle, the change in internal energy, Du is zero, and therefore from the first law of thermodynamics for a closed system undergoing a cycle is given by,

vÚ

W=

vÚ

Q

Wn = Q n (Wt – Wp) = (Qin – Qout) An automobile engine is another example of a heat engine. The burning of fuel in a combustion chamber is equivalent to a heat flow into a fluid from a heat source. As a result, work is produced to move the automobile vehicle. The heat is rejected to the heat sink (atmosphere) through the radiator and exhaust gases. When compared to the steam power plant, the automobile engine does

Second Law of Thermodynamics

177

not constitute a complete thermodynamic cycle since the exhaust gas never returns to its original state of fuel and air mixture. Hence, automobile engine cycles are treated as a mechanical cycle, not a thermodynamic cycle. For the analysis purposes, automobile engine cycle is approximated as a thermodynamic cycle.

4.2.1 Thermal Efficiency In general, thermal efficiency of any device or system is defined as follows. Efficiency =

Desired output Required input

Apply this definition to a heat engine. Here,

\

desired output Æ net work output required input Æ total heat input efficiency Æ thermal efficiency Net work output Wn = Thermal efficiencyHeat Engine = Total heat input Qin =

Qout Qin - Qout Q = 1 - out = 1 Qin Qin Qin

This is strictly a first law efficiency. Both | Qin | and | Qout | are absolute values. The above equation says that the thermal efficiency of a heat engine is always less than unity. Thermal efficiency is a measure of how efficiently a heat engine converts the heat into work. The purpose of heat engine is to convert heat into work. The aim of the engineer is to improve the efficiency of heat engine since increased efficiency means less fuel consumption and thus lowers fuel bills and less pollution. A fluid is used in a heat engine to receive and reject heat while undergoing a cycle. This fluid is called the working fluid.

4.3

REVERSED HEAT ENGINES

A reversed heat engine is a device which takes heat from a low temperature reservoir (sink) and rejects heat to a high temperature reservoir (source). In order to do so, special devices are required. These special devices are refrigerators and heat pumps. A reversed heat engine, refrigerator, is shown in Figure 4.9.

4.3.1

Refrigerator

Refrigeration cycle is used to produce a cooling effect. In this cycle also, working medium is required to receive and reject heat. The working fluid used in the refrigeration cycle is called a refrigerant. The commonly used refrigeration cycle is vapour compression refrigeration (VCR) cycle. A VCR cycle is shown in Fig. 4.10 which consists of four main components—a compressor, a condenser, an expansion valve and an evaporator.

178

Basic Thermodynamics Surroundings (atmosphere)

Heating space

Source High temperature

Source High temperature

QH

QH

R system

W

HP system

W

QL

QL

Sink Low temperature (cooling space)

Sink Surroundings (atmosphere)

Figure 4.9 Schematic diagram of reversed heat engine. W1 Closed system boundary

1

Compressor 2

Qin

3

Sink

Condenser

Condenser

Source

Qout

Expansion valve 4

Figure 4.10 Schematic arrangement of a refrigerator.

The refrigerant enters the compressor as a vapour and is compressed to the condenser pressure (1–2). It leaves the compressor at a relatively high temperature (more than atmospheric) and condenses in the condenser (2–3) by rejecting heat to the surrounding medium. The working medium enters a throttle valve where its pressure and temperature drop drastically (3–4). The low temperature-working medium then enters the evaporator, where the refrigerant evaporates by absorbing heat from the refrigerated space (4–1). The cycle is completed as soon as the refrigerant re-enters the compressor. In this case also, the combination of components and pipelines together is considered as a closed system. We know that for a closed system in a cycle, Du = 0 \

vÚ

Q=

vÚ

W

Second Law of Thermodynamics

179

(Qout – Qin) = Wt The first law efficiency of a refrigerator or heat pump cycle is usually called the co-efficient of performance (COP) of the device. The COP of a reversed heat engine is defined as follows: COP =

Desired effect Required input

The co-efficient of performance for cooling is denoted as COPC. Here, desired effect means cooling effect (Qin) and required input means turbine work input, Wt. \

COPC =

Qout Qin = (Qout - Qin ) Wt

The co-efficient of performance for heating is denoted by COPHP. Here, desired effect means heating effect (Qout) and required input means turbine work input Wt. \

COPHP =

Qout Qout Q - Qin + Qin Qin = = out =1+ (Qout - Qin ) (Qout - Qin ) (Qout - Qin ) Wt

= 1 + COPC = 1 + COPRefrigerator

4.4

HEAT RESERVOIRS (HR)

A heat reservoir (thermal energy reservoir) is a large body that may absorb or reject an unlimited quantity of heat without undergoing an appreciable change in temperature or in any other thermodynamic properties. Even if there is change in thermodynamic properties, it is an extremely small one and too small to be measured. Heat reservoirs are of two types: 1. Heat source 2. Heat sink

4.4.1 Heat Source A heat reservoir that supplies energy in the form of heat to a system is called a heat source. Example: A combustion chamber where fuel is continuously burned.

4.4.2 Heat Sink A heat reservoir that absorbs energy in the form of heat from a system is called a heat sink. Examples: Oceans, lakes, atmosphere, etc. Temperature of heat source (TH) > Temperature of heat sink (TS)

4.5

STATEMENTS OF SECOND LAW OF THERMODYNAMICS

4.5.1 Kelvin-Planck Statement “It is impossible to construct a device that operates on a cycle, receives heat from a single reservoir and produces a net amount of work.”

180

Basic Thermodynamics

“It is impossible to construct a device that operating continuously (in a cycle), will produce no effect other than the transfer of heat from a single thermal reservoir at a constant temperature and performing an equal amount of work.” According to the above statement, it is impossible to develop (device) a heat engine which operates in a cycle and receives Q amount of heat from a high temperature body and does an equal amount of work W. The efficiency of such an engine would be 100% [h = (W/Q) × 100 = (W/W) × 100 = 100%]. Such a device violates Kelvin-Planck statement even though it satisfies the first law of thermodynamics, i.e. principle of conservation of energy. A heat engine which violates Kelvin-Planck statement of the second law of thermodynamics is known as perpetual motion machine of the second kind (PMM-II), i.e. 100% efficient machine which is impossible to obtain in nature.

Figure 4.11 Kelvin-Planck statement.

Now consider a system as shown in Figure 4.11(b). Qin amount of heat is transferred to the heat engine from a heat source, a part of it is utilized to do work W and Qout amount of heat is rejected to the heat sink. This cycle (device) is possible and it does not violate the Kelvin-Planck statement. There is a degradation of energy in the process of producing mechanical work (W), from the heat supplied Machine (Qin). Therefore, the Kelvin-Planck statement of second (device) law of thermodynamics sometimes is known as the law W Q=0 PMM-I of degradation of energy. A machine which violates the first law of thermoFigure 4.12 Illustration of PMM-I. dynamics is known as perpetual motion machine of the first kind (PMM-I). It is a machine which produces work energy without consuming an equivalent amount of energy from other source (Figure 4.12). Such a machine is impossible in nature. In Kelvin-Planck statement, the word continuous or cycle has an important implication. Consider an example of an isothermal expansion of an ideal gas. The change in internal energy, DU, is zero (T = C). Therefore, heat absorbed by the gas is completely converted to work (against second law). It seems that it contradicts the second law; in fact, it is not so. In this process, the pressure of the gas decreases and the pressure of the gas would soon become equal to that of the

Second Law of Thermodynamics

181

surroundings, and further expansion would be impossible, i.e. production of work stops. In this process, continuous conversion of heat to work is impossible. But this is possible only in a cyclic process.

4.5.2 Clausius Statement “It is impossible to construct a device that operates on a cycle whose sole result is the transfer of heat from a cooler to a hotter body without the assistance of external work”. According to the above statement, it is clear that heat cannot flow itself from a lower temperature body to a higher temperature body without the assistance of an external work [Figure 4.13(a)]. This process is impossible and violates the Clausius statement of the second law of thermodynamics. Alternatively, heat can transfer from a low temperature body to a high temperature body with the help of an external work (W) [Figure 4.13(b)]. This process is possible. Heat source High temperature

Heat source High temperature

Qout

Qout

HP

Qin

Qin Heat sink Low temperature

Heat sink Low temperature

(a) Impossible

(b) Possible

Figure 4.13

4.6

W = (Qout – Qin)

HP

Clausius statement.

EQUIVALENCE OF KELVIN-PLANCK AND CLAUSIUS STATEMENTS

At first sight, the Kelvin-Planck and the Clausius statements appear to be quite unconnected, but we can show that the two statements of the second law are equivalent. The two statements are equivalent if the truth of the other, or if the violation of each statement implies the violation of the other. Consider a system as shown in Figure 4.14(a). High temperature reservoir

Q2

Q1

HP

Q1

Q2 HP

HE

HE

W = Q1 – Q2 PMM-II

Q2

Q2 Q2

Q2

Low temperature reservoir, TL (a)

Q1 – Q2

OR Combined system

W = Q1 – Q2

Low temperature reservoir, TL (b)

(c)

Figure 4.14 Equivalence of Clausius and Kelvin-Planck statements.

182

Basic Thermodynamics

In this system, a heat pump or a refrigerator operates in a cycle and transfers Q2 amount of heat from a low temperature (TL) reservoir to a high temperature reservoir (TH) without any external work done on the system. Then the system is known as PMM-II and violates the Clausius statement. Consider a system as shown in Figure 4.14(b). In this system, a heat engine operates in a cycle and takes Q1 amount of heat from a high temperature reservoir (TH) and rejects Q2 amount of heat to a low temperature reservoir (TL). In doing so, the engine does W amount of work. This process is possible as per the Kelvin-Planck statement. Now consider a system as shown in Figure 4.14(c). In this case, both the heat pump and heat engine are combined together to constitute a single system. Heat ‘Q2’ rejected by the heat engine is taken by the heat pump and rejects to the high temperature reservoir (TH). Therefore, the low temperature reservoir can be removed. Since there is no net heat transfer to the low temperature reservior, the low temperature reservior, the heat engine, and the refrigerator can be considered together as a device that operates in a cycle and produces net work output. The combined system receives (Q1 – Q2) an amount of heat from the high temperature reservoir and does an equivalent amount of work (W = Q1 – Q2). Hence, the combined system violates the KelvinPlanck statement. Thus, the violation of the Clausius statement leads to the violation of the Kelvin-Planck statement. Consider a system as shown in Figure 4.15(a).

Figure 4.15 Violation of Kelvin-Planck statement leads to voilation of Clausius statement.

In this system, a heat engine takes Q1 amount of heat from a high temperature reservoir and does an equal amount of work without rejecting heat to the low temperature reservoir. This system violates the Kelvin-Planck statement and it is known as PMM-II. Consider a system as shown in Figure 4.15(b). Here Q2 amount of heat is taken from the low temperature reservoir and transfer (W + Q2) to the high temperature reservoir with the help of an external work W. This process is possible as per the Clausius statement. Now consider a system as shown in Figure 4.15(a) and (b). In this case, both the heat engine and heat pump are combined together to form a combined system. The resulting device operates in a cycle and transfers heat Q2 from a low temperature (TL) reservoir to a higher temperature (TH) reservoir by using the work developed by the heat engine (not from the external agency). And the

Second Law of Thermodynamics

183

combined system takes Q1 quantity of heat from the reservoir and rejects (Q1 + Q2) amount of heat to the high temperature reservoir without any external work input. This violates the Clausius statement. Thus, a violation of the Kelvin-Planck statement leads to a violation of the Clausius statement.

4.7

REVERSIBLE AND IRREVERSIBLE PROCESSES

A reversible process is one that is carried out in such a way that at the conclusion of the process, both the system and the local surroundings are restored to their initial states without producing any change in the rest of the universe, i.e. process can be reversed without leaving any trace (change) on the local surroundings. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined process. If the driving force between the parts of the system or between the system and the surroundings is infinitesimal in magnitude, the process is reversible.

4.7.1 Conditions for Reversibility 1. The system and local surroundings must be thermally and mechanically equilibrium, i.e. the temperature and pressure of the system and surroundings must be same. 2. No dissipative effects, this is also referred as the condition for external reversibility. 3. System should be chemically equilibrium. 4. Both a system and its surroundings must pass through a series of equilibrium states, and therefore the changes in states must occur very slowly. 5. There should be no friction. 6. There should be no mixing. 7. There should be free or unresisted expansion. 8. Heat transfer should take place with infinitesimal temperature difference.

4.7.2 Internally and Externally Reversible Process Reversible processes can be classified as follows. 1. Internally reversible. 2. Externally reversible 3. Total reversible or simply reversible 4.7.2.1 Internally reversible An internally reversible process is one in which the system can be returned to its original state, i.e. no irreversibility occurs within the boundaries of the system during the process. Example: Quasi-equilibrium heat transfer process. Consider an examples as shown in Figure 4.16. Heat is transferred to the system from an external heat source that must be at a higher temperature than that of the system. The

System 30°C (T)

40°C (T + DT ) Thermal energy reservoir

Figure 4.16 Internally reversible.

184

Basic Thermodynamics

system can be returned to its original state by reversing the direction of the heat transfer. However, there is a net change in the surroundings, i.e. irreversibility in the surroundings. The heat transfer process of the above example is internally reversible but is not externally reversible. Modify the above example to show why it is not externally reversible, i.e. externally irreversible. Modified figure is shown in Figure 4.17.

Figure 4.17 Illustration of internally reversible and externally irreversible

Heat is transferred to the system from an external heat source that must be at a higher temperature than that of the system. There is an increase in the temperature of the system from 30°C to 35°C. The system can be returned to its original state by reversing the direction of the heat transfer (rejecting heat to the surroundings) [Figure 4.17(c)], by transferring energy to an external source that is at a lower temperature than that of the system. Thus, the system can be returned to its original state by reversing the process, but permanent changes are present in the surroundings. Because of the heat transfer processes, the higher-temperature source is at a lower-energy level and the lower-temperature source is at a higher-energy level. 4.7.2.2 Externally reversible An externally reversible process is one in which the surroundings are returned to their original state but a change is left within the system. 4.7.2.3 Totally reversible A totally reversible or simply reversible process is one in which the system and surroundings are returned to their original state. A totally reversible process involves no heat transfer through a finite temperature difference, no non-quasi-equilibrium change and no friction of other dissipative effects. Consider an example as shown in Figure 4.18. Heat is transferred to the system, undergoing a constant pressure (and constant temperature) phase-

System 30°C (T)

External source 300°C + ... + 0.01°C (T + dt)

Figure 4.18 Totally reversible process.

Second Law of Thermodynamics

185

change process, from an external source through an infinitesimal temperature difference (dt), i.e. 0.001°C; the heat transfer across a vanishingly small temperature difference is a totally reversible heat transfer process. For a completely reversible or quasi-equilibrium process, both a system and its local surroundings must pass through a series of equilibrium states, and therefore the changes in state must occur very slowly. Systems (devices) are not built to operate at extremely slow rates. We know that the rate of heat transfer is proportional to the temperature difference between the system and its surroundings as well as the surface area available for heat transfer. Then an infinitely large surface area would be required to transfer finite quantity of heat through an infinitesimally small temperature difference, attempting to do so would certainly not be economically feasible.

4.7.3 Factors That Support Reversible Processes 1. 2. 3. 4. 5. 6.

Frictionless movement Frictionless adiabatic expansion and compression (a fast process) Frictionless isothermal expansion and compression (a slow process) Phase change process (condensation or evaporation of liquids) Electric current flow through a zero resistance Expansion or compression with infinitesimal pressure difference

Characteristics of a Reversible Process 1. The driving and opposing forces are in exact balance. 2. A reversal process takes place if there is small change in the external conditions. 3. An infinite time is required to complete a reversible process (i.e. reversible process is essentially a slow process). 4. A reversible process produces maximum amount of work in case of work producing machine. 5. A reversible process consumes least amount of work in case of work absorbing machine. 6. It is an idealized and imaginary concept and may be used for the comparision of the performance of actual systems and to calculate deviation of efficiency of actual system.

4.7.4 Examples of Reversible Processes 4.7.4.1 Frictionless motion of solids (Figure 4.19) The gas (system) in the cylinder expands against the atmospheric pressure (Pa). The work developed by the system is (P – Pa) × A. The system is covered with an adiabatic cover, so that no heat transfer takes place. During the reversal process, the same quantity of work [(P – Pa)A] is required to bring back the system to its initial state. Both the system and the surroundings can be restored to the initial conditions. This process is a totally reversible process.

186

Basic Thermodynamics

M

M

Frictionless rollers

(a)

Figure 4.19

4.7.4.2

(b)

Motion of solids—(a) Motion without friction and (b) motion with friction.

Frictionless adiabatic process (Compression and Expansion) (Figure 4.20) F=0

Pa

System P Gas

F=0

Figure 4.20 Frictionless adiabatic process.

4.7.4.3

Frictionless isothermal process (Compression and Expansion) (Figure 4.21) F=0

P System Steam 1

Figure 4.21

C/S area A Pa

Frictionless isothermal process.

The steam present in the cylinder is compressed with the help of a frictionless piston, and takes place very slowly. This is an isothermal process; hence, there is an infinitesimal temperature difference that exists between the system and the surroundings. During the compression process, the steam condenses, liberating heat, and transferring it to the surroundings through a small temperature difference. The work done on the system is (P – Pa) × A, converted into heat and transferred to the surroundings. During the reversal process, the same quantity of heat is transferred from the

Second Law of Thermodynamics

187

surroundings to the system very slowly. At each equilibrium point, condensate inside the cylinder gets vapourized, increasing its specific volume, moves the piston outward, developing the work. The work developed by the system during the reversal process will be equal to (P – Pa) × A. Hence, both the system and the surroundings can be restored to the initial conditions. This process is a totally reversible process. 4.7.4.4 Expansion or compression with infinitesimal pressure difference The frictionless piston is loaded with a large number of small stones as shown in Figure 4.22. The system is in equilibrium with the external load. Remove the stones one by one from the piston and place on platforms, which are at the same level as the stone pieces. As soon as each stone is removed, there is a small pressure difference that exists between the system (gas) and the surroundings, and the gas expands. When the entire weight is removed from the piston, the gas expands, the work is done on the surroundings, the potential energy of the surroundings is increased, as is evident from the height of the pile obtained on the side. During the reversal process, piston should be loaded with stone, one by one. As soon as each stone is put on the piston, a small pressure difference exists between the system and the surroundings and the gas is compressed. During the reversal process, the work done on the system is exactly equal to the work done during the forward process, because the stone and the platform are at the same height. At the end of the reversal process, both the system and the surroundings can be restored to the initial state. Hence, this is a totally reversible process.

Heap of small stones

F=0 System System System

Figure 4.22 Expansion with infinitesimal pressure difference, a reversible process.

4.7.5 Irreversible Process An irreversible process is one that is carried out in such a way that the system and its local surroundings cannot be exactly restored to their respective initial states at the end of the reverse process, i.e. a net change occurs in the universe. If a process can proceed in one direction only and the reversal of the direction leads to a violation of the second law of thermodynamics, the process is called an irreversible process. If the driving force between the parts of the system or between the system and the surroundings is finite in magnitude, the process is irreversible. A high pressure gas is taken in a piston cylinder arrangement as shown in Figure 4.22(a). The pressure of the gas is balanced by the atmospheric pressure and the weight of the piston and

188

Basic Thermodynamics

weight is applied on the piston. Assume that there is no heat transfer between the system and the surroundings and no friction between the piston and the cylinder. Now, the weight ‘W’ is removed by sliding on to the platform. Piston moves up due to the force imbalance as shown in Figure 4.22. During the process, the work obtained against the surroundings is not available for restoring the gas to its initial position. In order to restore to the initial state (i.e. reversal process), work would be done on the system. There is a net change in the surroundings. Therefore, the process is clearly an irreversible one.

W

W

Platform

Pa Pa

Pa

Pa Piston

System

Piston System

Figure 4.22(a)

Expansion with finite pressure difference, an irreversible process.

4.7.6 Factors That Support Irreversible Processes 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

Friction due to movement—sliding friction as well as fluid friction Transfer of electricity through a resistor Inelastic deformation Spontaneous chemical reaction Heat transfer through a finite temperature difference Combustion (formation of new chemical constituents) Mixing of substances at different compositions Unrestrained expansion of a fluid (gas or liquid) to a lower pressure Magnetic hysteresis of a material Irregular stirring of a viscous liquid in contact with a reservoir Bring a rotating or vibrating liquid to stationary position Gas seeping through a porous plug (throttling expansion) Diffusion of two dissimilar inert ideal gases Sudden change of phase (condensation of superheated vapour)

4.7.7 Classification of Irreversible Process 1. Mechanical irreversibility 2. Thermal irreversibility

189

Second Law of Thermodynamics

3. Internal irreversibility 4. External irreversibility 5. Irreversibility due to dissipative effect 4.7.7.1 Mechanical irreversibility The dissipation of work (mechanical work) into internal energy either in a system or of a surroundings may be due to the phenomena as viscosity (fluid friction among the molecules), friction (mechanical friction between surfaces), inelasticity, electric resistance and magnetic hysteresis. 4.7.7.2 Thermal irreversibility Thermal irreversibility is associated with transfer of heat between a system and surroundings by virtue of a finite temperature difference. 4.7.7.3 Internal irreversibility Internal irreversibility is associated with fluid friction (rubbing of one layer of gas over the other and turbulence) and temperature difference variations within the fluid. Combustion and diffusion cause internal irreversibility. 4.7.7.4 External irreversibility External irreversibility is due to friction at the bearings and between the atmosphere and rotating members. All these absorb some work developed by the system. 4.7.7.5 Irreversibility due to dissipative effect Irreversibility of a process may be due to dissipative effects in which work done on a system, instead of increasing the P.E. and K.E. of the system (high grade energies) if it increases the molecular internal energy of the system, i.e. heat energy (considered as low grade energy) increasing the temperature of the system, then the work is said to be dissipated. Dissipative effect may be either internal or external. The energy which is dissipated cannot be recovered and becomes an unavailable energy. Dissipative effects may be due to viscocity, friction, inelasticity, electric resistance, etc. Due to dissipative effect, work is lost.

4.7.8 Examples of Irreversible Processes (External Irreversibility) 4.7.8.1 Mechanical friction (Figure 4.23) Let F = Force acting on the piston

F

dx = Displacement of the piston f = Frictional force between the piston and the surface A = Cross sectional area of the piston \

Work done on the piston from the surroundings = F dx

\

Effective work done on the system = (F – f) dx

P

System

f

Figure 4.23 Mechanical friction.

190

Basic Thermodynamics

The pressure of the system increases due to the effective work done on the system. During the reversal process, a force (F – f) acts on the inner surface of the piston. \ The effective force during the reversal process = [(F – f) – f] \ Effective work done during the reversal process = [(F – f) – f] dx = (F – 2f) dx Therefore, the work obtained from the system during the reversal process on the surroundings is less than the work done on the system from the surroundings, i.e. work done during compression process > work done during expansion process \ There is a net change in the surroundings. Hence, this process is irreversible. 4.7.8.2

Unrestrained expansion Pa

Pa

Pa

dx Pg = Pa

Pin

–Q

Gas at high pressure (Pg, T1)

Gas at high pressure (Pg, T1) (a)

(b)

(c)

Figure 4.24 Illustration of unrestrained process.

A high-pressure gas is enclosed within a piston and cylinder arrangement as shown in Figure 4.24. The piston is held in its position by a pin. The initial condition of the gas is Pg and T1. As soon as the pin is removed, the piston moves up due to the pressure difference between the gas pressure Pg and external pressure Pa. A sudden expansion takes place. Finally, the piston comes to an equilibrium condition as shown in Figure 4.24(b). The work done by the system =

Ú P dV a

dV = A dx A = Cross-sectional area of the piston During the reversal process, work is done on the system followed by heat rejection to the surroundings. Finally, the system is restored to the initial condition, i.e. Pg and T1. The work done on the system =

Ú P dV g

We know that Pg > Pa. Therefore, the work done on the system during the reversal process is greater than the work done by the system during the forward process. Therefore, some quantity of work has been lost by the surroundings and at the same time some quantity of heat is gained by the surroundings also. There is a net change in the surroundings. Hence, this process is an irreversible one.

191

Second Law of Thermodynamics

Apply the first law of thermodynamics. The work loss by the surroundings = the heat gain by the surroundings.

4.8

CARNOT CYCLE

The Carnot cycle is a hypothetical cycle developed by a French engineer, Sadi Carnot in 1824 for a heat engine or for a reversed heat engine. All the processes involved in the Carnot cycle are reversible, thereby providing the best possible device for power development purpose. The following are the assumptions made in the working of the Carnot engine. • No friction between the piston and the cylinder. • The walls of the piston and the cylinder are perfectly insulated. • The cylinder head is so arranged that it can be a perfect insulator or a perfect conductor as well. • A source and sink with infinite quantity of heat at high and low temperature, respectively. • A perfect gas is used as a working medium.

4.8.1 Working Principle of Carnot Cycle The working principle of Carnot cycle is shown in Figure 4.25. Process 1–2 (Isothermal compression process) T

P 3

T=

3

Tmax

C

4

T=C

Qin 4

Q=0

Q=0 g

PV = C g

PV =C

Tmin

2

1

2 T= C

H

3

I = Insulator C = Conductor

Piston

Sink

Source

1

2

Piston

3

C

1

Tout S

S I

T=C

4

Figure 4.25 Carnot cycle.

In the beginning, the piston is at point 1, i.e. the piston is at the outermost dead centre. The state of the working medium is represented by P1, V1 and T1. At this point, volume of the system is maximum (V1) (total volume). The sink is brought in contact with the head (H). The piston moves from state 1 to state 2, towards the inner dead centre compressing the gas isothermally. Whatever

192

Basic Thermodynamics

the work done on the system is converted into heat and rejected to the sink at constant temperature (Tmin). This process should take place as slow as possible in order to transfer heat, Qout, to maintain temperature constant, tmin = C. Qout = W1- 2 = P1V1 ln

V1 V = RT1 ln 1 V2 V2

Temperature remains constant Tmin = T1 = T2 Pressure increases from P1 to P2 Volume decreases from V1 to V2 Process 2–3 (Adiabatic compression process) Sink is removed and insulator is placed on the head (H). The piston is moved further till it reaches point 3. The volume at this point is called as clearance volume. This process should take place as fast as possible to avoid heat transfer. Temperature increases from T2 to T3 = Tmax Pressure increases from P2 to P3 Volume decreases from V2 to V3 Work done by the system W2–3 Heat interaction Q = 0 Change in entropy DS = S3 – S2 = 0 (S3 = S2) W2 -3 =

P3V3 - P2V2 RT3 - RT2 R (Tmax - Tmin ) = = -1 -1 -1

Process 3–4 (Isothermal expansion process) Insulator is removed and conductor is placed on the head (H). The source is brought in contact with the conductor. During this process, heat transfer takes place from the source to the system at constant temperature. The piston is moved from the inner dead centre to the outer dead centre expanding the gases. Temperature remains constant T3 = T4 = Tmax Pressure decreases from P3 to P4 Volume increases from V3 to V4 Work done by the system W3–4 W3 - 4 = Qin = P3V3 ln

V4 V = RTmax ln 4 V3 V3

Process 4–1 (Adiabatic expansion process) Source and conductor are removed and insulator is placed on the head (H). The piston moves from point 4 to point 1, i.e. it reaches the outermost dead centre, completing one thermodynamic cycle. During the process gases expand from state 4 to state 1. Temperature decreases from T4 to T1 = Tmin Pressure decreases from P4 to P1

Second Law of Thermodynamics

193

Volume increases from V4 to V1 Work done by the system W4–1 Heat interaction Q = 0 Change in entropy DS = S4 – S1= 0 (S4 = S1) W4 -1 =

P4V4 - P1V1 RT4 - RT1 R(Tmax - Tmin ) = = -1 -1 -1

Net work done during the cycle (Wn) Wn =

ÚW =W

1- 2

= - RTmin ln = RTmax ln

+ W2 -3 + W3- 4 + W4 -1

R(Tmax - Tmin ) V1 R(Tmin - Tmin ) V + RTmax ln 4 + V2 -1 V3 -1

V4 V - RTmin ln 1 V3 V2

Process 2–3

V3 È T2 ˘ =Í ˙ V2 Î T3 ˚

Process 4–1

1 -1

V4 È T1 ˘ =Í ˙ V1 Î T4 ˚

(4.1)

1 -1

1

È T ˘ -1 =Í 2˙ (4.2) Î T3 ˚ [∵ T1 = T2; T4 = T3]

Equating Eqs. (4.1) and (4.2), we have, V4 V1 = = RA = Adiabatic compression and expansion ratio V3 V2 V4 V1 = = RI = Isothermal compression and expansion ratio V3 V2

\

T2 Tmin È V3 ˘ = =Í ˙ T3 Tmax Î V2 ˚

Wn = RTmax ln

-1

= ( RA )

-1

V1 V V - RTmin ln 1 = R ln 1 [Tmax - Tmin ] V2 V2 V2

From the first law of thermodynamics,

Ú

W=

Ú

Wn = Qn

Q

194

Basic Thermodynamics

c

= Thermal efficiency =

=

Net work done Wn Qin - Qout = = = Qin Total heat input Qin

Tmax - Tmin T = 1 - min = 1 - ( R A ) Tmax Tmax

R ln

V1 (Tmax - Tmin ) V2 V R ln 4 Tmax V3

-1

We can write c

\

=

Qin - Qout Tmax - Tmin Q T = = 1 - out = 1 - min Qin Tmax Qin Tmax

(4.2a)

Qout Tmin = Qin Tmax

If Qout = Q2, Qin = Q1, Tmin = T2, Tmax = T1, then the above equation can be written as, Q2 T2 = Q1 T1

4.9 4.9.1

(4.3)

CARNOT’S THEOREM Corollary-I

It states that “no heat engine can be constructed to operate between only two heat reservoirs with a higher efficiency than a reversible heat engine operating between the same two reservoirs.” Let I in Figure 4.26(a) be the irreversible heat engine and R be the reversible heat engine. Both the engines operate between the same temperature limits of Tmax and Tmin. Both engines receive, QH, amount of heat from the higher temperature reservoir. WI and WR are the work developed by

Figure 4.26 Illustration of Carnot theorem: Corollary-I.

Second Law of Thermodynamics

195

the irreversible and reversible heat engines, respectively. Now we have to prove that efficiency of reversible engine (R) is more than that of irreversible engine (I). Let us assume that converse of this corollary-I is true, hI > hR

i.e.

WI W > R WH WH

\

WI > WH

Now, reversible heat engine can be operated as a reversible heat pump with the same QL and QH, i.e. just reverse the direction. The reversible heat pump (i.e. HP) receives QL amount of heat from the low temperature reservoir and rejects QH, at high temperature reservoir. Now this heat (QH) is supplied from reversible heat pump (i.e. HP) to irreversible heat engine (I) as shown in Figure 4.26(b). Doing so, high temperature reservoir can be eliminated. Now the combined system shown in Figure 4.26(b) can produce net work because WI is greater than WR. Thus, the system produces net work (Wn) while exchanging heat with a single reservoir. Such a system is perpetual motion machine of the second kind (PMM-II) and is a violation of the Kelvin-Planck statement of the second law of thermodynamics. Therefore, our initial converse assumption is not true, i.e. efficiency of I cannot be greater than that of R, and the corollary-I must be true.

4.9.2

Corollary-II

“All reversible engines operating between the same two reservoirs have the same efficiency.” Let R1 and R2 in Figure 4.27 be the two reversible engines. Both the engines operate between the same temperature limits of Tmax and Tmin. Both the engines receive QH amount of heat from the higher temperature reservoir. W1 and W2 are the work developed by the engines R1 and R2, respectively. Now we have to prove that efficiency of R1 and R2 is same. Let us assume that efficiency of R1 is greater than that of R2,

Figure 4.27 Illustration of Carnot theorem: Corollary-II.

196

Basic Thermodynamics

i.e.

R1

\

>

R2

W1 W > 2 WH WH

W1 > W2

Now, engine R2 can be operated as a reversible heat pump with the same QL and QH, i.e. just reverse in direction. The reversible heat pump (HP) receives QL amount of heat from the low temperature reservoir and rejects QH amount of heat to the high temperature reservoir. The rejected heat, QH, from the heat pump, R2, is supplied to the heat engine, R1, as shown in Figure 4.27(b). Doing so, high temperature reservoir can be eliminated. Now the combined system shown in Figure 4.27(b) can produce net work because W1 is greater than W2. Thus, the combined system produces net work (Wn) while exchanging heat with a single reservoir. Such a system is perpetual motion machine of second kind (PMM-II) and is a violation of the Kelvin-Planck statement of the second law of thermodynamics. Therefore, our initial assumption is not true, i.e. efficiency of R1 cannot be greater than that of R2, and the corollary-II must be true.

4.10 THERMODYNAMIC TEMPERATURE SCALE As per the corollary-II, the thermal efficiency of a reversible heat engine does not depend on the nature of the heat engine (type of the heat engine) or the working fluid. The thermal efficiency depends only on the temperature of the two reservoirs. Based on this fact one can establish an absolute thermodynamic temperature scale. This thermodynamic temperature scale is the one which does not depend on the properties of any particular substance, but is based only on the characteristics of reversible heat engines and constant temperature reservoirs.

4.10.1

Corollary-III

“A temperature scale is one which is independent of the properties of the substance used for measurement of temperature.” Consider the reversible heat engines shown in Figure 4.28. High temperature reservoir T1 Q1 Q1

W1 = Q1 – Q2

R2

W2 = Q1 – Q3

Q3 R1

T3

Intermediate

Q3 R3 Q2

W3 = Q3 – Q2

Q2 Low temperature reservoir T2

Figure 4.28 Reversible heat engines (R1, R2 and R3) to illustrate thermodynamic temperature scale.

Second Law of Thermodynamics

197

The reversible heat engine, R1, receives heat, Q1, from the high temperature reservoir at T1 and produces net work, W1, while rejecting heat, Q2, to the low temperature reservoir at T2. The reversible heat engine, R2, also receives heat, Q1, from the high temperature reservoir at T1 while producing net work, W2, and rejecting heat, Q3, to the intermediate temperature reservoir at T3. The reversible heat engine, R3, is designed in such a way that it receives heat, Q3, from the intermediate temperature reservoir at T3 and produces net work, W3, and rejecting heat, Q2, to the low temperature reservoir at T2. The first law of thermal efficiency of any heat engine is given by, =1-

Q2 Q1

Apply this equation to reversible heat engine R1, \

R1

=1-

Q2 Q1

(4.4)

As per the Carnot cycle, the efficiency of any reversible heat engine can be expressed as a function of temperatures of the high and low temperature reservoirs. \ Equation (4.4) can be written as, R1

=1-

Q1 T = 1 - 2 = F (t1 , t2 ) Q2 T1

[Eq. (4.2a)]

Q1 = F (t1 , t2 ) Q3

Similarly for reversible heat engines R2, R3, Q3 = F (t1 , t3 ) Q2

Q3 = F (t3 , t2 ) Q2

Now

Q1 Q /Q F (t1 , t2 ) = F (t1 , t3 ) = 1 2 = Q3 Q3 / Q2 F (t3 , t2 )

(4.5)

The temperature t1, t2 and t3 are arbitrarily chosen. As per corollary-II (Carnot cycle), the ratio Q1/ Q3 depends only on t1 and t3, and is independent of t2. So t2 will drop out from the right side of Eq. (4.5). After the cancellation of t2, we get Q1 (t ) = F (t1 , t3 ) = 1 Q3 (t2 )

(4.6)

Several functions, f(t) will satisfy this equation, and the choice is completely arbitrary. Lord Kelvin first proposed that f(t) = T to define a thermodynamic temperature scale.

198

Basic Thermodynamics

Q1 T1 = Q3 T3

(4.7)

Equation (4.7) is called temperature scale, Kelvin scale and the temperatures on this scale are called absolute temperatures. Equation (4.7) is having two temperatures. If one temperature is fixed and assigned a number, the other temperature can be determined. The fixed point or reference temperature is taken as triple point temperature of water and assigned a valve of 273.16 K. This is shown in Figure 4.29. Then, Eq. (4.7) reduces to \

T1 = T2

Figure 4.29

Schematic representation of equation 4.7 (Q3 = Qtp, T3 = Ttp).

Q1 Q Q = Ttp 1 = 273.16 1 Q2 Qtp Qtp

Comparison of Eqs. (4.3) and (4.7) shows that ideal gas temperature and the absolute thermodynamic temperature are identical.

4.11 ZERO TEMPERATURE ON THERMODYNAMIC TEMPERATURE SCALE Consider a series of reversible heat engines R1, R2, R3, ..., etc. extracting heat from a high temperature source T1 as shown in Figure 4.30. Following points are to be remembered for reversible heat engines in series. All reversible heat engines work between two heat reservoirs, higher and lower temperature. Sink of each engine will act as a source for the next engine. Apply the first law of thermodynamics. Wt = total work output = W1 + W2 + W3 + ... Total work cannot be greater than Q1. In the limiting case, Wt = Q1 In order to satisfy this condition, the last engine should reject zero heat. Therefore, the last engine in the series violates the second law of thermodynamics, i.e. Kelvin-Planck statement of the second law of thermodynamics. Hence, it is impossible. We know that heat is proportional to temperature, i.e.

Source, T1

Q1 T1 = Q2 T2

Therefore, as the number of engines in the series is increased and heat rejection of the last engine approaches zero, the Figure 4.30 temperature at which this heat is rejected approaches zero on the thermodynamic scale.

Q1 R1 Q2

W1 T2

Q2 R2 Q3

W2 T3

Q3 R3 Q4

W3 T4

Q4 R4 Q5

W4 T5

Reversible heat engines in series.

Second Law of Thermodynamics

199

Violation of the second law means rejection of zero heat from the last engine. Then, rejection of zero heat is impossible means attainment of zero temperature is also impossible. Thus, absolute zero is an idealized one and not practically attainable.

4.12 COROLLARY-I This can also be stated as, “no refrigerator (or heat pump) can be constructed to operate between only two heat reservoirs with a higher co-efficient of performance than a reversible refrigerator (or heat pump) operating between the same two reservoirs.” Let I in Figure 4.31(a) be the irreversible heat pump (or refrigerator). Both the heat pumps operate between the same temperature limits of Tmax and Tmin. Q1¢ and Q1 be the heat taken by the heat pumps I and R, respectively from the low temperature reservoir. WI and WR are the work input to the irreversible and reversible heat pumps (or refrigerator), respectively. Now, we have to prove that COP of reversible heat pump (R) is more than that of irreversible heat pump (I). Let us assume that converse of this corollary-I is true, i.e.

COPI > COPR Q2 Q2 > WI WR W1 WR < Q2 Q2

WI < WR

Figure 4.31 Illustration of Carnot theorem: Corollary-I.

Now, reversible heat pump can be operated as a reversible heat engine with the same amount of Q1 and Q2, just reverse the direction. The reversible heat pump (RHP) receives Q2 amount of heat from the high temperature reservoir and rejects heat, Q1, to low temperature reservoir. The

200

Basic Thermodynamics

rejected heat, Q2, from the irreversible heat pump is supplied to the reversible heat engine, RHE, as shown in Figure 4.31(b). Doing so, high temperature reservoir can be eliminated. Now the combined system shown in Figure 4.32(b) can produce net work because WR is greater than WI. Thus, the combined system produces net work (Wn) while exchanging heat with a single reservoir. Such a system is perpetual motion machine of second kind (PMM-II) and is a voilation of the KelvinPlanck statement of the second law of thermodynamics. Therefore, our initial assumption is not true, i.e. COP of irreversible heat pump (I) cannot be greater than COP of reversible heat pump and the corollary-I must be true.

4.13 COROLLARY-II This can be stated in other way, “all reversible refrigerators operating between the same two reserviors have the same co-efficient of performance.” Let R1 and R2 in Figure 4.32(a) be the two reversible refrigerators. Both the refrigertors operate between the same temperature limits, Tmax and Tmin. Both the refrigerators receive Q1 amount of heat from the low temperature reservior. W1 and W2 are the work input to the refrigerators R1 and R2, respectively. Now, we have to prove that COP of R1 and R2 is same. Let us assume that COP of R1 is greater than COP of R2, COPR1 > COPR2

i.e.

Q1 Q1 > W1 W2 W1 W2 < Q1 Q1

W1 < W2 High temperature reservoir source Tmax

Q1 R1 Ref

Q2 W2 W1

Q1

R2 Ref Q1

High temperature reservoir source Tmax

Q2

Q2 W1

R1 Ref

W2

Wn = W2

W1

Q1

Q1

Low temperature reservoir sink Tmin (a)

R2 Ref

(b)

Figure 4.32 Illustration of Carnot theorem: Corollary-II.

Second Law of Thermodynamics

201

Now, refrigerator, R2, can be operated as a reversible heat engine with the same amount of Q1 and Q2, just reverse the direction. The reversible heat engine, R2HE, receives Q1 amount of heat from the high temperature reservior and rejects Q1 amount of heat to the low temperature reservior. The rejected heat, Q1, from the engine, R2HE, is taken by the refrigerator, R1, as shown in Figure 4.32(b). Doing so, the low temperature reservior or high temperature reservior can be eliminated. Now, the combined system shown in Figure 4.32(b) can reduce net work (Wn) because W2 is greater than W1. Thus, the combined system produces net work, Wn, while exchanging heat with a single reservior. Such a system is perpetual motion machine of second kind (PMM-II) and is a violation of the Kelvin-Planck statement of the second law of thermodynamics. Therefore, our initial assumption is not true, i.e. COP of R1 cannot be greater than COP of R2 and the corollary-II must be true.

SOLVED EXAMPLES EXAMPLE 4.1 An inventor claims that his engine has the following data. Power developed = 30 kW Calorific value = 45000 kJ/kg

Fuel consumed = 4 kg/hr Operating temperature limits = 750°C, 30°C

Justify the claim. Solution: hR = Thermal efficiency of reversible heat engine =

Tmax - Tmin 750 - 30 ¥ 100 = ¥ 100 = 70.35% Tmax 750 + 273

hHE = Thermal efficiency of heat engine = =

W.D. Heat supplied

30 kW 30 kW = ¥ 100 = 60% 4 kg kJ FC ¥ CV ¥ 45000 3600 s kg

Thermal efficiency of heat engine is less than thermal efficiency of reversible heat engine. Hence, claim is justified. EXAMPLE 4.2 A heat engine receives heat at the rate of 2000 kJ/min and does a work output of 8.5 kW. Calculate (a) the thermal efficiency and (b) the rate of heat rejection. Solution: h) (a) The thermal efficiency (h =

Work done 8.5 kW ¥ 100 = ¥ 100 = 25.5% 2000 kJ Heat supplied 60 s

202

Basic Thermodynamics

(b) The rate of heat rejection (QR) QR = Heat supplied – Work output =

2000 kW - 8.5 kW = 24.83 kW 60

EXAMPLE 4.3 The evaporator temperature of a Carnot refrigerator is maintained at 5°C. The heat is rejected to the surrounding at a temperature of 30°C. Now the evaporator temperature is decreased to –13°C, the surrounding temperature remaining constant at 30°C. Calculate the percent increase in work input for the same quantity of heat (QR) removed from the low temperature reservoir. Solution:

Tmax = 30°C

QH = Heat rejected from the refrigerator QR = Heat supplied to the refrigerator

QH

Tmin1 = 5 + 273 = 278 K Tmin2 = –13 + 273 = 260 K

Ref

Tmax = 30 + 273 = 303 K We know that

W

QR

QH – QR = W

(1) Tmin

QH Tmax = QR Tmin

(2)

ÈT ˘ W = QR Í max - 1˙ Î Tmin ˚ Case 1, when the refrigerator operates between 5°C and 30°C

\

Figure E4.3

(3)

È 303 ˘ W1 = QR Í - 1˙ = 0.0899QR Î 278 ˚

\

Case 2, when the refrigerator operates between –13°C and 30°C È 303 ˘ W2 = QR Í - 1˙ = 0.1654QR Î 200 ˚

\

The percent increase in work input (W%) W% =

W2 - W1 0.1654QR - 0.0899QR ¥ 100 = = 83.98% W1 0.0899QR

EXAMPLE 4.4 A reversible heat engine takes heat at the rate of 500 kJ per second from a heat source at 700 K. The work done by the cyclic device is 200 kJ per second and rejects heat to two sinks at 400 K and 500 K, respectively. Calculate (a) the engine thermal efficiency and (b) the amount of heat rejected to each sink.

Second Law of Thermodynamics

203

Solution: h) (a) The engine thermal efficiency (h =

Work done 200 kW ¥ 100 = ¥ 100 = 40% Heat supplied 500 kW Source Tmax 700 K 500 kJ/s = QS HE

W = 200 kJ/s

300 kJ/s = QR QRA

QRB

Sink A 400 K

Sink B 500 K

Figure E4.4

(b) Amount of heat rejected to each sink (QRA, QRB) Heat rejected = Heat supplied – Work done = 500 – 200 = 300 kW Let heat rejected to sink A = QRA Then heat rejected to sink B = (300 – QRA) = QRB From the second law of thermodynamics, i.e. Clausius theorem,

vÚ

Q Q Q Q˘ 500 QRA 300 - QRA = 0 = R - RA - RB = T ˙˚ REV Tmin Tmin A Tmin B 700 400 500

\

QRA = 228.6 kW

\

QRB = 300 – 228.6 = 71.4 kW

EXAMPLE 4.5 A reversible heat engine rejects heat at the rate of 1200 kJ/min at 30°C to a river. The efficiency of this engine is 45%. Calculate (a) temperature of the source and (b) power output of the engine.

Tmax QH

Solution: (a) Temperature of the source (Tmax) hR = Thermal efficiency of reversible HE T - Tmin Tmax - 303 0.45 = max = Tmax Tmax

\

Tmax = 550.91 K or

277.91°C

HE

QL Tmin = 30°C

Figure E4.5

W

204

Basic Thermodynamics

(b) Power developed (W.D.) R

\

W.D. Qin - Qout 1200 = =1= 0.45 Qin Qin Qin

Qin = 2181.82 kJ/min = 36.36 kW R

\

=

=

W.D. W.D. = = 0.45 36.36 Qin

W.D. = 16.36 kW

EXAMPLE 4.6 There are three reservoirs at temperatures 827°C, 127°C and 27°C parallel. A reversible heat engine operates between 827°C and 127°C and a reversible refrigerator operates between 27°C and 127°C, respectively. 502 kJ of heat is extracted from the reservoir at 827°C by the heat engine and 251 kJ of heat is abstracted by the refrigerator from the reservoir at 27°C. Find the net amount of heat delivered to the reservoir at 127°C. Can the heat engine drive the refrigerator and still deliver some net amount of work? If so how much? (VTU MQP) Solution:

First consider heat engine operating between 827°C and 127°C

\

HE

=

T1 - T2 W 827 - 127 ¥ 100 = ¥ 100 = E ¥ 100 827 + 273 T1 Q1

\

HE

=

T1 - T2 W Work done ¥ 100 = ¥ 100 = E ¥ 100 Heat supplied T1 Q1

\

WE ¥ 100 502 WE = 319.42 kJ

\

Q2 = Q1 – WE = 502 – 319.42

63.63 =

= 182.58 kJ of heat is rejected to the sink at 127°C

Figure E4.6

Second Law of Thermodynamics

Now consider reversible refrigerator, COPR =

T3 27 + 273 = =3 T2 - T3 127 - 27

COPR =

Q3 251 = =3 Q4 - Q3 Q4 - 251

\

205

Q4 = 334.67 kJ

Net amount of heat rejected to 127°C reservoir (QR) QR = Q2 + Q4 = 182.58 + 334.67 = 517.25 kJ Work done on the refrigerator (WR) WR = Q4 – Q3 = 334.67 – 251 = 83.67 kJ \

WE > WR

Net work developed (Wn) Wn = WE – WR = 319.42 – 83.67 = 235.75 kJ EXAMPLE 4.7 A heat pump is to be 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. Heat transfer through the walls and roof is estimated to be 0.525 kJ/s per degree temperature difference between the inside and outside. (a) If the outside temperature in winter is 5°C, what is the minimum power required to drive the heat pump? (b) If the power requirement is same as in part (a) above, what is the maximum outside temperature for which the inside can be maintained at 20°C? (KUD II-98) Solution: (a) Minimum power required (Wp) Q1 = 0.525 COPP =

kJ kJ ¥ (293 - 278) K = 7.875 sK s

T1 Q1 Q 293 = = = 1 T1 - T2 Q1 - Q2 293 - 278 WP T3

293 K, T1 Q1 WP

HP Ref.

Q3 WR

Ref. Rev.

Q4

Q2 278 K, T2

293 K, T1

Figure E4.7

206

Basic Thermodynamics

293 7875 = 293 - 278 WP \

Wp = 0.4031 kW

(b) Maximum outside temperature (T3) Given condition WP = WR Given that Q4 = 0.525 × (T3 – T1) = 0.525(T3 – 293) Q4 293 = WR T3 - 293 0.525(T3 - 293) 293 = 0.4031 T3 - 293

\

T3 = 308 K

EXAMPLE 4.8 A reversible engine operates between heat reservoirs a, b and c. The engine receives equal quantities of heat from reservoirs a and b at temperatures Ta and Tb, respectively, and rejects heat to reservoir c at temperature Tc. If the efficiency of the above mentioned engine is a times the efficiency of another reversible engine operating between a and c only, prove that, Ta T = [2 - 1] + 2[1 - ] a Tb Tc

Solution:

(KUD II-98, I-96)

Efficiency (ha) of engine operating between Ta and Tc a

=

Ta - Tc Wa = Ta Q1

(1)

Efficiency (hb) of engine operating between Tb and Tc b

Ta

=

Tb - Tc Wb = Tb Q1 Tb Q1

Q1

E

Wa

Wb

Q2a

(2)

Ta Q1 E2

E Q2b

Q2

Tc

Tc

Engine 1

Engine 2

Figure E4.8

W2

Second Law of Thermodynamics

207

Efficiency of one more engine, i.e. E2, operating between Ta and Tc

Given condition,

2

=

Ta - Tc Ta

(3)

1

=

Wa + Wb 2Q1

(4)

h1 = a h2

(5)

Substitute Eqs. (1), (2), (3) and (4) in Eq. (5) 1

=

Wa + Wb 1 È (Ta - Tc ) (Tb - Tc ) ˘ = Í + ˙= 2Q1 2 Î Ta Tb ˚

(Ta - Tc ) Ta

(Ta - Tc ) (Tb - Tc ) 2 (Ta - Tc ) + = Ta Tb Ta 12-2 -

Tc T T +1- c = 2 - 2 c Ta Tb Ta Tc T T +2 c = c Ta Ta Tb

2(1 - ) - (1 - 2 )

Multiply by

Ta , Tc

\

2(1 - )

Ta T T T T - (1 - 2 ) c a = c a Tc Ta Tc Tb Tc

Ta = (2 Tb

\ Hence the proof.

Tc Tc = Ta Tb

- 1) + 2(1 - )

Ta Tc

EXAMPLE 4.9 A household refrigerator is maintained at a temperature of 2°C. Every time the door is opened, warm material is placed inside, introducing an average of 420 kJ, but making only small change in the temperature of the refrigerator. The door is opened 20 times a day and the refrigerator operates at 15% of the ideal COP. The cost of work is 32 paise per kWh. What is the monthly bill for this refrigerator? The atmosphere is at 30°C. (KUD II-97)

T1 = 30°C Q1 WR

R

Solution: COPc = Carnot COP COPc =

T2 275 = = 9.82 28 T1 - T2

Q2 = 420 kJ T2 = 2°C

Figure E4.9

208

Basic Thermodynamics

COPa = Actual COP COPa = 0.15 × COPc = 0.15 × 9.82 = 1.47 COPa = WR =

Q2 WR Q2 420 kJ ¥ 20 ¥ 24 hr ¥ 30 = COPa 1.47 day ¥ 24 hr ¥ 3600 s

WR = 47.62 kWh Monthly bill for this refrigerator (Bill) Cost of the work = 0.32 Rs./unit

1 unit = 1 kWh = 0.32 Rs

Bill = 47.62 kWh = 47.62 × 0.32 = Rs. 15.20 EXAMPLE 4.10 A heat engine is used to drive a heat pump. The heat transfers from the heat engine and from the heat pump are used to heat the water circulating through the radiators of a building. The efficiency of the engine is 27% and the co-efficient of performance of the heat pump is 4. Evaluate the ratio of the heat transfer to the circulating water to the heat transfer to the heat engine. (KUD I-96) Solution:

Figure E4.10

hE = Efficiency of the heat engine = 27% COPP = Co-efficient of performance of the pump = 4 g

\

=

Q1 =

COPP =

W Q1

W g

Q4 W

=

W = 3.704W 0.27

Second Law of Thermodynamics

\

Q4 = W × COPP = 4W

We know that

Q2 = Q1 – W = 3.704W – W = 2.704W

209

The ratio of the heat transfer to the circulating water to the heat transfer to the heat engine (R) R=

H.T. to the circulating water Q2 + Q4 2.704W + 4W = = = 1.81 Q1 H.T. to the heat engine 3.704W

EXAMPLE 4.11 A series combination of two Carnot engines operate between the temperature of 180°C and 20°C. Calculate the intermediate temperature of the engines producing equal amount of work. (KUD March 2001) Solution: Intermediate temperature (T2)

T1 = 180°C

ha = Efficiency of heat engine A

Q1

Wa T1 - T2 453 - T2 = = 453 Q1 T1

(1)

hb = Efficiency of heat engine B (2)

T2 Q2

Q1 = Wa + Q2 Q2 = Wb + Q3

(3)

Wa = Wb (given data)

(4)

Similar to Eq. (1), we can write

HE B

Wb

Q3 T3 = 20°C

Wa T1 - T2 = Q2 T1

Q2 =

b

\

Wa

Q2

Wb T2 - T3 T2 - 293 = = Q2 T2 T2

\

HE A

=

Figure E4.11

WaT1 T1 - T2

(5)

Wb Wb (T1 - T2 ) T2 - T3 = = Q2 WaT2 T2

(T1 – T2) = (T2 – T3) 2T2 = T1 + T3 = 453 + 293

\

T2 = 373 K or

100°C

EXAMPLE 4.12 An inventor claims to have developed an engine which takes 100 MJ of heat at a temperature of 327°C, rejects 48 MJ at a temperature of 27°C and delivers 15 kWh of mechanical work. Is his claim valid?

210

Basic Thermodynamics

Solution: hc = Efficiency of Carnot engine c

=

T1 - T2 327 - 27 ¥ 100 = ¥ 100 = 50% 327 + 273 T2

This is the maximum possible efficiency. HE

15 kJ-hr 15 kJ × 3600 ds = ¥ 100 = ¥ 100 100 MJ 100 s × 1000 kJ = 54%

The efficiency of the heat engine is greater than the efficiency of the Carnot engine, hence the claim of the inventor is wrong.

327°C = T1 Q1 = 100 MJ

HE

W = 15 kWh

Q2 = 48 MJ 27°C = T2

Figure E4.12

EXAMPLE 4.13 A domestic food freezer maintains a temperature of –15°C. The ambient air temperature is 30°C. If heat leaks into the freezer at the continuous of 1.75 kJ/s, what is the least power necessary to pump this out continuously? (KUD July 2002) Solution: The least power necessary to pump continuously (WR) COPR =

T2 258 = = 5.733 T1 - T2 303 - 258

COPR =

Q2 1.75 kW = WR WR kW

WR =

\

1.75 = 0.3052 kW 5.733

Or we know that Q2 Q1 = T2 T1

\

Q1 =

Figure E4.13

Q2T1 1.75 ¥ 303 = = 2.055 kW T2 258

WR = Q1 – Q2 = 2.055 – 1.75 = 0.305 kW EXAMPLE 4.14 Calculate the thermal efficiency of a heat engine working on the Carnot cycle between the temperatures of 300°C and 30°C. To improve thermal efficiency by 10% what must be the new source temperature when sink temperature is held at 30°C. (KUD I-91) Solution:

New source temperature when sink temperature is held at 30°C (TN) ho = Thermal efficiency of original engine =

T1 - T2 300 - 30 ¥ 100 = ¥ 100 = 47.1% T1 300 + 273

Second Law of Thermodynamics

211

Figure E4.14

hN = Thermal efficiency of new engine = (ho + 10) 57.1 =

\

TN - 300 ¥ 100 TN

TN = 706.29 K or

433.29°C

EXAMPLE 4.15 A reversible engine operates between temperature limits of T1 and T, where T1 is the higher temperature. The heat rejected by this engine is received by a second reversible engine at the same temperature T, which in turn rejects heat to a sink at temperature T2. (a) If the two engines have equal efficiencies, show that T = T1T2 , (b) If the two engines have equal work output, show that T = ½(T1 + T2). (KUD I-92 year course) Solution: (a) When both the engines have the same efficiency

T1 (Source)

h1 = h2 1=

Q1

Q Q =1- 2 Q1 Q

T Q

T T2 = T1 T

\

T = T1T2

HE2

T = T1T2

(b) If both the engines have same work transfer 1

=

W1

Q

T T =1- 2 1T1 T

2

HE1

W1 W1 T = =1Q1 W1 + Q T1

Q2 T2 (Sink)

Figure E4.15

W2

212

Basic Thermodynamics

È T˘ W1 = (W1 + Q) Í1 - ˙ T1 ˚ Î

\ We have

W1 Q1 - Q T1 - T È T1 ˘ = = = Í - 1˙ Q Q T T Î ˚ ÈT ˘ W1 = Q Í 1 - 1˙ T Î ˚

\

(1)

h2 = Efficiency of 2nd engine = \ Given that \

W2 È T2 ˘ = 1- ˙ Q ÍÎ T˚

T ˘ È W2 = Q Í1 - 2 ˙ T˚ Î W1 = W 2

(2)

Equating Eqs. (1) and (2),

ÈT ˘ È T ˘ Q Í 1 - 1˙ = Q Í1 - 2 ˙ T˚ ÎT ˚ Î

T1 T -1=1- 2 T T \

T=

T1 T2 + =2 T T

T1 + T2 2

EXAMPLE 4.16 A reversible heat engine operates between two reservoirs at temperatures of 600°C and 40°C. The engine drives a reversible refrigerator which operates between reservoirs at temperature of 40°C and –20°C. The heat transfer to the heat engine is 2000 kJ and the net work output of the combined engine refrigerator plant is 360 kJ. Evaluate the heat transfer to the refrigerant and net heat transfer to the reservoir at 40°C. (KUD I-97) Solution: HE

È 40 + 273 ˘ = Í1 ˙ ¥ 100 = 64.15% 600 + 273 ˚ Î

\

WE = hHE × Q1 = 0.6415 × 2000 = 1282.93 kJ

\

Q2 = 2000 – 1282.93 = 717.07 kJ COPR = Co-efficient of performance of refrigerator =

Given that

-20 + 273 = 4.22 40 - ( -20)

WE – WR = 360 kJ WR = 1282.93 – 360 = 922.93 kJ

Second Law of Thermodynamics

T3 = 20°C

T1 = 600°C

Q3

T1 = 200 kJ HE Rev.

213

WE

WR

Ref. Rev.

300 kJ Q2

Q4 T2 = 40°C

Figure E4.16

COPR =

Q3 WR

(a) Heat transfer to the refrigerant (Q3) \

Q3 = WR × COPR = 922.93 × 4.22 = 3894.7646 kJ

(b) Net heat transfer to the reservoir at 40°C (Qn) \

Q4 = Q3 + WR = 3894.76 + 922.93 = 4817.695 kJ Qn = Q2 + Q4 = 717.07 + 4817.695 = 5534.765 kJ

EXAMPLE 4.17 A Carnot refrigerator has to remove 1680 kJ/hr of heat from a region at –50°C and reject it to atmosphere at 5°C. A Carnot engine operating between 600°C and 5°C is driving a refrigerator. Estimate the rate of heat supply to Carnot engine at 600°C. Solution:

Rate of heat supply to the Carnot heat engine (QA) HE

=

COPR =

TA - TB 600 - 5 ¥ 100 = ¥ 100 = 68.16% 600 + 273 TA TC -50 + 273 = = 4.055 5 - ( -50) TB - TC TC = 50°C

TA = 600°C

QC = 1680 kJ/hr Ref R

WR

QA WE

Q2B

HE R Q1B

TB = 5°C

Figure E4.17

214

Basic Thermodynamics

From Clausius theorem, we have WR = Work done by Carnot engine =

WE TA - TB = QA TA

WE =

QA (TA - TB ) TA

HE

\

COPR =

\

(1)

QC TC = Q2 B - QC TB - TC

(Q2B – QC) = Work of Carnot refrigerators =

QC (TB - TC ) TC

(2)

Work done by the Carnot engine = Work done on the Carnot refrigeration (given). Hence, equating Eqs. (1) and (2), QA (TA - TB ) QC (TB - TC ) = TA TC

QC (TB - TC ) TC Q T (T - TC ) QA = = C A B TA - TB TC (TA - TB ) TA QA =

1680 (600 + 273) (278 - 223) kJ ¥ = 607.95 ( -50 + 273) (873 - 278) hr

EXAMPLE 4.18 It is claimed that an engine working on new heat engine cycle engine between temperatures 1400°C and 30°C receives 4.2 kJ/s of heat and develops a power of 3.675 kW. (a) Disprove the claim. (b) What change in condition would validate the claim and its magnitude? Solution: (a) To disprove the claim hc = Efficiency of Carnot engine =

Tmax - Tmin 1400 - 30 ¥ 100 = ¥ 100 = 81.9% Tmax 1400 + 273

hN = Efficiency of newly developed engine

=

W.D. of new engine 3.675 kW ¥ 100 = ¥ 100 = 87.5% kJ Heat input to new engine 4.2 s

Second Law of Thermodynamics

215

h N > h C ∵ it is impossible as per the corollary of the second law of thermodynamics, and disproved. (b) Change in condition that would validate the claims and its magnitude The engine can be validated as below, Case I—If work output decreased for the same heat input. Case II—If heat input increased for the same work output. Case I: Let DW be the decrease in work output \

N

=

C

=

W.D. – DW 3.675 - DW = HS 4.2

3.675 - DW 4.2 DW = 0.2352 kW

0.819 =

\

If work is decreased by 0.2352 kW, then h N = h C Case II: Let DQ be the amount of heat increased N

=

0.819 = \

C

Q + DQ

=

W.D. 3.675 = Q + DQ 4.2 + DQ

3.675 4.2 + DQ

DQ = 0.2872 kW

In order to get hN = hC, amount of heat increased by 0.2872 kW. EXAMPLE 4.19 The heat engine used in thermal power station is considered as a Carnot heat engine operating between 350°C and 70°C. If the maximum rise in the cooling water is restricted to 10°C, and the power output is 106 kW, calculate the water flow rate required. Solution: HE

\

=

Q1 =

T1 = 350°C Q1 HE Rev.

T1 - T2 350 - 70 ¥ 100 = ¥ 100 = 44.94% T1 350 + 273

WE HE

10 6 kW = = 2.225 ¥ 10 6 kW 0.4494

Q1 quantity of heat is transferred to the sink and carried by circulating water where the temperature of water is raised by 10°C.

6

WE = 10 kW

Q2 in Water circulation = m kg/s

T2 = 70°C out

Figure E4.19

216

Basic Thermodynamics

The water flow rate (m) \

Q1 = m Cp dT 2.225 ¥ 10 6 kW = m

kg kJ ¥ 4.187 ¥ 10 K s kg K

m = 53140.67

\

kg s

EXAMPLE 4.20 A heat engine is shown in the figure where 10000 kJ/hr of heat is supplied from source at 1400°C while the working fluid is at 540°C. 8000 kJ/hr of heat is rejected to a sink at temperature 5°C and working fluid is at 60°C. Calculate the following: (a) The actual efficiency of the engine (b) The fraction of actual efficiency of the internally reversible efficiency (c) The fraction of actual efficiency of the external reversible efficiency. Solution: TFH = Fluid temperature at which heat is supplied TFC = Fluid temperature at which heat is rejected h a) (a) The actual efficiency of the engine (h a

Source TH = 1400°C

W 2000 = E = = 0.2 Q1 10000

Q1 = 10000 kJ/hr

h FI) (b) Fraction of internally reversible efficiency (h

540°C

hI = Internal reversible efficiency È T = Í1 - FC TFH Î FI

=

a

=

1

˘ È 60 + 273 ˘ ˙ = Í1 ˙ = 0.59 540 + 273 ˚ ˚ Î

0.2 = 0.34 0.59

h FE) (c) Fraction of external reversible efficiency (h he = The external reversible efficiency =

\

FE

=

HE

WE 60°C Q2 = 8000 kJ/hr TC = –5°C

Figure E4.20

TH - TC 1400 - 5 = = 0.834 1400 + 273 TC a e

=

0.2 = 0.2399 0.834

EXAMPLE 4.21 An ice plant working on reversed Carnot cycle produces 5 tonnes of ice per day at 0°C from water at 0°C. Heat is rejected to the atmosphere at 30°C. Heat engine absorbs heat from the source maintained at 250°C by burning fuel of calorific valve of 45000 kJ/kg. Calculate the mass flow rate of fuel consumption.

Second Law of Thermodynamics

Solution: T3 = 30°C

T1 = 250°C

Fuel

QH

Q1 WE

HE Rev.

WR

Ref. Rev.

Q2

Q3 T3 = 0°C

T2 = 30°C

Figure E4.21

Q3 = Amount of heat removed from ice per kg of ice = Latent heat of fusion of ice = 335 kJ/kg m3 = mass of ice produced = Q3 = 335 COPR =

WR =

5 ¥ 1000 kg kg = 0.05787 24 ¥ 3600 s s

kJ kg kJ ¥ 0.05787 = 19.3865 kg s s

Q 0 + 273 = 9.1 = 3 30 - 0 WR

Q3 19.3865 = = 2.130 kW COPR 9.1

hHE = Efficiency of heat engine = 1 -

HE

\

=

Q1 =

T2 30 + 273 =1= 0.4206 T1 250 + 273

WE Q1 WE HE

=

2.130 = 5.064 kW 0.4206

[WR - We ]

Mass flow rate of fuel consumption (m) \

Q1 = Heat added to the heat engine = mass flow rate of fuel × C.V. of fuel = m 5.064 = m × 45000

\

m = 0.000112524

kg s

or 0.4051

kg hr

kg kJ × C.V. s kg

217

218

Basic Thermodynamics

EXAMPLE 4.22 A reversible heat engine operates between heat source (T1) and heat sink (T2) developing work WE. A reversible refrigerator takes heat from a cold chamber of T3 and rejects heat to heat sink (T2). The work is developed by the heat engine taken by the refrigerator. Derive a relation between Q1 and T1, T2 and T3. Solution: T3

T1

Q3

Q1 WE

HE R

WR

Q2

Ref. Rev. Q4

T2

Figure E4.22

hHE = Thermal efficiency of reversible heat engine =

\

Q1 =

WE T1 - T2 = Q1 T1

WE T1 T1 - T2

(1)

COPR = Co-efficient of performance of reversible refrigerator =

\

WR =

Q3 Q3 T3 = = WR Q2 - Q3 T2 - T3 Q3 (T2 - T3 ) T3

Substituting Eq. (2) in Eq. (1), we know that WE = WR Q1 =

Q3T1 (T2 - T3 ) T3 (T1 - T2 )

EXAMPLE 4.23 A fluid is compressed in an adiabatic steady state steady flow process as shown in the figure. If 14 kJ/s is the work done on the compressor, calculate the efficiency of the compressor for a mass flow rate of 0.5 kg/s. Given that h1 = 200.00 kJ/kg, h2 = 225 kJ/kg Solution: This is a steady flow steady state process \ m1 = m2 = m K.E. and P.E. changes are neglected.

Second Law of Thermodynamics

219

2

Compressor

Wa = 14 kJ/s

Wa

m = 0.5 kg/s

1

Figure E4.23

Assuming isentropic process, WI = Ideal work due to compression Wa = Actual work done on the compressor From SFEE,

WI = (h2 – h1) = (225 – 200) = 25 kJ/kg

14 kJ 14 kJ = = 28 kgs 0.5 kg m s h) Efficiency of the compressor (h Wa =

=

WI 25 ¥ 100 = ¥ 100 = 89.3% 28 Wa

EXAMPLE 4.24 A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle takes in Q1 heat units at T1 and rejects Q2 at T2. The heat pump abstracts Q4 from the sink at T4 and discharges Q3 at T3. Develop an expression for the ratio Q4/Q1, in terms of the four temperatures. Solution: T4

T1

Q4

Q1

HE R

WE

WR

Ref. Rev.

Q3

Q2

T3

T2

Figure E4.24

220

Basic Thermodynamics

hHE = Efficiency of reversible heat engine =

\

Q1 =

WE T1 - T2 = Q1 T1 WE T1 T1 - T2

(1)

COPR = Co-efficient of performance of a refrigerator =

\

Q4 =

Q4 Q4 T4 = = WR Q3 - Q4 T3 - T4 WR T4 T3 - T4

(2)

Now, divide Eq. (2) by Eq. (1), Q4 WR T4 (T1 - T2 ) = Q1 WE T1 (T3 - T4 )

\

Given that work developed by the heat engine is used to run the refrigerator, i.e. WR = WE Q4 T4 (T1 – T2 ) = Q1 T1 (T3 – T4 )

\

EXAMPLE 4.25 An S.I. engine has a thermal efficiency of 30% and develops 50 kW of power output. Calculate mass flow rate of fuel. Take calorific valve of fuel as 45000 kJ/kg. Solution: Mass flow rate of fuel ( m ) h = Thermal efficiency of S.I. engine 0.30 =

\

m=

W W = = Q1 m ¥ C.V.

50 kW m

kg kJ ¥ 45000 s kg

50 kg kg = 0.003703 or 13.33 0.3 ¥ 45000 s hr

Figure E4.25

EXAMPLE 4.26 Heat energy is generated by two methods. Method 1: By electricity, i.e. electric burner. Efficiency of electric burner is 75%. Consider a 2 kW electric burner where the unit cost of electricity is Rs. 0.09/kW h. Method 2: By using LPG gas (burner) energy. Efficiency of this unit is 40%. Cost of energy Rs. 0.6 per therm. Calculate (a) the rate of energy consumption by the burner and (b) the unit cost of utilized energy for both electric and LPG burners.

Second Law of Thermodynamics

221

Solution: (a) Rate of energy consumption (ee C) Electric burner: Q utilized = 2 kW × 0.75 = 1.5 kW \

eC = 2 kW to get 1.5 kW of heat energy

LPG burner: In order to get 1.5 kW of energy from LPG burner, heat input required = 1.5/0.4 = 3.75 kW \

eC = 3.75 kW to get 1.5 kW of heat energy

(b) Unit cost of utilized energy for both electric and gas burners Electric burner: Cost of electric energy = = LPG burner:

Cost of LPG energy = =

Cost of energy input Efficiency Rs. 0.09 Rs. 0.12 = kW h × 0.75 kW h

Cost of energy input Efficiency Rs. 0.6 Rs. 0.0512 = 29.3 kW h × 0.4 kW h [1 Therm = 29.3 kWh]

EXAMPLE 4.27 An inventor claims to have developed a refrigerator operating between a temperature limits of 5°C and 30°C and has a COP of 15.0. Is this claim possible? Solution: COPR = Co-efficient of performance of a reversible refrigerator =

5 + 273 = 11.12 30 - 5

COPNR = Co-efficient of performance of new refrigerator operating between same temperature limits = 15.00 (given data) Here COPNR > COPR This is impossible.

\

The claim is false.

EXAMPLE 4.28 A reversible heat engine converts one sixth of the heat input into work. When the temperature of the sink is reduced by 60°C, the efficiency gets doubled. Calculate source and sink temperature. Solution: h = Efficiency of the reversible heat engine = Q = Heat supplied

T1 - T2 T1

222

Basic Thermodynamics

Source temperature (T1) and sink temperature (T2) W = Work developed =

In the beginning,

W T1 - T2 1 = = Q T1 6

6T2 5 When the sink temperature is reduced by 60°C, efficiency is doubled \

T1 = 6T1 – 6T2

=

\

T1 =

T1 - (T2 - 60) 1 =2¥ T1 6

2T1 = 6T1 – 6T2 + 36

Substituting for T2 and simplifying, we get 6T2 = 1800 \

T2 = 300 K

and T1 = 360 K

EXAMPLE 4.29 A reversible refrigerator is used to produce 800 kg/hr of ice at 4°C from water at 20°C. The refrigerator operates between 20°C and –4°C. Take Cp ice = 2.0 kJ/kg K and latent heat of ice = 335 kJ/kg K. Calculate the power input to the engine. Solution: m = 800 kg/h of ice COPR = Co-efficient of the reversible refrigerator Q = Heat to be removed from water to get ice at –4°C = m[(Cp dT)water + (hfg)ice + (Cp dT)ice] = 800[4.187 × (20 – 0) + 335 + 2.0 × [0 – (–4)] = 341392 COPR =

kJ kJ or 94.8 s hr

Tmin -4 + 273 = = 11.21 Tmax - Tmin 20 - ( -4)

Power input to the machine (W.D.)

\

COPR =

Desired effect W.D.

W.D. =

Desired effect 94.8 kW = = 8.813 kW COPR 11.21

EXAMPLE 4.30 A light is provided in the refrigerator. It is designed in such a way that light will be switched on when the door of the refrigerator is opened. Due to malfunction of the switch, bulb remains ‘on’ continuously. Calculate (a) the increase in energy consumption of the refrigerator and (b) its cost per month.

Second Law of Thermodynamics

223

Take: Bulb – incandescent = 40 watts COP of refrigerator = 2 Cost of electricity = Rs. 3 per unit Assume refrigerator is opened 20 times per day for 30 seconds. Solution: If bulb is continuously ‘on’, 40 watts of energy is added to the refrigerator. This 40 watts is an additional heat load on the refrigerator. Therefore, work required to remove 40 watts. Heat to be removed COP = Work done (W ) W=

\

Heat to be removed 40 W = = 20 watts COP 2

Wa = Additional power consumption due to bulb ‘on’ continuously = 40 W + 20 W = 60 W or J/s (a) Increase in energy consumption (Wa) In general, when the refrigerator is opened, light will be ‘on’, hence the light will be ‘on’ for, Normal operating hours =

20 times 30 s h 30 days 5h ¥ ¥ ¥ = day times 3600 month month

The bulb is ‘on’ for the refrigerator given in this example for, monthly hour = \

30 days 24 h 720 h ¥ = month day month

The additional hour, the light remains on as a result of the malfunction becomes, Additional operating hours = Monthly hour – Normal operating hours = 720 - 5 = 715

Additional power consumption due to 715

h month

h 60 kW 715 h kWh = ¥ = 42.9 month 1000 month month

(b) Cost of the additional power at Rs. 3 per unit = Additional power consumption × Cost =

42.9 kWh 3 Rs. ¥ = Rs. 128.7 per month month kWh

EXAMPLE 4.31 Series combination of three Carnot engines A, B and C operate between temperatures of 1500 K and 300 K. If the amount of heat addition to each engine is in the ratio of 6:3:2, calculate the intermediate temperatures.

224

Basic Thermodynamics

Solution:

The efficiencies of the engines are given by 1

2

3

=

W1 1500 - T2 = Q1 1500

(1)

Q1

T - T3 W = 2 = 2 Q2 T2 =

T1 = 1500 K

(2)

W3 T3 - 300 = Q3 T3

A Q2

(3)

T2

We have

Q2

W1 T1 - T2 1500 - T2 = = Q2 T2 T2

W1 =

Q3

Q2 (1500 - T2 ) T2

T3

(4)

Q3

Substitute Eq. (4) in Eq. (1),

Q4 T4 = 300 K

Q 3 Given condition 2 = = 2 Q1 6

\

W3

C

Q2 (1500 - T2 ) 1500 - T2 = Q1T2 1500

\

W2

B

Intermediate temperatures (T2 and T3) \

W1

Figure E4.31

2 (1500 - T2 ) 1500 - T2 = T2 1500

T2 = 750 K

Substitute T2 in Eq. (2), \

W2 750 - T3 = Q2 750

(5)

W2 T2 - T3 = Q3 T3

(6)

We have

Substitute Eq. (6) in Eq. (5), (750 - T3 ) Q3 750 - T3 ¥ = T3 Q2 750

2 1 = 3T3 750

\

T3 = 500 K

È Q2 2 ˘ = given ˙ Í∵ Q3 3 Î ˚

Second Law of Thermodynamics

225

EXAMPLE 4.32 A reversible heat engine works between the temperature limits of 300°C and 80°C. Efficiency of the heat engine is to be increased either by increasing the source temperature to 350°C or by lowering the sink temperature to 30°C. Which is preferable? Solution: h = Efficiency of Carnot engine =

Tmin 300 - 80 = = 0.384 Tmax - Tmin 300 + 273

If the source temperature is increased to 350°C hmodified 1 =

350 - 80 = 0.433 300 + 273

If the sink temperature is lowered to 30°C modified 2

\

=

300 - 30 = 0.471 300 + 273

Lowering the sink temperature results in increased efficiency.

EXAMPLE 4.33 In a power plant cycle, temperature range is 200°C to 50°C. The upper temperature being maintained in the boiler where heat is received and the lower temperature being maintained in the condenser where heat is rejected. Turbine and pump are adiabatic devices. Specific enthalpies are given in figure. (a) Verify the Clausius inequality. (b) If this is irreversible, what suggestion would you recommend to make the cycle reversible? Solution: (a) Verify the Clausius inequality QS = Heat supplied to the boiler = h1 – h4 = 2800 – 700 = 2100 kJ/kg QR = Heat rejected by the condenser = h2 – h3= 2400 – 450 = 1950 kJ/kg

1 (tS) QS

h1 = 2800 kJ/kg

Boiler W Turbine h4 = 700 kJ/kg

h2 = 2400 kJ/kg

4 2 Condenser (tR)

Pump

Figure E4.33

3

h3 = 450 kJ/kg

226

Basic Thermodynamics

Remaining two devices are adiabatic. \

We have Q

ÂT \

=

QS QR 2100 1950 + = = - 1.5974 < 0 473 323 TS TR

The Clausius inequality is proved. Therefore, the cycle is irreversible.

(b) Suggestion to make the cycle reversible Case I: Increase the heat supply to the boiler at the same temperature (TS = 200°C), keeping the heat rejected in the condenser same. \

QS = ?; QR = 1950 kJ/kg

\

\

TE 323 =1= 0.3171 423 TS

R

=1-

R

= 0.3171 =

QS1 = 2855.6

QS - QR QS1 - 1950 W = 1 = QS1 QS1 QS1

kJ kg

Check

Â

Cycle

Q QS1 QR 2855.6 1950 = + = =0 473 323 T TS TR

The cycle is reversible. Case II:

Qs = 2100 kJ/kg, QR1 = ?

= 0.3171 = \

QR1 = 1434.1

QS - QR1 QS

=

2100 - QR1 QS

kJ kg

Check

Â

Cycle

Q QS1 QR1 2100 1434.1 = = =0 473 323 T TS TR

The cycle is reversible. EXAMPLE 4.34 The figure represents an imaginary ideal cycle. Assuming constant heat capacities, show that the thermal efficiency is, =1-

(V1 / V2 - 1) ( P3 / P2 - 1)

227

Second Law of Thermodynamics

Solution:

P

h = Efficiency of the imaginary cycle QE QS

(A)

QR = Heat rejected by the system QS = Heat supplied to the system

Adiabatic

V=C

=1-

3

2

P=C

1

Assume quasi-static operation QS = Cv (T3 – T2)

(1)

QR = Cp (T1 – T2)

(2)

Figure E4.34

V

Process 2–3 (V = C), P2V2 P3V3 = T2 T3 T2 P2 = T3 P3

(3)

Process 1–2 (P = C), P1V1 P2V2 = T1 T2 T1 V1 = T2 V2

(4)

Substitute Eqs. (1), (2), (3) and (4) in Eq. (A),

=1-

C p (T1 - T2 ) Cv (T3 - T2 )

=1-

(V1 /V2 – 1) (T1 / T2 - 1) =1( P3 /P2 – 1) (T3 / T2 - 1)

EXAMPLE 4.35 An ideal gas engine operates in a cycle which, when represented on a PV diagram, is a rectangle. Call P1 and P2 the lower and higher pressures, respectively and V1 and V2 the lower and higher volumes, respectively. (b) Indicate which parts of the cycle involve heat flow into the gas, and calculate the amount of heat flowing into the gas in one cycle. (b) Calculate the work done in one cycle. [Assume constant heat capacities.] Solution: (a) Indicate which parts of the cycle involve heat flow Process 1–2 (V = C) P1V1 P2V2 = T1 T2

228

Basic Thermodynamics

T1 P1 = T2 P2

Qs¢ = Cv [T2 – T1]

Process 2–3 (P = C), P2V2 P3V3 = T2 T3 T3 V3 V3 = = T2 V2 V1

Qs≤ = Cp [T3 – T2]

QT = Qs¢ + Qs≤ = Total heat supplied = Cv [T2 – T1] + Cp [T3 – T2] ÈÊ È ÈT ˘ ÊV ˆ˘ T ˘ P ˆ = CvT2 Í1 - 1 ˙ + C pT2 Í 3 - 1˙ = Cv T2 Í Á 1 – 1 ˜ + Á 3 – 1˜ ˙ T2 ˚ P2 ¯ Ë V1 ¯ ˙˚ ÍÎ Ë Î Î T2 ˚ (b) Work done (WT) WT = Area of P–V diagram = (P2 – P1)(V2 – V1)

EXERCISES 4.1 Is the perpetual motion of first kind possible? If yes, show. If not, why?

(KUD I-89)

4.2 Explain the second law of thermodynamics with reference to Kelvin-Planck statement and Clausius statement and hence prove that both the statements are equivalent to each other although they appear to be different. (KUD I-89) 4.3 Give some typical examples of irreversible processes. State the method by which we determine whether a given process is reversible or irreversible. (KUD I-89) 4.4 State the two statements of the second law of thermodynamics and using these statements define efficiency of a heat engine and COP of a heat pump. (KUD II-93) 4.5 What are the limitations of the first law of thermodynamics?

(KUD Dec. 94)

4.6 Explain the concept of reversibility. What is the importance of this concept in thermodynamics? State the condition of thermodynamic reversibility. (KUD Dec. 94) 4.7 Define the term COP as applied to refrigerator and heat pump. Show that COPHP = 1 + COPref

(KUD Dec. 94)

4.8 Show that two reversible engines operating between two reservoirs at constant temperature have the same efficiency. (KUD Dec. 94) 4.9 State the two statements of the second law of thermodynamics and show that the violation of one statement violates the other. (KUD I-95) 4.10 Write a note on perpetual motion machine.

(KUD I-95)

4.11 Define and differentiate between reversible and irreversible process. Give example for both. (KUD I-95)

Second Law of Thermodynamics

4.12 Explain how you arrive at the thermodynamic temperature scale.

229

(KUD I-95)

4.13 Substantiate the statement “steam power plant can be classified as a direct heat engine whereas domestic refrigeration plant as a reversed heat engine”. (KUD I-96) 4.14 A reversible engine operates between heat reservoirs a, b and c. The engine receives equal quantities of heat from reservoirs a and b at temperatures Ta and Tb, respectively and rejects heat to reservoir c at temperature Tc. If the efficiency of the above mentioned engine is ‘a’ times the efficiency of another reversible engine operating between a and c only, prove that, Ta T = (2 - 1) + 2(1 - ) a (KUD I-96) Tb Tc 4.15 Is it impossible to construct perpetual motion machine of II kind?

(KUD I-90)

4.16 Mention which of the following processes can be reversible.

(KUD I-90)

1. 2. 3. 4.

Isothermal evaporation of water at constant pressure Throttling process Sensible heating of water from 0°C to 100°C from a constant temperature source Isentropic process.

4.17 What are the factors that render the process as irreversible?

(KUD II-90)

4.18 Briefly explain the Carnot cycle with TS and PH diagram and derive efficiency relation. (KUD II-90) 4.19 Explain briefly two propositions regarding efficiency of Carnot cycle.

(KUD II-90)

4.20 Define the following terms: 1. 2. 3. 4. 5.

Thermal reservoirs Heat engines Heat pump Refrigerator PMM-I and PMM-II

(KUD I-95)

4.21 Differentiate between 1. Heat engine and heat pump 2. Heat pump and refrigerator 3. Reversible and irreversible process

(KUD I-93)

4.22 Explain, with the help of neat sketch and example, reversible and irreversible processes. 4.23 Classify the reversible process and explain briefly with example and neat sketch. 4.24 Explain the concept of thermodynamic temperature scale. 4.25 What practical difficulties prevent a real heat engine from working on Carnot cycle? (KUD I-91)

230

Basic Thermodynamics

4.26 A reversible engine works between temperature limits of 300°C and 80°C. Which of the following cases is preferable? 1. Raising the source temperature to 400°C. 2. Lowering the sink temperature to 30°C.

(KUD I-95)

4.27 An inventor claims to have designed an engine which receives 10 kJ of heat and produces 2.5 kJ of useful work when operating between source at 60°C and sink at –16°C. Is his claim valid? Is his claim valid if the engine produces 1.5 kJ of work? (KUD I-95) 4.28 Calculate the thermal efficiency of a heat engine, working on the Carnot cycle between the temperature of 300°C and 30°C. To improve this thermal efficiency by 10%, what must be the new source temperature, when sink temperature is held at 30°C. (KUD I-91) 4.29 A reversible heat engine drives a reversible refrigerator. The engine operates between 800°C and 50°C. The refrigerator operates between 50°C and –10°C. 3000 kJ of heat is transferred to the engine and the net work output of the combined system is 400 kJ. (a) Calculate the heat transferred to the refrigerant and the net heat transferred to the reservoir at 50°C. (b) Reconsider (a) given that the efficiency of the heat engine and the COP of the refrigerator are each 45% of their maximum possible valves. 4.30 A Carnot heat engine operates between two reservoirs, one is at 600 K and the other one is at T3 K. The work out from the engine is used to drive a Carnot refrigerator, which operates between 300°K and T3 K. Calculate (a) The temperature T3 such that heat supplied to the engine Q1 is equal to the heat absorbed by refrigerator Q2. (b) The efficiency of Carnot engine and COP of Carnot refrigerator. [Ans: 400 K, 3] 4.31 Heat 300 kJ/s is supplied to a heat engine at constant temperature 290°C. The heat rejection takes place at 8.5°C. The following results were obtained. (a) 220 kJ/s are rejected. (b) 155 kJ/s are rejected (c) 80 kJ/s are rejected Classify the reversible, irreversible and impossible. [Ans: irreversible, irreversible, impossible] 4.32 A power plant cycle is shown in the figure. It operates between 164°C and 51°C. Heat interactions take place in boiler and condenser and turbine and pump are adiabatically covered. Calculate heat rejected by the condenser to the surroundings, if the cycle is reversible. [Ans: 534.7 kJ/kg] 4.33 Three Carnot engines are connected in series and the combination of these series engines operates between 727°C and 27°C. The work produced by these engines are in the ratio of 5:4:3. Calculate the intermediate temperature. [Ans: 435.5°C, 202°C]. 4.34 Prove that two reversible adiabatic curves plotted on any thermodynamic co-ordinates cannot intersect with each other.

Second Law of Thermodynamics

231

4.35 A house is maintained at a temperature of 30°C by using a heat pump. Heat, 190 MJ/hr is supplied to the house from the atmosphere which is at a temperature of –6°C. Calculate the power required to drive heat pump and COP of heat pump. [Ans: 6.283 kW, 8.4] 4.36 A heat engine drives a refrigerator whose COP is 4.5. 1260 kJ/hr heat is removed from the cold body. The engine efficiency is 35%. Calculate the rate of heat supplied per hour to the heat engine. [Ans: 800 kJ/hr] 4.37 An ice plant is working on a reversed Carnot cycle. The ice at 0°C is formed from the water at 0°C. Heat is rejected to the atmosphere at 27°C. A heat engine drives the refrigerator. The heat engine is maintained at 227°C by burning fuel of CV 20935 kJ/kg. The work developed by the heat engine is 7.7 kW. Calculate the fuel consumption per hour and tonnes of ice produced. [Ans: 3.3 kg/s, 20 tonnes of ice]

5 CHAPTER

Entropy

5.1 INTRODUCTION The second law of thermodynamics states that all spontaneous processes are irreversible and are accompanied by a degradation. It also states that it is impossible to transfer energy from a given state to a higher state of availability for any self-acting machine. In order to make these two statements quantitative, some property is required. During a spontaneous process this property always changes in a certain way, which will characterize such a change. Internal energy gives quantitative significance to the first law of thermodynamics. In a spontaneous process internal energy does not change in any characteristic way and it will not help in any way in the development of the second law. The property which will change in a particular way in a spontaneous process is called as ENTROPY, introduced by Clausius in 1851. Entropy means transformability (change) in Greek. In this chapter it will be proved that all the spontaneous processes result in an increase in entropy and no process is possible that results in a decrease in entropy.

5.2

ENTROPY AND HEAT

According to the first law of thermodynamics, there is an exact quantative equivalence among the different forms of energy. However, the availability of these forms of energy for useful work is not same. Heat energy is the least ‘available’ form of energy and transformation of other forms (higher grade energy) into heat represents a degradation of energy. Hence, entropy can be used to measure the unavailability or degradation of energy (internal energy), i.e. increase in unavailability or the total energy of a system is quantitatively expressed by an increase in its entropy. 232

Entropy

5.3

233

CARNOT THEOREM

The Carnot cycle comprising two reversible isothermal and two reversible isentropic processes and operating between two temperature limits, T1 = Tmax and T2 = Tmin, is shown in Figure 5.1. Q1 = Heat absorbed by the Carnot engine from the source at temperature T1 = Tmax Q2 = Heat rejected by the Carnot engine to the sink at temperature T2 = Tmin = Efficiency = =1-

T2 T1

Q1 - Q2 Q =1- 2 Q1 Q1

[Eq. (4.6)]

(5.1) (5.2)

Figure 5.1 Carnot cycle (a) TS diagram, (b) Symbolic representation of Carnot cycle.

From Eqs. (5.1) and (5.2), we can write Q2 T2 = Q1 T1 Q1 Q2 = T1 T2

\

Q1 Q2 =0 T1 T2

(5.3) (5.4)

We know that in a Carnot cycle, heat transfer occurs only during the two isothermal processes, then Eq. (5.4) can be written as,

Ú

Q =0 T

Equation (5.5) states that cyclic integral of the quantity dQ/T for a reversible cycle is zero.

(5.5)

234

5.4

Basic Thermodynamics

CLAUSIUS THEOREM

Consider a system operating on reversible cycle P Adiabatics A-B-C-D as shown in Figure 5.2. This cycle is divided into a large number of small Carnot cycles by means of a large number of reversible B adiabatics and reversible isothermals as shown in 7 6 Isotherms Figure 5.2. 2 3 C The original cycle A-B-C-D is thus replaced A by small Carnot cycles as 1-2-3-4, 5-6-7-8, 9-104 8 11-12 and so on. Now consider first two small 5 Carnot cycles 1-2-3-4 and 5-6-7-8. Adiabatic D expansion (3-4) of cycle 1-2-3-4 is being cancelled with adiabatic compression (5-6) of cycle 5-6-7-8. V Similarly, adiabatic expansion of one small cycle is cancelled with adiabatic expansion of the next Figure 5.2 Illustration of clasius theorem. small cycle. For the first small Carnot cycle 1-2-3-4, heat dQ1 is absorbed by the system reversibly at T1 and heat dQ2 is rejected by the system reversibly at T2. Then according to Eq. (5.3), Q1 Q2 = T1 T2

(5.6)

Q2 Q2 + =0 T1 T2

(5.7)

We know that heat supplied is taken as +ve and heat rejected as –ve. Similarly, we can write for the next small cycle 5-6-7-8 as

Then,

Q3 Q4 + = 0, etc. T3 T4 Q3 Q1 Q2 Q4 + + + + T1 T2 T3 T4

È Q˘ ˙ =0 ˚R

Ú ÍÎ T

(5.8)

+=0

(5.9) (5.10)

Equation (5.10) is valid only for reversible cycles. The cyclic integral of (dQ/T) for a reversible cycle is equal to zero. This is known as Clausius theorem. The quality of energy and quantity of irreversibility of a process are very important in the analysis of a thermal system. A property used to measure the quality of energy and irreversibility of the process is known as ‘Entropy’. Therefore, (dQ/T) of reversible process is a point function and it is a property of the system. The property is called ‘Entropy’.

235

Entropy

5.5

ENTROPY—PROPERTY OF A SYSTEM

Consider Figure 5.3, in which three reversible process paths are shown between state 1 and state 2 on a PV diagram. The PV diagram is chosen arbitrarily. The system changes from state 1 to state 2 along two paths a and b, and then returns from state 2 to state 1 along path c. All three paths a, b and c are of reversible processes. Now consider a reversible cycle 1-a-2-c-1, from Cluasius theorem, i.e. from Eq. (5.5),

Ú

Q =0= T

Ú

2È

1

Q˘ Í T ˙ + Î ˚a

2 a b

c

1 V

Q˘ Í ˙ 2 Î T ˚c

Ú

P

1È

(5.11)

Figure 5.3 Entropy as a property.

Similarly, consider a reversible cycle 1-b-2-c-1

Ú

Q =0= T

Ú

2È

Q˘ Í T ˙ + Î ˚b

1

1È

Q˘ ˙ ˚c

Ú ÍÎ T 2

(5.12)

Subtracting Eq. (5.12) form Eq. (5.11),

Ú

2È

1

Q˘ Í T ˙ = Î ˚a

Ú

2È

1

Q˘ Í T ˙ Î ˚b

(5.13)

Equation (5.13) says that the Ú(dQ)/T is the same for all reversible paths between states 1 and 2. Therefore, this quantity [Ú(dQ)/T] is independent of the path and is a function of the end states only. When the quantity is independent of the path, and is therefore a property, the property is called Entropy and is denoted by S. S1 is the entropy at state 1 and S2 is the entropy at state 2. Then,

S2 - S1 =

Ú

2È

1

Q˘ Í T ˙ =0 Î ˚R

È Q˘ ds = Í ˙ proved Î T ˚R

(5.14) (5.15)

where ds is an exact differential because S is a point function and a property. One important point to note here is that the change in entropy between states 1 and 2 must be the same whether the process is reversible or irreversible, in view of the fact that entropy is a property.

5.6 CLAUSIUS INEQUALITY Consider a cycle as shown in Figure 5.4. Q1 = Heat absorbed by the system either reversibly or irreversibly

236

Basic Thermodynamics

Q2 = Heat rejected by the system reversibly Now, the efficiency of a general cycle will be equal to or less than the efficiency of a Carnot cycle. For general cycle, Source T1 = Tmin Q1 - Q2 = (5.16) Reversible or Q1 Q1

Q1 = Reversible or irreversible Q2 = Reversible For reversible cycle, Q - Q2 = 1 Q1 Q1 = Reversible Q2 = Reversible T - T2 = 1 T1

HEc

(5.17)

irreversible heat engine W

Q2 Sink T2 = Tmin

Figure 5.4 Illustration of Clausius inequality.

Suffix IR = irreversible, R = reversible 1-

Q2 Q1IR or R Q2 Q1IR or R Q2 Q1IR or R

Q1IR or R From Eq. (5.17), we have

Q2

Q2 Q1R

£1-

Q2 Q1R

£≥

£

Q1R Q2 R

Q2 Q1R

Q1R

(5.18)

Q2 =

T1 T2

(5.19)

Substitute Eq. (5.19) in Eq. (5.18) then,

Q1IR or R Q2 Q1IR or R T1

£ £

T1 T2 Q2 T2

Equation (5.20) is for any process, reversible or irreversible. \ Substitute Eq. (5.15) in Eq. (5.20), hence for any process, Q1IR or R £ ds T1

(5.20)

Entropy

237

In differential form, Q £ ds T Since entropy is a property and the cyclic integral of any property is zero,

Ú

Ú

Q £0 T

Ú

(5.21)

(5.21a)

Or from Eq. (5.20),

Q1IR or R

-

T1

Q2 £0 T2

(5.22)

In differential form, Q £0 T Equations (5.21a) and (5.23) are known as inequality of Clausius

Ú

Q =0 T

Ú

(5.23)

(5.24a)

For reversible cycle

Q1R

i.e.

T1

Q2 =0 T2

-

(5.24b)

We know that dQ/T = ds, therefore,

Ú ds = DS

cycle

=0

and

Ú i.e.

Q < 0 For irreversible cycle T

Q1R T1

-

Q2 =0 T2

(5.24c) (5.24d)

and Q >0 (5.24e) T cycle is impossible, and violates the second law of thermodynamics. The inequality of Clausius is a result of the second law of thermodynamics, and will be proved to be valid for all the possible cycles. This includes both reversible and irreversible heat engines and refrigerators. Whenever a system undergoes a complete cycle, the cyclic integral of (dQ/T) around the cycle is less than or equal to zero. This is known as Clausius inequality.

Ú

238

5.7

Basic Thermodynamics

ENTROPY CHANGE OF AN IRREVERSIBLE PROCESS OF A CLOSED SYSTEM

The relationship between the entropy change and heat transfer interactions during irreversible processes can be explained with the following example. Consider a cycle composed of two processes; one, which is internally reversible, and the other one is irreversible as shown in Figure 5.5. Apply Eq. (5.24b) to a reversible cycle 1-A-2-B1, then,

Ú Process

Q = T

Ú

1-A-2 2-B-1 2-C-1 Cycle 1-A-2-B-1 Cycle 1-A-2-C-1

2

1A

Q + T

Ú

1

2B

P

A B 1

Q =0 T

2

C

(5.25) V

= Reversible = Reversible = Irreversible = Reversible cycle = Irreversible cycle

Figure 5.5

Entropy change of an irreversible process in a closed system.

Similarly for an irreversible cycle, 1-A-2-C-1, Eq. (5.24d)

Ú

Q = T

Ú

2

1A

Q + T

Ú

1

2C

Q

Ú

1

2C

Q T

In general, we can write ds ≥

Q T

(5.28a)

239

Entropy

Ú

2

Q T

¸ Ô Ô Q where possible, reversible process Ô ds = Ô T ˝ Q possible, irreversible process Ô ds > Ô T Ô Q Ô impossible ds < ˛ T Equations (5.21) and (5.28) are same. \

5.8

DS = S2 - S1 ≥

1

(5.28b)

PRINCIPLE OF INCREASE OF ENTROPY

Isolated system is composed of system and surroundings of any open or closed system. The entropy change of the isolated system is the net entropy change of the closed or open system plus the net entropy change of the surroundings. This entropy change is called the total entropy change, i.e. dstot = dsnet sys + dsnet sur = dsnet iso [sys = system, sur = surrounding, iso = isolated, tot = total] Consider a system and surroundings as shown in Figure 5.6. A quantity of heat dQ is transferred from the system at temperature T to the surroundings at temperature T0. The work done by the system is dw. Apply Eq. (5.28) to system Q dssys ≥ (5.30) T [dQ, –ve, heat transferred from the system] Similarly, Eq. (5.28) can be applied to the surroundings

dssur =

Q T0

(5.29)

System T

dW

dstot = dsnet iso ≥ -

T > T0

Surroundings T0

(5.31) Figure 5.6

[dQ, +ve, heat transferred to the surrounding] Substitute Eqs. (5.30) and (5.31) in Eq. (5.29),

dW

Principle of increase of entropy.

Q Q + T T0

È1 1˘ ≥ QÍ - ˙ Î T0 T ˚ È1 1˘ We know that T > T0, the quantity Í - ˙ is positive. Î T0 T ˚

(5.32)

240

Basic Thermodynamics

\

dstot ≥ 0

(5.33)

The net entropy change of the system and surroundings associated with any process must be greater than or equal to zero. dstot = 0 both the system and surroundings are internally reversible. dstot > 0 both the system and surroundings are irreversible or any one is irreversible

(5.34)

Equation (5.33) is called increase in entropy principle. Equation (5.33) says that net entropy change of the universe is +ve or equal to zero. This does not mean that entropy cannot decrease during a process, but the sum of the entropy change of the system and the entropy change of its surroundings cannot be negative.

5.9

THE COMBINED FIRST AND SECOND LAW

Consider a simple system that undergoes a reversible process. From the first law of thermodynamics, dQ = dW + dE

(5.35)

The total energy of the system E is internal energy only (du). In a reversible process, work done can be expressed as PdV work only. \ Equation (5.35) can be written as dQR = PdV + dU

(5.36)

From second law of thermodynamics and Eq. (5.21) or (5.28), Q = dS T dQR = TdS

\

TdS = PdV + dU

(5.37)

The first and second laws can also be combined in terms of enthalpy, and then, H = U + PV dH = dU + PdV + VdP

(5.38)

Substitute Eq. (5.37) in Eq. (5.38), dH = TdS + VdP TdS = dH – VdP

(5.39)

Equations (5.37) and (5.38) are very useful in the thermal analysis of pure substances. Example: Consider a system as shown in Figure 5.7. The system is a pure substance, mixture of liquid and vapour states. Heat is transferred externally. The heating process is carried very slowly, so that the process can be considered as a quasi-equilibrium or internally reversible process. The liquid part of the mixture vaporizes at constant temperature and constant pressure. Heat is supplied until all liquid is converted into vapour.

241

Entropy

From Eq. (5.21) or (5.28) 2 2 È Q˘ dS = Í ˙ 1 1 Î T ˚R

Ú

Frictionless piston

Ú

ÈQ ˘ S2 - S1 = Í 1- 2 ˙ Î T ˚R

Vapour

System boundary

Liquid

Insulation

Q I

The process carried out in the system is considered as reversible and heat transfer from the surroundings to the system is irreversible. From Eq. (5.39),

\

Figure 5.7

dS =

dh VdP T T

dS =

dH T

Illustration of the change in entropy for a reversible process.

[P = C, T = C]

On integrating between 1 and 2,

Ú

\

2

1

dS =

S2 - S1 = T

Ú

2

1

Ú

2

1

dH 1 = T T

Ú

2

1

dH

H 2 - H1 T

dS = H 2 - H1

Q1–2 = H2 – H1

(5.40)

i.e. entropy change of a pure substance (P = C, T = C) can be obtained directly from the enthalpy change of the process.

5.10 ENTROPY CHANGE FOR AN IDEAL GAS Entropy property is determined by the end states. The change in entropy will be independent of whether the process is carried out in an open or closed system. The entropy change for an ideal gas can be determined from the ideal gas Eqs. (5.37), (5.38), du = Cv dT and dH = Cp dT. \ From Eq. (5.37), TdS = dU + PdV and dU = Cv dT \

TdS = CvdT + PdV

dS =

Cv dT PdV + T T

dS =

Cv dT RdV + T V

È P R˘ Í∵ T = V ˙ Î ˚

(5.41)

242

Basic Thermodynamics

Similarly from Eq. (5.39), TdS = dH – VdP and dH = Cp dT \

TdS = Cp dT – VdP dS =

dS =

C p dT T

-

VdP T

Cv dT RdV T V

È V R˘ Í∵ T = P ˙ Î ˚

(5.42)

Equations (5.41) and (5.42) can be written in integral form, S2 - S1 = Cv ln

T2 V + R ln 2 T1 V1

T P S2 - S1 = C p ln 2 + R ln 2 T1 P1

For solids and liquids DV = 0, DP = 0 Equation (5.43) reduces to S2 - S1 = Cv ln

¸ Ô Ô ˝ Ô Ô ˛

(5.43)

T2 T = C p ln 2 T1 T1

(5.44)

For solids and incompressible liquids, Cp = Cv = C \ Equation (5.44) reduces to S2 - S1 = C ln

T2 T1

(5.45)

5.11 CHANGE OF ENTROPY FOR DIFFERENT PROCESSES 1. Constant volume process (Figure 5.8) Substitute V2 = V1 in Eq. (5.43) \

S2 - S1 = Cv ln P

T2 T1

(5.46)

T 2

2

1

1

V

Figure 5.8 Constant volume process.

S

Entropy

243

2. Constant pressure process (Figure 5.9) P

T

2 1

2

1

V

S

Figure 5.9 Constant pressure process.

Substitute P2 = P1 in Eq. (5.43) \

S2 - S1 = C p ln

T2 T1

(5.47)

3. Isothermal process (Figure 5.10) P

T 1 1

2

1 V

S

Figure 5.10 Constant temperature process.

Substitute T1 = T2 in Eq. (5.43) V2 P P = - R ln 2 = R ln 1 V1 P1 P2 4. Adiabatic or isentropic process (Figure 5.11)

\

S2 - S1 = R ln

P

(5.48)

T 2

2

1

1 V

Figure 5.11

S

Isentropic process.

244

Basic Thermodynamics

From Eq. (5.43), S2 - S1 = C p ln

We know that

T2 P - R ln 2 T1 P1

P1V1 P2V2 = T1 T2

\

T P2V2 = T P1V1

\

S2 - S1 = C p ln

P2V2 P P V P - R ln 2 = C p ln 2 + C p ln 2 - R ln 2 P1V1 P1 P1 V1 P1

= (C p - R) ln

P2 V P V + C p ln 2 = Cv ln 2 + C p ln 2 P1 V1 P1 V1

For reversible adiabatic process, Q = 0 Cv ln

5. Reversible polytropic process Equation (5.43) holds good.

(5.49)

\ ds = 0

P2 V + C p ln 2 = 0 P1 V1

6. Adiabatic mixing process When two substances at different temperatures are mixed together adiabatically, both will attain an intermediate temperature, T. The change in entropy of each is calculated as DS =

Ú

dQR = T

Ú

T

T1

Cp

dT T

where T1 denotes its initial temperature. The total entropy change is then obtained by adding the individual changes. This is used in processes like mixing of two fluid streams or quenching of metalic bodies in liquids. 7. Isothermal mixing of ideal gases The change in entropy resulting from the irreversible process of mixing of ideal gases in their pure state at temperature, T, and pressure, P, to form the mixture at the same temperature (T) and pressure (P) can be computed by the following equation DS = –R S xi ln xi where

xi = Mole fraction of the components in the mixture.

5.12 ISENTROPIC PROCESS FOR SOLID OR INCOMPRESSIBLE FLUID From Eq. (5.37), TdS = dU + PdV

Entropy

245

The specific volume of a solid and an incompressible fluid is constant. \

PdV = 0

\

TdS = dU

or

dS =

dU T

dU T dU = 0

Since dS = 0 for an isentropic process, 0 = \

Cv dT = 0

i.e. \

dT = 0

(5.50)

Thus, an isentropic process of a solid or an incompressible liquid is essentially an isothermal process also. From Eq. (5.39), TdS = dH – VdP For isentropic process, dS = 0 \ On integrating,

dH = VdP H2 – H1 = V(P2 – P1)

(5.51)

Equation (5.51) is particularly useful for work done calculation in the pumping of liquids.

5.13 ISENTROPIC WORK IN A STEADY FLOW SYSTEM SFEE equation can be written as V = velocity

Q = dH + VdV + gdZ + W For a reversible process dQ = TdS \

TdS = dH + VdV + gdZ + W

Substitute Eq. (5.38) in the above equation \

dH - VdP = dH + VdV + gdZ + W -VdP = VdV + gdZ + W

On integration, -

Ú

2

1

VdP =

DV 2 + gDZ + W1- 2 2

If P.E. and K.E. changes are neglected, then W1- 2 = -

Ú

2

1

VdP

(5.52)

246

Basic Thermodynamics

dQ = 0, ds = 0, apply in Eq. (5.39), 0 = dH – VdP or dH = VdP On integration, H 2 - H1 =

Ú

2

1

VdP or H1 - H 2 = -

Ú

2

1

VdP

(5.53)

5.14 AVAILABLE AND UNAVAILABLE ENERGY Figure 5.12(a) and (b) shows the symbolic representation of Carnot cycle and its TS diagram. T

Reservoir, Tmax

Tmax

QS

1

2

W1

RHE

Tmin

QR

3

4

Surroundings, Tmin (T0) 6 (a)

Figure 5.12

5

S

(b)

Carnot heat engine cycle: (a) Symbolic representation of Carnot cycle and (b) TS diagram.

QS amount of heat (area 1–2–5–6–1) at Tmax (reservoir temperature) is supplied to the reversible heat engine. QR (area 4–3–5–6–4) amount of heat at temperature Tmin (T0) is rejected by the reversible heat engine to the surroundings. T0 is the lowest possible temperature at which heat can be rejected. The maximum net work that can be obtained is Wc. This maximum amount of work output obtainable from an engine is called the available energy (availability). The amount of heat rejected QR is referred to as the unavailable energy (i.e. which is not available for doing work or unavailable part of the energy supplied). WnI = WI = QS – QR = The maximum amount of work (Ideal work done or reversible work done) = Available energy QR = Unavailable energy Consider an actual heat engine as shown in Figure 5.13(a) and (b), which operates between the same reservoir temperature (Tmax) and surroundings temperatures (Tmin = T0) as that of reversible heat engine. QS (area 1–2–5–6–1) amount of heat is supplied (same as that of reversible heat engine) to the actual heat engine by a reservoir. The work done by the actual heat engine is Wa, which is less than WI of that of Carnot engine.

Entropy T

Reservoir, Tmax

Tmax QS

247

1

2

Actual heat engine Wa

HE

Tmin

QR

4

3

7

Surroundings, Tmin (T0) 6

5

(a)

Figure 5.13

(b)

8

S

DSnet

Actual heat engine: (a) Symbolic representation of actual heat engine and (b) TS diagram.

The heat rejected by the actual heat engine is QR¢ (area 4–7–8–6–4) which will be greater than the heat rejected by the reversible heat engine. Wna = Wa = QS – QR¢

\

The unavailable energy in case of the actual engine is QR¢ (area 4–7–8–6–4) which is greater than that of the reversible engine. The irreversibility (I) for the actual engine is defined as the difference between the reversible work and actual work. \

I = WI – Wa WI = QS – QR = Area (1–2–5–6–1) – Area (4–3–5–6–4) Wa = QS – QR¢ = Area (1–2–5–6–1) – Area (4–7–8–6–4)

\

I = WI – Wa = (Area 3–7–8–5–3) = T0 DSnet Energy supplied = Available energy + Unavailable energy QS = W.D. + QR

SOLVED EXAMPLES EXAMPLE 5.1 A perfect gas is heated from 60°C to 300°C at a constant pressure of 4 bar. The gas is then cooled to 60°C at constant volume. The mass of the gas is 5 kg. Calculate change in entropy. Take Cp = 1.0 kJ/kg, Cv = 0.72 kJ/kg K. Solution: D1S3) (a) The change of entropy (D D1S2 = change of entropy between 1 and 2 S2 - S1 = mC p ln

T2 573 kJ = 5 ¥ 1 ¥ ln = 2.714 333 K T1

D2S3 = change of entropy between 2 and 3

(+ve increase in entropy)

248

Basic Thermodynamics P

T

2

1

2 P=C V=C 1

3

3 V

S

Figure E5.1

S3 - S2 = mCv ln

T3 333 kJ = 5 ¥ 0.75 ¥ ln = - 1.954 573 K T2

(–ve decrease in entropy)

D1S3 = DSn = S3 – S1= D1S2 + D2S3 = 2.714–1.954 = +0.76 (+ve increase of entropy) EXAMPLE 5.2 2 kg of air at 10 bar absolute pressure and 600°C temperature expands isothermally to 5 times its original volume. Calculate the following: (a) The original volume (b) The final pressure (c) The change in entropy. Solution: (a) The original volume (V1) V1 =

mRT1 2 kg × 0.287 kN m × 573 K × m 3 = = 0.5011 m 3 P1 10 × 100 kN kg K

(b) The final pressure (P2) P2 =

P1V1 10 ¥ 0.5011 = = 2 bar V2 5 ¥ 0.5011

DS) (c) The change in entropy (D DS = mR ln

V2 kJ 5 kJ = 2 kg ¥ 0.287 ln = 0.9238 V1 kgK 1 K

EXAMPLE 5.3 A vessel of capacity 4 m3 contains air at pressure 2 bar and temperature 30°C. Then, additional air is pumped into the system until the pressure rises to 30 bar and the temperature rises to 70°C. Calculate the following: (a) The mass of air pumped in (b) Equivalent volume of air pumped in expressed at 1 bar and 30°C If the vessel is now allowed to cool until the temperature is again 30°C, calculate the following:

Entropy

249

(c) The pressure in the vessel (d) Quantity of heat transferred (e) The change of entropy of the gas during the cooling process Solution: (a) The mass of air pumped in (m) Assume rigid vessel, m1 =

P1V1 1 ¥ 100 ¥ 4 = = 9.199 kg RT1 0.287 ¥ 303

m1 = mass of air initially, i.e. before pumping m2 = mass of air after pumping additional air, i.e. final mass of air m2 =

\

P2V2 30 ¥ 100 ¥ 4 = = 121.9 kg RT2 0.287 ¥ 343

m = mass of air pumped in = m2 – m1 = 121.9 – 9.199 = 112.7 kg

(b) Equivalent volume (V) of air pumped in expressed at (P) 1 bar and (T) 30°C mRT 112.7 ¥ 0.287 ¥ 303 = = 98 m 3 P 100 (c) The pressure in the vessel (P3) If the vessel is cooled, pressure drops, temperature drops, but volume remains constant. V=

\

P3 = ?

T3 = 30°C V3 = 4 m3 P3 =

m3 = m2 = 121.9 kg

mRT3 121.9 ¥ 0.287 ¥ 303 = = 2650.14 kPa V3 4

Considering cooling process as constant volume, P3 =

P2V2 303 K = 30 bar ¥ = 26.5 bar or 2650 kPa V3 343 K

(b) Quantity of heat transferred (Q2–3) Q2–3 = m Cv(T3 – T2) = 121.9 × 0.714 × (70–30) = –3481.46 kJ (–ve, heat transferred to the surrounding) D2S3) (e) The change of entropy of the gas during the cooling process (D D 2 S3 = mCv ln

T2 303 = 121.9 ¥ 0.714 ¥ ln = - 10.79 kJ 343 T3

(–ve decrease in entropy)

EXAMPLE 5.4 Steam at 100°C is condensed at a constant pressure from saturated vapour to saturated liquid. During the process, the heat of condensation is rejected to the surroundings at 30°C. Calculate the entropy change of the universe per kg of steam condensed during the process.

250

Basic Thermodynamics

Solution:

DSu) Entropy change of the universe (D q1–2 = heat rejected during condensation = (h1 – h2) = (2675.7 – 419.06) = 2256.64 kJ/kg S2 – S1 = change of entropy during condensation process (system) DSsys= 1.3069 – 7.3545 = – 6.0476 kJ/kg (decrease)

From steam tables at 100°C (saturation) h1 = hg = 2675.7, h2 = hf = 419.06 kJ/kg s1 = sg = 7.3545, s2 = sf = 1.3069 kJ/kg K Ta = ambient temperature = 303 K Heat, q1–2, is added to the surrounding DSsur = change of entropy of the surrounding q1- 2 2256.64 kJ (+ve increase) = = 7.4477 303 kg K Ta

\

DSu = DSn = net change in entropy = DSsys + DSsur = (S2 – S1) + DSsur = – 6.0476 + 7.4477 = +1.4 kJ/kg K (increase of entropy)

Figure E5.4

EXAMPLE 5.5 An inventor claims that his newly designed engine develops 400 kW of work output. Water is supplied to the engine at 95°C, 1 bar and leaves the engine at 35°C and 1 bar. The engine continuously rejects heat to the surroundings at 1 bar and 25°C. Calculate (a) the heat rejection rate and (b) possibility of the claim. Assume water flow rate at 10 kg/s. Solution: (a) The heat rejection (QR) Apply SFEE, neglect DK.E., DP.E. First law analysis

251

Entropy

\ \

h1 = W + h2 + QR QR = (h1– h2) – W QR = 10(398 – 147) – 400 = 2110 kJ/s (b) Possibility of claim Second law analysis

95°C 1 bar water

New Engine 1

W QR

È Q ˘ Sgen C-V = m ¥ [ s2 - s1 ] - Í - R ˙ Î Ta ˚ = decrease in entropy of water between inlet and outlet + increase in entropy of the surroundings

= 10

2

35°C 1 bar water

Ta = 25°C Surroundings

Figure E5.5

kg kJ 2110 kJ ¥ [0.505 - 1.25] + = - 0.3695 s kg 298 K

–ve, entropy of the control volume decreases. This is impossible, hence the claim is not valid. [From steam tables at 95°C, h1 = hf1 = 398 kJ/kg, S1 = Sf1 = 1.25 kJ/kg K, at 35°C, h2 = hf2 = 147 kJ/kg, S2 = Sf2 = 0.505 kJ/kg K] EXAMPLE 5.6 Modify Example 5.5 in such a way that the new engine develops 200 kW of work output. Remaining conditions are same. Calculate the following: (a) Amount of heat rejected (b) Possibility of claim Solution:

95°C 1 bar water

New Engine 1

2

35°C 1 bar water

(a) The heat rejection (QR) Apply SFEE, neglect DK.E., DP.E. From the first law analysis, mh1 = W + mh2 + QR \ QR = m(h1 – h2) – W kg ¥ [398 - 147] - 200 = 2310 kJ s (b) Possibility of claim From the second law analysis, = 10

W = 200 kW QR

Ta = 25°C Surroundings

Figure E5.6

È Q ˘ Sgen C-V = m ¥ [ s2 - s1 ] - Í - R ˙ Î Ta ˚ kg kJ 2310 kJ = 10 ¥ [0.505 - 1.25] + = - 0.3016 s kg 298 kg

+ve, increase in entropy of the control volume, hence claim is possible.

252

Basic Thermodynamics

EXAMPLE 5.7 A device is supplied with 3 kg/s of air at 400 kPa and 300 K. Two separate streams of air leave the device. Stream 2 leaves at 100 kPa and 330 K and stream 3 leaves at 100 kPa and 270 K. Both streams are having the same mass flow rate. The atmospheric temperature is at 300 K. Prove the validity of the claim. Solution:

1

Stream 2, 100 kPa, 330 k m2

Stream 1, 400 kPa, 300 k, 3 kg/s m1 Stream 3, 100 kPa, 270 k m3 3

2

Figure E5.7

This is an open flow system W = 0, Q = 0 Entropy generation of the control volume (Sgen C-V)

Sgen C-V =

Âm s - Âm s e e

exit

i i

inlet

-

Q Ta

= m2s2 + m3s3 – m1s1 [∵ m1= m2 + m3] = m2s2 + m3s3 – (m2 + m3)s1 = m2s2 + m3s3 – m2s1 – m3s1 = m2[s2 – s1] + m3[s3 – s1] Now calculate change of entropy between 1 and 2 and 1 and 3.

\

S2 - S1 = C p ln

T2 P 330 100 kJ - R ln 2 = 1005 ¥ ln - 0.287 ¥ ln = 0.494 300 400 kgK T1 P1

S3 - S1 = C p ln

T3 P 270 100 kJ - R ln 3 = 1005 ¥ ln - 0.287 ¥ ln = 0.292 300 400 kgK T1 P1

Sgen C-V = (0.494 + 0.292)

kJ kg kJ kW ¥3 = 2.358 or kgK s sK K

EXAMPLE 5.8 A gas is contained in a piston and cylinder arrangement at 500 kPa, and 400 K. The gas is heated isothermally and expands until its pressure is reduced to 100 kPa. During the process the work done by the gas is 0.6 kJ. (a) Show the process is internally reversible or irreversible. (b) Calculate the entropy change.

Entropy

253

Solution: P 1

W = 0.6 kJ

Q 3

Gas 750 cm 500 kPa Æ 100 kPa 400 K Æ 400 K

2

V

Figure E5.8

(a) To show the process is internally reversible or irreversible W1- 2 int rev = P1V1 ln

V2 P 500 = P1V1 ln 1 = 500 ¥ 750 ¥ 10 -6 ¥ ln V1 P2 100

= 0.604 kJ (Work done by the system) Wa = Actual work done by the system = 0.6 kJ (given data) Wa < W1–2 int rev \

The actual process is irreversible.

(b) The change of entropy during the process (internal reversible) (S2 – S1) S2 - S1 =

Ú

2È

Q˘ 1 = Í T ˙ T Î ˚ int rev

1

Ú

2

1

[ Q]int rev

[∵ T = C]

(1)

[∵ Du = 0]

(2)

From the first law of thermodynamics Q1–2 int rev = W1–2 int rev + Du Q1–2 int rev = W1–2 int rev = 0.604 kJ \

Substitute Eq. (2) in Eq. (1), S2 - S1 =

Q1- 2 int rev T

=

0.604 kJ = 0.00151 K 400

(c) Change of entropy (actual) Q1–2a = W1–2a = 0.6 kJ

(given data)

We know that

Ú

2È

1

Q1- 2a 0.6 kJ Q˘ Í T ˙ = T = 400 = 0.0015 K Î ˚a

S2 - S1 >

Ú

2È

1

Q˘ Í T ˙ Î ˚a

This is possible and in accordance with the second law of thermodynamics.

254

Basic Thermodynamics

EXAMPLE 5.9 An iron rod of mass of 2 kg is initially at 800 K. It is allowed to cool by transferring heat to the surrounding at 300 K. Determine the change in entropy of the iron and the total change in entropy of the iron and surrounding air after the iron reaches thermal equilibrium. Cp of iron is 0.45 kJ/kg K. Solution:

Apply the first law of thermodynamics dq = dw + Du,

dw = 0

dq = Du = mCv dT = m C dT]iron

[For solids, liquids, Cp = Cv = C]

= 2 × 0.45 × (300 – 800) dq1–2 iron = – 450 kJ Change in entropy of iron (DSiron)

DSiron = (S2 - S1 )iron = mC ln

T2 ˘ kJ 300 = - 0.883 ˙ = 2 ¥ 0.45 ln K T1 ˚ w 800

Change of entropy of surrounding (DSs) D Ss = (S2 - S1 ) s =

Ú

Q1- 2 air Q1- 2 iron Q˘ = = Í T ˙ Tair Tair Î ˚ int rev

2È

1

450 kJ = 1.5 300 K

(S2 – S1)t = Total entropy change = (S2 – S1)iron + (S2 – S1)s = – 0.883 + 1.5 = 0.617

kJ K

The change in total entropy is positive in accordance with the increase-in-entropy principle, indicating that the process is irreversible. The irreversibility is a result of heat transfer through a finite temperature difference. EXAMPLE 5.10 Using the increase-in-entropy principle, show that the direction of heat transfer must be from a higher temperature body to a body at a lower temperature. Solution:

Total entropy change = entropy change of A + entropy change of B dstot =

QA QB Q˘ Q˘ = = + ˙ ˙ T ˚ A int rev T ˚ B int rev TA TB

Boundary Body A at TA

dQ

Body B at TB

Figure E5.10

Entropy

We know

255

(dQ)A = – dQ (dQ)B = dQ

È1 Q Q Q Q 1 ˘ + = = QÍ ˙ TA TB TB TA Î TB TA ˚ We know that principle of increase of entropy ≥ 0, in order to satisfy this condition, TA > TB, i.e. temperature of body A > temperature of body B and heat always transfers from higher temperature to lower temperature. dstot = -

\

EXAMPLE 5.11 Modify Example 5.8 in such a way that the actual work developed by the system during the process is 0.7 kJ. Remaining data are same. Show that the process is internally reversible. Solution:

To show that actual process is internally reversible W1–2 int rev = 0.604 kJ

Here, \

(from Example 5.8)

Wa = 0.7 kJ

(given in 5.11)

Wa > W1-2 int rev ,

(it is impossible)

Wa cannot be more than 0.604 kJ. This is against the second law of thermodynamics. Q1–2a = W1–2a = 0.7 kJ

Ú

2È

1

Q1- 2a 0.7 Q˘ kJ Í T ˙ = T = 400 = 0.00175 K Î ˚a

S2 - S1 = 0.0015

Here,

S2 - S1
Wa (6.40) \ Irreversibility (I) for a given process is given by I = Wrev – Wa (6.41) The reversible work is given by Eq. (6.8), i.e. È ˘ È ˘ È ˘ V2 V2 V2 Wrev = m1 Í h1 - T0 S1 + 1 + Z1 ˙ - m2 Í h2 - T0 S2 + 2 + Z 2 ˙ - d Íu + m + mZ - T0 S ˙ 2 2 2 ÍÎ ˙˚ S ÍÎ ˙˚ ÍÎ ˙˚ Wa = Actual work È ˘ È ˘ È ˘ V2 V2 V2 Wa = m1 Í h1 + 1 + Z1 ˙ - m2 Í h2 + 2 + Z 2 ˙ - d Íu + m + mZ ˙ + Q 2 2 2 ÎÍ ˚˙ S ÎÍ ˚˙ ÎÍ ˚˙

(6.41a)

290

Basic Thermodynamics

Substitute Wrev and Wa in Eq. (6.41) I = m1[–T0S1] – m2[–T0S2] – d[–T0S]S – Q

(6.42)

Case I: Steady flow process Assumptions: m1 = m2 = m

d(–T0S)S = 0

Apply the above assumptions to Eq. (6.42) Isteady flow = m(–T0S1) – m(–T0S2) – Q = m(T0S2 – T0S1) – Q = mT0(S2 – S1) – Q

(6.43)

Isteady flow = T0(S2 – S1) – Q

(6.44)

For per unit mass Case II: Non-flow process (closed system) Assumption: m1 = m2 = 0 Apply the above assumption to Eq. (6.42) \

Iclosed system = –d (–T0S)S – Q = d (T0 S)S – Q

\

6.6

Iclosed system = T0 (S2 – S1) – Q

(6.45)

AVAILABILITY FOR IDEAL GAS SYSTEMS

We can write relations for ideal gas as (u1 – u0) = m Cv(T1 – T0)

(6.46)

(V1 – V0) = m(v1 – v0) È RT RT ˘ = mÍ 1 - 0 ˙ P0 ˚ Î P1

\

(V1 - V0 ) =

mRT0 P0

È T1 P0 ˘ - 1˙ Í Î T0 P1 ˚

ÈT P ˘ P0 (V1 - V0 ) = mRT0 Í 1 0 - 1˙ Î T0 P1 ˚

∵

P1V1 P0V0 ; = T1 T0

V1 T1 P0 = V0 T0 P1

ÈV ˘ P0 (V1 - V0 ) = mRT0 Í 1 - 1˙ Î V0 ˚

˘ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙˙ ˚

(6.47)

Availability and Irreversibility

291

We know that entropy change for ideal gas,

È T V ˘ ˘ S1 - S0 = m ÍCv ln 1 + R ln 1 ˙ ˙ T0 V0 ˚ ˙ Î ˙ È T1 P0 ˘ ˙ T1 = m ÍCv ln + R ln ˙ T0 T0 P1 ˚ ˙ Î ˙ È T0 P1 ˘ ˙ T1 = m ÍCv ln - R ln ˙˙ T0 T1 P0 ˚ ˙ Î ˙ È T1 P1 ˘ ˙ = m ÍC p ln - R ln ˙ ˙ T0 P0 ˚ ˙˚ Î

(6.48)

f = (u1 – u0) + P0(V1 – V0) – T0(S1 – S0) [Eq. (6.26)] Substitute Eqs. (6.46), (6.47) and (6.48) in Eq. (6.26) ideal gas

ÈV ˘ È TP˘ T = mCv (T1 - T0 ) + mRT0 Í 1 - 1˙ - mT0 ÍCv ln 1 - R ln 0 1 ˙ T0 T1 P0 ˚ Î V0 ˚ Î = mCv T1 - mCvT0 + mRT0

TP V1 T - mRT0 - mT0 Cv ln 1 + mT0 R ln 0 1 V0 T0 T1 P0

ÈT ÈV T ˘ V ˘ = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 1 - 1 - ln 1 ˙ T0 ˚ V0 ˚ Î T0 Î V0 ÈT P ˘ T ˘ ÈP = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 0 - 1 - ln 0 ˙ T0 ˚ P˚ ÎP Î T0

È mRT0 ˘ Í We know that P0V0 = mRT0 ; \ V0 = ˙ P0 ˙ Í Í ˙ P P V V Í\ 1 = 1 = 0 = 0 ˙ P Í V0 mRT0 mRT0 ˙ Í ˙ mRT mRT T P 0 0 Í\ P = ˙ P1 = 0 1 = ÍÎ ˙˚ V1 mRT1 T1 Availability Eq. (6.49) for ideal gas system contains two terms. (a) Thermal availability term fideal gas (T) (b) Pressure availability term fideal gas (P) ideal gas system (T )

ÈT T ˘ = mCvT0 Í 1 - 1 - ln 1 ˙ T0 ˚ Î T0

ideal gas system (P )

P ˘ ÈP = mRT0 Í 0 - 1 - ln 0 ˙ P˚ ÎP

(6.49)

292

Basic Thermodynamics

If T1 = T0, the thermal term vanishes and if P¢ = P0, the pressure term vanishes Availability of ideal gas system based on flow system \

6.7

ideal gas system (flow)

ÈT Ê P ÈT ˆ P ˘ T ˘ = mC pT0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 1 Á 0 - 1˜ - ln 0 ˙ T0 ˚ P1 ˙˚ ¯ ÍÎ T0 Ë P1 Î T0

(6.50)

AVAILABILITY FOR SUBSTANCES IN THE SOLID AND LIQUID PHASES

Generally, solids and liquids can be regarded as incompressible, i.e. change in volume is negligible, i.e. (V1 – V0) ª 0. We know that For solids and liquids, \ \

Tds = du + Pdv Cp ª C v ª C du = CdT T ds ª du ds ª

du T dT T

ds = C

S1 - S2 =

Ú

2

1

mC

T dT = mC ln 1 T T2

Initial state is represented by 1. Final state is nothing but reference environment represented by 0. T S1 - S0 = mC ln 1 (6.51) \ T0 u1 – u0 = mC(T1 – T0)

(6.52)

Equation (6.26) is f = (V1 – V0) + P0(V1 – V0) – T0(S1 – S0) \ Substitute Eqs. (6.51) and (6.52) in Eq. (6.26) Availability of substances of solids and liquids solid, liq. system

= mC (T1 - T0 ) + 0 - T0 mC ln

T1 T0

ÈT T ˘ = mCT0 Í 1 - 1 - ln 1 ˙ T0 ˚ Î T0

6.8

(6.53)

AVAILABILITY ASSOCIATED WITH HEAT FLOW FROM SOURCE

Heat (Q) is transferred from the surroundings at a uniform temperature T1 to a system through the boundary as shown in Figure 6.3(a).

293

Availability and Irreversibility

This system can be re-written as shown in Figure 6.3(b). T

Heat Q(T1)

Con

T2

T1, Source or surrounding

Con

Q T1 RHE

We T0

Q0

System

Available energy

1

Unavailable

Reference environment Heat Q0(T0) (a)

Figure 6.3

S1

P0, T0

S2

ds

(b)

S

(c)

Availability associated with heat flow from source: (a) Schematic representation, (b) Symbolic representation and (c) TS diagram.

Q quantity of heat from source at T1 temperature is transferred to a reversible heat engine at T1 temperature. Q0 amount of heat is reflected to the sink at a temperature T0, i.e. reference environment (dead state). \

Availability associated with heat flow (Q) = maximum gross work

È T ˘ (6.54) Availability of a system with heat flow = Í1 - 0 ˙ Q T1 ˚ Î If heat is transferred from a higher temperature level to a lower temperature level, i.e. T1 to T0, then availability of the source decreases. If heat is transferred from a lower temperature level to a higher temperature level, then availability of that medium is increasing. Availability destruction in non-flow system (closed system) due to irreversibility.

Availability destruction = Irrversibility = T0dSuniverse – Q I = T0 [dSsystem + dSsurrounding] – Q

(6.55)

Infinitesimal (dQS) amount of heat is added to the system over the temperature range of T1 and T2. If T is the temperature at which heat addition takes place and T0 is the dead state temperature at which heat rejection takes place, the maximum work possible is given by,

È T ˘ È T ˘ Wmax = QS Í1 - 0 ˙ = ( AE ) = QS Í1 - 0 ˙ T˚ T˚ Î Î Integrating between 1and 2

Ú

2

1

È T ˘ QS Í1 - 0 ˙ = 1 T˚ Î 1AE2 = 1Q2 – T0(S2 – S1) ( AE ) =

Ú

2

Ú

2

1

QS - T0

Ú

2

1

Q T

294

Basic Thermodynamics

UE = Unavailable energy = QS – AE 1UE2 = 1Q2

6.9

– 1AE2 = T0 (S2 – S1)

SECOND LAW EFFICIENCY OR EFFECTIVENESS Effectiveness ˘ Increase in availability of the system ˙= (heating or expansion) ˚ Loss of availability of the surrounding Effectiveness ˘ Increase in availability of the surrounding ˙= (cooling or compression) ˚ Loss of availability of the system

SOLVED EXAMPLES EXAMPLE 6.1 Determine the availability of the following closed systems. Reference environment conditions are 27°C, 100 kPa. (a) 4 kg of water at P1 = 100 kPa, 80°C (b) 2 kg of ice at P1 = 100 kPa, –15°C (c) 0.5 kg of steam at P1 = 5 MPa, 500°C (d) 0.5 kg of wet steam at P1 = 10 kPa with quality 0.9 Solution: Given data

T0 = 27 + 273 = 300 K

P0 = 100 kPa

Availability of closed system is given by Eq. (6.26) f = (U1 + P0V1 – T0S1) – (U0 + P0V0 – T0S0) From steam tables at 27°C, properties correspond to the reference environment.

\

V0 = 113.1 kJ/kg

h0 = 113.2 kJ/kg

S0 = 0.396 kJ/kg K

u0 = 0.001004 m3/kg

(U0 + P0V0 – T0S0) = [113.1 + 100(0.001004) – 300 × 0.396] = –5.6 kJ/kg

(a) 4 kg of water at P1 = 100 kPa, 80°C From steam tables, at 80°C, V1 = 335 kJ/kg v0 = 1.029 ×

10–3

h1 = 335 kJ/kg m3/kg

S0 = 1.075 kJ/ kg K

\

(U1 + P0V1 – T0S1) = (h1 – T0S1) = 335 – 300(1.075) = 12.5 kJ/kg

\

f = 12.5 – (–5.6) = 18.1 kJ/kg

Total f = 4 × 18.1= 72.4 kJ (b) 2 kg of ice at P1 = 100 kPa, –15°C From properties of ice and water tables at –15°C h1 = –364 kJ/kg

S1 = –1.336 kJ/ kg K

Availability and Irreversibility

295

y = Availability of the system (closed) = (h1 – h0) – T0(S1 – S0) = (–364 – 113.2) – 300(–1.336 – 0.395) = 42.1 kJ/kg Total y = 2 kg × 42.1 kJ = 84.2 kJ Even though the temperature of the system is less than zero, the availability of the system is greater than zero and work could be obtained from the system. (c) 0.5 kg of steam at P1 = 5 MPa, 500° C From steam tables at 50 bar and 500°C

\

h1 = 3433.8 kJ/kg

u1 = 3091.0 kJ/kg

v1 = 0.0686 m3/kg

S1 = 6.976 kJ/ kg K

(u1 + P0V1 – T0S1) = [3091 + 100(0.0686) – 300 × 6.976] = 1005.06 kJ/kg

\

f = 0.5[1005.06 – (–5.6)] = 505.33 kJ

(d) From steam tables, P1 = 10 kPa, with x = 0.9 \

\

hf = 191.8 kJ/kg

hfg = 2392.8 kJ/kg

Vf = 192 m3/kg

Sf = 0.649 kJ/kg K

Sfg = 7.501 kJ/kg K

Vg = 2437 m3/kg

vf = 0.001010 m3/kg

Vfg = (14.67 – 0.001010) m3/kg

h1 = hf + x(hfg) = 191.8 + 0.9(2392.8) = 2345.32 kJ/kg u1 = uf + x(ufg) = 192 + 0.9(2437 – 192) = 2212.5 kJ/kg S1 = Sf + x(Sfg) = 0.649 + 0.9(7.501) = 7.3999 kJ/kg K

\

(u1 + P0V1 – T0S1) = (h1 – T0S1) = 2345.32 – 300(7.3999) = 125.35 kJ/kg

\

f = 125.35 – (–5.6) = 130.95 kJ/kg

Total f = 0.5 × 130.95 = 65.475 kJ EXAMPLE 6.2 Determine the exergy (availability) of 1 m3 of complete vacuum. Dead state conditions are 27°C and 100 kPa. Solution:

Exergy (availability) of 1 m3 of complete vacuum

For vacuum: v1 = 0, H1 = 0, S1 = 0, P1 = 0, V1 = 1 m3 Properties corresponding to dead state, i.e. 27°C and 100 kPa u0 = 0, h0 = 0, S0 = 0, V0 = 0, P0 = 100 kPa Availability of a closed system (f), from Eq. (6.26) f = (u1 – u0) + P0(u1 – u0) – T0(S1 – S0) = 0 + 100(1 – 0) – 300(0) = 100 kJ This shows that vacuum has availability, i.e. work capability. EXAMPLE 6.3 Calculate the availability of air in the following cases. Dead state conditions are 27°C and 100 kPa. R = 0.287 kJ/kg K g = 1.4 Cp = 1.005 kJ/kg K

296

Basic Thermodynamics

(a) (b) (c) (d) (e) (f)

0.1 kg of air at P1 = 10 bar and 300 K (T1 = T0) 0.5 kg of air at P1 = 1 bar (P1 = P0) and 500 K 0.5 kg of air at P1 = 1 bar (P1 = P0) and –125°C 0.5 kg of air at P1 = 0.5 bar and 15°C 0.5 kg of air at P1 = P0, T1 = T0 1 kg of air at P1 = 10 bars, 867°C

Solution: (a) 0.1 kg of air at P1 = 10 bar, 300 K (T1 = T0) By using Eq. (6.49) f = Availability of ideal gas system (closed) ÈT P ˘ T ˘ ÈP = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 0 - 1 - ln 0 ˙ T0 ˚ P¢ ˚ Î P¢ Î T0 È ˘ PT 10 ¥ 100 ¥ 300 Here T1 = T0, hence thermal term = 0 Í P ¢ = 1 0 = 1000 kPa ˙ T 300 1 Î ˚

\

100 ˘ È 100 = 0.1 ¥ 0.287 ¥ 300 Í - 1 - ln ˙ 1000 1000 Î ˚

Or by using Eq. (6.26) or substituting Eqs. (6.46), (6.47) and (6.48) in Eq. (6.26) T1 = T0 , f = (u1 – u0) + P0(V1 – V0) – T0(S1 – S0) = (u1 + P0V1 – T0S1) – (u0 + P0V0 – T0S0) È RT È RT ˘ TP˘ T = Cv (T1 - T0 ) + T0 Í 1 - 0 ˙ - T0 ÍC p ln 1 - R ln 0 1 ˙ P0 ˚ T0 T1 P0 ˚ Î P1 Î 1000 ˘ È 300 300 ˘ È = 0 + 100 ¥ 0.287 Í - 300 Í 0 - 0.287 ln = 120.76 kJ/kg ˙ 100 ˙˚ Î 1000 100 ˚ Î

Total f = 0.1 kg × 120.76 = 12.076 kJ (b) 0.5 kg of air at P1 = 1 bar (P1 = P0) and 500 K By using Eq. (6.49) f = Availability of an ideal gas system (closed system) ÈT P ˘ T ˘ ÈP = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 0 - 1 - ln 0 ˙ T T P P¢ ˚ ¢ Î 0˚ Î 0 È ˘ P1T0 100 ¥ 300 = = 60 kPa ˙ ÍP¢ = T1 500 Î ˚ 500 ˘ 100 ˘ È 500 È 100 = 0.5 ¥ 0.718 ¥ 300 Í - 1 - ln + 0.5 ¥ 0.287 ¥ 300 Í - 1 - ln ˙ 300 ˚ 60 ˙˚ Î 300 Î 60

= 16.784 + 6.71 = 23.49 kJ

Availability and Irreversibility

Or

297

f = (u1 – u0) + P0(u1 – u0) – T0(S1 – S0) È RT È RT ˘ TP˘ T = Cv (T1 - T0 ) + T0 Í 1 - 0 ˙ - T0 ÍC p ln 1 - R ln 0 1 ˙ P0 ˚ T0 T1 P0 ˚ Î P1 Î 500 100 ˘ È 500 300 ˘ È - 300 Í1.005 ln - 0.287 ln = 0.718(500 – 300) + 100 × 0.287 Í ˙ 300 100 ˙˚ Î 100 100 ˚ Î

= 46.986 kJ/kg Total f = 0.5 × 46.986 = 23.49 kJ (c) 0.5 kg of air at P1 = 1 bar, and –125°C ÈT P ˘ T ˘ ÈP = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 0 - 1 - ln 0 ˙ T T P P¢ ˚ ¢ Î 0˚ Î 0 È ˘ P1T0 100 ¥ 300 = = 202.7 kPa ˙ ÍP¢ = T1 148 Î ˚

148 ˘ 100 ˘ È 148 È 100 = 0.5 ¥ 0.718 ¥ 300 Í - 1 - ln ˙ + 0.5 ¥ 0.287 ¥ 300 Í 202.7 - 1 - ln 202.7 ˙ 300 300 Î ˚ Î ˚ = 21.53 + 8.61 = 30.14 kJ

(d) 0.5 kg of air at P1 = 0.5 bar, 15°C È ˘ P1T0 100 ¥ 300 = = 104.71 kPa ˙ ÍP¢ = T1 (273 + 15) Î ˚ Substitute in Eq. (6.49)

288 ˘ 100 ˘ È 288 È 100 = 0.5 ¥ 0.718 ¥ 300 Í - 1 - ln + 0.5 ¥ 0.287 ¥ 300 Í - 1 - ln ˙ 300 ˚ 104.71 ˙˚ Î 300 Î 104.71

= 0.0885 + 0.0354 = 0.12394 kJ (e) 0.5 kg of air at P1 = P0, T1 = T0 f = Availability of the given system = 0 (f) 1 kg of air P1 = 10 bar, 867°C Case I:

Using Eq. (6.49)

ÈT P ˘ T ˘ ÈP = mCv T0 Í 1 - 1 - ln 1 ˙ + mRT0 Í 0 - 1 - ln 0 ˙ T0 ˚ P¢ ˚ Î P¢ Î T0 1140 ˘ 100 ˘ È 1140 È 100 = 1 ¥ 0.718 ¥ 300 Í - 1 - ln + 1 ¥ 0.287 ¥ 300 Í - 1 - ln ˙ 300 ˚ 263.15 ˙˚ Î 300 Î 263.15

= 315.56 + 29.93 = 345.49 kJ/kg È ˘ P1T0 100 ¥ 300 = = 263.15 kPa ˙ ÍP¢ = T1 1140 Î ˚

298

Basic Thermodynamics

Case II:

Using Eqs. (6.26), (6.46), (6.47) and (6.48) f = (u1 – u0) + P0(V1 – V0) – T0(S1 – S0)

È 1140 300 ˘ = 0.718(140 - 300) + 100 ¥ 0.287 Í ˙ Î 1000 100 ˚ 1140 1000 ˘ È - 300 Í1.005 ln - 0.287 ln 300 100 ˙˚ Î = 603.12 + (–53.382) – 204.25 = 345.49 kJ/kg Case III: Using Eqs. (6.26), (6.46), (6.47) and (6.48) f = (u1 – u0) + P0(V1 – V0) – T0(S1 – S0) From properties of air tables at 300 K, S0 = 1.70203 kJ/kg K u0 = 214.07 kJ/kg S1 = 3.11883 kJ/kg K at 1140 K, u1 = 880.35 kJ/kg È 1140 300 ˘ = (880.35 - 214.07) + 100 ¥ 0.287 Í ˙ - 300[3.11883 - 1.70203] Î 1000 100 ˚

= 666.28 + (–53.382) – 425.04 = 187.86 kJ/kg It can be seen that availability of case I and case II is same, but in case of case III, it is not same. The reason is that, for very high temperature of the system, case I and case II should not be used, only case III method should be used. In case III, S1 and S0 values are taken from the property table. Correct formula for È P ˘ R T0 (S1 - S2 ) = T0 Í S (T1 ) - S (T0 ) ln 1 ˙ M P0 ˚ Î

È Ê 8.314 ˆ 1000 ˘ = 300 Í3.11883 - 1.70203 - Á ln ˙ = 226.8 kJ/kg Ë 28.97 ˜¯ 100 ˚ Î \

Rewriting the case III method f = (u1 – u0) + P0(u1 – u0) – T0(S1 – S0) È P ˘ R ln 1 ˙ = (u1 - u0 ) + P0 (V1 - V0 ) - T0 Í S (T1 ) - S (T0 ) M P0 ˚ Î = 666.28 + (–53.38) – 226.8 = 386.1 kJ/kg

case III(a) π case III(b) EXAMPLE 6.4 Calculate the availability of 5 kg of copper Cp = Cv = C = 0.380 kJ/kg K at 2 bar and 500 K. Dead state conditions are 27°C and 100 kPa.

Availability and Irreversibility

299

f) Solution: Availability of 5 kg of Copper (f By using Eq. (6.26) or (6.53) = (u1 – u0) + P0(V1 – V0) – T0(S1 – S0) or

ÈT T ˘ 500 ˘ È 500 = mCT0 Í 1 - 1 - ln 1 ˙ = 5 ¥ 0.38 ¥ 300 Í - 1 - ln = 88.83 kJ T0 ˚ 300 ˙˚ Î 300 Î T0

EXAMPLE 6.5 The following data is connected to chilling of beaf. Dead state conditions are 27°C, 100 kPa. Calculate availability of the system. Mass of beaf = 1 ton = 1000 kg Initial temp. = 30°C Final temp. = –20°C Initial pressure = 100 kPa (P1 = P0) Freezing point = –2.0°C Specific heat below freezing = Cp solid = 1.5 kJ/kg K Specific heat above freezing = Cp liq = 3.0 kJ/kg K Latent heat of fusion = 250 kJ/kg K DY ) Solution: Availability of the system (D By using Eq. (6.18) DY = (h2 – h1) – T0(S2 – S1) (h2 – h1) = (h2 – hb) + (hb – ha) + (ha – h1) = mCp solid (T2 – Tb) – latent heat of beaf × m + mCp liq (Ta – T1) (h2 – h1) = 1000 × 1.5 × [–20 – (–2)] – 250 × 1000 + 1000 × 3.0 × [–2.0 – 30] = –373000 kJ È T ˘ T (S2 - S1 ) = m(S2 - Sb ) + m(Sb - Sa ) + m(Sa - S1 ) = m ÍC p solid ln 2 - h fg + C p liq ln a ˙ Tb T1 ˚ Î 253 250 271 ˘ È = 1000 Í1.5 ln - 3 ln = - 1360.445 kJ 271 271 303 ˙˚ Î DY = (h2 – h1) – T0(S2 – S1) = –373000 – 300 (–1360.45)

= 35135 kJ or 35135 kWs or 35.135 MJ or 9.76 kW h T °C 1

30°C T1

Liquid fish b

–2° = T2

a Solid fish

–20°C T2

2 S

Figure E6.5

300

Basic Thermodynamics

EXAMPLE 6.6 The temperature of 1 kg of water initially at 30°C is raised to 80°C using an electric heating coil. Calculate the first law and second law of thermal efficiency of the process. Dead state conditions are 27°C and 100 kPa. Assume no heat losses to the surrounding. Solution: This is a closed system. Apply the first law of thermodynamics Q = (Wd + Ws) + (u2 – u1) Q = 0, Wd = P0(V2 – V1),

P0 = P1 = P2 = 100 kPa

Ws

– (Wd + Ws) = (u2 – u1) –WS = Wd + u2 – u1 = u2 – u1 + P0(V2 – V1) Q=0

= (u2 – P0V2) – (u1 – P0V1)

Figure E6.6

= (u2 – P2V2) – (u1 – P1V1) = h2 – h1 From steam tables at 30°C and 80°C at 30°C

h1= 125.8 kJ/kg

at 80°C h2 = 3349 kJ/kg

–WS = (334.9 –125.8) = 209.1 kJ/kg \

WS = –209.1 kJ/kg

–ve sign represents the work (shaft) input to the system. h 1) (a) First law efficiency (h 1

=

Desired output ¥ 100 Required input

Desired output means, in this case increase in internal energy Required input = Work input From steam tables, u1 = 125.8 kJ/kg 1

=

u2 = 334.9 kJ/kg

u2 - u1 334.9 - 125.8 = ¥ 100 = 100% 209.1 Ws

h 2) (b) Second law efficiency (h heat input = 209.1 kJ. The availability of the system (closed system) (Df) Df = (u2 + P0V2 – T0S2) – (u1 + P0V1 – T0S1) Df = (h2 – h1) – T0(S2 – S1) = 209.1 – 300(1.075 – 0.437) = 17.7 kJ/kg [∵ From stream tables, S1 = 0.437 kJ/kg K 2

=

Availability output 17.7 = ¥ 100 = 11.8% Availability input 209.1

S2 = 1.075 kJ/kg K]

Availability and Irreversibility

301

It means, nearly 88% of the availability supplied is destroyed at the time of converting into internal energy. EXAMPLE 6.7 Reconsider Example 6.6. The heating of water from 30°C to 80°C is achieved from a constant temperature source of (a) 1500 K, (b) 500 K and (c) 350 K. Determine the second law efficiency. Solution:

Heat supplied to the water to raise from 30° to 80°C, Q = mCp dt = 1 × 4.187 × (80 – 30) = 209.35 kJ/kg

Availability of a system associated with heat flow (fheat flow) Eq. (6.54) heat flow

È T ˘ = Í1 - 0 ˙ Q T1 ˚ Î

(a) Source temperature = 1500 K heat flow

300 ˘ È = Í1 ˙ 209.35 = 167.48 kJ Î 1500 ˚

Change in availability of the closed system (Df) Df = (u2 + P0V2 – T0S2) – (u1 + P0V1 – T0S1) = (h2 – T0S2) – (h1 – T0S1) = (h2 – h1) – T0(S2 – S1) From steam table at 30°C, at 80°C,

h1 = 125.8 kJ/kg

S1 = 0.437 kJ/kg K

h2 = 334.9 kJ/kg

S2 = 1.075 kJ/kg K

Df = (334.9 – 1258) – 300(1.075 – 0.437) = 17.7 kJ/kg 17.7 ¥ 100 = 11.8% 167.48 (b) Source temperature = 500 K 2

=

heat flow

300 ˘ È = Í1 209.35 = 83.64 kJ/kg 500 ˙˚ Î

17.7 ¥ 100 = 21.16% 83.64 (c) Source temperature = 350 K 2

=

300 ˘ È f heat flow = Í1 209.35 = 29.87 kJ/kg 350 ˙˚ Î 17.7 ¥ 100 = 53.3% 29.87 It can be concluded that, to minimize loss of availability, heating should be carried out with the heat source whose temperature is very near to the required hot water temperature, i.e. in this example, required water temperature = 80°C, source temperature = 80°C. 2

=

302

Basic Thermodynamics

EXAMPLE 6.8 5 kgs of water undergoes a process from initial state to final state. Initial state: Water is saturated at 130°C Velocity 30 m/s Elevation 5 m Final state: Water is saturated at 20°C Velocity 20 m/s Elevation 4 m Calculate (a) Availability at the initial state (b) Availability at the final state (c) The change in availability Dead state conditions are 27°C and 100 kPa. Solution: 1

Saturated water at 130°C; 30 m/s

2

Saturated water at 20°C; 20 m/s

5m 4m

Figure E6.8

Assumption: The water is in closed system Considering the effect of K.E. in Eq. (6.26) = (u1 - u0 ) - T0 (S1 - S0 ) + P0 (V1 - V0 ) +

From steam tables at 130°C saturation, P1 = 2.7 bar, uf1 = 546.02 kJ/kg, at 27°C, uf0 = 113.25 kJ/kg, at 20°C, uf2 = 1.0018 cm3/g,

V2 + gz 2

ug1 = 2539.9 kJ/kg vf1 = 668.5 cm3/g, vf0 = 1.0035 cm3/g, vf2 = 83.95 kJ/kg,

Sg1 = 7.0269 kJ/kg K Sf0 = 0.3954 kJ/kg K Sf2 = 0.2966 kJ/kg K

f 1) (a) Availability of a closed system at initial state (f 1

= [2539.9 - 113.25] + 100[668.5 - 1.0035]

cm 3 × 10 –6

10 –3 kg – 300[7.0269 – 0.3954]

= 503.95 kJ/kg f 2) (b) Availability of a closed system at final state (f f2 = m[(u2 – u0) + P0(V2 – V0) – T0(S2 – S0)] = [(83.95 – 113.25) + 100(1.0018 – 1.0035) × 10–3 – 300[0.296 – 0.3954] = 0.3383 (c) Decrease in availability or change in availability = f 1 – f 2 = Df Df = 503.95 – 0.3383 = 503.6117 kJ/kg = 5 × 503.612 = 2518.01 kJ

Availability and Irreversibility

303

EXAMPLE 6.9 A solid iron ingot is placed in a room. Dead state conditions are 27°C, 100 kPa. Initial temperature of ingot = 500°C Final temperature of ingot = 40°C Diameter of the iron ingot = 0.1 m Thickness of the ingot = 0.2 m Iron ingot density = 9000 kJ/kg K Calculate, (a) Heat transfer from the ingot to the room (b) Irreversibility of the process Assume Cv of iron ingot = 0.42 kJ/kg K Solution:

Apply the first law of thermodynamics

(a) Heat transfer from the ingot to the room (Q) Consider (assume) iron ingot alone as a system Q = W + du

Room

Q

[No work done, ∵ W = 0]

Iron ingot

Q = du = mCvdT = mCv(T2 – T1) m = mass of ingot =

Q

2

d ¥t¥ 4 [t = thickness = 0.2 m]

Q

Figure E6.9

(0.1)2 ¥ 0.2 ¥ 9000 = 14.14 kg 4 Q = 14.14 × 0.42 × (40 – 500) = – 2731.3 kJ

–ve indicates heat transfer from the system (iron ingot) to the surroundings (room). (b) Irreversibility of the process (I) The irreversibility for any process is given by Eq. (6.42) I = Âm2 T0S2 – Âm1 T0S1 + mT0[ds]s – Qc.v. This is a closed system \

m2 = m 1 = 0 I = mT0(dS)s – Qc.v. = mT0(S2 – S1) – Qc.v. È È T V ˘ T ˘ (dS )s = m ÍCv ln 2 + R ln 2 ˙ = m ÍCv ln 2 ˙ T V T1 ˚ 1 1˚ Î Î

(dS )s = 14.14 ¥ 0.42 ¥ ln

[∵ V2 = V1]

(40 + 273) = - 5.37 kJ/K (500 + 273)

I = 300(–5.37) – (–2731.3) kJ = 1120.56 kJ EXAMPLE 6.10 Two blocks of iron ingot are initially at 1500 K and 500 K, respectively. After the thermal communication between the two blocks, they attain the same temperature. Specific heat of iron ingot is 1.0 kJ/kg K. Calculate irreversibility. Assume mass of 1st block = 5 kg, 2nd block = 10 kg.

304

Basic Thermodynamics

Solution: Irreversibility (I) After attaining the thermal equilibrium, both of the blocks attain same temperature which is represented by Tm m1Cp (T1 – Tm) = m2Cp (Tm – T2) m1T1 – m1Tm = m2Tm – m2T2 m1T1 + m2T2 = Tm (m2 + m1) \

Tm =

m1T1 + m2T2 5 ¥ 1500 + 10 ¥ 500 = = 833.33 K = 560.33∞C 10 + 5 m4 + m2

Net entropy change = dSnet = dS1 + dS2 We know that, È È T P ˘ T ˘ dS = m ÍC p ln 2 - R ln 2 ˙ = m ÍC p ln 2 ˙ T1 P1 ˚ T1 ˚ Î Î

\

[∵ P2 = P1]

dS1 = m1C p1 ln

Tm 833.33 = 5 ¥ 1 ¥ ln = - 2.94 kJ/K 1500 T1

dS2 = m2C p ln

Tm 833.33 = 10 ¥ 1 ¥ ln = 5.108 kJ/K 500 T1

dSnet = –2.94 + 5.108 = 2.168 kJ/K \

I = T0 dsNet= 300 × 2.168 = 650.47 kJ

EXAMPLE 6.11 5 kg of iron ingot at 1500 K is dipped into a water bath at 50°C containing 10 kg of water. The C pi (iron) and C pw (water) are 0.4 and 4.2 kJ/kg K, respectively. Dead state conditions, are 27°C and 100 kPa. Calculate loss of availability after the materials reached a common temperature. Solution: Loss of availability (I) Hot iron ignot immensed into the water. Find the mean temperature (Tm) after attaining the thermal equilibrium condition. Suffix

i — for iron block w — for water T1 — for initial temperature of iron T2 — for initial temperature of water Tm — for mean temperature for both m1C pi (T1 - Tm ) = mwC pw (Tm - T2 ) mi C pi T1 - mi C pi Tm = mwC pw Tm - mwC pw T2 mi C pi T1 + mwC pw T2 = mwC pw Tm + mi C pi Tm = Tm (mwC pw + mi C pi )

Availability and Irreversibility

\

Tm =

mi C pi T1 + mwC pw T2 mi C pi + mwC pw

= 376.5 K

or

=

305

5 ¥ 0.4 ¥ 1500 + 10 ¥ 4.2 ¥ 323 5 ¥ 0.4 + 10 ¥ 4.2

103.5°C

Net entropy change = dSnet = dSi + dSw We know that È È T P ˘ T ˘ dS = m ÍC p ln 2 - R ln 2 ˙ = m ÍC p ln 2 ˙ T1 P1 ˚ T1 ˚ Î Î dSi = mi C pi ln

Tm 376.5 = 5 ¥ 0.4 ¥ ln = - 2.765 kJ/K 1500 T1

dSw = mwC pw ln

\

[∵ P2 = P1]

Tm 376.5 = 10 ¥ 4.2 ¥ ln = 6.44 kJ/K 323 T2

dSnet = dSi + dSw = –2.765 + 6.44 = 3.67 kJ/K

\

I = T0 dSnet = 300 × 3.67 = 1101.65 kJ/K

EXAMPLE 6.12 Air at 25 bar, 350 K and a velocity of 90 m/s is expanded in a gas turbine and leaves at 10 bars 280 K and 60 m/s. Assuming no heat losses and air to be an ideal gas with constant specific heats, calculate, (a) The reversible work, (b) The actual work Dead state conditions 27°C, 100 kPa. Solution: (a) The reversible work (Wrev) By using Eq. (6.14)

\

Wrev = (h1 - h2 ) - T0 (S1 - S2 ) +

V12 - V22 g + ( Z1 - Z 2 ) gc 2 gc

Wrev = (h1 - h2 ) - T0 (S1 - S2 ) +

V12 - V22 2 gc

dz = 0

È T P ˘ V 2 - V22 = C pa (T1 - T2 ) - T0 ÍC pa ln 1 - R ln 1 ˙ + 1 2 gc T2 P2 ˚ Î

623 25 ˘ 90 2 - 60 2 È = 1.005(350 - 280) - 300 Í1.005 ln - 0.287 ln ˙ + 553 10 ˚ 2 ¥ 1000 Î = 70.35 + 42.96 + 2.25= 115.56 kJ/K (b) Actual work (Wa) Steady flow energy Eq. (6.41a) Wa = (h1 - h2 ) +

V12 - V22 90 2 - 60 2 = 1.005[(350 - 280)] + = 72.6 kJ/kg 2 gc 2 ¥ 1000

306

Basic Thermodynamics

EXAMPLE 6.13 Air is to be cooled from 400 K to 280 K. Amt. of air is 60 kg/min. Dead state conditions are 400 K and 100 kPa. Calculate minimum power required to cool the air. Solution: Minimum power required to cool the air (Wrev) By using Eq. (6.14) Wrev = (h1 – h2) – T0(S1 – S2) = Cp(T1 – T2) – T0(S1 – S2) È T P ˘ = C p (T1 - T2 ) - T0 ÍC p ln 1 - R ln 1 ˙ T P 2 2˚ Î 400 ˘ È = 1.005(400 - 280) - 400 Í1.005 ln ˙ 280 Î ˚

[∵ P1 = P2]

= –22.8 kJ/kg –ve means, minimum work input EXAMPLE 6.14 Air from atmosphere is compressed by using centrifugal air compressor. The atmosphere conditions are 15°C. The air is compressed from 1 bar to 2 bar. The temperature increases from 15°C to 100°C. The flow rate of air is 30 kg/min. Calculate (a) Minimum work and power. (b) Actual work and power. Neglect K.E. and P.E. changes. Neglect the heat interaction between the compressor and surroundings. (c) Irreversibility. Solution: 1

2

Air 1 bar 15°C

Air 2 bar 100°C 1

2

Figure E6.14

(a) Minimum work (Wrev = Wmin) and power (Pmin) Equation (6.14) can be used to calculate minimum work input. È T P ˘ Wrev = ( h1 - h2 ) - T0 (S1 - S2 ) = C p (T1 - T2 ) - T0 ÍC p ln 1 - R ln 1 ˙ T2 P2 ˚ Î 288 1˘ È = 1.005(15 - 100) - 288 Í1.005 ln - 0.287 ln ˙ = - 67.86 kJ/kg 273 2˚ Î

–ve indicates work done on the system. Or

Y1 = Availability at inlet condition = (h1 – h0) – T0(S1 – S0) È T P ˘ = C p (T1 - T0 ) - T0 ÍC p ln 1 - R ln 1 ˙ = 0 - 0 ª 0 T P 0 0˚ Î

Availability and Irreversibility

307

Y2 = Availability at outlet condition = (h2 – h0) – T0(S2 – S0) È T P ˘ = C p (T2 - T0 ) - T0 ÍC p ln 2 - R ln 2 ˙ T0 P0 ˚ Î 373 2˘ È = 1.005(100 - 15) - 288 Í1.005 ln - 0.287 ln ˙ = 67.86 kJ/kg 288 1˚ Î

Wmin = Change in availability = Decrease in availability = Y1 – Y2 = 0 – 67.86 = – 67.86 kJ/kg Pmin = - 67.86 ¥

30 kJ kg = - 33.93 kW 60 kg s

(b) Actual work (Wa) and actual power (Pa) Wa = h1 – h2= Cp(T1 – T2) [Equation (6.41a), K.E. and P.E. are neglected, Q = 0] = 1.005(15 – 100) = 85.43 kJ/kg Pa = mWa = -

30 kg kJ ¥ 85.43 = - 42.71 kW 60 s kg

(c) Irreversibility (I) I = Wmin (Wrev) – Wa [From Eq. (6.44)] = – 67.86 – (– 85.43) = 17.57 kJ/kg =

30 kg kJ ¥ 17.57 ¥ = 8.785 kW 60 s kg

Or from Eq. (6.55) and dssurrounding = 0, Q = 0 È T P ˘ I = T0 dSsystem = T0 ÍC p ln 2 - R ln 2 ˙ T1 P1 ˚ Î 373 2˘ 30 È = 288 Í1.005 ln - 0.287 ln ˙ = 17.57 kJ/kg = ¥ 17.57 = 8.785 kW 288 1 60 Î ˚

EXAMPLE 6.15 Air is compressed from 27°C and 1 bar to 3 bar. The efficiency of the compressor is 80%. Calculate irreversibility.

P 3

2

Solution: Ideal path 1–2 Actual path 1–3 Assume inlet temperature of the air = dead state condition Here, dead state conditions are 27°C, 1 bar \

T1 = T0 = 27°C.

1

Figure E6.15

V

308

Basic Thermodynamics

Irreversibility (I) Consider the process 1–2 (Ideal)

ÊP ˆ T2 = T1 Á 2 ˜ ËP ¯

-1

1

c

\

= 0.8 =

Ê 3ˆ = 300 Á ˜ Ë 1¯

1.4 - 1 1.4

= 410 K

T2 - T1 410.6 - 300 = T3 - T1 T3 - 300

T3 = 438.3 K È T P ˘ I = T0 dSsystem = T0 (S3 - S1 ) = T0 ÍC p ln 3 - R ln 3 ˙ T1 P1 ˚ Î 438.3 3˘ È = 300 Í1.005 ln - 0.287 ln ˙ = 19.698 kJ/kg 300 1˚ Î

Wideal = Cp(T2 – T1) = 1.005[410.6 – 300] = 111.153 kJ/kg Wactual = Cp(T3 – T1) = 1.005[438.3 – 300] = 138.99 kJ/kg This gives an idea that, when 138.99 kJ/kg (Wactual) of energy is added to the air as work, 19.698 kJ/kg (I) of energy is wasted or cannot be reconverted to work. For the reversible process, 111.153 kJ/kg of work is required and the entire amount of energy can be reconverted into work by reversing the process. EXAMPLE 6.16 In a Babcox and Wilcox boiler, saturated steam is generated at 400° K. Hot gases from a fire having a specific heat of 1.005 kJ/kg K and a temperature of 1500° C transfer heat to the boiler. The final temperature of the gases is 800 K. Latent heat of water is 1950 kJ/kg, Calculate, (a) Increase in entropy of the combined system. (b) Increase in unavailable energy due to the irreversible heat transfer. (c) %age loss of available energy. Dead state conditions are 27°C and 1 bar. Solution: (a) Increase in entropy of the combined system (dsnet) Change of entropy of the gases (dsg) 800 = - 0.6318 kJ/kgK 1500 Change of entropy of the water (dsw) dsg = C pg ln

Latent heat 1950 = = 4.875 kJ/kg K 400 400 Net change in entropy due to heat transfer (dsnet) dsnet = – 0.6318 + 4.875 = 4.2432 kJ/kg K dsw =

309

Availability and Irreversibility

(b) Increase in unavailable energy due to the irreversible heat transfer = T0dsnet = 300 × 4.2432

T

= 1272.96 kJ/kg (c) %age loss of available energy Available energy from the gas due to this heat transfer

1500 K

s

Ga

800 K

Steam

400 K

= 1950 – 300 × 0.6318 = 1760.46 kJ/kg

Increase in unavailable energy

300 K

%age of loss of available energy =

dsg

S

dsnet dsw

1272.96 ¥ 100 = 72.31% 1760.46

Figure E6.16

EXAMPLE 6.17 Hot gases leaving from the turbine at 1000°K are used to heat water at constant pressure. Water enters at temperature 350 K. The exit temperature of gases is 450 K. The mass flow rates of gases and water are 25 kg/s and 30 kg/s, respectively. Calculate: (a) The available and (b) unavailable energies of the process. Solution:

T

Heat lost by gases = Heat gained by water

1000 K 450 K

mg Cpg dtg = mw Cpw dtw 25

kg kJ ¥1 (1000 - 450) K s kgK = 30

\

Available

kg kJ (Two - 350) K ¥ 4.2 s kgK

350 K Available

300 K Unavailable

Two = 459.13 K

Unavailable

Figure E6.17

f gas) For gas: Available energy from the gases (f fgas = Total heat transferred from the gases – unavailable energy \

Total heat transferred from the gases (Qg) Qg = mg Cpgdtg = 13750 kJ/s

Unavailable energy from the gases (UEg) È T P ˘ T UEg = T0 dsg = T0 ÍC p ln 1 - R ln 1 ˙ = T0 C p ln 1 T2 P2 ˚ T2 Î 1000 = 239.55 kJ/kg 450 fgas = 13750 – 239.55 × 25 = 7761.25 kJ/S = 300 ¥ 1 ¥ ln

\

[∵ P1 = P2]

S

310

Basic Thermodynamics

For water: Total heat energy transferred Qw = 13750 kJ/kg Unavailable energy from water (UEw) È T P ˘ T UEw = T0 dsw = T0 ÍC p ln 2 - R ln 2 ˙ = T0 C p ln 2 T P T1 1 1˚ Î

[∵ P1 = P2]

459.13 = 341.96 kJ/kg 350 fwater = Available energy from water = Qw – UEw = 300 ¥ 4.2 ¥ ln

= 13750 – 341.96 × 30 = 3491.2 kJ/s f t) and unavailable energies of the process The total available (f \

ft = – Available from gas + Available from water = –7761.3 + 3491.2 = – 4270.05 kJ/s

–ve sign indicates that, during this heat transfer process, the available energy decreased by 4270.05 kJ/s and unavailable energy increased by the same amount. \ Loss of available energy = 4270.05 kJ/s EXAMPLE 6.18 A reservoir at a temperature of 1000 K transfers heat at the rate of 2200 kJ/min to a system at 500 K. The low temperature reservoir is at a temperature of 300 K. Assume that the temperatures of system and sourse remain constant during heat transfer. Calculate: (a) The entropy produced during heat transfer, (b) The decrease in available energy after heat transfer. Solution:

Figure E6.18

Availability and Irreversibility

311

(a) The entropy produced during heat transfer (dsnet) Change in entropy of the source during heat transfer (dssource) dssource =

-Q -2200 kJ = = - 2.2 T1 1000 min K

Change in entropy of the system during the heat transfer (dssystem) dssystem =

Q 2200 kJ = = 4.4 T2 500 min K

The net change in entropy = dsnet = dssource + dssystem= -2.2 + 4.4 = 2.2

kJ min K

(b) The decrease in available energy after heat transfer [loss of available energy] = Available energy from the source – Available energy from the system = T0 dsnet Unavailable energy from the source = T0dssource = 300 × (–2.2) = – 660 kJ/min Unavailable energy from the system = T0dssystem = 300 × 4.4 = 1320 kJ/min Available energy from the source = Heat transferred from the source – Unavailable energy from the source = 2200 – 660 = 1540 kJ/min Available energy from the system = Heat transferred to the system – Unavailable energy from the system = 2200 –1320 = 880 kJ/min \

Decrease in available energy = Available energy from the source – Available energy from the system = 1540 – 880 = 660 kJ/min

Or decrease in available energy

= T0dsnet = 300 × 2.2 = 660 kJ/min

EXAMPLE 6.19 A heat source at a temperature of 675 K transfers heat to a gas of 1 kg initially at 300 K and 10 bar at constant pressure. The gas is then expanded reversibly and adiabatically to a final pressure of 1 bar and 300 K temperature C. Assume gas as ideal and constant specific heat. Calculate irreversibility Assume: ggas = 1.7 Cp = 1.2 kJ/kg K Cv = 0.7059 kJ/kg K Solution:

Irreversibility of the overall process = I1–3 = I1–2 + I2–3 I1–2 = T0ds1–2 – Q1–2

\

I2–3 = 0

I1–3 = T0ds1–2 – Q1–2 – 0 ds1- 2 = C p ln

T1 P - R ln 1 T2 P2

1–2 is a constant pressure process,

\

P1 = P2

[∵ 2–3 is a reversible process]

312

Basic Thermodynamics 10 bar

T

1 bar

2

P 1

10 bar

1

300 K

3

1 bar

3

2

V

S

Figure E6.19

ds1- 2 = C p ln

T2 T1

We know that 2–3 is a reversible adiabatic process.

ÈP ˘ T2 = T3 Í 2 ˙ Î P3 ˚

\

-1

È 10 ˘ = 300 Í ˙ Î1˚

1.4 - 1 1.4

= 774.26 K

774.26 = 1.13774 kJ/kg K 300 From the first law of thermodynamics

\

ds1- 2 = 1.2 ¥ ln

Q1-2 = du1–2 + w1–2 = Cv(T2 – T1) + P(V2 – V1) V1 =

mRT1 1 ¥ 0.4941 ¥ 300 = = 0.14823 m 3 /kg P1 10 ¥ 100

V2 = V1

T2 0.14823 ¥ 774.26 = = 0.3826 m 3 /kg T1 300

\

Q1–2 = 0.7059[774.26 – 300] + 10 × 100[0.3826 – 0.1482]= 569.142 kJ/kg

\

I1–3 = 300 × 1.1377 – 569.142 = – 227.83 kJ/kg

EXAMPLE 6.20 A closed container is having compressed air at 700 kPa and 27°C. The dead state conditions are 27°C and 100 kPa. Calculate work potential or availability. f) Solution: Work potential or availability (f From Eq. (6.26) f = (U1 + P0V1 – T0S1) – (U0 + P0V0 – T0S0)

(1)

= (U1 – U0) + P0(V1 – V0) – T0(S1 – S0) = Cp(T1 – T0) + P0(V1 – V0) – T0(S1 – S0) = 0 + P0(V1 – V0) – T0(S1 – S0) Here,

T1 = T0 = 27°C

(2)

Availability and Irreversibility

d s = (S1 - S2 ) = C p ln = - 0.287 ln

V0 =

RT0 P0

313

T1 P P - R ln 1 = 0 - R ln 1 T0 P0 P0

700 = - 0.5585 kJ/kg K 100

V1 =

RT1 P1

È1 1 ˘ 1 ˘ È 1 3 (V1 - V0 ) = RT0 Í - ˙ = 0.287 ¥ 300 Í ˙ = - 0.738 m /kg P P 700 100 Î ˚ 0˚ Î 1 Substitute the above data in Eq. (2)

\

\

f = 100(– 0.738) – 300(– 0.5585) = 93.75 kJ/kg

EXAMPLE 6.21 A battery is used to start a petrol car. 5.0 MJ of electrical energy is required to start the car. In order to get the same energy from compressed air of 8 MPa and 27°C, what will be the volume of the container of the compressor? Solution: First of all calculate availability of the compressed air at 8 MPa, 27°C. Surrounding (dead state) conditions are 27°C, 1 bar. \ From Eq. (6.26) f = (U1 – U0) + P0(V1 – V0) – T0 (S1 – S0) T1 = T0 = 27°C V1 =

RT1 0.287 ¥ 300 = = 0.010763 m 3 /kg P1 8000

V0 =

RT0 0.287 ¥ 300 = = 0.861 m 3 /kg P0 100

È T P ˘ = 0 + 100(0.010763 - 0.861) - 300 ÍC p ln 1 - R ln 1 ˙ T0 P0 ˚ Î

8000 ˘ È = 0 + ( -0.85024) - 300 Í 0 - 0.287 ln 100 ˙˚ Î

= – 0.85024 + 377.3 = 376.44 kJ/kg In order to deliver 5 MJ of electrical energy (work), the quantity of air required (V) ma =

5000 kJ kg = 13.28 kg 376.44 kg

V = ma v1 = 13.28 ¥ 0.010763 kg ¥

m3 = 0.14296 m 3 kg

314

Basic Thermodynamics

EXAMPLE 6.22 Steam at an initial condition of 10 bars and 250°C is contained in a piston and cylinder arrangement. The initial volume is 2500 cm3. The steam is then expanded to 2 bars and 150°C. During this process 0.5 kJ of heat is transferred to the system. Dead state conditions are 1 bar and 27°C. Calculate: (a) The actual useful work done by the system. (b) The change in availability during the process. Solution: Wd P1, V1, T1 Piston

P0, V0, T0

Ws

P1 = 10 bar P2 = 2 bar T1 = 250°C T2 = 150°C 3 V1 = 2500 cm T0 = 27°C P0 = 1 bar

Q = 0.5 kJ

Figure E6.22

(a) Actual useful work done by the system Displacement work associated with the system boundary against the pressure P0 is Wd Wa = Actual work = Wd + Ws Wd = Work done by the system on the surroundings Ws = Actual useful work Apply the first law of thermodynamics Q = (Wd + Ws) + du = P0(V2 – V1) + Ws + m(u2 – u1) Ws = Q – P0(V2 – V1) – m(u2 – u1) \

(1)

From steam tables at 10 bar, 250°C u1 = 2709.9 kJ/kg

h1 = 2942.6 kJ/kg

s1= 6.925 kJ/kg K v1 = 0.233 m3/kg

at 2 bar and 150°C u2 = 2576.9 kJ/kg h2 = 2768.8 kJ/kg s2 =7.279 kJ/kg K \

m= m=

V1 v1 V2 v2

=

2500 ¥ 10 6 m 3 0.233 m 3 /kg

v2 = 0.960 m3/kg

= 0.1073 kg

\ V2 = mv2 = 0.1073 kg × 0.96 = 0.103 m3

Substitute the above data in Eq. (1) Ws = 0.5 – 100[0.103 – 0.0025] – 0.1073[2576.9 – 2709.9] = 4.721 kJ

Availability and Irreversibility

Df ) (b) The change in availability during the process (D Df = m(u1 – u2) + P0(V1 – V2) – mT0(S1 – S2) = 0.1073(2709.9 – 2576.9) + 100(0.0025 – 0.103) – 0.1073 × 300(6.925 – 7.279) = 15.616 kJ This availability is also called as maximum useful work. 4.721 ¥ 100 15.616 = 30.23% of the maximum useful work.

The actual useful work is approximately =

EXAMPLE 6.23 The following data refers to a heat exchanger. (a) Parallel flow mode P1 = 50 kPa, T2 = 300°C, m1 = m2 = mw = 50 kg/S, T3 = 1000 K, T4 = 450 K

(b) Counter flow mode P1 = 50 kPa, T2 = 300°C, m3 = m4 = ma, T3 = 1000 K, T4 = 375 K

Dead state conditions are (T0) 270C, 1 bar Calculate: (i) Irreversibility (ii) Second law efficiency. Solution: (a) Parallel flow mode First law analysis mwh1 + mah3 = mwh2 + mah4 \

ma = m w

mw(h1 – h2) = ma(h4 – h3)

h1 - h2 h4 - h3

From steam tables at 50 kPa and saturated water h1 = 340.6 kJ/kg, at 50 kPa and 300°C, h2 = 3075.5 kJ/kg, \

ma =

s1 = 1.0912 kJ/kg K s2 = 8.537 kJ/kg K

50 (340.6 - 3075.5) = 247.39 kg/s 1.005(450 - 1000)

315

316

Basic Thermodynamics

(i) Irreversibility (I) From Eq. (6.43) for steady flow process

È I = T0 Í ÎÍ

˘

 ms -  ms ˙˙ - Q

˚ e = exit I = inlet Q = 0 steady flow, Adiabatic process e

i

I = T0 [mws2 + mas4 – (mws1 + mas3)] = T0 [mw(s2 – s1) + ma(s4 – s3)] (S4 - S3 ) = C p ln

T4 P - R ln 4 T3 P3

No pressure drop in the pipe line,

\ P4 = P3

500 = - 0.6966 kJ/kg K 1000 (S2 – S1) = (8.537 – 1.0912) = 7.446 kJ/kg K

(S4 - S3 ) = 1.005 ln

Substitute in Eq. (1) \

I = 300[50 × (7.446) + 247.39 × (–0.6966)] = 59990.44 kJ/S or kW = 59.99 MW= 60 MW

Or from availability relation I=

 mY -  mY = m i

w Y1

+ ma Y 3 - ( mw Y 2 + m a Y 4 )

e

= mw(Y1 – Y2) + ma(Y3 – Y4) = mw [(h1 – h2) – T0(s1 – s2)] + ma [(h3 – h4) – T0(s3 – s4)] = 50 [(340.6 – 3075.5) – 300(–7.466)] + 247.39{[1.005 (1000 – 450)] – 300(0.6966)} = 60290.26 kJ/s = 60.29 MW h 2) (ii) Second law efficiency (h 2

=

=

Availability production Loss availability

Increase in availability of cold cream = Loss of availability of hot stream or decrease in availability of hot stream

IC LH

fIC = mw [(h2 – h1) – T0(s2 – s1)] = 50[(3075.5 – 340.6) – 300(7.466)] = 24755 kJ/s fLH = ma[(h4 – h3) – T0(s4 – s3)] = 247.39[1.005(450 – 1000) – 300(– 0.6966)] = – 85045.26 kJ/s

(1)

Availability and Irreversibility

317

–ve sign means loss of availability or decrease in availability. \

2

(b) Counter flow mode ma =

\

=

24755 ¥ 100 = 29.11% 85045.26

50 (340.6 - 3075.5) = 217.7 kg/S 1.005(375 - 1000)

(i) Irreversibility (I) From Eq. (6.44) for steady flow process

È ˘ I = T0 Í ms ms ˙ - Q ÍÎ e ˙˚ i Q = 0, Adiabatic process

Â

\

Â

I = T0[mwS2 + maS4 – (mwS1 + maS3)] = T0[mw(S2 – S1) + ma(S4 – S3)] (S4 - S3 ) = C p ln

T4 P T - R ln 4 = C p ln 4 T3 P3 T3

(2) [∵ P4 = P3]

375 = - 0.9857 kJ/kg K 1000 (S2 – S1) = (8.537 – 1.0912) = 7.446 kJ/kg K = 1.005 ln

\

I = 300[50 (7.446) + 217.7(–0.9857)] = 47313.933 kJ/s = 47.314 MW

(ii) Second law efficiency fIC = mw[(h2 – h1) – T0(S2 – S1)]= 24755 kJ/s fLH = ma[(h3 – h4) – T0(S3 – S4)] = 217.7[1.005(1000 – 375) – 300(0.9857)] = 72366.75 kJ/s 2

=

24755 ¥ 100 = 34.21% 72366.75

EXAMPLE 6.24 A piston-cylinder contains 10 kg of water initially at 200°C, 1 MPa. During an internally reversible isothermal expansion process, the heat transfer to the water from a thermalenergy reservoir, 600°C, is 10000 kJ. The temperature and pressure of the environment are 298 K and 100 kPa, respectively. Determine (a) the work for this process, (b) the useful work, (c) the reversible useful work, (d) the irreversibility and (e) the second law efficiency. Solution:

The process is internally reversible, hence entropy change S2 - S1 =

Q1- 2 QR 10000 kJ = = (200 + 273) ¥ 10 kg K T1 T1

318

Basic Thermodynamics

P

T

600°C (TR) m1 = 10 kg T1 = 200°C P1 = 1 MPa

1

1

Reservoir

2

10000 kJ = QR T0 = 298 K P0 = 100 kPa

S

V

Figure E6.24

\

S2 = S1 +

10000 473 ¥ 10

From steam table at 10.00 bar, saturated S1 = 6.694 kJ/kg K S2 = 6.694 +

v1 = 0.206 m3/kg

u1 = 2621.9 kJ/kg

10000 = 8.8082 kJ/kg K 473 ¥ 10

From steam tables at 200°C and S2 = 8.8082 kJ/kg K u2 = 2659.45 kJ/kg

P2 = 0.625 bar

u2 = 3.628 m3/kg

(a) Actual work (Wa = W1–2) Wa = Q1–2 – (u2 – u1)m = 10000 – 10(2659.45 – 2621.9) = 9624.5 kJ (b) Actual useful work (Wa useful) Wa useful = Q1–2 – P0(V2 – V1)m = 9624.5 – 100 × 10(3.628 – 0.206) = 6202.5 kJ (c) Reversible useful work (Wrev useful) Wrev useful = (

1

-

2)

2

Ê T ˆ - QR Á 1 - 0 ˜ TR ¯ Ë

Ê T ˆ = (u1 - u2 ) + P0 (u2 - u1 ) - T0 (s1 - s2 ) - QR Á1 - 0 ˜ TR ¯ Ë

= (2621.9 – 2659.45) + 100(3.628 – 0.206) – 298(6.694 – 8.8082) 298 ˆ Ê – (–10000) Á 1 Ë 873 ˜¯ [Heat is rejected from the surrounding hence QR is –ve for the surrounding]

Availability and Irreversibility

319

Wrev useful = –37.55 + 342.2 – (–630.03) + 6586.48 = 7521.16 kJ/kg = 10 kg × 7521.16 kJ/kg = 75211.6 kJ (d) Irreversibility (I) I = Wrev useful – Wa useful = 75211.6 – 6202.5= 69009.1 kJ h 2) (e) Second law efficiency (h 2

= =

Wa useful Wrev useful

=

6202.5 ¥ 100 = 8.25% 75211.6

The ratio of the actual useful work produced to the reversible useful work that could have been produced can be used to measure the second law efficiency for the process. This ratio is called as effectiveness (e) of the process.

EXERCISES 6.1 Explain the term available and unavailable energy. When does the system become dead state? 6.2 Define the term availability. 6.3 Derive an expression for (a) Availability for non-flow system. (b) Availability for open system. 6.4 Derive the expression for (a) Reversible work done for non-flow system . (b) Reversible work done for open system. 6.5 Define the term maximum work and maximum useful work. 6.6 Define the term irreversibility. 6.7 Write the expression for irreversibility of (a) closed system (b) open system 6.8 Define second law efficiency for (a) Expansion process or heating. (b) Compression process or cooling. 6.9 A cylinder of an internal combustion engine contains 2450 cm3 of gaseous combustion products at a pressure of 7 bar and temperature of 867°C just before exhaust valve opens. Determine the specific availability of the gas, in kJ/kg. Ignore the effects of motion and gravity, and model the combustion products as air as an ideal gas. Take T0 = 27°C and P0 = 1 atm. [Ans: 368.91 kJ/kg] 6.10 Water initially at saturated liquid at 100°C is contained in piston cylinder assembly. The water undergoes a process to the corresponding saturated vapour state, during which piston

320

Basic Thermodynamics

moves freely in the cylinder. For each of the processes described below, determine on a unit mass basis the change in availability, the availability transfer accompanying work, the availability transfer accompanying heat, and the irreversibility, each in kJ/kg. Let T0 = 20°C. P0 = 1.014 bar. (a) The change in state is brought about by heating the water as it undergoes an internally reversible process at constant temperature and pressure. (b) The change in state is brought about adiabatically by the stirring action of a paddle wheel. [Ans: (a) 484 kJ/kg, 0, 484 kJ/kg, (b) 484 kJ/kg, –2257 kJ/kg, 1773 kJ/kg] 6.11 Steam enters a turbine with a pressure of 30 bars, a temperature of 400°C and a velocity of 160 m/s. Steam exists as saturated vapour at 100°C with a velocity of 100 m/s. At steady state, the turbine develops work at a rate of 540 kJ/kg of steam flowing through the turbine. Heat transfer between turbine and its surrounding occurs at an average surface temperature of 500 K. Determine the irreversibility per unit mass of steam flowing through the turbine in kJ/kg. Neglect the change in potential energy between inlet and exit. Let T0 = 25°C and [Ans: 142.71 kJ/kg] P0 = 1 atm. 6.12 Determine the availability of the following closed systems: (a) 5 kg of water at P = 100 kPa and 90°C (b) 2 kg of ice at P = 100 kPa and –10°C (c) 0.1 kg of steam at P = 4 MPa and 500°C (d) 0.5 kg of wet steam at P = 10 kPa with a quality of 0.85 [Ans: 122 kJ, 81.2 kJ, 98.7 kJ, 623 kJ] 6.13 Determine the availability of 1 m3 of complete vacuum.

[Ans: 100 kJ]

6.14 The motors used in mines are often designed to run on compressed air. In a particular design, an air motor requires a minimum of 7 bar air pressure to operate while delivering 105 kJ/kg of air supplied. A compressed air bottle of volume 1.66 m3 containing air at 20 bar and 300 K is available for operating the air motor. If the ambient pressure is 1 atm and the ambient temperature is 300 K, determine (a) the maximum work output of the air motor operated from one bottle of compressed air. (b) the second law efficiency achieved in the operation of the air motor. [Ans: 6740 kJ, 38.9%] 6.15 Determine the irreversibility per kg of Freon 12 flow, associated with an expansion valve. Freon = 12 enters at 1.2 MPa, 30°C and is expanded to 100 kPa. [Ans: 7.87 kJ/kg] 6.16 A 0.1 m diameter, 0.1 m high solid copper cylinder is initially at 180°C. It is then placed in a room and is allowed to cool to a final temperature of 30°C. Calculate the heat transfer and irreversibility. Dead state conditions are 25°C and 1 atm. Assume copper density = 8954 kJ/kg K. 6.17 The following data refers to a power plant Steam pressure entry into the turbine Steam pressure entry into the turbine

5 MPa 500°C

Availability and Irreversibility

321

Steam is extracted from the turbine at 1 MPa and 300°C, allowed into the high pressure heater. At some other plant the steam is bled at 100 kPa and 100°C and allowed into the low pressure heater. The remaining quantity of steam exits the turbine at 5 kPa with a enthalpy of 2200 kJ/kg. Assume m2 = 0.13, m3 = 0.1. m3 m2 = 0.13, = 0.1 m1 m1

let

m1 = Amount of steam entering into the turbine. m2 = entering into the high pressure feed water heater. m3 = Amount of steam entering into the low pressure water heater. m4 = Quantity of water existing the turbine end. Calculate the irreversibility. [Ans: 71.70 kJ/kg]

6.18 Heat is transferred from the hot gases in a boiler to generate saturated steam at 200°C. Find the increase in total entropy of the combined system of gas and water and increase in unavailable energy due to irreversible heat transfer. The gases are cooled from 1000°C to 500°C and all the heat from gases goes to water. Assume water enters the boiler at saturated water and leaves saturated steam. C pg = 1.0 kJ/kg, hfg of steam = 1940 kJ/kg, atmospheric temperature = 20°C. [Ans: 635.22 kJ/kg]

7 CHAPTER

Pure Substances

7.1

PURE SUBSTANCES—DEFINITION AND EXPLANATION

A pure substance is a substance that has a uniform chemical composition throughout the system. Examples: Water, N2, He and CO2.

O2

(a)

Air (Gaseous) (b)

H2O (vap)

H2O (liq) OR H2O (sol) OR H2O (vap)

CO2

(c)

(d)

H2O (liq) H2O (s) (e)

Figure 7.1 Pure substances.

Figure 7.1 shows the pure substances. A pure substance may be a single chemical element [Figure 7.1(a)] or a compound [Figure 7.1(c), (d)] or a mixture of various chemical elements or compounds [Figure 7.1(b)]. A system consisting of a mixture of various phases of a pure substance [Figure 7.1(e)] is still a pure substance because all the phases have the same chemical composition. Consider a system having a mixture of liquid air and gaseous air [Figure 7.2(a)]. Each substance (liquid air and gaseous air) is a pure substance separately. However, mixture is not a pure substance, because the composition of liquid air is different from the composition of gaseous air; hence the mixture is not chemically homogeneous. Similarly, a mixture of water and oil [Figure 7.2(b)] is not a pure substance because the composition of water (H2O) is different from the composition of oil. Hence, the mixture is not chemically homogeneous. 322

Pure Substances

323

A pure substance must be homogeneous in Oil chemical composition. Homogeneous in chemical Air (gaseous) Water Air (liquid) composition means, each part of the system must (liquid) have precisely the same chemical constituents, and (a) (b) the chemical elements must be combined chemically in the same way (chemical aggregation or Figure 7.2 Non-pure substances: (a) Pure homogeneous in chemical aggregation). substance, chemically aggregate Consider an example of a mixture of water and (b) Non-pure substance, chemically non-aggregate. (liquid) and water (vapour) as shown in Figure 7.3(a). This is a pure substance. All parts of the system of H2 + 0.5O2 H2O Vapour Figure 7.3(a) are having chemical composition of (mix. of gases) H2O [1 mole of H2 and 0.5 mole of O2] and then H2 Liquid chemical elements (H 2 and O 2 ) are combined H2O Liquid chemically in the same way in all parts of the system. (a) (b) Now consider an example of a mixture of water Figure 7.3 Illustration of chemical (liquid) and mixture of 1 mole of H2(g) and 0.5 mole aggregation. of O2 (g) as shown in Figure 7.3(b). This is not a pure substance. All parts of a system of Figure 7.3(b) are having chemical component of H2O (1 mole of H2 and 0.5 mole of O2) but the chemical elements (H2 + O2) are not combined chemically in the same way in all parts of the system. A pure substance must also be invariable in chemical composition with respect to time, i.e. chemical composition of the system should not change with respect to time. Consider a system as shown in Figure 7.4. The system is a mixture of 1 mole of C (g) and 1 mole of O2 (g). If this mixture is remained as a mixture over a period of time (period of interest) without undergoing a chemical reaction, then the system is called as a pure substance.

Over a period of time

Mixture of C + O2

Mixture of C + O2

(Period of interest)

(a)

(b)

Figure 7.4 Illustration of no change with respect to time.

Consider a system as shown in Figure 7.5. The system is a mixture of O2 (g) + Fuel vapour (g). This is a pure substance, until it is remained as a mixture of O2 + Fuel vapour over a period of time O2 + Fuel (vap)

with time

(a)

O2 + Fuel (vap)

over a period of time

(b)

over a period CO2 + H2O

of time

(c)

(d)

Chemical reaction Pure substance

Figure 7.5

non-pure substance

CO2 + H2O

Pure substance

Illustration of pure substance to non-pure substance.

324

Basic Thermodynamics

[Figure 7.5(a) to (b)]. If a chemical reaction takes place such that the chemical composition of the system becomes non-homogeneous over a period of time, then the system cannot be regarded as a pure substance for that period [Figure 7.5(b) to (c)]. However, the final product of the reaction can be regarded as a pure substance, if the composition of the mixture again becomes homogeneous throughout the system. This is shown in Figure 7.5(c) to (d). Table 7.1 shows the examples of systems as pure substance. Table 7.1

System

Examples of systems as pure substance

Homogeneous in chemical composition

Homogeneous in chemical aggregates

Pure substance

Yes Yes Yes Yes

Yes Yes Yes Yes

Yes Yes Yes Yes

Yes No Yes

No No Yes

No No No

Yes

Yes

Yes

O2 CO2 Steam + Water Air (g) H2O (liquid) + H2 (g) + 0.5 O2 (g) Water + Oil O2 + Fuel (vapour) Product of combustion CO2 + H2O

7.2

TWO PROPERTY RULE

The thermodynamic state of a pure substance of given mass can be fixed by any two independent intensive properties provided that 1. The system is in equilibrium and 2. The effects due to gravity, motion, surface tension, electricity and magnetism are negligible on the system.

7.3

FORMATION OF STEAM AT CONSTANT PRESSURE OR PHASES OF PURE SUBSTANCES

A mixture of more than one phase of a pure substance co-exists in equilibrium (example, water exists as a mixture of liquid and vapour in boiler and condenser). Similarly, there are many practical situations where more than two phases co-exist in equilibrium. In this section, the behaviour of the liquid and vapour phase and their mixtures is considered. Changes in temperature and specific volume of a pure substance such as water when it is heated at constant pressure can be studied by considering a simple experimental device, piston and cylinder arrangement as shown in Figures 7.6 and 7.7. Let us consider 1 kg of water at 25°C and 1 atmospheric pressure (101.32 kPa). Under these conditions, water exists in the liquid phase, and it is called a compressed liquid or a sub-cooled liquid, i.e. the system is not about to vaporize. The condition of water at 25°C is shown in Figure 7.6(1) and represented by state 1 in Figure 7.7.

Pure Substances

325

Figure 7.6 Process showing phase changes of a pure substance (H2O).

The water is heated from external source at constant pressure until its temperature rises to, say 50°C and at the same time, the water will expand. When there is increase in energy of the water at constant pressure, specific volume will also increase. Since the system is a closed one, the piston will move up slightly in order to accommodate the expansion. This is shown in Figure 7.6(1a) and is represented by state 1a in Figure 7.7. At this state, water is still a sub-cooled liquid, as it has not started to vaporize. If heating process is continued, the temperature will keep rising until it reaches 100°C. At this point the water is still completely a single phase (liquid phase). Further heat addition to the water will start vaporization and a mixture of liquid and vapour will present. A liquid phase that is about to vaporize is called a saturated liquid. This is shown in Figure 7.6(2) and is represented by state 2 in Figure 7.7. State 2 is called a saturated liquid state. Boiling of water starts on further heating. Then vaporization will commence and there is no increase in temperature of the two-phase mixture of liquid and vapour. As soon as the system reaches to point ‘2a’, approximately half of the mass of the water in the cylinder is liquid phase and the remaining half is in the vapour phase. Further heating increases the quantity of vapour and reduces the quantity of liquid. The vaporization process will continue until the last drop of liquid is vaporized. This is shown in Figure 7.6(3) and is represented by state ‘3’ in Figure 7.7. Eventually, at state 3 all of the water in the cylinder is in the vapour phase. Any heat loss from this vapour will cause some of the vapour to condense (change of phase from vapour to liquid) or any further

326

Basic Thermodynamics

Figure 7.7 TV and PV equilibrium diagram for the heating process of water at constant pressures.

Pure Substances

327

heating will cause the temperature to increase, again converting vapour to superheated vapour. A vapour that is about to condense at same temperature or about to increase in its temperature is called a saturated vapour (dry steam or saturated steam). Therefore, state ‘3’ is saturated vapour state. The mixture of liquid phase and vapour phase at states between 2 and 3 is called as a saturated liquid-vapour mixture or wet substance since the liquid and vapour phases co-exist in equilibrium in between these states. The temperature at which a pure substance (in this case water) changes its phase depends on the pressure acting on it. This temperature (corresponding to the line 2 to 3) is called as saturation temperature, TS. The saturation temperature (TS) is defined as the temperature at which a pure substance changes the phase at the stated pressure. Similarly at a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure PS. On further heating the saturated vapour at the same constant pressure increases its temperature above the saturation temperature, TS and also increases in specific volume. Any state to the right of state ‘3’ on constant pressure line is called a superheated vapour state (superheated steam). The temperature of the vapour (steam) above the saturation temperature at a given pressure is called superheated temperature Tsup. The process of heating right to the state 3 on constant pressure line is called a superheating. This is shown in Figure 7.6(4) and is represented by state 3 to 4 in Figure 7.7. Heat is either absorbed or rejected to melt a solid or vaporize a liquid. The amount of energy absorbed or released during a process is called the enthalpy. Sensible heat is the amount of heat required to raise the temperature of water from an initial temperature (T 1 ) to saturation temperature (TS). This is also known as the heat of the liquid or the enthalpy of the liquid, denoted by, hf (Figure 7.8). The amount of heat required to convert saturated liquid at saturated temperature (TS) into saturated vapour (saturated dry vapour) at the same saturated temperate (TS) at given constant pressure (saturated pressure, PS) is called latent heat of evaporation or latent heat Figure 7.8 Temperature–enthalpy diagram of vaporization or enthalpy of vaporization, for a pure substance (water). denoted by hfg. The reverse is called as latent heat of fusion (Figure 7.8). Total heat of evaporation (hg) is the amount of heat required to convert water (sub-cooled) at initial temperature (T1) into saturated vapour (TS) (dry saturated vapour) (Figure 7.8). hg = hf + hfg Enthalpy of wet steam (hw) (mixture of liquid and vapour) is the amount of heat required for the formation of wet steam of dryness fraction ‘X’ at saturated temperature (TS) from water at initial temperature (T1).

328

Basic Thermodynamics

hw = hf + X hfg Enthalpy of superheated steam (hsup) is the total amount of heat required for the formation of superheated vapour (superheated steam) at superheated temperature, Tsup, from the water (subcooled liquid) from the initial temperature of T1. hsup = hg + C pv (Tsup - TS ) = h f + h fg + C pv (Tsup - TS ) where

C pv = Mean specific heat of superheated vapour (Tsup – TS) = q = Degree of superheat

The mixture of saturated liquid-vapour states between 2 to 4 as shown in Figure 7.7 is called wet steam (wet substance). Therefore, wet steam is defined as a mixture of two phases where saturated water molecule and vapour are in thermal equilibrium at the saturation temperature corresponding to a given pressure. As the heating process continues, quantity of vapour increases and quantity of liquid decreases, i.e. quality of the mixture changes. Then the quality of the mixture is called as dryness fraction. The ratio of the mass of dry saturated vapour (dry steam) actually present to the total mass of the wet steam is called dryness fraction. It is denoted by X. X=

7.4

mg Mass of dry steam = Total mass of wet stream (sum of the mass of liquid and vapour) mg + m f

SPECIFIC VOLUME

It is the volume occupied by the unit mass of a substance. It is expressed as m3/kg.

7.4.1 Specific Volume of Saturated Water It is the volume occupied by the 1 kg of saturated water at the saturation temperature (TS) at a given pressure. This is denoted by Vf.

7.4.2 Specific Volume of Dry Steam or Saturated Vapour It is the volume occupied by the 1 kg of saturated vapour at the saturation temperature (TS) at a given pressure. This is denoted by Vg.

7.4.3 Specific Volume of Wet Steam It is the sum of volume of dry steam and the volume of water in the mixture of 1 kg of wet steam. This is denoted by VW. VW = (1 – X)Vf + X Vg m3/kg Since Vf 1. Or

dQ = dW + du =

( P1V1 - P2V2 ) ( P V - P2V2 ) R(T2 - T1 ) + Cv (T2 - T1 ) = 1 1 + n -1 n -1 -1

=

( P1V1 - P2V2 ) ( P2V2 - P1V1 ) ( P1V1 - P2V2 ) ( P1V1 - P2V2 ) + = n -1 -1 n -1 -1

=

( P1V1 - P2V2 ) n -1

=

( P1V1 - P2V2 ) È ( - n) ˘ Í ˙ n -1 Î ( - 1) ˚

=

( - n) ¥ the work done during the polytropic process ( - 1)

È (n - 1) ˘ ( P1V1 - P2V2 ) È ( - 1 - n + 1) ˘ Í1 ˙= Í ˙ - 1) ˚ -1 ( n -1 Î Î ˚

(8.46)

We know that h = u + PV dh = Cp dT h2 – h1 = Cp (T2 – T1) = (P2V2 + u2) – (P1V1 + u1) = (P2V2 – P1V1) + (u2 – u1) = (P2V2 – P1V1) + Cv (T2 – T1)

= ( P2V2 - P1V1 ) +

Cv È C ˘ ( P2V2 - P1V1 ) = ( P2V2 - P1V1 ) Í1 + v ˙ R R˚ Î

È È Cp ˘ Cv ˘ = ( P2V2 - P1V1 ) Í1 + ˙ = ( P2V2 - P1V1 ) Í ˙ C p - Cv ˙˚ ÍÎ ÍÎ C p - Cv ˙˚

Divide and multiply by Cv È ˘ dh = ( P2V2 - P1V1 ) Í ˙ 1 Î ˚ (e) The change in entropy is given by, dQ = dW + du = PdV + Cv dT

Ú

2

1

ds =

Ú

2

1

dQ = T

Ú

2

1

PdV + T

Ú

2

1

Cv

(8.47)

dT T

Ideal Gases and Mixtures of Ideal Gases

s2 - s1 = R

Ú

2

1

dV + Cv V

Ú

2

1

381

dT T

= R ln

V2 T + Cv ln 2 V1 T1

= R ln

V2 PV + Cv ln 2 2 V1 P1V1

= R ln

V2 P V + Cv ln 2 + Cv ln 2 V1 P1 V1

(8.48) È T2 P2V2 ˘ = Í∵ ˙ Î T1 P1V1 ˚

= Cv ln

P2 V + ( R + Cv ) ln 2 P1 V1

= Cv ln

P2 V + C p ln 2 P1 V1

= Cv ln

P2 PT + C p ln 1 2 P1 P2T1

= Cv ln

P2 P T - C p ln 2 + C p ln 2 P1 P1 T1

= C p ln

T2 P - (C p - Cv ) ln 2 T1 P1

= C p ln

T2 P - R ln 2 T1 P1

(8.49) È V2 P1T2 ˘ = Í∵ ˙ Î V1 P2T1 ˚

(8.50)

Polytropic process for different values of n is shown in Figure 2.10.

8.6.6 Specific Internal Energy In 1843, Joule showed that the specific internal energy u of an ideal gas is a function of temperature only. That is, an ideal gas has a definite specific internal energy u at a given temperature irrespective of the pressure.

8.6.7 Specific Enthalpy From the definition of the enthalpy and equation of state of an ideal gas, we have, h = u + PV = u + RT Since u is a function of temperature only and R is a constant, the enthalpy of an ideal gas is also a function of temperature only. Thus, we can write h = f (T)

382

Basic Thermodynamics

8.7 REVERSIBLE ADIABATIC PROCESS From the first and the second laws of thermodynamics, Tds = du + PdV From Eq. (8.13) [reversible adiabatic, \

Tds = Cv dT + PdV \

Q = 0]

(8.51)

0 = Cv dT = PdV

Differential form of ideal gas equation PV = RT dT =

is PdV + VdP = RdT

( PdV + VdP ) R

(8.52)

Substitute Eq. (8.52) in Eq. (8.51), 0 = Cv

Multiply the above equation by 0 = Cv

( PdV + VdP ) + PdV R

R PV

( PdV + VdP ) dV +R PV V

= Cv

dV dP dV dV È dV dP ˘ + Cv +R = Cv Í + + (C p - Cv ) ˙ V P V P ˚ V ÎV

= Cv

dP dV + Cp P V

Divide by Cv,

C p dV dP + =0 Cv V P dV dP + =0 V P

Integrating the above equation we get, g ln V + ln P = C1 ln PVg = C1 PV = eC1 = C This equation is the governing equation for the isentropic process. \

P1V1 = P2V2 = C

(8.53)

Ideal Gases and Mixtures of Ideal Gases

383

From the first law of thermodynamics, we have dQ = dW + dU dW = –dU = – Cv dT

(Isentropic process, Q = 0)

Pdv = – CvdT R dV dT =Cv V T

[∵ PV = RT]

È R ˘ = ( - 1) ˙ Í∵ Î Cv ˚ On intergration between the initial and final states, we get ( - 1)

dV dT =V T

ln

T2 V = ( - 1) ln 2 T1 V1 -1

T2 È V2 ˘ \ =Í ˙ T1 Î V1 ˚ By using ideal gas equation, we can write

From Eqs. (8.53a) and (8.53b), we get

From Eqs. (8.53a) and (8.53c), we get

(8.53a)

V2 P1T2 = V1 P2 T1

T2 È P2 ˘ =Í ˙ T1 Î P1 ˚

V1 È P2 ˘ =Í ˙ V2 Î P1 ˚

(8.53b) -1

(8.53c)

1

P1V1 = P2V2 PVg = Constant = C

(8.53d)

Equations (8.53) and (8.53d) are one and the same.

8.8

OTHER RELATIONS FOR AN ISENTROPIC PROCESS (a) Relation between T and P From Eq. (8.53), P1V1 = P2V2

V1 È P2 ˘ =Í ˙ V2 Î P1 ˚

1

(8.54a)

384

Basic Thermodynamics

P2 È V1 ˘ =Í ˙ P1 Î V2 ˚

(8.54b)

P1V1 P2V2 = T1 T2 V1 P2T1 = V2 P1T2

(8.55a)

P2 V1T2 = P1 V2T1

(8.55b)

Equating Eqs.(8.54a) and (8.55a)

P2T1 È P2 ˘ =Í ˙ P1T2 Î P1 ˚ T1 È P2 ˘ =Í ˙ T2 Î P1 ˚

1

1-

(8.56)

(b) Relation between T and V Equating Eqs. (8.54b) and (8.55b) V1T2 È V1 ˘ =Í ˙ V2T1 Î V2 ˚

T2 È V1 ˘ =Í ˙ T1 Î V2 ˚

8.9

-1

ANALYSIS OF THE STEADY FLOW PROCESS (OPEN SYSTEM)

For a reversible process, heat and work transfer is given by dq = dh – VdP dW = –VdP – d(K.E.) – d(P.E.)

Ú

W = - VdP - DK.E. - DP.E. 1. Constant volume process (isochoric process) [Figure 8.7(a)] (the specific volume at the inlet is equal to the specific volume at the exit) (a) The change in the internal energy is given by, Du = Cv dT = Cv (T2 – T1) (b) The change in the enthalpy is given by, Dh = Cp dT = Cp (T2 – T1)

(8.57)

Ideal Gases and Mixtures of Ideal Gases

385

(c) The work done is given by,

Ú

W.D. = - VdP - DK.E. - DP.E. Neglecting the P.E. and K.E. changes, W.D. = -

Ú

2

1

VdP

= –V(P2 – P1) = V(P1 – P2)

(8.58)

(d) The heat transfer is given by, dq = dh – VdP

Ú

q = Dh -

2

1

VdP

= Cp(T2 – T1) – V(P1 – P2)

(8.59)

(e) The change in the entropy is given by,

Ú

2

1

Ú

2

Ú

2

ds =

s2 - s1 =

1

1

dq T dh T

= C p ln

Ú

2

1

V

dP = T

Ú

2

1

Cp

dT T

Ú

2

1

R

dP P

T2 P - R ln 2 T1 P1

(8.60)

2. Constant pressure process (isobaric process) [Figure 8.7(b)] (a) The change in the internal energy is given by, Du = Cv dT = Cv (T2 – T1) (b) The change in the enthalpy is given by, Dh = Cp dT = Cp (T2 – T1) (c) The work done is given by,

Ú

W.D. = - VdP - DK.E. - DP.E. Neglecting P.E. and K.E. changes W.D. = -

Ú

2

1

VdP = V ( P1 - P2 ) = 0

[∵ P1 = P2]

(8.61)

(d) The heat transfer is given by, q = Dh -

Ú

2

1

VdP

= Cp(T2 – T1) – 0 = Cp(T2 – T1)

(8.62)

386

Basic Thermodynamics

(e) The change in the entropy is given by,

Ú

2

1

ds =

Ú

dq = T

2

1

Ú

2

1

dh = T

s2 - s1 = C p ln

Ú

2

1

Cp

dT T

T2 T1

(8.63)

3. Constant temperature process (isothermal process) [Figure 8.7(c)] (a) The change in the internal energy is given by, Du = Cv dT = Cv (T2 – T1) = 0 [∵ T1 = T2] (b) The change in the enthalpy is given by, Dh = Cp dT = Cp(T2 – T1) = 0 (c) The work done is given by,

Ú

W.D. = - VdP - DK.E. - DP.E. Neglecting P.E. and K.E. changes,

W.D. = -

2

Ú

1

VdP = -

= - P1V ln = P1V1 ln

Ú

2

1

C

dP P

C˘ È Í∵ PV = C , V = P ˙ Î ˚

P2 P1

P1 V V = P1V1 ln 2 = RT1 ln 2 P2 V1 V1

(8.64)

(d) The heat transfer is given by, dq = dh – VdP

Ú

q = Dh - VdP q = - RT1 ln

[∵ Dh = 0]

V2 V1

(8.65)

\ Heat transfer = Work done (e) The change in the entropy is given by,

Ú

2

1

ds =

Ú

dq =T

2

1

=-

Ú

2

1

R

Ú

2

1

V

dP T

P dP = - R ln 2 P P1

È V R˘ Í∵ T = P ˙ Î ˚

4. Isentropic process [Figure 8.7(d)], PVg = C (a) The change in the internal energy is given by, Du = Cv dT = Cv (T2 – T1)

(8.66)

Ideal Gases and Mixtures of Ideal Gases

387

(b) The change in the enthalpy is given by, Dh = Cp dT = Cp(T2 – T1) (c) The work done is given by,

Ú

W.D. = - VdP - DK.E. - DP.E. Neglecting P.E. and K.E. changes, W.D. = -

Ú

2

1

Ú

VdP = -

2ÈC ˘

1˘ È C Ê ˆ Í∵ PV = C, V = ˙ ÁË P ˜¯ ˙ Í ÍÎ ˚˙

1

Í P ˙ dP Î ˚

1

1 1 1˘ È - 1 +1 - +1 Í P1 ˙ 1 È - 1 +1 ˘ ( ) ( ) P V P P V 1 1 2 2 2 - +1 1 Í ˙ ÍP ˙ - P1 Î ˚ 2 Í ˙ = -C = 1 -1 Í ˙ - +1 ÍÎ ˙˚

= ( P1V1 - P2V2 ) = Cp

( - 1)

(T1 - T2 ) ( - 1)

= R

(T1 - T2 ) ( - 1)

È R ˘ Í∵ C p = ˙ ( - 1) ˚ Î

(d) The heat transfer is given by, dq = dh – VdP = Cp (T2 – T1) + Cp (T1 – T2) = Cp (T2 – T1) – Cp (T2 – T1) = 0 (e) The change in the entropy is given by,

Ú

2

1

ds =

5. Polytropic process [Figure 8.7(e)],

Ú

2

1

PVn

dq =0 T

=C

(a) The change in the internal energy is given by, Du = Cv dT = Cv (T2 – T1) (b) The change in the enthalpy is given by, Dh = Cp dT = Cp (T2 – T1) (c) The work done is given by,

Ú

W.D. = - VdP - DK.E. - DP.E.

(8.67)

388

Basic Thermodynamics

Neglecting P.E. and K.E. changes, W.D. = -

Ú

2

1

VdP = -

Ú

1˘ È Í∵ PV n = C , V = Ê C ˆ n ˙ Í ËÁ P ¯˜ ˙ ÍÎ ˙˚

1 2 ÈC ˘n

1

Í P ˙ dP Î ˚

1 1 1˘ È - 1 +1 - +1 n n n n n n 1 1 Í ˙n P ( P V ) P ( P V ) È ˘ 1 1 2 2 1 2 - +1 - +1 1 ˙˚ Í P2 n - P1 n ˙ ÍÎ = -Cn Í ˙= 1 n -1 Í ˙ - +1 n ÎÍ ˚˙

= ( P1V1 - P2V2 ) = Cp

(T - T2 ) n = nR 1 (n - 1) (n - 1)

È nR ˘ Í∵ C p = ˙ (n - 1) ˚ Î

(T1 - T2 ) (n - 1)

(8.68)

(d) The heat transfer is given by, dq = dh – VdP = C p (T2 - T1 ) + nR

È (T2 - T1 ) nR ˘ = (T2 - T1 ) ÍC p ˙ (n - 1) n - 1˚ Î

È Ï ( - 1) ¸˘ = (T2 - T1 ) ÍC p - ÌnCv ˝˙ (n - 1) ˛˚ Ó Î

[∵ R = Cv(g – 1)] P

P

P 2

P=C

1

2

1 T=C

W.D.

W.D.

2

1

V

(a)

V

(b)

P

(c)

V

P 1

1

g

PV = C

g

PV = C

W.D.

W.D. 2

(d)

Figure 8.7

(8.69)

2

V

(e)

V

PV diagram for flow processes: (a) Constant volume, (b) Constant pressure, (c) Isothermal, (d) Isentropic, (e) Polytropic process.

Ideal Gases and Mixtures of Ideal Gases

389

8.10 IDEAL GAS MIXTURE Mixtures of substances are commonly encountered in systems and processes of interest to the engineers [a mixture of two phases of the same substance, such as a mixture of liquid water and water vapour, pure substance has already been discussed]. If the mixture remains homogeneous in composition throughout the process, it may itself be considered as a pure substance. The mixtures of pure substances may be classified broadly into three groups. I. Mixtures consisting only of gases are considered as ideal gases. Example: dry air (H2, N2, O2, etc.) II. Mixtures consisting of ideal gases and vapour. Example: moist air (dry air + water vapour) III. Mixtures consisting of two phases of the same substance. Example: liquid water and water vapour The third type has already been discussed in the chapter ‘pure substances’. The first and second types are dealt in this chapter. When two or more ideal gases are mixed, the behaviour of a molecule normally is not influenced by the presence of other similar or dissimilar molecules, and therefore a non-reacting mixture of ideal gases also behaves as an ideal gas. Therefore, the basic thermodynamic principles presented so far are valid for non-reacting gas mixtures. Property tables are available for a mixture having fixed composition, for example air, water, etc. If the possible mixture composition is infinite, tabulation of the thermodynamic properties of such mixtures would be impractical. The alternative is to develop a procedure for predicting the mixture properties on the basis of the composition of the mixtures and the thermodynamic properties of the individual components of the mixtures.

8.11 PVT BEHAVIOUR OF AN IDEAL GAS MIXTURE 8.11.1

Mass Fraction

Consider a gas mixture as shown in Figure (8.8) composed of j components. The mass of the mixture (total mass of the mixture) m is the sum of the masses of the individual components. j

m = m1 + m2 + m3 +

+ mi =

Âm

(8.70)

i

i =1

A+B m1 + m2 = m (j = 2)

Figure 8.8

=

A m1

B m2

Schematic representation of mass fraction.

390

Basic Thermodynamics

The ratio of the mass of a component to the mass of the mixture is called the mass fraction, mf.

mi m The ratio (mi/m) is defined as the mass fraction of the ith component in the mixture. m fi =

j

i

mf =

\

(8.71)

mi

Âm = Â m fi

i =1

i =1

if j = 2, 2

\

Â

mf =

i =1

2

m fi =

mi

m1

Âm = m + i =1

2

mf =

or

Âm i =1

8.11.2

fi

= m f1 + m f1 =

m2 m1 + m2 m = = =1 m m m

(8.72)

m1 m2 m1 + m2 m + = = =1 m m m m

Mole Fraction

Consider a gas mixture as shown in Figure 8.9 composed of j components. The mole of the mixture (total mole of the mixture) n is the sum of the moles of the individual components. A+B n1 + n2 = n (j = 2)

Figure 8.9

A n1

=

B n2

+

Schematic representation of mole fraction. j

n1 = n1 + n2 + n3 +

+ ni =

Ân

(8.73)

i

i =1

2

if j = 2,

Ân = n i

1

+ n2

j =1

The ratio of the mole of a component to the mole of the mixture is called the mole fraction, nf.

ni n The ratio (ni/n) is defined as the mole fraction of the ith component in the mixture. \

n f1 =

(8.74)

j

n f = n f1 + n f2 + n f3 +

+ nfj =

Ân i =1

fi

=1

(8.75)

Ideal Gases and Mixtures of Ideal Gases

391

For the above example, if j = 2, 2

nf =

Ân i =1

= n f1 + n f2 =

fi

2

Â

or

i =1

2

n fi =

ni

n1 n2 n1 + n2 + = =1 n n n

n1

Ân = n + i =1

n2 n1 + n2 = =1 n n

The composition of the mixture of gases is usually specified either by the number of moles of each component in the mixture (a molar analysis) or by the mass of each component in the mixture (a gravimetric analysis).

8.11.3

Molecular Weight of the Mixture of Ideal Gases

The mass of a component is related to the number of moles and the molecular weight by the relationship m = nM ; for ith component (8.76) mi = niMi The molecular weight (M, apparent molecular weight) of a mixture is defined as the total mass of the mixture (m) divided by the total number of moles (n) in the mixture. j

m M= = n

Â

j

mi

i =1

n

Ân M i

=

i

i =1

j

=

n

ni Mi = n i =1

Â

j

Ân i =1

fi M i

(8.77)

R (8.78) M T, P where R = Apparent gas constant for the mixture m, n, V M = Apparent molecular weight of the mixture Gases 1, 2, etc. Figure 8.10 shows the homogeneous non-reacting mixture of ideal gases 1, 2, etc. at temperature T, pressure P and volume V. Figure 8.10 A mixture of Let n1, n2, etc. and m1, m2 etc. be their moles and masses of the ideal gases. individual gases, respectively. In the same manner we can write,

8.11.4

R=

Dalton’s Law of Partial Pressure or Dalton’s Law of Additive Pressure

It states that the total pressure (P) exerted by the mixture of gases is equal to the sum of the pressures exerted by each component of the mixture, if each component exists separately at the same temperature and volume as the mixture. If this component pressure is denoted by Pi, Dalton’s law can be written as, j

P=

ÂP

i

i =1

392

Basic Thermodynamics

Gas 1, Gas 2 T, P, V m, n

Gas 1 T, P1, V m 1, n 1

=

Gas 2 T, P2, V m2 , n 2

+

Figure 8.11 Schematic representation of Dalton’s law of additive pressure.

Figure 8.11 shows a schematic representation of Dalton’s law of additive pressures. Here, mixture is also assumed to behave as an ideal gas. The two ideal gases 1 and 2, both at absolute temperature T and volume V are equal to the temperature and volume of the mixture. Therefore, the final pressure of the mixture is the sum of partial pressures of individual gases. j

\

P = P1 + P2 +

+ Pi =

ÂP

(8.79)

i

i =1

This can be proved as below. Apply ideal gas equation to the mixture as well as to the individual components. Gas 1

Gas 2

Mixture

P1V1 = n1 RT1

P2V2 = n2 RT2

PV = nRT

PV n1 = 1 1 RT1

PV n2 = 2 2 RT2

n=

PV RT

We know that n = n1 + n2 ; V1 = V2 = V ; PV P1V1 P2V2 = + RT RT1 RT2

T1 = T2 = T

j

\

P = P1 + P2 +

+ Pi =

ÂP

i

i =1

Hence the proof.

8.11.5

Amagat’s Law of Partial Volume or Amagat’s Law of Additive Volume

It states that total volume (V) of a mixture of gases is equal to the sum of the volume, each component would occupy if each component exists separately at the same temperature and pressure as the mixture. If this component is denoted by Vi, Amagat’s law can be written as, j

V=

ÂV

i

i =1

Figure 8.12 shows the schematic representation of Amagat’s law of additive volume. Here, mixture is also assumed to behave as an ideal gas. The two ideal gases (gas 1, gas 2) both at temperature T and absolute pressure P are equal to the temperature and pressure of the mixture. Then the final volume of the mixture is the sum of the partial volume of the individual gases.

Ideal Gases and Mixtures of Ideal Gases

T, P, V Gas 1, Gas 2 m, n

Gas 1 T, P, V1 m1, n1

=

393

Gas 2 T, P, V1 m2 , n 2

+

Figure 8.12 Schematic representation of Amagat’s law of additive volume. j

V = V1 + V2 +

+ Vi =

ÂV

(8.80)

i

i =1

This can be proved as below. Gas 1

Gas 2

Mixture

P1V1 = n1 RT1

P2V2 = n2 RT2

PV = nRT

PV n1 = 1 1 RT1

PV n2 = 2 2 RT2

n=

PV RT

We know that n = n1 + n2 ;

P1 = P2 = P ;

T1 = T2 = T

PV P1V1 P2V2 = + RT RT1 RT2 j

V = V1 + V2 +

+ Vi =

ÂV

i

i =1

Hence the proof. For Figure 8.12, j = 2, then we can write, P2V = n2RT; P1V = n1RT;

P1V n1 RT = ; PV nRT

P2V n2 RT = ; PV nRT

P1 = n f1 ; P \

P1 P2 + + P P

P2 = n f2 ; P +

Pj P

PV = nRT

PV n RT i = i PV nRT Pi = n fi P

j

=

Pi = n f1 + n f2 + P i =1

Â

(8.81)

j

+ nfj =

Ân i =1

fi

\ Partial pressure ratio = mole fraction For Figure 8.13, j = 2, then we can write,

P1V1 = nRT1 ;

P2V2 = nRT2 ;

PV1 n1 RT ; = PV nRT

PV2 n2 RT = PV nRT

V1 n1 = ; V n

V2 n2 = V n

PV = nRT

394

Basic Thermodynamics

\

Volume fraction = mole fraction

Vi ni = = n fi V n j

Vj

j

Vi + = = n f1 + n f2 + V i =1 V

V1 V2 + + V V

\

(8.82)

Â

+ n f1 =

Ân i =1

fi

Comparing Eqs. (8.81) and (8.82), we can write

Pi Vi ni = = = n fi P V n

8.11.6

(8.83)

Characteristic Gas Constant of the Mixture of an Ideal Gas

The characteristic gas constant (apparent gas constant), R, and molecular weight, M, for the mixture can be determined by the following relation. m n If the molecular analysis of the mixture and the molecular weight of each gas in the mixture is given, then M=

From Eq. (8.76),

j

m M= = n

Â

j

i i

i =1

i =1

=

n

j

M=

ÂM n

mi

ni Mi = n i =1

Â

n j

Ân i =1

fi M i

=

m

1 j

mi mM i i =1

Â

=

j

mi

Ân  M i

i =1

=

m

=

j

i =1

i

1 j

(8.84)

m fi

ÂM i =1

i

We know that

R=

R M

(8.85)

Substitute Eq. (8.84) in Eq. (8.85) R= j

R =R 1 m fi

ÂM i =1

i

j

m fi

j

 M = Âm i =1

i

i =1

fi

R = Mi

j

Âm i =1

fi Ri

(8.86)

Ideal Gases and Mixtures of Ideal Gases

395

8.11.7 Internal Energy and Constant Volume Specific Heat of an Ideal Gas Mixture All extensive properties of a non-reacting gas mixture can be obtained by summing the contributions of individual gases of the mixture. For example, the internal energy of the mixture can be written as, j

U = U1 + U 2 +

+Uj =

ÂU

i

(8.87)

fi ui

(8.88)

i =1

Ui is the internal energy of ith gas. (a) Internal energy on mass basis U = m u; Ui = mi ui j

U=

Âm u

i i

i =1

u=

U = m

j

mi ui = m i =1

Â

j

Âm i =1

(b) Internal energy on molar basis j

u=

Ân i =1

fi ui

(8.89)

(c) Constant specific heat at constant volume on mass basis

È ∂u ˘ Cv = Í ˙ = Î ∂t ˚ v

j

È ∂u ˘ m fi Í i ˙ = Î ∂t ˚ v i =1

Â

j

Âm i =1

fi Cvi

(8.90)

fi Cvi

(8.91)

(d) Similarly on molar basis

È ∂u ˘ Cv = Í ˙ = Î ∂t ˚ v

8.11.8 Let

j

È ∂u ˘ n fi Í i ˙ = Î ∂t ˚ v i =1

Â

j

Ân i =1

Enthalpy and Constant Pressure Specific Heat of an Ideal Gas Mixture H = Enthalpy of the mixture U = Internal energy of the mixture P = Pressure of the mixture V = Total volume of the mixture

We know that H = U + Pu ; j

 i =1

Hi = Ui + PiVi

j

Hi =

 i =1

j

Ui + V

ÂP

i

i =1

396

Basic Thermodynamics

(a) Enthalpy on mass basis H = mh;

Hi = mihi j

H = H1 + H2 +

+ Hj =

Â

j

Hi =

i =1

Âm h

i i

(8.92)

i =1

j

\

H h= = m

Âm h

i i

j

i =1

m

j

mi hi = m i =1

Â

=

Âm i =1

fi hi

(8.93)

(b) Enthalpy on molar basis j

h=

Âm i =1

fi hi

(8.94)

(c) Constant specific heat at constant pressure on mass basis From Eq. (8.15) we have constant pressure specific heat of an ideal gas mixture,

È ∂h ˘ Cp = Í ˙ = Î ∂t ˚ p

j

È ∂h ˘ m fi Í i ˙ = Î ∂t ˚ p i =1

Â

j

Âm i =1

fi C pi

(8.95)

(d) Similarly on molar basis

È ∂h ˘ Cp = Í ˙ = Î ∂t ˚ p

8.11.9

j

È ∂h ˘ m fi Í i ˙ = Î ∂t ˚ p i =1

Â

j

Âm i =1

fi C pi

(8.96)

Entropy of an Ideal Gas Mixture j

s = s1 + s2 +

+ sj =

Âs

i

i =1

S = ms;

Si = mi s i

j

S=

Âm s

i i

i =1

S s= = m

j

mi si m i =1

Â

j

s=

Âm i =1

fi si

on mass basis

(8.97)

j

s=

Âm i =1

fi si

on molal basis

(8.98)

Ideal Gases and Mixtures of Ideal Gases

397

8.11.10 Change in Entropy Change in entropy is given by (a) On mass basis dTi dP - Ri i Ti Pi

dsi = C pi s2 - s1 =

Ú

2

1

ds =

Âm

fi

C pi

Ú

2

1

dTi Ti

Âm

fi

Ri

Ú

2

1

dPi Pi

(8.99)

For ith gas (component)

ÈT ˘ ÈP ˘ (s2 - s1 )i = s2i - s1i = m fi C pi ln Í 2 ˙ - m fi Ri ln Í 2 ˙ T Î 1 ˚i Î P1 ˚i

(8.100)

(b) Similarly on molar basis s2 - s1 =

Ú

2

1

ds =

Â

n fi C pi

Ú

2

1

dTi Ti

Â

n fi Ri

Ú

2

1

dPi Pi

ÈT ˘ ÈP ˘ ( s2 - s1 )i = s2i - s2i = n fi C pi ln Í 2 ˙ - n fi Ri ln Í 2 ˙ Î T1 ˚i Î P1 ˚i

(8.101)

SOLVED EXAMPLES EXAMPLE 8.1 A gas at 700 kPa absolute pressure, 2.0 m3, expands at constant temperature to a volume of 12 m3. Find the pressure of the gas after the expansion. Solution: P1 = 700 kPa,

V1 = 2.0 m3,

P2 = ?,

V2 = 12 m3,

T1 = T2,

P1V1 = P2V2

Final pressure (P2) \

P2 =

P1V1 700 kPa ¥ 2 m 3 = = 116.67 kPa V2 12 m 3

EXAMPLE 8.2 Heat is added at constant pressure to a gas at 50°C, 0.2 m3. The final temperature of the gas is 250°C. Find the final volume and increase in volume during the supply of heat. Solution: T1 = 50 + 273 = 323 K, We know that

V1 = 0.2 m3

T2 = 250 + 273 = 523 K, P1 = P2 P1V1 P2V2 = T1 T2

i.e.

V1 V2 = T1 T2

398

Basic Thermodynamics

(a) Final volume of the gas (V2) V2 =

V1T2 0.2 ¥ 523 = = 0.324 m 3 T1 323

(b) Increase in volume during the supply of heat (dV) dV = V2 – V1 = 0.324 – 0.2 = 0.124 m3 EXAMPLE 8.3 The density of a gas at NTP is 1.5 kg/m3. Find (a) Characteristic gas constant (b) Molecular weight (c) Specific volume of the gas at 15 bar and 40°C. Solution: NTP = 0°C, We know that P = rRT

\

760 mm Hg (101.325 kPa)

R=

P kN m2 1 kJ = 101.325 2 ¥ ¥ = 0.2474 T 1.5 kg 273 K kg K m

EXAMPLE 8.4 50°C air is contained in a spherical vessel of 2 m diameter. The vessel is evacuated till it becomes 70 cm of Hg. During the process, the temperature remains constant. Find (a) The mass pumped out (b) The pressure in the tank in cm of Hg, if the tank is cooled to 5°C Assume atmospheric pressure of 760 mm Hg. Solution: (a) The mass pumped out (m) m1 = Initial mass =

P1V1 RT1

4 3 4 r = ¥ ¥ 13 = 4.1888 m 3 3 3 P2 = Final pressure in the tank = 76 – 70 = 6 cm of Hg

V1 = Volume of the spherical vessel = 6 ¥ 101.325 = 7.999 kPa 76 101.325 ¥ 4.1888 m1 = = 4.578 kg 0.287 ¥ (50 + 273) =

\

P2V2 7.999 ¥ 4.1888 = = 0.3614 kg RT2 0.287(50 + 273) m = m1 – m2 = 4.578 – 0.3614 = 4.2166 kg

m2 = Final mass =

\

(b) The pressure (P3) in the tank in cm of Hg, if the tank is cooled to 5°C P3 = ? P2 = 7.999 kPa T3 = 5 + 273 = 278 K T2 = 50 + 273 = 323 K

Ideal Gases and Mixtures of Ideal Gases

\ \

P2V2 P3V3 = T2 T3 P2 =

399

But V2 = V3

P3T2 273 = ¥ 7.999 = 6.885 kPa T3 323

EXAMPLE 8.5 An ideal gas is heated from 20°C to 100°C. Assume 1 kg of mass, R = 0.2 kJ/kg K, g = 1.2. Find, (a) Specific heats (b) Change in internal energy (c) Change in enthalpy. Solution: (a) Specific heats (Cp, Cv) We know that

Cp

Cp – Cv = R \

Cp – Cv = 0.2

\

1.2Cv – Cv = 0.2

Cv Cp Cv

= = 1.2

Cv = 1 kJ/kg K

Cp = 1.2 × Cv = 1.2 × 1 = 1.2 kJ/kg K (b) Change in internal energy (du) du = mCv (T2 - T1 ) = 1 kg ¥ 1

kJ ¥ (100 - 20) K = 80 kJ kg K

(c) Change in enthalpy (dh) dh = mC p (T2 - T1 ) = 1 kg ¥ 1.2

kJ ¥ (100 - 20) K = 96 kJ kg K

EXAMPLE 8.6 A gas undergoes polytropic process from 100 N/m2 and 120°C to 20 N/m2, –20°C. Calculate polytropic index n. Solution:

Polytropic index (n) P1 = 100 N/m2

P2 = 20 N/m2

T1 = 120 + 273 = 393K

T2 = –20 + 273 = 253 K

We have

T1 È P1 ˘ =Í ˙ T2 Î P2 ˚

n -1 n

393 È 100 ˘ = = 253 ÍÎ 20 ˙˚

n -1 n

400

Basic Thermodynamics

1.553 = [5]

ln 1.553 =

n -1 n

n -1 ln 5 n

n = 1.3765

EXAMPLE 8.7 A 0.5 m3 capacity of oxygen at a pressure of 15 bar abs is stirred by internal paddle wheel until the pressure becomes 30 bar abs. The vessel is insulated. Find, (a) Change in internal energy (b) Work input (c) Heat transferred Assume Cv = 0.65 kJ/kg K; R = 0.265 kJ/kg K; m = 1 kg Solution: P1 = 15 bar

V1 = 0.5 m3

m = 1 kg

P2 = 30 bar

T1 =

P1V1 15 × 100 kN × 0.5 m 3 ×1 kg K = = 2830.2 K mR 1 kg × m 2 × 0.265 kN m

T2 =

P2V2 30 ¥ 100 ¥ 0.5 ¥ 1 = = 5660.37 K mR 1 ¥ 0.265

V2 = V1

(a) Change in internal energy (du) du = Cv (T2 - T1 ) = 0.65

kJ kJ ¥ (5660.37 - 2830.2) K = 1839.6 kg K kg

(b) Work input (W) W = du = 1839.6

kJ kg

(c) Heat transferred (Q) Q=0

[∵ insulated]

EXAMPLE 8.8 2 m3 of gas is heated at constant pressure from 30°C to 200°C. Find, (a) Characteristic gas constant (b) Ratio of specific heats (c) Heat added (d) Work done (e) Change in internal energy (f) Final volume (g) Initial pressure Assume: Cp = 0.98 kJ/kg K Cv = 0.714 kJ/kg K m = 1 kg Solution: (a) Characteristic gas constant (R) R = C p - Cv = 0.98 - 0.714 = 0.266

kJ kg K

Ideal Gases and Mixtures of Ideal Gases

401

(b) Ratio of specific heats (g)

= (c) Heat added (Q)

Cp Cv

Q = C p (T2 - T1 ) = 0.98

=

0.98 = 1.373 0.714

kJ kJ ¥ (200 - 30) K = 166.6 kg K kg

(d) Work done (W) W = Q - Du = Q - Cv (T2 - T1 ) = 166.6 - 0.714 (200 - 30) = 45.22

kJ kg

Du) (e) Change in internal energy (D Du = Cv (T2 - T1 ) = 0.714 (200 - 30) = 121.38

kJ kg

(f) Final volume (V2) P1V1 P2V2 = T1 T2

\

V2 =

[∵ Constant pressure]

T2V1 (200 + 273) = ¥ 2 = 3.122 m 3 T2 (30 + 273)

(g) Initial pressure (P1) mRT1 1 kg × 0.266 kN m × 303 K kN P1 = = = 40.299 2 3 V1 2m m EXAMPLE 8.9 2 kg of O2 from a pressure of 1 bar, 60°C is compressed to a final pressure of 5 bar along with a polytropic path for which PV1.3 = C. Calculate, (a) The heat transferred (b) The change of entropy Assume: R = 0.280 kJ/kg K Cp = 0.98 kJ/kg K. Solution: m = 2 kg P1 = 1 bar T1 = 60 + 273 = 333 K (a) The heat transferred (Q)

P2 = 5 bar

Cv = C p - R = 0.98 - 0.28 = 0.7 kJ/kg K

=

Cp Cv

=

0.98 = 1.4 0.70

ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚ Q = mCv

n -1 n

È5˘ = 333 Í ˙ Î1˚

( - n) (T2 - T1 ) (1 - n)

1.3 - 1 1.3

= 482.78 K

PV1.3 = C

402

Basic Thermodynamics

= 2 kg ¥ 0.7

kJ (1.4 - 1.3) ¥ (482.78 - 333) K = - 69.897 kJ kg K (1 - 1.3)

Or from the first law of thermodynamics W=

R(T2 - T1 ) 0.28(482.78 - 333) kJ = = 139.79 n -1 1.3 - 1 kg

Du = Cv (T2 - T1 ) = 0.7(482.78 - 333) = 104.85

kJ kg

Q = m[W + Du] = 2[–139.79 + 104.85] = – 69.88 kJ (b) The change of entropy (DS) DS = S2 - S1 = mCv

= 2 ¥ 0.7 ¥

T ( - n) ¥ ln 2 (1 - n) T1

(1.4 - 1.3) 482.78 ¥ ln = 0.1733 kJ (1 - 1.3) 333

EXAMPLE 8.10 A container contains 5 kg of an ideal gas, initially at a temperature of 100°C. A heat of 600 kJ is added to the gas. The gas then undergoes a reversible isobaric process. Find the final temperature of the gas. Assume: R = 0.28 kJ/kg K g = 1.2 Solution: Cv =

R 0.28 = = 1.4 kJ/kg K - 1 1.2 - 1

Cp = R + Cv = 0.28 +1.4 = 1.68 kJ/kg K Final temperature (T2) Q = m Cp(T2 – T1) 60 kJ = 5 kg ¥ 1.68 T2 = 380.14 K

kJ ¥ (T2 - 373) K kg K

EXAMPLE 8.11 An ideal gas initially at 80°C and 1 bar undergoes the following reversible processes. (a) The gas is compressed adiabatically to 120°C (1–2). (b) It is then cooled at constant pressure from 120°C to 80°C (2–3). (c) Finally the gas is expanded at constant temperature to a final pressure of 1 bar (3–1). Assume:

Cv =

5 kJ R 3 kg mole K

Determine, (a) Dh, (b) Du, (c) W and (d) q for the entire cycle on per kg mole basis.

Ideal Gases and Mixtures of Ideal Gases

403

Solution:

Figure E8.11(a)

R = Universal gas constant = 8.314 C p = R + Cv = R +

kJ kg mole K

5 8 8 kJ R = R = ¥ 8.314 = 22.171 3 3 3 kg mole K

Cv = C p - R = 22.171 - 8.314 = 13.86

=

Cp Cv

=

kJ kg mole K

22.171 = 1.599 13.86

T 2

(i) For a reversible adiabatic compression process q1–2 = 0 Du1- 2 = - W1- 2 = C v DT = 13.86 (393 - 353)

= 554.4

kJ kg mole K

1

3

[Work done on the system]

Dh1- 2 = C p DT = 22.171(393 - 353) = 886.84

Figure E8.11(b)

kJ kg mole K

(ii) For a constant pressure process q2 -3 = Dh2 -3 = C p DT = C p (T3 - T2 ) = - 886.84 Du2 -3 = Cv DT = Cv (T3 - T2 ) = - 554.4 From the first law of thermodynamics, q2–3 = w + Du

kJ kg mole K

kJ kg mole K

S

404

Basic Thermodynamics

w2 -3 = q2 -3 - Du = - 886.84 - ( -544.4) = - 332.44

kJ kg mole K

(iii) For an isothermal expansion process Du3–1 = 0

[∵ Temperature constant]

Dh3–1 = 0 q3 -1 = W3-1 = RT ln

ÈT ˘ P2 = P1 Í 2 ˙ Î T1 ˚ \

-1

P3 P = RT ln 2 P1 P1

È 393 ˘ = 1Í ˙ Î 353 ˚

1.6 - 1 1.6

= 1.33 bar

q3-1 = 8.314 ¥ 353 ¥ ln 1.33 = 836.96

(a) Cyclic integral of heat transfer

Úq=q

(b) Cyclic integral of work transfer 1- 2

kJ kg mole K

( Ú q)

+ q2 -3 + q3-1 = 0 + ( -886.84) + 836.96 = - 49.88

1- 2

Ú w =W

[∵ P3 = P2]

kJ kg mole K

( Ú w)

+ W2 -3 + W3-1 = - 554.4 - 332.44 + 836.96 = - 49.88

kJ kg mole K

Du) (c) Internal energy of the total cycle (D Du = Du1–2 + Du2–3 + Du3–1 = 554.4 – 554.4 + 0 = 0 Dh) (d) Enthalpy of the cycle (D Dh = Dh1–2 + Dh2–3 + Dh3–1 = 886.84 – 886.84 + 0 = 0 Since the initial and final states are same (cyclic process), Du and Dh are zero. As per the first law of thermodynamics, EXAMPLE 8.12 Solution:

Ú q = Ú w, hence proved.

Re-calculate Example 8.11 with an efficiency of 70% for each process.

Process 1–2 is no longer reversible adiabatic.

(a) W1-2a =

W1- 2 rev 0.7

=

-554.4 kJ = - 792 0.7 kg mole K

q1–2a = w1–2a + Du1–2 = –792 + 554.4 = –237.6 (b) W2 -3a =

W1-2 rev 0.7

=

-332.44 kJ = - 474.9 0.7 kg mole K

[∵ Du1–2 from Example 8.11]

405

Ideal Gases and Mixtures of Ideal Gases

q2 -3a = w2 -3a + Du2 -3 = - 474.9 - 554.44 = - 1029.34 (c) w3-1a = (w3-1 rev ) = 836.96 ¥ 0.7 = 585.87

kJ kg mole K

q3-1a = w3-1a + Du3-1 = 585.87 + 0 = 585.87 For the whole cycle, Du = 0, Dh = 0 \ (a) Cyclic integral of heat transfer

Úq

a

a

= q1- 2a + q2 -3a + q3-1a = - 237.6 - 1029.34 + 585.87 = - 681.07

Processes

a

Reversible Du 554.4 –554.4 0 0

Dh 886.84 –886.8 0 0

kJ kg mole K

(ÚW )

= W1- 2a + W2 -3a + W3-1a = - 792 - 474.9 + 585.87 = - 681.03

a

1–2 2–3 3–1 Cycle

kJ kg mole K

(Ú q )

(b) Cyclic integral of work transfer

ÚW

kJ kg mole K

kJ kg mole K

Irreversible Du

q

W

0 –886.84 836.96 –49.88

–554.4 –332.44 836.96 – 49.88

554.4 –554.4 0 0

Dh 886.84 –886.8 0 0

q –237.6 –1029.34 585.87 – 681.07

W –792 –474.9 585.87 –681.07

EXAMPLE 8.13 A gas initially at 273 K and 12 bar is continuously compressed to a final pressure of 150 bar. The path used to compress the gas is a reversible isothermal process. Determine, (a) The work per kg mole of the gas required to operate the compressor. (b) The heat removed from the compressor. Assume (i) The gas is an ideal (ii) The gas behaves according to Pu = RT – where, a = 10.5

K m3 kg mole

b = 0.25

m3 kg mole

R = 1.9

kJ kg mole K

Solution: This is a steady flow process (open system). We have SFEE, Eq. (3.50), È È V2 gz ˘ V2 gz ˘ h + + + q = h + + ˙ + w Í ˙ Í 2 gc gc ˙˚ 2 gc gc ˙˚ ÎÍ ÎÍ in out

a P + bP T

406

Basic Thermodynamics

dw

P 2

1

12 bar, 273 K GAS input

Compressor

150 bar

2

1 dq

V

Figure E8.13

Neglecting P.E. and K.E. changes, we have h1 + qq = h2 + qw Dh = h2 – h1 = qq – qw

Ds =

(1)

q Tsys

(2)

We have total derivative of enthalpy and entropy

[ Dh1- 2 ]T =

Ú

È ∂V ˘ V - T Í ˙ dP Î ∂T ˚ P

P2

P1

[ DS1- 2 ]T = -

Ú

P2

P1

(3)

È ∂V ˘ Í ∂T ˙ dP Î ˚P

(4)

(i) Ideal gas behaviour PV = RT

∂V ˘ R V = = ˙ ∂T ˚ P P T

(5)

È ∂V ˘ T Í ˙ =V Î ∂T ˚ P

(6)

Subsitute Eq. (6) in Eq. (3).

[ Dh1-2 ]T =

Ú

P2

P1

È ∂V ˆ ∂V ˆ ˘ -T ÍT ˙ dP = ˜ ∂T ˜¯ P ˚ Î ∂T ¯ P

Ú

P2

P1

0 dP = 0

(7)

Substitute Eq. (6) in Eq. (4) [ DS1- 2 ] T =

Ú

P2

P1

P 190 kJ R = - 4.799 dP = - R ln 2 = - 1.9 ln 12 kg mole K P P1

(8)

Ideal Gases and Mixtures of Ideal Gases

407

(a) The quantity of heat to be removed (q) Substitute Eq. (8) in Eq. (2) \

q = Tsys ( DS1- 2 )T = (2734)( -4.799)

kJ kJ = - 1310.095 kg mole K kg mole K

(9)

(b) The work done on the system (w) Substitute Eqs. (8) and (9) in Eq. (1) q–W=0 q = W = - 1310.095

kJ kg mole K

(ii) Real gas behaviour, according to È a ˘ PV = RT + Í - + b ˙ P T Î ˚ V=

RT È a ˘ + Í- + b˙ P Î T ˚

(10)

R a È ∂V ˘ Í ∂T ˙ = P + 2 T Î ˚P [ Dh1- 2 ]T =

Ú

P2

P1

(11)

2a ˘ 2a ˘ È È Í b - T ˙ dP = Í b - T ˙ ( P2 - P1 ) Î ˚ Î ˚

2 ¥ 10.5a ˘ kJ È = Í 0.25 ˙ (150 - 12) ¥ 100 = 2388.46 kg mole 273 Î ˚

(12)

Substitute Eq. (11) in Eq. (4) [ DS1- 2 ]T = -

Ú

P2

P1

È ˘ P2 a ˘ a ÈR Í P + 2 ˙ dP = - Í R ln P - 2 ( P2 - P1 ) ˙ T ˚ T Î 1 Î ˚

150 10.5 kJ È ˘ = - Í1.9 ln (150 - 12) ¥ 100 ˙ = 29.288 2 12 273 kg mole K Î ˚

(13)

(a) The quantity of heat removed (q) Substitute Eq. (13) in Eq. (2) q = 273 ¥ 29.388 = 8022.85

kJ kg mole

(b) The work done on the system (w) \

Dh = q – W W = q - Dh = 8022.85 - 2388.46 = 5634.39

kJ kg mole

(14)

408

Basic Thermodynamics

EXAMPLE 8.14 A gas is raised from 30°C to 120°C. Calculate, (a) Molar specific heat at constant pressure (b) Cp (c) Cv (d) Change in specific enthalpy Assume molecular weight of the gas as 40(M) and the gas follows a relation of Cp = 5/3R. Solution:

Assume the given gas behaves like an ideal gas.

(a) Molar specific heat at constant pressure (C p ) C p = MC p = M ¥

5 5 5 kJ R = R = ¥ 8.314 = 13.86 3 3 3 kg mole K

(b) Specific heat at constant pressure (Cp) Cp =

Cp M

=

kJ 13.86 kJ = 0.3464 kg kg mole K kg K 40 kg mole

(c) Specific heat at constant volume (Cv) R=

R 8.314 kJ = = 0.208 M 40 kg K

Cv = C p - R = 0.3464 - 0.208 = 0.1386

kJ kg K

Dh) (d) Change in specific enthalpy (D Dh = h2 - h1 = C p (T2 - T1 ) = 0.3464[120 - 30] = 31.176

kJ kg mole

EXAMPLE 8.15 Show that for an ideal gas, the slope of the constant volume process is greater than that of the slope of the constant pressure process on TS diagram. Solution: We have,

Assume 1 kg of gas. Tds = du + PdV = CvdT + PdV

\

T È ∂T ˘ Í ∂S ˙ = C Î ˚V v

T

V=C

P=C

Also we have Tds = dh – udp = CpdT – udp \

T È ∂T ˘ Í ∂S ˙ = C Î ˚P p

We know that Cp > Cv

A

Figure E8.15

S

Ideal Gases and Mixtures of Ideal Gases

\

T T > Cv C p

\

È ∂T ˘ È ∂T ˘ Í ∂S ˙ > Í ∂ S ˙ Î ˚V Î ˚ P

409

Hence proved.

EXAMPLE 8.16 Air at 70 kPa, 50°C and 0.5 kg is compressed reversibly and adiabatically to 0.5 MPa. Then it is expanded to the original volume at constant pressure. Calculate for the whole path, (a) Work transfer (b) The heat transfer Solution: P

T V=C 2

P=C

3

3 2

1 1 V

S

Figure E8.16

V1 = Volume of air at 1 =

mRT1 0.5 ¥ 0.287 ¥ 323 = = 0.66215 m 3 P1 70

For process 1–2,

ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚ \

W1- 2 =

-1

È 500 ˘ = 323 Í ˙ Î 70 ˚

1.4 - 1 1.4

= 566.5 K

P1V1 - P2V2 mR(T1 - T2 ) 0.5 ¥ 0.287(323 - 566.5) = = = - 87.35 kJ -1 -1 1.4 - 1

For adiabatic process, 1

1

ÈP ˘ È 70 ˘ 1.4 3 V2 = V1 Í 1 ˙ = 0.66215 Í ˙ = 0.1626 m P 500 Î ˚ Î 2˚ Q1–2 = 0 (∵ Adiabatic process) For process 2–3, W2–3 = P2(V1 – V2) = 500[0.6622 – 0.1626] = 249.81 kJ

410

Basic Thermodynamics

P2V2 P3V3 = T2 T3

\

T3 =

But P2 = P3

V3T2 0.6622 ¥ 566.5 = = 2307.53 K V2 0.1626

Q2–3 = m Cp(T2 – T1) = 0.5 × 1.005 × (566.5 – 323) = 122.4 kJ (a) Cyclic work transfer

(Ú W)

ÚW =W (b) Cyclic heat transfer

1–2 +

W2–3 = –87.35 + 249.81= 162.46 kJ

( Ú Q)

ÚQ=Q

1–2 +

Q2–3 = 0 + 122.4 = 122.4 kJ

EXAMPLE 8.17 Air at a pressure of 10 bar occupies a volume of 0.3 m3 and contains 2 kg. This air is then expanded to a volume of 1.4 m3. Find, (a) The final temperature (b) The work done (c) The heat absorbed or rejected by the air (d) The change in enthalpy for each of the following processes 1. Constant pressure 2. Isothermal 3. Adiabatic 4. Polytropic PV1.2 = C Solution: T1 =

P1V1 10 ¥ 100 ¥ 0.3 = = 522.65 K mR1 2 ¥ 0.287

1. Constant pressure process (a) The final temperature (T2) T2 =

P2V2 10 ¥ 100 ¥ 1.4 = = 2439.02 K mR2 2 ¥ 0.287

(b) The work done (W) W = P2(V2 – V1) = 10 × 100 × [1.4 – 0.3] = 1100 kNm or kJ (c) The heat transfer (Q) Q = W + Du = 1100 + m Cv(T2 – T1) = 1100 + 2 × 0.714(2439.02 – 522.65) = 3836.58 kJ Dh) (d) The change in enthalpy (D Dh = h2 – h1 = m Cp(T2 – T1) = 2 × 1.005(2439.02 – 522.65)=3851.9 kJ

Ideal Gases and Mixtures of Ideal Gases

2. Isothermal process (a) The final temperature (T2) T1 = T2 = 522.65 K (b) The work done (W) W = P1V1 ln

V2 1.4 = 10 ¥ 100 ¥ 0.3 ln = 1540.45 kJ V1 0.3

(c) The heat transfer (Q) Q = W + Du = 1540.45 + m Cv(T2 – T1)= 1540.45 kJ

(∵ T2 = T1)

Dh) (d) The change in enthalpy (D Dh = h2 – h1 = m Cp(T2 – T1) = 0

(∵ T2 = T1)

3. Adiabatic proces (a) The final temperature (T2) ÈV ˘ T2 = T1 Í 1 ˙ Î V2 ˚

We have

-1

1.4 - 1

È 0.3 ˘ = 522.65 Í ˙ Î 1.4 ˚

= 282.2 K

(b) The work done (W) 1.4

ÈV ˘ È 0.3 ˘ P2 = P1 Í 1 ˙ = 10 ¥ 100 Í ˙ Î 1.4 ˚ Î V2 ˚

W=

= 115.72

kN m2

P1V1 - P2V2 (100 ¥ 0.3 - 115.7 ¥ 1.4) = = 345.05 kJ -1 1.4 - 1

(c) The heat transfer (Q) Q=0

(∵ Adiabatic process)

(d) The change in enthalpy (Dh) Dh = h2 – h1 = m Cp(T2 – T1) = 2 × 1.005 × (282.2 – 522.65) = – 483.3 kJ 4. Polytropic process, PV1.2 = C (a) The final temperature (T2) ÈV ˘ T2 = T1 Í 1 ˙ Î V2 ˚ (b) The work done (W)

n -1

n

1.2 - 1

È 0.3 ˘ = 522.65 Í ˙ Î 1.4 ˚

= 384.07 K

1.2

ÈV ˘ È 0.3 ˘ P2 = P1 Í 1 ˙ = 10 ¥ 100 Í ˙ V Î 1.4 ˚ Î 2˚

= 157.47

kN m2

411

412

Basic Thermodynamics

W=

P1V1 - P2V2 (100 ¥ 0.3 - 157.47 ¥ 1.4) = = 397.71 kJ n -1 1.2 - 1

(c) The heat transfer (Q) Q = W + Du = W + m Cv(T2 – T1) = 397.71 + 2 × 0.714 × (384.07 – 522.65) = 199.81 kJ Dh) (d) The change in enthalpy (D Dh = h2 – h1 = m Cp(T2 – T1) = 2 × 1.005 × (384.07 – 522.65) = –278.55 kJ EXAMPLE 8.18 A vessel of 0.05 m3 capacity is filled with air. The pressure at the end of the pumping operation is 80 bar and temperature 50°C. The air is then cooled to the atmospheric temperature of 20°C. Then leakage occurred to 20 bar. When the leakage was stopped, the temperature of air being 5°C. Find, (a) How much heat was lost by the air in the vessel before the leakage began? (b) How much heat was transferred during leakage by the air remaining in the vessel? Assume index of expansion during leakage to be constant. R = 0.287 kJ/kg K Cv = 0.714 kJ/kg K Solution: m = mass of air in the vessel at the end of filling operation m=

P1V1 80 ¥ 100 ¥ 0.05 = = 4.315 kg RT1 0.287(50 + 273)

(a) Heat lost by air before the leakage (m) Q = m Cv(T2 – T1) = 4.315 × 0.714 × (50 – 20) = 92.43 kJ (b) The heat transferred during the leakage (QL). At the beginning of the leakage, the condition in the vessel, P2 = ?, V2 = 0.05 m3, T2 = 20 + 273 = 293 K \

P1 = 80 bar P1V1 P2V2 = T1 T2

\

P2 =

After the leakage,

V 1 = V2

T1 = 50 + 273 = 323 K

But V1 = V2

P1T2 80 ¥ 100 ¥ 293 kN = = 7256.97 2 T1 323 m

P3 = 20 × 100 kN V3 = 0.05 m3 m3 = mass of air remained =

T3 = 5 + 273 = 278K

m3 = ?

P3V3 20 ¥ 100 ¥ 0.05 = = 1.253 kg RT3 0.287(5 + 273)

We know that 4.315 kg of air is occupied in 0.05 m3 \

1.253 kg of air would have occupied

Finally it occupied 0.05 m3 at 20 bars.

0.05 ¥ 1.253 = 0.0145 m 3 at 80 bars 4.315

Ideal Gases and Mixtures of Ideal Gases

413

\ 80 bars and 0.0145 m3 air expanded to 0.05 m3, 20 bar expansion process being polytropic. With index n PbVb = PaVa a: after, b: before expansion n

n

Ê Vn ˆ Ê 0.05 ˆ Ê 80 ˆ ÁË V ˜¯ = ÁË 0.0145 ˜¯ = ÁË 20 ˜¯ b Q1 =

=

n = 1.1197

( - n) ( - n) ( PbVb - PaVa ) ¥ W.D. = ( - 1) ( - 1) (n - 1)

(1.4 - 1.11) (80 ¥ 0.0145 - 20 ¥ 0.05) ¥ 100 (1.4 - 1) (1.11 - 1)

= 105.45 kJ

Heat transfer to the surroundings.

EXAMPLE 8.19 A container is having air at a pressure of 2 bar and temperature 30°C. The volume of the container is 3 m3. Now additional air is pumped into the container until the pressure rises to 40 bar abs and the temperature rises to 65°C. Determine, (a) The mass of air pumped in (b) Express this mass as a volume at a pressure of 1 bar and 25°C. If the vessel is allowed to cool until the temperature is again 30°C, find, (c) the pressure in the vessel (d) the heat transferred during cooling process (e) the change of entropy during cooling process Neglect the heat capacity of the vessel. Assume air as ideal gas. Solution: m1 = initial mass in the container P1V1 2 ¥ 100 ¥ 3 = = 6.8997 kg RT1 0.287(30 + 273) m2 = mass of air after pumping operation =

=

P2V2 40 ¥ 100 ¥ 3 = = 123.7 kg RT2 0.287(65 + 273)

(a) The mass of air pumped (m) m = m2 – m1 = 123.7 – 6.8997 = 116.8 kg (b) Mass (m) in terms of volume (V) V=

mRT 116.8 ¥ 0.287 ¥ (25 + 273) = = 99.897 m 3 P 1 ¥ 100

(c) The pressure in the vessel after cooling (P3) V2 = V3 (cooling, const. vol. process) \

P3 =

P2T3 40 ¥ 100 ¥ (30 + 273) kN = = 3585.798 2 T2 (65 + 273) m

414

Basic Thermodynamics

(d) The heat transferred during the cooling (Qc) Qc = W + Du

(W = 0, const. vol. process)

Qc = Du = m Cv(T2 – T1) = 123.7 × 0.714 × (303 – 338) = –3091.26 kJ

(heat is rejected)

Ds) during const. volume process (e) The change in entropy (D Ds = mCv ln

= - 9.65

T2 303 = 123.70.714 ln 338 T1

kJ kg K

(decrease in entropy)

EXAMPLE 8.20 A rigid vessel is having two compartments. Both the compartments, A and B, are of a volume 0.25 m3. The pressure in A is 2 bar and that in B is 4 bar. Both the compartments are at the same temperature. When 40 kJ of heat is added, the partition wall is damaged. What is the final pressure when equilibrium is attained? R = 0.287 kJ/kg K. Solution: Final pressure when equilibrium is attained (Pf) ma = mass of air on compartment A

Similarly, \

=

PaVa 2 ¥ 100 ¥ 0.25 174.22 = = 0.287 Ta RTa Ta

mb =

PbVb 4 ¥ 100 ¥ 0.25 348.45 = = 0.287 Tb RTb Tb

m = Total mass = ma + mb

(174.22 + 348.43) 522.65 (∵ Ta = Tb = T1 ) = T1 T1 If 40 kJ of heat is added, at const. volume, then there is increase in the temperature. Q = m Cv DT =

40 kJ =

522.65 ¥ 0.714 ¥ (T2 - T1 ) T1

Pressure when the partition wall is damaged =

\ T2 = 1.1072 T1

2+4 = 3 bar. Since the volume remains constant, 2

final pressure is proportional to temperature.

Pf 3

=

1.1072 T1 T1

\ Pf = Final pressure = 3.32 bar

EXAMPLE 8.21 A balloon is filled with H2 at 20°C and atmospheric pressure. The surrounding air is at 15°C and barometer reads 75 cm of Hg. The diameter of the balloon is 20 m. Find what load the balloon will lift.

Ideal Gases and Mixtures of Ideal Gases

Solution: Vb = Volume of balloon =

d3 = 6

¥ 20 3 = 4188.79 m 3 6

8.314 kJ R = M H2 2 kg K

RH2 =

= 760 mm Hg Æ 101.325 kPa \

101.325 ¥ 750 = 99.99 kPa 760 = mass of H2 in the balloon

750 mm Hg Æ

mH2

=

PH2 V RH3 TH3

=

99.99 ¥ 4188.79 = 343.88 kg 4.157(20 + 273)

Volume of surrounding air displaced = volume of the balloon = 4188.79 m3 \ ma = mass of air displaced =

PaV 99.99 ¥ 4188.79 = = 5067.23 kg RaTa 0.287 (15 + 273)

The load which can be lifted by the balloon due to buoyancy force = 5067.23 – 343.88 = 4723.35 kg EXAMPLE 8.22 Air contains 21% of O2 and 79% of N2 by volume. Determine (a) Molecular weight of air, (b) R, (c) r at NTP Solution: NTP means 101.325 kPa, 0°C (a) Molecular weight of air (Mair) 1 Æ Oxygen, 2 Æ Nitrogen ; n = n1 + n2 2

Mair =

 i =1

2

n fi M i =

ni

ÂnM i =1

i

n Èn ˘ = Í i M1 + 2 M2 ˙ n În ˚

kg È 21 ¥ 32 79 ¥ 28 ˘ =Í + ˙ = 29.63 kg mole 100 100 Î ˚ (b) Characteristic gas constant of air (Rair) Rair =

R kJ kg mole kJ = 8.314 ¥ = 0.2997 M air kg mole K 29.63 kg kg K

r air) (c) Density (r air

=

Mair kJ kg mole kJ = 29.63 ¥ = 1.322 3 3 V kg mole 22.4 m m

415

416

Basic Thermodynamics

EXAMPLE 8.23 A gas mixture consists of 3 mole O2, 5.0 N2 and 2 mole of CO2. Determine (a) the mole fraction, (b) mass fraction of each component, (c) the apparent molecular weight of the mixture and (d) the apparent gas constant for the mixture. Solution:

We have mi = ni Mi mO2 = 3 kg mole ¥ 32

kg = 96 kg kg mole

mH2 = 5 kg mole ¥ 28

kg = 140 kg kg mole

mCO2 = 2 kg mole ¥ 44

kg = 88 kg kg mole

m = Total mass = mO2 + mN2 + mCO2 = 96 + 140 + 88 = 324 kg (a) The mole fraction of the components

n fi =

We have

n f O2 = n f H2 = n f CO2 =

ni n niO2 n niH2 n

=

3 = 0.3 10

=

5 = 0.5 10

niCO2

=

2 = 0.2 10

=

96 = 0.2963 324

=

140 = 0.4321 324

n (b) The mass fraction of the components We have

m fi = m f O2 = m f H2 =

mi m miO2 m miH2 m

miCO2

88 = 0.2716 m 324 (c) The apparent molecular weight of the mixture (M) m f CO2 =

M=

=

m 324 kg = = 32.4 n 10 kg mole

Ideal Gases and Mixtures of Ideal Gases

417

(d) The apparent gas constant for the mixture (R) R 8.314 kJ = = 0.2566 M 32.4 kg K

R=

EXAMPLE 8.24 A mixture of 0.5 kg of carbon dioxide and 0.3 kg of N2 is compressed from P1 = 1 atm, T1 = 20°C to P2 = 5 atm in a polytropic process for which n = 1.3. Find, (a) The final temperature (b) The work (c) The heat transfer (d) The change in entropy of the mixture Solution: P 2

h

PV = C

0.5 kg CO2 0.3 kg N2

1

V

Figure E8.24

(a) The final temperature (T2)

ÈP ˘ T2 = T1 Í 2 ˙ Î P1 ˚

We have

n -1 n

È5˘ = 293 Í ˙ Î1˚

1.3 - 1 1.3

= 424.78 K

(b) The work (W) nCO2 =

0.5 = 0.0114 kg mole 44

nH 2 =

0.3 = 0.0107 kg mole 28

M=

0.8 kg m = = 36.19 kg mole n 0.0114 + 0.0107 kJ ¥ (424.78 - 293) K kg mole K = 80.73 kJ kg ¥ 0.3 36.19 kg mole

0.8 kg ¥ 0.834

\

W=

Work done on the system.

418

Basic Thermodynamics

(c) The heat transfer (Q) Refer table, Du = nCO2 [uCO2 T2 - uCO2 T1 ] + nN2 [uN2 T2 - uN2 T1 ] = 0.0114[10714 - 6651] + 0.0107[8733 - 6021] = 75.34 \

kJ kg mole

Q = W + Du = –80.73 + 75.34 = –5.39 kJ [Heat is transferred to the surroundings]

DS) (d) The change in entropy (D Refer table, DS = nCO2 DSCO2 + nN2 DSN2 DSi = Si T2 - Si T1 - R ln

Pi2 Pi1

DSCO2 = 227.258 - 212.66 - 8.314 ln

5 = 1.217 kJ/kg mole K 1

DSN2 = 201.5 - 190.7 - 212.66 - 8.314 ln

5 = - 2.581 kJ/kg mole K 1

DS = 0.0114 ¥ 1.217 + 0.0107 ¥ ( -2.581) = - 0.01374

kJ K

EXAMPLE 8.25 A mixture of 1 kg of O2 and 3 kg of N2 at 30°C and 10 bar is compressed in a reversible adiabatic process to 60 bar. Calculate, (a) The final partial pressure of the constituents (b) The final temperature (c) The change in internal energy of the mixture Assume: CvN2 = 0.74 kJ/kg K, CvO2 = 3.5 kJ/kg K, C pN2 = 1.05 kJ/kg K,

C pO2 = 5.5 kJ/kg K

Solution: nN 2 =

n fN2 =

3 = 0.107 28

nN2 n

=

nO2 =

0.107 = 0.774 0.107 + 0.03125

n fO2 =

1 = 0.03125 32

nO2 n

=

0.03125 = 0.226 0.107 + 0.03125

(a) The final partial pressure of each components We know that

nN 2 È P2 ˘ Í ˙ = n f N2 = n Î P1 ˚ N 2

[ P2 ]N2 = [ P1n f ]N2 = 10 ¥ 0.774 = 7.74 bar [ P2 ]O2 = [ P1n f ]O2 = 10 ¥ 0.226 = 2.26 bar

Ideal Gases and Mixtures of Ideal Gases

419

(b) The final temperature of the mixture (T2) m f N2 =

mN 2 m

=

3 = 0.75 4

m f O2 =

mO2 m

=

1 = 0.25 4

[C p ]mix = [ m f C p ]N2 + [ m f C p ]O2 = 0.75 ¥ 1.05 + 0.25 ¥ 5.5 = 2.1625 [Cv ]mix = [ m f Cv ]N2 + [ m f Cv ]O2 = 0.75 ¥ 0.74 + 0.25 ¥ 3.5 = 1.43

kJ kg K

kJ kg K

È Cp ˘ 2.1625 =Í ˙ = = 1.512 C 1.43 Î v ˚ mix -1

1.512 - 1

ÈP ˘ È 60 ˘ 1.512 T2 = T1 Í 2 ˙ = 303 Í ˙ = 555.83 K Î 10 ˚ Î P1 ˚ Du) (c) The change in internal energy (D Du = U2 – U1 = m Cv(T2 – T1)mix = 4 × 1.43(555.83 – 303) = 1446.199 kJ EXAMPLE 8.26 A tank capacity of 5 m3 contains 70% H2 and 30% CH4 by volume at 100 kPa and 350 K. Calculate (a) the partial pressures and (b) the moles of H2 and CH4. Solution: n f H2 =

nH2 n

=

70 = 0.7 100

n f CH 4 =

nCH 4 n

=

30 = 0.3 100

(a) The partial pressures ( PH 2 , PCH 4 ) PH2 = n f H2 P = 0.7 ¥ 100 = 70 kPa PCH 4 = n f CH 4 P = 0.3 ¥ 100 = 30 kPa (b) The moles of H2 and CH4 ( nH 2 and nCH 4 )

nH2 = nCH4 =

PH2 V RT PCH 4 V RT

=

70 ¥ 5 = 0.1203 kg mole 8.314 ¥ 350

=

30 ¥ 5 = 0.0515 kg mole 8.314 ¥ 350

EXAMPLE 8.27 The compartments A, B and C are filled with O2, N2 and H2 as shown in the figure. Find, (a) Cv mix (b) Tmix (c) Rmix (d) Pmix (e) Partial pressure of individual component and (f) The entropy change.

420

Basic Thermodynamics

O2

N2

A(1)

H2

B(2)

n1 = 2

n2 = 3

n3 = 4

P1 = 4 bar

P2 = 5 bar

P3 = 6 bar

T1 = 120°C

T2 = 180°C

T3 = 300°C

Take CvO2 = 0.65,

CvN2 = 0.743,

C(3)

CvH2 = 0.5 kJ/kg K

Solution: 3

(a) Cv mix =

Âm i =1

fi Cvi

=

m m1 m Cv + 2 Cv2 + 3 Cv3 m 1 m m

m1 = n1 M1 = 2 kg mole × 32 = 64 kg m2 = n2 M2 = 3 kg mole × 28 = 84 kg m3 = n3 M3 = 4 kg mole × 2 = 8 kg 156 kg n = n1 + n2 + n3 = 2 + 3 + 4 = 9 Cv mix =

64 ¥ 0.65 84 ¥ 0.743 8 ¥ 0.5 kJ + + = 0.692 156 156 156 kg K

(b) Tmix Internal energy before mixing = Internal energy after mixing (mCvT )O2 + (mCvT ) N2 + (mCv T )H2 = (mCv T ) mix 64 × 0.65 × 120 + 84 × 0.743 × 180 + 8 × 0.5 × 300 = 156 × 0.692 × Tmix Tmix = 161.42°C 3

(c)

Rmix -

Âm i =1

fi Ri

=

m m1 m R1 + 2 R2 + 3 R3 m m m

=

64 8.314 84 8.314 8 8.314 kJ ¥ + ¥ + ¥ = 0.4797 156 32 156 28 156 2 kg K

M mix =

R 8.314 kJ = Rmix 0.4797 kg mole

(d) Pmix È nRT ˘ 2 ¥ 8.314 ¥ 393 = 16.336 m 3 VO2 = Í ˙ = ¥ P 4 100 Î ˚ O2 È nRT ˘ 3 ¥ 8.314 ¥ 453 = 22.599 m 3 VN2 = Í ˙ = ¥ P 5 100 Î ˚ N2

Ideal Gases and Mixtures of Ideal Gases

È nRT ˘ 4 ¥ 8.314 ¥ 573 = 31.75 m 3 VH2 = Í ˙ = 6 ¥ 100 Î P ˚ H2 70.685 m3 j

Pmix =

 i

PV i i = V

3

PV PV PV PV i i = 1 1 + 2 2 + 3 3 V V V V i =1

Â

400 ¥ 16.336 500 ¥ 22.599 600 ¥ 31.75 + + = 521.81 kPa 70.685 70.685 70.685 (e) Partial pressure of individual components =

Pi = n f1 P =

We have,

PO2 = PN2 = PH2 =

nO2 n nN2 n nH2 n

ni P n

P=

2 ¥ 5.22 = 1.16 bar 9

P=

3 ¥ 5.22 = 1.74 bar 9

P=

4 ¥ 5.22 = 2.32 bar 9 5.22 bar

(f) The change in entropy of the individual constituents From tables, C pN2 = 1.302 DSO2 =

Ú

T2

T1

C pH2 = 1.809 kJ/kg K

È dT P ˘ - R ln 2 ˙ ÍC p T P1 ˚ O Î 2

= 1.357 ln Similarly,

C pO2 = 1.357

434.42 8.314 1.16 kJ ln = 0.4576 393 32 4 kg K

DSN2 = 1.302 ln

434.42 8.314 1.74 kJ ln = 0.259 453 28 5 kg K

DSH2 = 1.809 ln

434.42 8.314 2.32 kJ ln = 3.449 573 28 6 kg K

EXAMPLE 8.28 A mixture of ideal gases is shown in the figure. Find, (a) The partial pressure of N2, O2 (b) The molecular weight of the mixture (c) Mass proportion of the mixture (d) The gas constant for the mixture (e) The volume per mole of the mixture

421

422

Basic Thermodynamics

N2

O2

H2

50% vol.

20% vol.

30% vol.

PN2 = ?

PO2 = ?

PH2 = 50 kPa

Temperature = 20°C

Solution:

We have Volume fraction = Mole fraction VN2

i.e.

=

nN2

= nf N2 =

V n (a) The partial pressure of N2, O2 We have, \

n f1 = P=

Pi P

50 = 0.5 100

n fH2 =

nfO2 = 0.2

nf H2 = 0.3

50 kPa P

50 kPa = 166.67 kPa 0.3

PN2 = n f N2 ¥ P = 0.5 ¥ 166.67 = 83.33 kPa PO2 = n f O2 ¥ P = 0.2 ¥ 166.67 = 33.33 kPa (b) The molecular weight of the mixture (M) 3

M=

Ân i =1

fi M i

= n f1 M1 + n f2 M2 + n f3 M3

= n f N2 M N2 + nH2 M H2 + nH2 M H2 = 0.5 ¥ 28 + 0.2 ¥ 32 + 0.3 ¥ 2 = 21 (c) The mass proportion of the mixture

m fi =

mi ni Mi M = = n fi i m nM M

m f N 2 = 0.5 ¥

28 = 0.667 21

m f O2 = 0.2 ¥

32 = 0.305 21

m f H 2 = 0.3 ¥

2 = 0.028 21 1.001

kg kg mole

Ideal Gases and Mixtures of Ideal Gases

423

(d) The gas constant for the mixture (R) R=

R 8.314 kJ = = 0.396 M 21 kg K

(e) The volume per mole of the mixture

( PV = nRT )mix \

V=

nRT 8.314 ¥ 293 = P 166.67

= 14.62

[∵ n = 1 kg mole]

m3 kg mole

EXAMPLE 8.29 A mixture occupies a volume of 1.5 m3 and contains 30 kg of O2 and 5 kg of N2 at a temperature of 300 K. Calculate (a) the specific volume, (b) the pressure, (c) the specific enthalpy, (d) the specific internal energy and (e) the specific entropy of the mixture and each component. Assume: C pO2 = 0.91, C pN2 = 1.05 kJ / kg K Solution: RO2 =

8.314 kJ = 0.259 32 kg K

RN2 =

8.314 kJ = 0.297 28 kg K

(a) The specific volume of each component We have

vi =

V mi

vO2 =

1.5 m3 = 0.5 3 kg

vN2 =

1.5 m3 = 0.3 3 kg

(b) The partial pressure of each component PO2 =

mRT 3 ¥ 0.259 ¥ 300 = = 155.4 kPa V 1.5

mRT 5 ¥ 0.297 ¥ 300 = = 297.0 kPa V 1.5 (c) The specific enthalpy of each component PN2 =

hO2 = C pO2 T = 0.9 ¥ 27 = 24.30 kJ/kg hN2 = C pN2 T = 1.05 ¥ 27 = 28.35 kJ/kg (d) The specific internal energy of each component (Cv )O2 = (C p - R)O2 = 0.91 - 0.259 = 0.651

kJ kg K

(Cv )N2 = (C p - R) N2 = 1.05 - 0.297 = 0.753

kJ kg K

424

Basic Thermodynamics

uO2 = (Cv T )O2 = 0.651 ¥ 27 = 17.58

kJ kg K

uO2 = (Cv T )O2 = 0.753 ¥ 27 = 20.33

kJ kg K

(e) The specific entropy of each component SO2 = C p ln

ÈP ˘ T2 - R ln Í 2 ˙ T1 Î P1 ˚ O

2

[Assumption: Entropy = 0 at 273 K (T1) and 100 kPa (P1)] SO2 = 0.91 ln

300 155.4 kJ - 0.259 ln = - 0.0284 273 100 kg K

SN2 = 1.05 ln

300 297 kJ - 0.259 ln = - 0.224 273 100 kg K

For the mixture (a)

1 = v

1

Âv

=

i

1 1 1 1 + = + = 5.333 vO2 vN2 0.5 0.3

\ v = 0.1875

m3 kg

(b) P =

ÂP = P

(c) h =

Âm

fi hi

= mO2 hO2 + mN2 hN2 =

3 5 kJ ¥ 24.3 + ¥ 28.35 = 26.83 8 8 kg

(d) u =

Âm

fi ui

= mO2 uO2 + mN2 uN2 =

3 5 kJ ¥ 17.58 + ¥ 20.33 = 19.298 8 8 kg

(e) s =

Âm

fi si

= mO2 sO2 + mN2 sN 2 =

3 5 ¥ ( -0.0284) + ¥ ( -0.234) 8 8

i

O2

= - 0.15065

+ PN2 = 155.4 + 297 = 452.4 kPa

kJ kg K

EXAMPLE 8.30 A gas mixture contains equal molal proportions of N2, O2 and H2. Calculate (a) gas constant, (b) g of the mixture and (c) work done on the mixture, if gas mixture is compressed from 100 kPa and 300 K to 600 kPa. F˘ È Assume isentropic compression process. Take C p = Í1 + ˙ R 3˚ Î where F = degree of freedom, F = 3 (mono atomic), 5 (diatomic) and 6 (polyatomic)

Ideal Gases and Mixtures of Ideal Gases

425

Solution: (a) The gas constant (R) Let us consider total mole = 1 =

Ân

i

nN 2 = nO2 = nH 2 =

R

R=

Ân

fi Ri

=

1 3

R

Â

nN2 n

RN2 +

nO2

RO2 +

n

nH2 n

RH2

8.314 kJ = 0.4022 1 1 1 kg K ¥ 28 + ¥ 32 + ¥ 2 3 3 3 (b) Ratio of the specific heats (gg ) =

We have

Cp =

Â

Cp M

5 ˘ 8.314 kJ È (C p ) N2 = Í1 + ˙ ¥ = 0.792 3˚ 28 kg K Î 5 ˘ 8.314 kJ È (C p )O2 = Í1 + ˙ ¥ = 0.693 3˚ 32 kg K Î È (C p ) H 2 = Í1 + Î

Cp =

5 ˘ 8.314 kJ ¥ = 11.085 3 ˙˚ 2 kg K

Âm C Âm i

pi

=

( m f C p ) N 2 + ( m f C p )O2 + ( m f C p ) H 2

Âm

i

=

i

(nMC p ) N2 + (nMC p )O2 + (nMC p ) H2

Ân M i

i

È1 ˘ È1 ˘ È1 ˘ Í 3 ¥ 28 ¥ 0.792 ˙ + Í 3 ¥ 32 ¥ 0.693˙ + Í 3 ¥ 2 ¥ 11.086 ˙ ˚ Î ˚ Î ˚ = 1.073 kJ =Î 1 1 kg K È1 ˘ Í 3 ¥ 28 + 3 ¥ 32 + 3 ¥ 2 ˙ Î ˚ Cv = C p - R = 1.073 - 0.40 = 0.672

=

Cp Cv

=

1.073 = 1.594 0.672

kJ kg K

426

Basic Thermodynamics

(c) Work done (W) Apply SFEE (This is an open system) Dh = Q – W [P.E. and K.E. changes are neglected]

Q=0

(h2 – h1) = –W W = (h1 - h2 ) = C p (T1 - T2 ) = 1.073 ¥ [300 - 615.3] = - 338.24 È È P2 ˘ Í ÍT2 = T1 Í P ˙ Î 1˚ Í Î

-1

È 600 ˘ = 300 Í ˙ Î 100 ˚

1.594 - 1 1.594

kJ kg

˘ ˙ = 615.3 K ˙ ˙ ˚

EXAMPLE 8.31 0.5 m3 H2 at 20 bar and 40°C and 0.6 m3 of O2 at 5 bar and 10°C are contained in a container with a valve. These two gases are mixed by opening the valve. Calculate heat transfer if the final temperature of the mixture is 30°C. Solution:

Figure E8.31

We have

MPV mRT m= RT M 20 ¥ 100 ¥ 0.5 ¥ 2 = = 0.7686 kg 8.314 ¥ 313

PV = mH 2

mO2 =

5 ¥ 100 ¥ 0.6 ¥ 32 = 4.08 kg 8.314 ¥ 283

m = mH2 + mO2 = 0.7686 + 4.08 = 4.8487 kg F˘ È C p = Í1 + ˙ R 2˚ Î

È (C p ) H2 = Í1 + Î

Cp =

Cp M

5˘ 7 R= R 2 ˙˚ 2

(C p ) H2 =

7 8.314 kJ ¥ = 14.55 2 2 kg K

5˘ 7 È (C p )O2 = Í1 + ˙ R = R 2˚ 2 Î

(C p )O2 =

7 8.314 kJ ¥ = 0.909 2 32 kg K

(Cv ) H2 = (C p - R)H2 = 14.55 -

8.314 kJ = 10.39 2 kg K

Ideal Gases and Mixtures of Ideal Gases

(Cv )O2 = (C p - R)O2 = 0.909 Cv = =

Âm

fi Cvi

427

8.314 kJ = 0.65 32 kg K

= ( m f C v ) H 2 + ( m f C v )O2

0.7686 ¥ 10.39 + 4.08 ¥ 0.65 kJ = 2.19 4.8487 kg K

Tmix = Temperature of the mixture, assuming adiabatic process (mCvT )H2 + (mCv T )O2 = (mCvT )mix 0.7686 × 10.39 × 40 + 4.08 × 0.65 × 10 = 4.8487 × 2.19 × Tmix \

Tmix = 32.58°C

Or temperature of the mixture can also be calculated from Tmix =

Ân

fi Ti

= ( n f T ) H 2 + ( n f T )O2

20 ¥ 100 ¥ 0.5 È PV ˘ nH 2 = Í ˙ = 8.314 ¥ 313 = 0.3843 RT Î ˚ H2

5 ¥ 100 ¥ 0.5 È PV ˘ nO2 = Í = = 0.1275 ˙ 8.314 ¥ 283 Î RT ˚ O2

n = nH2 + nO2 = 0.3843 + 0.1275 = 0.5118 Tmix =

0.3843 ¥ 40 0.1275 ¥ 10 + = 32.53°C 0.5118 0.5118

Heat transfer (Q) Q = mCv(T2 – T1) = 4.8487 × 2.19 × (32.53 – 30) = 26.82 kJ Therefore, heat is transferred from the system; hence our assumption of adiabatic process is not true. EXAMPLE 8.32 Mixing of CO2 and N2 takes place as shown in the figure. Determine for the mixture, when it is in equilibrium. (a) The temperature (b) Pressure (c) The change in entropy.

CO2

N2

2 moles 150 kPa 70°C

4 moles 250 kPa 60°C

Figure E8.32

428

Basic Thermodynamics

Solution: We have

m = nM mCO2 = (nM )CO2 = 2 ¥ 44 = 88 kg mN2 = (nM )N2 = 4 ¥ 28 = 112 kg 200 kg F˘ È C p = Í1 + ˙ R 2˚ Î

Cp =

Cp M

È (C p )CO2 = Í1 + Î

5˘ 7 R= R 2 ˙˚ 2

(C p )CO2 =

7 8.314 kJ ¥ = 0.6613 2 44 kg K

È (C p ) N2 = Í1 + Î

5˘ 7 R= R ˙ 2˚ 2

(C p )N2 =

7 8.314 kJ ¥ = 1.0393 2 28 kg K

(Cv )CO2 = (C p - R)CO2 = 0.6613 (Cv ) N2 = (C p - R)N2 = 1.0393 -

8.314 kJ = 0.4724 44 kg K

8.314 kJ = 0.742 28 kg K

RCO2 = (C p - Cv )CO2 = (0.6613 - 4724) = 0.1889

kJ kg K

RN2 = (C p - Cv )N2 = (1.0393 - 0.742) = 0.2973

kJ kg K

The Cv of the mixture Cv = =

Âm

fi Cvi

= (m f Cv )CO2 + (m f Cv )N2

88 ¥ 0.4724 + 112 ¥ 0.742 kJ = 0.623 200 kg K

(a) The temperature of the mixture (Tmix) (mCvT )CO2 + (mCvT ) N2 = (mCvT )mix 88 × 0.4724 × 70 + 112 × 0.742 × 60 = 200 × 0.623 × Tmix \

Tmix = 63.37°C

Or temperature of the mixture can also be calculated from Tmix = =

Ân

fi Ti

= (n f T )CO2 + (n f T ) N2

2 4 ¥ 70 + ¥ 60 = 63.33°C 6 6

Ideal Gases and Mixtures of Ideal Gases

429

(b) The pressure of the mix (Pmix) È mRT ˘ 88 ¥ 8.314 ¥ 343 = = 38.02 m 3 VCO2 = Í ˙ 44 ¥ 150 Î P ˚CO2 È mRT ˘ 112 ¥ 8.314 ¥ 333 = 44.30 m 3 VN2 = Í ˙ = 28 ¥ 250 Î P ˚ N2

Vmix = VCO2 + VN2 = 38.02 + 44.30 = 82.32 m 3 If total volume remains the same, then Pmix is given by, È nRT ˘ 6 ¥ 8.314 ¥ (63.33 + 273) Pmix = Í = = 203.8 kPa ˙ 82.32 Î V ˚ mix

The partial pressure of CO2 and N2 PCO2 = (n f P )CO2 =

2 ¥ 203.8 = 67.94 kPa 6

4 ¥ 203.8 = 135.88 kPa 6 DS)mix (c) The change in entropy of the mixture (D PN2 = ( n f P )N 2 =

È T P ˘ DS = m ÍC p ln 2 - R ln 2 ˙ T1 P1 ˚ Î

È (63.33 + 273) 67.94 ˘ kJ ( DS )CO2 = 88 Í 0.6613 ln - 0.1889 ln ˙ = 12.023 (70 + 273) 150 ˚ kg Î È (63.33 + 273) 135.83 ˘ kJ ( DS ) N2 = 112 Í1.0393 ln - 0.2973 ln ˙ = 21.46 (60 273) 250 kg + Î ˚ ( DS ) mix = ( DS )CO2 + ( DS )N2 = 12.023 + 21.46 = 33.48

kJ kg

EXAMPLE 8.33 Air is compressed polytropically with index n = 1.25 from 1 bar, 300 K to 5 bar in a single stage compressor. Assume mass of air compressed is 20 kg, Cpair = 27.0 + 6.0 × 10–3 T – 9.0 × 10–7 T2, where T in K, Cpair in kJ/kg mole K. Calculate (a) the work done by the compressor per cycle and (b) the amount of heat transferred to the surroundings. Solution: We know that, n=

m kg mole = 20 kg ¥ = 0.69 kg mole or 0.69 K mole Ma 29 kg

430

Basic Thermodynamics

We know that, V1 =

kN m m2 nRT = 0.69 kg mole ¥ 8.314 ¥ 300 K + = 17.21 m 3 P1 kg mole K 100 kN

We know that P1V1n = P2V2n

(Polytropic process)

1 bar × 17.211.25 m3 = 5 bar × V21.25 m3 \

V2 = 4.75 m3

(a) The work done by the compressor per cycle (W.D.) W.D. =

P2V2 - P1V1 (500 ¥ 4.5 - 100 ¥ 17.2) kN = ¥ m 3 = 2616 kJ 2 n -1 (1.25 - 1) m

Work done on the system. (b) The amount of heat transferred to the surroundings (Q) We know that P1V1 P2V2 = T1 T2

1 bar ¥ 17.21 m 3 5 bar ¥ 4.751 m 3 = 300 K T2 K T2 = 414 K The given substance is air, i.e. ideal gas, we have

Ú

Ú

Ú

Du = Cv dT = (C p - R) dT = (27 + 6 ¥ 10 -3 T - 9 ¥ 10 -7 T 2 - 8.314) dT = 18.69(414 - 300) +

= 2361.7

6 ¥ 10 -3 (4142 - 300 2 ) 9 ¥ 10 -7 (4143 - 3003 ) 2 3

kJ kg mole

DU = 2361.7 ¥ n = 2361.7

kJ ¥ 0.69 kg mole = 1629.6 kJ kg mole

Q = DU + W.D. = 1629.6 kJ – 2616 kJ = – 986.4 kJ Heat is liberated from the system and transferred to the surroundings.

EXERCISES 8.1 Define specific heat of gas.

(VTU Aug 2001)

8.2 What do you understand by isentropic process? Starting with the relation ds = dQ/T, derive (KUD II-98) the relation PV g = C for the above process of the ideal gas.

Ideal Gases and Mixtures of Ideal Gases

431

8.3 Define the terms as applied to a mixture of ideal gases (i) mass fraction (ii) mole fraction (iii) volume fraction. (KUD March 2001) 8.4 State Gibbs Dalton’s law.

(KUD March 2001)

8.5 Define a perfect gas. Show that for a perfect gas, Cp – Cv = R

(KUD Aug 2001)

8.6 State Gibbs Dalton’s law and derive an expresion for the molecular weight of a mixture of ideal gases. (VTU Aug 2002) 8.7 Prove that in a mixing process at constant temperature and pressure, the total entropy of a mixture of gases increases. (VTU Aug 2003) 8.8 Define (i) Ideal gas, (ii) Real gas and (iii) Universal gas constant.

(VTU Feb 2003)

8.9 Derive an expression for (i) Change in internal energy (ii) Change in enthalpy (iii) Work done (iv) Heat transfer (v) Change in entropy For the following processes of an open system and closed system with PV diagram. (a) Constant volume (b) Constant pressure (c) Isothermal (d) Polytropic (e) Isentropic 8.10 Enunciate the Boyle’s and Charle’s laws and deduce the equation of perfect gases. 8.11 A vessel of capacity 5 m3 contains 20 kg of an ideal gas having a molecular weight of 25. The temperature is 15°C. Find the pressure. [Ans: 3.84 bar] 8.12 The values of Cp and Cv for a gas are 0.2375 and 0.1691, respectively. Find the density of this gas at 15°C and 1.0 bar absolute pressure. [Ans: 1.23 kg/m3] 8.13 The values of specific heats at constant pressure and constant volume for an ideal gas are 0.235 and 0.174, respectively. Find the values of the characteristic gas constant R and the ratio of specific heats g for the gas, if 1 kg of this gas is heated at constant pressure from 25°C to 200°C. Calculate the heat added, ideal work done and change in internal energy. Also calculate the pressure and final volume if the initial volume was 2 m3. 8.14 Air at a pressure of 15 bar and a temperature of 250°C expands according to the law of PV1.25 = C to a pressure of 1.5 bar. Determine the work done, heat exchange and change in entropy of the system if it contains 0.9 kg of air. [Ans: 199 kJ, 74.3 kJ, 0.178 kJ/kg] 8.15 Air at 26°C and 1 bar is compressed according to the law PVn = C from a volume of 0.3 m3 to 0.07 m3. Determine the final pressure temperature, work done and changes in internal energy and enthalpy, given the index n takes values of 1.5, 1.4 and 1.0. [Ans: n = 1.5, 8.87 bar, 346°C, –64 kJ, 80.3 kJ, 346 kJ]

432

Basic Thermodynamics

8.16 0.5 kg of air is compressed reversibly and adiabatically from 80 kPa, 60°C to 0.4 MPa and is then expanded at constant pressure to the original volume. Calculate the heat transfer and work transfer for the whole path. [Ans: 527.85 kJ, 93.6 kJ] 8.17 A mixture of gases contains 60% N2, 30% O2 and 10% CO2 by mass. 1.5 kg of the mixture is compressed from 100 kPa and 300 K to 400 kPa, polytropically which follows the law PV1.2 = C. Determine the work done, heat transfer and change in entropy. [Ans: –161 kJ, –78 kJ, –0.23 kJ/k] 8.18 A gas mixture contains 3 kg of N2 and 5 kg of CO2 at a pressure of 3 bar and a temperature of 20°C. Find (a) the mole fraction of each constituent, (b) the molecular weight and the gas constants of the mixture, (c) the partial pressure and partial volume of CO2 and N2 in the mixture and (d) the volume and density of the mixture. [Ans: 0.485, 0.515, 228 kJ/kg K, 1.455 bar, 1.545 bar, 0.868 m3, 0.925 m3, 4.46 kg/m3] 8.19 A tank of volume 0.1 m3 contains 4.0 kg N2, 1.5 kg O2 and 0.75 kg CO2. If the temperature of the mixture is 20°C, determine the following: (a) density of the mixture (b) pressure of the mixture (c) gas constant of the mixture (d) relative molecular mass of the mixture (KUD II-98) 3 [Ans: 62.5 kg/m , 5035.6 kPa, 0.2749 kJ/kg K, 30.236] 8.20 Nitrogen expands through a turbine in a reversible adiabatic process. The gas enters at 1100 K and 550 kPa and exits at 100 kPa. The turbine produces 37 Mw. Determine the following: (a) exit temperature, (b) the flow rate required and (c) the exit specific enthalpy. Take Cp = 1.0399 kJ/kg K, Cv = 0.7431 kJ/kg K. (KUD II-98) [Ans: 676.69K, 84.05 kg/s, 703.69 kJ/kg] 8.21 1 kg of H2 undergoes a polytropic process from an initial condition of 1 bar and 30°C to a final condition of 5 bar and 3.5 m3. Find (a) the work done and (b) the change in entropy. Take Cp = 14.32 kJ/kg K, Cv = 10.28 kJ/kg K. (KUD March 2001) [Ans: –1845.18 kJ/kg, –1.384 kJ/kg K] 8.22 A vessel of 0.2 m3 capacity contains 2 kg of CO2 and 1.5 kg of N2 at 300 K. Determine (a) pressure in the vessel, (b) mole fraction of each constituent and (c) R and M of the mixture. (KUD March 2001) [Ans: 1235 kPa, 0.722, 0.278, 235 Nm/kg K, 27.8] 8.23 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 (a) the total pressure of each gas in the vessel and (b) the volume of the vessel. (KUD March 2001) [Ans: 100 kPa, 100 kPa, 200 kPa, 7.75 m3] 8.24 A mixture of ideal gases consists of 7 kg of N2 and 6.6 kg of CO2 at a pressure of 300 kPa and a temperature of 20°C. Find, (a) Mole fraction of each constituent (b) Equivalent molecular mass of the mixture

Ideal Gases and Mixtures of Ideal Gases

433

(c) Equivalent gas constant for the mixture (d) Total volume of the mixture (KUD Aug 2001) [Ans: 0.625, 0.375, 34 kJ/kg mole, 244.52 J/kg K, 3.248 m3] 8.25 A rigid chamber of 0.5 m3 capacity contains 2 kg of a gas (Cp = 1.968 and Cv = 1.507 kJ/ kg K) at 60°C. Heat is transferred to the gas until the temperature becomes 100°C. Find the work done, the heat transferred, the changes in internal energy, enthalpy and entropy. Find also gas constant and its molecular mass. (KUD Aug 2001) [Ans: 0, 141.658, 141.658 kJ/kg K, 184.992 kJ/kg K, 0.437 kJ/kg K] 8.26 A mixture of 1 kg of O2 and 2 kg of N2 occupies volume of 1 m3 at temperature 27°C. Assuming perfect gas behaviour, determine the following properties of the mixture: (a) specific volume, (b) pressure, (c) gas constant and (d) relative molecular mass. (KUD, I-96) È ˘ 1 m3 kN kJ , 256.2 2 ; 0.284 , 29.2 ˙ Í Ans: 3 kg kg K m ÎÍ ˚˙ 8.27 2 kg of an ideal gas is contained in a rigid cylinder, 21.1 kJ of heat is added to the gas which has an initial temperature of 32°C. Determine, (a) the final temperature (b) the change of enthalpy (c) the change of entropy Take R = 0.317 kJ/kg K, g = 1.26 (KUD I-96) [Ans: 40.650°C, 21.09 kJ, 0.068 kJ/kg K]

9 CHAPTER

Real Gases

9.1 INTRODUCTION Clerk Maxwell developed an ideal gas equation of state Pv = RT from the postulates of the kinetic theory of gases. A gas that follows the gas laws (Boyle’s and Charles’) at all range of pressure and temperature can be considered as an ideal gas. However, no such gas exists in nature. The real gases obey very closely the ideal gas equation when the pressure is very small or temperature is very large. However, as pressure increases the intermolecular forces of attraction and repulsion increases, and the volume of the molecules becomes appreciable compared to the total volume of the gases. Therefore, the real gases deviate considerably from the ideal gas equation. Then the question arises, what correction is to be done to the ideal gas equation?

9.2

VAN DER WAALS EQUATION OF STATE

In 1873, Van der Waals introduced two constant terms to ideal gas equation based on the laws of mechanics to individual molecules. Then the state equation for real gases, È a ˘ ÍP + ˙ (v - b) = RT ( v )2 ˚ Î

(9.1)

a˘ È Í P + 2 ˙ (v - b) = RT v ˚ Î

(9.2)

434

Real Gases

435

where P = Pressure of the gas, kPa R = Universal gas constant 8.314 kJ/kg mole K T = Temperature of the gas, K v = Molal volume of the gas, m3/k mol a, b = Constants, depend on gases

9.3

VAN DER WAALS CONSTANTS, a AND b

Van der Waals introduced two constants. These constants are determined from the behaviour of a substance at the critical point. The coefficient a accounts for the mutual attraction between the molecules. The term a/v2 accounts for the inter molecular forces (force of cohesion), and b accounts for the volume occupied by the gas molecules. The determination of the two constants in Eq. (9.1) is based on the observation that the critical isotherm on a PV diagram has a horizontal inflection point at the critical point as shown in Figure 9.1. \ At critical point, temperature, T = Tc

P Critical point

Tc Pc

Vc

Figure 9.1

V

Critical isotherm with an inflection at critical point.

È ∂P ˘ Í ∂v ˙ = 0 Î ˚Tc

È ∂2 P ˘ Í 2 ˙ =0 ÍÎ ∂v ˙˚T c Equation (9.2) can be written as

Differentiating Eq. (9.3),

È RT ˘ a P=Í ˙- 2 Îv - b˚ v RTc 2a È ∂P ˘ + 3 =0 Í ∂v ˙ = 2 vc Î ˚Tc (vc - b) RTc =

Differentiating Eq. (9.4a),

2a vc3

( vc - b)2

(9.3)

(9.4a) (9.4b)

È ∂2 P ˘ 2 RTc 6a - 4 =0 Í 2˙ = 3 vc ÍÎ ∂v ˙˚T (vc - b) c RTc =

3a vc4

(vc - b)3

(9.5)

436

Basic Thermodynamics

Equating Eqs. (9.4b) and (9.5) 2a vc3

(vc - b)2 =

3a vc4

(vc - b)3

vc = 3b

(9.6)

Substitute Eq. (9.6) in Eq. (9.4b)

RTc =

2a 27b

3

(3b - b)2 =

8a 27b

8a 27bR Substitute Eqs. (9.6) and (9.7) in Eq. (9.3)

Pc =

Tc =

(9.7)

R 8a a a ¥ - 2 = (3b - b) 27bR 9b 27b2

(9.8)

The critical constants vc, Tc and Pc are given by Eqs. (9.6), (9.7) and (9.8), respectively. From Eqs. (9.6) and (9.8) a = 3Pc vc2

(9.9)

From Eqs. (9.6) and (9.7) a=

9 RTc vc 8

(9.10)

a=

27 R 2Tc2 64 Pc

(9.11)

From Eqs. (9.7) and (9.8)

From Eqs. (9.9), (9.10) and (9.11) a = 3Pc vc2 =

9 27 R 2Tc2 RTc vc = 8 64 Pc

(9.12)

From Eqs. (9.6), (9.7) and (9.11) b=

vc 8 ¥ 27 ¥ R 2Tc2 RTc 8a = = = 3 27Tc R 27Tc R ¥ 64 Pc 8Pc

(9.13)

From Eqs. (9.9) and (9.10) R=

8Pc vc 3Tc

(9.14)

9.3.1 Non-dimensional Properties Non-dimensional properties (Pr, vr and Tr) (reduced properties) can be written as Pr =

P Pc

vr =

v vc

Tr =

T Tc

(9.15)

Real Gases

437

From Eqs. (9.6) and (9.15) (9.16)

v = 3bvr From Eqs. (9.7) and (9.15)

T=

8aTr 27bR

(9.17)

From Eqs. (9.8) and (9.15) P=

aPr

(9.18)

27b2 Substitute Eqs. (9.16), (9.17) and (9.18) in Eq. (9.2) È aPr R8aTr a ˘ + 2 2 ˙ (3bvr - b) = Í 2 27bR 9b vr ˚˙ ÎÍ 27b È Pr 8Tr 1˘ Í + 2 ˙ (3vr - 1) = 3 ÎÍ 3 vr ˙˚

(9.19)

Divide and multiply Eq. (9.19) by 3 È 3 ˘Ê 1 ˆ 8Tr Í Pr + 2 ˙ Á vr - ˜ = 3¯ 3 vr ˚˙ Ë ÎÍ

(9.20)

Equation (9.2) is clearly stated in terms of temperature and pressure and is implicit in volume. This equation is not accurate over a wide range of conditions. If the constants a and b are varied with respect to temperature, the resulting equation would become more accurate over a wide range of conditions. Tables 9.1 and 9.2 give the constants a and b and the critical constants of selected gases, respectively. Table 9.1 Constants for the Van der Waals equations of state

Substance

Van der Waals a bar

Air Butane (C4H10) Carbon dioxide (CO2) Carbon monoxide (CO) Methane (CH4) Nitrogen (N2) Oxygen (O2) Propane (C3H8) Refrigerant 12 Sulphur dioxide (SO2) Water (H2O)

(m3/kmol )2 1.368 13.86 3.656 1.474 2.293 1.366 1.393 9.349 10.49 6.883 5.531

b m3/kmol 0.0367 0.1162 0.0423 0.0395 0.0428 0.0386 0.0314 0.0901 0.0971 0.0569 0.0305

438

Basic Thermodynamics

Table 9.2 Critical constants

Substance Formula Ammonia Argon Benzene Carbon dioxide Carbon monoxide Chlorine Dichlorodifluoromethane (R-12) Dichlorofluoromethane (R-22) Ethane Ethyl alcohol Ethylene Helium Hydrogen (normal) Methane Neon Nitrogen Oxygen Propane Sulphur dioxide Water

9.4

NH3 Ar C6H6 CO2 CO Cl2

Molecular weight 17.03 39.948 78.115 44.01 28.01 70.906

TC Temp., K

Pressure, PC = MPa

Volume, m3 kg-mol

405.5 151 562 304.2 133 417

11.28 4.86 4.92 7.39 3.50 7.71

0.0724 0.0749 0.2603 0.0943 0.0930 0.1242

CCl2F2

120.91

384.9

4.13

0.2167

CHCl2F C2H6 C2H5OH C2H2 He H2 CH4 Ne N2 O2 C3H8 SO2 H2O

102.92 30.070 46.07 28.054 4.003 2.016 16.043 20.183 28.013 31.999 44.097 64.063 18.015

451.7 305.2 516 282.4 5.3 33.3 190.7 44.5 130.0 153.0 370 430.7 647.3

5.17 4.88 6.38 5.12 0.23 1.30 4.64 2.73 3.39 5.1 4.26 7.88 22.09

0.1973 0.1480 0.1673 0.1242 0.0578 0.0649 0.0993 0.0417 0.0899 0.0780 0.1998 0.1217 0.0560

COMPRESSIBILITY FACTOR (Z)

The ideal gas equation is very simple and is used conveniently. Real gases behave as ideal gases only in certain regions, i.e. the real gases behave like ideal gases as they approach zero pressure. This is shown in Figure 9.2.

Real gases

PÆ0 Or High temperature

Ideal gases

Figure 9.2 Real gases behave like ideal gases at zero pressure or elevated temperature.

Figure 9.3 shows the deviation of real gases from the ideal gas behaviour. Let us take an example of water-steam as a substance. The water vapour region shown in shaded area behaves like an ideal gas, i.e. saturated water vapour at very low pressure and superheated water vapour at very high temperature behave like an ideal gas. The steam at high pressure and saturated conditions and at critical conditions deviates significantly from the ideal gas behaviour. This deviation at a given temperature and pressure can be accounted by the introduction of a factor called compressibility factor, Z, defined as,

Real Gases

439

Figure 9.3 TV diagram shows the region where steam can be treated as an ideal gas.

Pv RT Pv = ZRT Z=

(9.21)

Compare Eq. (9.21) and state equation Pv = RT The compressibility factor, Z=1

for ideal gases

The value of Z can be obtained from experimentation and the chart is prepared. However, the chart is not available for many gases. In such cases the principle of corresponding states can be used to develop a generalized compressibility chart. For a given property (T, P) gases behave differently but behave very much same when the gases are at the same reduced pressure (Pr) and temperature (Tr). A reduced property is defined as the ratio of the property (P, T) to the value of the property at the critical state (Pc, Tc). The reduced pressure (Pr) is Pr =

P Pc

440

Basic Thermodynamics

Similarly the reduced temperature (Tr) is T Tc The value of Z factor for all the gases is approximately the same at the reduced pressure and temperature. This is called the principle of corresponding states. Figure 9.4 shows the generalized compressibility chart where experimentally determined Z values are plotted against the reduced pressure (Pr) and temperature (Tr) for different gases. Tr =

Figure 9.4

Generalized compressibility chart.

Then according to the principle of corresponding states, the compressibility factor for any gas is a function of only Pr and Tr. i.e. Z = f (Pr, Tr) Hence, compressibility factor for most of the gases is the same when the gases are at the same reduced pressure (Pr) and temperature (Tr). REAL As shown in Figure 9.5, Z = 1 for ideal gases and GASES 1 ≥ Z ≥ 1 for real gases. IDEAL >1 Sometimes P and v or T and v are known instead GAS Z =1 of P and T. Under this situation one more reduced Z=1 W2 \ Engine will deliver more power.

400°C

T1

100°C

Q2 25000 kW

Q1 12000 kW W1

HE Q2 20°C 1st HE

T1

W2

HE Q3

T2

20°C

T4

2nd HE

5(a) Derive Clasius inequality and prove that entropy is a property. Ans. Refer to Sections 5.5 and 5.6. Or the Carnot principle states that a reversible heat engine is more efficient than an irreversible engine. The efficiency of a reversible heat engine is given by, T2 T1 - T2 = R =1T1 T1 T1 = Source temperature T2 = Sink temperature Now consider an irreversible engine operating between the same thermal reservoir. Let dQ be the heat absorbed and dQ2 be the heat rejected by the engine. The efficiency of the engine, dQ2 dQ1 - dQ2 = IR = 1 dQ1 dQ1 Then by Carnot principle, dQ1 dQ2 dQ T dQ2 T T1, so there will be heat transfer. But no work across the boundary. \ W.D. = 0 (ii) In this case also the temperature inside the shelf may be high or low, according to the surrounding temperature. But there is no work done. \ W.D. = 0. (iii) Rigid container, containing uranium 235 decays to be lead will also not perform any kind of work. \ W.D. = 0. In all the three cases work done or supplied to or be the system is zero. \ W.D. = 0. 5(c)

A rigid tank contains air at 35°C and is stirred to paddle wheel which does 500 kJ of work on the air. During the stirring process the temperature of air remains constant because of heat transfer to surrounding at 15°C. Estimate the change in entrophy of air in the tank and the change in the entropy of the surroundings. Ans. T1 = 35°C, W.D. = 500 kJ Constant temperature process in tank and heat transfer to surrounding at T2 = 15°C. For constant temperature process in tank Change in entropy, S2 - S1 = mRa ln

V2 T + mCV air ln 2 V1 T1

W.D. = mRaT1 ln

V2 V1

500 = mRa (35 + 273) ln

V2 V1

(1)

460

Model Question Papers

\

mRa ln

V2 500 = = 1.6233 V1 308

From Eq. (1), S2 - S1 = 1.6233 + 1 ¥ 0.817 ln

T2 = 1.6233 kJ/kg°K T1

Now with the surrounding, heat lost by the tank Q==

Ú

Ú

T2

T1

T1

T2

mC p DT

(–ve, as heat is lost)

mC p DT = 1.0 ¥ 1.005(T1 - T2 ) = 1 ¥ 1.005 ¥ (35 - 15)

= 20.1 kJ/kg Change in entropy with surrounding ( DS )sur =

20.1 Q = = 0.06979 kJ/kg K T2 (15 + 273)

(DS)universe = (DS)tank + (DS)sur = 1.6233 + 0.06979 = 1.69309 6(a)

Briefly explain what is meant by (i) Available energy (ii) Dead state with respect to a system.

Ans. Available energy: When a system is subjected to a process from its initial state to a dead state (when the system is under thermodynamic equilibrium with the surrounding), the maximum amount of useful work that can be obtained under ideal conditions without dissipative effects is called available energy. The concept of availability is related to maximum amount of theoretical work that can be obtained from a system at a given state up to its dead state. The maximum useful work thus obtained under these conditions is the available energy. Dead state with respect to a system: When a system at state 1 defined by P, V, T, can be brought down up to surrounding condition (P0,V0, T0) and the system can no longer do or produce any work, the surrounding conditions are called the dead state with respect to the system. Usually at dead state P0 = 1 atm, T0 = 25°C. 6(b) Write a brief note on the law of degradation of energy. Ans. Whenever heat is transferred through a finite temperature difference, there is decrease in available energy for given temperatures T1 and T0. Q1 = T1DS

Q0 = T0DS

\

W = Ae = (T1 – T0)DS

If Q1 is transferred through a finite temperature difference from reservoir at T1 to the engine absorbing heat Q1 at T1¢.

Model Question Papers

T

T T1

Q1

Q1 Q1¢

T1

T1

T0

461

Q0

Q0¢

DS

DS ¢

T0

Q2 S

S

Q1 received by engine at T1¢ can be determined between temperature limits T1¢ and T0 as shown. Q1 = T1DS = T1¢DS¢ T 1 > T 1¢ \ DS¢ = DS (1) Q0 = T0DS and Q0¢ = T0DS¢ (2) From Eqs. (1) and (2) Q0¢ > Q0 W = Q1 – Q0 and W¢ = Q1 – Q0¢ W¢ < W [Q0¢ being greater than Q0] \ Loss of available energy W – W¢ = Q0¢ – Q0 = (Q1 – T0DS) – (Q1 – T0DS) = T0(DS¢ – DS) Greater the difference (T1 – T1¢), greater is heat rejecting Q0 and greater will be unavailable part of energy. Energy is degraded each time it is transferred through a finite temperature difference. For this reason the second law of thermodynamics is also called the law of degradation of energy. 6(c)

A turbine expands air adiabatically from 6 bar, 400 K, 200 m/s to 1 bar, 290 K, 50 m/s. Estimate the actual work output and the maximum possible work output for the above operating conditions. Also find the irreversibility of the process. Take surrounding atmosphere to be at 1 bar and 280 K. Ans. Initial conditions Final conditions Atmospheric conditions P2 = 1 bar P0 = 1 bar P1 = 6 bar T1 = 400 K T2 = 290 K T0 = 280 K C1 = 200 m/s C2 = 50 m/s C0 = 1.005 kJ/kg K Q=0 R = 0.287 kJ/kg K DP.E. = 0 Apply SFEE, Eq. (3.50) Wa - Q = (h1 - h2 ) + Actual work

C12 - C22 g( Z1 - Z 2 ) + 2 gc ¥ 1000 gc ¥ 1000

Wa = C p (T1 - T2 ) +

C12 - C22 2 gc ¥ 1000

= 1.005(400 - 290) +

200 2 - 50 2 kJ = 129.3 2 ¥ 1 ¥ 1000 kg

462

Model Question Papers

Maximum work

Wm = (h1 - h2 ) - T0 (S1 - S2 ) +

C12 - C22 2 gc ¥ 1000

Ê T P ˆ C 2 - C22 = C p (T1 - T2 ) - T0 Á C p ln 1 - R ln 1 ˜ + 1 T2 P2 ¯ 2 gc ¥ 1000 Ë

400 6 ˆ 200 2 - 50 2 kJ Ê = 1.005(400 - 290) - 280 Á1.005 ln - 0.287 ln ˜ + = 182.8 Ë 290 1 ¯ 2 ¥ 1 ¥ 1000 kg Irreversibility = Wm – Wa = 182.8 − 129.3 = 53.5

kJ kg

7(a)

Define the following (i) Pure substance (ii) Triple point (iii) Critical point Ans. Pure substance: It is a substance that has a uniform chemical composition throughout the system and it remains invariant during heat and work transfer with the surrounding. Refer to Section 7.1. Triple point: It is the point on which all the three phases of mixture, i.e. soild, liquid and vapour can co-exist in equilibrium. Critical point: As we keep on increasing the pressure, a state comes at which latent heat of vaporization becomes zero at a point; that point is called critical point at state. Properties of steam at critical point. Tc = 373.15°C, Vc = 0.000317 m3/kg Pc = 221.2, 7(b) Briefly explain what you understand by two property rule. Ans. Refer to Section 7.2. 7(c)

Define dryness fraction and briefly explain how one could estimate the same thing by separating and throttling calorimeter. Ans. Dryness fraction (x): It is defined as the ratio of dry saturated steam in a given mass of wet steam. mg x= mg + m f mg = Mass of dry saturated steam mf = Mass of saturated liquid. Combined separating and throttling calorimeter: Refer to Section 7.11.2. 7(d)

A rigid container is filled with steam at 600 kPa and 200°C. At what temperature does the steam begin to condense when it is cooled? Determine the corresponding pressure.

Ans. From: S.T. at 600 kPa and 200°C, V1 = 0.352 m3 /kg As it is rigid, V1 = V2 = 0.352 m3/kg; steam starts condensing from point 2. Therefore, pressure and temperature corresponding to V2 = 0.352 m3/kg is 5.3 bar.

Model Question Papers

P

463

T 1

600

P2

200°C

2

T2

1

2

S

V

8(a)

Clearly distinguish between ideal and real gases. Mention any two equations of state you know off. Ans. Refer to Sections 8.1, 9.1, 8.2.1 and 8.2.2. 8(b) Write a note on compressibility factor. Ans. Refer to Section 9.4. 8(c)

State Dalton’s law of partial pressure and derive an expression for all gas constants of a mixture of ideal gases. Ans. Refer to Sections 8.11.4 and 8.11.6. 8(d)

A gas mixture consists of 6 kmol of H2 and 4 kmol of N2. Determine the mass of each gas and the gas constant of the mixture.

Ans. Mass of each gas (mH2 , mN2 ) mi = ni Mi mH2 = 6 kg mole ¥ 2

kg = 12 kg kg mole

mN2 = 4 kg mole ¥ 28

kg = 112 kg kg mole 124 kg

Gas constant of the mixture (R)

R 8.314 kJ/kg mole K kJ = = 0.671 kg M kg K 12.4 kg mole Or refer to Example 8.23. R=

464

Model Question Papers

PAPER 2 1(a)

Explain with example the difference between a closed system, an open system and an isolated system. Ans. Closed system: It is a system where only energy (heat and work) flow takes place across the boundary but no mass flow takes place. Example: piston and cylinder arrangement. Open system: A system is called an open system when energy flow as well as mass flow takes place across the boundary. Example: diesel engine, air compressor. Isolated system: A system is called an isolated system when no flow of heat, work and mass takes place across the boundary. Example: universe, thermo flask. 1(b)

State Zeroth law of thermodynamics and explain how this law is used to measure temperature of a given system. Ans. Refer to Section 1.11.2. The readings tA and tB of two Celsius thermometers A and B agree at the ice point (0°C) and the steam point (100°C) but elsewhere are related by the equation ta = L + Mt B + Nt B2 , where L, M and N are constants. When both the thermometers are immersed in a well stirred oil bath, A registers 51°C, while B registers 50°C. (i) Determine the reading on B when A reads 25°C. (ii) Which thermometer (A or B) is correct? Ans. Refer to Example 1.16 (exactly same). tB = 24.96°C, when A reads 25°C, B reads 24.96°C. A has higher accuracy, so thermometer A is correct. 1(c)

2(a)

A simple magnetic substance is one involving only magnetic displacement work, i.e. a change in magnetization of the substance in the presence of a magnetic field. For such a substance undergoing a quasi-static process at constant volume, the displacement work is given by dW = C0HdM, where H = magnetic field intensity, M = magnetization and C0 = a proportionality constant. For a first approximation, assume that magnetization is proportional to the magnetic field divided by the temperature of the magnetic substance. Determine the work done in an isothermal process when the magnetization changes from M1 to M2.What would be the work done if temperature varies from T1 to T2 and the magnetic field intensity is constant?

Ans. Magnetization μ

Magnetic field Temperature

(given condition)

H KH ; M= (where K is a constant) (1) T T (i) Work done is an isothermal process when the magnetization changes from M1 to M2. Substitute Eq. (1) in work done equation Mμ

ÈT ˘ dW = - C0 M Í ˙ dM ÎK ˚

465

Model Question Papers

Integrate between M1 and M2 \

Ú

ÈT ˘ dW = W.D. = - C0 Í ˙ ÎK ˚

Ú

M2

M1

M

MdM = -

2 C0 T M 2 ˘ C0 T [ M12 - M22 ] ˙ = 2K K 2 ˙˚ M1

(ii) Work done, when temperature varies from T1 to T2, magnetic field intensity constant, i.e. H = C We have,

Mμ

dM =

H KH TM ; M= ; H= T T K

- KH

dT T2 Substitue H and dM in work done equation C MT È TM ˘ È - KH ˘ dW = - C0 Í dT = 0 dT ˙ Í ˙ 2 T Î K ˚Î T ˚

\

Ú dW = W.D. = C MH Ú 0

T2

T1

dT T = C0 MH ln T ]T2 = C0 MH ln [T2 - T1 ] 1 T

An engine cylinder has a piston of area 0.12 m2 and contains gas at a pressure of 1.5 MPa. The gas expands according to a process which is represented by a straight line on a PV diagram; the final pressure is 0.15 MPa. Calculate the W.D. by the gas on the piston if the piston stroke is 0.3 m. P2 = 0.15 MPa Ans. P1 = 1.5 MPa, P 2 c/s area = 0.12 m , stroke length = 0.3m 1 Work done by the gas on the piston W1–2 W1–2 = Area under the line 1–2 3 = Area 1–2–3–1 + Area 3–2–4–5–4 2

2(b)

=

1 × 0.12 × 0.3(1.5 – 0.15) × 103 kJ + 0.15 × 103 2 × 0.12 × 0.3 kJ = 24.3 + 5.4 = 29.7 kJ

5

4

V

2(c) Compare heat and work. Ans. Refer to Sections 2.1, 2.8.1 and 2.8.2. 3(a)

Starting from the first law of a closed system undergoing a cyclic process, obtain the first law equation for a closed system undergoing a non-cyclic process and show that energy is a property of a system. Ans. The first law of thermodynamics requires that the change in the total energy of the system be compensated by an equal but opposite change in the total energy of the surrounding, so that there is no net change in the total energy so the surrounding occurs only through the exchange of heat or work with the system. Then the change in the total energy of the surroundings, (DE)sur, must be equal to the energy transferred to or from it as heat and

466

Model Question Papers

work. Since Q is the heat transferred to the system and W is the work extracted from it during the process, (DE)sur = –Q + W (1) We know that, (DE)sys = DK.E. + DP.E. + DU (2) We know that, (DE)sur = –(DE)sys From Eqs. (1) and (2), we get, Q – W = DK.E. + DP.E. + DU (3) For a closed system, DK.E. and DP.E. are neglected. \ Q – W = DU For differential changes in the thermodynamics of a closed system, dQ − dW = dU To prove energy is a property, refer to Section 3.4. 3(b)

Ans.

A nozzle is a device for increasing the velocity of a steadily flowing stream. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the velocity is 60 m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The nozzle is horizontal and there is neglisible heat loss from it. (i) Find the velocity at exit section of the nozzle. (ii) If the inlet area is 0.1m2 and the specific volume at inlet is 0.187 m3/kg, find the mass flow rate. (iii) If the specific volume at the exit of the nozzle is 0.498 m3/kg, find the diameter at the exit section of the nozzle. (i) SFEE

C12 Z1 g C22 Z2 g + + h1 = + + h2 2 gc ¥ 1000 gc ¥ 1000 2 gc ¥ 1000 gc ¥ 1000

Conditions: Q = 0 W.D. = 0 DP.E. = 0 C22 60 2 + 3000 = + 2762 2 ¥ 1 ¥ 1000 2 ¥ 1 ¥ 1000

1

\ C2 = 692.53 m/s (ii) Inlet mass flow rate (m)inlet (m)inlet =

Area × Velocity at inlet 0.1 ¥ 60 kg = = 32.08 Specific volume at inlet 0.187 s

(iii) Diameter at the exit section of the nozzle (d2) È Area × Velocity ˘ (mass flow rate)inlet = (mass flow rate)exit = Í ˙ Î Specific volume ˚ exit 32.08 =

A ¥ 692.7 0.498

A = 0.0228 m 2 =

d22 4

d2 = 0.17 m

2

Model Question Papers

467

4(a)

Show that, of all the reversal heat engines working between the same two constants but different temperature reservoirs, the reversible heat engine will have the maximum coefficiency of performance. Ans. Refer to Section 4.12. 4(b) Demonstrate, using the second law, that free expansion is an irreversible process. 4(c)

A reversible engine works between three thermal reservoirs A, B and C. The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperatures TA and TB, respectively, and rejects heat to a thermal reservoir C kept at temperature TC.The efficiency of this engine is a times the efficiency of a reversible engine which works between the two reservoirs A and C. Show that TA T = (2 - 1) + 2(1 - ) A TB TB

Ans. Refer to Example 4.8. 5(a) State and prove Clasius inequality. Ans. Refer to Section 5.6 OR 5(a) of Paper 1. 5(b)

Show by invoking Kelvin-Planck statement of the second law that two reversible adiabatic paths cannot intersect each other.

Ans. Two reversible adiabatic paths cannot intersect each other. Process 1–2 = Reversible isothermal process 2–3 and 3–1 = Reversible adiabatic process Let two reversible adiabatic processes 1–3 and 2–3 intersect each other. Consider isothermal process 1–2 intersects two reversible adiabatic at points 1 and 2. Heat supplied in the cycle. Qnet = Q1–2 + Q2–3 + Q3–1 = Q1–2

[∵ Q2–3, Q3–1 = 0]

Work done is the area under 231 = Wnet According to the first law of the thermodynamics, Qnet = Wnet But here it is not true. Hence, two adiabatic paths cannot intersect. 5(c)

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 1680 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 temperature at which heat is rejected by engine A. (ii) W.D. by engines A and B. (iii) The heat rejected by engine B. (iv) If engines A and B deliver equal work, find the efficiencies.

468

Ans.

Model Question Papers

(i) Temperature at which heat is rejected by engine A (T) A

=1-

T T =1T1 1000

B

=1-

T1 = 1000 K

T2 100 =1T T

Q1 = 1680 kJ

hA = hB (Given condition) Q2

T 100 \ T = 316.22 K =1000 T (ii) The W.D. by engine A (WA) and B (WB) -

A

\

T Q2

W W T 316.22 =1 = 0.6832 = A = A T1 Q1 1680 1000

=1-

Q3

WA = 1148.75 kJ = Q1 – Q2 = 1680 – Q2

T3 = 100 K

Q2 = 531.25 kJ A

=

B

= 0.6837 =

WB WB = Q2 531.25

WB = 363.2 kJ

(iii) The heat rejected by engine B (Q3) B

=

Q2 - Q3 531.25 - Q3 = 0.6837 = Q2 531.25

\ Q3 = 168.02 kJ

(iv) Efficiencies of engines A and B, if W1 = W2(h1 = h2) For engine A,

=

ÊT -T ˆ W1 = Q1 Á 1 Ë T1 ˜¯

T1 - T W1 = T1 Q1

For engine B,

ÊT -T ˆ W2 = Q2 Á 2 Ë T2 ˜¯

Now if

ÊT -T ˆ ÊT -T ˆ = Q2 Á 2 W1 = W2, Q1 Á 1 ˜ T Ë 1 ¯ Ë T2 ˜¯

But

Q1 Q2 Q3 = = T1 T2 T3

\

T1 – T = T – T2 1

6(a)

1

=1-

(From Carnot principle)

1000 – T = T – 100

T 550 =1= 0.45 T1 1000

2

=1-

\

T = 550 K

T2 100 =1= 0.818 T 550

A rigid vessel contains 1 kg of wet steam at a pressure of 0.15 MPa. When the mixture is heated, the state passes through the critical point. Determine (i) The volume of the vessel. (ii) Mass of liquid and vapour initially in the vessel. (iii) Temperature of the mixture when the pressure has risen to 3 MPa. (iv) The heat transfer required to produce the final state.

Model Question Papers

Ans.

469

(i) The volume of the vessel (V1) From S.T. at critical condition, P2 = 221.2 bar, T2 = 374.5 K, v2 = 0.00317 m3/kg In a rigid vessel, volume remains constant. \

v1 = v2 = 0.00317 m3/kg

\

V1 = mv1 = 1 × 0.00317 = 0.00317 m3

(ii) Mass of liquid and vapour initially in the vessel (ml1 , mv1 ) v1 = v 2; From S.T. at 1.5 bar

vg1 = 1.159 m 3 /kg

v f1 = 0.001 m 3 /kg

x1vg1 + (1 - x1 ) v f1 = v2

x1 ¥ 1.159 + (1 - x1 ) 0.001 = 0.00317

x1 = 0.0031 x1 =

mv Mass of dry steam = 0.0031 = 1 Mass of wet steam 1

mv1 = 0.0031 kg;

ml1 = 1 - mv1 = 1 - 0.0031 = 0.9969 kg

(iii) Temperature of the mixture when the pressure has risen to 3 MPa (Tsat). At P = 3 MPa = 30 bar, Tsat = 233.8°C (from S.T) (iv) The heat transfer required to produce the final state (Q) State corresponding to 3 MPa is point 3 At 30 bar (3 MPa), vg = 0.0666 m3/kg,

v1 = v2 = x3 vg ˘˚

at point 3

vf = small

0.00317 = x3 ¥ 0.0666

At 30 bar,

hf = 1008.3 kJ/kg

hfg = 1794 kJ/kg

At 1.5 bar,

hf = 461.1 kJ/kg

hfg = 2226.3 kJ/kg

\

h3 = 1008.3 + 0.04 × 1794 = 1080.06 kJ/kg

\ x3 = 0.04

h1 = 467.1 + 0.0031 × 2226.3 = 474 kJ/kg \ 6(b)

Q = h3 − h1 = 1080.06 – 474 = 606.05 kJ/kg

Steam at 10 bar and 200°C undergoes a reversible adiabatic process to 1 bar in a turbine. Determine the final specific volume, the final temperature and the final specific entropy. If the mass flow rate of steam through the turbine is 2 kg/s, determine the work output from the turbine. Ans. At 1, 10 bar and 200°C, v1 = 0.2059 m3/kg, h1= 2826.8 kJ/kg, S1 = 6.692 kJ/kg At 2, 1 bar, vg = 1.694 m3/kg, hf = 417.5 kJ/kg

470

Model Question Papers P

T 200

1 bar

99.63

1

10

2

1

1

2 V

S

hfg = 2258 kJ/kg, Sf = 1.303 kJ/kg K, Sfg = 6.057 kJ/kg K

S1 = S2 = S f + x2 S fg ˘˚

x2 = 0.889

at point 2

h2 = hf + x2Sfg = 417.5 + 0.889 × 1.694 = 2424.86 kJ/kg (i) Final specific volume (v2) v2 = x2vg2 = 0.889 × 1.694 = 1.5059 m3/kg (ii) The final temperature (t2) From S.T. at 1.5 bar t2 = 99.63°C (iii) Work output (W) W = 2[h1 − h2] = 2[2826.8 – 2424.86] = 803.88 kW 7(a)

An ideal gas cycle consisting of three processes uses argon (molecular weight = 40) as working substance. Process 1–2 is a reversible adiabatic process from 0.014 m3, 700 kPa and 280°C to 0.056 m3, process 2–3 is a reversible isothermal process. Process 3–1 is an isobaric process. Sketch the cycle on PV and TS diagrams and find (i) the work transfer in process 1–2 (ii) work transfer in process 2–3 (iii) network output from the cycle. Assume change in enthalpy for each process.

Ans.

P1 = 700 kPa, PV = nRT

V1 = 0.014 m3, 7×

n = 0.0617 =

105

T1 = 280 + 273 = 553 K

× 0.014 = n × 287 × 553

mass mass = mole weight 40

\ mass = 2.469 kg

(i) Work transfer in process 1–2 (W1–2) T1 È V2 ˘ =Í ˙ T2 Î V1 ˚

-1

1.67 - 1

553 È 0.056 ˘ =Í ˙ T2 Î 0.014 ˚

T2 = 218.44 K

Model Question Papers 1.67

P1 È V2 ˘ =Í ˙ P2 Î V1 ˚ W1- 2 =

471

7 È 0.056 ˘ = P2 ÍÎ 0.014 ˙˚

P2 = 0.6912 bar

mR(T1 - T2 ) 2.469 ¥ 1.005(553 - 218.44) = = 353.83 kJ ( - 1)1000 (1.67 - 1)1000

dh = mCp(T2 – T1) = 2.469 × 1.005(218.44 – 553) = –8301.5 kJ (ii) Work transfer in process 2–3 (W2–3) T2 = T3 P2V2 = P3V3 W2 -3 =

V3 = 0.005529 m3

0.6912 × 0.056 = 7V3

È P ˘ 2.469 ¥ 287 ¥ 218.44 È 0.6912 ˘ mRT2 ln Í 2 ˙ = ln Í ˙ = - 358.36 kJ 1000 1000 Î 7 ˚ Î P3 ˚

dh = mCp(T3 – T2) = 0

[∵ T2 = T3]

(iii) Work transfer in process 3–1(W3–1) V3 V1 = T3 T1

P3 = P1

0.005529 0.014 = 553 T3

T3 = 218.44 K

W3–1 = P(V1 – V2) = 7 × 105(0.056 – 0.005529) = 5.92 kJ dh = mCp(T1 – T3) = 2.469 × 1.005(553 – 218.44) = 830.15 kJ Wn = W1–2 + W2–3 + W3–1 = 353.83 – 358.36 + 5.92 = 1.39 kJ 7(b)

The gas constant of the mixture, (ii) the molecular weight of the mixture, (iii) specific heat Cp and Cv of the mixture (iv) the change in enthalpy of the mixture if the mixture is heated at constant volume to a temperature of 100°C. Ans. m02 = 1 kg, mN 2 = 2 kg, P = 150 kPa, T = 20 + 273 = 293 K, O2 = 32 kg, N2 = 28 kg RO =

8314 J = 259.8 32 kg K

RN =

8314 J = 296.9 28 kg K

(i) The gas constant of the mixture (R)

R=

mO2 RO + mN2 RN m

=

1 ¥ 259.8 + 2 ¥ 296.9 J = 284.5 3 kg K

(ii) The molecular weight of the mixture (M) xO2 =

1 / 32 = 0.3043 1 / 32 + 2 / 28

x N2 =

2 / 28 = 0.6957 1/ 32 + 2 / 28

M = xO2 M O2 + xN2 M N2 = 0.3043 ¥ 32 + 0.6957 ¥ 28 = 29.216

kg kg mole

(iii) Specific heat Cp and Cv of the mixture. Both O2 and N2 are diatomic gas, gdiatomic = 1.4 For O2

Cp − Cv = 259.8,

Cp/Cv = 1.4,

Cv = 0.6495 kJ/kg K

Cp = 1.4Cv = 1.4 × 0.6495 = 0.9093 kJ/kg K

472

Model Question Papers

For N2

Cp − Cv = 296.9

Cp/Cv = 1.4

Cv = 0.742 kJ/kg K

Cp = 1.4Cv = 1.4 × 0.742 = 1.039 kJ/kg K mO2 C pO2 + mN 2 C pN 2

Cp =

mO2 + mN2

=

1 ¥ 0.9093 + 2 ¥ 1.039 kJ = 0.996 1+2 kg K

1 ¥ 0.6495 + 2 ¥ 0.942 kJ = 0.7112 1+2 kg K (iv) Change in entropy Cv =

P

At constant volume V1 = V2 P1 P2 = T1 T2

P 1.5 ¥ 10 5 = 2 293 373

2

1

\ P2 = 1.9 bar = 1.9 × 105 N/m2 DS = mR ln

P1 T + mC p ln 2 [As a mix O2 + N2 ª Air] P2 T1

= 3 ¥ 0.2845 ln

V

1.5 373 kJ + 3 ¥ 0.996 ln = 0.5196 1.9 293 kg K

8(a)

Write down the Van der Waals equation of state. How is it different from the ideal gas equation of state? Ans. Refer to Section 9.2 and 9.3. 8(b) Explain the law of corresponding states. Ans. For a given property (T, P) the gases behave differently but behave very much same when the gases are at the same reduced pressure (Pr) and temperature (Tr) . Pr = P/Pc, Tr = T/Tc. The value of z factor for all the gases is approximately the same at the reduced pressure and temperature; this is called the principle of corresponding states. 8(c)

Obtain expressions for the constants a, b and R in terms of critical properties for a Van der Waals gas. Ans. Refer to derivation up to Eq. (9.13).

473

Model Question Papers

PAPER 3 1(a)

Ans.

Define the following terms: (i) System (iii) Process (v) Thermal equilibrium

(ii) Cycle (iv) Property

(i) System: Refer to Section 1.3. (ii) Cycle: Refer to Section 1.7. (iii) Process: The transformation of a system from one state to another is called a process (also refer to Section 1.6). (iv) Property: Every system has certain characteristics by which is physical condition may be described, e.g. volume, temperature, pressure, etc. such characteristics are called properties of the system (also see Section 1.4). (v) Thermal equilibrium: (Refer to Section 1.9.1).

1(b)

State the similarities between heat and work. Explain an example to illustrate the difference between heat and work. Ans. Similarities between heat and work. Refer to Section 2.8.1. Difference between heat and work. Refer to Section 2.8.2. 1(c)

A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law of PV2 = C until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original position; heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 20 bar. Calculate the network done by the fluid for an initial volume of 0.5 m3. Ans. V1 = 0.5 m3 V2 = 2V1 = 1.0 m3 P P1 = 20 bar V3 = V1 = 0.5 m3 1 We know that 2

2

ÈV ˘ Ê 0.5 ˆ P2 = P1 Í 1 ˙ = 20 ¥ Á = 5 bar Ë V 1 ˜¯ Î 2˚ W.D.1- 2

3

( P V - P2V2 ) = 1 1 (n - 1)

(20 ¥ 0.5 - 5 ¥ 1) ¥ 100 = = 500 kN m (2 - 1)

W.D.2 -3 = P2 (V3 - V2 ) = 5 ¥ 100 W.D.3–1 = 0

[∵ V = C]

kN m2

¥ (0.5 - 1) m 3 = - 250 kN m

2

V

474

Model Question Papers

Net work done (Wn) Wn = W1–2 + W2–3 + W3–1 = 500 kN m – 250 kN m + 0 = 250 kN m 2(a)

Starting from a common state point draw the following expansion processes on the PV plane and write expression for the work in each case. (i) Isochoric process (ii) Isobaring process (iii) Isentropic process (iv) Isothermal process (v) Polytropic process if n > k, k = Cp/Cv P (ii) Ans. (i) W.D. Isochoric process = 0 P=C 1 2 (ii) W.D. Isobaric process = P(V2 – V1)

(iv) W.D. Isothermal process = P1V1 ln (v) W.D. Polytropic process = 2(b)

2

V=C

( P1V1 - P2V2 ) k -1

(iii) W.D. Isentropic process =

V2 V1

2 2

( P1V1 - P2V2 ) n -1

(iii)

2

(i)

(iv)

(v) V

Consider the system shown in the Figure of Q.2(b). Initial condition of the gas are V1 = 0.1 m3, P1 = 200 kPa. The ambient atmospheric pressure is 100 kPa. The spring exerts a force which is proportional to the displacement from its equilibrium position. The gas is heated until the volume is doubled at which point P = 600 kPa. Determine the work done by the gas.

kx1

Piston

P1 = 200 kPa 3

V1 = 0.1 m

A = Area of the piston Ps = kx1

100 kPa = Pa

x Q

Work done against gas (Wg) Ans. Force balance at any instant gives K - (V - V0 ) A where V0 is the volume of the gas when the spring is at it natural length PA = Pa A + kx1 = Pa A +

Wg =

Ú

2

1

PdV =

Wg = (V2 - V1 )

Ú

2

1

Pa +

˘ (V - V0 ) ˙ dV A ˚ k

2

( P1 + P2 ) (200 + 600) = (0.2 - 0.1) = 40 kJ 2 2

475

Model Question Papers

3(a) Write down SFEE and explain briefly the different terms in the equation. Ans. Refer to Section 3.9.1. 3(b)

A cylinder containing the air comprises the system. Cycle is completed as follows: (i) 82000 Nm of work is done by the piston on the air during compression stroke and 45 kJ of heat is rejected to the surroundings. (ii) During expansion stroke 100000 Nm of work is done by the air on the piston. Calculate the quantity of heat added to the system. Ans. (i) W1–2 = –82 kN m Q2–1 2 Q1–2 = –45 kJ W2–1 (ii) W2–1 = 100 kN m W1–2 Q2–1 = ? Apply the first law of thermodynamics to a cyclic process.

Ú

W=

Ú

Q1–2

Q

1

W1–2 + W2–1 = Q1–2 + Q2–1 – 82 kJ + 100 kJ = –45 kJ + Q2–1 kJ Q2–1 = 63 kJ

\ 3(c)

A stone of 20 kg mass and a tank containing 200 kg water comprise a system. The stone is 15 m above the water level initially. The stone and water are at the same temperature initially. If the stone falls into water then determine Du, DP.E., DK.E., Q and W when, (i) The stone is about to enter the water. (ii) The stone has come to rest in the tank. (iii) The heat is transferred to the surroundings in such an amount that the stone and water come to their initial temperature. Ans. mS = mass of stone = 20 kg mW = mass of water = 200 kg (i) When the stone is about to enter the water: Q = 0, W = 0, DU = 0

\

Q = (u2 - u1 ) + m (c22 - c12 ) + mg ( z2 - z1 ) + W DP.E. = –DK.E. = 20 × 9.81 × (–15) = –2943 J DK.E. = +2943 J DP.E. = –2943 J Stone

1 z1

0 = z2

15 m 2 Water

476

Model Question Papers

(ii) When the stone has come to rest in tank: DK.E. = 0, Q = 0, W = 0 0 = DU + 0 + DP.E. + 0 DU = –DP.E. = –(–2943) = 2943 J DP.E. = –2943 J This shows that the I.E. (temperature) of the system increases. (iii) When the water and the stone come to their initial temperature: W = 0, DK.E. = 0 Q = –DU = –2943 J –ve sign shows that heat is lost from the system to the surroundings. 4(a) Explain the various reasons for irreversibility. Ans. Refer to Section 4.7.7. 4(b)

Mention which of the following are reversible process. (i) Isentropic process (ii) Evaporation of water at constant pressure (iii) Throttling process without heat loss to surroundings (iv) Heat transfer through a finite temperature difference (v) Constant volume combustion with adiabatic wall. Ans. (i) Reversible process (ii) Reversible process (iii) Irreversible process (iv) Irreversible process (v) Irreversible process. 4(c)

A reversible heat engine operates between two reservoirs at temperature 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. (i) Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir at 50°C. (ii) Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir at 50°C given that the efficiency of the heat engine and the COP of the refrigerator are each 45% of their maximum possible values. Ans. (i) Heat transferred to the refrigerant (Q4) and net heat transfer to the reservoir at 50°C (Q2 + Q3) Wn = W1 – W2 = 400 kJ HE

È T ˘ 323 ˘ È = Í1 - 2 ˙ ¥ 100 = Í1 ¥ 100 = 66.8% T1 ˚ 973 ˙˚ Î Î

0.668 =

W1 W1 = Q1 2500 kJ

Model Question Papers

T3 = –25°C

T1 = 700°C

Q4

Q1 = 2500 kJ HE R

477

W1

W2

Ref.

Wn = W1 – W2 Q2

Q3 T2 = 50°C

\ \

W1 = 0.668 × 2500 = 1670 kJ W2 = Wn – W1 = 400 – 1670 = –1270 kJ COPRef = 3.306 =

\

\

T3 248 = = 3.306 T2 - T3 323 - 248 Q4 Q = 4 W2 1270

Q4 = 3.306 × 1270 = 4200 kJ Q2 = Q1 – W1 = 2500 – 1670 = 830 kJ Q3 = W2 + Q4 = 1270 + 4200 = 5470 kJ Q2 + Q3 = 830 + 5470 = 6300 kJ

(ii) If the efficiency of the heat engine and COP of the refrigerator each are 45% of their maximum possible values. ha = hHE × 0.45 = 0.668 × 0.45 = 0.3 0.3 =

\

W1 = 750 kJ Wn = W1 – W2 = 400 kJ W2 = W1 – 400 = 750 – 400 = 350 kJ COPa = 0.45(COP)R = 0.45 × 3.306 = 1.4877 1.4877 =

\

\

W1 W = 1 Q1 2500

Q4 Q = 4 W2 350

Q4 = 1.4877 × 350 = 520 kJ Q3 = W2 + Q4 = 350 + 520 = 870 kJ Q2 = Q1 – W1 = 2500 – 750 = 1750 kJ Q3 + Q2 = 870 + 1750 = 2620 kJ

478

Model Question Papers

5(a) Explain briefly available and unavailable energy. Ans. Refer to Section 5.14. 5(b) 300 kJ/s of heat is supplied at a constant fixed temperature of 290°C to a heat engine. The heat rejection takes at 85°C. The following results were obtained: (i) 215 kJ/s is rejected (ii) 150 kJ/s is rejected (iii) 75 kg/s is rejected Classify which of the results report a reversible cycle or irreversible cycle or impossible cycle. Ans. Case I dQ Q1 Q2 300 215 = + = = - 0.2309 < 0 T T1 T2 563 281.5

Ú

\

Cycle is irreversible.

Case II

Ú \

dQ 300 150 = = 0, T 563 281.5

Cycle is reversible.

Case III

\

Ú

Cycle is impossible.

dQ 300 75 = = 0.2664 > 0, T 563 281.5

5(c)

In a particular chemical plant 5000 kg vapour of a substance is produced at 1000°C. While a second unit in the same plant yields 5000 kg of the same substance at 600°C. These two products are mixed together in an insulated container and the mixture is used as a source for a heat engine till the temperature is reduced to 40°C. A recently recruited engineer suggested that if the two products are used separately as source for a heat engine, additional work can be obtained. Determine how much additional work can be obtained by adopting the suggestion. The ambient temperature is 25°C. Cp of the product is 1.0 kJ/kg K. Ans. Equilibrium temperature after mixing 1000 + 600 = 800°C = 1073.15 K 2 Available energy or the mixture (AEm) AEm = Heat supplied by the mixture – Unavailable energy = (mCpdT)mix – T0 | DS | =

kJ ¥ (1073.00 - 313) kg K kJ 1073 ln - 298 K ¥ 10 ¥ 10 -3 kg ¥ 1.0 kg K 313 6 AEm = 3.9278 × 10 kJ = 10 ¥ 103 kg ¥ 1.0

Model Question Papers

479

Available energy if these two used separately (AES) AE1 = Q – T0 | DS | = (mCpdT)1 – T0 | DS | AE1 = 5 ¥ 10 3 ¥ 1 ¥ (1000 - 40) - 298.15 ¥ 5 ¥ 103 ¥ 1 ¥ ln

1273.15 313.15

= 2.71 × 106 kJ AE2 = 5 ¥ 10 3 ¥ 1 ¥ (600 - 40) - 298.15 ¥ 5 ¥ 10 3 ¥ 1 ¥ ln

873.15 313.15

= 1.27 × 106 kJ AES = AE1 + AE2 = 2.71 × 106 + 1.27 × 106 = 3.98 × 106 kJ Additional work obtained by incorporating the suggestion (Wa) Wa = AES – AEm = 3.98 × 106 kJ – 3.93 × 106 kJ = 5 × 104 kJ 6(a) Draw a PT diagram for pure substance and indicate all necessary points on it. Ans. Refer to Figure 7.9. 6(b)

If a certain amount of steam is produced at a pressure of 8 bar and dryness fraction 0.8, calculate (i) External work done during evaporation (ii) Internal latent heat of the steam.

Ans. From steam table at 8 bar, Vg = 0.240 m3/kg, hfg = 2046.5 kJ/kg (i) External work done during evaporation (W.D.) W.D. = PxVp = 8 × 100 × 0.8 × 0.240 = 153.6 kJ (ii) Internal latent heat of steam (IL) IL = hw – W.D. = xhfg – W.D. = 0.8 × 2046.5 – 153.6 = 1483.6 kJ 6(c)

Ans.

Two boilers one with superheater and other without superheater are delivering equal quantities of steam into a common main. The pressure in the boiler and main is 20 bar. The temperature of the 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 steam supplied by the other boiler, take Cps = 2.25 kJ/kg.

480

Model Question Papers

hB1 = hg + CPS (tsup - tsat ) = 2797.2 + 2.25(350 – 212.4) = 3106.8 kJ/kg hB2 = h f + x B2 h fq = 908.6 + x B2 ¥ 1888.6 hmain = 2[hg + CPS (tsup – tsat)]at 20 bar & 250°C = 2[2797.2 + 2.25(250 – 212.4)] = 5763.6 kJ By energy balance. hB1 + hB2 = hmain 3106.8 kJ + (908.6 + x B2 1888.6) = 5763.6 kJ \

x B2 = 0.925

7(a) Define perfect gas. Ans. It is a gas having no forces of intermolecular attraction. The gases which follow the gas laws (Boyle’s and Charle’s) at all ranges of pressure and temperature are considered as ‘Ideal gases’. 7(b) Write a note on compressibility factor and compressibility chart. Ans. Refer to Section 9.4. 0.2 m3 of mixture of fuel and air at 1.2 bar and 60°C is compressed until its pressure becomes 12 bar and temperature becomes 270°C. Then it is ignited suddenly at constant volume and its pressure becomes twice the pressure at the end of the compression. Find the maximum temperature reached and change in internal energy. Also find the heat transfer during the compression process. C pm = 1.072 kJ/kg K, R = 0.2943 kJ/kg K, g = C p/Cv = 1.4. Ans. We know that, P 7(c)

1 ˘n

1 È 1.2 ˘ 1.25

ÈP V2 = V1 Í 1 ˙ = 0.2 Í ˙ Î 12 ˚ Î P2 ˚ We know that T1 È P2 ˘ =Í ˙ T2 Î P1 ˚ W.D.1- 2 = Q1- 2 =

1- n n

333 È 12 ˘ ; = 543 ÍÎ 1.2 ˙˚

3

= 0.0316 m 3 2 1- n n

\ n = 1.25

( P1V1 - P2V2 ) 10 5 (1.2 ¥ 0.2 - 12 ¥ 0.0316) = = - 55.68 kJ n -1 0.25 -n (1.4 - 1.25) ¥ W.D.1- 2 = ¥ ( -55.68) = - 33.408 kJ n -1 (1.25 - 1)

Maximum temperature (T3) We have that between 2–3 process,

1

V

Model Question Papers

481

T3 P3 2 P2 = = =2 T2 P2 P2

T3 = 2T2 = 2 × 543 = 1080 K m=

P1V1 1.2 ¥ 10 5 ¥ 0.2 = = 0.2449 kg RT1 0.2943 ¥ 10 3 ¥ 333

Cv = Cp – R = 1.072 – 0.2943 = 0.7777 kJ/kg K Change in internal energy (DQ1–3) Dv1–3 = mCv(T3 – T1) = 0.2449 × 0.7777 × (1086 – 333) = 143.415 kJ 8(a) Define specific heat of a gas. Ans. Specific heat of a gas at a constant volume Cv is defined as the rate of specific internal energy with respect to temperature when the volume is held constant, i.e. Cv =

∂u ˘ ∂T ˙˚V = C

Specific heat of a gas at a constant pressure Cp is defined as the rate of specific enthalpy with respect to temperature when the pressure is held constant, i.e. C p = 8(b) Write down Van der Waals equation and Beattie Bridgeman equation. Ans. Refer to Section 9.2. Beattie Bridgeman equation:

∂h ˘ ∂T ˙˚ P = C

RT (1 - )(V + B) A - 2 2 V V 3 8(c) A rigid tank of 2 m volume is tilled with dry saturated steam at 0.2 MPa. Owing to poor insulation of the tank, the pressure of the steam is found to be 0.1 MPa after some time. Determine the final state of the steam and the amount of energy transferred as heat to the surroundings. Ans. From steam table at 0.2 MPa h1 = hg = 2706.7 kJ/kg, vg = 0.8857 m3/kg = v1 P=

m=

V 2 = = 2.258 kg vg 0.8857

At 0.1 MPa, vf = 0.001043 m3/kg, vg = 1.694 m3/kg, hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg v2 = v1 = 0.8857 (volume held constant) = volume of dry steam + volume of water

482

Model Question Papers

0.8857 = 1.694x2 + (1 – v2) × 0.001043 x = 0.5225 h2 = h7 + xhfg h2 = 417.46 + 0.5225 × 2258 = 1597.27 kJ/kg Change in internal energy (DU) DU = Heat transferred for constant volume process = Q = m(u2 – u1) = m[(h2 – p2v2) – (h1 – p1v1)] = 2.258 [(1597.27 – 0.1 × 103 × 0.8857) – (2706.7 – 0.2 × 103 × 0.8857)] = –2273.26 kJ = Q

Model Question Papers

483

PAPER 4 1(a)

Define mean free path and hence differentiate between microscopic and macroscopic approaches. Ans. Refer to Q. No. 1a (Paper 1). 1(b)

Define: (i) Closed system (ii) Open system (iii) Boundary (iv) Point functions (v) Path functions. Ans. Refer Q. No. 1(b) (Paper 1). 1(c)

Classify the following properties as intensive or extensive: (i) Refractive index of a glass slab. (ii) Velocity of a bullet. (iii) Energy required to lift a bucket of water. (iv) Specific heat of a substance. Ans. (i) Intensive (ii) Extensive (iii) Extensive (iv) Intensive 2(a)

Clearly distinguish between heat and work from the point of view and in the following pair of interactions identify which is heat and which is not. A patient is sitting with a thermometer in his mouth and BP apparatus attached to his hand. The mercury column in both the instruments goes up. Ans. Refer to Sections 2.1 and 2.8.2. Thermometer Æ Heat interacting BP apparatus Æ Work interacting 2(b)

2(c)

In the following three interactions identify the common factor. The system is shown in italic. (i) A dentist grinding a damaged tooth (ii) Glowing of an incandescent lamp (iii) A gas being compressed by a compressor. Each of these represents interaction with the concerned system having both (i) the work and (ii) the heat interactions.

State Zeroth law of thermodynamics and extract the concept of temperature from it. The temperature T on a thermometric scale is defined as 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 0°C and 100°C, respectively. Calculate the temperature for a value of K = 2.42. Ans. Refer to Section 1.11.2. Refer to Example 1.9.

484

Model Question Papers

3(a)

Precisely state the first law of thermodynamics for a closed system undergoing a cycle and prove that internal energy is a property. Ans. Refer to Sections 3.2 and 3.4. 3(b) Write the steady flow energy equation and explain all the terms involved. Ans. Refer to Eq. (3.50). Refer to Section 3.8. 3(c)

A small turbine runs an aircraft refrigeration system. Air enters the turbine at 4 bar and 40°C with a velocity of 40 m/s. At the exit the air is at 1 bar, 2.5°C and having a velocity of 200 m/s. If the work output of a turbine is 52 kJ/kg of air, calculate the heat transferred per kg of air.

Ans. Data: P1 = 4 bar, T1 = 313 K, V1 = 40 m/s, P2 = 1 bar, T2 = 275.5 K,

V2 = 200 m/s, W = 52 kJ/kg, Q = ?, D2 = 0. h1 = C pT1 = 1.005 ¥ (273 + 40); h1 +

V12 40 2 kJ = 1.005 ¥ 313 + = 315.4 2 2 ¥ 100 kg

h2 +

V22 200 2 kJ = 1.005 ¥ 275.5 + = 297 2 2 ¥ 1000 kg

Q - W = (h2 - h1 ) +

(V22 - V12 ) g ( Z 2 - Z1 ) + g1 ¥ 1000 2 ¥ 1000

Q – 52 = (297 – 315.4) = –18.54 kJ/kg Q = 33.5 kJ/kg 4(a) Give clearly the Clausius statement of the second law of thermodynamics. Ans. Refer to Section 4.5.2. 4(b) 4(c)

Between any two heat reservoirs at temperatures T1 and T2 (T1 > T2) prove that no refrigerator can have a compression greater than a Carnot’s refrigerator. A reversible heat engine is operating between two thermal reservoirs at 800°C and 30°C, respectively. It drives a reversible refrigerator operating between –15°C and 30°C. The heat input to the heat engine is 1900 kJ and the net work output from the combined plant (engine and refrigerator both) is 290 kJ. Calculate the heat absorbed by the refrigerant and the total heat transferred to 30°C reservoir.

Ans.

HE

=1-

T2 303 =1= 0.718 T1 1078

WT = hHE × Q1 = 0.718 × 1900 = 1364.2 WT = WR + 290 = 1364.2 \

WR = 1074.2 kJ

Model Question Papers

485

Q2 = Q1 – WT = 1900 – 1364.2 = 535.8 kJ COPRE = =

\

T3 258 = = 5.7333 T2 - T3 45 Q4 Q4 = WR 1074.2

Q4 = 6158.74 kJ Q3 = WR + Q4 = 6158.74 + 1074.2 = 7232.7 kJ

Total heat rejected at T2 = 30°C (Q) Q = Q2 + Q3 = 7232.7 + 535.8 = 7768 kJ 5(a) Derive Clausius inequality and prove that entropy is a property. Ans. Refer to Section 5.6. Or The Carnot principle states that a reversible heat engine is more efficient than an irreversible engine. The efficiency of a reversible engine is given by R

dQ1

T - Tmin T = 1 - min = max Tmax Tmax

Now consider an irreversible engine operating between the same thermal reservoirs. Let dQ1 be the heat absorbed and dQ2 be the heat rejected by the engine. The efficiency of the engine is given by I

=

Tmax

HE

dQ2 Tmin

dQ1 - dQ2 dQ2 =1dQ1 dQ1

Then by Carnot principle, 1-

T T dQ1 dQ2 dQ2 dQ2

Ú

2

1

Q / T and thence show that

5(b) Write the two ‘Tds’ relations and identify each term in these relations. Ans. Tds = PdU + dU [Eq. (5.37)] Tds = dH – VdP [Eq. (5.39)] where P = Pressure dV = Change in volume du = Change in internal energy dH = Change in enthalpy, dP = Change in pressure V = Volume T = Abs. temperature dS = Change in entropy 5(c)

A 5 kg copper block at a temperature of 200°C is dropped into an insulated tank containing 100 kg oil at a temperature of 30°C. Find the increase in entropy of the universe due to this process when copper block and oil reach thermal equilibrium. Assume that the specific heats of copper and oil are respectively 0.4 kJ/kg K and 2.1 kJ/kg K. Ans. Let t be the equilibrium temperature. By energy balance [mCp dt]Cu = [mCp dt]oil 5 × 0.4 (200 – t) = 100 × 2.1 × (t – 30); t = 31.6°C or 304.6 K

DSCu = mC p ln

T˘ 304.6 kJ = - 0.88 ˙ = 5 ¥ 0.4 ¥ T1 ˚ Cu 473 K

DSoil = mC p ln

T˘ kJ 304.6 = 1.106 ˙ = 100 ¥ 2.1 ¥ T1 ˚oil 303 K

DSuniverse = DSCu + DSoil = –0.88 + 1.106 = 0.226 kJ/K

492

Model Question Papers

6(a)

Draw the PV diagram for water and show on it (i) the saturation liquid line, (ii) the saturated vapour line, (iii) the critical point, (iv) an isotherm passing through sub-cooled liquid, mixture and superheated regions and (v) constant dryness fraction lines. Ans. AB = Saturated liquid line BC = Saturated vapour line B = Critical point DBE, FBG etc. = Constant dryness fraction line HIJK = Isotherm line passing through

P

B

H

J

I

K

A

G E C

D F

V

6(b)

Steam at 1 bar and dryness fraction of 0.523 is heated in a rigid vessel until it becomes saturated vapour. Calculate the heat transferred per kg steam.

Ans.

v1 = xvg ˘˚

P

at P = 1 bar

m3/kg

= 0.523 × 1.694 = 0.8857 Heating at constant volume; v2 – v1 = 0.8857 m3/kg

2

At ‘2’ steam is dry saturated, v2 = vg ˘˚ P

2

From steam table, corresponding to v2 = vg ˘˚ P

1

2

V

= 0.8857 m3/kg and dry, pressure = P2 = 2 bar. u1 = h1 – P1v1 \

h1 = (1 - x )h f + xhg ˘˚

and

h1 = (1 – 0.523) × 417.46 + 0.523 × 2675.5 = 1598.4 kJ/kg u1 = 1598.4 -

10 5 ¥ 0.8857 = 1509.8 kJ/kg 10 3

u2 = h2 - P2 v2 = hg - P2 v2 ˘˚

6(c)

P =2

5

2 ¥ 10 ¥ 0.8857 = 2529.6 kJ/kg 103 Q = Du + W.D. = Du [W.D. = 0, rigid vessel] = u2 – u1 = 2529.6 – 1509.8 = 1059.8 kJ/kg. = 2706.7 -

\

P =1

2.5 kg steam at 10 bar and 400°C is cooled at constant pressure in a heat exchanger until it becomes saturated vapour. Find the available and unavailable parts of the energy from this steam. Assume the surroundings are at 27°C.

Ans. Steam is cooled at constant pressure of 10 bar from 400°C to t2 = t5 ]P =10 From S.T., h2 = hg ˘˚

P = 10

= 2778.1 kJ/kg

493

Model Question Papers

h1 = h]P =10 and 400°C = 3263.9 kJ/kg (from steam table) Heat transfer from cooling steam Q = m(h2 – h1) = 2.5 (2778.1 – 3263.9) = –1214.5 kJ

S1 = S ]P =10, t = 400 = 7.4651 kJ/kg

(from S.T.)

S2 = Sg ˘˚

(from S.T.)

P =10, sat dry

= 6.5865 kJ/kg

Unavailable energy = T0DS = T0 mDS = (273 + 27) × 2.5 × (7.4651 – 6.5865) = 659 kJ \

Available energy = | Q | – unavailable energy = 1214.5 – 659 = 555.5 kJ

7(a) Distinguish between universal gas constant and particular gas constant. Ans. R = Universal gas constant = 8.314 kJ/k mole K, Remains same for an substances. R = Particular gas constant = R / M M = Molecular weight of a particular gas. 7(b) 7(c)

Ans.

For question and answer, refer to Section 8.6.5(d), Eq. (8.45). 2 kg air (Cp = 1.005 kJ/kg K, Cv = 0.718 kJ/kg K) is compressed reversibly according to PV1.3 = C from 1 bar 37°C to 5 bar. (i) Find the increase in internal energy, (ii) use the relation in 7(b) and calculate the heat transfer, (iii) using the results in (i) and (ii) calculate the magnitude and direction of work, (iv) show the initial and final states and the process path on TS diagram. (i) Increase in internal energy (Du) n -1 ˘ n

T2 P2 = ˙ T1 P1 ˚

P 2

; T2 = 310 ¥

0.3 1.3 (5)

= 449.4 K

Du = mCv(T2 – T1) = 2 × 0.718 × (449.4 – 310) = 200.18 kJ

1

V

494

Model Question Papers

(ii) Heat transfer (Q) Q=

nCv (T2 - T1 ) -1

1.005 0.718 ¥ 0.718 (49.310) = 1.3 - 1 = –33.36 kJ/kg; –33.36 × 2 = –66.72 kJ (iii) Magnitude of work and direction From the first law of thermodyamics W = Q – Du = –66.72 – 200.18 = –266.9 kJ (iv) Path diagram on PV diagram.

T

1.3 -

P=2 449.4

2

P=1 310

1

S

8(a) Define mass fractions and mole fractions of the constituents of an ideal gas mixture. Ans. Refer to Sections 8.11.1 and 8.11.2. Find the gas constant and apparent molar mass of a mixture of 2 kg of O2 and 3 kg of N2 given that universal gas constant is 8314.3 J/k mole K, molar masses of O2 and N2 are respectively 32 and 28. Ans. (i) Gas constant (R)

8(b)

RN2 =

8314.3 R J = = 296.94 28 M N2 kg K

RO2 =

8314.3 J = 259.82 32 kg K

Mass fraction of O2 and N2 (m f O2 and m f N2 ) m f O2 =

mO2 mO2 + mN2

=

2 = 0.4 2+3

m f N2 = 1 - m f O2 = 1 - 0.4 = 0.6 R=

Âm

fi

Ri = ÈÎ m f O2 RO2 + m f N2 RN2 ˘˚

= 0.4 × 259.82 + 0.6 × 296.94 = 282 J/kg K (ii) Apparent molar mass (M) M=

8(c)

R 8314.3 kg = = 29.48 R 282 kg K

Write short notes on (i) Van der Waals equation of state, (ii) Reduced properties and (iii) Compressibility chart. Ans. Refer to Section 9.3 and 9.3.1, Figure 9.4.

Model Question Papers

495

PAPER 6 1(a)

With suitable examples, distinguish between (i) closed system and open system, (ii) point function and path function, (iii) intensive and extensive properties. Ans. (i) Closed and open system: Refer to Q.1(a) of Paper 2. (ii) Point function: When the characteristics of property depend on the start and end state (or starting point and end point) then that system is called system depending on point function. Example: pressure, temperature, volume. Path function: When the characteristics of a property depend on the path followed by the system and not depend on end states then that system is called system depending on path function. Example: heat and work. (iii) Intensive property: The properties whose value does not depend on the mass of the system are known as intensive properties. Example: pressure, temperature, velocity. Extensive property: The value of some of the properties depends on the mass of the system. They are known as extensive properties. Example: total volume, weight, surface area, enthalpy. 1(b)

With a neat diagram explain the use of constant volume gas thermometer for temperature measurement. Mention how it can be used as constant pressure gas thermometer. Which is preferable and why? Ans. There is a fixed marking ‘M’ over the glass tube. The other Pa limb of the Hg manometer is open to the atmosphere and can be moved vertically to adjust the Hg levels so that the Hg just touches neck ‘M’ of the capillary. The pressure in h the bulb is used as a thermometric property and is given by, M B P = Pa + (h r g); r = density of Hg Initially the bulb B is kept in melting ice and reservoir Bulb level is suitably adjusted so that Hg level is at mark M. Tube Then the ice point pressure (Pi) shall be, Pi = Pa + (hi r g) = P0 (Pressure at 0°C) hi = Height, when bulb kept in melting ice Similarly, the boiling water (steam point) (PS) shall be, PS = Pa + hs r g = P100 (Pressure at 100°C) hs = Height, when bulb kept in boiling water. When the bulb is brought in contact with the system whose temperature is to be measured, the bulb, in course of time, comes in thermal equilibrium with the system. Pt = Pa + ht r g = Pressure at unknown temperature. \

t=

V Pt R

496

Model Question Papers

In a constant pressure gas thermometer, the Hg levels have to be adjusted to keep ‘h’ constant, and the volume of gas V which would vary with the temperature of the system, becomes the thermometric property.

Pt =V R The constant volume gas thermometer is, however, mostly in use, since it is simpler in construction and easier to operate. t=

1(c)

A new scale N of temperature is devised in which the ice point is assigned 100°N and the steam point is assigned 400°N. Establish the relationship between the N scale and the Celsius scale. At what temperature will both the Celsius and the new thermometer readings would be identical numerically? Ans. Refer to example 1.6. 2(a)

Write the thermodynamic definition of work. With a suitable example, explain how it is more general than the definition of work in mechanics. Ans. Definition of work: Work is said to be done by a system if the sole effect of the system on things external to the system, i.e. surroundings, can be reduced to the raising of a weight. Refer to Section 2.1.2, Figure 2.2 2(b)

Show that the displacement work done by a closed system consisting of an ideal gas undergoing an isothermal expansion from state 1 to state 2 is given by W1–2 = mRT1 ln V2/V1, where m is the mass of the gas in kg. V1 and V2 are the initial and final volumes and R is the characteristics gas constant. Ans. Refer to Section 2.5. 2(c)

A fluid contained in a horizontal cylinder with a frictionless leak-proof piston is continuously agitated by means of a stirrer passing through the cylinder cover. The cylinder diameter is 0.4 m. During the stirring process lasting 10 min, the piston slowly moves out a distance of 0.485 m against the atmospheric pressure of 101 kPa. The net work done by the fluid during the process is 2 kJ. The speed of the electric motor driving the stirrer is 840 rpm. Determine the torque in the shaft and the power output of the motor. Wdisp =

Ans.

=

Ú

2

1

PdV = P (V2 - V1 ) = P 3

101 ¥ 10 ¥

4

d2L

2

¥ 0.4 ¥ 0.485 = 6.15 ¥ 10 3 J 4

Wnet = Wdisp – Wst Wst = Wdisp – Wnet = 6.15 × –2 = 4.15 kJ = 2p NT \

T=

2¥

4.15 = 7.8 ¥ 10 -5 kN m ¥ 840 ¥ 10

Wst = Stirring work

Model Question Papers

497

0.485 m

d

Motor

Pa = 101 kPa (1)

(2)

Power output of the motor (P) P=

4.15 = 0.00692 kW 10 ¥ 60

2(d)

A steady electric current from a battery flows through resistor immersed in 10 kg of water contained in a rigid insulated vessel. As a result the temperature of water rises. State whether the heat (Q) and the work (W) are positive, negative , are zero for each of the following systems: (i) Resistor (ii) Water. Ans. (i) Resistor : Q = –ve; W = –ve (ii) Water : Q = +ve; W = 0 3(a) What is meant by law of conservation of energy? Explain. Ans. Refer to Section 3.2. 3(b)

Show that ‘heat’ is a path function.

Ans.

Q1- 2 =

Ú

2

1

Tds

Area under 1–2¢ π 1–2. Hence, heat, Q1–2 is a path function. 3(c)

The properties of a certain fluid are related as follows: u = 196 + 0.718t and Pv = 0.287 (t + 273), where u is specific internal energy in kJ/kg, t is in °C, P is pressure in kN/m2, v is specific volume in m3/k. A closed system consisting of 2 kg of this fluid expands in an irreversible adiabatic process in which the initial and final states are related by Pv1.2 = C. The initial condition are 1 MPa and 200°C and the final pressure is 100 kPa. Determine the work transfer and the change in internal energy for the process. Evaluate also the value of ÚPdV and comment on the results. Ans. Data: u = 196 + 0.718t, Pv = 0.287(t + 273), m = 2 kg, P1v11.2 = P2v 11.2, P1 = 1 MPa, P2 = 100 kPa, t1 = 200°C, Q1–2 = 0, P1v1 = 0.287(t1 + 273) 1000 × v1 = 0.287(200 + 273); \ 1

v1 = 0.135 m3/kg 1

È 1000 kPa ˘ 1.2 P ˘ 1.2 m3 v2 = 1 ˙ ¥ v1 = Í ¥ = 0.135 0.92 ˙ P2 ˚ kg Î 100 kPa ˚ P2v2 = 0.287(t2 + 273)

498

Model Question Papers

100 × 0.92 = 0.287(t2 + 273); \

t2 = 47.4°C

(u2 – u1) = 0.718(t2 – t1) = 0.718(200 – 47.4) = –89 kJ/kg (i) Work transfer (W1–2) 1

Q2

0

= 1W2 = Du = u2 – u1

–W1–2 = –89.0; W1–2 = 89 kJ/kg; W1–2 = 178 kJ (ii) Pdv work (ÚPdv)

Ú

2

1

Pdv =

P1V1 - P2V2 = 430 kJ (n - 1)

Comment on the results: The first process is a non-quasi-static process. The relationship P1v11.2 = P2v21.2 holds good only for the end states. Hence, ÚPdv is not valid and cannot be used. 3(d)

Air flows steadily at the rate of 0.4 kg/s through an air compressor, entering at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg and leaving at 4.5 m/s with a pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The internal energy of the air leaving is 88 kJ/kg greater than that of air entering. The air looses heat at the rate of 59 W to the surroundings as it flows through the compressor. Calculate the power in kW required to drive the compressor. Neglect the change in evaluation between inlet and exit. Ans. Data: m = 0.4 kg/s, C1 = 6 m/s, P1 = 1 × 105 N/m2, v1 = 0.85 m3/kg, C2 = 4.5 m/s, P2 = 6.9 × 105 N/m2 1 2 v2 = 0.16 m3/kg, Du = u2 – u1 = 88 kJ/kg Comp. Q = –59 kW, Dz = 0 Power input (W) C22 - C12 g( z2 - z1 ) Q -W = (u2 - u1 ) + ( P2 v2 - P1 v1 ) + + m 2 ¥ 1000 ¥ gc gc ¥ 1000

4.52 - 62 -59 - W = 88 + (6.9 ¥ 0.16 - 1 ¥ 0.85) ¥ 10 2 + +0 0.4 2 ¥ 1000 ¥ 1 \

–W = 113.392 kJ/kg; W = –113.392 kJ/kg

4(a) Write the Kelvin-Planck and the Clausius statements of the second law of thermodynamics. Ans. Refer to Sections 4.5.1 and 4.5.2. 4(b)

Show that the violation of Clausius statement of second law leads to the possibility of a PMM of the second kind. Ans. Refer to Section 4.6.1. 4(c)

A heat engine is used to drive a heat pump. The heat transfers from the heat engine and the heat pump are used to heat the water circulating through the radiators of a building. The efficiency of the heat engine is 27% and the COP of the heat pump is 4. Evaluate the ratio of the heat transfer to the circulating water in the radiator to the heat transfer to the heat engine. Ans. Refer to Example 4.10.

499

Model Question Papers

4(d)

There are two ways of increasing the efficiency of a Carnot heat engine. (i) Lowering the temperature T2 of the low temperature reservoir by DT, while keeping the temperature T1 on the high temperature reservoir a constant. (ii) Increasing the temperature T1 by DT, while keeping the temperature T2 a constant. Which is more effective? Prove your answer. Ans. Let T2 be decreased by DT keeping T1 a constant T1 T2 + DT (Case I) I

=

T1 - (T2 - DT ) T1 - T2 + DT = T1 T1

(1)

Now let T1 be increased by DT, keeping T2 a constant (Case II) II

=

T1 + DT - T 2 (T1 + DT )

(2)

Q1

Q1

Carnot engine

Carnot engine

Q2

Q2

T2 – DT

T2

Case I

Case II

Comparing Eqs. (1) and (2), it can be seen that the numerators are the same in both the equations. But the denominators of Eq. (2) is greater than the denominator of Eq. (1). Hence, hI > hII, i.e. lowering the temperature of LTR is the more effective way of increasing the Carnot cycle efficiency. 5(a) Prove that entropy is a property of the system. Ans. Refer to Section 5.5. 5(b)

A reversible heat engine shown in the Figure of Q.5(b) operates between three constant temperature reservoirs A, B and C at 600 K, 400 K and 300 K, respectively. It receives 2500 kJ of energy as heat from the reservoir at 600 K and does 1000 kJ of work. Determine the magnitude and direction of heat interactions with the reservoirs B and C. Ans. (i) Q2 and Q3 A B C From the first law of thermodynamics 600 K 400 K 300 K SQ = SW Q2 Q3 i.e. Q1 + Q 2 + Q 3 = W 2500 + Q2 + Q3 = 1000; HE Q1 = 2500 kJ \ Q2 + Q3 = –1500 kJ (1) W = 1000 kJ From Clausius inequality for a reversible cycle, Q1 Q2 Q3 + + =0 T1 T2 T3

Q 2500 Q2 + + 3 =0 600 400 300 From Eqs. (1) and (2), Q2 = –1000 kJ, Q3 = –500 kJ, i.e. heat is rejected to reservoir B and C.

500

Model Question Papers

5(c)

0.5 kg of ice block at –10°C is brought into contact with 5 kg copper block at 80°C in an insulated container. Determine the change in entropy of (i) ice block, (ii) copper block, (iii) the universe. Given specific heat of ice = 2 kJ/kg K, specific heat of water = 4.2 kJ/kg K, specific heat of copper = 0.5 kJ/kg K, enthalpy of fussion of water at 0°C = 334 kJ/kg. Ice Copper Ans. When the ice and copper block are brought into 0.5 kg 1 kg contact, let the equilibrium temperature attained –10°C 80°C be ‘t’. Heat lost by copper = Heat gained by ice

m C p DT ˘˚

copper

= m C p DT ˘˚

ice

5 × 0.5 × (80 – t) = 0.5{C Pice [0 - ( -10)] + 334 + C PH \

2O

(t - 0)}

t = 5°C DScopper = mCP ln

T 278 kJ = 5 ¥ 0.5 ¥ ln = 0.597 T1 353 K

È T ˘ 273 h fg DSice = mice ÍC Pice ln + + CPH O ln ˙ 2 263 273 273 ˚ Î 273 334 278 ˘ kJ È = 0.5 Í 2 ln + + 4.2 ln = 0.687 ˙ 263 273 273 ˚ K Î kJ K 6(a) Sketch TP phase diagram for water. Mark on it the following regions of solid, liquid and vapour, triple point and critical point. Ans. Refer to Figure 7.10.

\

6(b)

DSuniverse = DScopper + DSice = –0.597 + 0.687 = 0.0899

State whether the following samples of steam are wet, dry or superheated. Justify your answer: (i) Pressure = 1 MPa abs, enthalpy = 2880 kJ/kg (h) (ii) Pressure = 500 kPa abs, volume = 0.35 m3/kg (v) (iii) Temperature = 200°C, pressure = 1.2 MPa (iv) Temperature = 100°C, entropy = 6.88 kJ/kg K (s) (v) Pressure = 10 kPa, enthalpy = 2584.8 kJ/kg (h) Ans. From steam table (i) hg = 2776.2 kJ/kg, \ 2880 > hg, steam is superheated (ii) vg = 0.3746 m3/kg, v < vg, \ steam is wet (iii) tsat = 187.96°C, 200°C > 187.96°C, steam is superheated (iv) sg = 7.3554 kJ/kg K, s < sg, steam is wet (v) hg = 2584.8 kJ/kg, h = hg, steam is dry saturated

Model Question Papers

501

6(c)

Dry steam at 1 MPa absolute expands in a horizontal adiabatic nozzle to 10 kPa. Find the velocity at steam issuing out of the nozzle, assuming frictionless flow. Neglect initial velocity of steam. Ans. From steam table at inlet (dry saturated) hg1 = 2776.2 kJ/kg, sg1 = 6.5828 kJ/kg 1 2 At exit, at 10 kPa, hf2 = 191.8 kJ/kg, hfg2 = 2392.9 kJ/kg, sf2 = 0.6493 kJ/kg K, sfg2 = 7.5018 kJ/kg K Adiabatic expansion in the nozzle, \ S1 = S2 Sf2 + x2Sfg2 = S1 = Sg1 0.6493 + x2 + 7.5018 = 6.5828; x2 = 0.79 h2 = hf2 + x2hfg2 = 191.8 + 0.79 × 2392.8 = 2082.2 kJ/kg Velocity at exit (C2) if C1 = 0 C2 = 2(h1 - h2 ) + C12 = 2 ¥ (2776.2 - 2082.2) ¥ 103 = 1178 m/s 6(d)

Wet steam at 1 MPa enters a throttling calorimeter and is throttled to 100 kPa. What is the minimum value of dryness fraction that can be measured using the throttling calorimeter under these conditions? How can the dryness fraction be measured, if it is less than this value?

Ans.

x min =

hg 2 - h f 1 h fg1

=

2675.4 - 762.6 = 0.95 2013.6

from S.T., h f 1 = 762.6, h fg1 = 2013.6 ˘ È P1 = 1 MPa, Í ˙ hg 2 = 2675.4 ÎÍ P2 = 100 kPa, ˚˙ If the dryness fraction of the steam entering the throttling calorimeter is less than this, a separator has to be used to ensure that steam gets superheated after throttling. 7(a)

Starting from the relation Tds = du + Pdv, show that for an ideal gas undergoing a reversible adiabatic process, the law for the process is given by Tvg–1 = C. Ans. Refer to Section 8.7. 7(b)

A mass of 0.25 kg of an ideal gas has a pressure of 300 kPa, a temperature of 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 and Cv of the gas and the increase in entropy of the gas. Also find the molecular weight of the gas.

Ans.

R=

P1V1 kN 1 1 kJ = 300 2 ¥ 0.07 m 3 ¥ ¥ K = 0.238 mT1 0.25 kg (273 + 80) kg K m

T2 =

P2V2 kN 1 1 kg K = 300 2 ¥ 0.1 m 3 ¥ ¥ = 505 K mR 0.25 kg 0.238 kJ m

502

Model Question Papers

From the first law of thermodynamics, Q1- 2

0

= W1- 2 + Du = mCv (T2 - T1 ) + W1- 2

0 = 0.25 kg ¥ Cv

kJ (505 - 353) K - 25 kJ kg K

kJ kg K CP = CV + R = 0.658 + 0.238 = 0.896 kJ/kg K

\

CV = 0.658

DS = S2 - S1 = m Cv ln

= 0.25 ¥ 0.896 ln

P2 v + m C P ln 2 P1 v1

0.10 kJ = 0.08 0.07 kg K

Molecular weight (M) M=

R kJ 1 kg K kg = 8.314 ¥ = 34.9 R kg mole K 0.238 kJ kg mole

7(c)

A mixture of ideal gases contains 3 kg of N2 and 5 kg of CO2. The partial pressure of CO2 in the mixture is 155 kPa. Find (i) Partial pressure of N2, (ii) Gas constant for the mixture and (iii) Molecular weight of the mixture. 3 = 0.1071 Ans. Number of moles of N2, N2 = 28 5 = 0.1130, CO2 = T = N 2 + CO2 = 0.1071 + 0.113 = 0.221 44 Mole fraction of N2, xN2 =

xCO2 =

N2

= 0.485,

T CO2

= 0.515 nT Partial pressure ratio of CO2 = mole fraction of CO2 PCO2 = xCO2 = 0.515 PT \ Total pressure of the mixture, PT = 301 kPa Partial pressure of N2, PN2 = PT × mole fraction of N2 = 301 × 0.485 = 146 kPa Molecular weight of the mixture (M) M=

mT 3+5 8 = = = 36.2 0.221 0.221 nT

Characteristic gas constant (R) R=

R 8.314 kJ = = 0.229 M 36.2 kg K

Model Question Papers

503

Or R = m f CO2 ¥ RCO2 + m f N2 ¥ RN2 = 5 ¥ M=

8(a)

R 8.314 kg = = 36.2 R 0.229 kg mole

Write the Van der Waals equation of state. In what ways, is it an improvement over the ideal gas equation of state?

Ans. 8(b) Ans.

8(c)

8.314 8.314 K +3¥ = 0.229 44 28 kg

a ˘ È Í P + 2 ˙ [ v - b] = RT . Refer to Section 9.1. u ˚ Î

Explain: (i) Reduced properties, (ii) Compressibility factor, (iii) Law of corresponding states, (iv) Generalised compressibility chart. P T V , Tr = , Vr = Pc Tc Vc (ii) Compressibility factor: Real gases behave as ideal gases only in certain regions, i.e. the real gas behaves like an ideal gas as it approaches zero pressure. The deviation at a given temperature and pressure can be accounted by the introduction of a factor called compressibility factor, Z, defined as Z = PV/RT, Z = 1 for ideal gases. (iii) Law of corresponding states: Refer to Q.8(b) of Paper 2. (iv) Generalized compressibility chart: It is a chart in terms of reduced properties as shown in Figure 9.4. It is very useful in predicting the properties of substances for which more precise data are not available. (i) Reduced properties: Pr =

1 kg of propane (C3H8) is at a pressure of 7 MPa and a temperature of 150°C. The critical properties of propane are Pc = 4.26 MPa, Tc = 370 K and Vc = 0.0054 m3/kg, compressibility factor = 0.54. Calculate. (i) The reduced pressure, volume and temperature (ii) Specific volume of propane using ideal gas equation. P 7 Pr = = = 1.64 Pc 4.26 T 423 Tr = = = 1.144 Tc 370 R 8.314 kJ = = 0.1889 M 44 kg K 0.54 ¥ 0.1889 ¥ 423 v= = 6.16 ¥ 10 -3 m 3 /kg 3 7 ¥ 10 R=

vr = From ideal gas equation v=

v 6.16 ¥ 10 -3 = = 1.36 vc 0.0054 m3 RT 0.1889 ¥ 423 = = 0.0114 kg P 7 ¥ 10 3

504

Model Question Papers

PAPER 7 1(a)

Define open, closed and isolated system. Draw the boundary of the following systems and label their type: (i) A wind mill, (ii) mixture of ice and water in a metal container, (iii) thermo flask filled with hot tea, (iv) pressure cooker, (v) storage cell producing electricity. Ans. Refer to Sections 1.3.1, 1.3.2 and 1.3.3. (i) Open system, (ii) Closed system, (iii) Isolated system, (iv) Closed system, (v) Closed system. 1(b)

Define Zeroth law of thermodynamics, steam point and ice point. The temperature scale of a certain thermometer is given by the relation t = a ln (x) + b, where a and b are constants and x is the thermometric property of the fluid in the thermometer. If at the ice and steam points the thermometric properties are found to be 1.5 and 7.5, respectively, what will be the temperature corresponding to the thermometric property 3.5? Ans. Refer to Section 1.11.2 and Example 1.9. 2(a)

Define heat and thermodynamic definition of work. The initial pressure and volume of a mass of gas in a cylinder fitted with a moveable piston are P1 bar and V1 m3, respectively. Calculate the work done by the system when the gas expands reversibly to a volume V2 m3 during the following processes: (i) Isothermal process and (ii) Polytropic process. Ans. Heat: Refer to Section 2.7. Work: Refer to Section 2.1.2. Isothermal process: Refer to Section 2.5(ii). Polytropic process: Refer to Section 2.5(iv). 2(b) Define (i) Equilibrium state and (ii) Quasi-static process. Ans. Refer to Sections 1.6.1 and 1.6.2. 2(c)

A perfect gas is undergoing a process in which T μ V–2/5. Calculate the work done by the gas in going from state 1 in which pressure is 100 bar and volume is 4 m3 to the state 2 in which volume is 2 m3. Also calculate the final pressure.

Ans.

PV = mRT,

T μ V–2/5,

mR = K

mRT KT = = KT V -1 = KV -2 / 5V -1 = KV -7 / 5 V V K = PV7/5 = 100 × 47/5 = 696.44

P=

\

Work done (W.D.) W.D. =

Ú

2

1

P dV =

Ú

W.D. = –319.66 kJ

2

1

KV -7 / 5 dV = - K

5 2

È 1 1 ˘ Í 2/5 - 2/5 ˙ V1 ˙˚ ÍÎ V2

Model Question Papers

505

Define enthalpy and show that H = U + PV with usual notations. A system contains 0.15 m3 of air at a pressure of 3.8 bar and 150°C. It is expanded adiabatically till the pressure falls to 1 bar. The gas is then heated at a constant pressure till its enthalpy increases by 70 kJ. Determine the total work done. Ans. The total enthalpy of a substance is given by

3(a)

H = U + PV We know that at constant pressure, DH = Q dH = dQ dH = d(U + PV) = dU + d(PV) = dU + PdV [∵ P = C] H = U + PV

T2 È P2 ˘ =Í ˙ T1 Î P1 ˚

-1

0.4

È 100 ˘ 1.4 ; T2 = 423 ¥ Í ˙ = 288.86 K Î 380 ˚ 1

È 380 ˘ 1.4 3 V2 = 0.15 Í ˙ = 0.3892 m Î 100 ˚ m=

P1V1 380 ¥ 0.15 = = 0.4695 kg RT1 0.287 ¥ 423

Process 2–3 DH = mCP(T3 – T2) 70 = 0.4695 × 1.005 × (T3 – 288.84) ; T3 = 437.43 K V3 =

T3 437.43 ¥ 0.3892 V2 = = 0.58944 m 3 T2 288.86

Total work done (W) È P V - P2V2 ˘ W = W1- 2 + W2 - 3 = Í 1 1 ˙ + P2 [V3 - V2 ] -1 ˚ Î (380 ¥ 0.15 - 100 ¥ 0.3892) + 100[0.58944 - 0.3892] 0.4 = 65.8275 kJ

=

3(b)

Define steady flow process with an example. A fluid flows through a steady flow system at the rate of 3 kg/s. The inlet and outlet conditions are P1 = 5 bar, V1 = 150 m/s, u1 = 2000 kJ/ kg and P2 = 1.2 bar, V2 = 80 m/s, u2 = 1300 kJ/kg. The change in specific volume is from 0.4 m3/kg to 1.1 m3/kg. The fluid loses 25 kJ/kg heat during the process. Neglecting potential energy, determine power output of the system. Ans. Refer to Sections 3.9 and 3.10.1.1.

506

Model Question Papers

È Ve2 - Vi2 ˘ È g ( Z e - Z i ) ˘ Q - W = [ he - hi ] + Í ˙+Í ˙ ÎÍ 2 ¥ 1000 ¥ gc ˚˙ Î gc ¥ 10000 ˚

[Eq. (3.50)]

È V 2 - Vi 2 ˘ Q - W = (u2 - u1 ) + ( P2V2 - P1V1 ) + Í e ˙ ÎÍ 2 ¥ 1000 ¥ 1 ˚˙ È V 2 - Ve2 ˘ W = (u1 - u2 ) + ( P1V1 - P2V2 ) + Í 1 ˙+Q ÎÍ 2 ¥ 1000 ¥ 1 ˚˙

= (2000 - 1300) + (500 ¥ 0.4 - 120 ¥ 1.1) +

150 2 - 80 2 + 25 2 ¥ 1000 ¥ 1

= 801.05 kJ/kg Power output (P) P = m ¥W =3

kg kJ ¥ 801.05 = 2404 kW s kg

4(a)

Define the two statements of the second law of thermodynamic. Show that the violation of Clausius statement of the second law of thermodynamics violates the Kelvin-Planck statement of the second law of thermodynamics. Ans. Refer to Sections 4.5.1, 4.5.2 and 4.6.1.

Define heat engine and heat pump and show that COP ]HP = 1 + COP ]R where HP stands for heat pump and R stands for refrigerator. The minimum power required to drive an HP which maintains a house 20°C is 3 kW. If the outside temperature is 3°C, estimate the amount of heat which the house loses per minute. Ans. Heat engine: Refer to Section 4.2 and Figure 4.7. Heat pump: Refer to Section 4.3 and Figure 4.9(b). Derivation: Refer to Section 4.3.1.

4(b)

COPHP =

\

TH Q 293.15 = = 17.24 = H TH - TL 293.15 - 276.15 Wn

QH = Wn × COPHP = 17.24 × 3 = 51.72 kJ/s

Heat lost per minute = 51.72 × 60 = 3103.2 kJ TH = Source temperature, TL = Sink temperature, QH = Heat lost to source. 5(a)

Define inequality of Clausius and entropy of a system. Show that for an irreversible process dS ≥ dQ/T. Ans. Whenever a system undergoes a complete cycle, the cyclic integral of dQ/T around the cycle is less than or equal to zero. This is known as Clausius in equality. The quality of energy and quantity of irreversibility of a process are very important in the analysis of a thermal system. A property used to measure the quality of energy and irreversibility of the process is known as entropy (dQ/T). For derivation, refer to Section 5.7, Eq. (5.28).

Model Question Papers

507

5(b)

Define the principle of increase of entropy. A heat engine is supplied with 278 kJ/s of heat at a constant fixed temperature of 283°C and the heat rejections take place at 5°C. The following results were reported: (i) 208 kJ/s of heat rejected, (ii) 139 kJ/s of heat rejected and (iii) 70 kJ/s of heat rejected. Classify which of the results report a reversible cycle, irreversible cycle or impossible cycle. Ans. Principle of increase of entropy. Refer to 5.8. For solution, refer to Example 5.38. 6(a)

Define critical temperature and pressure. Draw a neat sketch of temperature volume diagram for water showing liquid and vapour phases. Mark all the salient points on the diagram. Ans. Refer to Figure 7.7. 6(b)

Define dryness fraction. Determine the dryness fraction of the steam sample which is tested in a separating and throttling calorimeter and the following data were obtained: (i) Pressure of steam sample = 15 bar (ii) Pressure of steam at exit = 1 bar (iii) Temperature of steam at exit = 150°C (iv) Water collected from separating calorimeter = 0.2 kg/min (v) Discharge collected at exit = 10 kg/min. x1 =

Ans.

W2 10 = = 0.98 (W1 + W2 ) (0.2 + 10)

From S.T. at 15 bar, hf = 844.67, at 1 bar and 150°C, hsup = 2776.4

hfg = 1945.7

[ h f + x2 h fg ]P=15bar = hsup ˘˚

1bar and150°C

844.67 + x2 × 1945.7 = 2776.4 \ x2 = 0.99 Dryness fraction (x) x = x1 × x2 = 0.98 × 0.99 = 0.97 7(a)

Define mass fraction and mole fraction. A gas mixture consists of 0.5 kg of carbon monoxide and 1 kg of CO2. Determine: (i) Mass fraction of each component (ii) Mole fraction of each component (iii) The average molar mass (iv) The gas constant of the mixture.

Ans. Mass fraction: Refer to Section 8.11.1. Mole fraction: Refer to Section 8.11.2. Mass fraction (m f CO , m f CO2 )

m f CO =

mCO 0.5 = = 0.3333, m 1.5

[ m = mCO + mCO2 ]

508

Model Question Papers

m f CO2 =

1 = 0.6667 1.5

Mole fraction of each component (n f CO , n f CO2 ) N CO =

N CO2 =

mCO 0.5 kg = = 0.0179 kmole M CO 28 kg/kmol

mCO2 M CO2

Total number of mole, N = f CO

f CO2

= =

=

f CO

+

1 kg = 0.0227 kmole 44 kg/kmol f CO2

= 0.0406 kmole

N CO 0.0179 = = 0.441 0.0406 n N CO2 n

=

0.0227 = 0.559 0.0406

Average molar mass (M) M=

m 1.5 = = 36.95 kg mole N 0.0406

R=

R 8.314 kJ = = 0.225 M 36.95 kg mole K

Gas constant (R)

7(b)

Define the term ideal gases and real gases. One kg of air at a pressure of 8 bar and temperature of 100°C undergoes a reversible polytropic process following the law PV1.2 = C. If the final pressure is 1.8 bar, determine. (i) Final specific volume, temperature and increase in entropy, (ii) Work done and heat transfer. Ans. Ideal gas: Refer to Section 8.1. Real gas: Refer to Section 9.1.

v1 =

RT1 0.287 ¥ 373 m3 = = 0.1338 P1 800 kg

Final specific volume (v2), temperature (T2) 1

1

È P ˘ 1.2 m3 È 8 ˘ 1.2 v2 = v1 Í 1 ˙ = 0.1338 ¥ Í ˙ = 0.4637 kg Î 1.8 ˚ Î P2 ˚

T2 =

P2V2 1.2 ¥ 100 ¥ 0.4637 = = 290.8 K, 17.8°C R 0.287

Model Question Papers

509

Increase in entropy (DS) DS = CV ln

T2 V 290.8 0.4637 + R ln 2 = 0.718 ln + 0.287 ln 273 0.1338 T1 V1

DS = 0.1782 kJ/kg K Work done (W) W=

R (T1 - T2 ) 0.287(100 - 17.8) kJ = = 117.96 ( n - 1) 0.2 kg

Heat transfer (Q) Du = Cv(T2 – T1) = 0.718(290.8 – 373) = –58.94 kJ/kg Q = Du + W = –58.94 + 117.96 = 59.02 kJ/kg 8(a)

Write short notes on: (i) Van der Waals equation of state (ii) Law of corresponding states Ans. (i) Refer to Sections 9.2 and 9.3. (ii) For a given property (P, T) gases behave differently but behave very much same when the gases are at the same reduced properly (Pr, Tr). The value of Z factor for all the gases is approximately the same at the reduced pressure and temperature. This is called the principle of corresponding states. 8(b)

Show that entropy of an ideal gas is given by the equation S2 - S1 = C p ln

V2 P + Cv ln 2 V1 P1

Starting from the general property relations Tds = du + PdV and Tds = dh – VdP Ans. Tds = du + PdV ds =

du PdV dT R + = Cv + dv T T T v

È du = Cv dT , ˘ Í Pv = RT ˙ Î ˚

Integrating this equation between 1 and 2 S2 - S1 = Cv

Ú

2

1

dT +R T

Ú

2

1

T v dv = Cv ln 2 + R ln 2 v T1 v1

T2 P2 v2 = T1 P1 v1

\

S2 - S1 = Cv ln = Cv ln

P2 v2 v P v v + R ln 2 = Cv ln 2 + Cv ln 2 + R ln 2 P1v1 v1 P1 v1 v1 P2 v P v + ( R + Cv ) ln 2 = Cv ln 2 + C p ln 2 P1 v1 P1 v1

Bibliography

Bejan, A. (2006): Advanced Engineering Thermodynamics, 3rd ed., John Wiley & Sons, New York. Black, W.Z. and Hartley, J.G. (1991): Thermodynamics, 2nd ed., Harper Collin Publisher, New York. Cengel, Y.A. and Boles, M.A. (2008): Thermodynamics—An Engineering Approach, 6th ed., Tata McGraw-Hill, New Delhi. Engel, T. and Reid, P. (2007): Thermodynamics, Statistical Thermodynamics and Kinetics, Pearson Education, New Delhi. O’Connell, J.P. and Haile, J.M. (2006): Thermodynamics Fundamentals for Applications, Cambridge University Press, New York. Rao, Y.V.C. (1993): An Introduction to Thermodynamics, Wiley Eastern Limited, New Delhi. Wylen, G.V. and Sonntag, R.E. (1989): Fundamentals of Classical Thermodynamics, Wiley Eastern Limited, New Delhi. Zemansky, M.W., Abbott, M.M., and Van Ness, H.C. (1975): Basic Engineering Thermodynamics, McGraw-Hill, New York.

453

Index

Amagot’s law, 392 Availability, 281, 284 heat flow, 292 liquids, 292 non-flow, 287 reversible work, 284, 286 solids, 292 steady flow work, 285 Avogadro’s hypothesis, 370 Calorimeter combined, 338 throttling, 337 Carnot cycle, 191 Carnot theorem, 194, 233 corollary-I, 194, 199 corollary-II, 195, 200 corollary-III, 196 Characteristics gas constant, 394 Clausis, 181 inequality, 235 theorem, 234 Compressibility factor, 438

Constants characteristics, 394 Van der Waals, 435 Control surface, 4 volume, 3 Corollary I, 194, 199 II, 195, 200 III, 196 Cycle, 9 Dalton’s law, 391 Diathermic wall, 11 Effectiveness, 294 Energy, 91 internal, 91 Enthalpy, 92, 371, 381 latent heat, 327 sensible, 327 superheat, 327

511

512

Index

Entropy, 232 change, 241 irreversible, 238 principle of increase, 239 process, 243 adiabatic, 243 adiabatic mixing, 244 constant pressure, 243 constant temperature, 243 constant volume, 242 isothermal mixing, 244 polytrophic, 244 property, 235 Equality of temperature, 11 Equilibrium, 8 Equivalence of second law, 181 Expansion, unrestrained, 190 External work dry steam, 329 superheat, 329 wet steam, 329

First law of thermodynamics, 88, 240 Formation of steam, 324 saturated steam, 327 superheated steam, 327 Free expansion process, 42

Gases, 434 real, 434

Heat, 35, 43, 44 Heat engine, 175 reversed, 177 refrigerator, 177 Heat reservoir sink, 179 source, 179

Ideal gas, 368 equation, 369 mixture, 368, 389 constant specific heat, 395 enthalpy, 395 Internal energy, 91, 371, 381 steam, 330 dry, 330

superheated, 330 wet, 330 Internal reversible, 9 International practical temperature scale, 15 Irreversibility, 281, 289 non-flow, 290 steady flow, 290 Kelvin-Planck, 179 Latent heat, 327 Law Amagot’s, 392 Boyle’s, 369 Charles’, 369 Dalton’s, 391 first law of thermodynamics, 88, 240 closed system, 89 adiabatic, 95 closed, 97 constant pressure, 94 constant temperature, 95 constant volume, 94 isentropic, 96 limitation, 171 open system, 97 cycle, 88 process, 89 second law efficiency, 294 second law effectiveness, 294 second law of thermodynamics, 179, 240 Clasius, 181 equivalence, 181 Kelvin-Planck, 179 zeroth law of thermodynamics, 12, 198 Macroscopic, 2 Mass fraction, 389 Maximum shaft, 283 work, 283 Microscopic, 2 Mole fraction, 390 Molecular weight, 391 Non-dimensional property, 436 Non-flow process, 374 adiabatic, 377

Index

constant pressure, 375 constant temperature, 376 constant volume, 374 polytrophic, 378

Real gases, 434 Reservoir sink, 179 source, 179 Reversible process, 9

Open system, 384 Path, 6 Path function, 7 PdV work, 37 constant pressure, 39 constant temperature, 39 constant volume, 38 polytropic, 39 Point function, 7 Process, 6 adiabatic, 243 adiabatic mixing, 244 constant pressure, 243 constant temperature, 243 constant volume, 242 external, 183, 184 internal, 183 irreversible, 183, 187, 188, 189, 238 dissipative, 189 external, 189 mechanical, 189 thermal, 189 isothermal mixing, 244 polytrophic, 244 quasi-static, 8 reversible, 9, 183, 185 total, 184 Property entropy, 235 extensive, 6 intensive, 5 internal energy, 91, 93 non-dimensional, 436 Pure substance, 322 constant pressure, 332 constant temperature, 333 constant volume, 332 isentropic, 334 throttling, 335 Quasi-static process, 8

Second law effectiveness, 294 Second law efficiency, 294 Second law of thermodynamics, 179, 240 Clasius, 181 equivalence, 181 Kelvin-Planck, 179 Sensible heat, 327 SFEE, 97 Sink, 179 Source, 179 Specific heat, 372 constant pressure, 93 constant volume, 93 liquids, 94 solids, 94 Specific volume dry steam, 328 saturated water, 328 superheat, 329 Standard scale, 13 State, 6 Steam dry, 330 superheated, 330 table, 336 wet, 330 Steam formation, 324 saturation temperature, 327 superheated steam, 327 Superheat, 327 System, 3 closed, 3 isolated, 3 open, 3 Temperature concepts, 11 equality, 11 measurements, 14 scale, 12, 196 Thermodynamic cycle, 9

513

514

Index

equilibrium, 9 chemical, 11 mechanical, 10 thermal, 9 path, 6 process, 6 property, 5 extensive, 6 intensive, 5 state, 6 system, 3 closed, 3 isolated, 3 open, 3 total reversible, 9 Thermometer, 12 Thermometric property, 12 Two property rule, 324 Unrestrained expansion, 190

Work, 35, 36, 40, 44 changing area, 42 constant pressure, 39 constant temperature, 39 constant volume, 38 electrical, 40 external dry steam, 329 superheat, 329 wet steam, 329 paddle, 41 PdV, 37 constant pressure, 39 constant temperature, 39 constant volume, 38 polytropic, 39 polytropic, 39 reversible, 284, 286 shaft, 41 steady flow, 285 straining, 42

Van der Waals, 434 constants, 435

Zeroth law of thermodynamics, 12