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Engineering Thermodynamics

Engineering Thermodynamics [For the Students of all branches of B.E./B.Tech.]

Er. S.K. Gupta

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© 2013, Er. S.K. Gupta

All rights reserved. No part of this publication may be reproduced or copied in any material form (including photo copying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, tribunals and forums of New Delhi, India only.

First Edition 2013 ISBN : 81-219-4270-5 printed in india

Code : 10 589

By Rajendra Ravindra Printers Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055 and published by S. Chand & Company Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055.

Preface This standard treatise titled as ‘Engineering Thermodynamics’ is intended for the use of students of B.E./B.Tech. of all Indian and Foreign Universities. The subject matter is presented in the most concise, to-the-point and lucid manner. The book contains wellgraded examples, most of which are taken from the recent examination papers of Indian and Foreign Universities. In order to make more useful for the students, Highlights for Quick Revision before examination and Objective Type Questions with Answers are added, at the end of each chapter. My sincere thanks are due to the Management Team and Editorial Staff of S. Chand & Co. Pvt. Ltd., New Delhi, for taking their keen interest in bringing the book in a very short time. Though utmost care has been taken to check mistakes, yet it is difficult to claim perfection. The author will, therefore, acknowledge any errors, omissions and suggestions for the improvement of the book and will be incorporated in the next edition.

Er. S.K. Gupta

Disclaimer : While the author of this book has made every effort to avoid any mistake or omission and has used his skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and S. Chand do not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chand and the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

Contents

1. BASIC CONCEPTS

1-18

1.1 Introduction

1

1.2 Scope and Applications of Thermodynamics

1

1.3 Working Substance

2

1.4 Pure Substance

2

1.5 Macroscopic and Microscopic Approach

2

1.6 Mass and Weight

2

1.7 Force

3

1.8 Thermodynamic System

3

1.9 Types of Thermodynamic System

3

1.10 Concept of Continuum

4

1.11 Phase

4

1.12 Homogeneous and Hetrogeneous System

4

1.13 Thermodynamic State and Path

4

1.14 Thermodynamic Property

5

1.15 Thermodynamic Process

5

1.16 Thermodynamic Cycle

5

1.17 Thermodynamic Equilibrium

5

1.18 Quasi-Static Process

6

1.19 Reversible Process

6

1.20 Irreversible Process

6

1.21 Heat

6

1.22 Specific Heat

7

1.23 Thermodynamic and Mechanical Work

7

1.24 Point Function

8

1.25 Path Function

8

1.26 Comparison of Heat and Work

9

1.27 Power

9

8 ♦ Contents

1.28 Energy

9

1.29 Different Forms of Stored Energy

9

1.30 Law of Conservation of Energy

10

1.31 Specific Volume and Density

10

1.32 Pressure

11

1.33 Atmospheric Pressure, Gauge Pressure and Absolute Pressure

11

1.34 Measurement of Pressure

12

1.35 Normal Temperature and Pressure

13

1.36 Standard Temperature and Pressure

13

1.37 Laws of Thermodynamics

13

2. ZEROTH LAW OF THERMODYNAMICS

19-24

2.1 Introduction

19

2.2 Concept of Temperature

19

2.3 Equality of Temperature

19

2.4 Measurement of Temperature

19

2.5 Temperature Measuring Scales

21

3. LAWS OF IDEAL AND REAL GASES

25-44

3.1 Introduction

25

3.2 Ideal or Perfect Gas

25

3.3 Boyle’s Law

25

3.4 Charles’ Law

26

3.5 Combination of Boyle’s and Charles’ Law—General Gas Equation

28

3.6 Joule’s Law

28

3.7 Characteristic Equation of a Gas

29

3.8 Universal Gas Constant

29

3.9 Avogadro’s Hypothesis

29

3.10 Specific Heats of a Gas

33

3.11 Enthalpy of a Gas

34

3.12 Regnault’s Law

34

3.13 Difference Between Two Specific Heats

34

3.14 Molar Specific Heats of a Gas

35

3.15 Ratio of Specific Heats

35

3.16 Deviation of Real Gas from Ideal Gas

38

Contents ♦ 9

4. FIRST LAW OF THERMODYNAMICS

4.1 Introduction

4.2 First Law of Thermodynamics for a Closed or

45-89 45

Non-flow System Undergoing a Cycle

45

4.3 First Law of Thermodynamics for a Closed or

Non-flow System Undergoing a Change of State

46

4.5 Limitations of First Law of Thermodynamics

47

4.6 Classification of Thermodynamic Process

49

4.7 Workdone for a Closed or Non-flow Process

49

4.8 Types of Non-flow Processes

51

4.9 Constant Volume Process (Isometric or Isochoric Process)

52

4.10 Constant Pressure Process (or Isobaric Process)

54

4.11 Constant Temperature Process (or Isothermal Process)

56

4.12 Adiabatic Process (or Isentropic Process)

61

4.13 Relation Between Pressure, Volume and Temperature

During an Adiabatic Change

64

4.14 Polytropic Process

68

4.15 Irreversible Non-flow Process-Free Expansion Process

83

5. FIRST LAW APPLIED TO FLOW PROCESSES

90-115

5.1 Introduction

90

5.2 Assumptions in Steady Flow Processes

90

5.3 Flow Energy

91

5.4 First Law of Thermodynamics as Applied to Steady Flow Process

91

5.5 Workdone for a Steady Flow Process

94

5.6 Steady Flow Energy Equation Applied to Various Processes

95

5.7 Throttling Process-Joule Thomson Porous Plug Experiment

97

5.8 Engineering Applications of Steady Flow Energy Equation

98

6. SECOND LAW OF THERMODYNAMICS

116-139

6.1 Introduction

116

6.2 Statements of Second Law of Thermodynamics

116

6.3 Equivalence of Kelvin-Planck and Clausius Statements

117

6.4 Perpetual Motion Machine of Second Kind (PMM-II)

118

6.5 Important Terms

118

6.6 Carnot Cycle

120

6.7 Efficiency of Carnot Cycle

123

10 ♦ Contents

6.8 Practical Difficulties (or Impractibility) of Carnot Cycle

124

6.9 Carnot’s Theorem

124

6.10 Corollaries of Carnot’s Theorem

125

7. ENTROPY, AVAILABILITY AND IRREVERSIBILITY 140-186

7.1 Introduction

140

7.2 Temperature-Entropy (T-S) Diagram

141

7.3 Clausius Inequality

141

7.4 Principle of Increase of Entropy

144

7.5 Change of Entropy for Ideal Gases

145

7.6 Change of Entropy During Thermodynamic Processes

148

7.7 Change of Entropy During Constant Volume (or Isochoric) Process

148

7.8 Change of Entropy During Constant Pressure (or Isobaric) Process

150

7.9 Change of Entropy During Constant Temperature (or Isothermal) Process 152

7.10 Change of Entropy During Reversible Adiabatic (or Isentropic) Process n

154

7.11 Change of Entropy During Polytropic (or pv = constant) Process

156

7.12 Available and Unavailable Energy

162

7.13 Loss in Available Energy

166

7.14 Availability

170

7.15 Availability of a Closed System

170

7.16 Availability in a Steady Flow Process

170

7.17 Helmholtz Function

171

7.18 Gibbs Function

173

7.19 Irreversibility

173

8. PROPERTIES OF PURE SUBSTANCES

187-218

8.1 Introduction

187

8.2 Phases of Water

188

8.3 Process of Formation of Steam

188

8.4 Graphical Representation of Formation of Steam

189

8.5 Pressure-Volume (p -v) Diagram for Water

189

8.6 Temperature-Entropy (T- s) Diagram for Water and Steam

190

8.7 Pressure-Specific Volume-Temperature (p-v -T) Surface

of a Pure Substance

192

8.8 Important Terms as Applied to Steam

193

8.9 Use of Steam Tables

195

8.10 Enthalpy-Entropy (h-s) Diagram or Mollier Chart for Water and Steam

195

Contents ♦ 11

8.11 External Workdone During Evaporation

198

8.12 Internal Energy of Steam

199

8.13 Entropy of Water

200

8.14 Entropy of Evaporation

200

8.15 Entropy of Wet and Dry Steam

201

8.16 Entropy of Superheated Steam

201

8.17 Measurement of Dryness Fraction or Quality of Steam

206

8.18 Barrel or Tank Calorimeter

206

8.19 Separating Calorimeter

208

8.20 Throttling Calorimeter

209

8.21 Separating and Throttling Calorimeter

210

9. VAPOUR PROCESSES

9.1 Introduction

219

9.2 Constant Volume (or Isochoric) Process

219

9.3 Constant Pressure (or Isobaric) Process

225

9.4 Constant Temperature (or Isothermal) Process

228

9.5 Hyperbolic (or pv = C ) Process

230

9.6 Reversible Adiabatic (or Constant Entropy) Process

233

n

219-244

9.7 Polytropic (or pv = C ) Process

236

9.8 Throttling (or Constant Enthalpy) Process

240

10. AIR STANDARD CYCLES

245-287

10.1 Introduction

245

10.2 Assumptions in Air Standard Cycle

246

10.3 Important Terms used in Air Standard Cycle

246

10.4 Otto Cycle (Constant Volume Cycle)

247

10.5 Compression Ratio for Maximum Output of an Otto Cycle

249

10.6 Mean Effective Pressure for Otto Cycle

250

10.7 Diesel Cycle (Constant Pressure Cycle)

258

10.8 Mean Effective Pressure for Diesel Cycle

260

10.9 Dual Cycle

267

10.10 Mean Effective Pressure for Dual Cycle

269

10.11 Brayton Cycle

271

10.12 Difference Between Otto Cycle and Diesel Cycle

274

10.13 Comparison Between Otto, Diesel and Dual Cycle

275

10.14 Four Stroke Cycle Petrol Engine

276

12 ♦ Contents

10.15 Two Stroke Cycle Petrol Engine

277

10.16 Diesel Engine or Compression Ignition Engine

278

10.17 Four Stroke Cycle Diesel Engine

278

10.18 Two Stroke Cycle Diesel Engine

280

10.19 Comparison Between Two Stroke and Four Stroke Cycle Engines

281

11. MIXTURE OF GASES

288-311

11.1 Introduction

288

11.2 Mass Fraction

288

11.3 Mole Fraction

289

11.4 Volume Fraction

289

11.5 Dalton’s Law of Partial Pressure

290

11.6 Amagat’s Leduc Law

291

11.7 Properties of Gas Mixtures

292

11.8 Change of Entropy During Mixing Process

295

11.9 Volumetric Analysis and Gravimetric Analysis

306

12. VAPOUR POWER CYCLES

312-350

12.1 Introduction

312

12.2 Carnot Vapour Cycle

313

12.3 Limitations of Carnot Vapour Cycle

314

12.4 Rankine Cycle

315

12.5 Difference Between Rankine Cycle and Carnot Cycle

316

12.6 Rankine Cycle Using Superheated Steam

322

12.7 Modified Rankine Cycle

327

12.8 Methods of Improving the Efficiency of the Rankine Cycle

334

12.9 Reheat Cycle

334

12.10 Regenerative Cycle

338

12.11 Regenerative Cycle with an Open Feed Water Heater

338

12.12 Regenerative Cycle with a Closed Water Heater

340

12.13 Comparison Between Open and Closed Feed Water Heaters

340

12.14 Regenerative Cycle with Two Feed Water Heaters

343

Table 1: Properties of Saturated Water and Steam (Pressure based)

352

Table 2: Properties of Saturated Water and Steam (Temperature based)

356

Table 3: Properties of Superheated Steam

359

INDEX 367-369

1 BASIC CONCEPTS 1.1 Introduction 1.2 Scope and Applications of Thermodynamics 1.3 Working Substance 1.4 Pure Substance 1.5 Macroscopic and Microscopic Approach 1.6 Mass and Weight 1.7 Force 1.8 Thermodynamic System 1.9 Types of Thermodynamic System 1.10 Concept of Continuum 1.11 Phase 1.12 Homogeneous and Heterogeneous System 1.13 Thermodynamic State and Path 1.14 Thermodynamic Property 1.15 Thermodynamic Process 1.16 Thermodynamic Cycle 1.17 Thermodynamic Equilibrium 1.18 Quasi-Static Process 1.19 Reversible Process

1.20 Irreversible Process 1.21 Heat 1.22 Specific Heat 1.23 Thermodynamic and Mechanical Work 1.24 Point Function 1.25 Path Function 1.26 Comparison of Heat and Work 1.27 Power 1.28 Energy 1.29 Different Forms of Stored Energy 1.30 Law of Conservation of Energy 1.31 Specific Volume and Density 1.32 Pressure 1.33 Atmospheric Pressure, Gauge Pressure and Absolute Pressure 1.34 Measurement of Pressure 1.35 Normal Temperature and Pressure 1.36 Standard Temperature and Pressure 1.37 Laws of Thermodynamics

1.1 INTRODUCTION The subject *Thermodynamics is defined as the science which deals with the study of energy in its various forms or types. It may also be defined as the science which deals with the conversion of heat into mechanical energy. The science of thermodynamics is used often by engineers and technologists in very practical design problems and in problems of the operation of large or complicated systems.

1.2 SCOPE AND APPLICATIONS OF THERMODYNAMICS The scope of applied thermodynamics is restricted to the study of heat and work and the conversion of one into the other. The applications of thermodynamics are found in the measurement of temperature and humidity in the air, as in the design of a heating or air conditioning system. The engines of automobiles, trucks and tractors are designed by using the concept of thermodynamics.

* The literal meaning of thermodynamics is heat-force action and it deals with matter and interaction between quantities of matter.

2 Engineering Thermodynamics Jet engines and rockets are analysed by using the principles of thermodynamics. The central thermal plants, captive power plants based on coal, nuclear power plants, gas turbine plants and chemical process plants are fully dependent on the Laws of Thermodynamics. Thus, a good knowledge in thermodynamics is very necessary for an engineer so that he can use fundamental ideas of thermodynamics and applications of those ideas to engineering problems.

1.3 WORKING SUBSTANCE A substance through which the conversion of heat into work and vice versa takes place, is known as working substance. It should be capable of changing its volume, so the working substances are, in general, fluids. Air and steam are the commonly used working substances.

1.4 PURE SUBSTANCE A substance whose chemical composition is both homogeneous and constant is known as pure substance. In other words, it is a homogeneous substance whose molecular structure does not vary. A pure substance exists in three phases, i.e. solid, liquid and gas. Water or steam or a mixture of water and steam are examples of pure substances as they have the same molecular or chemical structure through its mass. Air in liquid and gaseous form is another example of pure substance.

1.5 MACROSCOPIC AND MICROSCOPIC APPROACH The behaviour of a matter may be studied by the following two approaches: 1. Macroscopic approach. The term macroscopic is used in regard to large units which is visible to the naked eye. In the macroscopic approach, a certain quantity of matter is considered without taking into consideration the events occuring at the molecular level. In other words, macroscopic approach to thermodynamics is concerned with gross or overall behaviour of matter. This is called classical thermodynamics. In order to understand clearly the macroscopic approach of thermodynamics, let us consider the pressure that a gas exerts on the walls of its container. This pressure is average value of pressures exerted due to the collision of all the molecules made on a unit area. We are not concerned with the action of individual molecules from the macroscopic point of view. We measure the force on a given unit area by using a pressure gauge. There are certain properties like temperature and volume which can be expressed as macroscopic quantities. In the microscopic approach, the matter is considered to be composed of particles and each particle having a certain position, velocity and energy, at a given instant. Such a study which is concerned directly with the structure of the matter, is known as statistical thermodynamics. It may be noted that all the results of classical or macroscopic thermodynamics may be derived from the statistical or microscopic study of matter.

1.6 MASS AND WEIGHT The mass of a system is a measure of quantity of matter as well as property of matter called inertia. In S.I. system of units, the mass is expressed in kilogram (briefly written as kg). Though the mass is related to weight, yet both these terms are different. Weight refers to the force exerted by gravity on mass. It may be noted that mass of a body or system does not change but the weight of a body changes from place to place. In S.I. system of units, the weight is expressed in newton (briefly written as N). Mathematically, the relation between the mass (m) and the weight (W) of the body is m = W/g or W = m.g where m is in kg, W is in newton and g is the acceleration due to gravity whose value is taken as 9.81 m/s2. Note: From the above discussion, we see that weight of a body of mass m kg, at a place where gravitational acceleration is ‘g’ m/s2 is mg newton.

Basic Concepts 3

1.7 FORCE The force is well defined with the aid of Newton’s Second Law of Motion which states that force is proportional to the product of mass (m) and acceleration (a). Mathematically, F m a or F = k m a where k is a constant of proportionality. In S.I. system of units, the force is in newton (N), mass in kilogram (kg) and acceleration in metre per second per second (m/s2). We know that F = m a 1m or 1 N = 1 kg × 2 = 1 kg-m/s2 s From the above, we may define a newton as the force while acting upon a mass of 1 kg produces an acceleration of 1 m/s2 in the direction of which the force acts.

1.8 THERMODYNAMIC SYSTEM A thermodynamic system is defined as any region of space or a finite quantity that occupies a volume and has a boundary. For example, the gas in a cylinder of a Boundary reciprocating engine is a finite portion of matter and, therefore, it can be considered as a system. Anything external to the system is called surroundings. The system is separated from the surroundings by an imaginary envelope, which is known as boundary of the system as shown System in Fig. 1.1. Actually, the boundaries are the limits of the system. This may be fixed or movable. The system and its surroundings together constitute Surroundings a universe.

1.9 TYPES OF THERMODYNAMIC SYSTEM

Fig. 1.1. Thermodynamic system.

Following are the four types of thermodynamic systems: 1. Closed system. It is also known as non-flow system. In this system, the mass (gas) within the boundary remains constant. A closed system can permit exchange of heat and work with the surroundings, but does not permit any mass transfer across the boundary. The cylinder and piston of a reciprocating engine containing gas in the cylinder, is an example of a closed system, as shown in Fig. 1.2. Another simple example of a closed system is a tea kettle in which heat is supplied to the kettle but the mass of water remains constant.

Fig. 1.2. Closed system or non-flow system.

2. Open system. It is also known as flow system. In this system, both mass and energy (i.e. heat and work) cross the boundary. The mass within the system may not be constant. An open system

4 Engineering Thermodynamics is also called control volume. A particular example of an open system is that of a reciprocating compressor, as shown in Fig. 1.3.

Fig. 1.3. Open system or flow system.

Other examples of an open system are boilers, turbines and heat exchangers in which the fluid flows through them and heat or work is taken out or supplied to them. 3. Isolated system. In an isolated system, there is no interaction between the system and the surroundings. It is of fixed mass and energy and there is no mass or energy transfer across the system boundary. The universe which constitute a system and its surrounding taken together, is an example of isolated system. Another example of isolated system is a thermos flask containing a liquid. It may be noted that an isolated system is not at all influenced by its surroundings. 4. Adiabatic system. A system which is thermally insulated from its surroundings, is known as adiabatic system. It can only exchange work and not heat with the surroundings.

1.10 CONCEPT OF CONTINUUM The system is regarded as a continuum, i.e. the system is assumed to contain continuous distribution of matter. Thus, from the continuum point of view, the matter is seen as being distributed through space and treats the substance as being continuous, disregarding the action of individual molecules. There are no voids and cavities. The pressure, temperature, density and other properties are the average values of action of many molecules and atoms.

1.11 PHASE A quantity of matter which is homogeneous (i.e. uniform) throughout in physical structure and chemical composition is called a phase. Every substance can exist in any one of the three phases, i.e. solid, liquid and gas.

1.12 HOMOGENEOUS AND HETEROGENEOUS SYSTEM A system which consists of a single phase is known as homogeneous system. The examples of this system are water, ice, mixture of air and water vapour, dry saturated steam, mixture of ammonia in water, water plus nitric acid, octane plus heptane etc. A system which consists more than one phase is called a hetrogeneous system. The examples of this system are water plus steam, ice plus water, mixture of water and oil etc. Notes: (a) A mixture of a liquid and gas is a two phase system. (b) A mixture of water, ice and steam forms a three phase system.

Basic Concepts 5

1.13 THERMODYNAMIC STATE AND PATH The condition of physical existence of a system at any instant of time is called a state. The state of a system may be described by certain macroscopic properties such as pressure, temperature, volume etc. A system may undergo a change of state without undergoing a change of phase. A thermodynamic system passing through a series of states constitutes a path.

1.14 THERMODYNAMIC PROPERTY Every system has some characteristic by which its physical condition may be described such as pressure, volume, temperature etc. Such characteristics are called properties of the system. A property is identified by the fact that it is dependent upon the state only and is independent of its path, by which the state has been brought about. Each of the properties of the system has a definite unique value at a state. A thermodynamic property may be divided into the following two types: 1. Intensive property; and 2. Extensive property. The properties which are independent of the mass of the system, are known as intensive properties. The intensive properties include pressure and temperature and also those properties which are based on unit mass such as specific volume, specific energy, density etc. The properties which depends upon the mass of the system, are known as extensive properties. The extensive properties include volume, total energy etc. Notes: 1. Any extensive property per unit mass becomes the intensive property. 2. The properties are co-ordinates to describe the state of a system. When all the properties of a system have definite values, the system is said to exist at a definite state.

1.15 THERMODYNAMIC PROCESS When any property of a system changes, there is a change of state and the system is then said to have undergone a thermodynamic process. The expansion of a gas as it flows through a turbine to perform work is an example of a process. Thus, when a system undergoes a certain process, work transfer takes place across the boundary of the system. There are following three commonly used processes: 1. Isometric or Isochoric process. This process takes place at constant volume. 2. Isopiestic or Isobaric process. This process takes place at constant pressure. 3. Isothermal process. This process takes place at constant temperature.

1.16 THERMODYNAMIC CYCLE

1.17 THERMODYNAMIC EQUILIBRIUM

A 1

Pressure

When a process or processes are performed on a system in such a way that the final state is identical with the initial state, it is then known as a thermodynamic cycle or cyclic process. In Fig. 1.14, A–1–B and A–2–B are the processes where as A–1–B–2–A is a thermodynamic cycle or a cyclic process.

2

A system is said to be in thermodynamic equilibrium when B it does not tend to undergo any further change of its own accord. Volume Any further change must be produced by external means. A system will be in a state of thermodynamic equilibrium if the Fig. 1.4. Thermodynamic cycle. following three types of equilibrium are satisfied: 1. Mechanical equilibrium. When there is no unbalanced forces on any part of the system or between the system and the surroundings, then the system is said to be in mechanical equilibrium. If an unbalanced force exists, then either the system alone or both the system and the surroundings will undergo a change of state till mechanical equilibrium is attained. For example, if the pressure is

6 Engineering Thermodynamics not uniform throughout the system, then internal changes in the state of the system will take place until the mechanical equilibrium is reached. 2. Chemical equilibrium. When there is no chemical reaction or transfer of matter from one part of the system to another such as diffusion of solution, then it is said to exist in a state of chemical equilibrium. For example, a mixture of oxygen and gasoline is not in chemical equilibrium once the mixture is ignited. 3. Thermal equilibrium. When there is no temperature difference between the parts of the system or between the system and the surroundings, it is then said to be in thermal equilibrium. Equilibrium states 1

A quasi-static process is one in which the system deviates from the thermodynamic equilibrium state by only infinitesimal amount throughout the entire process. This process may, practically, be considered as a series of equilibrium states and its path may be represented graphically as a continuous line on a state diagram, as shown in Fig. 1.5. Note: A quasi-static process is an ideal concept that is applicable to all thermodynamic systems. It may be noted that conditions for such a process can never be satisfied rigorously in practice.

Pressure

1.18 QUASI-STATIC PROCESS

Quasi-static process

2 Volume

Fig. 1.5. Quasi-static process.

1.19 REVERSIBLE PROCESS

1.20 IRREVERSIBLE PROCESS A process in which the system passes through a sequence of *non-equilibrium states, is known as irreversible process. This process will not retrace the reverse path to restore the original state, as shown in Fig. 1.7. The heat transfer by convection, combustion of air and fuel, plastic deformation etc. are few examples of irreversible process.

Pressure

1

2 Volume

Fig. 1.6. Reversible process.

1

1

Pressure

A process which can be reversed in direction and the system retraces the same continuous series of equilibrium states, is said to be a reversible process, as shown in Fig. 1.6. A quasi-static process is also known as a reversible process. It may be noted that the process will be reversible only when there is no friction in the system. A reversible process should be carried out with absolute slowness, so that the system is always in mechanical, chemical and thermal equilibrium. There should also be no heat transfer across finite temperature difference. In actual practice, a reversible process can not be attained, but it can be approximated as closely as possible. For example, a gas confined in a cylinder with a well lubricated piston can be made to undergo an approximately reversible process by pushing or pulling the piston in the slow motion.

2 Volume

Fig. 1.7. Irreversible process.

* The non-equilibrium state means that the properties such as pressure, volume and temperature are not uniform throughout the system.

Basic Concepts 7

1.21 HEAT The heat is an energy interaction between the system and the surrounding. It may be defined as the energy transferred, without the transfer of mass, across the boundary of a system due to the temperature difference between the system and the surrounding. The system for heat transfer is usually such that its boundaries are fixed. For example, raising steam in a boiler. The heat is transferred across the walls of the tube or the boiler shell, whose boundaries are fixed. It is expressed in joules (briefly written as J) and the rate of heat transfer is given in kW. It may be noted that heat is not a thermodynamic property whereas temperature and pressure are thermodynamic properties. Following are three ways through which the heat may be transferred: 1. Conduction. The process of heat transfer from one particle to another particle of the body, in the direction of fall of temperature, is called conduction. For example, heat transfer through solids is by conduction. In this case, the particles themselves remain in fixed position relative to each other. 2. Convection. The process of heat transfer from one particle to another particle of the body by convection currents, is called convection. For example, heat transfer through fluids is by convection. In this case, the particles of the body move relative to each other. 3. Radiation. The process of heat transfer from a hot body to a cold body in a straight line without affecting the intervening medium, is called radiation. The main characteristics of heat are as follows: (a) The heat flows from a system or source at a higher temperature to a system at a lower temperature. (b) The heat exists only during transfer of energy into or out of a system. (c) The heat when it flows into the system is taken as positive, and when it flows out of the system or when the heat is rejected by the system, then it is taken as negative.

1.22 SPECIFIC HEAT The amount of heat required to raise the temperature of a unit mass of a substance through one degree, is known as specific heat. The unit of specific heat is kJ/kg K. Mathematically, heat required to raise the temperature of a body or system, Q = mc (T2 – T1) in kJ where m = Mass of the substance in kg, c = Specific heat in kJ/kg K, T1 = Initial temperature in degrees centigrade (°C) or Kelvin (K), and T2 = Final temperature in degrees centigrade (°C) or Kelvin (K) It may be noted that solids and liquids have only one specific heat. But the specific heat of gases depend upon the process of heating the gas, i.e. whether it is heated at constant pressure or at a constant volume. Thus, we have the following two types of specific heats for gases, i.e. 1. Specific heat at constant pressure (denoted by cp ), and 2. Specific heat at constant volume (denoted by cv ). Notes: (a) The value of cp is always greater than cv. (b) The ratio of cp /cv is known as ratio of specific heats and it is denoted by a Greek letter gamma (g).

1.23 THERMODYNAMIC AND MECHANICAL WORK The thermodynamic work may be defined as the energy transferred (without the transfer of mass) across the boundary of a system because of an intensive property difference other than temperature that exists between the system and surroundings.

8 Engineering Thermodynamics For the work transfer, the system has to be such selected that its boundaries must move. There can not be work transfer in a closed system, without moving the system boundaries. In a cylinderpiston arrangement, the top of the piston is the moving system boundary, and the work is transferred by the movement of the piston. It may be noted that work done by a system is considered to be positive and work done on the system as negative. The following points are worthnoting: 1. In thermodynamics, work is energy in transition, i.e. a form of energy interaction between the two systems. It is a path function. 2. The work is a product of an intensive property and a change in extensive property. 3. In engineering practice, the intensive property is the pressure difference. The pressure difference (between the system and the surroundings) gives rise to a force and the action of this force over a distance is termed as mechanical work. For example, if F is the force and x is the distance moved in direction of the force, then Workdone = F × x The unit of work depends upon the unit of force and the distance moved. If the force is in newton and the distance moved is in metre, the unit of work will be newton-metre (briefly written as N-m) or joule (briefly written as J). It may be noted that 1 N-m = 1 J. Note: The mechanical work is said to be positive when both the force and displacement are in the same direction. A negative work is obtained when the force and displacement are in the opposite direction.

1.24 POINT FUNCTION

2

∫1 dv = [v]1 = v

2

2

p1

1

Pressure

When the thermodynamic property has a definite value for a given state, it is called a point function. For example, pressure, temperature, volume etc. are point functions. These functions depend upon the initial and final states of the system. It may be noted that the differentials of point function are exact or perfect differentials. Let the system undergoes a change in volume from point 1 to point 2. This change may take place through path A or path B as shown in Fig. 1.8. Thus, this property does not depend upon the path, so it is a point function or a state function. Mathematically,

B A

p2

2 v1

v2 Volume Fig. 1.8. Point function.

– v1

1.25 PATH FUNCTION

2

∫1 ´ Q = 2

∫1

and

Similarly

2

∫1 δW

B

A 2

[Q]12 = Q1 – 2 or 1Q2

δW = [W ]12 = W1 – 2 or 1W2

It may be noted that

1

Pressure

The quantities which are not thermodynamic properties, are known as path functions. For example, heat and work are path functions because they depend upon the path taken from one state (say state 1) to another state (say state 2), as shown in Fig. 1.9. The heat and work are inexact differentials and are written as dQ and dW respectively. On integrating, we have

2

∫1 δQ

Volume Fig. 1.9. Path function.

is not equal to (Q2 – Q1), because heat is not a point function.

is not equal to (W2 – W1), because work is not a point function. Thus heat and work

are not properties of the system.

Basic Concepts 9 Note: From Fig. 1.9, we see that the system is taken from an initial equilibrium state 1 to a final equilibrium state 2 by two different paths 1–A–2 and 1–B–2. The processes are quasi-static. Since the area under the pressure-volume ( p – v) diagram represents the workdone during the process, therefore, the areas under these curves are different and the quantities of workdone are also different. Thus, the work can not be expressed as a difference between the values of some property of the system in the two states. Thus it is not correct to represent

∫

2

1

δW = W2 − W1 .

1.26 COMPARISON OF HEAT AND WORK There are many similarities and differences between heat and work, as discussed below: Similarities 1. The heat and work are both boundary phenomena. They are observed at the boundary of the system. 2. The heat and work are both transient phenomena. The systems do not possess heat or work. 3. When a system undergoes a change, heat transfer or workdone may occur. 4. The heat and work represent the energy crossing the boundary of the system. 5. The heat and work are path functions and depend upon the process. Hence they are not thermodynamic properties and are inexact differentials. They are written as dQ and dW. Differences 1. In a stable system (when volume of the system is constant and there is no movement of the boundary), there cannot be reversible work transfer. However, there is no restriction for the transfer of heat. 2. For the transfer of heat, the temperature difference is needed. 3. The sole effect on the surroundings, in case of work transfer, can be reduced to the raising of a weight. But in case of heat transfer, other effects are also observed.

1.27 POWER The rate of doing work is termed as power. In other words, power is the workdone per unit time. In S.I. system of units, the unit of power may be expressed either in watt (briefly written as W) or kilowatt (briefly written as kW, such that 1 kW = 1000 W).

1.28 ENERGY The energy is defined as the capacity to do work. In a broad sense, energy is classified as stored energy and energy in transition. The energy that remains within the boundary of the system is called stored energy, e.g. potential energy, kinetic energy and internal energy etc. The energy that crosses the system boundaries is known as energy in transition, e.g. heat, work and electricity etc. Notes: 1. The stored energy is a thermodynamic property whereas the energy in transition is not a thermodynamic property as it depends upon the path. 2. The unit of energy is same as that of work i.e. N-m.

1.29 DIFFERENT FORMS OF STORED ENERGY The different forms of stored energy are as follows: 1. Potential energy. The energy possessed by a body or system by virtue of its position above the datumn or ground level, is known as potential energy. For example, a body resting on the top of a building has stored up potential energy because it can do some work by falling on the ground level. Mathematically, potential energy, P.E. = W × h = m g h where W = Weight of the body in N,

10 Engineering Thermodynamics m = Mass of the body in kg , g = Acceleration due to gravity = 9.81 m/s2, and h = Distance through which the body falls in m. The unit of potential energy is N-m, as discussed below: We know that P.E. = W × h = N × m = N-m or J m 1kg × m = 1 N = m g h = kg × 2 × m = N-m or J ... 2 s s 2. Kinetic energy. The energy possessed by a body or system by virtue of its mass and velocity of motion, is known as kinetic energy. Mathematically, kinetic energy, 1 K.E. = × mV 2 2 where m = Mass of the body in kg, and V = Velocity of the body in m/s. The unit of kinetic energy is N-m, as discussed below: We know that kinetic energy 1 K.E. = × mV 2 2 = kg ×

m2 2

= kg ×

m 2

× m = N-m or J... 1 kg ×

m = 1 N s2

s s 3. Internal energy. The energy possessed by a body or a system by virtue of its molecular arrangement and motion of molecules, is called internal energy. The change in temperature causes the change in internal energy. It is usually denoted by U. Note: The sum of the above three types of energies is the total energy of the system. Mathematically, total energy of the system, 1 E = P.E. + K.E. + U = m g h + × mV 2 + U 2 When the system is stationary and the effect of gravity is neglected, then P.E. = 0, and K.E. = 0. Thus E = U i.e. the total energy of the system is equal to the internal energy of the system.

1.30 LAW OF CONSERVATION OF ENERGY This law states that energy can neither be created nor destroyed, though it can be transformed from one form to another form, in which the energy can exist. In other words, the total energy possessed by a body remains constant.

1.31 SPECIFIC VOLUME AND DENSITY The space occupied by a substance is called its volume. It is measured in m3 or litre, such that 1 m3 = 1000 litres = 1 kilo litre The volume per unit mass of a substance is termed as specific volume. It is usually expressed in m3 / kg. The mass per unit volume of a substance is called density. It is generally stated in kg / m3. It is usually denoted by a Greek letter rho (r). From the above, we find that the density of any substance is the reciprocal of the specific volume of the substance. Similarly, specific volume is the reciprocal of density of a substance.

Basic Concepts 11 Example 1.1. A fan is to accelerate quiescent air to velocity of 10 m/s delivering the air at the rate of 4 m3/s. If the density of air is 1.18 kg/m3, determine the minimum power that must be supplied to the fan. Solution. Given: Velocity,

V = 10 m/s

Volume,

v = 4 m3 /s

Density,

r = 1.18 kg/m3

We know that mass of air supplied by the fan,

m = Volume × Density = 4 × 1.18 = 4.72 kg/s

\ Kinetic energy imparted by the fan to the air, 1 1 = × mV 2 = × 4.72 (10)2 = 236 N-m/s = 236 J/s ...(Q 1 N-m = 1 J) 2 2 \ Minimum power that must be supplied to the fan = 236 J/s = 236 W Ans.

...(Q 1 J/s = 1 W)

1.32 PRESSURE The pressure is defined as the force per unit area. In S.I. system of units, the unit of pressure is Pascal (briefly written as Pa), such that 1 Pa = 1 N/m2 The pressure, depending upon its magnitude, may be expressed in bar, kilo-pascal (kPa) or Megapascal (MPa), such that 1 bar = 1 × 105 N/m2 = 100 × 103 N/m2 = 100 kPa = 0.1 MN/m2 Sometimes, the pressure is also expressed in terms of the height of a liquid column e.g. mm of Hg or mm of water.

1.33 ATMOSPHERIC PRESSURE, GAUGE PRESSURE AND ABSOLUTE PRESSURE The atmospheric pressure is the pressure due to the weight of the column of air above the earth’s surface. It is usually measured by a mercury barometer and is expressed in mm of Hg. The standard atmospheric pressure is 760 mm of Hg at 0°C (or 1.013 bar) at sea level. In other words, Atmospheric pressure = 760 mm of Hg = 1.013 bar = 1.013 × 105 N/m2 ...(Q 1 bar = 1 × 105 N/m2) = 101.3 × 103 N/m2 = 101.3 kPa 101.3 × 103 = 133.3 N/m2 760 Most pressure measuring instruments are so designed that zero point on their dials is the atmospheric pressure. When an instrument records pressure above the atmospheric pressure, it is called a pressure gauge and the pressure recorded is gauge pressure. If the instrument is designed to record pressure below the atmospheric pressure, it is a vacuum pressure (also called negative gauge pressure). In thermodynamic calculations, we use the actual pressure called absolute pressure, such that Absolute pressure = Gauge pressure + Atmospheric pressure = Atmospheric pressure – Vacuum pressure \

1 mm of Hg =

12 Engineering Thermodynamics This relation is shown in Fig. 1.10. Gauge pressure Vacuum pressure

Absolute pressure

Atmospheric pressure

Atmospheric pressure

Absolute pressure

Fig. 1.10. Relation between absolute pressure, gauge pressure and vacuum pressure.

1.34 MEASUREMENT OF PRESSURE The pressure is measured with a device known as pressure gauge. Manometers are the simplest types of pressure gauges that are widely used. A manometer is a U-tube containing a fluid to measure a pressure in a tube. The fluids most often used are water and manometer fluids. Most manometer fluids are slightly lighter than water, having specific gravity of about 0.8 to 0.95. Mercury is also used in manometers and barometers. A U-tube manometer containing mercury is shown in Fig. 1.11 (a). One end of the tube is open to the atmosphere while the other end is connected to a system or vessel. If the pressure ( p) is higher than the atmospheric pressure ( pa), then the mercury is forced up the tube that opens to the atmosphere, as shown in Fig. 1.11 (a). On the other hand, if the pressure is the vessel ( p) is lower than the atmospheric pressure ( pa), then the mercury is forced into the tube connected to the vessel, as shown in Fig. 1.11 (b). The pressure higher than the atmospheric pressure is known as gauge pressure while the pressure lower than the atmosphere pressure is called vacuum pressure (or negative gauge pressure). Vessel

pa

p

Vessel

pa

p

hg hv

(a)

(b)

Fig. 1.11. U-tube manometer.

If h (in metres) is the difference in height of the fluid column in the two limbs of a U-tube, r is the density of the fluid (in kg/m3) and ‘g’ is the acceleration due to gravity in m/s2, then according to principle of hydrostatics, gauge pressure ( pg) is given by pg = r . h . g (in N/m2)

...

kg m kg − m 1 1kg − m ×m × 2 = × 2 = N / m 2 ... = 1 N m3 s s2 m s2

Note: Very low pressures are generally measured by the height of water column. This is convenient because the height of water column is 13.6 times greater than that of mercury column for any given pressure, i.e. 1 mm of Hg = 13.6 mm of water

Basic Concepts 13

1.35 NORMAL TEMPERATURE AND PRESSURE The conditions for normal temperature and pressure (N.T.P.) are 0°C and 760 mm of Hg respectively.

1.36 STANDARD TEMPERATURE AND PRESSURE The conditions for standard temperature and pressure (S.T.P.) are 15°C and 760 mm of Hg respectively. Example 1.2. If a mercury barometer reads 720 mm of Hg, find the atmospheric pressure in kPa. Solution. Given: Barometer reading = 720 mm of Hg We know that 1 mm of Hg = 133.3 N/m2 \ Atmospheric pressure = 720 × 133.3 = 95 976 N/m2 or Pa = 95 976/1000 = 95.976 kPa Ans. Example 1.3. The steam leaving a turbine is at a pressure of 70 kPa vacuum. Find the absolute pressure in bar, if the atmospheric pressure is 101 kPa. Solution. Given: Vacuum pressure = 70 kPa Atmospheric pressure = 101 kPa We know that Absolute pressure = Atmospheric pressure – Vacuum pressure = 101 – 70 = 31 kPa = 31/100 = 0.31 bar ... (Q 1 bar = 100 kPa) Example 1.4. The pressure of the gas supplied to an engine is measured as 76.2 mm of water gauge. If the barometer reads 730 mm of Hg, what is the absolute pressure of the gas in mm of Hg and kPa. Solution. Given: Gauge pressure of gas = 76.2 mm of water 76.2 = = 5.6 mm of Hg ...(Q 1 mm of Hg = 13.6 mm of water) 13.6 Barometer reading or Atmospheric pressure = 730 mm of Hg We know that absolute pressure of the gas = Gauge pressure + Atmospheric pressure = 5.6 + 730 = 735.6 mm of Hg Ans. = 735.6 × 133.3 = 98 055 N/m2 or Pa ...(Q 1 mm of Hg = 133.3 N/m2) = 98.055 kPa Ans.

1.37 LAWS OF THERMODYNAMICS

Following are the three laws of thermodynamics : 1. Zeroth law of thermodynamics ; 2. First law of thermodynamics ; and 3. Second law of thermodynamics. These laws are discussed, in detail, in the subsequent chapters.

14 Engineering Thermodynamics

HIGHLIGHTS 1. The science which deals with the conversion of heat into mechanical energy is called Thermodynamics. 2. A substance through which the conversion of heat into work and vice versa takes place, is known as working substance. 3. A substance whose chemical composition is both homogeneous and constant, is known as pure substance. 4. In thermodynamics, macroscopic approach is concerned with gross or overall behaviour of matter. This is called classical thermodynamics. 5. In the microscopic approach, the matter is considered to be composed of particles and each particle having a certain position, velocity and energy, at a given constant. Such a study which is concerned directly with the structure of the matter is known as statistical thermodynamics. 6. The mass (m) of a system is a measure of quantity of matter as well as property of matter called inertia. The weight (W) refers to the force exerted by gravity on mass. Mathematically, m = W/g or W = m g where m is in kg and W is in newton and g is acceleration due to gravity (9.81 m /s2). 7. The force is proportional to the product of mass (m) and acceleration (a). Its unit is newton. 8. Any region of space or finite quantity that occupies a volume and has a boundary, is known as thermodynamic system. Anything external to the system is called surroundings. The system is separated from the surroundings by an envelope, known as boundary. 9. A thermodynamic system is of four types, i.e. closed system (or non-flow system); open system (or flow system); isolated system and adiabatic system. 10. In a closed system, mass within the boundary remains constant. In other words, there is no mass transfer across the boundary. 11. In an open system, both mass and energy cross the boundary. 12. In an isolated system, there is no interaction between the system and the surroundings. 13. In an adiabatic system, only work can be exchanged and not heat with the surroundings. 14. A quantity of matter which is homogeneous throughout in physical structure and chemical composition is called a phase. 15. A system which consists of a single phase is known as homogeneous system and a system which consists more than one phase is called hetrogeneous system. 16. The condition of physical existence of a system at any instant of time is called a state and a system passing through a series of states constitutes a path. 17. Every system has some characteristic by which its physical condition may be described such as pressure, volume, temperature etc. Such characteristics are called properties of the system. 18. The properties which are independent of the mass of the system, are known as intensive properties. 19. The properties which are dependent upon the mass of the system, are known as extensive properties. 20. Any extensive property per unit mass becomes intensive property.

Basic Concepts 15 21. When any property of a system changes, there is a change of state and the system is then said to have undergone a thermodynamic process. 22. When a process or processes are performed on a system is such a way that the final state is identical with the initial state, it is then known as a thermodynamic cycle or cyclic process. 23. A system is said to be in thermodynamic equilibrium when it does not tend to undergo any further change of its own accord. 24. A quasi-static process is one in which the system deviates from the thermodynamic equilibrium state by only infinitesimal amount throughout the entire process. 25. A process which can be reversed in direction and the system retraces the same continuous series of equilibrium states, is said to be a reversible process. 26. A process in which the system passes through a sequence of non-equilibrium states, is known as irreversible process. 27. The heat may be defined as the energy transferred, without the transfer of mass, across the boundary of a system due to the temperature difference between the system and the surrounding. The unit of heat is joules. 28. The amount of heat required to raise the temperature of a unit mass of a substance through one degree is known as specific heat. The unit of specific heat is kJ/kg K. 29. The thermodynamic work may be defined as the energy transferred (without the transfer of mass) across the boundary of a system because of an intensive property difference other than temperature that exists between the system and the surroundings. 30. The mechanical work is defined as the product of the force (F ) and the distance moved (x) in the direction of the force. The unit of work depends upon the unit of force and distance moved. The practical unit of work is N-m. 31. When the thermodynamic property has a definite value for a given state, it is called a point function. For example, pressure, volume, temperature etc. are point functions. 32. The quantities which are not thermodynamic properties, are known as path functions. For example, heat and work are path functions. 33. The rate of doing work (or workdone per unit time) is termed as power. The unit of power is watt (W) or kilowatt (kW). 34. The capacity to do work is called energy. The energy that remains within the boundary of the system is called stored energy, e.g. potential energy, kinetic energy and internal energy. 35. The energy possessed by a body by virtue of its position above the datumn or ground level, is known as potential energy. 36. The energy possessed by a body by virtue of its mass and velocity of motion, is known as kinetic energy. 37. The energy possessed by a body by virtue of its molecular arrangement and motion of molecules, is called internal energy. 38. The law of conservation of energy states that ‘Energy can neither be created nor destroyed, though it can be transformed from one form to another form, in which the energy can exist. 39. The volume per unit mass of a substance is termed as specific volume. 40. The mass per unit volume is called density. 41. The pressure is defined as the force per unit area. The unit of pressure is Pascal (Pa) or bar, such that 1 bar = 1 × 105 Pa = 1 × 105 N/m2.

16 Engineering Thermodynamics 42. In thermodynamics, we use the actual pressure called absolute pressure, such that Absolute pressure = Gauge pressure + Atmospheric pressure = Atmospheric pressure – Vacuum pressure 43. The conditions for normal temperature and pressure (N.T.P.) are 0°C and 760 mm of Hg respectively. Similarly, conditions for standard temperature and pressure (S.T.P.) are 15°C and 760 mm of Hg respectively.

EXERCISES 1. A pressure gauge on a steam boiler reads 0.8 N/mm2. Find the corresponding absolute pressure of steam in the boiler, if the barometer reads 720 mm of Hg. [Ans. 0.896 N/mm2] 2. A vacuum gauge on the condenser reads 620 mm of mercury and at the same time barometer reads 740 mm of mercury. Find the absolute pressure in kPa. [Ans. 15.996 kPa] 3. What is the vacuum pressure (in mm of Hg) inside the combustion chamber of a gasoline engine at position when the pressure is 1 bar? The atmospheric pressure is 1.033 bar. [Ans. 24.75 mm of Hg] 4. A turbine is supplied with steam at a gauge pressure of 1.4 MPa.The steam after expansion in the turbine, flows into a condenser maintained at a vacuum of 710 mm of Hg. The barometric pressure is 772 mm of Hg. Express the inlet and exhaust steam pressure in pascal (absolute). Take density of mercury as 13 600 kg/m3. [Ans. 1.503 MPa; 8.272 kPa]

[Hint: Atmospheric pressure = r . g . h = 13 600 × 9.81 × 0.772

= 103 × 103 N/m2 or Pa

...( h = 772 mm = 0.772 m)]

QUESTIONS 1. Define thermodynamics. State the scope and applications of thermodynamics. 2. What is the difference between working substance and pure substance? 3. Explain the difference between macroscopic and microscopic approach. 4. Differentiate between statistical and classical thermodynamics with suitable examples. 5. What is a thermodynamic system? Discuss its types with suitable examples. 6. Explain the concept of continuum and its relevance in thermodynamics. 7. Explain the following terms as related to thermodynamics : (a) System; (b) State; (c) Property; (d) Process; and (e) Cycle 8. What is thermodynamic equilibrium? Explain mechanical, chemical and thermal equilibrium. 9. What do you mean by cyclic and quasi-static process? 10. What do you understand by reversible and irreversible processes? Give some causes of irreversibility. 11. Define heat transfer in thermodynamics. 12. State the thermodynamic definition of work. What is positive work and negative work. 13. What are the similarities and differences between heat and work? 14. What is meant by thermodynamic work and mechanical work? How does the thermodynamic work differ from mechanical work? 15. What do you understand by path function and point function? Show that work is a path function. 16. Define energy. Explain the different forms of energy and pin point the modes of work. 17. Define pressure. How it is measured?

Basic Concepts 17

OBJECTIVE TYPE QUESTIONS 1. Any region of space or a finite quantity that occupies a volume and has a boundary, is known

as a (a) thermodynamic cycle (b) thermodynamic system (c) thermodynamic process (d) none of these 2. A system in which the mass of the working substance remains constant, is known as a (a) closed system (b) open system (c) isolated system (d) none of these 3. In an open system, which of the following crosses the boundary of the system? (a) heat (b) work (c) mass (d) heat and work 4. A closed system is also known as .................... system. (a) flow (b) non-flow 5. In an isolated system, (a) both mass and energy (i.e. heat and work) crosses the boundary of the system (b) neither mass nor energy transfer takes place across the boundary of the system

(c) only mass crosses the boundary of the system (d) only energy crosses the boundary of the system 6. Thermos flask containing a liquid is an example of a (a) closed system (b) open system (c) isolated system (d) adiabatic system 7. A mixture of ammonia in water is a .................... system. (a) homogeneous (b) heterogeneous 8. In an extensive property of a thermodynamic system (a) extensive energy is utilised (b) extensive work is done (c) extensive heat is transferred (d) none of these 9. A system consisting of a single phase is called a (a) closed system (b) open system (c) homogeneous system (d) heterogeneous system 10. Which of the following is an intensive property of a thermodynamic system? (a) Pressure (b) Volume (c) Temperature (d) Energy 11. Which of the following is an extensive property of a thermodynamic system? (a) Pressure (b) Volume (c) Temperature (d) Energy 12. A series of operations, which take place in a certain order and restore the initial condition is known as (a) reversible cycle (b) irreversible cycle (c) thermodynamic cycle (d) none of these 13. When there are no unbalanced forces on any part of the system, then the system is said to be in (a) mechanical equilibrium (b) chemical equilibrium (c) thermal equilibrium (d) none of these 14. Which of the following is a path function? (a) Pressure

(b) Volume

(c) Temperature

(d) Heat

18 Engineering Thermodynamics 15. The change in .................... causes the change in internal energy. (a) pressure (b) volume (c) temperature 16. The total energy of the system (E) is given by (a) P.E. + K.E. + U (b) P.E. + K.E. + U (c) K.E. – P.E. – U (d) P.E. – K.E. – U 17. The unit of pressure is (a) Pascal (b) bar (c) mm of Hg (d) all of these 18. One atmospheric pressure is equal to (a) 760 mm of Hg (b) 1.013 bar (c) 101.3 kPa (kilo-pascal) (d) all of these 19. Which of the following is correct? (a) Absolute pressure = Gauge pressure + Atmospheric pressure (b) Vacuum pressure = Atmospheric pressure – Absolute pressure (c) Atmospheric pressure = Absolute pressure + Gauge pressure (d) Gauge pressure = Atmospheric pressure – Absolute pressure 20. The normal temperature and pressure is (a) 15°C and 760 mm of Hg (b) 0°C and 760 mm of Hg (c) 15°C and 760 mm of water (d) 0°C and 760 mm of water

ANSWERS

1. (b) 6. (c) 11. (b) 16. (a)

2. (a) 7. (a) 12. (c) 17. (d)

3. (d) 8. (d) 13. (a) 18. (d)

4. 9. 14. 19.

(b) (c) (d) (a), (b)

5. 10. 15. 20.

(b) (a) (c) (b)

2 ZEROTH LAW OF THERMODYNAMICS 2.1 Introduction 2.2 Concept of Temperature 2.3 Equality of Temperature

2.4 Measurement of Temperature 2.5 Temperature Measuring Scales

2.1 INTRODUCTION

A

B

A

B

When two systems are each in thermal equilibrium with C a third system, then they are also in thermal equilibrium with each other. This statement is known as Zeroth law of C thermodynamics. For example, let a body A is in thermal equilibrium with a body B, and also separately with a body Fig. 2.1. Zeroth law of thermodynamics. C, then according to zeroth law of thermodynamics, bodies B and C shall also be in thermal equilibrium with each other, as shown in Fig. 2.1. This law is the basis of concept of temperature as well as all temperature measurements.

2.2 CONCEPT OF TEMPERATURE The temperature is an important thermodynamic property (intensive property) of the system, because it is independent of mass. It is also a macroscopic property, like pressure. The temperature is not a function of path, but a function of state of the system, i.e. it is a point function. The temperature may be defined as the degree of hotness or coldness of a body or environment. When two bodies are brought in contact, the heat will flow from the hot body at a higher temperature to a cold body at a lower temperature. In other words, temperature is a thermal potential causing the flow of heat energy.

2.3 EQUALITY OF TEMPERATURE When two bodies at different temperatures (say one hot and the other cold) are brought into contact, the cold body becomes warmer and the hot body becomes colder. After sometime, these bodies attain an equal temperature and this is a state of thermal equilibrium. It, thus, follows that the two systems have equal temperature and there is no change in their properties when they are brought in thermal contact with each other.

2.4 MEASUREMENT OF TEMPERATURE The branch of heat relating to the measurement of a body is called thermometry. A *thermometer is an instrument to measure the temperature of a body. Some commonly used thermometers are discussed, briefly, as follows:

* The low temperatures are measured with thermometers while the high temperatures are measured by instruments known as thermocouples or pyrometers.

20 Engineering Thermodynamics 1. Mercury-in-glass thermometers. The mercury-in-glass thermometers, as shown in Fig. 2.2, are most commonly used for the measurement of temperature. These are based upon the principle of change in volume of a mercury with change in temperature. In a mercury-in-glass thermometer, the upper limit of the temperature is determined by the boiling temperature of mercury (357°C at atmospheric pressure) or by the temperature at which the glass softens. The lower temperature is the freezing temperature of mercury (–39°C). The range can be extended upto 600°C by filling the space above the mercury with a gas, like nitrogen under pressure, thereby increasing the boiling point of mercury. 2. Bimetallic thermometers. The bimetallic thermometers are based upon the principle of expansion of solids, and consists of two solid metal strips of different metals (i.e. having different coefficient of expansion), bonded together as shown in Fig. 2.3. One end of the bimetallic strip is fixed and the other end is attached to a long pointer. By suitable calibration, a scale can be used to record the bending caused by a change in temperature. The bimetallic thermometers are used in metrology for recording changes in temperature during the day. They are also used to measure temperatures at high altitudes. 3. Electrical resistance thermometers (Thermistors). The electrical resistance thermometers are based on the principle of change in resistance with the change in temperature, such as platinum resistance thermometer. These thermometers give accurate measurement of temperature. Thermistor is a semi-conductor having negative temperature coefficient in contrast with positive temperature coefficient of most metals. It is a very sensitive device and can be used to measure temperature accurately.

Safety bulb

Capillary tube

Stem

Temperature sensing bulb

Mercury

Fig. 2.2. Mercury-in-glass thermometer.

Temperature scale

Metal B

Metal A

Pointer

Fig. 2.3. Bimetallic thermometer.

4. Thermocouples or Pyrometers. The thermocouples or pyrometers are commonly used for measuring large temperature differences or high temperatures. These devices are made of two different metal wires joined together at each end, and a temperature difference exists between the two junctions and an electromotive force (e.m.f.) will exist between the two junctions. This e.m.f. is directly proportional to the temperature difference and if one temperature (cold junction temperature or reference temperature) is known, then the measurement of the e.m.f. will give the hot junction temperature which is to be measured. The cold junction A, as shown in Fig. 2.4, is placed in a mixture of ice and distilled water (0°C) and this is the reference temperature. The hot junction B is placed at a point where temperature is to

Zeroth Law of Thermodynamics 21 be measured. The circulating current due to the e.m.f. developed can be measured by a milliammeter and can be calibrated to indicate the hot junction temperature. When great accuracy is not required, as in measuring high temperature (such as furnace temperature), the cold junction reference temperature may be the room temperature. It may be noted that a little variation of the room temperature will not affect the temperature measured much. Such devices are often referred Fig. 2.4. Thermocouple. to pyrometers. The range of temperature to be measured depends upon the metals used in the thermocouple. Some of the commonly used thermocouple wires combination are copper-constantan, iron-constantan, and chromel-alumel.

2.5 TEMPERATURE MEASURING SCALES Following are the important scales commonly used for measuring the temperature of a body: 1. Celsius and Fahrenheit scales. The Celsius (also known as centigrade scale) and Fahrenheit scales are the two commonly used scales for measuring the temperature of a body. These scales are based on two fixed points known as freezing point of water (or ice point) under atmospheric pressure and the boiling point of water (or steam point) as shown in Fig. 2.5. In *Celsius scale, the freezing point of water (lower fixed point) is marked as zero and the boiling point of water (upper fixed point) as 100. The interval between the two fixed points is divided into 100 equal parts, and each part represents one degree celsius (briefly written as 1°C).This scale is mostly used by scientists and engineers. In **Fahrenheit scale, the freezing point of water is marked as 32 and the boiling point of water as 212. The interval between these two points is divided into 180 equal parts and each part represents one degree fahrenheit (briefly written as 1°F). The relation between the Celsius scale and Fahrenheit scale is given byt Fig. 2.5. Celsius and Fahrenheit scales. C F − 32 C F − 32 = or = 100 180 5 9 It may be noted that Celsius and Fahrenheit scales show the same reading at – 40°, i.e. – 40°C = – 40°F (See Example 2.1.) 2. Kelvin and Rankine absolute scales. Whenever the value of temperature is used in thermodynamic calculations, then the value of temperature,whose reference point is true zero or absolute zero is used. The temperature, below which the temperature of any substance can not fall, is known as absolute zero temperature. In case of Celsius scale, the absolute zero temperature is taken as – 273°C for all sorts of calculations. The absolute temperature in Celsius scale is known as Kelvin (briefly written as K), such that Kelvin temperature = Celsius temperature + 273 or K = °C + 273

* The Celsius scale was suggested by Celsius in 1742. ** The Fahrenheit scale was suggested by Fahrenheit in 1720.

22 Engineering Thermodynamics Similarly, in case of Fahrenheit scale, the absolute zero temperature is taken as – 460°F. The absolute temperature in Fahrenheit scale is known as Rankine (briefly written as R), such that Rankine temperature = Fahrenheit temperature + 460 or R = °F + 460 Example 2.1. At which temperature the Celsius and Fahrenheit scales coincide? Solution. Let x = Temperature at which the Celsius and Fahrenheit scales coincide. C F − 32 x x − 32 We know that = or = 5 9 5 9 9x = 5x – 5 × 32 9x – 5x = –160 or 4x = –160 −160 \ x = = – 40° Ans. 4 Example 2.2. Find the temperature at which the Kelvin and Fahrenheit scales coincide. Solution. Let x = Temperature at which the Kelvin and Fahrenheit scales coincide. We know that K = °C + 273 or °C = K – 273 C F − 32 K − 273 F − 32 = We also know that = or 5 9 5 9 x − 273 x − 32 = 5 9 9x – 9 × 273 = 5x – 5 × 32 9x – 5x = 9 × 273 – 5 × 32 = 2457 – 160 or 4x = 2297 2297 \ x = = 574.25° Ans. 4 Example 2.3. Express the temperature of 86°F and – 40°C into the following units: 1. Celsius absolute; and 2. Fahrenheit absolute Solution. Given: Fahrenheit temperature = 86°F Celsius temperature = – 40°C 1. Temperature in Celsius absolute (i.e. Kelvin) First of all, let us convert Fahrenheit temperature into Celsius temperature. We know that C F − 32 86 − 32 = = =6 5 9 9 \ C = 6 × 5 = 30°C We also know that Celsius absolute (i.e. Kelvin), K = °C + 273 = 30 + 273 = 303 K Ans. For – 40°C ; K = – 40 + 273 = 233 K Ans. 2. Temperature in Fahrenheit absolute (i.e. Rankine) We know that Fahrenheit absolute (i.e. Rankine), R = °F + 460 = 86 + 460 = 546 R Ans. \

Zeroth Law of Thermodynamics 23 Now converting – 40°C into degrees Fahrenheit. We know that C F − 32 − 40 F − 32 = = or 5 9 5 9 or \ and

F – 32 =

− 40 × 9 = – 72 5

F = – 72 + 32 = – 40°F R = °F + 460 = – 40 + 460 = 420 R Ans.

HIGHLIGHTS 1. Zeroth law of thermodynamics states that when two systems are each in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other. 2. The temperature may be defined as the degree of hotness or coldness of a body or environment. 3. The temperature is measured by an instrument called thermometer. 4. The thermocouples or pyrometers are commonly used for measuring large temperature differences or high temperatures. 5. In Celsius scale, the freezing point of water is zero and the boiling point of water is 100.

6. In Fahrenheit scale, the freezing point of water is 32 and the boiling point of water is 212. 7. The relation between the Celsius scale and Fahrenheit scale is

C F − 32 C F − 32 = or = 100 180 5 9

8. The temperature, below which the temperature of any substance cannot fall, is known as absolute zero temperature. 9. The absolute temperature in Celsius scale is called Kelvin (briefly written as K), such that Kelvin temperature = Celsius temperature + 273 or

K = °C + 273

10. The absolute temperature in Fahrenheit scale is called Rankine (briefly written as R), such that Rankin temperature = Fahrenheit temperature + 460 or R = °F + 460

EXERCISES 1. The temperature of steam in a boiler is 200°C. What is its temperature in degrees Fahrenheit and degrees Kelvin? [Ans. 392°F; 473 K] 2. The temperature of a steam in a boiler is 343 K. What will be its temperature in degrees Celsius? [Ans. 70°C] 3. The normal boiling point of liquid oxygen is –183°C. What is the temperature on Kelvin scale and Rankine scale? [Ans. 90 K; 162.6 R] 4. The boiling point of liquid hydrogen is 20.2 K. Convert the temperature into degrees Rankine. [Ans. 36.6 R]

24 Engineering Thermodynamics

QUESTIONS

1. State and explain zeroth law of thermodynamics. 2. How the zeroth law of thermodynamics helps to introduce the concept of temperature? 3. What is the equality of temperature? 4. State the principle of thermometry. How it is used for the measurement of temperature? 5. Explain the working principle of Mercury-in-glass thermometer. 6. Explain the working principle of a thermocouple. 7. Give a relation between Celsius and Fahrenheit scales. 8. What is Kelvin and Rankine scales?

OBJECTIVE TYPE QUESTIONS 1. When two systems are each in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other. This statement is known as (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Kelvin-Plank’s law 2. The temperature is an .................... property of the thermodynamic system. (a) extensive (b) intensive 3. The measurement of a temperature is based upon (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) none of these 4. The relation between the Celsius scale and Fahrenheit scale is (a)

C + 32 F C − 32 F = (b) = 9 5 9 5

(c)

C F − 32 C F + 32 = = (d) 5 9 5 9

5. The absolute zero temperature is taken as (a) – 237°C (b) 237°C (c) 273°C

(d) – 273°C

ANSWERS

1. (a)

2. (b)

3. (a)

4. (c)

5. (d)

3 IDEAL AND REAL GASES 3.1 Introduction 3.2 Ideal or Perfect Gas 3.3 Boyle’s Law 3.4 Charles’ Law 3.5 Combination of Boyle’s and Charles’ LawGeneral Gas Equation 3.6 Joule’s Law 3.7 Characteristic Equation of a Gas 3.8 Universal Gas Constant

3.9 Avogadro’s Hypothesis 3.10 Specific Heats of a Gas 3.11 Enthalpy of a Gas 3.12 Regnault’s Law 3.13 Difference Between Two Specific Heats 3.14 Molar Specific Heats of a Gas 3.15 Ratio of Specific Heats 3.16 Deviation of Real Gas from Ideal Gas

3.1 INTRODUCTION A gas may be defined as a substance of which the vaporisation from the liquid state is complete. Such substances are oxygen, nitrogen, hydrogen, air etc. A vapour may be defined as a partially evaporated liquid and consists of pure gaseous state but with some suspension of liquid particles of the gases. It does not behave in the same way as a gas because it is further liable to evaporate consequently by increasing and decreasing the pressure and temperature. The vapours commonly used in engineering practice are steam, carbon dioxide, sulphur dioxide and ammonia.

3.2 IDEAL OR PERFECT GAS An ideal or perfect gas is one which strictly obeys the gas laws under all conditions of temperature and pressure. For engineering purposes, an ideal or perfect gas is a substance which remains in the gaseous state during the whole thermodynamic cycle in an engine. In fact, no real gas behaves exactly as an ideal or perfect gas, but gases like hydrogen, oxygen, nitrogen and even air may be regarded as ideal or perfect gases.

3.3 BOYLE’S LAW This law was discovered by Robert Boyle in 1662 and found experimentally that when a gas is heated at constant temperature, the pressure multiplied by the volume remains constant. In other words, it states that ‘The volume of a given mass of a perfect gas varies inversely as the absolute pressure, when the temperature is constant.’ Let p = Absolute pressure of the gas, and v = Volume of the gas occupied at pressure p.

26 Engineering Thermodynamics Now according to Boyle’s law, 1 or pv = constant p

The Boyle’s law may be represented on pressure volume ( p - v) diagram as discussed below: Let us consider a certain quantity of gas in the cylinder at point 1, and it changes its state from point 1 to point 2, as shown in Fig. 3.1. Let p1, v1 = Pressure and volume of gas at state 1, and p2, v2 = Pressure and volume of gas at state 2. Since pv = constant, therefore anywhere on the curve 1–2, p1 v1 = p2 v2 = ... = constant This is the most useful from of the Boyle’s law.

3.4 CHARLES’ LAW

Pressure

v

1

p1

pv = Constant

2

p2 v1

v2 Volume

Gas 1 Cylinder Piston

2

Fig. 3.1. Boyle’s law.

This law was discovered by a Frenchman Jacques A.C. Charles in about 1787, which states that ‘The volume of a perfect gas varies directly as its absolute temperature when the absolute pressure remains constant.’ Mathematically, v v T or = constant T or

v v1 v = 2 = 3 = ... = constant T1 T2 T3

Charles’ law also states that ‘The absolute pressure of a perfect gas varies directly as its absolute temperature when the volume remains constant.’ Mathematically, p p T or = constant T or

p p1 p = 2 = 3 = ... = constant T1 T2 T3

Further investigations by Gay Lussac and Regnault states that ‘All perfect gases expand by the same fraction of the volume they occupy at freezing temperature when their temperature is raised by one degree provided the pressure remains constant.’ This fraction is found to be 1/273 when Celsius scale is used. Let v0 = Volume of a given mass of gas at freezing temperature (i.e. 0°C), and vt = Volume of the same mass of gas at temperature t °C. Now as per the above statement, 1 t vt = v0 + × v0 × t = v0 1 + 273 273 T 273 + t = v0 = v0 × 273 T 0

Ideal and Real Gases 27 or where

vt T = v0 T0 T = Absolute temperature corresponding to t°C, and T0 = Absolute temperature corresponding to 0°C.

Note: All the solids and liquids expand at different rates when heated, but all the gases expand at the same rate when heated and contracts at the same rate when cooled. It may be noted that a perfect gas contracts 1/273 part of its volume at 0°C. This means that if a gas is cooled 273° below 0°C (i.e. – 273°C), the volume of the gas will become zero. In other words, the temperature at which the volume of gas becomes zero, is called absolute zero temperature.

Example 3.1. A cylinder contains 0.28 m3 of oxygen at 3.5 bar. What will be its volume when expanded to 1.5 bar at constant temperature. Solution. Given: Initial volume, v1 = 0.28 m3 Initial pressure, p1 = 3.5 bar Final pressure, p2 = 1.5 bar Let v2 = Final volume. We know that p1v1 = p2v2 ...(Boyle’s law)

p1v1 3.5 × 0.28 = = 0.653 m3 Ans. 1.5 p2 Example 3.2. A closed vessel contains gas at 1.5 N/mm2 and 20°C. The gas is compressed till it acquires a temperature of 280°C. Determine the pressure in MPa at the end of compression. Solution. Given: Initial pressure of the gas, p1 = 1.5 N/mm2 Initial temperature, T1 = 20°C = 20 + 273 = 293 K Final temperature, T2 = 280°C = 280 + 273 = 553 K Let p2 = Final pressure at the end of compression. p1 p We know that = 2 ...(Charles’ law) T1 T2 \

v2 =

p1 × T2 1.5 × 553 = = 2.83 N/mm2 293 T1 = 2.83 MPa Ans. ...(Q 1 N/mm2 = 1 MPa) Example 3.3. A gas with an initial pressure of 4 bar and initial temperature of 40°C is heated at constant volume until the final pressure becomes 12 bar. Calculate the final temperature of the gas. Solution. Given: Initial pressure, p1 = 4 bar = 0.4 × 106 N/m2 ...(Q 1 bar = 0.1 × 106 N/m2) Initial temperature, T1 = 40°C = 40 + 273 = 313 K Final pressure, p2 = 12 bar = 1.2 × 106 N/m2 Let T2 = Final temperature of the gas. p1 p We know that = 2 ...(Charles’ law) T1 T2 \

p2 =

\

T2 =

p2 × T1 1.2 × 106 × 313 = = 939 K p1 0.4 × 106

= 939 – 273 = 666°C Ans.

28 Engineering Thermodynamics

3.5 COMBINATION OF BOYLE’S LAW AND CHARLES’ LAW-GENERAL GAS EQUATION We have already discussed the Boyle’s law (Art. 3.3) and Charles’ law (Art. 3.4) for a unit mass of gas. We know that 1 v , when T is constant (Boyle’s law) p and v T, when p is constant (Charles’ law) From above, we see that 1 T v and T both or v p p \ pv T or pv = C × T where C is constant. This expression may be written as pv = C ...(i) T This shows that the product of the absolute pressure and volume divided by the absolute temperature, for a given mass of perfect gas is always constant. The equation (i) may also be expressed in the following form: pv p1v1 pv = 2 2 = 3 3 = ... = constant T1 T2 T3 Note: The value of constant (C ) in equation (i) depends upon the mass and the properties of the gas concerned.

Example 3.4. A gas at a temperature of 20°C and a pressure of 1.5 bar occupies a volume of 0.1 m3. If the gas is compressed to a pressure of 7.5 bar and a volume of 0.04 m3, what will be the final temperature of the gas. Solution. Given: Initial temperature, T1 = 20°C = 20 + 273 = 293 K Initial pressure, p1 = 1.5 bar = 0.15 × 106 N/m2 Initial volume, v1 = 0.1 m3 Final pressure, p2 = 7.5 bar = 0.75 × 106 N/m2 Final volume, v2 = 0.04 m3 Let T2 = Final temperature of the gas. p1v1 pv We know that = 2 2 ...(General gas equation) T1 T2 p2 v2T1 0.75 × 106 × 0.04 × 293 = = 586 K p1v1 0.15 × 106 × 0.1 = 586 – 273 = 313°C Ans. \

T2 =

3.6 JOULE’S LAW This law states that ‘the changes in internal energy (dU) of a perfect gas is directly proportional to the change of temperature (dT).’ Mathematically, dU dT or dU = mc (T2 – T1) where m is the mass of gas, and c is the constant of proportionality, known as specific heat of a gas.

Ideal and Real Gases 29

3.7 CHARACTERISTIC EQUATION OF A GAS When 1 kg of gas is taken, then the general gas equation pv = CT may be written as pv = RT ...(i) and for m kg of gas, the above equation (i) becomes pv = m RT ...(ii) where R is known as characteristic gas constant or simply gas constant and the equation (ii) is known as characteristic equation of a gas. When p is in N/m2 ; v in m3 ; m in kg and T in kelvin (K), then from equation (ii)

R =

pv N × m3 N-m = 2 = mT kg K m × kg × K

= N-m/kg K or J/kg K ...(Q 1 N-m = 1 J) The value of R depends upon the type of gas. For atmospheric air, the value of R is taken as 287 J/kg K or 0.287 kJ/kg K.

3.8 UNIVERSAL GAS CONSTANT The product of the characteristic gas constant (R) and the molecular mass (M) of an ideal gas is known as universal gas constant (also known as molar constant). It is usually denoted by Ru. Mathematically, Ru = R M If R1, R2, R3 etc. are the characteristic gas constants of different gases and M1, M2, M3 etc. are their molecular masses respectively, then Ru = R1 M1 = R2 M2 = R3 M3 = ... The value of universal gas constant (Ru) is same for all gases. Its value is taken as 8314 J/kg-mole K or 8.314 kJ/kg-mole K. In terms of molecular mass ( M ), the characteristic gas equation may be written as pv = MRT and pv = n Ru T where n = Number of kg moles. The following table shows the values of molecular mass for some important gases: Table 3.1. Molecular mass of important gases. Gas Oxygen (O2) Nitrogen (N2) Carbon monoxide (CO)

Molecular mass 32 28 28

Gas Hydrogen (H2) Carbon dioxide (CO2) Sulphur dioxide (SO2)

Molecular mass 2 44 64

3.9 AVOGADRO’S HYPOTHESIS According to Avogadro’s hypothesis ‘Equal volumes of all gases at the same pressure and temperature contain equal number of molecules.’ Therefore, equal volumes of different gases have a mass in direct proportion to mass of the molecules of which they consist. For example, the molecular mass of hydrogen is 2 and that of oxygen is 32. This means that an oxygen molecule is 16 times heavier than a hydrogen molecule. Since a given volume of both gases contains the same number of molecules, the mass of this volume of oxygen will be 16 times the corresponding mass of hydrogen. Thus, the density of any two gases is directly proportional to their molecular masses, if the gases are

30 Engineering Thermodynamics at the same temperature and pressure. Mathematically, for the two gases, ρ1 M = 1 ρ2 M2 Since the specific volume is the reciprocal of the density, therefore v2 M = 1 or M1 v1 = M2 v2 v1 M2 This equation states that the product of the molecular mass and the specific volume (i.e. M . v) for all gases is constant at the same temperature and pressure. The term (M . v) is called molar volume and is denoted by vM. The molecular mass expressed in any unit of mass is called a mole. For example, molecular mass expressed in gram is written as gram-mole (briefly written as g-mole). Similarly, molecular mass expressed in kg is written as kg-mole. It may be noted that 1 g-mole of all gases occupies a volume of 22.4 litres at normal temperature and pressure (briefly written as N.T.P.). In other words, the volume of 1 kg-mole of any gas is 22.4 m3 at N.T.P. Example 3.5. Determine the volume occupied by 2 kg of an ideal gas at 120 kPa and 300 K. Assume suitable value of the gas constant. Solution. Given: Mass of the gas, m = 2 kg Pressure, p = 120 kPa = 120 × 103 Pa = 120 × 103 N/m2 Temperature, T = 300 K Let v = Volume occupied by the gas. Taking the value of gas constant (R) as 287 J / kg K, we have pv = m R T m RT 2 × 287 × 300 = \ v = = 1.435 m3 Ans. p 120 × 103 Example 3.6. A vessel of capacity 3 m3 contains air at a pressure of 1.5 bar and a temperature of 25°C. Additional air is now pumped into the system until the pressure rises to 30 bar and temperature rises to 60°C. Determine the mass of air pumped in and express the quantity as a volume at a pressure of 1.02 bar and a temperature of 20°C. If the vessel is allowed to cool until the temperature is again 25°C, calculate the pressure in the vessel. Solution. Given: Initial volume, v1 = 3 m3 Initial pressure, p1 = 1.5 bar = 150 kPa = 150 × 103 N/m2 ...(Q 1 bar = 100 kPa) Initial temperature, T1 = 25°C = 25 + 273 = 298 K Pressure of air after additional air is pumped, p2 = 30 bar = 30 × 100 kPa = 3 × 106 N/m2 Temperature of air, T2 = 60°C = 60 + 273 = 333 K Mass of air pumped in Let the vessel is initially filled with m1 kg of air and m2 kg be the mass of air in the vessel after pumping. We know that p1 v1 = m1 R T1

p1v1 150 × 103 × 3 = = 5.26 kg 287 × 298 RT1 ...(Q R for air = 287 J/kg K) \

m1 =

Ideal and Real Gases 31 Similarly,

p2 v2 = m2 R T2

p2 v2 3 × 106 × 3 = = 94.17 kg ...(Q v2 = v1) 287 × 333 RT2 We know that mass of air pumped in, m = m2 – m1 = 94.17 – 5.26 = 88.91 kg Ans. Volume of air pumped in Let v3 = Volume of air pumped in, p3 = Pressure of air = 1.02 bar = 102 × 103 N/m2 ...(Given) T3 = Temperature of air = 20°C = 20 + 273 = 293 K ...(Given) We know that p3v3 = m RT3 \

m2 =

\

v3 =

m RT3 88.91 × 287 × 293 = = 73.3 m3 Ans. p3 102 × 103

Pressure in the vessel after cooling Let p4 = Pressure in the vessel after cooling, T4 = Temperature after cooling = 25°C = 25 + 273 = 298 K ...(Given) Since the cooling takes place at constant volume, therefore according to Charles’ law, p2 p = 4 T2 T4 \

p4 =

p2 × T4 3 × 106 × 298 = = 2.68 × 106 N/m2 333 T2

= 26.8 × 105 N/m2 = 26.8 bar Ans. Example 3.7. A spherical vessel of 1.5 m diameter, containing air at 40°C is evacuated till the vacuum inside the vessel becomes 735 mm of Hg. Calculate the mass of the air pumped out. If the tank is then cooled to 3°C, what will be the final pressure in the tank? Take atmospheric pressure as 760 mm of Hg. Solution. Given: Diameter of spherical vessel, d = 1.5 m \ Volume,

v1 =

πd3 π (1.5)3 = = 1.767 m3 6 6

Temperature of air, T1 = 40°C = 40 + 273 = 313 K Vacuum pressure, pv = 735 mm of Hg = 735 × 133.3 = 98 × 103 N/m2 ...(Q 1 mm of Hg = 133.3 N/m2) Final temperature of air in the tank, T3 = 3°C = 3 + 273 = 276 K Atmospheric pressure, p1 = 760 mm of Hg = 760 × 133.3 = 101.3 × 103 N/m2 Mass of air pumped out Let m1 = Initial mass of air in the tank at atmospheric pressure.

32 Engineering Thermodynamics We know that or

p1 v1 = m1 R T1 m1 =

p1v1 101.3 × 103 × 1.767 = = 1.993 kg 287 × 313 RT1

...(Q R for air = 287 J/kg K) Pressure of air in the tank after evacuation p2 = Atmospheric pressure – Vacuum pressure = 101.3 × 103 – 98 × 103 = 3.3 × 103 N/m2 Let m2 = Mass of air in the tank after evacuation. We know that p2 v2 = m2 R T2

p2 v2 3.3 × 103 × 1.767 = = 0.065 kg 287 × 313 RT2 ...(Q v2 = v1; and T2 = T1) \ Mass of air pumped out, m = m1 – m2 = 1.993 – 0.065 = 1.928 kg Ans. Final pressure in the tank Let p3 = Final pressure in the tank. Since the cooling takes place at constant volume, therefore according to Charles’ law, p3 p = 2 T3 T2 or

\

m2 =

p3 =

=

p2 × T3 3.3 × 103 × 276 = = 2910 N/m2 T2 313

...(Q T2 = T1)

2910 = 21.83 mm of Hg Ans. 133.3

Example 3.8. A tank has a volume of 5 m3 and contains 20 kg of an ideal gas having a molecular mass of 25. The temperature is 15°C. What is the pressure? Solution. Given: Volume of the tank, v = 5 m3 Mass of an ideal gas, m = 20 kg Molecular mass of the gas, M = 25 Temperature of the gas, T = 15°C = 15 + 273 = 288 K Let p = Pressure of the gas. We know that characteristic gas constant, R 8.314 R = u = = 0.33 256 kJ / kg K ...(Q Ru = R . M) M 25 = 332.56 J / kg K ...(Q Ru for all gases = 8.314 kJ / kg-mole K) The general gas equation is pv = m R T m RT 20 × 332.56 × 288 \ p = = = 383 × 103 N/m2 v 5 = 383 kN/m2 Ans.

Ideal and Real Gases 33 Example 3.9. Nitrogen is to be stored at 140 bar and 27°C in a steel flask of 0.05 m3 volume. The flask is to be protected against excessive pressure by a fusible plug which will melt and allow the gas to escape, if the temperature rises too high. Find: 1. How many kg of nitrogen will the flask hold at the designed conditions? Take molecular mass of nitrogen as 28. 2. At what temperature must the fusible plug melt in order to limit the pressure of the full flask to a maximum of 168 bar. Solution. Given: Initial pressure of nitrogen, p1 = 140 bar = 14 × 106 N/m2 Initial temperature, T1 = 27°C = 27 + 273 = 300 K Initial volume, v1 = 0.05 m3 1. Mass of nitrogen Let m = Mass of nitrogen which the flask will hold in kg. Since the molecular mass (M ) of nitrogen is given as 28, therefore characteristic gas constant, R Universal gas constant = u R = Molecular mass M = We know that \

8314 = 297 J/kg K 28

...(Q Ru for all gases = 8314 J/kg K)

p1 v1 = m R T1 m =

p1v1 14 × 106 × 0.05 = = 7.856 kg Ans. 297 × 300 RT1

2. Temperature at which fusible plug melt Let T2 = Temperature at which the fusible plug melts, and p2 = Maximum pressure of the flask = 168 bar = 16.8 × 106 N/m2 ...(Given) The rise of pressure will cause rise in temperature. Since the process is a constant volume, therefore according to Charles’s law, p1 p = 2 T1 T2 \

T2 =

p2 × T1 16.8 × 106 × 300 = = 360 K p1 14 × 106

= 360 – 273 = 87°C Ans.

3.10 SPECIFIC HEAT OF A GAS We have already discussed that the specific heat of any substance is defined as the quantity of heat required to raise the temperature of unit mass of the substance by one degree. The solids and liquids have only one specific heat while the gases are considered to have two specific heats as follows: 1. Specific heat at constant volume (cv). It is defined as the amount of heat required to raise the temperature of unit mass of gas, by one degree, while the volume is kept constant and the pressure increases.

34 Engineering Thermodynamics If m kg of the gas is heated in a closed vessel, its temperature and pressure will increase, but the volume remains constant because of closed vessel. The total heat supplied to the gas at constant volume is given by Q1 – 2 = Mass × Specific heat at constant volume × Rise in temperature = m cv dT = m cv (T2 – T1) where dT = Rise in temperature = T2 – T1, T1 = Initial temperature of the gas, and T2 = Final temperature of the gas. Since the gas is being heated at constant volume, therefore there will be no workdone by the gas. The heat supplied to the gas remains within the body of the gas which is equivalent to change in internal energy of the gas. This statement is in agreement with Joule’s law. 2. Specific heat at constant pressure (cp). It is the amount of heat required to raise the temperature of unit mass of gas, by one degree, when it heated at constant pressure. If m kg of the gas is heated at constant pressure, its volume and temperature will increase. The total heat supplied to the gas at constant pressure is given by, Q1–2 = Mass × Specific heat at constant pressure × Rise in temperature = m cp dT = m cp (T2 – T1) It may be noted that the heat supplied to the gas at constant pressure is utilised to increase the internal energy of the gas (dU) and to do some external work (W1–2) in expanding the gas. Thus Heat supplied to the gas = Increase in internal energy + Workdone by the gas or Q1–2 = dU + W1 – 2 = dU + p dv = (U2 – U1) + p dv = mcv (T2 – T1) + p (v2 – v1) where p = Constant pressure at which the gas is heated, v1 = Initial volume of the gas, and v2 = Final volume of the gas.

3.11 ENTHALPY OF A GAS The enthalpy of a gas ( H ) is equal to the internal energy of the gas (U) plus the product of pressure and volume ( p . v). Mathematically, enthalpy of a gas, H = U + p . v Since (U + p. v) is made up entirely of properties, therefore enthalpy of a gas (H) is also a property. We have discussed above, that dU = U2 – U1, and W1 – 2 = p dv = p (v2 – v1) \ Q1–2 = dU + W1 – 2 = U2 – U1 + p (v2 – v1) = (U2 + pv2) – (U1 + pv1) = H2 – H1

3.12 REGNAULT’S LAW This law states that ‘The specific heats (cp and cv ) remain constant for all temperatures and pressures’. This is not strictly true as the specific heats are found to change with temperature.

3.13 DIFFERENCE BETWEEN TWO SPECIFIC HEATS Consider that m kg of gas is heated at constant pressure from temperature T1 to temperature T2. We have discussed in Art. 3.10, that heat supplied to the gas, Q1–2 = dU + W1 – 2 = dU + p dv

Ideal and Real Gases 35 or m cp (T2 – T1) = mcv (T2 – T1) + p (v2 – v1) m (T2 – T1) [cp – cv] = pv2 – pv1 ...(i) From the characteristic gas equation, pv = mRT, we have pv1 = m RT1, and pv2 = mRT2 Now from equation (i), m (T2 – T1) [cp – cv] = mRT2 – mRT1 = m (T2 – T1) R \ cp – cv = R This shows that the difference of the two specific heats (cp – cv) is equal to the characteristic constant of a gas (R).

3.14 MOLAR SPECIFIC HEATS OF A GAS When the specific heat a gas (c) is multiplied by its molecular mass (M), then it is known as molar specific heat (cm) of a gas. It may be defined as the amount of heat required to raise the temperature of a unit mole of gas through one degree. Mathematically, molar specific heat, cm = c . M \ Molar specific heat at constant volume, cvm = cv . M and specific heat at constant pressure, cpm = cp . M Note: The difference of two molar specific heats (cpm – cvm) is equal to universal gas constant (Ru), i.e. cpm – cvm = Ru

3.15 RATIO OF SPECIFIC HEATS The ratio of the two specific heats (i.e. cp /cv) is an important factor in the thermodynamics of gases. It is usually denoted by a Greek letter gamma (g ). We have already discussed that cp – cv = R c p cv R − = or ... (Dividing throughout by cv) cv cv cv

g – 1 =

\

R cv

g = 1 +

cp

cv

...

= γ

R cv

Notes: 1. Since cp is greater cv , therefore the value of g is always greater than unity.

2. The value of cp and cv is different for different gases, therefore the value of g is also different for

different gases. For example, the value of cp for air is 1 kJ / kg K; and cv = 0.72 kJ / kg K. Therefore the value of g for air is cp / cv = 1/0.72 = 1.4.

Example 3.10. If a gas has cp = 1.97 kJ/kg K and cv = 1.5 kJ/kg K; determine its molecular mass and characteristic gas constant. Solution. Given: Specific heat at constant pressure, cp = 1.97 kJ / kg K

36 Engineering Thermodynamics Specific heat at constant volume, cv = 1.5 kJ / kg K Characteristic gas constant We know that characteristic gas constant, R = cp – cv = 1.97 – 1.5 = 0.47 kJ/kg K Ans. Molecular mass Let M = Molecular mass. Since universal gas constant (Ru) is same for all gases and its value is 8.314 kJ/kg-mole K, therefore R 8.314 M = u = = 17.689 Ans. ...(Q Ru = R . M) R 0.47 Example 3.11. A closed vessel contains 2 kg of CO2 at temperature of 20°C and pressure of 0.7 bar. The heat is supplied to the vessel till the gas acquires a pressure of 1.4 bar. Calculate the value of heat supplied. Solution. Given: Mass of gas, m = 2 kg Initial temperature, T1 = 20°C = 20 + 273 = 293 K Initial pressure, p1 = 0.7 bar = 70 × 103 N/m2 Final pressure, p2 = 1.4 bar = 140 × 103 N/m2 Let T2 = Final temperature of the gas. Since the gas is heated in a closed vessel i.e. at a constant volume, therefore according to Charles’ law, p1 p = 2 T1 T2 T2 =

or

p2 × T1 140 × 103 × 293 = = 586 K p1 70 × 103

We know that heat supplied at constant volume, Q = m cv (T2 – T1) = 2 × 0.72 (586 – 293) = 422 kJ Ans. ... (Taking cv = 0.72 kJ/kgK) Example 3.12. One kg of ideal gas is heated from 18.3°C to 93.4°C. Assuming R = 0.287 kJ/kg K and g = 1.18 for the gas, find out: 1. Specific heats; 2. Change in internal energy; and 3. Change in enthalpy. Solution. Given: Mass of the gas, m = 1 kg Initial temperature, T1 = 18.3°C = 18.3 + 273 = 291.3 K Final temperature, T2 = 93.4°C = 93.4 + 273 = 366.4 K 1. Specific heats Let cv = Specific heat at constant volume, and cp = Specific heat at constant pressure. We know that cp – cv = R Dividing throughout by cv ,

cp cv

− 1 =

R cv

Ideal and Real Gases 37 \ and

g – 1 = cv =

R cv

...

cp

= γ cv

R 0.287 = = 1.594 kJ / kg K Ans. γ − 1 1.18 − 1

cp

= g cv cp = cv × g = 1.594 × 1.18 = 1.88 kJ / kg K Ans.

\ 2. Change in internal energy We know that change in internal energy, dU = m cv (T2 – T1) = 1 × 1.594 (366.4 – 291.3) = 119.7 kJ Ans. 3. Change in enthalpy We know that change in enthalpy, dH = m cp (T2 – T1) = 1 × 1.88 (366.4 – 291.3) = 141.2 kJ Ans. Example 3.13. A vessel of 2.5 m3 capacity contains 1 kg-mole of nitrogen ( N2 ) at 100°C. Evaluate the specific volume and pressure. If the gas is cooled to 30°C, calculate the final pressure, change in specific internal energy, and specific enthalpy. The ratio of specific heats is 1.4. One kg-mole of nitrogen is 28 kg. Solution. Given: Volume of vessel, v1 = 2.5 m3 Temperature, T1 = 100°C = 100 + 273 = 373 K Specific volume of the gas Since 1 kg-mole of N2 is 28 kg [i.e. molecular mass (M) of nitrogen is 28], therefore specific volume of gas, v 2.5 vs = 1 = = 0.893 m3/kg Ans. 28 M Pressure of the gas Let p1 = Pressure of the gas. We know that universal gas constant (Ru) for all gases is 8.314 kJ/kg mole/K, therefore characteristic gas constant, R 8.314 R = u = = 0.297 kJ/kg K M 28 We also know that p1 v1 = M RT1 MRT1 28 × 0.297 × 373 \ p1 = = = 1240 kN/m2 2.5 v1 = 1240 × 103 N/m2 = 12.4 bar Ans. Final pressure Given: Final temperature, T2 = 30°C = 30 + 273 = 303 K Let p2 = Final pressure. p1v1 pv p1 p = 2 We know that = 2 2 or T1 T2 T1 T2

...(1 bar = 1×105 N/m2)

...(Q v1 = v2)

38 Engineering Thermodynamics \

p2 =

p1 × T2 12.4 × 303 = = 10.07 bar Ans. T1 373

Change in specific internal energy We know that ratio of specific heats, cp = g = 1.4 or cp = 1.4 cv ...(Given) cv and cp – cv = R = 0.297 kJ/ kg K or 1.4 cv – cv = 0.297 0.297 \ cv = = 0.7425 kJ/kg K 0.4 and cp = 1.4 cv = 1.4 × 0.7425 = 1.04 kJ/kg K We know that change in specific internal energy, dU = cv (T2 – T1) = 0.7425 (303 – 373) = – 524 kJ/kg Ans. The negative sign indicates that there is a decrease in specific internal energy. Change in specific enthalpy We know that change in specific enthalpy, dH = cp (T2 – T1) = 1.04 (303 – 373) = – 72.8 kJ/kg Ans. The negative sign indicates that there is a decrease in specific enthalpy.

3.16 DEVIATION OF REAL GAS FROM IDEAL GAS We have already discussed the ideal gas obeying the law pv = RT. There is, however, no actual or real gas for which this equation holds accurately over the entire range of temperature and pressure except at the pressure approaching zero (i.e p 0). The ideal gas law is simple and hence simple relationships can be derived from it which may be applied to real gases under limited conditions. Therefore, an assumption is made that a particular gas under certain conditions behaves as an ideal gas or perfect gas. For example, gases which are ordinarily difficult to liquify such as oxygen, nitrogen, air or hydrogen may be treated as ideal gases at temperatures that are high relative to the critical pressure. Most of the known gases can be liquified and even solidified. Hence under certain conditions, a real gas remains no more in gaseous phase but changes its state, at the same time it does not accurately obey the ideal gas law. A distinction is, therefore, made between an ideal gas and a real gas on the basis whether it obeys ideal gas law or not. It will be observed that assumption of ideal gas behaviour for the case of real gases is not sufficiently accurate at very high pressures and low temperatures. Under these conditions, real gases show marked deviation from the ideal gas law. On the basis of kinetic theory of gases, it is possible to analytically derive the ideal gas equation pv = RT, provided the following two important assumptions are made: 1. There is little or no attraction between the molecules of the gas. 2. The molecules of a gas are mere mass-points occupying no space. In other words, the volume occupied by the molecules themselves is negligibly small compared to the volume of the gas. In actual practice, the molecules of all actual or real gas do occupy some space and do attract each other. Hence no real gas conforms to the perfect gas equation pv = RT. The Dutch physicist J.D. Vander Waals was the first to correct this equation by applying corrections for the above two

Ideal and Real Gases 39 factors. Thus, the perfect gas equation known as Vander Waals equation for the real or actual gas for one mole of the gas becomes, a p + 2 (v – b) = Ru T v where p = Pressure of the gas in N/m2; v = Molar volume in m3/kg-mole ; T = Temperature in K ; Ru = Universal gas constant = 8314 J/kg mole K ; a = Coefficient for intermolecular force in N-m4/(kg mole)2 ; b = Coefficient for volume molecules in m3/kg mole 2 The term a/v is called force of cohesion. The following table gives the values of a and b for some important substances: Table 3.2. Coefficients a and b for Vander Waals equation. Substance

a (N-m4/(kg-mole)2

Air Oxygen Hydrogen Water (vapour) Mercury (vapour) Carbon dioxide

135.332 × 103 139.25 × 103 25.105 × 103 551.13 × 103 2031.94 × 103 362.85 × 103

b m3/kg

mole 0.0362 0.0314 0.0262 0.0300 0.0657 0.0423

Example 3.14. Determine the pressure exerted by 1 kg of carbon dioxide (CO2 ) at 100°C by using Vander Waals equation. The specific volume is 1 m3/kg. The molecular mass of CO2 in 44. Compare the result obtained, if CO2 is treated as an ideal gas. Solution. Given: T = 100°C = 100 + 273 = 373 K Since the molecular mass of CO2 is 44 and the specific volume is 1 m3/kg, therefore molar volume, vM = Molecular Mass × Specific volume = 44 × 1 = 44 m3/kg mole From Table 3.2, we find that the coefficients a and b for carbon dioxide (CO2)are as follows: a = 362.85 × 103 N-m4/(kg mole)2; and b = 0.0423 m3/kg mole Pressure of CO2 by Vander Waal’s equation Let p = Required pressure in N/m2. We know that a p + 2 (v – b) = Ru T v 362.85 × 103 p + (44 – 0.0423) = 8314 × 373 (44) 2

...[Q R = 8314 N-m/kg mole K]

40 Engineering Thermodynamics

( p + 187.4) (43.9577) = 3 101 122 3 101 122 p + 187.4 = = 70 548 43.9577

\ p = 70 548 – 187.4 = 70 360.6 N/m2 Ans. Pressure of CO2 by using general gas equation pv = Ru T RT 8314 × 373 \ p = u = = 70 480 N/m2 Ans. v 44

HIGHLIGHTS 1. An ideal or perfect gas is one which strictly obeys the gas laws under all conditions of temperature and pressure. 2. Boyle’s law states that ‘The volume of a given mass of a perfect gas varies inversely as the absolute pressure, when the temperature is constant.’ In other words 1 v or p v = constant or p1 v1 = p2 v2 = ... = constant p 3. Charles’ law states that ‘The volume of a perfect gas varies directly as its absolute temperature, when the absolute pressure remains constant.’ In other words v v v v T or = constant or 1 = 2 = ... = constant T T1 T2 Charles’ law also states that ‘The absolute pressure of a perfect gas varies directly as its absolute temperature when the volume remains constant’. In other words p1 p p = 2 = ... = constant p T or = constant or T T1 T2 3. Gay Lussac and Regnault states that ‘All perfect gases expand by the same fraction of the volume they occupy at freezing temperature when their temperature is raised by one degree provided the pressure remains constant.’ This fraction is 1/273 when Celsius scale is used. In other words, vt T = v0 T0 where v0 and vt = Volume of the gas at 0°C and t°C respectively, T0 and T = Absolute temperature of the gas at 0°C and t°C respectively. 4. When the gas is cooled 273° below 0°C (i.e. – 273°C), the volume of the gas will become zero. In other words, the temperature at which the volume of gas becomes zero, is called absolute zero temperature. 5. The general gas equation is p1v1 pv = 2 2 = ... = constant T1 T2 6. The characteristics equation for m kg of a gas is p v = m R T where R = Characteristic gas constant. 7. Joule’s law states that ‘The change in internal energy (dU) of a perfect gas is directly proportional to the change of temperature (dT). In other words, dU dT or dU = mc (T2 – T1) where m is the mass of gas and c is a constant, known as specific heat of a gas.

Ideal and Real Gases 41 8. When the characteristic gas constant (R) is multiplied by the molecular mass (M) of the gas, then the product is known as universal gas constant (Ru). In other words, Ru = R . M The value of Ru is 8314 J/kg-mole K for all gases. 9. The characteristic gas equation may also be written as p v = m R T = Ru T or p v = n Ru T where n = Number of kg moles. 10. According to Avogadro’s hypothesis, ‘Equal volumes of all gases at the same pressure and temperature contain equal number of molecules’. The product of the molecular mass (M) and the specific volume (v) is constant for all gases and is known as molar volume. 11. The specific heat at constant volume (cv) is defined as the amount of heat required to raise the temperature of unit mass of gas by one degree, while the volume is kept constant and the pressure increases. For example, if m kg of gas is heated at constant volume and its temperature rises from T1 to T2, then heat supplied to the gas is given by Q1–2 = m cv (T2 – T1) Since the volume remains constant, therefore no work will be done by the gas. The heat energy stored in the gas and used for raising the temperature of the gas is known as internal energy of the gas (dU), such that dU = U2 – U1 = m cv (T2 – T1) 12. The specific heat at constant pressure (cp) is defined as the amount of heat required to raise the temperature of unit mass of gas by one degree, when it is heated at constant pressure. For example, if m kg of gas is heated at constant pressure, its volume will increase from v1 to v2 and temperature from T1 to T2. The heat supplied to the gas is given by Q1–2 = m cp (T2 – T1) and workdone by the gas, W1–2 = p (v2 – v1) The heat supplied to the gas at constant pressure is utilised to increase the internal energy of the gas (dU) and to do some external work (W1 – 2). In other words, Q1–2 = dU + W1 – 2 = (U2 – U1) + p (v2 – v1) = U2 – U1 + pv2 – pv1 = (U2 + pv2) – (U1 + pv1) 13. The enthalpy of a gas (H) is given by H = U + p . v We know that dU = U2 – U1; and W1–2 = pdv = p (v2 – v1) \ Q1–2 = dU + W1–2 = (U2 – U1) + p(v2 – v1) = (U2 + pv2) – (U1 + pv1) = H2 – H1 14. The difference between two specific heats (cp – cv) is equal to the characteristic constant of the gas (R). In other words, cp – cv = R 15. The ratio of specific heats (cp /cv) is denoted by a greek letter gamma (g), i.e. cp g = cv

42 Engineering Thermodynamics We know that \

cp – cv = R or g = 1 +

cp cv

−1 =

R cv

R cv

16. The Vander Waals equation for the real or actual gas is given by a p + 2 (v – b) = Ru T v where p = Pressure of the gas in N/m2, v = Molar volume in m3/ kg-mole, T = Temperature in K, Ru = Universal gas constant = 8314 J/kg-mole K; a = Coefficient for intermolecular force in N-m4/(kg-mole)2, and b = Coefficient for volume molecules in m3/ kg-mole. 2 The term a/v is called force of cohesion.

EXERCISES m3

1. A cylinder contains 0.28 of oxygen at 31.5 bar. What will be its volume when expanded to 15 bar at constant temperature. [Ans. 0.588 m3] 2. The volume of a certain mass of gas is 145 cm3 at 17°C and under a pressure of 725 mm of Hg. How will the volume be affected if the temperature falls to 7°C? If the pressure further falls to 700 mm of Hg, what will be the final volume? [Ans. 140 cm3; 150.18 cm3] 3. A gas occupies a volume of 0.105 m3 at a temperature of 20°C and a pressure of 0.15 MPa. Find the final temperature of the gas, if it is compressed to a pressure of 0.75 MPa and occupies a volume of 0.04 m3. [Ans. 285.1°C] 4. Find the mass of a gas which occupies 5.6 m3 at 7 bar and 200°C. Take characteristic gas constant as 287 N-m/ kg K. [Ans. 28.876 kg] 5. An insulated rigid tank with zero heat capacity contains 5 kg of air at 2 bar and 35°C. It is quickly filled with air at 12 bar and 50°C. Determine the total mass of air in the tank and the final temperature of air. [Ans. 33.48 kg ; 274.75 K] 6. A high altitude chamber, the volume of which is 30 m3, is put into operation by reducing the pressure from 1.013 bar to 0.35 bar and temperature from 27°C to 5°C. How many kilogram of air must be removed from the chamber during the process. Express this mass as volume measured at 1.013 bar and 27°C. Take R = 287 J/ kg K for air. [Ans. 22.136 kg ; 18.8 m3] 7. A certain gas occupies 0.15 m3 at a temperature of 20°C and a pressure of 120 kPa. If the mass of the gas is 0.2 kg ; calculate : 1. Value of the gas constant ; and 2. Molecular mass of the gas. [Ans. 307.17 J/ kg K ; 27.07] 8. 12 kg-mole of gas occupies 724 m3 of volume at a temperature of 140°C. If the density of gas under these conditions is 0.65 kg/m3. Find the pressure, gas constant and molecular mass of the gas. [Ans. 0.57 bar ; 212.5 J/kg K ; 39.12] 9. Calculate the volume of 6 kg of air at a pressure of 5 bar and temperature of 50°C.Take specific heat of air at constant pressure and volume as 1.005 kJ/kg K and 0.72 kJ/kg K respectively. [Ans. 1.1 m3] 10. A certain gas has cp = 1.97 kJ/ kg K and cv = 1.507 kJ/kg K. Find its molecular mass and the gas constant. Take universal gas constant as 8.315 kJ/kg K [Ans. 17.96 ; 0.463 kJ/ kgK] 11. The gas constant for atmospheric air is 287 J/kg K. If the specific heat at a constant volume is 0.72 kJ/ kg K ; find the ratio of specific heats and the value of specific heat at constant pressure. [Ans. 1.4, 1.007 kJ/kg K] 12. The heat required to heat a certain perfect gas at constant pressure from 16°C to 96°C is 1230 kJ/kg. When the gas is heated at constant volume between the same temperatures, the heat required is 795 kJ/kg. Find

Ideal and Real Gases 43 the values of specific heats at constant pressure and constant volume respectively. Also determine the molecular mass of the gas. [Ans. 15.375 kJ/kg K ; 9.9375 kJ/kg K ; 1.529 kg] 13. 0.14 m3 of air at 100°C and at a pressure of 1.5 bar is compressed at constant pressure until the volume is 0.112 m3. What is the final temperature? How much heat is given out by the air? If the specific heat at constant pressure is 1.005 kJ/ kg K, find the value of specific heat at constant volume. Take gas constant as 0.287 kJ/ kg K. [Ans. 25.4°C ; 14.7 kJ ; 0.717 kJ/kg K] 14. 3 kg of an ideal gas is expanded from a pressure of 7 bar and volume 1.5 m3 to a pressure of 1.4 bar and volume of 4.5 m3. The change in internal energy is 525 kJ. The specific heat at constant volume for the gas is 1.047 kJ/ kg K. Calculate: 1. Gas constant ; 2. Change in enthalpy ; and 3. Initial and final temperatures.

[Ans. 0.838 kJ / kg K; 945 kJ (decrease); 144°C, – 23.14°C]

QUESTIONS 1. What do you understand by the ideal gas? 2. Define Boyles’ law, Charles’ law and Gay Lussac law. pv 3. Prove the general gas equation, = C. T 4. Differentiate between characteristic gas constant and universal gas constant. 5. Write a short note on Avogadro’s hypothesis. 6. State specific heat of a gas. Why there are two types of specific heats. 7. Define enthalpy of a gas. 8. Derive the relation, cp – cv = R where cp = Specific heat at constant pressure, cv = Specific heat at constant volume, and R = Characteristic gas constant. 9. What do you mean by molar specific heats of a gas? 10. Explain how real gases deviate from ideal gas at elevated pressure and temperature. 11. Write the Vander Waals equation with the meaning of different symbols.

OBJECTIVE TYPE QUESTIONS 1. The volume of a given mass of a perfect gas varies inversely as the absolute pressure, when the temperature is constant. This statement is according to (a) Boyle’s law (b) Charles’ law (c) Gay-Lussac law (d) none of these 2. According to Charles’ law, the volume of a perfect gas varies .................... as its absolute temperature, when the absolute pressure remains constant. (a) directly (b) indirectly 3. All perfect gases change in volume by 1/ 273 of its original volume at 0°C for every 1°C change in temperature, when the absolute pressure remains constant. This statement is according to (a) Boyle’s law (b) Charles’ law (c) Gay-Lussac law (d) Joule’s law 4. The temperature at which the volume of a gas becomes zero is called (a) absolute temperature (b) absolute scale of temperature (c) absolute zero temperature (d) none of these 5. The change in internal energy of a perfect gas is directly proportional to the change of temperature. This statement is called (a) Boyle’s law (b) Charles’ law (c) Gay-Lussac law (d) Joule’s law

44 Engineering Thermodynamics 6. The general gas equation is (a) pv = RT m (b) pv = mR T (c) pv m = C (d) pv = (RT)m 7. The value of gas constant (R) for atmosperic air is taken as (a) 2.87 J/kg K (b) 28.7 J/kg K (c) 287 J/kg K (d) 2870 J/kg K 8. The universal gas constant (Ru) is given by (a) R . M (b) M × cp (c) M × cv (d) M/R where R = Gas constant, M = Molecular mass of the gas cp = Specific heat at constant pressure, and cv = Specific heat at constant volume. 9. The value of universal gas constant (Ru) is (a) 8.314 J/kg-mole K (b) 83.14 J/kg-mole K (c) 831.4 J/kg-mole K (d) 8314 J/kg-mole K 10. According to Avogadro’s law, the density of any two gases is .................... their molecular masses, if the gases are at the same temperature and pressure. (a) equal to (b) directly proportional to (c) inversely proportional to 11. The volume of 1 kg-mole of any gas at normal temperature and pressure is (a) 0.224 m3 (b) 2.24 m3 (c) 22.4 m3 (d) 224 m3 12. The value of specific heat at constant volume (cv) is .............. that of at constant pressure (cp) (a) less than (b) equal to (c) greater than 13. When the gas is heated at constant volume, then (a) no external work is done (b) decreases the temperature of the gas (c) increases the internal energy of the gas (d) all of the above 14. The enthalpy (H ) of a gas is given by (a) H = U – pv (b) H = U + p v (c) H = U × pv (d) H = U/pv where U = Internal energy of the gas, p = Pressure, and v = Volume. 15. The gas constant (R) is equal to (a) cp – cv (b) cp + cv (c) cp /cv (d) cp × cv

ANSWERS

1. (a) 6. (b) 11. (c)

2. (a) 7. (c) 12. (a)

3. (b) 8. (a) 13. (a), (c)

4. (c) 9. (d) 14. (b)

5. (d) 10. (b) 15. (a)

4 FIRST LAW OF THERMODYNAMICS 4.1 Introduction 4.2 First Law of Thermodynamics for a Closed or Non-flow System Undergoing a Cycle 4.3 First Law of Thermodynamics for a Closed or Non-flow System Undergoing a Change of State 4.4 Limitations of First Law of Thermodynamics 4.5 Perpetual Motion Machine of First Kind (PMM-I) 4.6 Classification of Thermodynamic Processes 4.7 Workdone for a Closed or Non-flow Process

4.8 Types of Non-flow Processes 4.9 Constant Volume Process (Isometric or Isochoric Process) 4.10 Constant Pressure Process (or Isobaric Process) 4.11 Constant Temperature Process (or Isothermal Process) 4.12 Adiabatic Process (or Isentropic Process) 4.13 Relation Between Pressure, Volume and Temperature During an Adiabatic Change 4.14 Polytropic Process 4.15 Irreversible Non-flow Process-Free Expansion Process

4.1 INTRODUCTION It was established by an English Scientist James Prescott Joule (1819-83) in 1843, that heat and mechanical energies are mutually convertible. He established through experiments that whenever a certain amount of work is done, a definite quantity of heat is produced. In other words, there is a numerical relation between the unit of work and unit of heat. This relation is denoted by J (named after Joule) and it is known as Joule’s equivalent or mechanical equivalent of heat. Since the unit of work is joule or kilo-joule and the unit of heat is also joule or kilo-joule, therefore we can convert straightway heat units into mechanical units and vice versa. The relationship between heat and work is the basis of First Law of Thermodynamics. It is one of the most fundamental laws of energy and is of special importance to engineers. It is also called law of conservation of energy.

4.2 FIRST LAW OF THERMODYNAMICS FOR A CLOSED OR NON-FLOW SYSTEM UNDERGOING A CYCLE We have already discussed that according to first law of thermodynamics, the heat and mechanical work are mutually convertible. According to this law, when a closed or non-flow system undergoes a thermodynamic cycle, the net heat transfer is equal to the net work transfer. Mathematically,

∫ δQ = ∫ δW

... (i)

i.e. the cyclic integral of heat transfer is equal to the cyclic integral of work transfer. It may be noted that the symbol ∫ stands for cyclic integral (i.e. integral around a complete cycle), dQ and dW are

46 Engineering Thermodynamics small elements of heat and work transfers respectively and have same units. The equation (i) may be written as ∫ (δQ − δW ) = 0 It follows from this equation that “if a system is taken through a cycle of processes, so that it returns to the same state or condition from which it started, then the sum of heat and work will be zero.” It may be noted that dQ and dW are both path functions, but their difference i.e. (dQ – dW) is a point function as the cyclic integral of (dQ – dW) is zero, i.e. ∫ (δQ − δW ) = 0.

4.3 FIRST LAW OF THERMODYNAMICS FOR A CLOSED OR NON-FLOW SYSTEM UNDERGOING A CHANGE OF STATE We have already discussed that according to first law of thermodynamics, the energy can neither be created nor destroyed though it can be transformed from one form to another. According to this law, when a closed or non-flow system undergoes a change of state or a thermodynamic process, then both heat transfer and work transfer takes place. The net energy transfer is stored within the system and is called stored energy or total energy of the system. Mathematically, dQ – dW = dE or dQ = dW + dE ...(i) The symbol d is used for a quantity which is *inexact differential and symbol d is used for a quantity which is an *exact differential. The quantity E is an extensive property and represents the total energy of the system at a particular state. On integrating the above equation (i) for a change of state from state 1 to state 2, we have

2

2

∫1 (δQ − δW ) = ∫1 dE

or Q1–2 – W1–2 = E2 – E1 ...(The units of Q, W and E are same) ...(ii) where Q1–2 = Heat transferred to the system during the process 1–2, i.e. from state 1 to state 2, W1–2 = Work done by the system on the surrounding during the process 1–2, and E1 = Total energy or **stored energy of the system at state 1, and E2 = Total energy of the system at state 2. We have already discussed that total energy of the system at state 1, 1 E1 = PE1 + KE1 + U1 = mgz1 + mV1 2 + U1 2 Similarly, total energy of the system at state 2, 1 E2 = PE2 + KE2 + U2 = mgz2 + mV2 2 + U2 2 Now from equation (ii), Q1–2 – W1–2 = (PE2 + KE2 + U2) – (PE1 + KE1 + U1) = (PE2 – PE1) + (KE2 – KE1) + (U2 – U1) ...(iii) 1 1 2 2 = (mgz2 – mgz1) + × mV2 − × mV1 + (U2 – U1) 2 2

* We have already discussed that heat and work are path functions and the differentials of path function are inexact differential. Similarly, energy is a point function and the differentials of a point function are exact or perfect differential. ** For a non-flow process, the stored energy is the internal energy only.

First Law of Thermodynamics 47 = mg (z2 – z1) +

m 2 V2 − V12 2

(

) + (U

2

– U1) ...(iv)

The following points are worthnoting: 1. When PE1 = PE2, i.e. when there is no change of potential energy,then equation (iii) may be written as Q1–2 – W1–2 = (KE2 – KE1) + (U2 – U1) ...(v) 2. When PE1 = PE2, and also there is no flow of mass into or out of the system (i.e. KE1 = KE2), then Q1–2 – W1–2 = U2 – U1 = dU ...(vi) or Q1–2 = dU + W1–2 where dU = Change in internal energy = U2 – U1 The above expression is called non-flow energy equation. 3. Since for an isolated system, Q1–2 = 0 and W1–2 = 0,therefore from equation (ii), we have E2 = E1. This shows that the first law of thermodynamics is the law of conservation of energy.

4.4 LIMITATIONS OF FIRST LAW OF THERMODYNAMICS Following are the important limitations of first law of thermodynamics: 1. We have discussed in Art. 4.2, that when a closed or non-flow system undergoes a thermodynamic cycle, the net heat transfer is equal to the net work transfer. But this statement does not specify the direction in which the heat and work flows. It also does not give any condition under which these transfers take place. 2. The heat energy and mechanical work are mutually convertible. Though the mechanical energy can be fully converted into heat energy,but only a part of heat energy can be converted into mechanical work. Thus, there is a limitation on the conversion of one form of energy into another form of energy.

4.5 PERPETUAL MOTION MACHINE OF FIRST KIND (PMM-I) A perpetual motion machine of the first kind (PMM-I), as shown in Fig. 4.1, is a machine producing a continuous supply of work without absorbing energy from the surroundings. The first law of thermodynamics implies that a perpetual motion machine of the first kind (PMM-I) is *impossible, because if a net amount of heat is not supplied by the surroundings during a cycle, no net amount of work can be delivered by the system. Heat

Heat

Work (a) PMM-I.

Work (b) Reversed PMM-I.

Fig. 4.1. Perpetual motion machine of first kind.

A machine which continuously consumes work and produces no other form of energy is called reversed PMM-I. Such a machine is also not feasible.

* We can also say that a machine which violates first law of thermodynamics (i.e. energy can neither be created nor destroyed, but can be changed from one form to another) is known as PMM-I.

48 Engineering Thermodynamics Example 4.1. A closed system consists of water contained in a cylinder and being stirred by a paddle wheel. During the process, 35 kJ/h of work was imparted to the system and the internal energy is increased to 145 kJ from an initial value of 120 kJ during one hour of stirring. Sketch the process with the help of neat diagram of the arrangement and determine the heat transfer. Is the temperature of the system rising or falling. Solution. Paddle wheel Given: Work imparted to the system, W1-2 W1–2 = 35 kJ / h Final internal energy, U2 = 145 kJ Closed Initial internal energy, U1 = 120 kJ System \ Increase in internal energy in one hour of stirring, Water dU = U2 – U1 = 145 – 120 = 25 kJ / h The arrangement of the process is shown in Fig. 4.2. Now according to first law of thermodynamics, heat transfer, Fig. 4.2 Fig. 4.2 Q1–2 = dU + W1–2 = 25 + 35 = 60 kJ / h Ans. Since there is a increase in internal energy, therefore the temperature will rise. Ans. Example 4.2. The gas is compressed from an initial state of 0.35 m3 and 105 kPa to a final state of 0.14 m3 and to the same pressure. Determine change in internal energy of the gas which transfers 38 kJ of heat. Solution. Given: Initial volume, v1 = 0.35 m3 Initial pressure, p1 = 105 kPa = 105 × 103 N/m2 Final volume, v2 = 0.14 m3 Final pressure, p2 = p1 Heat transfer, Q1–2 = – 38 kJ ...(Negative sign due to compression) Let dU = U2 – U1 = Change in internal energy. We know that workdone on the gas due to compression, W1–2 = p1 (v2 – v1) = 105 × 103 (0.14 – 0.35) N-m = – 22.05 × 103 N-m or J = – 22.05 kJ Now using the relation, Q1–2 = dU + W1–2 – 38 = dU – 22.05 \ dU = – 38 + 22.05 = – 15.95 kJ Ans. The negative sign indicates that there is a decrease is internal energy. Example 4.3. A cylinder with air comprises the system. The cycle is completed as follows: (a) 82 kN-m work is done by the piston on the air during compression and 45 kJ of heat is rejected to the surroundings. (b) During expansion, 100 kN-m of work is done by the air on the piston. Determine the quantity of heat added to the system. Solution. Given: Work done on the air, W1–2 = 82 kN-m = 82 kJ ...(Q 1 N-m = 1 J) Heat rejected to the surrounding, Q1–2 = 45 kJ

First Law of Thermodynamics 49 Workdone by the air, W2–1 = 100 kN-m = 100 kJ Consider that the process 1–2 i.e. from state 1 to state 2 represents the compression and the process 2-1 i.e. from state 2 to state 1 represents the expansion. The following points may be noted: 1. During compression, the work is done on the system (air) and the heat is rejected to the surroundings. Therefore, both are negative, i.e. W1–2 and Q1–2 are negative. 2. During expansion, the work is done by the system (air) and the heat is added to the system. Thus both are positive, i.e. W2–1 and Q2–1 are positive. Now according to first law of thermodynamics, change in internal energy during compression, U2 – U1 = Q1–2 – W1–2 = – 45 – (– 82) = 37 kJ or U1 – U2 = – 37 kJ and change in internal energy during expansion, U1 – U2 = Q2–1 – W2–1 or – 37 = Q2–1 – 100 \ Q2–1 = 100 – 37 = 63 kJ Ans.

4.6 CLASSIFICATION OF THERMODYNAMIC PROCESSES Following are the two types of thermodynamic processes: 1. Closed or Non-flow processes. The closed or non-flow processes are those processes which occur in a closed system where there is no transfer of mass across the boundaries. In such processes, energy in the form of heat and work crosses the boundary of the system. 2. Open or Flow processes. The open or flow processes are those processes which occur in an open system which permit the transfer of mass to and from the system. In such processes, mass enters the system and leaves after exchanging the energy. The flow processes may be steady flow processes (such as flow through nozzles, turbines, compressors etc.) and non-steady flow processes (such as filling or evacuation of vessels).

4.7 WORKDONE FOR A CLOSED OR NON-FLOW PROCESS Consider a system (gas) enclosed in a cylinder and piston arrangement as shown in Fig. 4.3. Let the system is initially in equilibrium state when the piston is at position 1, where the pressure is p1 and volume is v1. The pressure of the gas causes movement of the piston to position 2, where the pressure is p2 and volume is v2. This variation of the pressure and volume is drawn on the pressure-volume (briefly written as p-v) diagram, as shown in Fig. 4.3. If p is the net pressure of the gas acting on the piston of cross-sectional area A, then the force acting on the piston is given by F = p . A Let this force causes the piston to move through a small distance dx. Thus workdone by the system, dW = F × dx = p . A dx = p . dv ...(i) where dv = Change in volume = A . dx Now when the system (gas) expands from state 1 to state 2, then workdone for a non-flow process is given by

W1–2 =

2

∫1 δW

=

2

∫1

p ⋅ dv ...(ii)

This expression shows that the workdone by the system (gas) from state 1 to state 2 is equal to the area

Fig. 4.3. Workdone for a closed or non-flow process.

50 Engineering Thermodynamics under the p-v diagram. We have already discussed that the workdone by the system is considered as positive and workdone on the system is considered as negative. It may be noted that the equation (i) is only applicable to a reversible process. For an irreversible process, dW p . dv, because the path of the process is not represented truly on the p – v diagram due its non-equilibrium states in the process. Example 4.4. A perfect gas expands in such a way that its pressure varies in a linear relationship with volume p = av + b where a and b are constants. If the initial and final states of the gas are 4 bar, 0.1 m3 and 2 bar, 0.2 m3, determine the work interaction. Solution. Given: Initial pressure, p1 = 4 bar = 400 × 103 N/m2 = 400 kN/m2 Initial volume, v1 = 0.1 m3 Final pressure, p2 = 2 bar = 200 × 103 N/m2 = 200 kN/m2 Final volume, v2 = 0.2 m3 For the initial state, p1 = av1 + b or 400 = a × 0.1 + b = 0.1 a + b ...(i) and for the final state, p2 = av2 + b 200 = a × 0.2 + b = 0.2 a + b ...(ii) Substracting equation (ii) from equation (i), 400 – 200 = 0.1 a – 0.2 a = – 0.1 a \ a = – 2000 Substituting the value of a in equation (i), we have 400 = 0.1 (– 2000) + b \ b = 600 We know that work interaction, dW = pdv Integrating this expression for the initial and final states, we have \

2

v2

∫1 δW = ∫v

1

p dv =

v2

∫v v

1

(av + b) dv v

− 2000 v 2 2 2 a v2 + 600 v + bv = W1–2 = 2 v1 2 v1 0.2

2 = − 1000v + 600 v 0.1

= [– 1000 (0.2)2 + 600 × 0.2] – [– 1000 (0.1)2 + 600 × 0.1] = (– 40 + 120) – (– 10 + 60) = 80 – 50 = 30 kNm = 30 kJ Ans. Example 4.5. 150 kJ of work is supplied to a closed system. If the initial volume is 0.5 m3 and pressure of the system changes as p = 8 – 4 v, where p is in bar and v is in m3, determine the final volume and pressure of the system. Solution. Given: Work supplied, W1–2 = 150 kJ = 150 kN-m = 150 × 103 N-m Initial volume, v1 = 0.5 m3

First Law of Thermodynamics 51 Final volume of the system Let v2 = Final volume of the system. We know that work supplied, W1–2 = ∫

v2

v1

pdv =

v2

∫v

1

(8 − 4v) × 105 dv ...(Q 1 bar = 1 × 105 N/m2)

v

150 ×

103 =

2 4v 2 5 8v − × 10 2 0.5

4v 2 4(0.5) 2 = 8v2 − 2 − 8 × 0.5 + × 105 2 2 = (8v2 × 2v 22 – 4 + 0.5) × 105 = (8v2 – 2v 22 – 3.5) × 105 or \ or

150 × 103 105

2v 22 v 22 –

= 8v2 – 2v 22 – 3.5

1.5 = 8v2 – 2v 22 – 3.5 – 8v2 + 5 = 0 4v2 + 2.5 = 0

4 ± 42 − 4 × 1 × 2.5 4 ± 16 − 10 4± 6 = = 2 ×1 2 2 4 ± 2.45 = = 3.225 m3 or 0.775 m3 2 when v2 = 3.225 m3, then p = 8 – 4 × 3.225 = 8 – 12.9 = – 4.9 bar when v2 = 0.775 m3, then p = 8 – 4 × 0.775 = 8 – 3.1 = 4.9 bar Since v2 = 3.225 m3 gives final pressure – 4.9 bar which is not feasible, therefore the final volume = 0.775 m3, and final pressure = 4.9 bar. Ans. \

v2 =

4.8 TYPES OF NON-FLOW PROCESSES The heating (or cooling) and expansion (or compression) of a gas may be performed in the following ways: 1. Reversible non-flow processes. These processes include (a) Constant volume process (isometric or isochoric process); (b) Constant pressure process (isopiestic or isobaric process); (c) Constant temperature process (or isothermal process or hyperbolic process); (d) Adiabatic process (or Isentropic process); and (e) Polytropic process. In all these processes, we shall determine the expressions for (i) Workdone by the gas ; (ii) Change in internal energy ; (iii) Heat supplied or heat transfer ; and (iv) Change in enthalpy 2. Irreversible non-flow process. The free expansion process (or unrestricted process) is an irreversible non-flow process. We shall now discuss the above mentioned processes, in detail, in the following pages:

52 Engineering Thermodynamics

4.9 CONSTANT VOLUME PROCESS (ISOMETRIC OR ISOCHORIC PROCESS) When a gas is heated in a fixed enclosed chamber, the volume of the gas will remain constant. The heat supplied to the gas will increase the pressure and temperature. The constant volume process is represented by a vertical line on the p-v diagram, and by an inclined line on p-T diagram as shown in Fig. 4.4 (a) and (b) respectively. Since there is no change in volume, therefore no external work is done by the gas. All the heat supplied is stored within the gas in the form of internal energy. Now consider m kg of a certain gas being heated at constant volume from an initial state 1 to the final state 2.

Fig. 4.4. Constant volume process.

Let p1, v1, T1 = Pressure, volume and temperature at state 1, and p2, v2, T2 = Corresponding values at state 2. Since the volume is constant (v1 = v2), therefore according to Charles’ law, p1 p = 2 T1 T2 The workdone and other changes during the process are as follows: 1. Workdone by the gas We know that, workdone, dW = p dv Integrating this equation from state 1 to state 2,

2

2

∫1 δW = ∫1

2

p dv = p ∫ dv 1

or W1–2 = p (v2 – v1) = 0 ...(Q v1 = v2) This shows that work done during constant volume process 1-2 (i.e. from state 1 to state 2) is zero. 2. Change in internal energy We know that change in internal energy, dU = m cv dT Integrating this equation from state 1 to state 2,

2

2

∫1 dU = ∫1 m cv dT

2

= m cv ∫ dT 1

or U2 – U1 = m cv (T2 – T1) 3. Heat transferred or heat supplied We know that according to first law of thermodynamics, dQ = dU + dW

First Law of Thermodynamics 53 Integrating this expression from state 1 to state 2,

2

2

∫ δQ = ∫1 dU + ∫1 δW

or Q1–2 = (U2 – U1) + W1 – 2 Since W1–2 = 0, therefore heat transferred (or heat supplied) during the process 1–2, (i.e. from state 1 to state 2), Q1–2 = U2 – U1 = m cv (T2 – T1) 4. Change in enthalpy We know that change in enthalpy, dH = dU + d ( pv) Integrating this expression from state 1 to state 2,

2

2

2

∫1 dH = ∫1 dU + ∫1 d ( pv)

or H2 – H1 = (U2 – U1) + ( p2 v2 – p1 v1) Since U2 – U1 = m cv (T2 – T1) ; p1 v1 = m R T1 ; and p2 v2 = m R T2 , therefore H2 – H1 = m cv (T2 – T1) + (m R T2 – m R T1) = m cv (T2 – T1) + mR (T2 – T1) = m (T2 – T1) (cv + R) = m cp (T2 – T1) ... (Q cp – cv = R or cp = cv + R) Notes: 1. During expansion and heating, W1 – 2 , dU and Q1 – 2 are positive.

2. During compression and cooling, W1 – 2 , dU and Q1 – 2 are negative.

Example 4.6. A closed vessel contains 2 kg of CO2 at a temperature of 20°C and a pressure of 0.7 bar. Heat is supplied to the vessel till the gas acquires a pressure of 1.4 bar. Calculate the value of heat supplied. Take cv = 0.657 kJ/kg K Solution. Given: Mass of CO2, m = 2 kg Initial temperature, T1 = 20°C = 20 + 273 = 293 K Initial pressure, p1 = 0.7 bar Final pressure, p2 = 1.4 bar Specific heat at constant volume, cv = 0.657 kJ/kg K First of all, let us find the final temperature (T2). Since the vessel is closed, therefore, the volume is constant. We know that for constant volume, \

p1 p = 2 T1 T2 p ×T 1.4 × 293 T2 = 2 1 = = 586 K 0.7 p1

... (Charles’ law)

and heat supplied, Q1 – 2 = m cv (T2 – T1) = 2 × 0.657 (586 – 293) = 385 kJ Ans. Example 4.7. A constant volume chamber of 0.3 m3 capacity contains 2 kg of gas at 5°C. Heat is transferred to the gas until the temperature is 100°C. Find the workdone, the heat transferred, the changes in internal energy and enthalpy. Take cp = 1.97 kJ/kg K, and cv = 1.51 kJ/kg K.

54 Engineering Thermodynamics Solution. Given: *Constant volume = 0.3 m3 Mass of the gas, m = 2 kg Initial temperature, T1 = 5°C = 5 + 273 = 278 K Final temperature, T2 = 100°C = 100 + 273 = 373 K Workdone Since the volume is constant, therefore workdone is zero. Ans. Heat transferred We know that the heat transferred, Q1–2 = m cv (T2 – T1) = 2 × 1.51 (373 – 278) = 287 kJ Ans. Change in internal energy We know that change in internal energy, dU = m cv (T2 – T1) = Q1 – 2 = 287 kJ Ans. Change in enthalpy We know that change in enthalpy, dH = m cp (T2 – T1) = 2 × 1.97 (373 – 278) = 374.3 kJ Ans.

Sliding piston

1

2

Pressure p1 = p 2

Weight

Pressure p1 = p 2

4.10 CONSTANT PRESSURE PROCESS (OR ISOBARIC PROCESS) 1

2

Cylinder v1 v2 Volume Q

Fig. 4.5. Heating at constant pressure.

(a) p –v diagram.

T1 T2 Temperature (b) p –T diagram.

Fig. 4.6. Constant pressure process.

When the gas is heated in an enclosed cylinder under a loaded sliding piston to produce a desired pressure p on the gas as shown in Fig. 4.5. The heating will thus be effected under a constant pressure p. The heat supplied to the gas at a constant pressure will increase the volume and temperature. The p-v and p-T diagrams are shown in Fig. 4.6 (a) and (b) respectively. Due to the increase in volume, external work is done and due to increase in temperature, internal energy of the gas increases. In other words, heat supplied to the gas is utilised is doing some external work and in increasing the internal energy of the gas. Now consider m kg of a certain gas being heated at constant pressure from an initial state 1 to the final state 2. Let p1, v1, T1 = Pressure, volume and temperature at state 1, and p2, v2, T2 = Corresponding values at state 2. Since the pressure is constant (i.e. p1 = p2), therefore, according to Charles’ law, v1 v = 2 T1 T2 The workdone and other changes during the process are as follows:

* Superfluous data.

First Law of Thermodynamics 55 1. Workdone by the gas We know that, workdone, dW = p dv Integrating this equation from state 1 to state 2,

2

2

∫1 δW = ∫1

2

p dv = p ∫ dv 1

or W1–2 = p (v2 – v1) Since pv1 = m R T1, and pv2 = m R T2, therefore the above relation may be written as W1–2 = mR (T2 – T1) 2. Change in internal energy We know that change in internal energy, dU = m cv dT Integrating this equation from state 1 to state 2,

2

2

∫1 dU = ∫1 m cv dT

or U2 – U1 = m cv (T2 – T1) 3. Heat transferred or heat supplied We know that according to first law of thermodynamics, dQ = dU + dW Integrating this expression from state 1 to state 2,

2

... (same as before)

2

∫1 δQ = ∫1 dU + ∫ δW

Q1–2 = (U2 – U1) + W1–2 = (U2 – U1) + p (v2 – v1) We have discussed above that (U2 – U1) = m cv (T2 – T1) and p (v2 – v1) = mR (T2 – T1), therefore the above equation may be written as Q1 – 2 = m cv (T2 – T1) + mR (T2 – T1) = m (T2 – T1) (cv + R) = m cp (T2 – T1) ...(Q cv + R = cp) 4. Change in enthalpy We know that change in enthalpy, dH = dU + d ( pv) Integrating this expression from state 1 to state 2,

2

2

2

∫1 dH = ∫1 dU + ∫1 d ( pv)

or H2 – H1 = (U2 – U1) + p (v2 – v1) = Q1–2 = m cp (T2 – T1) This shows that the heat transferred is equal to the change in enthalpy. Example 4.8. One kg of air is expanded at a constant pressure of 2.5 bar from a volume of 0.3 m3 to a volume of 0.45 m3. Find : 1. external workdone by the gas; 2. internal energy of the gas; and 3. heat transferred during the process. Assume R = 287 J/kg K ; cv = 0.72 kJ/kg K ; and cp = 1.005 kJ/kg K for air. Solution. Given: Mass of air, m = 1 kg Pressure, p = 2.5 bar = 250 × 103 N/m2 Initial volume, v1 = 0.3 m3 Final volume, v2 = 0.45 m3 Gas constant, R = 287 J/kg K

56 Engineering Thermodynamics Specific heat at constant volume, cv = 0.72 kJ/kg K Specific heat at constant pressure, cp = 1.005 kJ/kg K First of all, let us find the initial and final temperature of the gas. Let T1 and T2 = Initial and final temperature of the gas respectively. We know that p1 v1 = m R T1 pv 250 × 103 × 0.3 \ T1 = 1 1 = = 261 K ...(Q p1 = p) 1 × 287 mR Similarly p2 v2 = m R T2 pv 250 × 103 × 0.45 \ T2 = 2 2 = = 392 K ...(Q p2 = p) 1 × 287 mR 1. External workdone by the gas We know that external workdone by the gas, W1–2 = p (v2 – v1) = 250 × 103 (0.45 – 0.3) = 37 500 N-m = 37.5 kJ Ans. 2. Internal energy of the gas We know that internal energy of the gas, dU = m cv (T2 – T1) = 1 × 0.72 (392 – 261) = 94.32 kJ Ans. 3. Heat transferred We know that heat transferred, Q1–2 = m cp (T2 – T1) = 1 × 1.005 (392 – 261) = 131.6 kJ Ans. Example 4.9. 0.14 m3 of air at 100°C and at a pressure of 1.5 bar is compressed at constant pressure until the volume is 0.112 m3. Find: 1. Final temperature of the air; 2. Mass of the air compressed; 3. Workdone in compressing the air; 4. Heat given out by the air; and 5. Specific heat at constant volume. Take R = 287 J/kg K; and cp = 1 kJ/kg K Solution. Given: Initial volume, v1 = 0.14 m3 Initial temperature, T1 = 100°C = 100 + 273 = 373 K Pressure (constant), p = 1.5 bar = 150 × 103 N/m2 Final volume, v2 = 0.112 m3 Gas constant, R = 287 J/kg K = 0.287 kJ/kg K Specific heat at constant pressure, cp = 1 kJ/kg K 1. Final temperature of the air Let T2 = Final temperature at the air. v1 v We know that = 2 T1 T2 v ×T 0.112 × 373 \ T2 = 2 1 = = 298.4 K 0.14 v1 = 298.4 – 273 = 25.4°C Ans.

First Law of Thermodynamics 57 2. Mass of the air compressed Let m = Mass of the air compressed. We know that p1 v1 = m R T1 \

m =

p1v1 150 × 103 × 0.14 = = 0.196 kg Ans. 287 × 373 RT1

...(Q p1 = p)

Pressure

3. Workdone in compressing the air We know that workdone is compressing the air, W1 – 2 = p (v1 – v2) ...(Q of compression) = 150 × 103 (0.14 – 0.112) = 4200 J = 4.2 kJ Ans. 4. Heat given out by the air We know that heat given out by the air (in compressing), Q1 – 2 = m cp (T1 – T2) = 0.196 × 1 (373 – 298.4) = 14.62 kJ Ans. 5. Specific heat at constant volume We know that specific heat at constant volume, cv = cp – R = 1 – 0.287 = 0.713 kJ/kg K Ans. ...(Q cp – cv = R) Example 4.10. The values of specific heats at constant pressure and constant volume for an ideal gas are 0.984 kJ/kg K and 0.728 kJ/kg K respectively. Find the values of the characteristic gas constant (R) and ratio of specific heat (g ) for the gas. If one kg of this gas is heated at constant pressure from 25°C to 200°C, estimate the heat added, ideal workdone and change in internal energy. Also calculate the pressure and final volume, if the initial volume was 2 m3. Solution. 1 2 Given: Specific heat at constant pressure, cp = 0.984 kJ/kg K Specific heat at constant volume, cv = 0.728 kJ/kg K Mass of the gas, m = 1 kg Initial temperature of the gas, v1 v2 Volume T1 = 25°C = 25 + 273 = 298 K Fig. Fig. 4.7 4.7 Final temperature of the gas, T2 = 200°C = 200 + 273 = 473 K Initial volume of the gas, v1 = 2 m3 The heating of gas at constant pressure is shown in Fig. 4.7. Characteristic gas constant We know that characteristic gas constant, R = cp – cv = 0.984 – 0.728 = 0.256 kJ/kg K = 256 J/kg K Ans. Ratio of specific heats We know that ratio of specific heats, cp 0.984 = g = = 1.35 Ans. cv 0.728

58 Engineering Thermodynamics Heat added We know that heat added, Q1–2 = m cp (T2 – T1) = 1 × 0.984 (473 – 298) = 172.2 kJ Ans. Ideal workdone We know that ideal workdone, W1–2 = mR (T2 – T1) = 1 × 0.256 (473 – 298) = 44.8 kJ Ans. Change in internal energy We know that change in internal energy, dU = U2 – U1 = m cv (T2 – T1) = 1 × 0.728 (473 – 298) = 127.4 kJ Ans. Pressure and final volume of the gas Since the gas is heated at constant pressure, therefore initial and final pressure of the gas are same (i.e. p1 = p2 = p). Let v2 be the final volume of the gas. We know that p1 v1 = m R T1 m RT1 1 × 256 × 298 = \ p1 = = 38 144 N/m2 ...(R is taken in J/kg K) 2 v1 = 38.144 kN/m2 Ans. For constant pressure process, v1 v = 2 T1 T2 v ×T 2 × 473 \ v2 = 1 2 = = 3.1745 m3 Ans. 298 T1

4.11 CONSTANT TEMPERATURE PROCESS (OR ISOTHERMAL PROCESS) When the heat is supplied to a gas in such a manner that its temperature remains constant during expansion, it is then known as constant temperature process (also called isothermal process). It may be noted that in a constant temperature process, the gas will do some external work equal to the heat supplied. We know that when the gas is heated at constant temperature, the product of the pressure and volume remains constant, i.e. p.v = constant, which is nothing but a Boyle’s law. The p-v diagram for an isothermal process is hyperbolic as the pressure (p) varies inversely as the volume (v). This is shown in Fig. 4.8. 1 pv = C (Hyperbolic curve)

p p2

2 v1

dv Volume

v2

(a) p –v diagram.

p1

1

p2

2

Pressure

Pressure

p1

T1 = T2 Temperature (b) p –T diagram.

Fig. 4.8. Constant temperature or isothermal process (or hyperbolic process).

Consider m kg of a certain gas being heated at constant temperature from an initial state 1 to the final state 2.

First Law of Thermodynamics 59 Let p1, v1, T1 = Pressure, volume and temperature at state 1, and p2, v2, T2 = Corresponding values at state 2. Since the temperature is constant (T1 = T2), therefore according to Boyle’s law, p1 v1 = p2 v2 or pv = constant The workdone and other changes during the process are as follows: 1. Workdone by the gas We know that workdone, dW = p dv ...(i) Since the expansion of the gas is hyperbolic, i.e. pv = constant, therefore pv pv = p1 v1 or p = 1 1 v Substituting the value of p in equation (i), we have workdone during expansion, pv dv dW = 1 1 dv = p1v1 v v Integrating this equation from state 1 to state 2, 2 v2 dv ∫1 δW = p1v1 ∫v1 v or

v v W1–2 = p1 v1 [ log e v ]v2 = p1v1 [log e v2 − log e v1 ] = p1v1 log e 2 1 v1

v = 2.3 p1 v1 log 2 = 2.3 p1 v1 log r v1 The term v2 /v1 is termed as expansion ratio (r). 2. Change in internal energy We know that change in internal energy, dU = m cv dT Integrating this equation from state 1 to state 2,

2

2

∫1 dU = ∫1 m cv dT

or U2 – U1 = m cv (T2 – T1) Since the process is a constant temperature process, i.e. T1 = T2, therefore U2 – U1 = 0 or U1 = U2 3. Heat transferred or heat supplied We know that the heat transferred or heat supplied, dQ = dU + dW Integrating this expression from state 1 to state 2,

2

2

2

∫1 δQ = ∫1 dU + ∫1 δW

or Q1–2 = (U2 – U1) + W1–2 = W1–2 ... ( U2 = U1) It means that the heat transferred or heat supplied to the gas is equal to the workdone by the gas.

* For compression,

v W1 – 2 = 2.3 p1 v1 log 1 , where v1/v2 is termed as compression ratio. v2

60 Engineering Thermodynamics 4. Change in enthalpy We know that change in enthalpy, dH = dU + d ( pv) Integrating this expression from state 1 to state 2,

2

2

2

∫1 dH = ∫1 dU + ∫1 d ( pv)

or H2 – H1 = (U2 – U1) + p (v2 – v1) = p (v2 – v1) ...(Q U2 = U1) = pv2 – pv1 = m R T2 – m R T1 ...(Q pv = m R T) = mR (T2 – T1) = 0 ...(Q T2 = T1) \ H2 = H1 From the above, we see that when the gas is heated at constant temperature, there is no change in internal energy and enthalpy. It may be noted that during compression of gas, the work is done on the gas and the heat is rejected from the gas. Example 4.11. 0.1 m3 of air at 6 bar is expanded isothermally to 0.5 m3. Calculate the final pressure and the heat supplied during the expansion process. Solution. Given: Initial volume, v1 = 0.1 m3 Initial pressure, p1 = 6 bar = 600 × 103 N/m2 Final volume, v2 = 0.5 m3 Final pressure Let p2 = Final pressure in bar. We know that for an isothermal process, p1 v1 = p2 v2 pv 6 × 0.1 \ p2 = 1 1 = = 1.2 bar Ans. 0.5 v2 Heat supplied We know that workdone during the isothermal expansion,

v W1–2 = 2.3 p1 v1 log 2 v1

0.5 = 2.3 × 600 × 103 × 0.1 log 0.1 = 96 600 N-m = 96.6 kJ We also know that heat supplied during the isothermal process is equal to the workdone. Therefore heat supplied, \ Q1–2 = W1–2 = 96.6 kJ Ans. Example 4.12. 0.4 kg of air at a pressure of 1 bar and a temperature of 20°C is compressed isothermally until its volume is 0.08 m3. Determine the workdone on the air during this compression process. Take R = 287 J/kg K Solution. Given: Mass of air, m = 0.4 kg Pressure, p1 = 1 bar = 100 × 103 N/m2 Temperature, T1 = 20°C = 20 + 273 = 293 K Final volume, v2 = 0.08 m3 Gas constant, R = 287 J/kg K

First Law of Thermodynamics 61 First of all, let us find the initial volume (v1) of the air. We know that p1 v1 = m R T1 m R T1 0.4 × 287 × 293 = \ v1 = = 0.336 m3 p1 100 × 103 We know that workdone on the air, v 0.336 W1–2 = *2.3 p1 v1 log 1 = 2.3 × 100 × 103 × 0.336 log 0.08 v2

Pressure

= 48 160 N-m = 48.16 kN-m = 48.16 kJ Ans. Example 4.13. Air initially at 75 kPa pressure, 1000 K temperature and occupying a volume of 0.12 m3 is compressed isothermally until the volume is halved and subsequently it undergoes further compression at constant pressure till the volume is halved again. Sketch the process on p-v diagram and find the work transfer. Solution. 3 p=C 2 p 2 = p3 3 2 Given: Initial pressure, p1 = 75 kPa = 75 × 10 N/m Initial temperature, T1 = 1000 K pv = C Initial volume, v1 = 0.12 m3 v 0.12 1 p1 Final volume, v2 = 1 = = 0.06 m3 2 2 Volume after compression at constant pressure, v3 v1 v2 v2 0.06 Volume = v3 = = 0.03 m3 Fig. 4.9 2 2 The p-v diagram of the process is shown in Fig. 4.9. The curve 1–2 represents the compression of air isothermally (i.e. according to pv = C) and the horizontal line 2–3 represents compression at constant pressure (i.e. p = C) Considering the isothermal compression 1–2, we have p1 v1 = p2 v2 pv 75 × 0.12 \ p2 = 1 1 = = 150 kPa = 150 × 103 N/m2 0.06 v2 and workdone on the air, v 0.12 W1–2 = 2.3 p1 v1 log 1 = 2.3 × 75 × 103 × 0.12 log 0.06 v2 = 6231 N-m or J = 6.231 kJ Now considering the constant pressure process 2–3, we have Workdone on the air, W1–2 = p2 (v2 – v3) = 150 × 103 (0.06 – 0.03) N-m = 4500 N-m or J = 4.5 kJ \ Net work transfer, W = W1–2 + W2–3 = 6.231 + 4.5 = 10.731 kJ Ans.

4.12 ADIABATIC PROCESS (OR ISENTROPIC PROCESS) When a process is carried out in such a manner that there is no heat transfer into or out of the system (i.e. Q = 0), then the process is said to be **adiabatic. This process may be reversible or ***irreversible. Such a process is not really possible in practice, though it can be closely approached. v1 . v2 ** A frictionless adiabatic process is known as isentropic process. *** Throttling is an irreversible adiabatic process.

* For compression process, Compression ratio is

62 Engineering Thermodynamics

Pressure

This will happen when the system is sufficiently thermally insulated, so that no (or negligible) heat transfer takes place during 1 p1 Adiabatic process the process. For an ideal adiabatic process, the following three conditions must be satisfied: 1. No heat is supplied or rejected during the process. pv = C 2. Some work must be done by the gas in expanding or work p2 should be done on the gas during its compression. 2 3. The process must be frictionless. v1 v2 Now consider m kg of gas being heated adiabatically from an Volume initial state 1 to a final state 2, as shown in Fig. 4.10. Fig. 4.10. Adiabatic process. Let p1, v1, T1 = Pressure, volume and temperature at state 1, and p2, v2, T2 = Corresponding values at state 2. According to first law of thermodynamics, we have dQ = dU + dW ...(i) Since no heat transfer takes place in an adiabatic process, therefore dQ = 0. Now equation (i) may be written as dU + dW = 0 m cv dT + p dv = 0 − p dv or dT = ...(ii) m cv We know that the general gas equation is pv = m R T Differentiating this equation, we have p dv + v dp = m R dT Substituting the value of dT in this expression from equation (ii), − p dv − p dv = R× p dv + v dp = mR × m cv cv − p dv = (cp – cv) cv or

− cp p dv + v dp = + 1 = – g + 1 p dv cv

... (Q R = cp – cv) ...(Q

cp cv

= g, the ratio of specific heats)

v dp × = – g + 1 dv p v dp × or g = − dv p dv dp γ× = − v p \ Integrating both sides, dv dp ∫ γ × v = − ∫ p g loge v = – loge p + constant = – loge p + loge C or loge p + g loge v = loge C loge p + loge vg = loge C

1+

First Law of Thermodynamics 63 loge ( pv)g = loge C or pvg = C ...(iii) This is a general equation for a reversible non-flow adiabatic process using a perfect gas in the system. The equation (iii) may be expressed in the following form: p1 v1g = p2 v2g = ... = constant C γ

v p1 = 2 ...(iv) p2 v1

or

The workdonw and other changes during the process are as follows : 1. Workdone by the gas We have already discussed that workdone, dW = p . dv ...(v) Since the expansion of the gas is adiabatic and follows the law pvg = constant, therefore v1γ

= p1 × v1γ × v − γ vγ Substituting the value of p in the above equation (v), we have dW = p1 v1g v – g dv Integrating this expression from state 1 to state 2, pvg = p1 v1g or p = p1 ×

2

2

∫1 dW = ∫1

p1v1γ v − γ dv = p1v1γ

2 −γ

∫1 v

dv

2

or

g

W1–2 = p1 v1

=

v− γ + 1 p1v1γ (v2 )1− γ − (v1 )1− γ = 1− γ − γ + 1 1

p1v1γ (v2 )1− γ − p1v1γ (v1 )1− γ 1− γ

Since p1 v1g = p2 v2g, therefore, the above equation may be written as

p2 v2γ (v2 )1− γ − p1v1γ (v1 )1− γ 1− γ p2 v2 − p1v1 p v − p2 v2 = 11 = ...(vi) 1− γ γ −1 We know that p1 v1 = m R T1, and p2 v2 = m R T2. Therefore equation (vi) becomes m RT1 − m RT2 m R (T1 − T2 ) = W1–2 = ...(vii) γ −1 γ −1 For compression, equations (vi) and (vii) may be written as p v − p1v1 m R (T2 − T1 ) W1–2 = 2 2 = γ −1 γ −1 2. Change in internal energy We know that change in internal energy, dU = m cv dT Integrating this equation from state 1 to state 2

or

W1–2 =

2

2

∫1 dU = ∫1 m cv dT U2 – U1 = m cv (T2 – T1)

64 Engineering Thermodynamics 3. Heat transferred or heat supplied Since the heat transferred or heat supplied is zero in an adiabatic process, therefore Q1–2 = 0 4. Change in enthalpy We know that change in enthalpy, dH = dU + dW Integrating this expression from state 1 to state 2,

2

2

2

∫1 dH = ∫1 dU + ∫1 δW

or H2 – H1 = (U2 – U1) + W1 – 2 = (U2 – U1) + p (v2 – v1) We have already discussed that, (U2 – U1) = m cv (T2 – T1), and p (v2 – v1) = mR (T2 – T1), therefore H2 – H1 = m cv (T2 – T1) + mR (T2 – T1) = m (T2 – T1) (cv + R) = m cp (T2 – T1) ...(cp = cv + R)

4.13 RELATION BETWEEN PRESSURE, VOLUME AND TEMPERATURE DURING AN ADIABATIC CHANGE First of all, let us determine the relation between pressure and temperature during an adiabatic change. We know that 1

p γ v p1 v1g = p2 v2g or 1 = 2 ...(i) v2 p1

From the general gas equation, p1v1 pv v pT = 2 2 or 1 = 2 1 ...(ii) T1 T2 v2 p1T2 From equations (i) and (ii), 1

or

p2 γ p2T1 p = p T 1 1 2 p T2 p2 / p1 = = 2 1/ γ T1 p1 ( p2 / p1 )

1−

1 γ

p = 2 p1

γ −1 γ

Now let us determine the relation between the volume and temperature during an adiabatic γ change. We know that v2 p1 g g = p1 v1 = p2 v2 or ...(iii) p2 v1 and from the general gas equation, p1v1 pv = 2 2 or T1 T2

p1 v T = 2 × 1 ...(iv) p2 v1 T2

From equations (iii) and (iv), γ

\

v2 v2 T1 = v 1 v1 T2 v T1 = 2 T2 v1

γ −1

v T or 2 = 2 T1 v1

− ( γ − 1)

1− γ

v = 2 v1

First Law of Thermodynamics 65 Example 4.14. An ideal gas of mass 0.25 kg has a pressure of 3 bar, a temperature of 80°C and a volume of 0.07 m3. The gas undergoes an *irreversible adiabatic process to a final pressure of 3 bar and a final volume of 0.1 m3, during which the workdone on the gas is 25 kJ. Evaluate cp and cv of the gas. Solution. Given: Mass of the gas, m = 0.25 kg Initial pressure, p1 = 3 bar = 300 × 103 N/m2 Initial temperature, T1 = 80°C = 80 + 273 = 353 K Initial volume, v1 = 0.07 m3 Final pressure, p2 = 3 bar = 300 × 103 N/m2 Final volume, v2 = 0.1 m3 Workdone on the gas, W1–2 = 25 kJ Let R = Characteristic gas constant, and T2 = Final temperature of the gas. We know that p1 v1 = m R T1 pv 300 × 103 × 0.7 \ R = 1 1 = = 238 J/kg K 0.25 × 353 mT1 and p2 v2 = m R T2

p2 v2 300 × 103 × 0.1 = = 504 K 0.25 × 238 mR Since the gas undergoes an irreversible adiabatic process, therefore there is no heat transfer during the process, i.e. Q1–2 = 0 We also know that heat transfer, Q1–2 = dU + W1–2 = m cv (T2 – T1) + W1–2 0 = 0.25 cv (504 – 353) – 25 = 37.75 cv – 25 \

T2 =

...(W1–2 is taken –ve, because work is done on the gas)

25 \ cv = = 0.662 kJ / kg K Ans. 37.75 and cp – cv = R = 238 J / kg K = 0.238 kJ / kg K \ cp = R + cv = 0.238 + 0.662 = 0.9 kJ / kg Ans. Example 4.15. 0.25 m3 of the gas at 288 K and 100 kPa is compressed adiabatically to 700 kPa. Calculate : 1. The final temperature of the gas; and 2. The workdone on the gas. Take cp = 1.001 kJ/kg K; cv = 0.715 kJ/kg K for the gas. Solution. Given: Initial volume, v1 = 0.25 m3 Initial temperature, T1 = 288 K Initial pressure, p1 = 100 kPa = 100 kN/m2 Final pressure, p2 = 700 kPa = 700 kN/m2 Specific heat at constant pressure, cp = 1.001 kJ / kg K

* When friction is involved in the process,then the adiabatic process is said to be irreversible.

66 Engineering Thermodynamics Specific heat at constant volume, cv = 0.715 kJ/kg K 1. Final temperature of the gas Let T2 = Final temperature of the gas. We know that adiabatic index, We also know that

g =

cp cv

=

p T2 = 2 T1 p1

1.001 = 1.4 0.715 γ −1 γ

700 = 100

1.4 − 1 1.4

= (7)0.286 = 1.7446

\ T2 = T1 × 1.7446 = 288 × 1.7446 = 502.4 K Ans. 2. Workdone on the gas First of all, let us find the final volume of the gas after compression (i.e. v2). We know that γ

v p p1 v1g = p2 v2g or 2 = 1 p2 v1 1

1

\

p γ v2 100 1.4 = 1 = (0.143)0.714 = 0.2494 v1 p 700 2

and

v2 = v1 × 0.2494 = 0.25 × 0.2494 = 0.0624 m3

We know that workdone on the gas, p v − p1v1 700 × 0.0624 − 100 × 0.25 43.68 − 25 W1–2 = 2 2 = = γ −1 1.4 − 1 0.4 = 46.7 kN-m = 46.7 kJ Ans. Example 4.16. One kg of a gas expands reversibly and adiabatically. Its temperature during the process falls from 240°C to 115°C while its volume is doubled. The gas does 90 kJ of work in this process. Find : 1. The value of cp and cv ; and 2. Molecular mass of the gas. Solution. Given: Mass of a gas, m = 1 kg Initial temperature, T1 = 240°C = 240 + 273 = 513 K Final temperature, T2 = 115°C = 115 + 273 = 388 K Final volume, v2 = 2 × Initial volume = 2v1 Workdone, W1–2 = 90 kJ 1. Value of cp and cv We know that during adiabatic process, the heat transferred is zero, i.e. Q1–2 = 0. We also know that change in internal energy, dU = m cv (T1 – T2) = 1 × cv (513 – 388) = 125 cv Now using the relation for adiabatic process, \

dU = W1–2 or 125 cv = 90 cv =

90 = 0.72 kJ/kg K Ans. 125

First Law of Thermodynamics 67 Let We know that

g = Ratio of specific heats = v T1 = 2 T2 v1

γ −1

2v = 1 v1

γ −1

cp cv

.

= (2)g – 1

513 = (2)g – 1 or 1.322 = (2)g – 1 388 Taking log on both sides, we have log (1.322) = (g – 1) log 2 0.1212 = (g – 1) 0.301 0.1212 or g – 1 = = 0.4026 0.301 \ g = 1 + 0.4026 = 1.4026 cp We know that = g = 1.4026 cv

\ cp = 1.4026 × cv = 1.4026 × 0.72 = 1.01 kJ / kg K Ans. 2. Molecular mass of the gas Let M = Molecular mass of the gas. We know that gas constant, R = cp – cv = 1.01 – 0.72 = 0.29 kJ / kg K and universal gas constant, Ru = R × M R 8.314 \ M = u = = 28.67 Ans. ...(Q Ru for all gases = 8.314 kJ / kg K) R 0.29 Example 4.17. A 0.568 m3 capacity insulated vessel of oxygen at the pressure of 12.6 bar is stirred by the internal paddle until the pressure becomes 21 bar. Find out: 1. Heat transferred; 2. Work input; and 3. Change in internal energy per kg. Take cv = 0.658 kJ/kg K; and R = 260 J/kg K. Solution. Given: Initial volume, v1 = 0.568 m3 Initial pressure, p1 = 12.6 bar = 1260 × 103 N/m2 Final pressure, p2 = 21 bar = 2100 × 103 N/m2 Specific heat at constant volume, cv = 0.658 kJ/kgK Gas constant, R = 260 J/kg K = 0.26 kJ/kg K 1. Heat transferred Since the vessel is insulated, therefore no heat is transferred i.e. Q1–2 = 0. Ans. 2. Work input We know that specific heat at constant pressure, cp = R + cv = 0.26 + 0.658 = 0.918 ...( R = cp – cv) cp 0.918 = \ Adiabatic index, g = = 1.395 cv 0.658

68 Engineering Thermodynamics and

p1 v1g = p2 v2g γ

or \

v2 p1 = p2 v1 p v2 = v1 1 p2

1 γ

1

12.6 1.395 = 0.568 = 0.568(0.6)0.717 21

= 0.568 × 0.693 = 0.394 m3 We know that work input,

W1–2 =

=

p1v1 − p2 v2 1260 × 103 × 0.568 − 2100 × 103 × 0.394 = 1.395 − 1 γ −1 715.7 × 103 − 827.4 × 103 = – 283 × 103 N-m = – 283 kJ 0.395

The negative sign indicates the compression of gas. 3. Change in internal energy Let dU = Change in internal energy. We know that according to first law of thermodynamics, Q1–2 = dU + W1–2 \ dU = – W1–2 = – (– 283) = 283 kJ Ans. Note: The value of dU may also be obtained by using the following relation: dU = U2 – U1 = m cv (T2 – T1) Since p1 v1 = m R T1 and p2 v2 = m R T2, therefore pv pv T1 = 1 1 and T2 = 2 2 mR mR Substituting these values in the above expression, we have

p v pv dU = U2 – U1 = m cv 2 2 − 1 1 mR mR c 0.658 = v [ p2 v2 – p1 v1] = [2100 × 103 × 0.394 – 1260 × 103 × 0.568] R 0.26 = 283 × 103 J = 283 kJ Ans.

4.14 POLYTROPIC PROCESS

Pressure

When expansion or compression of a gas takes place according to a general law pv n = constant, then it is called a polytropic process as shown in Fig. 4.11. The index n is known as polytropic index, whose value varies from zero to infinity, depending upon the manner in which the gas is expanded or compressed. It may 1 p1 be noted that the various results derived for adiabatic process may n pv = C be used for polytropic process by simply changing the adiabatic index g to polytropic index n. Thus, the following relations are used: p2 p1 v1n = p2 v2n = ... = constant or

v p1 = 2 p2 v1

2

n

v1

v2

Volume Fig. 4.11. Polytropic process.

First Law of Thermodynamics 69 1. Workdone by the gas The workdone by the gas during a polytropic process is given by p v − p2 v2 m R (T1 − T2 ) = W1–2 = 1 1 n −1 n −1 2. Change in internal energy We know that change in internal energy, dU = U2 – U1 = m cv (T2 – T1) 3. Heat transfer or heat supplied We know that heat transfer or heat supplied, Q1–2 = dU + W1–2 m R(T1 − T2 ) = m cv (T2 – T1) + ...(i) n −1 Since cp – cv = R and cp /cv = g, therefore c p − cv cp R R −1 = = or . cv cv cv cv \

g – 1 =

R R or cv = cv γ −1

Substituting the value of cv in equation (i), we have m R (T1 − T2 ) R (T2 − T1 ) + Q1–2 = m × n −1 γ −1 1 1 + = mR (T1 – T2) − γ − 1 n − 1 − n + 1 + γ − 1 = mR (T1 – T2) ( γ − 1) (n − 1) γ−n = mR (T1 – T2) ( γ − 1) (n − 1) =

γ − n m R (T1 − T2 ) γ −1 n −1

From the above, we see that heat transfer or heat supplied, γ −n Q1–2 = × Workdone during a polytropic process γ −1 4. Change in enthalpy We know that change in enthalpy, dH = H2 – H1 = m cp (T2 – T1) Note: During compression, if p1 and v1 refer to initial state, then workdone will be negative and heat is rejected by the gas. This will happen only if n < g.

Example 4.18. 0.016 m3 of gas at constant pressure of 2055 kN/m2 expands to a pressure of 215 kN/m2 by following the law pv1.35 = C. Determine the workdone by the gas during expansion process. Solution. Given: Initial volume, v1 = 0.015 m3

70 Engineering Thermodynamics Initial pressure, p1 = 2055 kN/m2 Final pressure, p2 = 215 kN/m2 Polytropic index, n = 1.35 First of all, let us find the final volume of the gas after expansion (i.e. v2). We know that \

v p1 v1n = p2 v2n or 2 v1 p v2 = 1 v1 p2

1 n

n

=

p1 p2

1

2055 1.35 = = (9.56)0.74 = 5.315 215

and v2 = v1 × 5.315 = 0.016 × 5.315 = 0.085 m3 We know that workdone by the gas, p v − p2 v2 2055 × 0.016 − 215 × 0.085 W1 – 2 = 1 1 = 1.35 − 1 n −1 32.88 − 18.27 = = 41.74 kN-m = 41.74 kJ Ans. 0.35 Example 4.19. Gas at 1.5 bar and 20°C in a closed vessel is compressed to 10 bar. Its temperature then becomes 180°C. If the compression follows the law pv n = C, find the value of n. Solution. Given: Initial pressure, p1 = 1.5 bar Initial temperature, T1 = 20°C = 20 + 273 = 293 K Final pressure, p2 = 10 bar Final temperature, T2 = 180°C = 180 + 273 = 453 K We know that n −1 p2 n T2 = T1 p1 Taking log on both sides, we have \

p T n −1 log 2 log 2 = n p1 T1 n −1 453 10 log log = n 293 1.5 n −1 log (1.546) = log (6.667) n n −1 (0.8239) n 0.1892 n = 0.8293 n – 0.8239 0.8239 0.8239 = n = = 1.298 Ans. 0.8239 − 0.1892 0.6347 0.1892 =

Example 4.20. 0.2 kg of air at 1.5 bar and 27°C is compressed to 15 bar according to pv1.25 = constant. Determine: 1. Initial and final parameters of the air; 2. Workdone on or by the air; and 3. Heat flow to or from the air.

First Law of Thermodynamics 71 Solution. Given: Mass of air, m = 0.2 kg Initial pressure, p1 = 1.5 bar = 150 × 103 N/m2 Initial temperature, T1 = 27°C = 27 + 273 = 300 K Final pressure, p2 = 15 bar Polytropic index, n = 1.25 1. Initial and final parameters of the air First of all, let us find the final temperature of the air (i.e. T2). We know that

p T2 = 2 T1 p1

n −1 n

15 = 1.5

1.25 − 1 1.25

= (10)0.2 = 1.585

\ T2 = T1 × 1.585 = 300 × 1.58 = 475.5 K = 475.5 – 273 = 202.5°C Ans. Let v1 and v2 be the initial and final volume of the air respectively. Using the relation, taking gas constant (R) for air = 287 J/kg K, we have p1 v1 = m R T1 m RT1 0.2 × 287 × 300 = \ v1 = = 0.1148 m3 Ans. p1 150 × 103 n

We know that

v p p1 v1n = p2 v2n or 2 = 1 p2 v1 1

or \ 2. Workdone on or by the air We know that workdone,

1

p n 1.5 1.25 v2 = 1 = = (0.1)0.8 = 0.1585 15 v1 p2 v2 = v1 × 0.1585 = 0.1148 × 0.1585 = 0.0182 m3 Ans.

m R (T1 − T2 ) 0.2 × 287 (300 − 475.5) = = – 40 295 J 1.25 − 1 n −1 = – 40.295 kJ Ans. The minus sign means that the work is done on the air. 3. Heat flow to or from the air First of all, let us find the change in internal energy (dU) by assuming the value of cv for air as 0.712 kJ/kg K. We know that dU = m cv (T2 – T1) = 0.2 × 0.712 (475.5 – 300) = 254 kJ \ Heat flow, Q1–2 = dU + W1–2 = 25 – 40.295 = – 15.295 kJ Ans. The minus sign means that the heat flows from the air. Example 4.21. A gas mixture obeying perfect gas law has a molecular mass of 26.7. The gas mixture is compressed through a compression ratio of 12 according to the law pv1.25 = constant, from the initial conditions of 0.9 bar and 60°C. Assuming a mean molar specific heat at constant volume of 21.1 kJ/kg K, find per kg of gas, the workdone and the heat flow across the cylinder walls. For the above gas, determine the values of characteristic gas constant, molar specific heat at constant pressure and the ratio of specific heats.

W1–2 =

72 Engineering Thermodynamics Solution. Given: Molecular mass,

v1 / v2 = 12

Polytropic index,

n = 1.25

Initial pressure,

p = 0.9 bar = 900 × 103 N/m2

Initial temperature,

T = 60°C = 60 + 273 = 333 K

p2

Mean molar specific heat at constant volume,

\

pv

1.25

=C

p1

1

cvm = 21.1 kJ/kg K

The compression process on p-v diagram is shown in Fig. 4.12. First of all, let us find the final temperature (T2) of the gas and the characteristic gas constant (R). We know that

2

Pressure

Compression ratio,

M = 26.7

v T1 = 2 T2 v1

n −1

v1

v2 Volume Fig. 4.12 Fig. 4.12

1.25 − 1

1 = 12

= 0.537

T2 =

T1 333 = = 620 K 0.537 0.537

R =

Universal gas constant ( Ru ) 8314 = = 311.4 J/kg K 26.7 Molecular mass ( M )

and characteristic gas constant,

...(Q Ru for all gases = 8314 J/kg K) Workdone per kg of gas We know that workdone on the gas (during compression),

W1–2 =

m R(T2 − T1 ) 1 × 311.4 (620 − 333) = = 357 500 N-m 1.25 − 1 n −1

= 357.5 kJ Ans. ...(m = 1 kg) Heat flow across the cylinder walls We know that specific heat at constant volume,

cv =

cvm 21.1 = = 0.79 kJ / kg K M 26.7

\ Change in internal energy, dU = m cv (T2 – T1) = 1 × 0.79 (620 – 333) = 226.7 kJ Heat flow across the cylinder walls, Q1–2 = dU – W1–2 ...(Minus sign due to compression) = 226.7 – 357.5 = – 130.8 kJ Ans. The negative sign shows that the heat is rejected through the cylinder walls. Characteristic gas constant We have already calculated that R = 311.4 J/kg K = 0.3114 kJ/kg K Ans.

First Law of Thermodynamics 73 Molar specific heat at constant pressure Let cpm = Molar specific heat at constant pressure. We know that R = cp – cv or cp = R + cv = 0.3114 + 0.79 = 1.1014 kJ / kg K \ cpm = cp × M = 1.1014 × 26.7 = 29.4 kJ / kg K Ans. Ratio of specific heats We know that ratio of specific heats, cp 1.1014 = g = = 1.394 Ans. cv 0.79

Pressure

Example 4.22. 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. It is then ignited suddenly at constant volume and its pressure becomes twice the pressure at the end of compression. Calculate the maximum temperature reached and change in internal energy. Also compute the heat transfer during compression process. Consider the mixture as a perfect gas and take cp = 1.072 kJ/kg K; and R = 294 J/kg K. Solution. p3 3 Given: Initial volume, v1 = 0.2 m3 Initial pressure, p1 = 1.2 bar = 120 × 103 N/m2 2 p2 Initial temperature, T1 = 60°C n = 60 + 273 = 333 K pv = C Pressure after polytropic compression, p1 1 p2 = 12 bar Temperature, T2 = 270°C = 270 + 273 v1 v2 = v3 = 543 K Volume Final pressure after ignition at constant volume, Fig.4.13 4.13 Fig. p3 = 2p2 = 2 × 12 = 24 bar Specific heat at constant pressure, cp = 1.072 kJ/kg K Gas constant, R = 294 J/kg K = 0.294 kJ/kg K Maximum temperature reached The p-v diagram is shown in Fig. 4.13. The curve 1-2 represents polytropic compression according to the law pv n = C and the vertical line 2-3 represent constant volume process. Let T3 be the temperature at end of constant volume process, i.e. the maximum temperature reached. We know that for a constant volume process 2-3, p p2 = 3 T3 T2 \ Change in internal energy Let We know that \

T3 =

p3 × T2 24 × 543 = = 1086 K = 1086 – 273 = 813°C Ans. p2 12

cv = Specific heat at constant volume, and m = Mass of the mixture compressed. R = cp – cv cv = cp – R = 1.072 – 0.294 = 0.778 kJ/kg K

74 Engineering Thermodynamics p1 v1 = m R T1

and

p1v1 120 × 103 × 0.2 = = 0.245 kg 294 × 333 RT1 We know that change in internal energy, dU = m cv (T3 – T1) = 0.245 × 0.778 (1086 – 333) = 143.53 kJ Ans. Heat transfer during compression process Since the compression process is polytropic, i.e. according to the law pv n = C, therefore, first of all, let us find the value of the polytropic index n. We know that m =

\

p T2 = 2 T1 p1

p T n −1 log 2 ...(Taking log on both sides) log 2 = n p1 T1

or

n −1 n

n −1 543 12 log log = n 333 1.2

\

n −1 log (10) n n −1 0.2122 = ...(Q log 10 = 1) n 1 1 = n = = 1.27 1 − 0.2122 0.7878

log (1.63) =

and workdone during the process 1-2

W1–2 =

m R(T1 − T2 ) 0.245 × 0.294 (333 − 543) = = –56 kJ 1.27 − 1 n −1

We know that change in internal energy during the process 1-2, dU = U2 – U1 = m cv (T2 – T1) = 0.245 × 0.778 (543 – 333) = 40 kJ \ Heat transfer during the compression process 1-2, Q1–2 = dU + W1–2 = 40 – 56 = – 16 kJ Ans. The –ve sign means heat is rejected during the process 1–2. Note: The value of heat transfer (Q1–2) may also be obtained as discussed below: We know that ratio of specific heats, cp 1.072 = g = = 1.378 cv 0.778 \ Heat transfer,

Q1–2 =

=

γ − n m R (T1 − T2 ) γ −1 n −1 1.378 − 1.27 0.245 × 0.294 (333 − 543) 1.378 − 1 1.27 − 1

= 0.2857 (– 56) = – 16 kJ Ans.

First Law of Thermodynamics 75

Pressure

Example 4.23. A mass of air initially at 206°C is at a pressure of 7 bar and has a volume of 0.03 m3. The air is expanded at constant pressure to 0.09 m3, a polytropic process with n = 1.5 is then carried out, followed by a constant temperature process which completes the cycle. All the processes are reversible. Sketch the cycle on pressure-volume diagram and find the heat received and heat rejected in the cycle. Take R = 287 J/kg K and cv = 0.713 kJ/kg K. Solution. Given: Initial temperature, T1 = 206°C = 206 + 273 = 479 K Initial pressure, p1 = 7 bar = 700 × 103 N/m2 Initial volume, v1 = 0.03 m3 1 p=c 2 Volume after constant pressure, p1=p2 v2 = 0.09 m3 n pv = C Gas constant, R = 287 J / kg K pv = C = 0.287 kJ / kg K 3 p Specific heat at constant volume, 3 cv = 0.713 kJ / kg K v1 v3 v2 Heat received Volume The p-v diagram is shown in Fig. 4.14. The horizontal Fig. 4.14 line 1-2 represents constant pressure process, curve 2-3 Fig. 4.14 represents polytropic process with n = 1.5 and curve 3-1 represents constant temperature process. Since process 1-2 is a constant pressure process, therefore v1 v = 2 T1 T2 \

T2 =

v2 × T1 0.09 × 479 = = 1437 K 0.03 v1

The process 3-1 is a constant temperature process. \ T3 = T1 = 479 K For a polytropic process 2-3, or

p T3 = 3 T2 p2

n −1 n

n

1.5

T n −1 479 1.5 − 1 = 7 p3 = p2 3 = 7(0.333)3 = 0.26 bar T 1437 2

= 0.26 × 105 = 26 × 103 N/m2 Now for a constant temperature process 3-1, p3 v3 = p1 v1 pv 7 × 0.03 \ v3 = 1 1 = = 0.81 m3 0.26 p3 Let m = Mass of the gas. Using the general gas equation, p1 v1 = m R T1

... ( 1 bar = 1 × 105 N/m2)

76 Engineering Thermodynamics p1v1 700 × 103 × 0.03 = = 0.153 kg 287 × 479 RT1 We know that cp – cv = R \ cp = R + cv = 0.287 + 0.713 = 1 kJ / kg K The heat is received during the constant pressure process 1-2. \ Heat received, Q1 – 2 = m cp (T2 – T1) = 0.153 × 1 (1437 – 479) = 146.6 kJ Ans. Heat rejected We know that workdone during polytropic process 2-3, \

m =

W2–3 =

p2 v2 − p3v3 700 × 103 × 0.09 − 26 × 103 × 0.81 = n −1 1.5 − 1

63 × 103 − 21.06 × 103 = 84 × 103 N-m 0.5 = 84 kJ and change in internal energy during polytropic process 2-3, dU = U3 – U2 = m cv (T3 – T2) = 0.153 × 0.713 (479 – 1437) = – 104.5 kJ \ Heat rejected during the process 2-3, Q2–3 = dU + W2 – 3 = – 104.5 + 84 = – 20.5 kJ Let us now consider a constant temperature process 3-1. We know that workdone during constant temperature process 3-1, =

...(Q p2 = p1)

...(i)

v 0.03 W3–1 = 2.3 p1 v1 log 1 = 2.3 × 700 × 103 × 0.03 log 0.81 v3

= 48 300 log (0.037) = 48 300 (– 1.4318) = – 69 156 N-m = – 69.156 kJ Since the change in internal energy during constant volume process is zero, therefore heat rejected during process 3-1, Q3–1 = dU + W3–1 = 0 + W3–1 = – 69.156 kJ \ Total heat rejected during the cycle = Q2–3 + Q3–1 = – 20.5 – 69.156 = – 89.656 kJ Ans. The negative sign means that the heat is rejected. Example 4.24. A gas initially at 14.3 bar and 360°C is expanded isothermally to a pressure of 2.24 bar. It is then cooled at constant volume till the pressure falls to 1.02 bar. Finally, an adiabatic compression brings the gas back to the initial state. The mass of the gas is 0.23 kg and cp = 1.005 kJ/kg K. Draw the p-v diagram and determine: 1. The value of the adiabatic index of compression; and 2. The change of internal energy of the gas during the adiabatic process. Solution. Given: Initial pressure, p1 = 14.3 bar Initial temperature, T1 = 360°C = 360 + 273 = 633 K Pressure after isothermal expansion, p2 = 2.24 bar

First Law of Thermodynamics 77

Pressure

Pressure after constant volume cooling, p3 = 1.02 bar 1 p1 Mass of the gas, m = 0.23 kg Specific heat at constant pressure, pv = C cp = 1.005 kJ/kg K pv = C 1. Value of adiabatic index p2 Let g = Adiabatic index. The p-v diagram is shown in Fig. 4.15. The curve 1-2 p3 shows the isothermal (constant temperature) expansion (i.e. according to pv = C ) process ; vertical line 2-3 shows v1 constant volume cooling (v2 = v3) and the curve 3-1 Volume shows the adiabatic compression process i.e. according to g Fig. 4.15 4.15 Fig. pv = C. First of all, let us consider the isothermal expansion process 1-2. We know that p1 v1 = p2 v2 v2 p 14.3 or = 1 = = 6.384 2.24 v1 p2

2

3 v2 = v3

...(i)

Since v2 = v3, therefore equation (i) may be written as v3 = 6.384 v1 Now for the adiabatic process 3-1, p1 v1g = p3 v3g

v p1 = 3 p3 v1

γ

or

...(ii)

14.3 = (6.384)g 1.02

Taking log on both sides, 14.3 log = g log 6.384 or 1.1467 = g × 0.805 1.02 \

g =

1.1467 = 1.424 Ans. 0.805

2. Change of internal energy Let T3 = Temperature at the end of constant volume process. For the constant volume process 2-3, we have p p2 = 3 T3 T2 \

T3 =

= and specific heat at constant volume,

cv =

p3 × T2 p ×T = 3 1 ...(... T2 = T1) p2 p2 1.02 × 633 = 288.24 K 2.24 cp γ

=

1.005 = 0.706 kJ/kg K 1.424

cp

cv

...

= γ

78 Engineering Thermodynamics We know that change of internal energy, dU = m cv (T1 – T3) = 0.23 × 0.706 (633 – 288.24) = 0.1624 × 344.76 = 56 kJ Ans. Example 4.25. A system contains 0.15 m3 of air at 4 bar and 150°C. A reversible adiabatic expansion takes place till the pressure falls to 1 bar. The air is then heated at constant pressure till enthalpy increases by 67 kJ. Determine the total workdone. If these processes are replaced by a single polytropic process giving the same work between the same initial and final states, determine the index of expansion. Take cp = 1.009 kJ/kg K, and R = 0.287 kJ/kg K. Solution. Given: Initial volume, v1 = 0.15 m3 Initial pressure, p1 = 4 bar = 400 × 103 N/m2 Initial temperature, T1 = 150°C = 150 + 273 = 423 K Final pressure, p2 = 1 bar = 100 × 103 N/m2 Increase in enthalpy, dH = 67 kJ Specific heat at constant pressure, cp = 1.009 kJ/kg K Gas constant, R = 0.287 kJ/kg K = 287 J/kg K The p-v diagram is shown in Fig. 4.16, in which the curve 1-2 represents reversible adiabatic expansion of air according to pvg = C ; horizontal line 2-3 represents heating at constant pressure and the curve 1-3 represents a polytropic process according to pv n = C. Total workdone The total workdone is the sum of workdone during process 1-2 (i.e. W1 –2) and workdone during process 2-3 (i.e. W2 –3). First of all, let us find the mass of air (m). We know that p1 v1 = m R T1 \

m =

p1v1 400 × 103 × 0.15 = 287 × 423 RT1

Fig. 4.16

= 0.494 kg Let v2 and T2 = Volume and temperature at the end of adiabatic expansion, and v3 and T3 = Volume and temperature at the end of constant pressure process. Now for the adiabatic process 1-2, p1 v1g = p2 v2g γ

or

v2 p1 = p2 v1 1

\

1

p γ 4 1.4 v2 = v1 1 = 0.15 = 0.15 (4)0.714 = 0.4036 m3 1 p2

...(Taking g for air = 1.4)

First Law of Thermodynamics 79 and \

p2 v2 = m R T2 T2 =

p2 v2 100 × 103 × 0.4036 = = 284.7 K 0.494 × 287 mR

We know that workdone during adiabatic process 1-2, p1v1 − p2 v2 400 × 103 × 0.15 − 100 × 103 × 0.4036 = 1.4 − 1 γ −1 = 49 100 N-m or J = 49.1 kN-m or kJ We know that increase in enthalpy during the process 2-3, dH = m cp (T3 – T2) 67 = 0.494 × 1.009 (T3 – 284.7)

or

W1–2 =

T3 – 284.7 =

67 = 134.4 0.494 × 1.009

\ T3 = 134.4 + 284.7 = 419.1 K and for constant pressure process 2-3, \

v v2 = 3 T3 T2 v ×T 0.4036 × 419.1 v3 = 2 3 = = 0.594 m3 T2 284.7

We know that workdone during constant pressure process 2-3,

W2–3 = p2 (v3 – v2)

= 100 × 103 (0.594 – 0.4036) = 19040 N-m or J = 19.04 kJ \ Total workdone,

W1–2–3 = W1–2 + W2–3 = 49.1 + 19.04 = 68.14 kJ Ans.

Index of expansion It is given that the workdone by a single polytropic process 1-3 (W1–3) is same as the total workdone during the processes 1-2 and 2-3, i.e. Let

W1–3 = 68.14 kJ = 68 140 J or N-m n = Index of expansion

We know that workdone during polytropic process (1-3), p1v1 − p3v3 n −1 400 × 103 × 0.15 − 100 × 103 × 0.594 68 140 = n −1 60 × 103 − 59.4 × 103 600 = = n −1 n −1 600 or n – 1 = = 0.0088 68140

W1–3 =

...(Q p3 = p2)

n = 1 + 0.0088 = 1.0088 Ans.

Example 4.26. One kg of air at 1 bar and 27°C is compressed isothermally to one-fifth of the original volume. It is then heated at constant volume to a condition such that isentropic expansion

80 Engineering Thermodynamics from that state will return the system to the original state. Determine the pressure and temperature at the end of constant volume heating. Represent the process on the pressure-volume diagram. Determine, also, the net workdone during the cycle. Solution. Given: Mass of air, m = 1 kg Initial pressure, p1 = 1 bar Initial temperature, T1 = T2 = 27°C = 27 + 273 = 300 K Final volume (i.e. volume at the end of isothermal compression) 1 v2 = v3 = v1 = 0.2 v1 5 Pressure and temperature at the end of constant volume heating The p-v diagram is shown in Fig. 4.17. The curve 1-2 represents the isothermal compression i.e. pv = C, vertical Fig. 4.17 line 2-3 represents heating at constant volume i.e. v = C and the curve 3-1 represents the isentropic expansion i.e. pvg = C. Let p3 and T3 = Pressure and temperature at the end of constant volume heating or at the start of isentropic expansion. p2 = Pressure at the end of isothermal compression. We know that for isothermal process 1-2, p1 v1 = p2 v2 pv 1 × v1 p2 = 1 1 = = 5 bar v2 0.2 v1 Now considering isentropic expansion 3-1. We know that p1 v1g = p3 v3g γ

\

γ

v v p3 = p1 1 = p1 1 ...(Q v3 = v2) v2 v3 1.4

v = 1 1 0.2v1

= (5)1.4 = 9.52 bar Ans.

...(Q g for air = 1.4)

Now using the relation for constant volume process 2-3, p3 p = 2 T3 T2 p ×T 9.52 × 300 \ T3 = 3 2 = = 571.2 K p2 5 = 571.2 – 273 = 29.82°C Ans. Net workdone during the cycle We know that workdone during isothermal compression 1-2,

v v W1–2 = 2.3 p1 v1 log 2 = 2.3 m R T1 log 2 v1 v1

...(Q T2 = T1)

First Law of Thermodynamics 81 0.2 v1 = 2.3 × 1 × 287 × 300 log v1

Pressure

= 198030 × – 0.7 = – 138 621 N-m or J = – 138.621 kJ ...(–ve sign due to compression) Workdone during the constant volume process 2-3 is zero, i.e. W2–3 = 0 and workdone during isentropic expansion 3-1, p v − p1v1 m R (T3 − T1 ) = W3–1 = 3 3 γ −1 γ −1 1 × 287 (571.2 − 300) = = 194.6 × 103 N-m or J = 194.6 kJ 1.4 − 1 \ Net workdone during the cycle, W1–2–3 = W1–2 + W2–3 + W3–1 = – 138.621 + 0 + 194.6 = 55.979 kJ Ans. Example 4.27. Air undergoes a cyclic process in a cylinder and piston arrangement. One kg of atmospheric air at 1 bar and 300 K is first compressed adiabatically to 10 bar, then expanded isothermally upto the initial pressure and lastly the air is brought to initial conditions under constant pressure. Calculate for each process and for the complete cycle: 1. Change in internal energy ; 2. Change in enthalpy ; 3. Heat transfer ; and 4. Work transfer. Assume for air, cp = 1 kJ/kg K and g = 1.4. Solution. Given: Mass of air, m = 1 kg Initial pressure, p1 = 1 bar = 100 × 103 N/m2 Initial temperature, T1 = 300 K Final pressure, p2 = 10 bar = 1000 × 103 N/m2 Specific heat at constant pressure, cp = 1 kJ/kg K cp cp 1 = Ratio of specific heats, g = = 1.4 or cv = = 0.714 cv γ 1.4 We know that gas constant, R = cp – cv = 1 – 0.714 = 0.286 kJ/kg K = 286 J/ kg K If the p-v diagram, as shown in Fig. 4.18, 2 the process 1-2 represents adiabatic compression p2 (i.e. according to pvg = C ); process 2-3 represents isothermal expansion (i.e. according to pv = C) and the process 3-1 represents compression at constant pv = C pressure (i.e. p = C). Considering adiabatic compression process 1-2 pv = C First of all, let us find the initial and final volumes (v1 and v2 respectively) for the adiabatic 3 p1 = p 3 compression. 1 p=C We know that p1 v1 = m R T1 \

v2

m RT1 v1 = p1

=

1× 286 × 300 100 × 103

= 0.858 m3

v1 Volume Fig. 4.18 4.18 Fig.

v3

82 Engineering Thermodynamics Now using the relation, p1 v1g = p2 v2g γ

or

v2 p1 v = p 1 2 1

1

\

p γ 1 1.4 v2 = v1 1 = 0.858 = 0.858 × 0.1932 = 0.166 m3 10 p2

Let

T2 = Temperature at the end of adiabatic compression (i.e. at state 2).

We know that \

p T2 = 2 T1 p1

γ −1 γ

10 = 1

1.4 − 1 1.4

= (10)0.286 = 1.932

T2 = T1 × 1.932 = 300 × 1.932 = 579.6 K

1. Change in internal energy We know that change in internal energy,

dU = U2 – U1 = m cv (T2 – T1)

= 1 × 0.714 (579.6 – 300) = 199.6 kJ Ans. 2. Change in enthalpy We know that change in enthalpy,

dH = H2 – H1 = m cp (T2 – T1)

= 1 × 1 (579.6 – 300) = 279.6 kJ Ans. 3. Heat transfer Since the heat transfer during adiabatic process is zero, therefore

Q1–2 = 0 Ans.

4. Work transfer According to first law of thermodynamics, work transfer,

W1–2 = Q1 – 2 – dU = 0 – 199.6 = – 199.6 kJ Ans.

Considering isothermal expansion process 2-3 1. Change in internal energy We know that change in internal energy for isothermal process,

dU = U3 – U2 = 0 Ans.

2. Change in enthalpy The change in enthalpy for isothermal process is also zero, i.e.

dH = H3 – H2 = 0 Ans.

3. Heat transfer First of all, let us find the volume (v3) at the end of isothermal expansion. Using the relation, \

p2 v2 = p3 v3 v3 =

p2 v2 10 × 0.166 = = 1.66 m3 ...(Q p3 = p1) 1 p3

First Law of Thermodynamics 83 and workdone,

v W2–3 = 2.3 p2 v2 log 3 v2

1.66 = 2.3 × 1000 × 103 × 0.166 log N-m 0.166 = 381 800 N-m or J = 381.8 kJ We know that heat transfer, Q2–3 = dU + W2 – 3 = 0 + W2 – 3 = 381.8 kJ Ans. 4. Work transfer We have calculated above that the work transfer, W2–3 = 381.8 kJ Ans. Considering constant pressure process 3-1 1. Change in internal energy, dU = U1 – U3 = m cv (T1 – T3) = m cv (T1 – T2) ...(T3 = T2) = – 199.6 kJ ... (as calculated above) 2. Change in enthalpy, dH = H1 – H3 = m cp (T1 – T3) = m cp (T1 – T2) = – 279.6 kJ ... (as calculated above) 3. Heat transfer, Q3–1 = m cp (T1 – T3) = m cp (T1 – T2) = – 279.6 kJ ... (as calculated above) 3 4. Work transfer, W3–1 = p3 (v1 – v3) = 100 × 10 (0.86 – 1.66) N-m = – 80 000 N-m or J = – 80 kJ Considering the complete cycle 1-2-3 1. Change in internal energy = (U2 – U1) + (U3 – U2) + (U1 – U3) = 199.6 + 0 – 199.6 = 0 Ans. 2. Change in enthalpy = (H2 – H1) + (H3 – H2) + (H1 – H3) = 279.6 + 0 – 279.6 = 0 Ans. 3. Heat transfer = Q1–2 + Q2–3 + Q3–1 = 0 + 381.8 – 279.6 = 102.2 kJ Ans. 4. Work transfer = W1–2 + W2–3 + W3–1 = – 199.6 + 381.8 – 80 = 102.2 kJ Ans.

4.15 IRREVERSIBLE NON-FLOW PROCESS-FREE EXPANSION PROCESS Most of the processes dealt in thermodynamics become irreversible due to the presence of friction, turbulence etc.The effect of irreversibility is loss of work, i.e. in compression process more work is to be supplied than the reversible work, whereas in expansion process, less work is obtained from the system. The important non-flow irreversible process is free-expansion (also called unresisted expansion) process. It occurs when a fluid is allowed to expand suddenly into a vacuum chamber through an orifice of large size. In this process, no heat is supplied or rejected and no external work is done.Hence the enthalpy of the fluid remains constant. This type of expansion may also be called as constant enthalpy process. It is thus obvious that for a free expansion process, Q1–2 = 0 ; W1–2 = 0 ; and dU = 0 Note 1. The free expansion process differs from isothermal process, because in this case, no external work is done. It is also not a true adiabatic process. 2. It can not be truely represented on p-v diagram. Hence the workdone in the free expansion process is not equal to

∫ p ⋅ dv .

84 Engineering Thermodynamics

HIGHLIGHTS 1. According to First Law of Thermodynamics: (a) Heat and mechanical work are mutually convertible; and (b) Energy can neither be created nor destroyed, though it can be transformed from one form to another. 2. When a closed or non-flow system undergoes a change of state or a thermodynamic process, then according to first law of thermodynamics; both heat transfer and work transfer takes place. For a change of state, from state 1 to state 2,

2

2

∫1 (δQ − δW ) = ∫1 dE

or Q1–2 – W1–2 = E2 – E1 = (PE2 + KE2 + U2) – (PE1 + KE1 + U1) When the potential energy (PE) and kinetic energy (KE) are neglected, then Q1–2 – W1–2 = U2 – U1 = dU or Q1–2 = dU + W1 – 2 where dU = Change in internal energy = U2 – U1 3. The various relations for the reversible non-flow processes are as follows: Process

p-v-T relationship

Workdone (W1–2 )

Change in internal energy (dU = U2 – U1 )

Heat Supplied (Q1–2 ) = W1–2 + dU

Change in enthalpy (dH = H2 – H1)

(a) Constant volume (or isochoric) process

p1 p = 2 T1 T2

0

m cv (T2 – T1)

m cv (T2 – T1)

m cp (T2 – T1)

(b) Constant pressure (or isobaric) process

v1 v2 = T1 T2

p (v2 – v1) or mR (T2 – T1)

m cv (T2 – T1)

m cp (T2 – T1)

m cp (T2 – T1)

(c) Hyperbolic or Constant tempera ture or Isothermal) process

p1 v1 = p2 v2

v p1v1 loge 2 v1

0

mR T loge v2 v 1

1

(e) Polytropic (pvn = Constant) process

p1 v1g = p2 v2g

p1 v1n = p2 v2n

0

or

or v mRT loge 2 v

(d) Adiabatic (or Isentropic) process

p1 v1 loge v2 v1

p1v1 − p2v2 γ −1 or mR(T1 − T2 ) γ −1 p1v1 − p2v2 n −1 or mR(T1 − T2 ) n −1

m cv (T2 – T1)

0

m cp (T2 – T1)

m cv (T2 – T1)

p1v1 − p2v2 n −1 + mcv (T2 – T1) γ −n × or γ −1 mR (T1 − T2 ) n −1

m cp (T2 – T1)

First Law of Thermodynamics 85 4. Workdone for a closed or non-flow process from state 1 to state 2 is given by 2

W1–2 = ∫ p dv

1

where p = Net pressure, and dv = Change in volume. 5. A machine which continuously consumes work and produces no other form of energy is called a reversed perpetual motion machine of first kind (PMM-I). Such a machine is not feasible. 6. The various relations between pressure, volume and temperature during an adiabatic or isentropic change are as follows: γ −1

γ −1

p γ v T T (a) 2 = 2 ; and (b) 1 = 2 T1 T2 p1 v1 where p1, v1 and T1 = Pressure, volume and temperature respectively at state 1, and

p2, v2, and T2 = Pressure, volume and temperature respectively at state 2.

7. In an irreversible non-flow process (i.e. free expansion process), Q1–2 = 0 ; W1–2 = 0 ; and dU = 0.

EXERCISES

(a) How much heat flows into the system along the path ADB, if the workdone is 10 J.

C

B

Pressure

1. A system is taken from state A to the state B, along the path ACB as shown in Fig. 4.19. During this process, 80 J of heat flows into the system and the system does 30 J of work.

(b) The system is returned from the state B to the state A along the curved path. The workdone on the system is 20 J. Does the system absorb or liberate heat and how much?

A

D Volume Fig. 4.19 Fig. 4.19

(c) If the internal energy at A (UA) is zero and the internal energy at D (UD) is 40 J; find the heat absorbed in the process AD and DB. [Ans. 60 J; – 70 J; 50 J, 10 J] 2. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship, p = a + bv, where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.20 m3 and 1.20 m3. The specific internal energy of the gas is given by the relation, u = (1.5 pv – 85) kJ/kg, where p is in kPa and v in m3/kg. Calculate the maximum internal energy of gas attained and the net heat transfer during expansion. [Ans. 90 kJ ; 690 kJ] 3. The internal energy of a certain substance is given by the following equation:

u = 3.56 pv + 84

where u is given in kJ/kg ; p is in kPa and v is in m3/kg. A system composed of 3 kg of this substance expands from an initial pressure of 500 kPa and a volume of 0.22 m3 to a final pressure of 100 kPa in a process in which pressure and volume are related by pv1.2 = constant. If the expansion is quasi-static, find the internal energy and the heat transfer during the process. [Ans. – 92.56 kJ; 37.44 kJ] 4. A fluid is confined in a cylinder by a spring loaded, frictionless piston so that the pressure in the fluid is a linear function of the volume ( p = a + bv). The internal energy of the fluid is given by the following equation:

U = 34 + 3.15 pv

86 Engineering Thermodynamics where U is in kJ ; p is in kPa and v in m3. If the fluid changes from an initial state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06 m3, with no work other than done on the piston, find the direction and magnitude of the work and heat transfer. [Ans. 8.55 kJ; 68.09 kJ] 5. 0.3 m3 of gas is filled in a closed tank at an initial condition of 2.75 bar and 40°C. The gas is heated until the pressure in the tank becomes 4 bar. Find the change in internal energy of the gas and heat added. Assume R = 287 J/ kg K and cv = 0.72 kJ/ kg K. [Ans. 93.86 kJ; 93.86 kJ] [Hint: First find the final temperature and mass of the gas by applying

p1 p2 = , and p1 v1 = m R T1] T1 T2

6. A quantity of air at a temperature of 20°C and a pressure of 2.15 bar is found to occupy 0.2 m3. It is heated at a constant volume until the temperature is 60°C. Find: (a) pressure at the end of the process; (b) mass of the gas; (c) change in internal energy; and (d) change in enthalpy during the process. Take cp = 1 kJ/ kg K ; cv = 0.72 kJ/ kg K ; and R = 287 J/kg K [Ans. 2.44 bar ; 0.5 kg ; 14.4 kJ ; 14.4 kJ] 7. Two kg of air at 7 bar and having a volume of 0.3 m3 is expanded to a volume of 1.5 m3. If the expansion takes place at a constant pressure, find the final temperature, the workdone and the heat absorbed or rejected by the air during the process. Assume R = 0.287 kJ/kg K; and cv = 0.72 kJ/kg K for air. [Ans. 1557°C ; 840 kJ ; 2948.5 kJ] 8. A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1.05 bar to a final state of 0.15 m3 at constant pressure during the process. There is a transfer of 36 kJ of heat from the gas during the process. How much does the internal energy of gas change? [Ans. 15.75 kJ ; 20.25 kJ] 9. It is desired to compress 10 kg of gas from 1.5 m3 to 0.3 m3 at a constant pressure of 15 bar. During this compression process, the temperature rises from 20°C to 150°C and the increase in internal energy is 3250 kJ. Calculate the workdone, heat interaction and the change in enthalpy during the process. Also work out the average value of specific heat at constant pressure. [Ans. – 1800 kJ; 1450 kJ; 1450 kJ; 1.1154] 10. Determine the volume and heat supplied to one kg of air at a pressure of 5 bar and temperature of 100°C. The air is allowed to expand four times to its initial volume at constant temperature. [Ans. 0.214 m3; 148 kJ] 11. 0.2 kg of air at 1.1 bar and 15°C is compressed isothermally to 5.5 bar. Calculate the final volume and the heat rejected during the compression process. Take R = 287 J/kg K for air. [Ans. 0.03 m3; 26.565 kJ] 12. A quantity of air has a volume of 56.5 litres and a pressure of 7 bar. It is expanded in a cylinder to a pressure of 1 bar. Compare the workdone if the expansion is as follows: 1. Hyperbolic; 2. Adiabatic; and 3. pv1.2 = C. Take g = 1.4 [Ans. 76.874 kJ ; 42.2 kJ ; 54.75 kJ] 13. 1.5 kg of air at an initial condition of 3.5 bar and 180°C expands adiabatically and without friction to a final condition of 1.3 bar. Calculate the amount of workdone and the change in internal energy of the gas during expansion. Assume g = 1.4 and R = 287 J/kg K for air. [Ans. 120 kJ; – 120 kJ] 14. 0.3 m3 of gas at 15°C and 1 bar is compressed adiabatically to 7 bar. Calculate: 1. mass of the gas ; 2. the final temperature of the gas ; and 3. workdone on the gas. Take cp = 1.005 kJ/kg K ; and cv = 0.71 kJ/kg K. [Ans. 0.353 kg ; 229.45°C ; – 55.83 kJ] 15. 1.4 m3 of a gas at a pressure of 1.26 bar is compressed to a volume of 0.28 m3. The final pressure is 7 bar. Assuming the compression to be polytropic, calculate the heat transfer and change in internal energy. Assume g = 1.4. [Ans. – 257 kJ; 49 kJ] 16. One kg of air at pressure of 7 bar and a temperature of 90°C undergoes a reversible polytropic process which may be represented by pv1.1 = C. The final pressure is 1.4 bar. Evaluate:

First Law of Thermodynamics 87 1. The final specific volume, temperature and increase in internal energy. 2. The workdone and heat transfer during the process. Assume R = 0.287 kJ/kg K, and g = 1.4

[Ans. 0.648 m3/kg, 40.6°C, 35.44 kJ; 141.8 kJ, 177.24 kJ]

17. An ideal gas of molecular mass 30 and specific heat ratio 1.38 is compressed according to the law pv1.25 = constant, from a pressure of 1 bar and 60°C to a pressure of 16 bar. Calculate the temperature at the end of compression, the heat received or rejected and workdone by the gas during the process. Assume 1 kg mass of the gas. Use only calculated values of cp and cv. [Ans. 306.4°C; – 93.37 kJ; – 273 kJ] 18. 3 g of nitrogen gas at 6 atmospheres and 160°C in a frictionless piston-cylinder device is expanded adiabatically to double its volume, then compressed at constant pressure to its initial volume and then compressed again at constant volume to its initial state. Calculate the net workdone on the gas. [Ans. 87.3 J] 19. One kg of air is compressed adiabatically from initial conditions of 100 kN/m3 and 27°C to a final condition of 800 kN/m2. It is then cooled at constant pressure to its initial temperature of 27°C. Calculate: 1. The final temperature and workdone in the adiabatic process; and 2. The amount of heat rejected during constant pressure process. [Ans. 543.75 K; 175 kJ; 243.75 kJ]

QUESTIONS 1. State the First Law of Thermodynamics and prove that for a closed or non-flow process, it leads Q1–2 – W1–2 = dU 2. What are the limitations of First Law of Thermodynamics? 3. Show that the workdone for a closed or non-flow process is pdv. 4. Name the different reversible and irreversible non-flow processes. 5. Why an isothermal process is also known as hyperbolic process? Derive an expression for the workdone during this process. 6. What is an adiabatic process? Derive an expression for the workdone during adiabatic expansion of an ideal gas? 7. Explain the difference between an adiabatic process and an isentropic process. 8. What is a polytropic process? Prove that the heat absorbed during a polytropic process on a perfect gas is given by γ−n × Workdone γ −1 where g is the ratio of specific heats and n is the polytropic index. 9. Explain how the free expansion makes the process irreversible?

OBJECTIVE TYPE QUESTIONS 1. According to First Law of Thermodynamics (a) heat and mechanical work are mutually convertible (b) energy can neither be created or destroyed though it can be transformed from one form to another (c) the cyclic integral of heat transfer is equal to cyclic integral of work transfer (d) all of the above

88 Engineering Thermodynamics 2. The process in which the mass enters the system and leaves after enhancing the energy is known as .................... process. (a) flow (b) non-flow 3. In a closed process, (a) no transfer of mass takes place across the boundaries (b) heat and work crosses the boundary of the system (c) permit the transfer of mass to and from the system (d) none of the above 4. Which of the following is a reversible non-flow process? (a) Isochoric process (b) Isobaric process (c) Isentropic process (d) all of these 5. The free expansion process is .................... non-flow process. (a) reversible (b) irreversible 6. The heating of a gas at constant volume is governed by (a) Boyle’s law (b) Charles’ law (c) Joule’s law (d) Gay-Lussac law 7. The heat energy stored in the gas and used for raising the temperature of the gas is known as (a) kinetic energy (b) molecular energy (c) internal energy (d) external energy 8. The heating of a gas at constant pressure is governed by (a) Boyle’s law (b) Charles’ law (c) Joule’s law

(d) Gay-Lassac law

9. When a gas is cooled at constant pressure, (a) its temperature increase but volume decreases (b) its volume increases but temperature increases (c) both temperature and volume increases (d) both temperature and volume decreases 10. A process, in which the temperature of the working substance remains constant during expansion or compression, is called (a) isothermal process (b) hyperbolic process (c) adiabatic process (d) polytropic process 11. An adiabatic process is one in which (a) no heat enters or leaves the gas (b) the temperature of the gas changes (c) the change in internal energy is equal to the workdone (d) all of the above 12. The general law for the expansion or compression of gases is (a) pv = C (b) pv = mR T (c) pv n = C (d) pv g = C

First Law of Thermodynamics 89 1 3. When the expansion or compression takesplace according to the law pv n = C, the process is known as (a) isothermal process (b) adiabatic process (c) hyperbolic process (d) polytropic process 14. If the value of n = 1 in the equation pv n = C, then the process is called (a) constant volume process (b) constant pressure process (c) isothermal process (d) adiabatic process

15. In a reversible adiabatic process, the ratio of T1 /T2 is given by

p (a) 2 p1

γ −1 γ

p (b) 1 p2

γ −1 γ

v (c) 2 v1

γ −1 γ

v (d) 1 v2

16. The free expansion process is a (a) constant volume process (c) constant enthalpy process

(b) constant pressure process (d) none of these

17. In a free expansion process, (a) W1 –2 = 0 (b) Q1–2 = 0

(c) dU = 0

γ −1 γ

(d) all of these

ANSWERS

1. (d ) 6. (a) 11. (d ) 16. (c)

2. (a) 7. (c) 12. (c) 17. (d )

3. (a), (b) 8. (b) 13. (d )

4. (d ) 9. (d ) 14. (c)

5. (b) 10. (a) 15. (b)

5 FIRST LAW APPLIED TO FLOW PROCESSES 5.1 Introduction 5.6 Steady Flow Energy Equation Applied to Various Processes 5.2 Assumptions in Steady Flow Process 5.7 Throttling Process-Joule Thomson Porous 5.3 Flow Energy Plug Experiment 5.4 First Law of Thermodynamics as Applied 5.8 Engineering Applications of Steady Flow to Steady Flow Process Energy Equation 5.5 Workdone for a Steady Flow Process

5.1 INTRODUCTION The flow processes are those processes which occur in an open system (also called control volume) which permit the transfer of mass as well as energy to and from the system i.e. across its boundaries. In such processes, mass enters the system and leaves after enhancing the energy. Following are the two types of flow processes: 1. Steady flow process. In a steady flow process, the state of the working substance in the neighbourhood of a given point remains constant with time. The flow rate in and out of the system is equal and remains constant with time. Thus, during this process, there is no change of stored energy within the system. 2. Unsteady flow process. In an unsteady flow process, the state of the working substance at the boundary of the system varies with time and mass inflow and outflow are not balanced. There is a change of energy stored within the system during this process. The filling of a tank is an example of unsteady flow process. Since we are mainly concerned with the steady flow processes, therefore only these processes are discussed, in detail, in this chapter.

5.2 ASSUMPTIONS IN STEADY FLOW PROCESS

The following assumptions are made in the steady flow process: 1. The mass flow through the system remains constant. 2. The rate of heat and work transfer is constant. 3. The working substance is uniform in composition. 4. The state of the working substance at any point remains constant with time. 5. In dealing with the flow processes, potential, kinetic and flow energies are considered. First of all, let us discuss the flow energy as follows:

First Law Applied to Flow Processes 91

5.3 FLOW ENERGY The energy required to flow or move the working substance across the open system, is termed as flow energy or displacement energy. It is also known as flow work. Work p

Piston

A

Piston

L (b )

(a )

Fig. 5.1. Flow energy.

Let the working substance with pressure p ( in N/m2 ) pushes the piston of cross-sectional area A (in m2) through a distance L (in metres), as shown in Fig. 5.1. The magnitude of the flow energy must be exactly equal to the workdone by the piston. \ Flow energy = Workdone by the system = Force × Distance moved = pA × L ( in N-m or J ) = p v where v = Volume of the working substance. For 1 kg mass of the working substance, Flow energy = p vs (in N-m / kg or J / kg) where vs = Specific volume of the working substance in m3 / kg.

5.4 FIRST LAW OF THERMODYNAMICS AS APPLIED TO STEADY FLOW PROCESS

Fig. 5.2. Steady flow process.

Consider an open system (control volume) as shown in Fig. 5.2. Let the working substance enter the system at section 1-1 with a velocity V1 and pressure p1. Let the specific volume and specific internal energy at section 1-1 be vs1 and u1 respectively. The working substance leaves the system at section 2-2 with a velocity V2 and pressure p2. Let specific volume and specific internal energy at this section 2-2 be vs2 and u2 respectively. In a flow process, the energy contained within the open system is influenced by the following energy transfers:

92 Engineering Thermodynamics

(a) Heat transfer (q1–2), (b) Work transfer (w1–2), and (c) Energy flow accompanying mass flow at inlet and outlet of the system. It is, therefore, necessary to determine the energy flow per unit mass flowing at various positions of the open system. The energy accompanying unit mass flow consists of the following parts: 1. Potential energy, i.e. gz (in N-m / kg or J / kg), where g is the acceleration due to gravity in m / s2 and z is the height above the datumn plane, in metres.

V2 (in J / kg), where V is the velocity in m / s. 2 3. Internal energy, i.e. u (in J / kg), and 4. Flow energy, p vs (in N- m / kg or J / kg) Thus, for one kg of the working substance entering the system at section 1-1, total energy, V2 e1 = gz1 + 1 + u1 + p1 vs1 + q1–2 (in N-m / kg or J / kg) 2 Similarly, total energy leaving the system at section 2-2, 2. Kinetic energy, i.e.

V22 + u2 + p2 vs2 + w1–2 (in N-m / kg or J / kg) 2 Now according to First Law of Thermodynamics (i.e. Law of conservation of energy), the total energy entering the system is equal to the total energy leaving system. \ e1 = e2 ...(Neglecting any loss of energy)

gz1 +

or

e2 = gz2 +

V12 V2 + u1 + p1 vs + q1–2 = gz2 + 2 + u2 + p2 vs + w1–2 1 2 2 2

...(i)

*h1 = Enthalpy of the working substance entering the system = u1 + p1 vs1 (in J / kg) h2 = Enthalpy of the working substance leaving the system = u2 + p2 vs2 (in J / kg) Now the equation (i) may be written as Let

and

V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2 This equation is called steady flow energy equation (briefly written as SFEE). The above equation may also be written as

gz1 +

V 2 − V12 q1–2 – w1–2 = g (z2 – z1) + 2 2

+ (h2 – h1)

= ( pe2 – pe1) + (ke2 – ke1) + (h2 – h1) In the differential form, this equation may be written as dq – dw = d ( pe) + d (ke) + dh

...(ii)

...(iii) ...(iv)

* When u1 and u2 are in kJ / kg , then p1 and p2 should be in kPa or kN / m2 so that the flow energy p1 vs1 and p2 vs2 are in kJ / kg, as discussed below. We know that

kN m3 kN - m × = or kJ / kg m 2 kg kg \ h1 = u1 + p1 vs (in kJ / kg) 1 and h2 = u2 + p2 vs2 (in kJ / kg) Refer Example 5.1

Flow energy = p × vs =

First Law Applied to Flow Processes 93 V 2 − V12 Note: In the above equation, change in potential energy [g (z2 – z1)] and change in kinetic energy 2 2 are in J/kg. These terms may be divided by 1000 to obtain the result in kJ / kg, i.e. Change in potential energy =

g ( z2 − z1 ) kJ / kg 1000

and

V22 − V12 kJ / kg 2 × 1000

Change in kinetic energy =

Example 5.1. In a certain steady flow process, the properties of the fluid at inlet and outlet are as follows: At inlet: P ressure = 1.5 bar; density = 26 kg / m3; velocity = 110 m / s; internal energy = 910 kJ / kg. At exit: P ressure = 5.5 bar; density = 5.5 kg / m3; velocity = 190 m / s; internal energy = 710 kJ/kg. During the process, the fluid rejects 55 kJ / s of heat and rises through 55 metres. The mass flow rate of the fluid is 10 kg / min. Determine: 1. Change in enthalpy; and 2. Power developed during the process. Solution. Given: Properties of the fluid At inlet: Pressure, p1 = 1.5 bar = 150 × 103 N/m2 = 150 kN/m2 Density, r1 = 26 kg / m3 1 1 = or Specific volume, vs = = 0.038 m3 / kg 1 ρ1 26 V1 = 110 m / s u1 = 910 kJ / kg

Velocity, Internal energy,

2 55 m 1

System

q1-2

w1-2

Fig. 5.3

Properties of the fluid at outlet Pressure, p2 = 5.5 bar = 550 × 103 N / m2 = 550 kN/m2 Density, r2 = 5.5 kg / m3 1 1 = or Specific volume, vs2 = = 0.182 m3 / kg ρ2 5.5 Velocity, V2 = 190 m / s Internal energy, u2 = 710 kJ / kg Heat rejected by fluid, Q1–2 = 55 kJ / s Rise in elevation = 55 m

94 Engineering Thermodynamics Mass flow rate = 10 kg / min =

10 1 = kg / s 60 6

The system is shown in Fig. 5.3. 1. Change in enthalpy We know that enthalpy at inlet, h1 = u1 + p1 vs1 = 910 + 150 × 0.038 = 915.7 kJ / kg and enthalpy at outlet, h2 = u2 + p2 vs2 = 710 + 550 × 0.182 = 810.1 kJ / kg \ Change in enthalpy, dh = h2 – h1 = 810.1 – 915.7 = – 105.6 kJ / kg Ans. 2. Power developed during the process Let w1–2 = Workdone or power developed during the process in kJ / kg. We know that heat rejected by the fluid Q − 55 q1–2 = 1− 2 = = – 55 × 6 = – 330 kJ/kg m 1/ 6 The –ve sign is due to heat rejected. Using the steady flow energy equation, we have

V 2 − V12 q1–2 – w1–2 = *g (z2 – z1) + * 2 + (h2 – h1) 2 – 330 – w1–2 =

9.81 (55 – 0) + 1000

(190) 2 − (110) 2 + (– 105.6) 2 × 1000

= 0.54 + 12 – 105.6 = – 93.06 \ w1–2 = – 330 + 93.06 = – 236.94 kJ / kg The –ve sign indicates that work is done on the system. Since the mass flow rate is

1 kg / s, therefore workdone or power developed during process, 6

W1–2 = m × w1–2 =

1 (236.94) = 39.5 kJ/s = 39.54 kW Ans. 6

5.5 WORKDONE FOR A STEADY FLOW PROCESS We have already discussed that for a non-flow process, the workdone is 2

2

∫1

p dv , as shown in

Fig. 5.4 (a). But in case of steady flow process, the workdone is − ∫ v dp instead of 1

2

∫1

p dv , as

discussed below. The steady flow energy equation (SFEE) in the differential form, for a unit mass flow, is given by dq – dw = d ( pe) + d (ke) + dh ...(i) and enthalpy or total heat,

*

V22 − V12 is in J / kg or 2

h = u + pvs = u + pv V22 − V12 kJ / kg 2 × 1000

Similarly, g (z2 – z1) is in J / kg or

g (z – z ) kJ / kg 1000 2 1

...(Q For unit mass, vs = v)

First Law Applied to Flow Processes 95 Differentiating this expression, we have dh = du + d (pv ) = du + pdv + vdp

...(ii)

Fig. 5.4

We know that for a closed system, dq = du + pdv ...(According to first law of thermodynamics) Substituting for du + pdv = dq in equation (ii), we have dh = dq + vdp Now the equation (i) may be written as dq – dw = d( pe) + d (ke) + dq + vdp – dw = d( pe) + d(ke) + vdp In case the changes in potential and kinetic energies are negligible, then d (ke) = 0 and d ( pe) = 0. Now the above equation becomes – dw = vdp or dw = – vdp On integrating from state 1 to state 2,

2

2

∫1 δw = − ∫1 v dp

The p-v diagram for a steady flow process is shown in Fig. 5.4 (b).

5.6 STEADY FLOW ENERGY EQUATION APPLIED TO VARIOUS PROCESSES We have already discussed the various non-flow processes. Now let us determine the expressions for the workdone during various steady flow processes,like non-flow processes. 1. Constant volume process We have discussed in the previous article that the workdone in a steady flow process from state 1 to state 2 is 2 w1–2 = − ∫ v dp 1

2

= – v ∫ dp = – v ( p2 – p1 ) = v ( p1 – p2 ) 1

2. Constant pressure process We know that, workdone in a steady flow process from state 1 to state 2 is

2

w1–2 = − ∫ v dp = – v ( p2 – p1 ) = v ( p1 – p2 ) 1

Since the pressure is constant, therefore p1 = p2, and w1–2 = 0

96 Engineering Thermodynamics 3. Constant temperature or Isothermal process We know that, workdone in a steady flow process from state 1 to state 2 is 2

w1–2 = − ∫ v dp

...(i)

1

For a constant temperature process, pv = p1 v1 = p2 v2 = ... = constant pv or v = 1 1 p Substituting the value of v in equation (i), we have 2 pv 2 dp w1–2 = − ∫ 1 1 dp = − p1v1 ∫ 1 1 p p = – p1 v1 [ log e p ]1 = – p1 v1 [loge p2 – loge p1] 2

p = p1 v1 [loge p1 – loge p2] = p1 v1 loge 1 p2 p = 2.3 p1 v1 log 1 p2 p1 v = 2 , therefore equation (ii) may also be written as Since p1 v1 = p2 v2, or p2 v1 v w1–2 = 2.3 p1 v1 log 2 v1

...(ii)

4. Adiabatic process We know that workdone in a steady flow process from state 1 to state 2 is 2

w1–2 = − ∫ v dp

...(i)

1

For an adiabatic process,

p vg = p

g

1 v1

g

= p2 v2 = ... = constant

γ

or

v p1 p1 v = p or v = v1 p 1

1 γ

Substituting the value of v in equation (i), we have

w1–2 = − ∫

2

1

p v1 1 p

1 γ

1

dp = − v1 p1γ

2 −1 +1 1 γ p γ = − v1 = − v1 p1

− 1 + 1 γ 1

1 γ − v1 p1γ = γ − 1

2

∫1

( p)

−

1 γ

dp

2

1 p1γ

−1 + γ p γ −1 + γ γ 1

2

γ −1 p γ 1

First Law Applied to Flow Processes 97 1 γ −1 γ −1 γ γ γ p − v1 p1 − p1 γ = 2 γ −1 1 γ −1 1 γ −1 γ γ γ − v1 p1 p2 + v1 p1γ p1 γ = γ −1

=

1 γ −1 γ − v1 p1γ p2 γ + v1 p1 γ −1

...(ii)

Since p1 v1g = p2 v2g, therefore v p1 = 2 p2 v1 or

1 γ

γ

1

( p1 ) = ( p2 ) γ ×

p or 1 p2

1 γ

=

v2 v1

v2 v1

1

Now substituting the value of ( p1 ) γ in equation (ii)

w1–2 =

= 5. Polytropic process

γ −1 1 v γ − v1 × ( p2 ) γ × 2 × p2 γ + v1 p1 γ −1 v1

γ [ p v – p2 v2 ] γ −1 1 1

For a polytropic process, we have p v n = p1 v1n = p2 v2n = ... = constant Thus the workdone in a steady flow process from state 1 to state 2 for a polytropic process may be obtained by substituting n for g in adiabatic process. \

w1–2 =

n (p v – p v ) n −1 1 1 2 2

5.7 THROTTLING PROCESS-JOULE THOMSON POROUS PLUG EXPERIMENT A process that takes place in such a way that the fluid expands through a *minute aperature such as a narrow throat or a slightly opened valve in the line of flow, is known as throttling process as shown in Fig. 5.5 (a). During this process, the expansion takes place from a high pressure to a low pressure without doing any work and there is no change in the potential and kinetic energy of the fluid. Also, no transfer of heat occurs and in the case of a perfect gas, there is a no change in temperature. In other words, we may say that in a throttling process, w1–2 = 0 and q1–2 = 0

* When a fluid expands suddenly into a vacuum chamber through a orifice of large dimensions, it is then said to be free expansion.

98 Engineering Thermodynamics It may be noted that this process is an adiabatic process as no heat flows from and to the system, but it is not reversible. It is an irreversible steady flow expansion process. Such a process occurs frequently in a flow through a porous plug. The Joule-Thompson porous plug experiment is shown in Fig. 5.5 (b).

Fig. 5.5

Let p1, V1 and T1 = Pressure, velocity and temperature respectively of the gas at inlet, and p2, V2 and T2 = Corresponding values of the gas as outlet. A stream of gas at pressure p1 and temperature T1 is forced continuously through a porous plug which is thermally insulated from the surroundings and comes out of the porous plug at a lower pressure p2 and temperature T2. Now under steady flow conditions, equation (ii) of Art. 5.4 is applicable, i.e. or

g z1 +

V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2g 2

pe1 + ke1 + h1 + q1–2 = pe2 + ke2 + h2 + w1–2

...(i) ...(ii)

Since there is no change in potential energy and kinetic energies, therefore the above steady flow energy equation may be written as h1 + q1–2 = h2 + w1–2 ...(iii) We also know that during a throttling process, there is no heat transfer and work transfer, i.e. q1–2 = 0 and w1–2 = 0. Therefore equation (iii) becomes,

h1 = h2

This shows that during a throttling process,the enthalpy or total heat remains constant. Thus, this process is also known as constant enthalpy process which is an irreversible process.

5.8 ENGINEERING APPLICATIONS OF STEADY FLOW ENERGY EQUATION The steady flow energy equation (SFEE) applies to flow processes in many of engineering systems, as discussed below: 1. Nozzle. A nozzle is a passage of varying cross-section by means of which the pressure energy of the flowing fluid is converted into kinetic energy. The main use of the nozzle, as shown in Fig. 5.6, is to produce a jet of high velocity to drive a turbine and to produce thrust. The steady flow energy equation for a unit mass flow of fluid is given by

g z1 +

V2 V12 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2g

First Law Applied to Flow Processes 99 Since the nozzle is horizontal, therefore z1 = z2. Further, there is no heat transfer and work transfer, i.e. q1–2 = 0, and w1–2 = 0. Thus, the above expression becomes

V12 V2 + h1 = 2 + h2 2 2 2 V1 + 2h1 = V22 + 2 h2

2 \ V2 = V1 + 2 (h1 − h2 )

In case V1 is very small as compared to V2, then V1 may be neglected. In that case

V2 = 2 (h1 − h2 ) Control volume

Control surface

Flow 1 in

2 Flow out w1-2 = 0

q1-2 = 0

Convergent

Divergent

Fig. 5.6. Flow through a nozzle.

Let A1 and A2 = Cross-sectional area of the nozzle at inlet and outlet respectively, V1 and V2 = Velocity of the fluid at the inlet and outlet of the nozzle respectively, vs1 and vs2 = Specific volume of the fluid at inlet and outlet respectively. We know that for the continuous steady flow, mass flow rate, A1V1 A V = 2 2 m = vs1 vs 2 2. Diffuser. A diffuser is a passage of varying cross-section by means of which the kinetic energy of the flowing fluid is changed into pressure energy. The energy equation for steady flow may be applied in the similar way as for the nozzle. 3. Steam or Gas turbine. A turbine is used to convert the heat energy of steam or gas into useful work. In actual practice, the turbine as shown in Fig. 5.7, is fairly insulated so that there is no transfer of heat (i.e. q1–2 = 0). In other words, the flow through a turbine is adiabatic. In this system, the changes in potential energy and kinetic energy are negligible, i.e. z1 = z2 and V1 = V2. Steam or Gas in 1

Control surface q1-2 = 0

Turbine

w1-2

Control volume Exhaust

2

Fig. 5.7. Flow through a turbine.

Applying steady flow energy equation per unit mass, we have V2 V2 g z1 + 1 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2

100 Engineering Thermodynamics Since z1 = z2 , V1 = V2 , and q1–2 = 0, therefore, h1 = h2 + w1–2 or w1–2 = h1 – h2 4. Boiler. A boiler is used to generate steam from feed water by heating due to burning of a fuel. The steam may be used to drive steam engine or a steam turbine. The flow through a boiler is shown in Fig. 5.8. Control surface

Steam out 2 w1-2

Steam

Feed water in 1

Water

Control volume

Flue gases q1-2

Fig. 5.8. Flow through a boiler.

Applying steady flow energy equation for a unit mass, we have V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2 In a boiler system, there is no change in potential and kinetic energies. Also, no work is done in a boiler. In other words, z1 = z2 ; V1 = V2 ; and w1–2 = 0 Now the above equation for a boiler system may be written as h1 + q1–2 = h2 or q1–2 = h2 – h1 5. Reciprocating compressor. A reciprocating compressor is used to compress air or gas from low pressure to high pressure with the help of work input (w1–2). In this system, heat is absorbed or rejected to cooling water from the air. A reciprocating compressor, as shown in Fig. 5.9, includes a receiver which is sufficiently large so that there is no fluctuation of flow in the control volume.

g z1 +

Control surface Air in

1

2 Air out

Piston

Air receiver

Cylinder

q1-2

w1-2

Control volume

Fig. 5.9. Flow through a reciprocating compressor.

Applying steady flow energy equation for a unit mass, we have

g z1 +

V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2

First Law Applied to Flow Processes 101 Since there is no change in potential and kinetic energies, therefore z1 = z2, and V1 = V2. Also the heat is rejected from the system, therefore q1–2 is negative. Now the above equation for a reciprocating compressor may be written as h1 – q1–2 = h2 + w1–2 or w1–2 = (h1 – h2) – q1–2 Since h2 is greater than h1 due to increased pressure of air, therefore, w1–2 is negative, i.e. the work is done on the system. Air out 6. Rotary compressor. A rotary compressor, is used 2 to compresses air or gas with the help of a rotating element called rotor followed by a stator. This increases Rotary compressor the energy level of the working fluid in the direction w1-2 of flow. A rotary compressor, as shown in Fig. 5.10, is employed where high efficiency, medium pressure rise Control (upto 10 bar) and large flow rate (50 m3/s) are primary Control surface 1 volume considerations. It may be noted that the flow through Air in q1-2 = 0 the rotary compressor is adiabatic. Thus there is no heat Fig. 5.10. Flow through a rotary compressor. transfer takes place (i.e. q1–2 = 0). Applying steady flow energy equation per unit mass, we have V2 V2 g z1 + 1 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2 Neglecting the changes is potential and kinetic energies (i.e. z1 = z2 , and V1 = V2), the above equation may be written as h1 + q1–2 = h2 + w1–2 or w1–2 = h1 – h2 ...(Q q1–2 = 0) Since h2 is greater than h1, therefore w1–2 is negative, i.e. work is done on the system. 7. Condenser. A condenser (or a heat exchanger) is a device used to condense steam by rejecting heat from the steam to the cooling water. In a condenser, as shown in Fig. 5.11, there is no change in potential and kinetic energies, i.e. z1 = z2 and V1 = V2. Also, the workdone is zero, i.e. w1–2 = 0. 1

Steam in

Cooling water out

w1-2 = 0 Cooling water in Control volume

q1-2

Control surface

2 Condensate out

Fig. 5.11. Flow through a condenser.

Applying steady flow energy equation for a unit mass flow, we have V12 V2 + h1 – q1–2 = g z2 + 2 + h2 + w1–2 2 2 ...(q1–2 is taken negative as heat is rejected) or h1 – q1–2 = h2 and q1–2 = h1 – h2 8. Evaporator. An evaporator is the component of a refrigeration system to take away heat from the refrigerator to maintain low temperature. The refrigerating liquid enters the evaporator, g z1 +

102 Engineering Thermodynamics absorbs latent heat from the chamber at constant pressure and leaves as a vapour refrigerant. In the evaporator, as shown in Fig. 5.12, there is no change in potential and kinetic energies, i.e. z1 = z2 and V1 = V2. Also the workdone is zero, i.e. w1–2 = 0. q1-2

Refrigerant liquid in

Evaporator

w1-2 = 0

Referigerant vapour out 2

1 Control surface

Control volume

Fig. 5.12. Flow through an evaporator.

Applying steady flow energy equation for a unit mass flow, we have

g z1 +

V12 V2 + h1 + q1–2 = q z2 + 2 + h2 + w1–2 2 2

or h1 + q1–2 = h2 and q1–2 = h2 – h1 9. Centrifugal pump. A centrifugal pump is used to pump water from a lower level to a higher level, as shown in Fig. 5.13. The work required to run the pump is supplied from an external source such as an electric motor or a diesel engine. In this system, there is no change in temperature, i.e. change in internal energy is zero (or u1 = u2).

Control surface z2

Overhead tank

Control volume

Datumn

w1-2

Pump

Electric motor

q1-2 = 0

z1

Water

Fig. 5.13. Flow through a centrifugal pump.

Also q1–2 = 0 and w1–2 is taken negative as work is done on the system. Applying steady flow energy equation, we have V2 V2 – g z1 + 1 + u1 + p1vs + q1–2 = gz2 + 2 + u2 + p2 vs – w1–2 1 2 2 2 V12 V22 – g z1 + + p1vs = g z2 + + p2 vs – w1–2 1 2 2 2 z1 is taken negative because it is below the datumn.

First Law Applied to Flow Processes 103 10. Water turbine. The water turbine, as shown in Fig. 5.14, is used to generate electric power by supplying water from a reservoir to the turbine. The potential energy of the water is converted into kinetic energy when it enters into the turbine and part of it, is converted into useful work which is used to generate electricity. In this case, u1 = u2 and q1–2 = 0.

Reservoir

Control volume

Applying steady flow energy equation, we have

z1

Control surface

Datumn

V2 g z1 + 1 + u1 + p1 vs + q1–2 1 2

Generator z2 Turbine

V2 = g z2 + 2 + u2 + p2 vs + w1–2 2 2 2 V or g z1 + 1 + p1 vs 1 2 V2 = – g z2 + 2 + p2 vs + w1–2 2 2

Tail race

Fig. 5.14. Flow through a water turbine.

z2 is taken as negative because it is below the datumn. Example 5.2. A perfect gas flows through a nozzle where it expands in a reversible adiabatic manner. The inlet conditions are 22 bar, 500°C and 38 m / s. At exit, the pressure is 2 bar. Determine the exit velocity and exit area, if the flow rate is 4 kg / s. Take R = 190 J / kg K and g = 1.35. Solution. Given: Inlet pressure,

p1 = 22 bar = 2200 × 103 N/m2

Inlet temperature,

T1 = 500°C = 500 + 273 = 773 K

Inlet velocity,

V1 = 38 m / s

Exit pressure,

p2 = 2 bar = 200 × 103 N/m2

Mass,

m = 4 kg /s

Gas constant,

R = 190 J/ kg K

Ratio of specific heats,

g =

cp cv

= 1.35

Exit Velocity Let

V2 = Exit velocity in m/s, and

T2 = Exit temperature.

We know that for a reversible adiabatic process, \

p T2 = 2 T1 p1

γ −1 γ

2 = 22

1.35 −1 1.35

1 = 11

0.259

T2 = T1 × 0.537 = 773 × 0.537 = 415.1 K

=

1 = 0.537 1.86

104 Engineering Thermodynamics and change in enthalpy from inlet to exit, *R γ 190 × 1.35 (T1 – T2) = (773 – 415.1) γ −1 1.35 − 1 = 262.3 × 103 J / kg Using the steady flow energy equation, we have

h1 – h2 = cp (T1 – T2) =

2 V2 = V1 + 2(h1 − h2 )

2 3 = (38) + 2(262.3 × 10 ) = 725.3 m/s Ans. Exit area Let A2 = Exit area in m2. First of all, let us find the specific volume of the gas (in m3 / kg) at exit (i.e. vs ). We know that 2

p v = m R T or p vs = RT

...(Q

v = vs) m

p2 vs2 = RT2 RT 190 × 415.1 or vs2 = 2 = = 0.394 m3 / kg p2 200 × 103 We know that for the continuous steady flow, A V m = 2 2 vs 2 m × vs2 4 × 0.394 = \ A2 = = 2.173 × 10–3 m3 Ans. V2 725.3 \

Example 5.3. In a gas turbine, the gas enters at the rate of 5 kg / s with a velocity of 50 m / s and enthalpy of 900 kJ / kg and leaves the turbine with a velocity of 150 m / s and enthalpy of 400 kJ / kg. The loss of heat from the gases to the surroundings is 25 kJ / kg. Assume for gas, R = 0.285 kJ / kg K and cp = 1.004 kJ/kg K. The inlet condition is at 100 kPa and 27°C. Determine the power output of the turbine and diameter of inlet pipe. Solution. Given: Mass of the gas m = 5 kg / s Inlet velocity, V1 = 50 m / s Enthalpy at inlet, h1 = 900 kJ / kg Exit velocity, V2 = 150 m / s Enthalpy at exit, h2 = 400 kJ / kg Loss of heat to the surrounding, q1–2 = – 25 kJ / kg (– sign due to loss of heat) Gas constant, R = 0.285 kJ / kg K Specific heat at constant pressure, cp = 1.004 kJ / kg K Inlet pressure, p1 = 100 kPa Inlet temperature, T1 = 27°C = 27 + 273 = 300 K c p − cv

R R or g – 1 = cv cv cv R Rγ Multiplying by cp on both sides, cp (g – 1) = × cp = R g or cp = cv γ −1

* We know that cp – cv = R or

=

First Law Applied to Flow Processes 105 Power output of the turbine Let w1–2 = Workdone or power output of the turbine. Using the steady flow energy equation for a unit mass, we have V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2 V12 V22 or + h1 + q1–2 = + h2 + w1–2 ...( Taking z1 = z2 ) 2 2 (50) 2 (150) 2 * + 900 + (– 25) = + 400 + w1–2 2 × 1000 2 × 1000 1.25 + 900 – 25 = 11.25 + 400 + w1–2 876.25 = 411.25 + w1–2 \ w1–2 = 876.25 – 411.25 = 465 kJ/kg Since the mass of gas is 5 kg / s; therefore power output of the turbine, W1–2 = 5 × 465 = 2325 kJ/s = 2325 kW Ans. ...(Q 1 kJ/s = 1 kW) Example 5.4. In a steam plant, 1 kg of water per second is supplied to the boiler. The enthalpy and velocity of water entering the boiler are 800 kJ / kg and 5 m / s. The water receives 2200 kJ / kg of heat in the boiler at constant pressure. The steam after passing through the turbine comes out with a velocity of 50 m / s and its enthalpy is 2520 kJ / kg. The inlet is 4 m above the turbine exit. The heat losses from the boiler and turbine to the surroundings are 20 kJ / s. Calculate the power developed by the turbine considering boiler and turbine as single system. Solution. Given: Mass of water, m = 1 kg / s Enthalpy of water, h1 = 800 kJ / kg Velocity of water, V1 = 5 m / s Heat supplied to water, q1 = 2200 kJ / kg Velocity of steam, V2 = 50 m / s Enthalpy of steam, h2 = 2520 kJ / kg Heat losses from the boiler and turbine, q2 = 20 kJ/s = 20 × 1 = 20 kJ / kg ...(Q m = 1 kg/ s) Since the inlet is 4 m above the turbine exit, therefore z1 – z2 = 4 m We know that total heat transfer from the boiler to turbine, q1– 2 = q1 – q2 = 2200 – 20 = 2180 kJ / kg Power developed by the turbine Let W1 – 2 = Work or power developed by the turbine in kW. Considering the boiler and turbine as a single system, the steady flow energy equation is given by

g z1 +

g z1 +

or

* Kinetic energy

V12 V2 + h1 + q1 – 2 = g z2 + 2 + h2 + w1–2 2 2 V 2 − V22 w1 – 2 = g (z1 – z2) + 1 2 V2 V2 is in J / kg or kJ/ kg 2 2 × 1000

+ (h1 – h2) + q1–2

106 Engineering Thermodynamics 52 − (50) 2 = 9.81 × 4 + + (800 – 2520) + 2180 2 × 1000 25 − 2500 = 39.24 + + [–1720] + 2180 2000 = 39.24 – 1.2375 – 1720 + 2180 = 498 kJ / kg Since the mass of water (m) in 1 kg / s, therefore power developed, W1–2 = m × w1–2 = 1 × 498 = 498 kJ / s = 498 kW Ans. Example 5.5. Air enters an air compressor at 8 m/ s velocity, 100 kPa pressure and 0.95 m3 / kg volume. It flows steadily at the rate of 0.6 kg / s and leaves at 6 m/s, 700 kPa and 0.19 m3 / kg. The internal energy of the air leaving is 90 kJ / kg greater than that of air entering. The cooling water in the compressor jackets absorbs heat from the air at the rate of 60 kW. Find: 1. The ratio of the inlet pipe diameter to outlet pipe diameter; and 2. The rate of shaft work input to the air in kW. Solution. Given: Inlet velocity, V1 = 8 m / s Inlet pressure, p1 = 100 kPa = 100 kN / m2 Inlet specific volume, vs1 = 0.95 m3 / kg Mass of air, m = 0.6 kg / s Exit velocity, V2 = 6 m / s Exit pressure, p2 = 700 kPa = 700 kN / m2 Exit specific volume, vs2 = 0.19 m3 / kg Internal energy at exit, u2 = 90 + Internal energy at inlet, u1 or u2 – u1 = 90 kJ / kg Heat absorbed from the air, 60 q1–2 = 60 kW = 60 kJ / s = = 100 kJ / kg 0.6 1. Ratio of inlet pipe diameter to outlet diameter Let d1 = Inlet pipe diameter, and d2 = Outlet pipe diameter. A V A V We know that m = 1 1 = 2 2 vs1 vs 2 \

or

\

v A1 V 6 × 0.95 = 2 × s1 = = 3.75 A2 Vs 2 V1 0.19 × 8 π (d1 ) 2 4 = 3.75 π (d 2 )2 4 d1 = 3.75 = 1.936 Ans. d2

...(Q Mass of air = 0.6 kg /s)

First Law Applied to Flow Processes 107 2. Rate of shaft work input to the air We know that flow energy at inlet = p1 vs1 = 100 × 0.95 = 95 kJ / kg and flow energy at outlet = p2 vs2 = 700 × 0.19 = 133 kJ / kg Now applying the steady flow energy equation for a unit mass flow, we have V12 V2 + u1 + p1 vs1 – q1–2 = g z + 2 + u2 + p2 vs2 – w1–2 2 2 2 In the above equation, q1–2 and w1–2 is taken as negative, because the heat is rejected by the air and work is done on the air. Neglecting the potential energy, the above equation may be written as g z1 +

V 2 − V12 w1–2 – q1–2 = 2 2

6 2 − 82 w1–2 – 100 = + 90 + (133 – 95) = 128 2 × 1000

+ (u2 – u1) + ( p2 vs2 – p1 vs1)

\ w1–2 = 128 + 100 = 228 kJ / kg Since the mass of air (m) is 0.6 kg / s, therefore rate of shaft-work, W1–2 = m × w1–2 = 0.6 × 228 = 136.8 kJ/s = 136.8 kW Ans. Example 5.6. Air at a temperature of 20°C passes through a heat exchanger at a velocity of 40 m / s,where its temperature is raised to 820°C. It then enters a turbine with the same velocity of 40 m / s and expands till the temperature falls to 620°C. On leaving the turbine, the air is taken at a velocity of 55 m / s to a nozzle where it expands until the temperature has fallen to 510°C. If the air flow rate is 2.5 kg / s; calculate: 1. Rate of heat transfer to the air in the heat exchanger; 2. The power output from the turbine, assuming no heat loss; and 3. The velocity at exit from the nozzle, assuming no heat loss. Take enthalpy of air as h = cp dT, where cp is the specific heat at constant pressure and taken as 1.005 kJ / kg K and dT = T2 – T1, is the change in temperature. Solution. Given: Temperature of air entering the heat exchanger, T1 = 20°C = 20 + 273 = 293 K Velocity of air, V1 = 40 m / s Temperature of air leaving the heat exchanger, T2 = 820°C = 820 + 273 = 1093 K Velocity of air entering the turbine, V2 = V1 = 40 m / s Temperature of air leaving the turbine, T3 = 620°C = 620 + 273 = 893 K Velocity of air leaving the turbine or entering the nozzle, V3 = 55 m /s Temperature of air leaving the nozzle, T4 = 510°C = 510 + 273 = 783 K Air flow rate, m = 2.5 kg / s

108 Engineering Thermodynamics 1. Rate of heat transfer to the air in the heat exchanger The system is shown in Fig. 5.15 2

1

Heat exchanger

2 w

Turbine

3 4 Nozzle 3

Fig. 5.15

First of all, let us consider heat transfer through a heat exchanger. Applying the steady flow energy equation, we have V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2 Since z1 = z2 , V1 = V2 , and w1–2 = 0 for a heat exchanger, therefore q1–2 = h2 – h1 = cp (T2 – T1) = 1.005 (1093 – 293) = 804 kJ/kg Since the air flow rate (m) is 2.5 kg / s, therefore rate of heat transfer, Q1–2 = m × q1–2 = 2.5 × 804 = 2010 kJ / s or kW Ans. ...(Q 1 kJ /s = 1 kW) 2. Power output from the turbine Let W2–3 = Workdone or power output from the turbine in kW. Applying the steady flow energy equation, we have

g z1 +

V2 V22 + h2 + q2–3 = g z3 + 3 + h3 + w2–3 2 2 Since z2 = z3, and q2–3 = 0, therefore

or

g z2 +

V2 V22 + h2 = 3 + h3 + w2–3 2 2 V 2 − V32 w2–3 = 2 + (h2 – h3) = 2

V22 − V32 2

+ cp (T2 – T3)

(40) 2 − (55) 2 = + 1.005 (1093 – 893) 2 × 1000 = – 0.7125 + 201 = 200.2875 kJ/ kg Since the air flow rate (m) is 2.5 kg / s, therefore power output from the turbine. W1–2 = m × w1–2 = 2.5 × 200.2875 = 500.72 kJ / s or kW Ans. 3. Velocity at exit from the nozzle Let V4 = Velocity at exit from the nozzle in m / s.

First Law Applied to Flow Processes 109 Applying the steady flow energy equation, we have V32 V2 + h3 + q3–4 = g z4 + 4 + h4 + w3–4 2 2 Since z3 = z4 , q3–4 = 0, and w3–4 = 0, therefore

g z3 +

V2 V2 3 + h3 = 4 + h4 2 2 or

2 V4 = V3 + 2(h3 − h4 ) =

V32 + 2c p (T3 − T4 )

2 = (55) + 2 × 1.005 (893 − 783)

= 3025 + 221110 = 473.43 m /s Ans. ...[Q Here cp is taken in J/ kg K] Example 5.7. A centrifugal pump delivers water at the rate of 45.5 kg / s by increasing the pressure from 80 kN / m2 to 280 kN / m2. The suction is 2 m below the centre of the pump and delivery is 5 m above the centre of the pump. The suction and delivery pipe diameters are 150 mm and 100 mm respectively. Determine the power required to drive the pump. Solution. Delivery 2 Given: Mass of water delivered, pipe m = 45.5 kg / s Initial pressure or suction pressure, p1 = 80 kN / m2 = 80 × 103 N / m2 5m Pump Final pressure or delivery pressure, p2 = 280 kN / m2 Motor = 280 × 103 N / m2 Distance of suction from the centre of the pump, w1-2 2m z1 = 2 m Suction pipe Distance of delivery from the centre of the 1 pump, z2 = 5 m Fig. 5.16 Diameter of suction pipe, Fig. 5.16 d1 = 150 mm = 0.15 m Diameter of delivery pipe, d2 = 100 mm = 0.1 m The centrifugal pump is shown in Fig. 5.16. We know that cross-sectional area of suction pipe, π π A1 = (d1)2 = (0.15)2 = 17.7 × 10–3 m2 4 4 and cross-sectional area of delivery pipe, π π A2 = (d2)2 = (0.1)2 = 7.855 × 10–3 m2 4 4

110 Engineering Thermodynamics Since the density (r) of water is 1000 kg / m3, therefore specific volume of water, 1 1 = vs1 = vs2 = = 1 × 10–3 m3 / kg ρ 1000 Let V1 and V2 be the velocity of water (in m / s) at the inlet and outlet respectively. We know that AV AV m = 1 1 = 2 2 vs1 vs 2 \ and

V1 =

m × vs1 45.5 × 1 × 10 −3 = = 2.57 m/s A1 17.7 × 10 −3

V2 =

m × vs 2 45.5 × 1 × 10 −3 = = 5.79 m/s A2 7.855 × 10 −3

Now using the steady flow energy equation for a centrifugal pump, we have V12 V2 + u1 + p1 vs + q1–2 = q z2 + 2 + u2 + p2 vs2 – w1–2 1 2 2 Since u1, u2 and q1–2 is zero, therefore – g z1 +

– 9.81 × 2 +

(2.57) 2 + 80 × 103 × 1 × 10–3 2

(5.79) 2 + 280 × 103 × 1 × 10–3 – w1–2 2 – 19.62 + 3.3 + 80 = 49.05 + 16.762 + 280 – w1–2 63.68 = 345.812 – w1–2 \ w1–2 = 345.812 – 63.68 = 282.132 J / kg Since the mass (m) is given as 45.5 kg /s, therefore power required to drive the pump, W1–2 = m × w1–2 = 45.5 × 282.132 = 12 837 J / s or W = 12.837 kW Ans. Example 5.8. In a water turbine, the water head measured from the centre of the turbine is 1500 m and the flow rate is 500 kg / s. The tail race is 3 m below the turbine centre line and outlet velocity is 10 m / s. Determine the power developed by the turbine. Solution. Given: Distance of water head from the centre of the turbine, z1 = 1500 m Flow rate, m = 500 kg / s Distance of tail race from the turbine centre line, z2 = 3 m Outlet velocity, V2 = 10 m/s Applying steady flow energy equation, we have

gz1 + or

= 9.81 × 5 +

V12 V2 + (u1 + p1 vs ) + q1–2 = g z2 + 2 + (u2 + p2 vs ) + w1–2 1 2 2 2 g z1 +

V12 V2 + h1 + q1–2 = – g z2 + 2 + h2 + w1–2 2 2

...(z2 is taken negative as it is below the centre line of the turbine)

First Law Applied to Flow Processes 111 Neglecting V1, h1, q1–2 and h2, we have

V22 + w1–2 2 V2 (10) 2 \ w1–2 = g (z1 + z2) – 2 = 9.81 (1500 + 3) – 2 2 = 14744.43 – 50 = 14694.43 J / s Since the flow rate is 500 kg / s, therefore power developed by the turbine, W1–2 = m × w1–2 = 500 × 14694.43 = 7 347 215 J / s or W = 7347.215 kW Ans.

g z1 = – g z2 +

HIGHLIGHTS 1. The steady flow energy equation (SFEE) per unit mass (i.e. for m = 1 kg) is given by g z1 + or

V12 V2 + u1 + p1 vs1 + q1–2 = g z2 + 2 + u2 + p2 vs2 + w1–2 2 2 V 2 − V12 q1–2 – w1–2 = g (z2 – z1) + 2 + [(u2 + p2 vs ) – (u1 + p1 vs1)] 2 2

V 2 − V12 = g (z2 – z1) + 2 2

+ (h2 – h1)

where q1–2 = Heat transfer in J / kg, w1–2 = Work transfer in J / kg, z1 and z2 = Height above datumn at inlet and outlet respectively, in metres, V1 and V2 = Velocity at inlet and outlet respectively, in m / s, h1 and h2 = Enthalpy at inlet and outlet respectively, in J / kg. In using the steady flow energy equation (SFEE), the unit of each term should be same. In the above equations, g z1 and g z2 = Potential energy in J / kg, when g is in m /s2, z1 and z2 is in metres.

V12 V2 and 2 = Kinetic energy in J / kg, when V1 and V2 is in m / s. 2 2

p1 vs1 and p2 vs2 = Flow energy in J / kg, when p1 and p2 is in N / m2, vs1 and vs2 is the specific volume is m3 / kg. 2. The workdone for a steady flow process is given by

2

w1–2 = − ∫ v dp 1

The workdone for various steady flow processes, like non-flow processes are as follows: (a) For constant volume process, w1–2 = v ( p1 – p2 ) (b) For constant pressure process, w1–2 = 0 (c) For constant temperature process,

v w1–2 = 2.3 p1 v1 log 2 v1

112 Engineering Thermodynamics (d) For adiabatic or isentropic process, γ w1–2 = ( p v – p2 v2 ) γ −1 1 1 (e) For polytropic process, n ( p v – p2 v2 ) n −1 1 1 3. A process that takes place in such a way that the fluid expands through a minute aperture such as a narrow throat or a slightly opened valve in the line of flow, is known as throttling process. During this process, q1–2 = 0 ; and w1–2 = 0 4. The steady flow energy equation is given by

g z1 +

w1–2 =

V12 V2 + h1 + q1–2 = g z2 + 2 + h2 + w1–2 2 2

The application of this equation to many of the engineering systems is as follows: (a) For a nozzle, z1 = z2 , q1–2 = 0 and w1–2 = 0

\

or

and mass flow rate,

V12 V2 + h1 = 2 + h2 2 2 2 V2 = V1 + 2(h1 − h2 )

m =

A1 V1 A V = 2 2 vs1 vs 2

(b) For a steam or gas turbine, z1 = z2, V1 = V2 and q1–2 = 0

\

w1–2 = h1 – h2

(c) For a boiler, z1 = z2 , V1 = V2 and w1–2 = 0

\

q1–2 = h2 – h1

(d) For a reciprocating compressor, z1 = z2 , V1 = V2 and q1–2 is negative.

\

or

h1 – q1–2 = h2 + w1–2 w1–2 = (h1 – h2) – q1–2

Since h2 > h1, therefore work is done on the system. (e) For a rotary compressor, z1 = z2 ; V1 = V2 and q1–2 = 0

\

w1–2 = h1 – h2

Since h2 > h1, therefore work is done on the system. ( f ) For a condenser, z1 = z2 , V1 = V2 and w1–2 = 0

\

or

h1 – q1–2 = h2 q1–2 = h1 – h2

(g) For an evaporator, z1 = z2, V1 = V2 and w1–2 = 0

\

q1–2 = h2 – h1

...(q1–2 is –ve because heat is rejected)

First Law Applied to Flow Processes 113 (h) For a centrifugal pump, internal energy is zero (i.e. u1 = u2). Also q1–2 = 0 and w1–2 is negative. In this case, the equation is V12 V2 + p1 vs = g z2 + 2 + p2 vs – w1–2 1 2 2 2 (i) For a water turbine, u1 = u2 and q1–2 = 0. In this case, the equation is – g z1 +

V12 V2 + p1 vs1 = – g z2 + 2 + p2 vs2 + w1–2 2 2 The value of z2 is taken –ve as it is below the datumn.

g z1 +

EXERCISES 1. 12 kg of fluid per minute passes through a reversible steady flow process. The fluid enters at a pressure of 1.4 bar, density 25 kg / m3, velocity 120 m / s, and internal energy 920 kJ / kg. The properties of the fluid at outlet are pressure 5.6 bar, density 5 kg / m3, velocity 180 m / s and internal energy 720 kJ / kg. During the process, the fluid rejects 60 kJ / s of heat and rises through 60 m. Determine the workdone during the process in kW. [Ans. 43.2 kW] 2. The enthalpy of the fluid at the inlet to a nozzle is 2500 kJ / kg and the velocity is 50 m /s. The enthalpy at the discharge end is 2400 kJ / kg. Assuming the nozzle to be horizontal and negligible loss from it, determine: 1. Velocity at exit of the nozzle; 2. Mass flow rate, if the inlet area is 900 cm2 and the specific volume at inlet is 0.187 m3 / kg; 3. Exit area of the nozzle, if the specific volume at the nozzle exit is 0.498 m3 / kg.

[Ans. 450 m / s; 24.06 kg / s; 266 cm2]

3. In an isentropic flow through a nozzle, air flows at the rate of 600 kg / h. At inlet to nozzle, pressure is 2 MPa and temperature is 127°C. The exit pressure is 0.5 MPa. If the initial velocity is 300 m / s, determine : 1. Exit velocity of air ; and 2. Inlet and exit area of nozzle. [Ans. 594.4 m / s; 31.9 mm2, 43.2 mm2] p T [Hint: (a) First find the temperature of air at exit of nozzle (i.e. T2) by using the relation 2 = 2 T1 p1

γ −1 γ

,

where g for air = 1.4. (b) Change in enthalpy, h2 – h1 = cp (T2 – T1), taking cp for air = 1.005 kJ / kg K] 4. In a gas turbine, the mass flow rate of gas through the turbine is 17 kg / s. The power developed by the turbine is 1400 kW. The enthalpies of the gas at inlet and outlet are 1200 kJ / kg and 360 kJ / kg respectively. The velocity of the gas at inlet and outlet are 60 m / s and 150 m /s respectively. Calculate the rate at which the heat is rejected from the turbine. Also calculate the area of the inlet pipe, if specific volume of the gas at inlet is 0.5 m3 / kg. [Ans. – 12 719 kJ / s ; 0.1417 m2] 5. A steam turbine operating under steady flow conditions, receives 60 kg / min. The steam enters the turbine at a velocity of 80 m / s, an elevation of 10 m and specific enthalpy of 3276 kJ/ kg. It leaves the turbine with a velocity of 150 m / s, an elevation of 3 m and specific enthalpy of 2465 kJ / kg. The heat losses from the turbine to the surroundings amounts to 10 kJ / s. Estimate the power output of the turbine. [Ans. 792.88 kW] 6. Air flows in a compressor at the rate of 0.7 kg / s. The air enters at 5 m / s velocity, 100 kPa pressure, 0.85 m3 / kg volume and leaving at 3 m / s, 700 kPa and 0.17 m3 / kg. The internal energy of the air leaving is 80 kJ / kg greater than that of air entering. Cooling water in the compressor jackets absorb heat from the air at the rate of 60 kW. Determine the work input to the compressor and the ratio of inlet pipe diameter to outlet pipe diameter. [Ans. – 139.8 kW; 1.732]

114 Engineering Thermodynamics 7. In a centrifugal compressor, the suction and delivery pressure are 1 bar and 5.5 bar respectively. The compressor draws 0.4 m3 / s of air which has a specific volume of 0.8 m3 / kg. The specific volume of air at delivery point is 0.2 m3 / kg. The compressor is driven by a 40 kW motor and during passage of air through the compressor, the heat lost to the surroundings is 30 kJ / kg of air. Neglecting changes in potential and kinetic energies, determine the increase in internal energy per kg of air. [Ans. 20 kJ / kg] 8. A centrifugal pump delivers 3000 kg of water per minute from the initial pressure of 1 bar to a final pressure of 4.2 bar. The suction is 2.2 m below and the delivery is 8.5 m above the centre line of the pump. If the suction and delivery pipes are of 200 mm and 100 mm diameter respectively, find the power required to run the pump. [Ans. 22.2 kW]

QUESTIONS 1. Define steady flow process. Write the steady flow energy equation (SFEE) and explain the significance of each term. 2. Derive steady flow energy equation in a control volume. 3. In which system the control volume approach is applied? Derive an expression for calculating steady flow energy equation on per unit mass basis while highlighting assumptions. 4. Prove that for a steady flow process, the workdone is given by − ∫ v dp . Plot the result of steady flow work and non-flow work on p-v diagram. 5. (a) Write a short note on throttling process. (b) Using energy equation, show that the enthalpy of fluid before throttling is equal to that after throttling. 6. Obtain the expression for the energy equation of a steady flow open system applicable to nozzle and diffuser. 7. Write the simplified steady flow energy equation for a unit mass flow for the following: (a) Turbine ; (b) Reciprocating compressor ; (c) Boiler ; (d) Condenser ; (e) Evaporator ; and ( f ) Centrifugal pump.

OBJECTIVE TYPE QUESTIONS 2. The steady flow energy equation (SFEE) for a unit mass flow rate is (a) q1–2 – w1–2 = ( pe2 – pe1 ) + (ke2 – ke1) + (h2 – h1) (b) q1–2 + w1–2 = ( pe2 + pe1 ) + (ke2 + ke1) + (h2 + h1) (c) q1–2 – w1–2 = ( pe2 – pe1) – (ke2 – ke1) – (h2 – h1) (d) q1–2 + w1–2 = ( pe2 + pe1) – (ke2 + ke1) – (h2 + h1) where q1–2 and w1–2 = Heat and work transfer respectively, pe1 and pe2 = Potential energy at inlet and outlet respectively, ke1 and ke2 = Kinetic energy at inlet and outlet respectively, and h1 and h2 = Enthalpy at inlet and outlet respectively.

First Law Applied to Flow Processes 115 2. In a steady flow process, the ratio of (a) heat transfer is constant (b) work transfer is constant (c) mass flow at inlet and outlet is constant (d) all of these 3. The workdone in a steady flow process is given by

(a)

2

∫1 v dp

2

(b) − ∫ v dp 1

(c)

2

∫1

2

(d) − ∫ p dv

p dv

1

4. The throttling process is a .................... process. (a) non-flow (b) steady flow (c) non-steady flow 5. Which of the following statement is correct? (a) A nozzle is a passage of varying cross-section by means of which pressure energy of the flowing fluid is converted into kinetic energy. (b) A diffuser is a passage of varying cross-section by means of which the kinetic energy of the flowing fluid is converted into pressure energy. (c) A turbine is used to convert heat energy of steam or gas into useful work. (d) all of the above

ANSWERS

2. (a)

1. (d)

3. (b)

4. (b)

5. (d)

6 SECOND LAW OF THERMODYNAMICS

6.1 Introduction 6.2 Statements of Second Law of Thermodynamics 6.3 Equivalence of Kelvin-Planck and Clausius Statements 6.4 Perpetual Motion Machine of Second Kind (PMM-II)

6.5 Important Terms 6.6 Carnot Cycle 6.7 Efficiency of Carnot Cycle 6.8 Practical Difficulties (or Impractiability) of Carnot Cycle 6.9 Carnot’s Theorem 6.10 Corollaries of Carnot’s Theorem

6.1 INTRODUCTION We have already discussed in Chapter 4 (Art. 4.4), the limitations of First law of thermodynamics. According to the first statement of the first law, a closed system undergoing a thermodynamic cycle, the net heat transfer is equal to the net workdone. But this does not specify the direction in which the heat and work flows. It also does not give any condition under which these transfers take place. According to the second statement of the first law, the heat energy and mechanical work are mutually convertible. Though the mechanical energy can be fully converted into heat energy, but only a part of heat energy is converted into mechanical work. The second law of thermodynamics put the restriction that the processes proceed in a certain direction and not in the opposite direction. According to the second law, the whole of heat energy can not be converted into work and part of the energy, therefore, must be rejected to the surroundings. In other words, the energy is degraded in the process of producing mechanical energy from the heat supplied. Thus, the second law of thermodynamics provides the means for measuring the energy degradation, so it is also known as law of degradation of energy.

6.2 STATEMENTS OF SECOND LAW OF THERMODYNAMICS Following are the two common statements of the second law of thermodynamics: 1. Kelvin-Planck statement. The Kelvin-Planck statement of the second law of thermodynamics is as follows: “It is impossible to construct an engine which, while working in a cycle, produces no effect other than to extract heat from a single thermal reservoir and perform an equivalent work.” This statement relates to a heat engine and it means that the complete conversion of heat energy into work is impossible. In other words, the *efficiency of a heat engine cannot be 100 percent.

* The efficiency of an engine is defined as the ratio of the output (workdone) to the input (heat supplied).

Second Law of Thermodynamics 117 2. Clausius Statement. The clausius statement of the second law of thermodynamics is as follows: “It is impossible to construct an engine which, while working in a cycle, produces no effect other than to transfer heat from a low temperature reservoir to a high temperature reservoir without the aid of an external agency.” This statement relates to a heat pump or refrigerator and it means that external work is required to pump heat from a low temperature reservoir to a high temperature reservoir. Alternately, it may be stated that the coefficient of performance of a heat pump or a refrigerator is finite. Though the above two statements appear to be different, yet they are equivalent. It may be proved that violation of either statement leads to violation of the other.

6.3 EQUIVALENCE OF KELVIN-PLANCK AND CLAUSIUS STATEMENTS The Kelvin Planck and Clausius statements are two parallel statements of the second law of thermodynamics, and are equivalent is all respects. The equivalence of the two statements is established by showing that any device that violates one statement leads to the violation of second statement and vice versa. This is discussed as follows: 1. Violation of Kelvin Planck statement leads to violation of Clausius statement. Consider that a heat engine or a 100 percent efficient machine (PMM-II) is constructed which operates from a single heat reservoir (source) at temperature T1. Let it receives heat Q1 from this reservoir and converts it completely into work (i.e. W = Q1) without rejecting any heat (i.e. Q2 = 0) to the low temperature reservoir at temperature T2, as shown in Fig. 6.1 (a). This is in violation of Kelvin Planck statement. This work W is used to drive a heat pump or refrigerator. The heat pump receives heat Q2 from the low temperature reservoir at temperature T2 and supply it along with Q1 (i.e. total Q1 + Q2) to the high temperature reservoir. The heat pump or refrigerator operates in confirmity with Clausius statement. Reservoir at T1

Reservoir at T1

Q1 Heat engine

Q1 + Q2 W = Q1 E

Heat pump

P

PMM-II

Composite system

W=0 Q2

Q2 = 0

Q2

Reservoir at T2

Reservoir at T2

(a )

(b )

Fig. 6.1. Violation of Kelvin-Planck statement leads to violation of Clausius statement.

Now consider that the heat engine and the heat pump (or refrigerator) are coupled together to form a single system, as shown in Fig. 6.1 (b). This single system, operating in a cycle, constitutes a device which has no effect on the surrounding other than the transfer of heat from the low temperature reservoir at T2 to the high temperature reservoir at T1 without any work input from an external source. This is in violation of Clausius statement. Thus the violation of Kelvin-Planck statement violates the Clausius statement.

118 Engineering Thermodynamics 2. Violation of Clausius statement leads to violation of Kelvin-Planck statement Consider a heat pump or refrigerator that operates in a cycle and transfers heat (Q2) from a low temperature reservoir at temperature T2 to a high temperature reservoir at temperature T1, without any work input from an external agency (surroundings), as shown in Fig. 6.2 (a). This is in violation of Clausius statement. Reservoir at T1

Reservoir at T1

W=0

Q1

Q1

Q1 Heat pump

Heat engine P

W = Q1 – Q2

W = Q1 – Q2

E

Composite system Q2

Q2

Reservoir at T2 (a )

(b )

Fig. 6.2. Violation of Clausius statement leads to violation of Kelvin-Planck statement.

Now consider that a heat engine is working between the same two reservoirs in such a way that it rejects heat Q2 to a low temperature reservoir and takes heat Q1 from the high temperature reservoir doing external work, W = Q1 – Q2. It is in accordance with the Kelvin-Planck statement. Fig. 6.2 (b) shows, when the two systems (i.e. heat pump or refrigerator and heat engine) are coupled together. This composite system constitutes a self-acting machine, that receives (Q1 – Q2) heat from the high temperature reservoir and converts it completely into an equal amount of work, W = Q1 – Q2, without rejecting any heat to the low temperature reservoir. This operation of the composite system violates the Kelvin-Planck statement. Thus violation of Clausius statement leads to the violation of Kelvin-Planck statement.

6.4 PERPETUAL MOTION MACHINE OF SECOND KIND (PMM-II) A heat engine which violates the Kelvin-Planck statement of the second law of thermodynamics (a heat engine which converts whole of the heat energy into mechanical work) is called a perpetual motion machine of the second kind (PMM-II). It is impossible to construct such a machine (i.e. 100 percent efficient machine) because, in actual practice, no machine can convert whole of the heat energy supplied to it, into its equivalent amount of work.

6.5 IMPORTANT TERMS The following are the important terms widely used in connection with second law of thermodynamics: 1. Thermal or heat reservoir. The thermal or heat reservoir is a system or body of extremely large heat capacity, capable of absorbing or rejecting finite amount of heat without any change in temperature. For example, atmosphere, rivers, seas are reservoirs from which we can extract or dump any amount of heat without changing its temperature.

Second Law of Thermodynamics 119 2. Heat source. The heat source (or simply known as source) is a reservoir at a higher temperature from which heat is extracted without change of its temperature. The examples of a heat source are sun, boiler furnaces, combustion chamber, nuclear reactor etc. 3. Heat sink. The heat sink (or simply known as sink) is a reservoir at a low temperature. It is capable of absorbing any amount of heat without change in its temperature. The examples of a heat sink are atmosphere or surroundings. 4. Heat engine. A heat engine is a thermodynamic system used for converting heat into work, while operating in a cycle between the high temperature source and a low temperature sink. The principle of a heat engine is shown in Fig. 6.3. Let the heat energy (Q1) is supplied to the system (heat engine) from the heat source at temperature T1. Some of the heat energy (Q2) is rejected to the sink and the remaining heat (Q1 – Q2) is converted into workdone by the system. The performance of a heat engine is measured by the efficiency which may be defined as the ratio of the net workdone by the system (or output of the engine) to the total heat supplied. Mathematically, efficiency of a heat engine, hE =

Since for a reversible engine,

W Q − Q2 Q Net workdone = E = 1 =1− 2 Q1 Q1 Total heat supplied Q1

Q2 T = 2 , therefore the above expression may be written as, Q1 T1

hE = 1 −

Source T1

Q2 T = 1− 2 Q1 T1 High temperature body T1 > T2

T1

T1 > T2

Q1 = Q2 + WP Q1

Heat pump

System

WP

P (Heat engine)

E WE = Q1 – Q2 Q2 Q2 T2

Sink T2

Fig. 6.3. Principle of a heat engine.

Atmosphere

Low temperature body

Fig. 6.4. Principle of a heat pump.

5. Heat pump. A heat pump is a thermodynamic system used for extracting heat from a low temperature body and delivers it to a high temperature body. The principle of a heat pump is shown in Fig. 6.4. Let the heat energy Q2 from the atmosphere or surrounding at a low temperature T2 and supply heat Q1 to a body at a higher temperature T1 by using external energy in the form of work input (WP). The performance of a heat pump is measured by the coefficient of performance (briefly written as C.O.P.) which is defined as the ratio of heat supplied to the external work supplied (work input) to obtain the desired effect. Thus, coefficient of performance [also known as energy performance ratio (EPR)] is given by

120 Engineering Thermodynamics

(C.O.P.)P =

=

Q Heat supplied = 1 External work supplied WP Q1 T1 = T1 − T2 Q1 − Q2

... ( WP = Q1 – Q2)

Notes: 1. The heat pump is used to keep the rooms warm is winter. 2. The C.O.P. is the reciprocal of efficiency of a heat engine. It is, thus obvious that the value of C.O.P. is always greater than unity.

6. Refrigerator. A refrigerator is a device similar to a heat pump (i.e. reversed heat engine) which extracts heat from a low temperature body and delivers it to a high temperature body. The principle of a refrigerator is shown in Fig. 6.5. Let the heat Q2 is extracted from the space to be cooled (i.e. refrigerated space) at temperature T2 which is lower than the temperature of the surroundings and reject heat Q1 to the surroundings at temperature T1 which is equal to the atmospheric temperature. The refrigerator, while operating in a cyclic process, require an input work (WR) to transfer heat from a lower temperature to a higher temperature. The performance of a refrigerator is also measured by the coefficient of performance (C.O.P.). In this case, it is defined as the ratio of the *heat extracted (Q2) to the work input (WR). Fig. 6.5. Principle of a refrigerator. Thus, the coefficient of performance of refrigerator is given by Q Q2 Heat extracted (C.O.P.)R = = 2 = W Work input Q1 − Q2 R

...(According to first law of thermodynamics, WR = Q1 – Q2)

T = 2 T1 − T2 Adding 1 to both sides, we have or

(C.O.P.)R + 1 =

T2 T + T1 − T2 T1 +1 = 2 = = (C.O.P.)P T1 − T2 T1 − T2 T1 − T2

(C.O.P.)P = (C.O.P.)R + 1

6.6 CARNOT CYCLE This cycle was suggested by a French engineer Nicolas Leonard Sadi Carnot in 1824, which works on reversible cycle. The Carnot cycle consists of an alternate series of two reversible isothermal processes and two reversible adiabatic processes, as shown on the p-v and T-s diagram in Fig. 6.6 (a) and (b) respectively. The Carnot engine has air as it working substance, enclosed in a cylinder in which a frictionless piston moves. The walls of the cylinder and piston are insulated. The cylinder head can be covered

* The amount of heat extracted (Q2) is also termed as refrigerating effect produced or capacity of a refrigerator.

Second Law of Thermodynamics 121 with an insulating cap. The engine works between the hot body and cold body. It consists of the following four processes: 1. Process 1-2 (Isothermal expansion). The hot body is brought in contact with the head of the cylinder. The air expands at constant temperature (isothermally) T1 = T2 from volume v1 to v2. The heat supplied to the air is utilised in doing the external work. The curve 1-2 on the p-v diagram represents the isothermal expansion. According to the first law of thermodynamics, Q1–2 = dU + W1–2 Since the temperature is constant, therefore there is no change in internal energy, i.e. dU = 0, or Q1–2 = W1–2 1 T1 = T2 2

Pressure

p2

p4

4 3

p3

v1

v4

v2

T3 = T4

v3

3

4 s1 = s4

s2 = s3

(b ) T-s diagram.

(a ) p-v diagram.

Piston

Insulating cover 1 Cold body

2

Entropy

Volume Hot body

1

Temperature

p1

3 Cylinder

Fig. 6.6. Carnot cycle.

We know that work done by the air during isothermal expansion process 1–2, where

v W1–2 = 2.3 p1v1 log 2 = 2.3 mRT1 log re v1 v re = 2 = Expansion ratio. v1

... ( p1v1 = mRT1)

\ Heat supplied to the air during isothermal expansion process 1–2, Q1–2 = W1–2 = 2.3 p1v1 log (re ) = 2.3 mRT1 log (re ) 2. Process 2–3 (Reversible adiabatic expansion or isentropic expansion). The hot body is removed from the cylinder head and the insulating cover is put on. Thus no heat transfer takes place (i.e. Q2–3 = 0). The air expands reversibly and adiabatically, from volume v2 to v3. The temperature drops from T2 to T3. The curve 2–3 on the p-v diagram represents the reversible adiabatic (or isentropic expansion). According to the first law of thermodynamics, Q2–3 = dU + W2–3 or – dU = W2–3 ...( Q2–3 = 0) The – ve sign indicates that there is a decrease in internal energy.

122 Engineering Thermodynamics 2–3,

We know that work done by the air during reversible adiabatic (or isentropic) expansion process

W2–3 =

=

p2 v2 − p3v3 mRT2 − mRT3 = γ −1 γ −1 m R (T2 − T3 ) m R (T1 − T3 ) = γ −1 γ −1

... ( T2 = T1)

\ Decrease in internal energy,

dU = W2–3 =

m R (T1 − T3 ) γ −1

3. Process 3–4 (Isothermal compression). The insulating cover is removed and the cold body is brought in contact with the head of the cylinder. The air is compressed at constant temperature (T3 = T4) from volume v3 to v4. The heat rejected by the air to the cold body is equal to the workdone on the air. The curve 3–4 on the p-v diagram represents the isothermal compression. According to the first law of thermodynamics, Q3–4 = dU – W3–4 The –ve sign indicates that during compression, work is done on the air. Since the temperature is constant, therefore there is no change in internal energy, i.e. dU = 0, or Q3–4 = – W3–4 We know that workdone on the air during isothermal compression process 3–4, where

v W3–4 = 2.3 p3v3 log 3 = 2.3 mRT3 log (rc ) ... ( p3v3 = mRT3) v4 v rc = 3 = Compression ratio. v4

\ Heat rejected by the air during isothermal compression process 3–4, Q3–4 = W3–4 = 2.3 p3v3 log (rc ) = 2.3 mRT3 log (rc ) 4. Process 4–1 (Reversible adiabatic compression or isentropic compression). The cold body is removed from the cylinder head and the insulating cover is put on. Thus no heat transfer takes place (i.e. Q4–1 = 0). The air compresses reversibly and adiabatically from volume v4 to v1. The temperature increases from T4 to T1. The curve 4–1 on the p-v diagram represents the reversible adiabatic compression or isentropic compression. According to first law of thermodynamics, Q4–1 = dU – W4–1 The –ve sign indicates that during compression, work is done on the air. Since no heat transfer takes place, therefore Q4–1 = 0, or dU = W4–1 We know that workdone on the air during reversible adiabatic or isentropic compression process 4–1, p v − p1v1 m RT4 − m RT1 = W4–1 = 4 4 γ −1 γ −1 =

m RT3 − m RT1 m R(T3 − T1 ) = γ −1 γ −1

... ( T4 = T3)

Second Law of Thermodynamics 123 \ Increase in internal energy, m R (T3 − T1 ) γ −1

dU = W4–1 =

6.7 EFFICIENCY OF CARNOT CYCLE The efficiency of the Carnot cycle is defined as the ratio of the workdone per cycle to the heat supplied per cycle. Mathematically, efficiency, Workdone per cycle h = Heat supplied per cycle We know that the workdone per cycle, W = W1 − 2 + W2 −3 + W3− 4 + W4 −1 = 2.3 mRT1 log (re ) +

m R(T1 − T3 ) m R(T3 − T1 ) – 2.3 mRT3 log (rc ) + γ −1 γ −1

= 2.3 mRT1 log (re ) +

m R (T1 − T3 ) m R (T1 − T3 ) – 2.3 mRT3 log (rc ) – γ −1 γ −1

= 2.3 mRT1 log (re ) – 2.3 mRT3 log (rc ) = Heat supplied – Heat rejected The expansion ratio (re ) and the compression ratio (rc ) must be equal, otherwise the cycle will not close. Taking re = rc = r, we have W = Heat supplied – Heat rejected = 2.3 mRT1 log r – 2.3 mRT3 log r = 2.3 mR log r (T1 – T3) 2.3 m R log r (T1 − T3 ) Workdone (W ) = \ Efficiency, h = 2.3 m RT1 log r Heat supplied (Q1 − 2 ) =

T1 − T3 T = 1− 3 T1 T1

Notes: 1. We know that for the process 2–3 (Reversible adiabatic or isentropic expansion), v T2 = 3 T3 v2

γ −1

or

T1 v3 = T3 v2

γ −1

... ( T2 = T1)

Similarly, for the process 4–1 (Reversible adiabatic or isentropic compression)

v T1 = 4 T4 v1

γ −1

γ −1

γ−1

γ −1

or

T1 v4 = T3 v1

or

v3 v v v = 4 or 2 = 3 = r v2 v1 v1 v4

From above, we have where \

v3 v2

v = 4 v1

r = Ratio of expansion or compression. T1 T 1 = r or 3 = T1 r T3

... ( T4 = T3)

124 Engineering Thermodynamics We know that efficiency of the Carnot cycle, 1 T h = 1 − 3 = 1 − γ − 1 T1 (r ) Since T3 is less than T1, therefore efficiency is less than 100%. The efficiency increases when T1 is increased or T3 is decreased.

6.8 PRACTICAL DIFFICULTIES (or IMPRACTICABILITY) OF CARNOT CYCLE Though the Carnot cycle is the most efficient, yet it has some practical difficulties as mentioned below: 1. All the four processes are not reversible due to internal friction between the piston and cylinder. 2. During isothermal process, the piston must move very slow and during adiabatic process, it should move very fast. This is practically not possible to have such variation of speed during the same cycle. 3. The isothermal heat addition is not possible in actual practice because the temperature is bound to increase. 4. It is impossible to transfer the heat without finite temperature difference. It is thus obvious that in actual practice it is impossible to realise Carnot’s engine. But it is used as a ultimate standard of comparison of all heat engines.

6.9 CARNOT’S THEOREM Source (T1) According to Carnot’s theorem ‘No heat engine operating in a cycle between two constant temperature reservoirs can be more efficient than a reversible engine Q1 Q1 operating between the same two reservoirs.’ Let EI be an irreversible heat engine and ER be any Reversible WI reversible heat engine, operate between the given source heat engine ER EI at temperature T1 and the given sink at temperature T2, WR as shown in Fig. 6.7. Irreversible We have to prove that the efficiency of a reversible heat engine Q2 Q2 engine (hR) is more than the efficiency of an irreversible engine (hI), i.e. hR > hI. Sink (T2) Assume that this is not true and let hI > hR. Let the irreversible engine EI absorbs heat Q1 from Fig. 6.7. Carnot’s theorem. the high temperature reservoir or source at temperature T1 and rejects heat Q2 to the low temperature reservoir or sink at temperature T2. Similarly, let the reversible engine ER absorbs heat Q1 from the high temperature reservoir or source at temperature T1 and rejects heat Q2 to the low temperature reservoir or sink at temperature T2. Let WI and WR be the work produced by the engines EI and ER respectively. Since we have assumed that hI > hR

or

WI W > R Q1′ Q1

Let

Q1 = Q1 = Q

\

WI > WR

Second Law of Thermodynamics 125 Now consider that the reversible engine ER is reversed so that it absorbs heat Q2 from the low temperature reservoir and reject heat Q1 to high temperature reservoir, as shown in Fig. 6.8. The values of heat (Q1 and Q2) and work transfer (WR) remain same. Since WI > WR, therefore some part of WI (equal to WR) may be fed to drive the reversed heat engine ER. Since Q1 = Q1 = Q, therefore the heat Q1 discharged by engine ER may be supplied to engine EI. The heat source may, therefore, be eliminated, as shown in Fig. 6.9. The engines EI and ER together constitute a heat engine operating on a single reservoir (sink) at temperature T2 and produces a net work (WI – WR) as WI > WR. This violates the Kelvin-Planck statement of the second law of thermodynamics. Thus, the assumption that the efficiency of an irreversible engine (EI) is more than the efficiency of a reversible engine (i.e. hI > hR) working between the same temperature limits is wrong. Therefore, the efficiency of a reversible engine operating between two constant temperatures is maximum of all others, operating between the same temperature limits, i.e. hR > hI. Source (T1)

Q1

Q1 WI

WR

EI

Q2

Sink (T2)

Q = Q1

Q = Q1

ER

EI

Q2

Q2

WR

WI – WR

ER

Q2

Sink (T2)

Fig. 6.8. Heat engine ER is reversed.

WI

Fig. 6.9. Combined engines EI and ER. Heat source eliminated.

6.10 COROLLARIES OF CARNOT’S THEOREM Following are the two important statements about the reversible cycles. The statements are also called corollaries of the Carnot’s theorem. Corollary 1. The efficiency of all reversible heat engines operating between the two constant temperature reservoirs have the same efficiency. Consider that both the engines EI and ER, as shown in Fig. 6.7, are reversible. Let us assume that the efficiency of the engine EI (hI) is greater than the efficiency of the engine ER (hR). Proceeding in the similar way as discussed in the precious article (Art. 6.9), if the engine ER is reversed to run as a heat pump using some part of the work output (WI) of the engine EI, then the combined system of the heat engine EI and heat pump ER, becomes a perpetual motion machine of the second kind (PMM–II). Thus, efficiency of the engine EI (hI) can not be more than efficiency of the engine ER (hR). Therefore hI = hR. Corollary 2. The efficiency of any reversible heat engine operating between the two constant temperature reservoirs is independent of the nature of the working substance undergoing the cycle and depends solely on the temperature of the reservoirs. Example 6.1. In a heat engine, the temperature of the source and sink are 700°C and 50°C respectively. The heat supplied is 5 MJ/min. Find the power developed by the engine. Solution. Given. Source temperature, T1 = 700°C = 700 + 273 = 973 K

126 Engineering Thermodynamics Sink temperature, Heat supplied,

T3 = 50°C = 50 + 273 = 323 K 5 × 1000 Q1 = 5 MJ/min = = 83.33 kJ/s 60

We know that efficiency of the engine, T − T3 973 − 323 = hE = 1 = 0.668 T1 973 We also know that efficiency of the engine, Workdone hE = Heat Supplied \ Workdone or power developed by the engine, WE = hE × Heat supplied = 0.668 × 83.33 = 55.67 kJ/s or kW Ans. Example 6.2. A refrigerating machine works on a reversed Carnot cycle. It consumes 6 kW and the refrigerating effect is 1000 kJ/min. The sink temperature is – 40°C. Determine: 1. Source temperature; and 2. C.O.P. of the refrigerating machine. Solution. Given: Power consumed by the refrigerating machine or work input, WR = 6 kW = 6 kJ/s = 6 × 60 = 360 kJ/min Refrigerating effect or heat extraction capacity of the refrigerator, Q2 = 1000 kJ/min Sink temperature, T2 = – 40°C = – 40 + 273 = 233 K Source T1 The system is shown in Fig. 6.10. 1. Source temperature Let T1 = Source temperature. Q1 We know that coefficient of performance of refrigerating machine, Q 1000 WR = 6 kW (C.O.P.)R = 2 = = 2.78 R 360 WR We also know that or

(C.O.P.)R = 2.78 = T1–233 =

T2 T1 − T2 233 T1 − 233 233 = 83.8 2.78

Q2 = 1000 kJ/min

Sink T2 = 233 K

Fig. 6.10

\ T1 = 233 + 83.8 = 316.8 K = 316.8 – 273 = 43.8°C Ans. 2. C.O.P. of the refrigerating machine We have calculated above that C.O.P. of the refrigerating machine is (C.O.P.)R = 2.78 Ans. Example 6.3. An inventor claims to have developed a refrigerating machine which operates between – 20°C and 30°C and consumes 1 kW of power. The machine gives a refrigerating effect of 21.6 MJ/h. Comment on the claim of the inventor.

Second Law of Thermodynamics 127 Solution. Given: Sink temperature, T2 = – 20°C = – 20 + 273 = 253 K Source temperature, T1 = 30°C = 30 + 273 = 303 K Power consumed, WR = 1 kW = 1 kJ/s Refrigerating effect or heat extraction capacity of the refrigerating machine, Q2 = 21.6 MJ/h =

T1 = 303 K

Q1

21.6 × 103 = 6 kJ /s = 6 kW 3600

R

The system is shown in Fig. 6.11. We know that coefficient of performance of the refrigerating machine, T2 253 = (C.O.P.)R = = 5.06 T1 − T2 303 − 253

Q2 = 21.6 MJ/ h

T2 = 253 K

According to the inventor,

WR = 1 kW

(C.O.P.)R =

Q2 6 = =6 WR 1

Fig. 6.11

Since this is greater than 5.06, therefore the inventor’s claim is false. Ans. Example 6.4. A reversible heat engine operates with two environments. In the first, it draws 12 000 kW from a source at 400°C and in the second, it draws 25 000 kW from a source at 100°C. In both the operations, the engine rejects heat to a thermal sink at 20°C. Determine the operation in which the engine delivers more power. Solution. First Second Given: Heat drawn during first operation, Source Source T1 T3 Q1 = 12 000 kW = 12 000 kJ/s Temperature of first source, Q1 Q3 T1 = 400°C = 400 + 273 = 673 K Heat drawn during second operation, WE1 WE2 E E Q3 = 25 000 kW = 25 000 kJ/s Temperature of second source, Q2 T3 = 100°C = 100 + 273 = 373 K Q4 Temperature of the sink T2 T4 T2 = T4 = 20°C = 20 + 273 = 293 K Sink Sink The system is shown in Fig. 6.12. (a ) (b ) We know that efficiency of the engine, during first Fig. 6.12 operation, W T − T2 673 − 293 = hE1 = E1 = 1 = 0.5646 673 Q1 T1 \ Work obtained by the engine from the first source, WE1 = hE1 × Q1 = 0.5646 × 12 000 = 6775 kJ/s Similarly, efficiency of the engine during second operation,

hE2 =

WE2 T3 − T4 373 − 293 = = = 0.2145 373 Q3 T3

128 Engineering Thermodynamics \ Work obtained by the engine from the second source, WE2 = hE2 × Q3 = 0.2145 × 25 000 = 5362 kJ/ s Since WE1 is greater then WE2, therefore the first operation delivers more power. Ans. Example 6.5. A reversible engine receives heat from a reservoir at 700°C and rejects heat at a temperature T2 . A second reversible engine receives the heat rejected by the first engine and rejects to a sink at a temperature of 37°C. Calculate the temperature T2 for (a) equal efficiency of both the engines; and (b) equal work output of both the engines. Solution. Reservoir Given: Temperature of reservoir, T1 = 973 K T1 = 700°C = 700 + 273 = 973 K Temperature of sink, T4 = 37°C = 37 + 273 = 310 K Q1 The system is shown in Fig. 6.13. (a) Temperature T2 for equal efficiency of both the engines E1 T2 WE1 We know that efficiency of the first engine (E1), T − T2 973 − T2 Q2 = Q1 – WE1 = hE1 = 1 ...(i) T1 973 Similarly, efficiency of the second engine (E2), T − T4 T2 − T4 T2 − 310 = = hE2 = 3 ...(ii) T3 T2 T2 For equal efficiency, equating equations (i) and (ii), 973 − T2 T − 310 = 2 973 T2 or \

...( T3 = T2)

T3

E2

WE2 Q3 = Q2 – WE2

T4 = 310 K Sink Fig. 6.13

T2 (973 – T2) = 973(T2 – 310) 973T2 – T22 = 973T2 – 973 × 310

T2 = 973 × 310 = 549.2 K Ans.

(b) Temperature T2 for equal work output of both the engines We know that work output of the first engine (WE1) is directly proportional to (T1 – T2), i.e. WE1 T1 – T2 or WE1 = K (T1 – T2) ....(iii) where K is a constant of proportionality. Similarly, work output of the second engine, WE2 = K (T3 – T4) ...(iv) For equal work output, equating equations (iii) and (iv), K (T1 – T2) = K (T3 – T4) 973 – T2 = T2 – 310 973 + 310 \ T2 = = 641.5 K Ans. 2 Example 6.6. A house is to be maintained at 25°C in summer as well as winter. For this purpose, it is proposed to use a reversible device as a refrigerator in summer and a heat pump in winter. The ambient temperature is 40°C in summer and 3°C in winter. The energy losses as heat from the roof

Second Law of Thermodynamics 129 and the walls are estimated at 5 kW per degree celsius temperature difference between the room and the ambient conditions. Calculate the power required to operate the device in summer and winter. Solution. Given: Temperature of house to be maintained in summer and winter, T1 = 313 K T3 = 298 K T2 = T3 = 25°C = 25 + 273 = 298 K Ambient temperature in summer, Q1 Q3 TS = 40°C WR Ambient temperature in winter, R P WP TW = 3°C Heat loss, Q2 = Q3 = 5 kW/°C temperature difference Q2 Q4 Power required to operate the device in summer In summer, a reversible device as a refrigerator is T2 = 298 K T4 = 276 K used, as shown in Fig. 6.14 (a). We know that coefficient of performance of a (a ) In summer as (b ) In winter as referigerator. heat pump. refrigerator, Fig. 6.14 T 298 (C.O.P.)R = 2 = = 19.87 T1 − T2 313 − 298 and heat loss from the roof and walls in summer, Q2 = 5 kW/°C temperature difference = 5(TS – T2) = 5(40 – 25) = 75 kW = 75 kJ/s Now using the relation, Heat supplied or Heat loss (Q2 ) (C.O.P.)R = Workdone or power required \ Power required to operate the device Q2 75 = = = 3.77 kJ/ s = 3.77 kW Ans. (C.O.P.) R 19.87 Power required to operate the device in winter In winter, a reversible device as a heat pump is used, as shown in Fig. 6.14 (b). We know that coefficient of performance of a heat pump, T3 298 = (C.O.P.)P = = 13.55 T3 − T4 298 − 276 and heat loss from the roof and walls in winter, Q3 = 5 kW/°C temperature difference = 5 (T2 – TW) = 5 (25 – 3) = 110 kW = 110 kJ/s Now using the relation, Heat loss (Q3 ) (C.O.P.)P = Workdone or power required \ Power required to operate the device Q3 110 = = = 8.12 kJ/ s = 8.12 kW Ans. (C.O.P.) P 13.55

130 Engineering Thermodynamics Example 6.7. A reversible heat engine receives heat from two thermal reservoirs maintained at constant temperatures of 750 K and 500 K. The engine develops 100 kW and rejects 3600 kJ / min of heat to a heat sink at 250 K. Determine the heat supplied by each thermal reservoir and thermal efficiency of the engine. First Second Solution. reservoir reservoir T1 T3 Given: Temperature of first reservoir, = 750 K = 500 K T1 = 750 K Temperature of second reservoir, Q1 Q3 T3 = 500 K Work or power developed by the engine, WE WE = 100 kW = 100 kJ/s Heat engine Heat rejected to sink, Q2 + Q4 = 3600 kJ/min = 60 kJ/s Q2 Q4 Temperature of sink, T2 = T4 = 250 K The system is shown in Fig. 6.15. Sink Heat supplied by each thermal reservoir T2 = T4 = 250 K Let Q1 and Q3 = Heat supplied by the Fig. 6.15 first and second reservoir respectively. When heat is supplied by the first reservoir, then efficiency of the heat engine, W Q − Q2 T1 − T2 750 − 250 = hE1 = E1 = 1 = = 0.67 750 T1 Q1 Q1 \ Work obtained by the engine from the first reservoir, WE1 = Q1 – Q2 = hE1 . Q1 = 0.67 Q1 and heat rejected to the sink, Q2 = Q1 – WE1 = Q1 – 0.67Q1 = 0.33Q1 Similarly, when heat is supplied by the second reservoir, then efficiency of the engine, W Q − Q4 T3 − T4 500 − 250 = hE2 = E3 = 3 = = 0.5 Q3 Q3 500 T3 \ Work obtained by the engine from the second reservoir, WE3 = Q3 – Q4 = hE2 . Q3 = 0.5Q3 and heat rejected to the sink, Q4 = Q3 – WE3 = Q3 – 0.5Q3 = 0.5Q3 We know that total work developed by the engine (WE), 100 = WE1 + WE3 = 0.67Q1 + 0.5Q3 and total heat rejected to the sink, 60 = Q2 + Q4 = 0.33Q1 + 0.5Q3 Subtracting equation (ii) from equation (i), 40 = 0.34Q1 or Q1 = 40/0.34 = 117.65 kJ Ans. Substituting the valve of Q1 is equation (i) we have 100 = 0.67 × 117.65 + 0.5Q3 = 78.8 + 0.5Q3 100 − 78.8 \ Q3 = = 42.4 kJ Ans. 0.5

... (i) ... (ii)

Second Law of Thermodynamics 131 Thermal efficiency of the engine We know that thermal efficiency of the engine, WE Work obtained 100 = hE = = = 0.625 or 62.5 % Ans. Heat supplied Q1 + Q3 117.65 + 42.4 Example. 6.8. A refrigerator working between 25°C and – 15°C is driven by an electric motor rated at 2 kW. What quantity of ice at 0°C can be obtained from water at 20°C in 10 hours. Take cp for ice = 2.1 kJ/kg K and cp for water = 4.2 kJ/kg K. Solution. Given: Source temperature, T1 = 25°C = 25 + 273 = 298 K Sink temperature, T2 = – 15°C = – 15 + 273 = 258 K Power of motor or work input, WR = 2 kW = 2 kJ/s Specific heat at constant pressure for ice, cpi = 2.1 kJ/ kg K Specific heat at constant pressure for water, cpw = 4.2 kJ / kg K Temperature of water, Tw = 20°C = 20 + 273 = 293 K Temperature of ice, Ti = 0°C = 0 + 273 = 273 K We know that heat removed from 1 kg of water at 20°C (293 K) to produce 1 kg of ice at 0°C (273 K) is Q = m cpw (Tw – Ti ) + Latent heat of ice = 1 × 4.2 (293 – 273) + 335 = 419 kJ/ kg ... ( Latent heat of ice = 335 kJ/kg) and coefficient of performance of the refrigerator, T2 258 = (C.O.P.)R = = 6.45 T1 − T2 298 − 258 \ Total heat removed, Q2 = (C.O.P.)R × WR = 6.45 × 2 = 12.9 kJ/s = 12.9 × 3600 = 46 440 kJ / h We know that ice produced per hour Total heat removed per hour 46 440 = = = 110.8 kg / h Heat removed per kg 419 and ice produced in 10 hours = 110.8 × 10 = 1108 kg Ans. Example 6.9. A heat engine operating between two reservoirs at 1000 K and 350 K is used to drive a heat pump which extracts heat from the reservoir at 350 K, at a rate twice that at which the engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and the coefficient of performance of the heat pump is 50% of the maximum possible, make calculations for the temperature of the reservoir to which the heat pump rejects heat. Also work out the rate of heat rejection from the heat pump, if the rate of supply of heat to the engine is 50 kW. Solution. Given: Temperature of first reservoir (source), T1 = 1000 K Temperature of second reservoir (sink), T2 = 350 K

132 Engineering Thermodynamics Efficiency of the engine, hE = 40% of hmax Coefficient of performance of heat pump, (C.O.P.)P = 50% of (C.O.P.)max Heat supplied to the engine, Q1 = 50 kW = 50 kJ/s The system is shown in Fig. 6.16. Heat rejected by the heat pump We know that maximum efficiency of the heat engine, T 350 hmax = 1 − 2 = 1 − = 0.65 T1 1000

First reservoir T1 = 1000 K

T3

Q4 = Q3 + WP

Q1 WE

E

WP

Heat pump

P

Q1 – Q2

Heat engine

Q3 = 2Q2

Q2

T = 350 K

2 Since the efficiency of the engine is 40% of the maximum efficiency, therefore actual efficiency of Second reservoir the heat engine, Fig. 6.16 hE = 40% of hmax = 0.4 × 0.65 = 0.26 We also know that actual efficiency of the heat engine, Q − Q2 Heat supplied − Heat rejected Workdone = hE = = 1 Heat supplied Q1 Heat supplied \ Workdone by the heat engine, WE = Q1 – Q2 = hE × Q1 = 0.26 × 50 = 13 kJ and heat rejected by the heat engine, Q2 = Q1 – WE = 50 – 13 = 37 kJ We know that heat removed by the heat pump, Q3 = 2Q2 = 2 × 37 = 74 kJ and heat rejected by the heat pump, Q4 = Q3 + WP = 74 + 13 = 87 kJ Ans. ... ( WP = WE) Temperature of the reservoir to which the heat pump rejects heat Let T3 = Temperature of the reservoir to which the heat pump rejects heat. We know that maximum coefficient of performance of the heat pump,

(C.O.P.)max =

Q Heat supplied 87 = 4 = = 6.7 External work supplied 13 WP

Since the actual C.O.P. of the heat pump is 50% of the maximum C.O.P., therefore (C.O.P.)P = 50% of (C.O.P.)max = 0.5 × 6.7 = 3.35 T3 We know that (C.O.P.)P = T3 − T2 or \

3.35 =

T3 T3 − 350

3.35 (T3 – 300) = T3 or 3.35 T3 – T3 = 3.35 × 350 3.35 × 350 T3 = = 499 K Ans. 2.35

Second Law of Thermodynamics 133 Example 6.10. A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C and delivers heat to a reservoir at 60°C. The heat pump is driven by a reversible heat engine which takes in heat from a reservoir at 840°C and rejects heat to a reservoir at 60°C. The reversible heat engine also drives a machine and absorbs 30 kW. If the heat pump extracts 17 kJ/s from the 5°C reservoir, determine the rate of heat supply from the 840°C source. Solution. Given: Temperature at which heat is taken by a heat pump, T3 = 5°C = 5 + 273 = 278 K Temperature at which heat is delivered from the heat pump T2 = 60°C = 60 + 273 = 333 K Temperature at which heat is taken by a heat engine, T1 = 840°C = 840 + 273 = 1113 K Temperature at which heat is rejected by a heat engine T2 = 60°C = 60 + 273 = 333 K Net work output, W = WE – WP = 30 kW = 30 kJ/s Heat extracted by heat pump, Q4 = 17 kJ/ s The combined heat engine and heat pump system is shown in Fig. 6.17. T3 = 278 K

T1 = 1113 K

Q4 = 17 kJ/s

Q1

E Heat engine

WE

WP

30 kW Q2 = Q1 – WE

Heat pump P

Q3 = Q4 + WP

T2 = 333 K Reservoir Fig. 6.17

We know that efficiency of the heat engine, T 333 hE = 1 − 2 = 1 − = 1 – 0.28 = 0.72 1113 T1 Let Q1 = Heat supplied from the reservoir at 840°C. The efficiency of the heat engine is also given by Work done (WE ) Q − Q2 = 1 hE = Q1 Heat supplied (Q1 ) \ Workdone, WE = Q1 – Q2 = hE × Q1 = 0.72 Q1 ...(i) Since the net workput of the combined heat engine and heat pump is W = WE – WP = 30 kJ/s, therefore work required for the heat pump, WP = WE – W = 0.72 Q1 – 30 ...(ii)

134 Engineering Thermodynamics We know that C.O.P. of a heat pump,

(C.O.P.)P =

T2 333 = = 6.05 333 − 278 T2 − T3

We also know that C.O.P. of a heat pump, Q Heat supplied = 4 (C.O.P.)P = Work supplied WP \

WP =

Q4 17 = = 2.81 (C.O.P.) P 6.05

...(iii)

From equations (ii) and (iii), 0.72 Q1 – 30 = 2.81 2.81 + 30 \ Q1 = = 45.6 kJ/s Ans. 0.72 Example 6.11. Two Carnot refrigerators A and B operate in series. The refrigerator A absorbs energy at the rate of 1 kJ/s from a body at temperature 300 K, and rejects energy as heat to a body at a temperature T. The refrigerator B absorbs the same quantity of energy which is rejected by the refrigerator A from the body at temperature T, and rejects energy as heat to a body at temperature 1000 K. If both the refrigerators have the same C.O.P., calculate: 1. The temperature T of the body; 2. The C.O.P. of the refrigerators ; and 3. The rate at which energy is rejected as heat to the body at temperature 1000 K. Solution. Given: Heat absorbed by the refrigerator A, Q1 = 1 kJ/s Temperature at which heat is received by the refrigerator A, T1 = 300 K T4 = 1000 K Temperature of a body at which heat is rejected by the refrigerator B, T4 = 1000 K Q4 = Q3 + WB (C.O.P.)RA = (C.O.P.)RB The system is shown in Fig. 6.18. B WB 1. Temperature T of the body We know that coefficient of performance of the refrigerator A, Q3 T 300 T < T4 (C.O.P.)RA = 1 = ...(i) T T − T1 T − 300 Similarly, coefficient of performance of the refrigerator B,

(C.O.P.)RB =

T T = (ii) T4 − T 1000 − T

Since (C.O.P.)RA = (C.O.P.)RB, therefore equating equations (i) and (ii),

300 T = T − 300 1000 − T

Q2 = Q1 + WA A WA Q1 T1 = 300 K Fig. 6.18 Fig. 6.18

T1 < T

Second Law of Thermodynamics 135 or

300 (1000 – T) = T (T – 300)

30 × 104 – 300 T = T 2 – 300 T T 2 = 30 × 104

4 T = 30 × 10 = 547.72 K Ans.

\ 2. C.O.P. of the refrigerators

We know that C.O.P. of the refrigerator A,

(C.O.P.)RA =

300 300 = = 1.211 Ans. T − 300 547.72 − 300

and C.O.P. of the refrigerator B,

(C.O.P.)RB =

T 547.72 = = 1.211 Ans. 1000 − T 1000 − 547.72

3. Rate at which energy is rejected to a body at temperature 1000 K We know that heat absorbed by the refrigerator A,

Q1 = 1 kJ/s

...(Given)

and work input to the refrigerator A,

WA =

Heat abosrbed (Q1 ) 1 = = 0.826 kJ/ s 1.211 (C.O.P.) RA

\ Heat supplied to a body at temperature T,

Q2 = Q1 + WA = 1 + 0.826 = 1.826 kJ/ s

= Q3, the heat supplied to the refrigerator B Work input to the refrigerator B,

WB =

Heat supplied to refrigerator B (Q3 ) 1.826 = = 1.508 1.211 (C.O.P.) RB

\ Rate at which energy is rejected to a body at temperature 1000 K,

Q4 = Q3 + WB = 1.826 + 1.508 = 3.334 kJ/s Ans.

HIGHLIGHTS 1. The second law of thermodynamics provides the means for measuring the energy degradation, so it is also known as law of degradation of energy. 2. According to Kelvin-Planck statement of the second law of thermodynamics, ‘It is impossible to construct an engine which, while working in a cycle, produces no effect other than to extract heat from a single thermal reservoir and perform an equivalent work.’ 3. According to Clausius statement of the second law of thermodynamics. ‘It is impossible to construct an engine which, while working in a cycle, produces no effect other than to transfer heat from a low temperature reservoir to a high temperature reservoir, without the aid of an external agency.’ 4. Though the Kelvin Planck and Clausius statements appear to be different, yet they are equivalent in the sense that violation of one statement leads to the violation of second statement and vice-versa.

136 Engineering Thermodynamics 5. A machine which violates the Kelvin Planck statement of the second law of thermodynamics is known as perpetual motion machine of the second kind (PMM-II). It is impossible to construct such a machine (i.e. 100 per cent efficient machine). 6. A heat engine is a thermodynamic system used for converting heat into work, while operating in a cycle, between the high temperature source (T1) and a low temperature sink (T2).The efficiency of the heat engine is given by T − T2 T = 1− 2 hE = 1 T1 T1 7. A heat pump is a thermodynamic system used for extracting heat from a low temperature body (T2) and delivers it to a high temperature body (T1). The C.O.P. of a heat pump is given by

(C.O.P.)P =

T1 T1 − T2

8. A refrigerator is a device similar to a heat pump which extracts heat from a low temperature body (T2) and delivers it to a high temperature body (T1). The C.O.P. of a refrigerator is given by

(C.O.P.)R =

T2 T1 − T2

9. The C.O.P. of a heat pump is greater than C.O.P. of a refrigerator by unity, i.e. (C.O.P.)P = (C.O.P.)R + 1 10. The efficiency of Carnot cycle is given by

h = 1 −

T3 1 = 1− T1 ( r ) γ −1

where T3 = Lowest temperature, T1 = Highest temperature, r = Ratio of expansion or compression, and g = Ratio of specific heats or adiabatic index. 11. According to Carnot’s theorem, ‘No heat engine operating in a cycle between two constant temperature reservoirs can be more efficient than the reversible engine operating between the same two reservoirs.’ 12. The efficiency of a reversible engine operating between two constant temperatures is maximum of all others operating between the same temperature limits. In other words, the efficiency of a reversible engine (hR) is greater than the efficiency of an irreversible engine (hI). 13. The efficiency of all reversible heat engines operating between the two constant temperature reservoirs have the same efficiency. 14. The efficiency of any reversible heat engine operating between the two constant temperature reservoirs is independent of the nature of the working substance undergoing the cycle and depends solely on the temperature of the reservoirs.

EXERCISES 1. A refrigerator transfers 120 kJ of heat from a cold space. It requires 40 kJ of work input. Calculate its coefficient of performance. [Ans. 3] 2. A Carnot refrigerator working between 27°C and – 13°C is driven by an electric motor rated at 1.5 kW. What quantity of ice at 0°C can be obtained from water at 20°C in 10 hours. Take cp for ice = 2.1 kJ / kg K, and cp for water = 4.2 kJ / kg K. [Ans. 838 kg]

Second Law of Thermodynamics 137 3. A reversible engine is supplied with heat from two constant temperature sources at 900 K and 600 K and rejects heat to a constant temperature sink at 300 K. The engine develops work equivalent to 90 kW and rejects heat at the rate of 56 kJ / s. Determine: (a) Heat supplied by each source; and (b) Thermal efficiency of the engine. [Ans. 100 kJ / s , 46 kJ / s; 61.6%] 4. A reversible engine operates between a source temperature of 250°C and a sink temperature of – 15°C. If the system receives 90 kJ from the source, find: 1. efficiency of the system, 2. net work transfer; and 3. heat rejected to the sink. [Ans. 50.67% ; 45.6 kJ ; 44.4 kJ] 5. A reversible heat engine receives heat from a high temperature reservoir at temperature T K and rejects heat to a low temperature sink of 800 K. A second reversible engine receives the heat rejected by the first engine at 800 K and rejects to a cold reservoir at 280 K. Make calculations for temperature T for (a) equal thermal efficiencies of the two engines; and (b) the two engines to deliver the same amount of work. [Ans. 2285.7 K ; 1320 K] 6. A cold storage is to be maintained at –5°C while the surroundings are at 35°C. The heat leakage from the surroundings to the cold storage is estimated to be 29 kW. The actual C.O.P. of the refrigeration plant is onethird of an ideal plant working between the same temperatures. Find the power required to drive the plant. [Ans. 12.987 kW] 7. A heat pump is used for heating the interior of a house in a cold climate. The ambient temperature is –5°C and the desired interior temperature is 25°C. The compressor of the heat pump is to be driven by a heat engine working between 1000°C and 25°C. Treating both cycles as reversible, calculate the ratio in which the heat pump and the heat engine share the heating load. [Ans. 7.66] 8. 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 the reservoirs at temperatures of 40°C and – 20°C. The heat transfer to the engine is 2 MJ and the net work output of the combined engine and refrigerator plant is 360 kJ. Find the heat transfer to the refrigerant and the net heat transfer to the reservoir at 40°C. Also find these values if the efficiency of the heat engine and C.O.P. of the refrigerator are each 40% of their maximum possible values. [Ans. 3892.3 kJ ; 5532.3 kJ ; 258.4 kJ ;1898.4 kJ]

QUESTIONS 1. State the second law of thermodynamics. 2. Give the Kelvin-Planck and Clausius statements of the second law of thermodynamics and prove the equivalence of the two statements. 3. What is perpetual motion machine of second kind ? Why such a machine can not be constructed in actual practice? 4. What is a heat pump? How does it differ from a refrigerator? Explain. 5. Show that the coefficient of performance of a heat pump is greater than the coefficient of performance of a refrigerator by unity. 6. What is a Carnot cycle? Represent it on p-v and T-s diagrams and derive an expression for the efficiency of a Carnot cycle. 7. Explain the practical difficulties in constructing a Carnot engine. 8. State and prove Carnot’s theorem. 9. Show that the efficiency of a reversible engine operating between two given constant temperatures is maximum. 10. What are the corollaries of Carnot’s theorem ?

OBJECTIVE TYPE QUESTIONS 1. The law of degradation of energy is also known as (a) zeroth law of thermodynamics (b) first law of thermodynamics (c) second law of thermodynamics (d) none of these

138 Engineering Thermodynamics 2. It is impossible to construct an engine while working in a cycle, produces no effect other than to extract heat from a single thermal reservoir and perform an equivalent work. This statement refers to (a) Clausius statement (b) Kelvin-Planck statement (c) Carnot’s theorem (d) PMM-II 3. Which of the following statement is correct according to Clausius statement? (a) It is impossible to transfer heat from a body at a lower temperature to a body at a higher temperature, without the aid of an external source. (b) It is possible to transfer heat from a body at a lower temperature to a body at a higher temperature. (c) It is possible to transfer heat from a body at a lower temperature to a body at a higher temperature by using the refrigeration cycle. (d) none of the above 4. According to Kelvin-Planck’s statement, a perpetual motion of the ................. is impossible. (a) first kind (b) second kind 5. There is a definite amount of mechanical energy, which can be obtained from a given quantity of heat energy. This statement is according to (a) zeroth law of thermodynamics (b) first law of thermodynamics (c) second law of thermodynamics (d) none of these 6. The heat flows from a cold body to a hot body with the aid of an external source. This statement is given by (a) Clausius (b) Kelvin (c) Joule (d) Gay-Lussac 7. The coefficient of performance of a heat pump is given by T − T2 T1 T2 T − T2 (a) 1 (b) (c) (d) 1 T1 T1 − T2 T1 − T2 T2 where T1 is the temperature of a hot body and T2 is the temperature of a cold body. 8. The coefficient of performance of a refrigerator is given by T − T2 T1 T2 T − T2 (a) 1 (b) (c) (d) 1 T1 T1 − T2 T1 − T2 T2 9. The coefficient of performance of a heat pump is greater than the coefficient of performance of a refrigerator by (a) 1 (b) 2 (c) 2.5 (d) none of these 10. Carnot cycle consists of (a) two constant volume and two isentropic processes (b) two isothermal and two isentropic processes (c) two constant pressure and two isentropic processes (d) one constant volume, one constant pressure and two isentropic processes 11. The efficiency of the Carnot cycle is T T T T (a) 1 – 1 (b) 1 – 1 (c) 1 – 2 (d) 1 + 1 T2 T2 T1 T2 where T1 and T2 = Highest and lowest temperature during the cycle.

Second Law of Thermodynamics 139 12. Carnot cycle has maximum efficiency for (a) petrol engine (b) diesel engine (c) reversible engine (d) irreversible engine 13. The efficiency of a Carnot cycle may be increased by (a) increasing the highest temperature (b) decreasing the highest temperature (c) increasing the lowest temperature (d) keeping the lowest temperature constant 14. No heat engine operating in a cycle between two constant temperature reservoirs can be more efficient than a reversible engine operating between the same two reservoirs. This statement is according to (a) Kelvin Planck statement (b) Carnot’s theorem (c) Clausius statement (d) PMM-II 15. The efficiency of a reversible engine operating between two constant temperatures is .................... of all others operating between the same temperature limit. (a) maximum (b) minimum

ANSWERS

1. (c) 6. (a) 11. (c)

2. (b) 7. (b) 12. (c)

3. (a) 8. (c) 13. (a)

4. (b) 9. (a) 14. (b)

5. (c) 10. (b) 15. (a)

7 ENTROPY, AVAILABILITY AND IRREVERSIBILITY 7.1 Introduction 7.2 Temperature-Entropy (T-S) Diagram 7.3 Clausius Inequality 7.4 Principle of Increase of Entropy 7.5 Change of Entropy for Ideal Gases 7.6 Change of Entropy During Thermodynamic Processes 7.7 Change of Entropy During Constant Volume (or Isochoric) Process 7.8 Change of Entropy During Constant Pressure (or Isobaric) Process 7.9 Change of Entropy During Constant Temperature (or Isothermal) Process

7.10 Change of Entropy During Reversible Adiabatic (or Isentropic) Process 7.11 Change of Entropy During Polytropic (or pv n = constant) Process 7.12 Available and Unavailable Energy 7.13 Loss in Available Energy 7.14 Availability 7.15 Availability of a Closed System 7.16 Availability in a Steady Flow Process 7.17 Helmholtz Function 7.18 Gibbs Function 7.19 Irreversibility

7.1 INTRODUCTION The second law of thermodynamics leads to a property of a system which is termed as entropy. The literal meaning of entropy is transformation and it was first introduced by Clausius. He regarded entropy as a measure of the unavailability of heat energy for transformation into work. It is difficult to directly define the term entropy, but the change of entropy can be easily defined. Let a small amount of heat dQ be added to a working substance at an absolute temperature T. In a reversible process, over a small range of temperature, the quantity of heat added or rejected divided by the absolute temperature of the working substance, gives the increase or decrease of entropy. In other words, change of entropy. δQ dS = or dQ = T × dS T where dS is the increase or decrease of entropy. It is used by engineers as a measure of providing quick solution for problems dealing with reversible adiabatic process. It is usually denoted by the letter S. When the entropy of the working substance is divided by its mass, than it is called specific entropy. In S.I. units, the specific entropy is expressed in kJ/ kg K. Notes: 1. The entropy remains constant during reversible adiabatic process and increases during an irreversible adiabatic process such as throttling process. 2. It increases when heat is supplied and decreases when heat is removed. 3. The entropy is an extensive property of the system.

Entropy, Availability and Irreversibility 141

7.2 TEMPERATURE-ENTROPY (T-S) DIAGRAM

Temperature (T )

The change of entropy with temperature, as shown in Fig. 7.1, is known as temperature-entropy (T-S ) diagram. 2 In this diagram, the base represents the units of T2 entropy (S ) and the vertical ordinate represents the absolute temperature (T ). Let the point 1 represents the absolute temperature (T1) A and entropy (S1) of a certain mass of gas. Now imagine that T S2 this gas is receiving heat in some manner. As the heating 1 T1 proceeds, the absolute temperature and entropy increases as shown by the curve 1-2 on the T-S diagram. The point 2 represents the final condition of the gas at which the absolute temperature is T2 and entropy is S2. S1 dS S2 Consider any point A on the curve 1-2. Let the absolute Entropy (S ) temperature at this point is T, and a small addition of heat Fig. 7.1. Temperature-Entropy (T-S ) dQ under reversible conditions increases entropy by dS. Now diagram. according to the definition of entropy, dQ = T × dS ...(i) From Fig. 7.1, we see that T × dS is the area under the curve during this change of entropy dS, as shown shaded in the figure. Thus, the total heat supplied or absorbed during the reversible process 1-2 may be obtained by integrating the equation (i), i.e.

2

2

∫1 δQ = ∫1 T × dS

7.3 CLAUSIUS INEQUALITY The Clausius inequality states that whenever a closed system undergoes a cyclic process and during which it exchanges heat dQ with a number of reservoirs at respective temperatures T, then the cyclic integral of dQ/T is less than or equal to zero, i.e. δQ ∫ T 0 This equation is known as Clausius inequality. The equal sign holds if the cycle is reversible and the inequality sign holds, if the cycle is irreversible. In other words for a reversible cyclic process, δQ ∫ T = 0 and for an irreversible cyclic process, δQ ∫ T < 0 The Clausius inequality not only gives mathematical expression to the second law of thermodynamics, but it also gives the quantitative measure of irreversibility of the system. The above relation may be proved as discussed below: Consider a Carnot cycle operating in the temperature limits of T1 and T2 absorbing heat Q1 and rejecting heat Q2. The efficiency for such a cycle is given by Q − Q2 Q =1− 2 h = 1 Q1 Q1

142 Engineering Thermodynamics We know that for a reversible engine, Q1 Q = 2 T1 T2 Since Q2 is the heat rejected, therefore the above equation may be written as or

Q1 − Q2 − = 0 T1 T2

...(Negative sign to Q2 is due to heat rejection)

Q1 Q2 + = 0 T1 T2

Thus, the summation of the ratio of the heat transfer to the absolute temperature for a reversible cycle is equal to zero. In other words for a reversible engine, the cyclic integral of dQ/ T is equal to zero, i.e. δQ ∫ T = 0 When we consider irreversibilities, then an irreversible engine absorbing heat Q1 from reservoir at temperature T1 will reject more heat to low temperature reservoir at temperature T2 than the reversible engine. In other words, the efficiency of the irreversible engine will be lower. Let the heat rejected by the irreversible engine is Q 2. We know that, efficiency for an irreversible engine, Q − Q2′ T1 − T2 < h = 1 T1 Q1 or

1−

Q2′ T < 1 − 2 Q1 T1

Considering the heat rejected Q 2 as negative, we have \

Q1 (− Q2′ ) − < 0 T1 T2 Q1 Q2′ + < 0 T1 T2

Thus for an irreversible engine, the cyclic integral of dQ/T is less than zero, i.e. δQ ∫ T < 0 The equation for the irreversible cyclic process may be written as δQ ∫ T + I = 0 where I represents the amount of irreversibility of a cyclic process. When I = 0, the cyclic process is reversible. δQ If ∫ > 0, the cycle is impossible because it violates the second law of thermodynamics. T Example 7.1. A heat engine is supplied with 2512 kJ/min of heat at 650°C. The heat rejection takes place at 100°C. Specifiy which of the following heat rejections represents a reversible, irreversible or impossible result: (a) 867 kJ/min; (b) 1015 kJ/min ; and (c) 1494 kJ/min.

Entropy, Availability and Irreversibility 143 Solution. Given: Heat supplied, Q1 = 2512 kJ/min Temperature at which heat is supplied, T1 = 650°C = 650 + 273 = 923 K Temperature at which heat is rejected, T2 = 100°C = 100 + 273 = 373 K (a) When heat rejection is 867 kJ/min (i.e. when Q2 = 867 kJ/min) According to Clausius inequality, Q Q δQ 2512 867 ∫ T = T11 − T22 = 923 − 373 = 2.72 – 2.32 = 0.4 Since it is greater than zero, therefore the Clausius inequality does not hold good. Hence the heat rejection of 867 kJ/min is impossible. Ans. (b) When the heat rejection is 1015 kJ/min (i.e. when Q2 = 1015 kJ/min) According to Clausius inequality, Q Q δQ 2512 1015 ∫ T = T11 − T22 = 923 − 373 = 2.72 – 2.72 = 0 Since it is equal to zero, therefore the heat rejection of 1015 kJ/min is reversible. Ans. (c) When the heat rejection is 1494 kJ/min (i.e. when Q2 = 1494 kJ/min) According to Clausius inequality, Q Q δQ 2512 1494 ∫ T = T11 − T22 = 923 − 373 = 2.72 – 4.005 = – 1.285 Since it is less than zero, therefore the heat rejection of 1494 kJ/min is irreversible. Ans. Example 7.2. A reversible engine working in a cycle takes 4800 kJ of heat per minute from a source at 800 K and develops 20 kW power. The engine rejects heat to two reservoirs at 300 K and 360 K. Determine the heat rejected to each sink. Solution. Given: Heat supplied from source, Q1 = 4800 kJ/min = 80 kJ/s Temperature of source, T1 = 800 K Power developed, W = 20 kW = 20 kJ/s The system is shown in Fig. 7.2. Let Q2 = Heat rejected to sink at temperature, T2 = 300 K ... (Given) Q3 = Heat rejected to sink at temperature, T3 = 360 K ... (Given) According to the first law of thermodynamics, or \

∫ δQ = ∫ δW Q1 + Q2 + Q3 = W 80 + Q2 + Q3 = 20 Q2 + Q3 = 20 – 80 = – 60 kJ Q3 = – 60 – Q2 = – (Q2 + 60)

...(i)

144 Engineering Thermodynamics Now from the Clausius inequality, Q1 Q2 Q3 + + = 0, for reversible cycle T1 T2 T3 \

Source 800K Q1

Q Q 80 + 2 + 3 = 0 800 300 360 0.1 +

Reversible engine

Q2 Q2 + 60 − = 0 300 360

Power = 20 kW

0.1 × 300 × 360 + 360Q2 − 300 (Q2 + 60) = 0 300 × 360

Q2

10 800 + 360 Q2 – 300 Q2 – 18 000 = 0 60 Q2 = 18 000 – 10 800 = 7200 \ Q2 = 7200/60 = 120 kJ/s = 7200 kJ/min Substituting the value of Q2 in equation (i), we have

Q3

Sink 300K

Sink 360K

Fig. 7.2.

Q3 = – (Q2 + 60) = – (120 + 60) = – 180 kJ/s = – 10 800 kJ/min From above we see that 7200 kJ/min is rejected to sink at 300 K and 10 800 kJ/min is rejected to sink at 360 K. Ans.

7.4 PRINCIPLE OF INCREASE OF ENTROPY Consider a given quantity of heat energy dQ lost from a hot body at temperature T1 and gained by the cold body at temperature T2. We know that 2 δQ Loss of entropy by the hot body = T1 δQ T2

A

Pressure

and gain of entropy by the cold body =

B

Since T1 is greater than T2, therefore the gain of entropy by the cold body is greater than the loss of entropy from 1 the hot body. Thus, as the temperature falls in a system (i.e. irreversible process), the entropy increases. This condition can be extended that in any isolated system (say universe), the heat Volume exchange takes place in irreversible manner. Thus, the entropy Fig. 7.3. Principle of increase of of an isolated system (i.e. universe) increases. entropy. The principle of increase of entropy may also be discussed as follows: We know that change of entropy in a reversible process is given by δQ dS = T In order to find the change of entropy in an irreversible process, consider a closed system under going a change from state 1 to state 2 along a reversible process 1-A-2 and returns from state 2 to the initial state 1 along an irreversible process 2-B-1, as shown on the p-v diagram in Fig. 7.3. Since entropy is a thermodynamic property, therefore, we can write

2

1

∫ dS = ∫1A (dS )R + ∫2B (dS )I

= 0

where subscript R represents reversible process and I stands for irreversible process.

...(i)

Entropy, Availability and Irreversibility 145 Now, for a reversible process 1-A-2,

2

2

δQ T R

∫1A (dS )R = ∫1A

Substituting this value in equation (i), we have 2 δQ 1 ∫1A T R + ∫2B (dS )I = 0

...(ii)

Since the process 1-A-2 and 2-B-1 together form an irreversible cycle, therefore from Clausius inequality, 2 δQ 1 δQ δQ ...(iii) ∫ T = ∫1A T R + ∫2B T I < 0 Now subtracting equation (iii) from equation (ii), we get 1 1 δQ ∫ (dS ) I > ∫ , which for the infinitesimal changes in states, 2B 2B T I may be written as

δQ (dS)I > T I

...(iv)

δQ This states that the change in entropy in an irreversible process is greater than . Combining T equations (iii) and (iv), we can write the equation in general form as δQ dS T where equality sign stands for the reversible process and inequality sign stands for the irreversible process. Now considering an isolated system in which matter, heat or work can not cross the boundary of the system. Thus dQ = 0. Therefore for an isolated system, equation (iii) may be written as dS 0 From this expression, it follows that dS = 0 or entropy is constant for a reversible cyclic process, and for an irreversible process, dS > 0. Since, all the processes in nature are irreversible, therefore the entropy of such a system like universe goes on increasing. This is known as the principle of increase of entropy.

7.5 CHANGE OF ENTROPY FOR IDEAL GASES Consider a certain mass (say m) of an ideal gas is being heated by any thermodynamic process from state 1 (having pressure p1, temperature T1 and volume v1) to state 2 (having pressure p2, temperature T2 and volume v2). The change in entropy during the process may be obtained in the following three ways: 1. In terms of volume and temperature According to the first law of thermodynamics, for reversible and irreversible process and for a small change in the state of the working substance, we know that dQ = dU + dW ...(i) where dQ = Heat transferred or heat supplied, dU = Change in internal energy = m cv dT, and dW = Workdone = pdv

146 Engineering Thermodynamics Now the equation (i) may be written as dQ = m cv dT + pdv m cv dT pdv δQ + = T T T

or

...(Dividing throughout by T ) ...(ii)

δQ p mR = = dS and p v = mR T or T T v

We know that

Substituting these values in equation (ii), we have dT dv dS = m cv × + mR × T v Now integrating this equation for the state 1 to state 2, 2 2 dT 2 dv ∫1 dS = m cv ∫1 T + mR ∫1 v [S]12 = m cv [ log e T ]1 + mR [ log e v ]1 2

2

S2 – S1 = m cv [loge T2 – loge T1] + mR [loge v2 – loge v1]

or

T v = m cv log e 2 + mR log e 2 T1 v1

...(iii)

T v = 2.3m cv log 2 + R log 2 T1 v1

...(iv)

T = 2.3m cv log 2 T1

...(v)

v2 + (c p − cv ) log v1

2. In terms of pressure and volume p1 v1 pv T pv We know that = 2 2 or 2 = 2 2 T1 T2 T1 p1v1 Substituting the value of

... [Q R = cp – cv] ...[General gas equation]

T2 in equation (v), we have T1

pv v S2 – S1 = 2.3m cv log 2 2 + (c p − cv ) log 2 p v 11 v1

p v v v = 2.3m cv log 2 + cv log 2 + c p log 2 − cv log 2 p1 v1 v1 v1

p v = 2.3m cv log 2 + c p log 2 p1 v1 3. In terms of pressure and temperature p1v1 pv v p T We know that = 2 2 or 2 = 1 2 T1 T2 v1 p2 T1

...(vi)

Entropy, Availability and Irreversibility 147 Substituting the value of

v2 in equation (v), we have v1

T S2 – S1 = 2.3 m cv log 2 T1

p1 T2 × + (c p − cv ) log p2 T1

T p T = 2.3 m cv log 2 + c p log 1 + c p log 2 T p 1 2 T1

p1 T2 − cv log − cv log p2 T1

T p = 2.3 m c p log 2 + (c p − cv ) log 1 T1 p2

...(vii)

T p = 2.3 m c p log 2 + R log 1 T1 p2

...(viii)

Notes: 1. We have already discussed that when heat is supplied, there is an increase of entropy and when heat is removed, there is a decrease in entropy. 2. The equation, (viii) may be written as

−1 T p S2 – S1 = 2.3 m c p log 2 + R log 2 T1 p1

T2 T1

= 2.3 m c p log

p2 − R log p1

Example 7.3. 2 kg of air at 150°C and at 8 bar expands to a temperature of 50°C and pressure of 1 bar. Calculate the change in entropy. Given: R = 0.287 kJ/kg K; and cv = 0.707 kJ/kg K. Solution. Given: Mass of air, m = 2 kg Initial temperature, T1 = 150°C = 150 + 273 = 423 K Initial pressure, p1 = 8 bar Final temperature, T2 = 50°C = 50 + 273 = 323 K Final pressure, p2 = 1 bar Gas constant, R = 0.287 kJ / kg K Specific heat at constant volume, cv = 0.707 kJ/kg K We know that specific heat at constant pressure, cp = R + cv = 0.287 + 0.707 = 0.994 kJ/kg K \ Change in entropy (in terms of pressure and temperature),

T p S2 – S1 = 2.3m c p log 2 + R log 1 T1 p2

323 8 = 2.3 × 2 0.994 log + 0.287 log 423 1 = 4.6 (– 0.1164 + 0.2592) = 0.657 kJ/K Ans. The positive sign indicates that there is an increase in entropy.

148 Engineering Thermodynamics Note: The change in entropy is also given by

T p S2 – S1 = 2.3m c p log 2 − R log 2 T 1 p1 323 1 = 2.3 × 2 0.994 log − 0.287 log 423 8

= 4.6 [– 0.1164 – (– 0.2592)] = 4.6 [– 0.1164 + 0.2592] = 0.657 kJ/K

7.6 CHANGE OF ENTROPY DURING THERMODYNAMIC PROCESSES We have already discussed that the various thermodynamic processes are as follows: 1. Constant volume (or Isochoric) process; 2. Constant pressure (or Isobaric) process; 3. Constant temperature (or isothermal) process; 4. Reversible adiabatic (or isentropic) process; and 5. Polytropic (or pv n = constant) process. The change of entropy during the above mentioned processes are discussed, in detail, in the following pages:

7.7 CHANGE OF ENTROPY DURING CONSTANT VOLUME (OR ISOCHORIC) PROCESS The constant volume process on the temperature-entropy (T-S) diagram is shown in Fig. 7.4. The initial and final states of the working substance are shown by points 1 and 2. We have already discussed in the previous article that the change of entropy in terms of volume (v) and temperature (T ) is given by T S2 – S1 = 2.3 m cv log 2 T1

v2 + (c p − cv ) log v1

We know that

T S2 – S1 = 2.3 m cv log 2 T1 p1v1 pv T = 2 2 or 2 T1 T2 T1

v=C T1

1

S2 Entropy (S)

=

2

T2

S1

Since the volume is constant, i.e. v2 = v1, and log 1 = 0, therefore the above equation becomes

Temperature (T )

Fig. 7.4. Constant volume process

...(i) p2 p1

...( v1 = v2)

Now the above expression may also be written as

p S2 – S1 = 2.3 m cv log 2 p1

Notes: 1. The equation (i) may be derived as follows: We know that heat supplied at constant volume, dQ = m cv dT δQ m cv dT or = T T dT ∴ dS = m cv T

...(ii)

...( Dividing throughout by T )

Entropy, Availability and Irreversibility 149

Integrating this expression within limits of initial and final states,

∫

dS = m cv

∫

2

1

dT T

[S]12 = m cv [log e T ]1

2

T T S2 – S1 = m cv loge 2 = 2.3 m cv log 2 T 1 T1

or

2

1

2. We have seen above that

dT dS = m cv T

dS = cv ×

For a unit mass,

or

The term

dT T

dT T = dS cv dT is termed as slope of the curve 1-2 on the T-S diagram as shown in Fig. 7.4. dS

Example 7.4. A vessel of capacity 3 m3 contains air at a pressure of 1.5 bar and a temperature of 25°C. Additional air is now pumped into the system until the pressure rises to 30 bar and the temperature rises to 60°C. Determine the mass of air pumped in. In the vessel is allowed to cool at constant volume until the temperature is again 25°C, calculate the pressure in the vessel. Determine the quantity of heat transferred and change of entropy of the gas during cooling process only. Neglect the effect of heat capacity of the vessel. Assume air as an ideal gas. Solution. Given: Volume of vessel, v1 = 3 m3 Initial pressure, p1 = 1.5 bar = 150 × 103 N/m2 Initial temperature, T1 = 25°C = 25 + 273 = 298 K Final pressure, p2 = 30 bar = 3000 × 103 N/m2 Final temperature, T 2 = 60°C = 60 + 273 = 333 K Mass of air pumped in Let m1 and m2 be the mass of air before and after the air is pumped into the vessel. Now applying the general gas equation to the initial conditions of air, we have or

p1 v1 = m1 RT1 m1 =

p1v1 150 × 103 × 3 = = 5.26 kg 287 × 298 RT1

...(Q R for air = 287 J/ kg)

Similarly after the air is pumped in, p2 v2 = m2 RT2 or

m2 =

p2 v2 3000 × 103 × 3 = = 94.17 kg 287 × 333 RT2

...(Q v2 = v1)

150 Engineering Thermodynamics \ Mass of air pumped in, m = m2 – m1 = 94.17 – 5.26 = 88.91 kg Ans. Heat transferred during cooling Since the vessel is allowed to cool at constant volume from temperature T2 = 60°C = 333 K to temperature T3 = 25°C = 25 + 273 = 298 K, therefore quantity of heat transferred (dQ) when the process takes place at constant volume is given by dQ = Change in internal energy (dU) = m2 cv (T3 – T2) = 94.17 × 0.712 (298 – 333) ...(Taking cv for air = 0.712 kJ/ kg K) = – 2346.7 kJ Ans. The negative sign indicates that the heat is rejected by the air Change of entropy We know that change of entropy during the constant volume cooling process,

T 298 S2 – S1 = 2.3 m2 cv log 3 = 2.3 × 94.17 × 0.712 log T 333 2

= 156 (– 0.0482) = – 7.44 kJ/K Ans. The negative sign indicates that there is a decrease in entropy.

7.8 CHANGE OF ENTROPY DURING CONSTANT PRESSURE (OR ISOBARIC) PROCESS The constant pressure process on the temperature-entropy (T-S) diagram is shown in Fig. 7.5. The initial and final states of the working substance are shown by points 1 and 2. We have already discussed in the previous article that the change of entropy in terms of pressure ( p) and temperature (T ) is given by T p S2 – S1 = 2.3 m c p log 2 + (c p − cv ) log 1 T1 p2 Since the pressure is constant, i.e. p1 = p2 and log 1 = 0, therefore the above equation becomes, We know that

T S2 – S1 = 2.3 m cp log 2 T1 p1v1 pv = 2 2 T1 T2

...(i)

v S2 – S1 = 2.3 m cp log 2 v1

...(ii)

T2 v = 2 ...(Q p1 = p2) T1 v1 Now the above expression may be written as

or

Notes: 1. The equation (i) may be derived as follows: We know that heat supplied at constant pressure,

dQ = m cp dT or

\

dT dS = m cp T

Fig. 7.5. Constant pressure process.

m c p dT δQ = T T

...(Dividing throughout by T )

Entropy, Availability and Irreversibility 151

Integrating this expression within the limits of initial and final states,

∫

2

1

[S]12 = m cp [log e T ]1

2

T T S2 – S1 = m cp loge 2 = 2.3 m cp log 2 T1 T1

or

2. We have seen above that dT dS = m cp T

dT For a unit mass, dS = cp or T

The term

dT T = dS cp

...(iii)

dT is the slope of the curve 1-2 on the T-S diagram as shown in Fig. 7.5. dS

We have already discussed in constant volume process that

2 dT dS = ∫ m c p 1 T

dT T = dS cv

Since cv < cp or 1/cv > 1/cp, therefore

...(iv)

T T > . This shows that the slope of the constant volume process cv c p

(curve 1-2΄) is higher than the slope of the constant pressure process (curve 1 – 2).

Example 7.5. 0.28 m3 of gas at a pressure of 10.5 bar and temperature of 538°C is expanded at a constant pressure to a volume of 0.34 m3. Determine the change of entropy, assuming cv = 0.69 kJ/kg K and R = 287 J/kg K. Solution. Given: Initial volume,

v1 = 0.28 m3

Initial pressure,

p1 = 10.5 bar = 1050 × 103 N/m2

Initial temperature,

T1 = 538°C = 538 + 273 = 811 K

Final volume,

v2 = 0.34 m3

Specific heat at constant volume,

cv = 0.69 kJ/ kg K

Gas constant,

R = 287 J/kg K = 0.287 kJ/kg K

First of all, let us find the mass (m) and final temperature (T2) of the gas. We know that \

p1 v1 = mR T1 m =

p1v1 1050 × 103 × 0.28 = = 1.263 kg 287 × 811 RT1

Since the gas is expanded at constant pressure, therefore v1 v = 2 T1 T2

152 Engineering Thermodynamics \ We know that or \ Change of entropy,

T2 =

v2 T1 0.34 × 811 = = 985 K 0.28 v1

R = cp – cv cp = R + cv = 0.287 + 0.69 = 0.977 kJ/ kg K T S2 – S1 = 2.3 m cp log 2 = 2.3 × 1.263 × 0.977 log T1

985 811

= 2.838 × 0.0844 = 0.24 kJ/K Ans. Example 7.6. 0.34 m3 of a perfect gas at constant pressure of 2.8 bar is heated from 100°C to 300°C and it is then cooled at constant volume to its initial temperature. Calculate the overall change of entropy. Given that cp = 1.05 kJ/kg K and cv = 0.75 kJ/kg K. Solution. Given: Initial volume, v1 = 0.34 m3 Initial pressure, p1 = 2.8 bar = 280 × 103 N/m2 Initial temperature, T1 = 100°C = 100 + 273 = 373 K Final temperature, T2 = 300°C = 300 + 273 = 573 K Specific heat at constant pressure, cp = 1.05 kJ/ kg K Specific heat at constant volume, cv = 0.75 kJ/ kg K We know that gas constant, R = cp – cv = 1.05 – 0.75 = 0.3 kJ/ kg K = 300 J/kg K First of all, let us find the mass (m) of the gas. We know that p1 v1 = m R T1 \

m =

p1v1 280 × 103 × 0.34 = = 0.85 kg 300 × 373 RT1

We know that change of entropy when gas is heated at constant pressure,

T 573 S2 – S1 = 2.3 m cp log 2 = 2.3 × 0.85 × 1.05 log 373 T1

= 2.053 × 0.1864 = 0.383 kJ/K When the gas is cooled at constant volume to its initial temperature (i.e. T3 = 100°C = 373 K), then change of entropy,

T S2 – S1 = 2.3 m cv log 3 = 2.3 × 0.85 × 0.75 log T2

373 573

= 1.466 (– 0.1864) = – 0.273 kJ/K \ Overall change of entropy = 0.383 – 0.273 = 0.11 kJ/K Ans.

7.9 CHANGE OF ENTROPY DURING CONSTANT TEMPERATURE (OR ISOTHERMAL) PROCESS The constant temperature (or isothermal) process on the temperature entropy (T-S) diagram is shown in Fig. 7.6. The initial and final states of the working substance are shown by points 1 and 2. We have already discussed in the previous article that the change of entropy in terms of volume and temperature (T ) is given by

Entropy, Availability and Irreversibility 153 T v S2 – S1 = 2.3 m cv log 2 + (c p − cv ) log 2 T1 v1

Since the temperature is constant, i.e. T2 = T1 and log 1 = 0, therefore the above equation becomes v S2 – S1 = 2.3 m (cp – cv ) log 2 v1 p1v1 pv = 2 2 T1 T2

We know that

v2 p = 1 v1 p2 Now the above expression may be written as

or

...(i) Fig. 7.6. Constant temperature process.

...(Q T1 = T2)

p S2 – S1 = 2.3 m (cp – cv) log 1 p2

...(ii)

Note: The equation (i) may be derived as follows: We know that the heat supplied at constant temperature (isothermal) process is equal to the workdone by the working substance, i.e.

v v Q1 – 2 = W1 – 2 = m R T loge 2 = 2.3 mRT log 2 v 1 v1

and change of entropy, S2 – S1 =

v v Q1− 2 2.3 mRT = log 2 = 2.3 mR log 2 T T v 1 v1

v = 2.3 m (cp – cv ) log 2 v1

...(Q R = cp – cv)

p = 2.3 m (cp – cv ) log 1 p2

v p ... 2 = 1 v p 1 2

Example 7.7. A piston cylinder machine contains 0.05 m3 of nitrogen at 1 bar and 300 K. Determine the change in entropy and workdone during isothermal reversible compression of the gas during which pressure rises to 4 bar. Solution. Given: Initial volume, v1 = 0.05 m3 Initial pressure, p1 = 1 bar = 100 × 103 N/m2 Initial temperature, T1 = 300 K Final pressure, p2 = 4 bar = 400 × 103 N/m2 Since the molecular mass (M) of the nitrogen is 28, therefore mass of nitrogen m = Molecular mass × Volume of nitrogen = 28 × 0.05 = 1.4 kg and characteristic gas constant, Universal gas constant ( Ru ) 8.314 R = = = 0.297 kJ/kg K 28 Molecular mass ( M )

...(Q Ru for all gases = 8.314 kJ/kg K)

154 Engineering Thermodynamics Change in entropy We know that change in entropy during isothermal process,

v p S2 – S1 = 2.3 m R log 2 = 2.3 m R log 1 v1 p2

...

v2 p1 = v1 p2

1 = 2.3 × 1.4 × 0.297 log 4

= 0.956 (– 0.6) = – 0.5736 kJ/K The negative sign indicates that there is a decrease in entropy. Workdone v p We know that workdone, W = 2.3 m R T1 log 2 = 2.3 m R T1 log 1 v 1 p2 1 = 2.3 × 1.4 × 0.297 × 300 log = 287 (– 0.6) = – 172.14 kJ 4 The negative sign indicates that work is done on the gas.

7.10 CHANGE OF ENTROPY DURING REVERSIBLE ADIABATIC (OR ISENTROPIC) PROCESS

Temperature (T )

The reversible adiabatic (or isentropic) process on the temperature-entropy (T-S ) diagram is shown in Fig. 7.7. The initial and final states of the working substance are shown by points 1 and 2. We know that in a reversible adiabatic process, no heat enters or leaves the system, i.e. dQ = 0 2 2 T2 We also know that change of entropy, δQ dS = = 0 ...(Q dQ = 0) T 1 or S2 – S1 = 0 or S2 = S1 = S T1 Thus during a reversible adiabatic process, the entropy of S1 = S2 a system remains constant. In other words, reversible adiabatic Entropy (S) process is said to isentropic process (i.e. frictionless adiabatic Fig. 7.7. Adiabatic process process). If the adiabatic process is irreversible, as shown by 1-2 in Fig. 7.7, then due to internal friction, the heat absorbed by the gas will be more than the reversible process. If dQ is the heat absorbed, then dQ/T will also be more. In other words, the entropy (dS ) is more at point 2. Thus an irreversible process always results in increase in entropy, i.e. dS > 0. Example 7.8. An ideal gas of mass 0.25 kg has a pressure of 3 bar, a temperature of 80°C and a volume of 0.07 m3. The gas undergoes irreversible adiabatic process to a final pressure of 3 bar and a final volume of 0.10 m3, during which the workdone on the gas is 25 kJ. Evaluate cp and cv of the gas and the increase in entropy of the gas. Solution. Given: Mass of the gas, m = 0.25 kg Initial pressure, p1 = 3 bar = 300 × 103 N/m2 Initial temperature, T1 = 80°C = 80 + 273 = 353 K Initial volume, v1 = 0.07 m3

Entropy, Availability and Irreversibility 155 Final pressure, p2 = 3 bar = 300 × 103 N/m2 Final volume, v2 = 0.10 m3 Workdone on the gas (i.e. workdone during compression process), W1–2 = – 25 kJ Values of cp and cv of the gas Let R = Characteristic gas constant, and T2 = Final temperature of the gas. We know that p1 v1 = m R T1 \ and \

R =

p1v1 300 × 103 × 0.07 = = 238 J/kg K = 0.238 kJ/ kg K 0.25 × 353 mT1

p2 v2 = mRT2 T2 =

p2 v2 300 × 103 × 0.10 = = 504 K 0.25 × 238 mR

Since the gas undergoes irreversible adiabatic process, therefore the heat transfer (Q1 – 2) is zero. We know that change in internal energy, dU = m cv (T2 – T1) = 0.25 cv (504 – 353) = 37.75 cv \ Heat transfer, Q1 – 2 = dU + W1 – 2 or

0 = 37.75 cv – 25 25 cv = = 0.662 kJ/kg K Ans. 37.75

We know that R = cp – cv \ cp = R + cv = 0.238 + 0.662 = 0.9 kJ/ kg K Ans. Increase in entropy of the gas Since the adiabatic process takes place at constant pressure (i.e. p1 = p2 = 3 bar), therefore increase in entropy of the gas,

T S2 – S1 = 2.3 m cp log 2 = 2.3 × 0.25 × 0.9 log T1

504 353

= 0.5175 × 0.1547 = 0.08 kJ/K Ans. Example 7.9. A mass m kg of a gas at temperature T1 is isobarically and adiabatically mixed with an equal mass of the same gas at a temperature T2 (T1 > T2 ). Show that the change in entropy of the universe during the process is given by

T + T2 (dS )universe = 2 m cp loge 1 2 T T 1 2

.

Solution. Consider m kg of gas at temperature T1 in the first compartment is mixed with the same mass m kg of gas at temperature T2 in the second compartment as shown in Fig. 7.8. Let T3 is the common temperature of the gas after mixing. This temperature (T3) is less than T1 and greater than T2. We know that Heat lost by the gas at temperature T1 = Heat gained by the gas at temperature T2 m cp (T1 – T3) = m cp (T3 – T2) ...(where cp is the specific heat at constant pressure)

156 Engineering Thermodynamics or

T1 – T3 = T3 – T2

T1 + T2 ...(i) 2 We know that change of entropy of the gas in the first compartment at constant pressure, \

T3 =

T dS1 = m cp loge 3 T1

m kg Gas at T1

Since T3 is less than T1, therefore dS1 will be negative. Similarly, change of entropy of the gas in the second compartment at constant pressure,

T dS2 = m cp loge 3 T2

1

T1 > T2

m kg Gas at T2

2

Insulation

Fig. 7.8

Since T3 is greater than T2, therefore dS2 will be positive. \ Change of entropy of the universe,

T (dS)universe = dS1 + dS2 = m cp loge 3 + m cp loge T1

T T = m cp log e 3 + log e 3 T1 T2

T3 T2

Substituting the value of T3 from equation (i), we have

T + T2 T1 + T2 (dS)universe = m cp log e 1 + log e 2 T 2T2 1

T + T2 T1 + T2 = m cp log e 1 log e 2T1 2T2

...[Q loge x + loge y = loge (x × y)]

(T + T ) 2 2 = m cp loge 1 4 (T T ) 2 1 2 T + T2 = m cp loge 1 2 T1 T2

2

= 2 m cp loge

T1 + T2 Ans. 2 T1T2

7.11 CHANGE OF ENTROPY DURING POLYTROPIC (OR pv n = CONSTANT) PROCESS We have already discussed in Chapter 4 (Art. 4.14) that during a polytropic process, the heat transfer, γ−n γ−n dQ = × Workdone = × pdv γ −1 γ −1 or

δQ γ − n p dv × = T γ −1 T

...(Dividing throughout by T )

Entropy, Availability and Irreversibility 157 δQ p mR = = dS and pv = mR T or T T v Substituting these values in the above equation, we have γ−n dv × mR dS = γ −1 v We know that

Integrating this expression within the limits of initial and final states, 2 2 dv γ−n ∫1 dS = γ − 1 × mR ∫1 v γ−n 2 × mR [ log e v ]1 [S]12 = γ −1 v γ−n γ−n × mR [ log e v2 − log e v1 ] = × mR log e 2 γ −1 γ −1 v1 v γ−n = × m (cp – cv) 2.3 log 2 γ −1 v1

S2 – S1 =

c p − cv = 2.3 m cv

...(i)

...( R = cp – cv)

γ −n v2 cv log ...(Multiplying and dividing by cv) γ − 1 v1

v2 γ − n = 2.3 m (g – 1) cv log γ − 1 v1

cp v v − n log 2 = 2.3 m cv (g – n) log 2 = 2.3 m cv v1 v1 cv

cp

cv

...

= γ

...(ii)

v = 2.3 m (cp – n . cv) log 2 v1 1

p n v Since, p1 v1n = p2 v2n, or 2 = 1 , therefore equation (ii) may be written as v1 p2 1

p n S2 – S1 = 2.3 m cv (g – n) log 1 p2 p1 γ − n = 2.3 m cv log n p2

...(iii)

1

T n −1 v2 We also know that = 1 v1 T2 Now the equation (ii) may also be written as

1

T n −1 S2 – S1 = 2.3 m cv (g – n) log 1 T2

T1 γ − n = 2.3 m cv log n −1 T2

...(iv)

158 Engineering Thermodynamics Notes: 1. The equations (iii) and (iv) may also be expressed in the following forms: We know that

p1 γ −n S2 – S1 = 2.3 m cv log n p2 −1

= 2.3 m ×

= 2.3 mR ×

R ... c p − cv = R or γ − 1 = c v

p2 R γ −n log γ − 1 n p1 p n−γ log 2 n ( γ − 1) p1

2. Now from equation (iv),

T1 T2 γ −n R γ −n S2 – S1 = 2.3 m cv log = 2.3 m × log 1 1 − γ − n − n 1 T 2 T1

−1

T2 n−γ = 2.3 m R log . ( γ − 1)(n − 1) T1

Example 7.10. The initial pressure, volume and temperature of a gas are 28 bar, 0.04 m3 and 2200 K respectively. The gas expands in an engine cylinder to a volume of 0.17 m3 and a pressure of 4.2 bar. Find mean index of expansion, the heat flow to or from the gas and the change of entropy. Take gas constant, R = 0.3 kJ/kg K and cv = 0.795 kJ/kg K. Solution. Given: Initial pressure, p1 = 28 bar = 2800 × 103 N/m2 Initial volume, v1 = 0.04 m3 Initial temperature, T1 = 2200 K Final volume, v2 = 0.17 m3 Final pressure, p2 = 4.2 bar = 420 × 103 N/m2 Gas constant, R = 0.3 kJ/kg K = 300 J/kg K Specific heat at constant volume, cv = 0.795 kJ/kg K The polytropic process on the p-v and T-S diagram is shown in Fig. 7.9. Mean index of expansion Let n = Mean index of expansion. n

v p p1 v1 = p2 v2 or 1 = 2 p1 v2

We know that

n

n

Taking log on both sides,

v p n log 1 = log 2 or n log v2 p1

n (– 0.628) = – 0.824

\

n =

− 0.824 = 1.312 Ans. − 0.628

0.04 4.2 = log 0.17 28

Entropy, Availability and Irreversibility 159 Heat flow First of all, let us find the temperature (T2) of the gas at the end of expansion. We know that p1v1 pv = 2 2 T1 T2 T2 =

\

1 n

pv = C

Pressure

1

T1

Temperature

p1

p2 v2T1 420 × 103 × 0.17 × 2200 = = 1402.5 K p1v1 2800 × 103 × 0.04

p2

v1

2

T2

2 v2

S1

Volume (a) p -v diagram.

S2 Entropy (b) T-S diagram.

Fig. 7.9

and mass of the gas in the engine cylinder,

m =

p1v1 2800 × 103 × 0.04 = = 0.17 kg 300 × 2200 RT1

We know that heat flow during the polytropic process, mR(T1 − T2 ) Q1 – 2 = m cv (T2 – T1) + n −1 = 0.17 × 0.795 (1402.5 – 2200) +

...( p1v1 = m R T1)

0.17 × 300 (2200 − 1402.5) 1.372 − 1

= – 107.8 + 130.4 = 22.67 kJ Ans. The positive sign of Q1 – 2 indicates that the heat flows into the gas. Change of entropy We know that change of entropy (in terms of volume and temperature),

T v S2 – S1 = 2.3 m cv log 2 + R log 2 T1 v1

1402.5 0.17 = 2.3 × 0.17 0.795 log + 0.3 log 2200 0.04

= 0.391 (– 0.1554 + 0.1885) = 0.013 kJ/K Ans. Example 7.11. 0.2 kg of air with pressure 1.5 bar and temperature 27°C is compressed to a pressure of 15 bar according to the law pv1.25 = constant. Determine: 1. Initial and final parameters of air; 2. Workdone on or by the air ; 3. Heat flow to or from the air; 4. Change of entropy stating whether it is an increase or decrease.

160 Engineering Thermodynamics Solution. Given: Mass of air, m = 0.2 kg Initial pressure, p1 = 1.5 bar = 150 × 103 N/m2 Initial temperature, T1 = 27°C = 27 + 273 = 300 K Final pressure, p2 = 15 bar = 1500 × 103 N/m2 Polytropic index, n = 1.25 1. Initial and final parameters of air Let v1 and v2 be the initial and final volume of air. The final temperature (T2) of the air is given by \

p T2 = 2 T1 p1

n −1 n

15 = 1.5

1.25 −1 1.25

= (10)0.2 = 1.585

T2 = T1 × 1.585 = 300 × 1.585 = 475.5 K

= 475.5 – 273 = 202.5°C Ans. We know that p1 v1 = mRT1 \

v1 =

mRT1 0.2 × 287 × 300 = = 0.1148 m3 Ans. p1 150 × 103

Similarly

...(Taking R for air = 287 J/ kg K) mRT2 0.2 × 287 × 475.5 v2 = = = 0.0182 m3 Ans. p2 1500 × 103

2. Workdone on or by the air We know that workdone during polytropic process,

W1–2 =

mR (T1 − T2 ) 0.2 × 287 (300 − 475.5) = = – 40.3 kJ Ans. 1.25 − 1 n −1

The negative sign shows that the work is done on the air. 3. Heat flow We know that heat flow during polytropic process, γ−n Q1–2 = × Workdone n −1 1.4 − 1.25 = × (– 40.3) = – 24.18 kJ Ans. 1.25 − 1 The negative sign shows that the heat flows from the air. 4. Change of entropy We know that change of entropy during polytropic process

...(Taking g for air = 1.4)

T1 γ − n S2 – S1 = 2.3 m cv log n −1 T2 1.4 − 1.25 300 = 2.3 × 0.2 × 0.712 log 1 . 25 − 1 475.5 = 0.1965 × (– 0.2) = – 0.0393 kJ/K Ans. The negative sign shows that there is a decrease in entropy.

...(Taking cv = 0.712 kJ/kg K)

Entropy, Availability and Irreversibility 161 Example 7.12. An ideal gas of molecular mass 30 and specific heat ratio as 1.38, is compressed according to the law pv1.25 = constant, from a pressure of 1 bar and 60°C to a pressure of 16 bar. Find the change in entropy, assuming 1 kg mass of the gas. Use only calculated values of cp and cv. Solution. Given: Molecular mass of gas, M = 30 cp Specific heat ratio, g = = 1.38 cv Polytropic index,

n = 1.25

Initial pressure,

p1 = 1 bar = 100 × 103 N/m2

Initial temperature,

T1 = 60°C = 60 + 273 = 333 K

Final pressure,

p2 = 16 bar = 1600 × 103 N/m2

First of all, let us find the values of cp and cv Since the universal gas constant (Ru) for all gases is 8.314 kJ/kg K, therefore characteristic gas constant, R 8.314 R = u = = 0.277 kJ/kg K = 277 J/kg K M 30 We know that and

cp – cv = R = 0.277 kJ/kg K cp

...(i)

= g = 1.38

cv

...(ii)

From equations (i) and (ii), cv = 0.729 kJ/kg K ; and cp = 1.006 kJ/ kg K Let v1 = Initial volume of the gas, v2 = Final volume of the gas, and T2 = Final temperature of the gas. We know that p1 v1 = mR T1 mRT1 1 × 277 × 333 = \ v1 = = 0.9224 m3 p1 100 × 103 n

and

p1 v1 = p2 v2 n

n

v p or 1 = 2 p1 v2 1

\ or

1

p n 16 1.25 v1 = 2 = = (16)0.8 = 9.19 v2 1 p1 v2 =

v1 0.9224 = = 0.1004 m3 9.19 9.19

Now using the relation, \

p T2 = 2 T1 p1

n −1 n

16 = 1

1.25 −1 1.25

= (16)0.2 = 1.741

T2 = T1 × 1.741 = 333 × 1.741 = 580 K

162 Engineering Thermodynamics We know that change of entropy,

T1 γ − n S2 – S1 = 2.3 m cv log n − 1 T2

1.38 − 1.25 333 = 2.3 × 1 × 0.729 log 580 1.25 − 1 = 0.872 (– 0.241) = – 0.21 kJ/K Ans. The negative sign indicates that there is a decrease in entropy.

7.12 AVAILABLE AND UNAVAILABLE ENERGY

Temperature

According to second law of thermodynamics,when heat dQ is transferred to a system operating in a cycle, only a portion of heat is available for work. Thus, a part of the heat energy which can be converted into work under ideal conditions, is called available energy. The part of the heat energy which can not be converted into work and is rejected to the sink, is known as unavailable energy. The state of the sink thus puts an upper limit to the available energy of the system. The available energy of the system becomes maximum when its state is brought to the state of surroundings (or what is called a dead state), by reversible processes. From above, we have that the total heat energy or heat supplied to the system, 1 2 T1 dQ = Available heat energy (A.H.E.) Available heat + Unavailable heat energy (U.H.E.) energy = Workdone + Heat rejected When considering available and unavail4 able parts of energy, the following two cases may 3 T0 arise: Unavailable heat energy Case 1. Heat withdrawn from an infinite reservoir, i.e. when heat is withdrawn at constant temperature. Consider a heat reservoir at constant A B dS temperature T1, from which a quantity of heat dQ Entropy is withdrawn. The maximum work which it can Fig. 7.10. Available and unavailable heat energy. produce is through a reversible cycle i.e. a Carnot cycle. The heat rejection is generally to the *atmosphere at temperature T0, as shown in Fig. 7.10. We know that the efficiency of the reversible engine (Carnot engine) operating between the temperature limits of T1 and T0 is given by T − T0 T =1− 0 h = 1 ...(i) T1 T1 We also know that efficiency,

h =

From equations (i) and (ii)

1−

Maximum work obtained (δW ) Heat supplied (δQ)

T0 δW = T1 δQ

* The sink is generally the atmosphere.

...(ii)

Entropy, Availability and Irreversibility 163 \ Maximum work obtained or available heat energy (A.H.E.), T T dW = Area 1-2-3-4 = dQ 1 − 0 = dQ – dQ × 0 = dQ – T0 . dS T1 T1 T or dQ = dW + dQ × 0 = dW + T0 . dS ...(iii) T1 We have already discussed above that dQ = A.H.E. + U.H.E. ...(iv) From equations (iii) and (iv), we have Unavailable heat energy, T δQ U.H.E. = Area A-4-3-B = dQ × 0 = T0 × = T0 × dS T1 T1 where dS = Change of entropy. From above, we see that unavailable heat energy (U.H.E.) is the product of lowest temperature (T0) of heat rejection and the change of entropy of the system (dS) during the process of supplying heat. It may be noted that during irreversible process, the entropy of system increases or less energy is available for work. Case 2. Heat withdrawn from a finite reservoir, i.e. when heat withdrawn is not at a consant temperature. In a finite reservoir, the temperature of the reservoir would change as heat is withdrawn and hence the heat is supplied at varying temperature as shown in Fig. 7.11. In order to get maximum work from this process, we will have to assume a series of infinitesimal Carnot cycles getting heat dQ at temperature T (different for each Carnot cycle) and discarding heat at the sink temperature Fig. 7.11. Available and unavailable T0 (constant for all such cycles). The maximum work (or heat energy. available heat energy) for this case is given by T δQ A.H.E. = Wmax. = ∫ δQ 1 − 0 = ∫ δQ − T0 ∫ = ∫ δQ − T0 × dS T T

Example 7.13. 2 MJ of heat is withdrawn from a heat reservoir at 300°C. If the temperature of the surroundings is 15°C, find the available and unavailable part of the heat energy. Solution. Given: Heat withdrawn from a heat reservoir, dQ = 2 MJ Temperature (constant) at which heat is withdrawn, T1 = 300°C = 300 + 273 = 573 K Temperature of the surroundings, T0 = 15°C = 15 + 273 = 288 K We know that available heat energy, δQ δQ A.H.E. = dQ – T0 dS = dQ – T0 × ... dS = T1 T1 2 = 2 – 288 × = 2 – 1.005 = 0.995 MJ = 995 kJ Ans. 573

164 Engineering Thermodynamics and unavailable part of heat energy,

U.H.E. = dQ ×

T0 288 = 2× = 1.005 MJ = 1005 kJ Ans. T1 573

Note: Unavailable part of heat energy is also given by U.H.E. = dQ – A.H.E. = 2 – 0.995 = 1.005 MJ = 1005 kJ.

Example 7.14. In a certain process, vapour is condensed at 500°C by transferring energy to water, which in turn is vaporised at 250°C. The resulting water vapour is used in a Carnot engine with the ambient atmosphere at 25°C as its sink. Determine the fraction of the available energy lost due to the irreversible energy transfer. Solution. Given: Temperature at which vapour is condensed, T1 = 500°C = 500 + 273 = 773 K Temperature at which water is vaporised, T2 = 250°C = 250 + 273 = 523 K Ambient temperature, T0 = 25°C = 25 + 273 = 298 K Let dQ be the heat energy transferred by condensing vapour. \ Availability of heat dQ at temperature of 500°C,

T 298 (A.H.E.)500 = dQ 1 − 0 = dQ 1 − = 0.6145 dQ T1 773

...(i)

Similarly, availability of same heat dQ after being transferred to water vapour at temperature of 250°C,

T (A.H.E.)250 = dQ 1 − 0 = dQ T2

298 1 − = 0.43 dQ 523

\ Fraction of the availability lost =

(A.H.E.)500 − (A.H.E.) 250 (A.H.E.)500

0.6145 δQ − 0.43 δQ = 0.3 Ans. 0.6145 δQ

=

Example 7.15. 4 kg of water at 50°C is mixed with 6 kg of water at 80°C in a steady flow process. Determine: 1. The temperature of the resulting mixture; 2. Is the mixing process isentropic? If not, what is change in entropy; and 3. Unavailable energy with respect to the surroundings at 50°C. Solution. Given: Mass of water at temperature, t1 = 50°C (or T1 = 50 + 273 = 323 K), m1 = 4 kg Mass of water at temperature, t2 = 80°C (or T2 = 80 + 273 = 353 K), m2 = 6 kg 1. Temperature of the resulting mixture Let t3 be the temperature after mixing of (m1 + m2) of water. According to first law of thermodynamics, m1 h1 + m2 h2 = (m1 + m2) h3

Entropy, Availability and Irreversibility 165 Assuming that the specific heat of water (cpw) is constant, therefore

*m1 cpw (t1 – 0) + m2 cp (t2 – 0) = (m1 + m2) cp (t3 – 0) t3 =

\ or

m1t1 + m2 t2 4 × 50 + 6 × 80 = = 68°C 4+6 m1 + m2

T3 = 68 + 273 = 341 K Ans.

2. Change in entropy During mixing of water, pressure remains constant. We know that change in entropy due to the mixing process,

T T dS = 2.3 m1 cp w log 3 + 2.3 m2 cpw log 3 T1 T2

341 341 = 2.3 × 4 × 4.187 log + 2.3× 6 × 4.187 log 323 353

= 38.52 × 0.0237 + 57.78 (– 0.015)

= 0.913 – 0.868 = 0.045 kJ/K Since the change in entropy is greater than zero, therefore the mixing process is not an isentropic process. It is an irreversible process which results in increase in entropy. 3. Unavailable energy Given: Temperature of surroundings, T0 = 50°C = 50 + 273 = 323 K We know that unavailable energy, U.H.E. = T0 × dS = 323 × 0.045 = 14.535 kJ Ans. Example 7.16. 1.5 kg of gas flows through gas turbine unit from its initial pressure and temperature 600 kN/m2 and 1300 K respectively and exhausts at a pressure of 102 kN/m2 and a temperature of 600 K to the atmosphere. The atmospheric pressure and temperature are 100 kN/m2 and 298K. Calculate availability at the entrance to the gas turbine and exhaust of the gas turbine. Take necessary assumptions. Solution. Given: Mass of the gas, m = 1.5 kg Initial pressure, p1 = 600 kN/m2 Initial temperature, T1 = 1300 K Exhaust pressure, p2 = 102 kN/m2 Exhaust temperature, T2 = 600 K Atmospheric pressure, p0 = 100 kN/m2 Atmospheric temperature, T0 = 298 K * This expression may be written as m1 cp (T1 – 273) + m2 cp (T2 – 273) = (m1 + m2) cp (T3 – 273)

\

T3 – 273 =

m1 (T1 − 273) + m2 (T2 − 273) m1 + m2

166 Engineering Thermodynamics Availability at the entrance to the gas turbine We know that total heat energy at the entrance of the turbine, Q1 = m cp (T1 – T0) = 1.5 × 1.004 (1300 – 298) = 1509 kJ and change of entropy in terms of pressure and temperature,

...(Taking cp = 1.004 kJ/kg K)

T p dS = 2.3 m c p log 1 + R log 0 p1 T0

1300 100 = 2.3 × 1.5 1.004 log + 0.287 log 298 600 ...(Taking R = 0.287 kJ/kg K) = 3.45 (0.6423 – 0.2233) = 1.4455 kJ/K \ Availability at the entrance of the gas turbine = Q1 – *T0 × dS = 1509 – 298 × 1.4455 = 1509 – 431 = 1078 kJ Ans. Availability at exhaust of the gas turbine We know that total heat energy at the exhaust of the turbine, **Q2 = m cp (T2 + T0) = 1.5 × 1.004 (600 + 298) = 1352 kJ T and change of entropy, dS = 2.3 m c p log 2 T0

p0 + R log p2

600 100 = 2.3 × 1.5 1.004 log + 0.287 log 298 102 = 3.45 (0.3038 – 0.00252) = 1.04 kJ/K \ Availability at exhaust of the gas turbine = Q2 – T0 × dS = 1352 – 298 × 1.04 = 1042 kJ Ans.

7.13 LOSS IN AVAILABLE ENERGY We have already discussed that when heat flows from a higher temperature to a lower temperature, it is an irreversible process and there is a loss of available energy. For example, let a quantity of heat Q flows from a hot body at temperature T1 to a cold body at temperature T2 through a conducting material say a metal bar, as shown in Fig. 7.12. If T0 is the temperature of the atmosphere, then Available energy of heat Q at temperature T1 T = Q 1 − 0 T1 Similarly available energy of same heat Q after conduction at temperature T2 T = Q 1 − 0 T 2

* T0 × dS is the unavailable part of heat energy.

** Total temperature at exhaust of the gas turbine = T2 + T0

Entropy, Availability and Irreversibility 167 \ Loss of available energy T = Q 1 − 0 T1

= Q – Q ×

T0 − Q 1 − T2

T0 T −Q +Q× 0 T1 T2

Q Q = T0 − = T0 (S2 – S1) T2 T1 = Unavailable heat energy Q where S2 = = Gain of entropy by the cold body, and T2

S1 =

Fig. 7.12. Loss in available energy.

Q = Loss of entropy by the hot body. T1

Note: When an irreversible process takes place, then the entropy of the system increases and the available energy decreases. The larger the temperature difference, greater is the irreversibility and hence greater is the loss of available energy. It is, therefore, desirable that heat should be transferred with as small a temperature difference as possible.

Example 7.17. A system at 450 K receives 225 kJ/s of heat energy from a source of 1500 K and the temperature of both the system and source remain constant during the heat transfer process. Draw the process on temperature-entropy diagram and determine: 1. Net change in entropy; 2. Available energy of heat source and system; and 3. Decrease in available energy. Take atmospheric temperature as 300 K. Solution. Given: Temperature of the system, T2 = 450 K Heat received by the system, Q = 225 kJ/s Temperature of the source, T1 = 1500 K Atmospheric temperature, T0 = 300 K The process on the temperature-entropy diagram is shown in Fig. 7.13. 1. Net change in entropy We know that loss of entropy by the source during heat transfer, Q 225 S1 = = = 0.15 kJ/s-K T1 1500 and gain in entropy by the system during heat transfer, Q 225 S2 = = = 0.5 kJ/s-K T2 450

Fig. 7.13

168 Engineering Thermodynamics \ Net change in entropy, Snet = S2 – S1 = 0.5 – 0.15 = 0.35 kJ/s-K Ans. 2. Available energy of heat source and system We know that available energy with the heat source = (T1 – T0) S1 = (1500 – 300) 0.15 = 180 kJ/s Ans. Similarly, available energy with the system = (T2 – T0) S2 = (450 – 300) 0.5 = 75 kJ/s Ans. 3. Decrease in available energy We know that decrease in available energy = Available energy with the heat source – Available energy with the system = 180 – 75 = 105 kJ/s Ans. Note: The decrease or loss in available energy is also given by T0 (S2 – S1) i.e. 300 (0.5 – 0.15) = 105 kJ/s Ans.

Example 7.18. In a parallel flow heat exchanger, water enters at 60°C and leaves at 80°C while oil of specific gravity 0.8 enters at 250°C and leaves at 100°C. The specific heat of oil is 2.5 kJ/kg K and the surrounding temperature is 300 K. Determine the loss in availability per kg of oil flow per second. Solution. Given: Temperature of water entering, Tw1 = 60°C = 60 + 273 = 333 K Temperature of water leaving, Tw2 = 80°C = 80 + 273 = 353 K Specific gravity of oil, = 0.8 Temperature of oil entering, T01 = 250°C = 250 + 273 = 523 K Temperature of oil leaving, T02 = 100°C = 100 + 273 = 373 K Specific heat of oil, c0 = 2.5 kJ/kg K Temperature of surroundings, T0 = 300 K 250°C Oil The parallel flow heat exchanger is shown in Fig. 7.14. 100°C Considering one kg of oil (m0 = 1 kg) and let mw be the mass of water. We know that 80°C Heat loss by oil = Heat gained by water Water 60°C m0 . c0 (T01 – T02) = mw . cw (Tw2 – Tw1) Fig. 7.14 1 × 2.5 (523 – 373) = mw × 4.18 (353 – 333) 375 = 83.6 mw ...(Taking cw for water = 4.18 kJ/kg K) \ mw = 375/83.6 = 4.48 kg We know that change of entropy of water,

T (dS )w = 2.3 mw cw log w2 Tw1 353 = 2.3 × 4.48 × 4.18 log = 1.09 kJ/K 333

Entropy, Availability and Irreversibility 169 and change of entropy of oil,

T (dS)0 = 2.3 m0 c0 log 02 T01

373 = 2.3 × 1 × 2.5 log = – 0.844 kJ/K 523 We know that change in availability of water = mw cw (Tw2 – Tw1) – T0 (dS)w = 4.48 × 4.18 (353 – 333) – 300 × 1.09 = 374.5 – 327 = 47.5 kJ and change in availability of oil = m0 c0 (T02 – T01) – T0 (dS)0 = 1 × 2.5 (373 – 523) – 300 (– 0.844) = – 375 + 253.2 = – 121.8 kJ \ Change in availability = 47.5 – 121.8 = – 74.3 kJ Ans. The negative sign indicates the loss in availability. Example 7.19. Exhaust gases leave an internal combustion engine at 800°C and 1 atmosphere, after having done 1050 kJ of work per kg of gas in the engine (cp of gas = 1.1 kJ/kg K). The temperature of surroundings is 30°C. Determine: 1. How much available energy per kg of gas is lost by throwing away the exhaust gases? 2. What is the ratio of the lost available energy to the engine work? Solution. Given: Temperature of exhaust gases, TG = 800°C = 800 + 273 = 1073 K Pressure, pG = 1 atm = 1.013 bar = 101.3 × 103 N/m2 Workdone, W = 1050 kJ/ kg cp of gas, = 1.1 kJ/kg K Temperature of surroundings, T0 = 30°C = 30 + 273 = 303 K 1. Available energy lost per kg of gas We know that total heat energy of the exhaust gases, Q1 = m cp (TG – T0) = 1 × 1.1 (1073 – 303) = 847 kJ/kg and change of entropy during constant pressure process,

T 1073 dS = 2.3 m cp log G = 2.3 × 1 × 1.1 log 303 T0 = 1.39 kJ/kg K \ Unavailable heat energy, Q2 = T0 × dS = 303 × 1.39 = 421.2 kJ/kg We know that available energy lost by throwing away the exhaust gases or net available heat energy, A.H.E. = Wmax = Q1 – Q2 = 847 – 421.2 = 425.8 kJ/kg Ans. 2. Ratio of the lost available energy to the engine work We know that ratio of the lost available energy to the engine work A.H.E. 425.8 = = = 0.405 Ans. W 1050

170 Engineering Thermodynamics

7.14 AVAILABILITY The availability of a system is defined as the maximum possible useful work that can be obtained in constant environmental conditions (i.e. pressure p0 and temperature T0). The system should come to equilibrium with its surroundings. The availability is thus a composite property depending upon the state of both the system and the surroundings. It is briefly written by the letter A and its unit is Joule.

7.15 AVAILABILITY OF A CLOSED SYSTEM When a system is in equilibrium with the atmosphere, it will not be able to do any more work. Consider a closed system with initial values of pressure p, volume v, temperature T, entropy S and internal energy U. Let the system is surrounded by a local environment of constant temperature T0 and constant pressure p0. Since the system and the environment are not in equilibrium, therefore these two may interact and produce work. The total useful work transfer (WT) is made up of the maximum useful work (WU) and the workdone on the surroundings. A part of workdone by the system is spent in overcoming the resistance of the atmosphere. This part of the work is p0 dv, where p0 is the pressure of the atmosphere and dv is the change in volume of the system. Thus, the total useful work, WT = WU + p0 dv We know that according to first law of thermodynamics, Q = dU + WT = dU + WU + p0 dv or T0 × dS = dU + WU + p0 dv ...(Q Q = T0 × dS) where dS is the change in entropy of the system. \ Maximum useful work, WU = T0 dS – dU – p0 dv = T0 (S0 – S) – (U0 – U) – p0 (v0 – v) = (U + p0 v – T0 × S) – (U0 + p0 v0 – T0 × S0 ) = f – f0 Let f = U + p0 v – T0 × S where f is known as the availability function. Since the availability (A) of a closed system is defined as the maximum useful work, therefore A = WU = f – f0 Thus the maximum useful work is equal to the decrease in availability function during non-flow process. Let a closed system be taken from an initial equilibrium state 1 to a second equilibrium state 2. Let the corresponding availability is A1 and A2 respectively. The end state 2 is not in equilibrium with the environment. Now the maximum useful work that can be obtained in the process is given by WU = A1 – A2 = (f1 – f0) – (f2 – f0) = f1 – f2 The expression (A1 – A2) represents the loss of work potential during the process 1-2.

7.16 AVAILABILITY IN A STEADY FLOW PROCESS We have already discussed that the steady flow energy equation for a unit mass flow is given by q1 – 2 – w1 – 2 = (pe2 – pe1) + (ke2 – ke1) + (h2 – h1) Consider a system comprising mass m of the working substance flowing through a device in a steady flow. Assuming the changes in the potential energy and kinetic energy as negligible, the steady flow energy equation for the maximum useful work may be written as

Entropy, Availability and Irreversibility 171 Q – WU = H2 – H1 or Q = WU + (H2 – H1) where H1 and H2 is the enthalpy at the inlet and outlet of the system. Let the system at outlet, be in equilibrium with the environment at pressure p0 and temperature T0. Let the symbol without subscript represents the inlet condition of the system. \ Q = WU + (H0 – H) or T0 × dS = WU + (H0 – H) ... (Q Q = T0 × dS) and WU = T0 × dS – (H0 – H) = T0 (S0 – S) – (H0 – H) = (H – T0 × S) – (H0 – T0 × S0) Let B = H – T0 × S where B is called the availability function for steady flow. The function B, like the function f, is a composite property of the system and its environment, involving two extensive properties of the system (i.e. H and S) and one intensive property of the environment (i.e. T0). The availability (A) of the steady flow process is defined as A = B – B0 The maximum work obtainable from a system at the entrance of a device, when the pressure and temperature at the output are those of the environment is called the availability of the system in steady flow, and is given by A = WU = B – B0 In case the pressure and temperature at the outlet of the steady flow device are not those of the environment, then the maximum useful work (WU) is the difference between the availabilities at the inlet and outlet, i.e. A = A1 – A2 = (B1 – B0) – (B2 – B0) or A = WU = B1 – B2 Note: The alternative name for availability is exergy.

7.17 HELMHOLTZ FUNCTION Helmholtz function is the property of a system and is given by the equation, F = U – T × S ...(i) where F = Helmholtz function, U = Internal energy, T = Absolute temperature, and S = Entropy. Since (U – T × S) is made up entirely of properties, therefore Helmholtz function (F ) is also a property. In practice, the primary function of heat engines and other devices is to perform mechanical work. From the first law of thermodynamics, for a system working between two equilibrium states, dQ = dU + dW or dW = dQ – dU ...(ii) It means that the energy converted into work is provided partly by the heat reservoir with which the system is in contact and which gives up a quantity of heat dQ and partly by the system whose internal energy decreases by (– dU). When a system undergoes a process between two equilibrium states, the maximum workdone may be obtained, assuming that the system exchanges heat energy only with a single heat reservoir at a temperature T0.

172 Engineering Thermodynamics Let dS = Increase in entropy of the system, dS0 = Increase in entropy of the surroundings (reservoir), dQ = Heat absorbed by the system from the reservoir, and T0 = Temperature of the reservoir. From the principle of increase of entropy, we know that the sum of the increase in entropy of the system and that of the surrounding is equal to or greater than zero, i.e. dS + dS0 0 ...(iii) − δQ For a reservoir, dS0 = T0 Since the reservoir has given out heat, therefore dQ is taken negative. Now, the equation (iii) is written as δQ dS – 0 or T0 . dS dQ ...(iv) T0 From equations (ii) and (iv), we have dW T0 . dS – dU ...(v) The quantities U and S are the properties of the system and T0 is a constant. Therefore, for a finite process between state 1 and state 2, integrating equation (v),

2

2

∫1 δW ∫1

2

T0 ⋅ dS − ∫ dU 1

or W1 – 2 T0 (S2 – S1) – (U2 – U1) (U1 – U2) – T0 (S1 – S2) ...(vi) In case the initial and the final temperatures are equal and are the same as that of the heat reservoir, then T = T0 where T is the temperature of the system. Now the equation (vi) may be written as (WT ) 1 – 2 (U1 – U2 ) T – T (S1 – S2 ) T ..(vii) Helmholtz function is given by F = U – T . S \ For the two equilibrium states 1 and 2 at the same temperature T, F1 – F2 = (U1 – T . S1) – (U2 – T . S2 ) = (U1 – U2) – T (S1 – S2 ) or (F1 – F2 ) T = (U1 – U2 ) T – T (S1 – S2 ) T ...(viii) Now from equations (vii) and (viii), (WT)1 – 2 (F1 – F2) T ...(ix) From this expression, we find that the maximum work that can be done in any process between two equilibrium states at the same temperature is equal to the decrease in Helmholtz function of the system. Here the system exchanges heat with a single heat reservoir at the same temperature. Moreover, maximum work is done when the process is reversible. If the process is irreversible, the work done is less than the maximum. It may be noted that the maximum work that can be done is equal to the decrease in the Helmholtz function of a system but the energy converted into work is provided partly by the system and the remaining by the heat taken from the heat reservoir. For maximum work, equation (ix) may be written as (WT )max = (WT )reversible = (F1 – F2 )T

Entropy, Availability and Irreversibility 173

7.18 GIBBS FUNCTION The Gibbs function of a system is given by where

G = U – T . S + p . v = H – T . S ...(Q H = U + pv) G = Gibbs function, U = Internal energy, T = Temperature, S = Entropy, p = Pressure, v = Volume, and H = Enthalpy = U + pv

...(i)

Since (H – T . S) is made up entirely of properties, therefore Gibbs function (G) is also a property. Consider a system that can do other forms of work in addition to ( pdv) like electrical work, magnetic work. If dW0 is some other form of work in addition to pdv, then

dW = pdv + dW0

...(ii)

Integrating this equation from state 1 to state 2, considering a process where the system works at constant pressure p0 , we have or

2

2

∫1 δW = ∫1

2

pdv + ∫ δW0 1

W1 – 2 = p0 (v2 – v1) + W0 = – p0 (v1 – v2) + W0

...(iii)

For a system that exchanges heat with a reservoir temperature T0, the workdone W1 – 2 (U1 – U2) – T0 (S1 – S2) Substituting the value of W1 – 2 from equation (iii), we get – p0 (v1 – v2) + W0 (U1 – U2) – T0 (S1 – S2) or W0 (U1 – U2) – T0 (S1 – S2) + p0 (v1 – v2) ...(iv) Consider a specific process, where the initial and the final states of the system and the surroundings are at the same temperature (T0 ) and pressure ( p0 ), then T0 = T ; and p0 = p Now, the equation (iv) may be written as (W0)p, T (U1 – U2)p, T – T (S1 – S2)p, T + p (v1 – v2)p, T ...(v) Gibbs function is given by G = U – T . S + pv \ For the two equilibrium states at the same pressure and temperature, (G1 – G2)p, T = (U1 – U2)p, T – T (S1 – S2)p, T + p (v2 – v1)p, T ...(vi) From equations (v) and (vi), (W0)p, T (G1 – G2)p,T ...(vii) Thus, the difference between Gibbs function of a system between the two equilibrium states sets the maximum limit to the work in addition to pdv work, provided the initial and the final states are at the same pressure and temperature and the system exchanges heat with a single heat reservoir. The workdone will be maximum when the process is reversible. If the process is irreversible, workdone will be less than the maximum.

7.19 IRREVERSIBILITY It is defined as the decrease or loss in the available energy due to irreversible processes. The causes of irreversibilities are as follows:

174 Engineering Thermodynamics (a) Heat transfer through finite temperature difference, (b) Friction, (c) Mixing of two fluids, and (d) Free expansion Since the actual workdone by the system is always less than the maximum workdone, therefore the difference between the two is called irreversibility. Mathematically, Irreversibility*, I = Maximum workdone – Actual workdone = Wmax – Wact We have already discussed in Art. 7.16 that the maximum workdone, for a steady flow process is Wmax = Decrease in availability function = (H1 – T0 S1) – (H2 – T0 S2) According to first law of thermodynamics, actual workdone by the system (neglecting changes in potential and kinetic energies) is given by **Wact = – Q – (H2 – H1) Since heat is rejected by the system, therefore Q is taken as negative. Thus, irreversibility, I = Wmax – Wact = [(H1 – T0 . S1) – (H2 – T0 . S2)] – [– Q – (H2 – H1)] = H1 – T0 . S1 – H2 + T0 . S2 + Q + H2 – H1 = T0 (S2 – S1) + Q = T0 (dS)system + Q ...(i) Now, change of entropy of the environment (surroundings) due to the receipt of heat Q at T0 is Q (dS)surroundings = or Q = T0 (dS)surroundings T0 Substituting this value of Q in equation (i), we have Irreversibility, I = T0 (dS)system + T0 (dS)surroundings = T0 (dS)universe Notes: 1. The same expression for irreversibility applies to both flow and non-flow processes. 2. The quantity [T0 (dS)system + T0 (dS)surroundings] represents an increase in unavailable energy (also known as anergy).

Example 7.20. One kg of air is compressed polytropically from 1 bar pressure and temperature of 300 K to a pressure of 6.8 bar and temperature of 370 K. Determine the irreversibility if the sink temperature is 293 K. Assume R = 0.287 kJ/kg K; cp = 1.004 kJ/kg K and cv = 0.716 kJ/kg K. Solution. Given: Mass of air, m = 1 kg Initial pressure, p1 = 1 bar Initial temperature, T1 = 300 K Final pressure, p2 = 6.8 bar Final temperature, T2 = 370 K Sink temperature, T0 = 293 K Gas constant, R = 0.287 kJ/kg K Specific heat at constant pressure, cp = 1.004 kJ/kg K Specific heat at constant volume, cv = 0.716 kJ/kg K * The term irreversibility is also known as lost work or degradation. ** The steady flow energy equation for mass m (neglecting potential and kinetic energies) is Q – W = H2 – H1 or W = Q – (H2 – H1)

Entropy, Availability and Irreversibility 175 We know that change in internal energy, dU = U2 – U1 = m cv (T2 – T1) = 1 × 0.716 (370 – 300) = 50.12 kJ/kg T p and change in entropy, dS = S2 – S1 = 2.3 m c p log 2 + (c p − cv ) log 1 T1 p2 370 1 = 2.3 × 1 1.004 log + (1.004 − 0.716) log 300 6.8 = 2.3 (0.0914 – 0.239) = – 0.3413 kJ/ kg \ Maximum workdone during a reversible process from state 1 to state 2, Wmax = (U1 – T0 . S1) – (U2 – T0 . S2) = (U1 – U2) + T0 (S2 – S1) = – 50.12 + 293 (– 0.3413) = – 150.12 kJ/kg The –ve sign indicates that the work is done on the air. In order to find the actual workdone during polytropic compression, first of all, let us find the value of polytropic index (n). We know that Taking log on both sides,

p T2 = 2 T1 p1

n −1 n

p T n −1 log 2 log 2 = n p1 T1

n −1 370 6.8 log log = n 300 1

n −1 × (0.8325) n 0.091 n = 0.8325 n – 0.8325 \ n = 1.227 We know that actual workdone during polytropic process, mR (T1 − T2 ) 1 × 0.287 (300 − 370) = Wact = = – 163.7 kJ/kg 1.227 − 1 n −1 The –ve sign indicates that during compression, work is done on the air. \ Irreversibility, I = Wmax – Wact = – 150.12 – (– 163.7) = 13.58 kJ/kg Ans. Example 7.21. A rigid tank of volume 2.5 m3 contains air at 200 kPa and 300 K. The air is heated by supplying heat from a reservoir at 600 K until temperature reaches 500 K. The surrounding atmosphere is at 100 kPa and 300 K. Determine the maximum useful work and the irreversibility associated with the process. Solution. Given: Volume of tank, v1 = 2.5 m3 Initial pressure of air, p1 = 200 kPa = 200 × 103 N/m2 Initial temperature of air, T1 = 300 K Temperature of reservoir, T2 = 600 K

0.091 =

176 Engineering Thermodynamics Final temperature, T3 = 500 K Pressure of atmosphere, p0 = 100 kPa = 100 × 103 N/m2 Temperature of atmosphere, T0 = 300 K Maximum useful work First of all, let us find the mass of air (m). We know that p1 v1 = mR T1 \

m =

p1v1 200 × 103 × 2.5 = = 5.8 kg 287 × 300 RT1

...(Taking R for air = 287 J/ kg K) Since the tank is a rigid one, therefore heat is supplied at constant volume. We know that heat supplied at constant volume, Q = Change in internal energy = cv (T3 – T1) = 0.712 (500 – 300) = 142.4 kJ/kg ... (Taking cv for air = 0.712 kJ/kg K) Change of entropy for air,

T 500 (dS)air = 2.3 cv log 3 = 2.3 × 0.712 log = 0.3647 kJ/kg K 300 T1

We know that maximum useful work or available energy = Heat supplied – Unavailable heat energy = Q – T0 (dS )air = 142.4 – 300 × 0.3647 = 33 kJ/kg = 33 × Mass of air = 33 × 5.8 = 191.4 kJ Ans. Irreversibility We know that change in entropy for surroundings, Q 142.4 (dS)surroundings = = = 0.237 kJ/kg K T2 600 \ Irreversibility, I = m T0 [(dS)air + (dS)surroundings] = 5.8 × 300 (0.3647 – 0.237) = 222 kJ Ans. ...(The –ve sign is given as the entropy of surroundings decrease due to heat transfer) Example 7.22. Two kg of air at 500 kPa, 80°C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100 kPa and 5°C. For this process, determine: 1. The maximum work; 2. The change in availability; and 3. The irreversibility. For air, take cv = 0.718 kJ/kg K; and R = 0.287 kJ/kg K. Solution. Given: Mass of air, m = 2 kg Initial pressure, p1 = 500 kPa = 500 × 103 N/m2 Initial temperature, T1 = 80°C = 80 + 273 = 353 K Final volume, v2 = 2 × Initial volume = 2v1 Final pressure, p2 = p0 = 100 kPa = 100 × 103 N/m2

Entropy, Availability and Irreversibility 177 Final temperature, T2 = T0 = 5°C = 5 + 273 = 278 K Specific heat at constant volume, cv = 0.718 kJ/kg K Gas constant, R = 0.287 kJ/kg K 1. Maximum work First of all, let us find the change in entropy of air between the initial and final states. We know that change of entropy in terms of volume and temperature is

T S2 – S1 = 2.3 m cv log 2 T1

v2 + R log v1

2v1 278 = 2.3 × 2 0.718 log + 0.287 log 353 v1

= 4.6 [0.718 (– 0.1037) + 0.287 (0.301)] = 4.6 (– 0.0744 + 0.0864) = 0.055 kJ/K and change in internal energy, U1 – U2 = m cv (T1 – T2) = 2 × 0.718 (353 – 278) = 107.7 kJ \ Maximum work, *Wmax = (U1 – U2) – T0 (S1 – S2) = (U1 – U2) + T0 (S2 – S1) = 107.7 + 278 × 0.055 = 107.7 + 15.29 = 122.99 kJ Ans. 2. Change in availability First of all, let us find the initial volume (v1). We know that p1 v1 = mR T1 \

v1 =

mRT1 2 × 0.287 × 353 = = 0.405 m3 500 p1

We know that the change in availability = f1 – f2 = Wmax + p0 (v1 – v2) = Wmax + p0 (v1 – 2v1) = 122.99 + 100 (0.405 – 2 × 0.405) = 122.99 – 40.5 = 82.49 kJ Ans. 3. Irreversibility According to first law of thermodynamics, dQ = dU + dW We know that for an adiabatic process, dQ = 0, therefore actual workdone, dW = – dU = – (U2 – U1) = U1 – U2 = 107.7 kJ (Calculated above) \ Irreversibility, I = Maximum workdone – Actual workdone = 122.99 – 107.7 = 15.29 kJ Ans. Note: Irreversibility (I ) is also given by I = T0 (S2 – S1) + Q = T0 (S2 – S1) = 278 × 0.055 = 15.29 kJ Ans.

...(For adiabatic process, Q = 0)

Example 7.23. Air expands through a turbine from 500 kPa and 520°C to 100 kPa and 300°C. During expansion, 10 kJ/kg of heat is lost to the surroundings which is at 98 kPa and 20°C. Neglecting changes in potential and kinetic energies, determine per kg of air..

* Refer equation (vi) of Helmholtz function.

178 Engineering Thermodynamics 1. Decrease in availability; 2. Maximum work; and 3. Irreversibility. Take cp = 1.005 kJ/kg K. Solution. Given: Initial pressure, p1 = 500 kPa Initial temperature, T1 = 520°C = 520 + 273 = 793 K Final pressure, p2 = 100 kPa Final temperature, T2 = 300°C = 300 + 273 = 573 K Heat lost to surroundings, Q = 10 kJ/kg Pressure of surroundings, p0 = 98 kPa Temperature of surroundings, T0 = 20°C = 20 + 273 = 293 K Mass of air, m = 1 kg Specific heat at constant pressure, cp = 1.005 kJ/kg K 1. Decrease in availability We know that change in entropy of air during expansion (in terms of pressure and temperature),

T S2 – S1 = 2.3 m c p log 2 T1

p1 + R log p2

573 500 = 2.3 × 1 1.005 log + 0.287 log 793 100 ...(Taking R for air = 0.287 kJ/kgK] = 2.3 [1.005 (– 0.1411) + 0.287 (0.7)] = 2.3 (– 0.1418 + 0.201) = 0.136 kJ/kg K We know that decrease in availability, = (H1 – T0 . S1) – (H2 – T0 . S2) = (H1 – H2) – T0 (S1 – S2) = m cp (T1 – T2) – T0 (S1 – S2) = 1 × 1.005 (793 – 573) + 293 × 0.136 = 221.1 + 39.85 ...(Q S2 – S1 = 0.136 or S1 – S2 = – 0.136) = 260.95 kJ/kg Ans. 2. Maximum workdone Since the maximum workdone is equal to the decrease in availability, therefore maximum workdone, Wmax = 260.95 kJ/kg Ans. 3. Irreversibility According to steady flow energy equation, workdone, W = H1 – H2 = m cp (T1 – T2) = 1 × 1.005 (793 – 573) = 221.1 kJ/kg Since heat lost to the surroundings is 10 kJ/kg, therefore actual workdone, Wact = 221.1 – 10 = 211.1 kJ/kg

Entropy, Availability and Irreversibility 179 \ Irreversibility, I = Wmax – Wact = 260.95 – 211.1 = 49.85 kJ/kg Ans. Note: The irreversibility (I) may also be obtained by using I = T0 (S2 – S1) + Q = 293 × 0.136 + 10 = 49.85 kJ/kg Ans.

Example 7.24. Air enters a compressor in steady flow at 140 kPa, 17°C and 70 m/s and leaves it at 350 kPa, 127°C and 110 m/s. The environment is at 100 kPa, and 7°C. Calculate per kg of air, 1. The actual amount of work required, 2. The minimum work required, and 3. The irreversibility of the process. Solution. Given: Initial pressure, p1 = 140 kPa = 140 × 103 N/m2 Initial temperature, T1 = 17°C = 17 + 273 = 290 K Initial velocity, V1 = 70 m/s Final pressure, p2 = 350 kPa = 350 × 103 N/m2 Final temperature, T2 = 127°C = 127 + 273 = 400 K Final velocity, V2 = 110 m/s Environment pressure, p0 = 100 kPa = 100 × 103 N/m2 Environment temperature, T0 = 7°C = 7 + 273 = 280 K Mass of air, m = 1 kg 1. Actual amount of work required per kg of air We know that work required due to change in enthalpy, w1 = m cp (T2 – T1) = 1 × 1.005 (400 – 290) = 110.55 kJ/kg ...(Taking cp for air = 1.005 kJ/kg K) and work required due to change in velocity,

V 2 − V12 (110) 2 − (70) 2 w2 = m 2 = 1 = 3.6 kJ/kg 2000 2 × 1000

\ Actual amount of work required per kg of air Wact = w1 + w2 = 110.55 + 3.6 = 114.15 kJ/kg Ans. It may be noted that this amount of work is done on the air. 2. Minimum work required We know that change of entropy in terms of pressure and temperature,

T p S2 – S1 = 2.3 m c p log 2 + R log 1 T1 p2 400 140 = 2.3 × 1 1.005 log + 0.287 log 290 350

...(Taking R for air = 0.287 kJ/kg K) = 2.3 [1.005 × 0.1396 + 0.287 (– 0.398)] = 2.3 (0.1403 – 0.1142) = 0.06 kJ/kg K and work required due to change of entropy, w3 = T0 (S2 – S1) = 280 × 0.06 = 16.8 kJ/kg

180 Engineering Thermodynamics \ Minimum work required (For a steady flow process, reversible work, wrev = wmin),

wmin = w1 + w2 – w3 = wact – w3

= 114.15 – 16.8 = 97.35 kJ/kg Ans. 3. Irreversibility of the process We know that irreversibility of the process = wmin – wact = – 97.35 – (– 114.15) = 16.8 kJ/kg Ans. The –ve sign to wmin and wact is due to workdone on the air.

HIGHLIGHTS 1. Entropy is regarded as a measure of the unavailability of heat energy for transformation into work. It increases when heat is supplied and decreases when heat is removed. 2. The increase or decrease of entropy (dS) when multiplied by the absolute temperature (T) gives the heat absorbed or rejected (dQ) by the working substance. Mathematically, δQ dQ = T × dS or = dS T 3. The entropy is an extensive property of the system and remains constant during reversible adiabatic process. 4. The Clausius inequality states that when a closed system undergoes a cyclic process and during which it exchanges heat dQ with a number of reservoirs at respective temperatures T, then the cyclic integral of dQ/T is less than zero or equal to zero. This holds good for both reversible and irreversible cycles. Mathematically, δQ ∫ T 0 The equal sign holds for reversible cycle and inequality sign for irreversibile cycle. 5. For adiabatic reversible or irreversible adiabatic process, dQ = 0. Hence change of entropy, dS 0 i.e. change in entropy is zero (i.e. dS = 0) for reversible adiabatic process and more than zero (i.e. dS > 0) for irreversible adiabatic process. Since all the processes,in nature, are irreversible, therefore entropy of such a system like universe goes on increasing. This is known as principle of increase of entropy. 6. The general expressions for the change of entropy of an ideal gas is given as follows: (a) In terms of volume and temperature

T S2 – S1 = 2. 3 m cv log 2 T1

v2 + R log v1

T v = 2.3 m cv log 2 + (c p − cv ) log 2 T1 v1 (b) In terms of pressure and volume

p v S2 – S1 = 2.3 m cv log 2 + c p log 2 p1 v1

Entropy, Availability and Irreversibility 181 (c) In terms of pressure and temperature T S2 – S1 = 2.3 m c p log 2 T1

p1 + R log p2

T p = 2.3 m c p log 2 + (c p − cv ) log 1 T 1 p2 7. The change of entropy during constant volume (or isochoric) process is given by p T S2 – S1 = 2.3 m cv log 2 = 2.3 m cv log 2 p 1 T1 8. The change of entropy during constant pressure (or isobaric) process is given by v T S2 – S1 = 2.3 m cp log 2 = 2.3 m cp log 2 v1 T1 9. The change of entropy during constant temperature (or isothermal) process is given by v p S2 – S1 = 2.3 m R log 2 = 2.3 m R log 1 v1 p2 v p = 2.3 m (cp – cv) log 2 = 2.3 m (cp – cv) log 1 v1 p2 10. The change of entropy during polytropic process (or pv n = constant) is given by T1 v2 γ − n S2 – S1 = 2.3 m cv log = 2.3 m cv (g – n) log n −1 T2 v1 = 2.3 m (cp – n cv) log

p1 v2 γ −n = 2.3 m cv log n v1 p2

11. A part of heat energy which can be converted into work under ideal conditions, is called available energy. The part of heat energy which cannot be converted into work and is rejected to the sink, is known as unavailable energy. 12. The maximum work obtained (dW) or available heat energy (A.H.E.) is given by dW = dQ – T0 × dS and unavailable heat energy (U.H.E.) = T0 × dS where dQ = Heat supplied, T0 = Lowest temperature of heat rejection, and dS = Change of entropy during the process of supplying heat. 13. When heat (Q) flows from a higher temperature (say T1) to a lower temperature (say T2), then it is an irreversible process and there is a loss of available energy. If T0 is the temperature of the atmosphere, then loss of available energy Q Q = T0 − = T0 ( S2 − S1 ) T2 T1 where

Q = Gain of entropy by the cold body, and T2 Q S1 = = Loss of entropy by the hot body. T1 S2 =

182 Engineering Thermodynamics 14. The availability of a system is defined as the maximum possible useful work that can be obtained in constant environmental conditions (i.e. pressure, p0 and temperature, T0). 15. The availability (A) of a closed system is defined as the maximum useful work. Mathematically, A = WU = f – f0 where f = (U + p0 v – T0 × S) and f0 = (U0 + p0 v0 – T0 S0) The term f is known as availability function. 16. The availability (A) of the steady flow process is given as A = B – B0 where B = H – T0 × S and B0 = H0 – T0 × S0 The term B is called the availability function for steady flow. 17. Helmholtz function (F) is a property of the system and is given by F = U – T × S where U = Internal energy, T = Absolute temperature, and S = Entropy. 18. Gibbs function (G ) of a system is given by G = U – T × S + p . v = H – T . S where H = Enthalpy = U + p . v p = Pressure, and v = volume. 19. The irreversibility is defined as the decrease or loss in the available energy due to irreversible processes. Since the actual workdone (Wact) by the system is always less than the maximum workdone (Wmax), therefore the difference between the two is called irreversibility. 20. For a steady flow process, irreversibility (I ) is given by I = Wmax – Wact = [(H1 – T0 × S1) – (H2 – T0 × S2)] – [– Q – (H2 – H1)] = T0 (S2 – S1) + Q = T0 (dS)system + Q = T0 (dS)system + T0 (dS)surrounding where

(dS)surroundings =

Q T0

EXERCISES 1. A heat engine operates between two thermal reservoirs which are at 900 K and 300 K. The heat engine receives 500 kJ of heat from the source and rejects 300 kJ of heat to the sink at 300 K. Determine if this heat engine violates the second law of thermodynamics on the basis of Clausius inequality. [Ans. Satisfies Clausius inequality] 2. A heat engine receives 125 kJ of heat per cycle from a reservoir at 282°C and rejects heat to a reservoir at 5°C in the hypothetical amounts of (a) 94 kJ/cycle; (b) 62.5 kJ/cycle; and (c) 31 kJ/cycle. Which of these cases represent a reversible cycle, an irreversible cycle, and an impossible cycle.

Entropy, Availability and Irreversibility 183

[Ans. (a) Irreversible; (b) reversible; (c) impossible]

3. An inventor claims that the engine designed by him absorbs 300 kJ of energy from a reservoir at 325 K and delivers 75 kJ of work. He also states that the engine rejects 125 kJ of heat to a reservoir at 300 K and 100 kJ to a reservoir at 275 K. State whether his claim is true or false. [Ans. False] 4. 0.05 m3 of air at a pressure of 800 kPa and temperature of 550 K expands until its temperature becomes 300 K. If the final volume of the air becomes eight times its original volume, find the change of entropy during the process. Take specific heat at constant pressure as 1.005 kJ/kg K and gas constant as 0.290 kJ/kg K. [Ans. 0.0425 kJ/K] 5. A certain gas cp = 1.97 kJ/kg K and cv = 1.51 kJ/kg K. Find its molecular mass and the gas constant. A constant volume chamber of 0.3 m3 capacity contains 2 kg of this gas at 5°C. The heat is supplied to the gas until its temperature rises to 100°C. Find the change in enthalpy and entropy. [Ans. 18; 0.46 kJ/kg K; 374.3 kJ; 0.8863 kJ/K] 6. 0.34 m3 of a perfect gas at constant pressure of 2.8 bar is heated from 100°C to 300°C. It is then cooled at constant volume to its initial temperature. Determine the change of entropy during each process stating whether it is an increase or decrease. Also determine the overall change of entropy. Take cp = 1.05 kJ/kg K and cv = 0.75 kJ/kg K.

[Ans. 0.3834 kJ/K (increase); – 0.274 kJ/K (decrease); 0.109 kJ/K]

7. 0.05 m3 of a gas contained in a cylinder at 1 bar and 290 K is compressed isothermally and reversibly until the pressure is 5 bar. Calculate change of entropy and the heat transferred.

[Ans. – 27.7 J/K ; – 8.033 kJ]

8. A closed system contains 0.04 m3 of air at a pressure of 1 bar and a temperature 298 K. The system undergoes a thermodynamic cycle as follows: (a) heat is added at constant volume till the pressure becomes 5 bar; (b) cooling at constant pressure; and (c) isothermal heating to initial state. Calculate the change in entropy for each process. Take cp = 1.004 kJ/kg K ; and cv = 0.717 kJ/kg K. [Ans. 0.054 kJ/K ; – 0.755 kJ/K ; 0.216 kJ/K] 9. A 0.568 m3 capacity insulated vessel of oxygen at the pressure of 12.6 bar is stirred by internal paddle until the pressure becomes 21 bar. Find out: (a) Heat transferred; (b) Work input; and (c) Change in entropy per kg. Take cv = 0.658 kJ/kg K ; R = 0.260 kJ/kg K.

[Ans. (a) zero (b) – 283 kJ ; zero]

10. 0.2 m of air at 10 bar and 65°C is compressed to 0.02 m according to the law pv1.3 = constant. The ratio of specific heats is 1.4 and the gas constant is equal to 0.2927 kJ/kg K. The heat is now added at constant volume until its pressure is 70 bar. Determine the heat exchange with the cylinder walls during compression and change of entropy during each process. [Ans. 165.825 kJ ; – 0.34 kJ/K ; – 1.546 kJ/K] 3

3

11. 9 kg of air at 1.75 bar and 286 K is compressed to 24.5 bar according to the law pv1.32 = constant and then cooled at constant volume to 288 K. Determine: (a) the volume and temperature at the end of compression; and (b) change of entropy during compression and during constant volume cooling. [Ans. (a) 0.5714 m3, 542.26 K (b) – 1.024 kJ/K, – 4.05 kJ/K] 12. A mass of air of volume 0.04 m3 at 500 K and 6 bar is expanded at constant pressure to 0.12 m3. The gas is then expanded according to the law pv1.35 = constant, followed by a constant temperature process to complete the cycle. All the processes are reversible. Calculate the change of entropy during each process and over the cycle. Take cv = 0.717 kJ/kg K and cp = 1.004 kJ/kg K. [Ans. 0.184 kJ/K ; 0.0188 kJ/K ; – 0.203 kJ/K ; zero] 13. 2 kg of water at 90°C is mixed with 3 kg of water at 10°C in an insulated system. Calculate the change in entropy due to mixing process. Assume the specific heat of water as 4.2 kJ/kg K. [Ans. 0.179 kJ/K]

184 Engineering Thermodynamics 14. 15 MJ of heat is withdrawn from a thermal reservoir at 220°C. If the temperature of the heat sink is 15°C, determine the available and unavailable heat energy. [Ans. 6.24 MJ ; 8.76 MJ] 15. In a certain process, a vapour while condensing at 420°C, transfers heat at water evaporating at 250°C. The resulting steam is used in a power cycle which rejects heat at 35°C. What is the fraction of the available energy in the heat transferred from the process vapour at 420°C that is lost due to the irreversible heat transfer at 250°C? [Ans. 0.26] 16. 2 kg of water at 50°C is mixed with 3 kg of water at 100°C in a steady flow process. Determine the temperature of the resulting mixture. State whether the mixing process is isentropic? If not, what is the entropy change? Also find the unavailable energy with respect to the surrounding at 50°C. [Ans. 80°C ; 0.05 kJ/K ; 16.15 kJ] 17. A system at 600 K receives 8200 kJ/min heat from a source of 1000 K. The temperature of the atmosphere is 300 K. Assume that the temperature of system and the source remain constant during heat transfer. Determine: 1. The entropy produced during heat transfer; and 2. The decrease in available energy after heat transfer. [Ans. 5.47 kJ/min K ; 1639 kJ/min] 18. In an heat exchanger (parallel flow type), water enters at 50°C and leaves at 70°C while oil enters at 250°C and leaves at 80°C. If the surrounding temperature is 27°C, determine the loss in availability on the basis of one kg of oil per second. Take specific gravity of oil as 0.82 and specific heat as 2.6 kJ/kg K. [ Ans. – 90 kJ] 19. 1 kg of air is compressed polytropically from a pressure of 100 kN/m2 and a temperature of 17°C to a pressure of 500 kN/m2 and a temperature of 127°C. If the temperature and pressure of the surroundings is 15°C and 100 kN/m2 respectively ; determine the maximum workdone and the irreversibility. [Ans. – 119.15 kJ/kg ; 7.15 kJ/kg] 20. Air expands in a turbine adiabatically from 500 kPa ; 400 K and 150 m/s to 100 kPa ; 300 K and 70 m/s. The environment is at 100 kPa, and 17°C. Calculate per kg of air ; (a) the maximum work output; (b) the actual work output ; and (c) the irreversibility. For air, take cp = 1.005 kJ/kg K ; and R = 0.297 kJ/kg K. [Ans. 164 kJ/kg ; 109.3 kJ/kg ; 54.7 kJ/K]

QUESTIONS 1. What is entropy? Prove that entropy is a property of the system. 2. Derive an equation for inequality of Clausius. 3. State and explain the principle of increase of entropy and show that the entropy of universe is increasing. 4. Prove that for an irreversible process, δQ ∫ dS > ∫ T 5. Show that the specific entropy change of an ideal gas in a process is given by

v p s2 – s1 = cp loge 2 + cv loge 2 v 1 p1

where subscripts 1 and 2 relate to initial and final states and other symbols have their usual meaning. 6. Show that for an ideal gas, the slope of the constant volume line on a temperature-entropy diagram is higher than that of constant pressure line. 7. Show that the entropy change between state 1 and state 2 in a polytropic process ( pv n = constant) is given by the following relations:

Entropy, Availability and Irreversibility 185

T2 n−γ S2 – S1 = 2.3 mR log ( γ − 1) (n − 1) T1

p2 n−γ = 2.3 mR log n ( γ − 1) p1 8. Explain the following: (a) Available energy; (b) Unavailable energy; and (c) Loss in available energy. 9. Write down the availability function. Is it a property? Explain. 10. Discuss briefly Helmholtz function and Gibbs function. 11. Explain irreversibility. What are the causes of irreversibility? 12. Prove that for a steady flow process, the irreversibility is given by I = T0 (dS)universe

OBJECTIVE TYPE QUESTIONS 1. The second law of thermodynamics defines (a) heat (b) work (c) enthalpy

(d) entropy

2. The property of a working substance which increases or decreases as the heat is supplied or removed in a reversible manner, is known as (a) enthalpy (b) entropy (c) internal energy (d) external energy 3. The area under the temperature-entropy curve of any thermodynamic process represents (a) heat absorbed (b) heat rejected (c) either (a) or (b) (d) none of these 4. The entropy is an .................... property of the system. (a) intensive (b) extensive 5. The entropy during adiabatic reversible process (a) increases (b) decreases (c) remains constant (d) first increases and then decreases 6. The entropy .................... in an irreversible process. (a) increases (b) decreases (c) remain constant 7. The condition for a reversible cyclic process, δQ δQ (a) ∫ > 0 (b) ∫ < 0 (c) T T δQ 8. If ∫ > 0, then the cyclic process is T (a) reversible

(b) irreversible

9. In an irreversible process, there is a (a) loss of heat (b) gain of heat

∫

δQ = 0 T

(d)

∫

δQ =1 T

(c) impossible (c) no loss of work

(d) no gain of heat

10. A part of heat which is rejected to the sink or cold reservoir, is called (a) available energy (b) work output (c) unavailable energy (d) none of these

186 Engineering Thermodynamics 11. The available energy of the system becomes .................... when its state is brought to the state of the surrounding by reversible processes. (a) maximum (b) minimum (c) zero 12. If T0 is the lowest temperature of heat rejection and dS is the change of entropy of the system during the process of supplying heat, then unavailable heat energy is given by (a) T0 – dS (b) T0 + dS (c) T0 × dS (d) T0 /dS 13. When a closed system with initial values of pressure p, volume v, temperature T, entropy S and internal energy U, is surrounding by a local environment of constant temperature T0 and constant pressure p0, then maximum useful work is given by (a) WU = T0 dS + dU – p0 dv (b) WU = T0 dS – dU – p0 dU (c) WU = T0 dS + dU + p0 dv (d) WU = dU – T0 dS + p0 dU 14. The main causes for the irreversibility is (a) mechanical and fluid friction (b) free expansion (c) heat transfer through finite temperature difference (d) all of the above 15. The irreversibility is defined as the ..................... in the available energy due to the irreversible processes. (a) decrease (b) increase

ANSWERS 1. (d) 6. (a) 11. (a)

2. (b) 7. (c) 12. (c)

3. (c) 8. (c) 13. (b)

4. (b) 9. (a) 14. (d)

5. (c) 10. (c) 15. (a)

8 PROPERTIES OF PURE SUBSTANCES

8.1 Introduction 8.2 Phases of Water 8.3 Process of Formation of Steam 8.4 Graphical Representation of Formation of Steam 8.5 Pressure-Volume (p-v) Diagram of Water 8.6 Temperature-Entropy (T-s) Diagram for Water and Steam 8.7 Pressure-Specific Volume and Temperature (p-v-T) surface of a Pure Substance 8.8 Important Terms as Applied to Steam 8.9 Use of Steam Tables 8.10 Enthalpy-Entropy (h-s) Diagram or Mollier Chart for Water and Steam

8.11 External Workdone during Evaporation 8.12 Internal Energy of Steam 8.13 Entropy of Water 8.14 Entropy of Evaporation 8.15 Entropy of Wet and Dry Steam 8.16 Entropy of Superheated Steam 8.17 Measurement of Dryness Fraction or Quality of Steam 8.18 Barrel or Tank Calorimeter 8.19 Separating Calorimeter 8.20 Throttling Calorimeter 8.21 Separating and Throttling Calorimeter Combined

8.1 INTRODUCTION A pure substance is a single substance which is uniform is chemical composition and does not change even though there may be a change of phase. For example, water is a pure substance. It exists as a solid (ice), liquid (water) and as a vapour (steam) or any combination of these. This is due to the fact that the chemical combination of *water is same in all the three phases. The other examples of pure substances are oxygen, carbon dioxide, ammonia, freon, mercury etc. On the other hand, an equilibrium mixture of liquid and gaseous air is not a pure substance, as the system is not uniform in composition, because the liquid will be richer in nitrogen than the gaseous state. However, the atmospheric air may be treated as a pure substance, as long as it remains is the gaseous state. From the engineering point of view, steam is one of the most important pure substance and therefore it is discussed, in detail, in this chapter. It is used as a working fluid in steam boilers, steam engines, steam turbines, steam condenser and steam power plants. When heat energy is transferred to water, its enthalpy and its physical state changes. As heating takes place, the temperature of the water rises and generally its density increases. The vapour thus formed is steam and it possesses properties like those of gases.

* The iced-water is a pure substance as the chemical composition of ice and water is same.

188 Engineering Thermodynamics

8.2 PHASES OF WATER We have discussed in the previous article that a pure substance exists in three phases, i.e. solid phase (e.g. ice), liquid phase (e.g. water) and gaseous or vapour phase (e.g. steam). These phases change from one phase to another by the addition or removal of heat. The process by which the phase changes from solid to liquid (e.g. ice to water), is known as melting or fusion. The process by which the liquid changes to gaseous state (e.g. water to steam), is called evaporation or boiling. Similarly, the process of changing the phase from the gaseous state to liquid (e.g. steam to water) is called condensation. When solid is directly changed to gaseous state, the process is then known as sublimation.

8.3 PROCESS OF FORMATION OF STEAM Consider a cylinder, containing 1 kg of ice at –20°C, with a frictionless tight piston of negligible weight to represent atmospheric pressure, as shown is Fig. 8.1(a). When heat is supplied to the cylinder, there is a gradual rise in temperature of ice. The heat added during this period is known as sensible heat of ice. When the temperature reaches 0°C, the temperature remains constant until whole of the ice has converted into liquid water, as shown in Fig. 8.1(b). When further heat is supplied to the water, the temperature rises and continues rising until the boiling point is reached. It may be noted that the boiling point of water depends upon the pressure. Under normal atmospheric pressure (i.e. at 1.013 bar or 101.3 kPa), the boiling point of water is 100°C, but as the pressure increases, the temperature also increases correspondingly. The temperature at which the water begins to boil, is known as saturation temperature and the heat absorbed by the water from the freezing point (i.e. 0°C) to boiling point (i.e. 100°C) is known as sensible heat of water.

Fig. 8.1. Formation of steam.

When the boiling point is reached, the temperature remains constant (even though the heating continues) and the water evaporates, thus pushing the piston up against the constant pressure p. At this stage, whole of the water will not be converted into steam, but is will have some particles of water in suspension and the steam thus formed is termed as wet steam, as shown in Fig. 8.1(c). The heat supplied during this time is known as latent heat of vaporisation. When further heat is added to this wet steam, the temperature remains constant and the suspended particles of water in the wet steam commence to evaporate till the last particle of water has evaporated and the steam is known as dry or saturated steam as shown in Fig. 8.1(d). At this stage. whole of the latent heat has been absorbed. The dry steam is no longer a vapour and behaves like a perfect gas.

Properties of Pure Substances 189 Now, if the dry steam is further heated at the same constant pressure, the temperature starts rising according to Charles’ law. This heating of dry steam above the saturation temperature is known as superheating and the steam is thus termed as superheated steam, as shown is Fig. 8.1(e).

8.4 GRAPHICAL REPRESENTATION OF FORMATION OF STEAM The graphical representation of formation of steam on the temperature-heat added (or enthalpy per kg) is shown in Fig. 8.2. The horizontal ordinate (or abscissa) represents the total heat (h) and the vertical ordinate represents the temperature in degrees centigrade (t°C). The line OA represents the rise in temperature of ice during heating and the heat added during this period is known as sensible heat of ice. When the temperature reaches at 0°C (i.e. at point A), the additional heat supplied does not immediately accomplish any increase of temperature and the ice begins to melt i.e. the change of phase starts. The temperature remains constant until whole of the ice has converted into liquid water as shown by the line AB in Fig. 8.2. When further heat is added at constant pressure, the temperature of water increases and it continues till the temperature is reached upto the boiling point C (i.e. 100°C). Vapour phase E

g

tsup

O

te wa of g

in at

Solid phase

Solid + Liquid phase

r

Liquid phase

0°C

D Vaporisation

rh

C

pe

Saturation temperature

Su

ts = 100°C

He

Temperature (t )

ea

tin

Liquid + vapour

A B Melting

S.H. of ice L.H. of fusion

S.H. of water (hf)

L.H. of vaporisation Heat of (hfg) superheat

Heat added (h)

(hsup)

Fig. 8.2. Graphical representation of formation of steam.

The temperature at which the water starts boiling is known as saturation temperature, ts (i.e. 100°C). The heat supplied to water from 0°C to 100°C is known as sensible heat of water or enthalpy of liquid or total heat of water and is denoted by hf. After reaching the boiling point, the steam formation takes place at constant temperature until the last particle of water is evaporated into steam. Such a steam is dry saturated steam and the heat supplied during this period i.e. during the evaporation process CD, is called latent heat of vaporisation and is denoted by hfg. When the dry saturated steam is further heated, the temperature of steam begins to rise again. The heating of steam above its saturation temperature is known as superheating as shown by the line DE and the heat supplied during superheating is known as heat of superheat or enthalpy of superheat or total heat of superheated steam and it is denoted by hsup.

8.5 PRESSURE-VOLUME (p-v) DIAGRAM FOR WATER The process of formation of steam from a unit mass of water, at constant pressure p, is shown on the pressure-volume ( p-v) diagram by the line abcde in Fig. 8.3. The state a shows a unit mass of water at room temperature ta and volume va. When heat is added, there is a rise in temperature and

190 Engineering Thermodynamics increase in volume as shown by the line a–b. At state b on the saturated liquid line, the condition of water is saturated liquid and the corresponding temperature is saturation temperature. The saturated liquid at b begins to boil on further addition of heat, but the temperature during boiling remains constant. This process of boiling at constant temperature will continue till state d (on the saturated vapour line). At this point, the whole mass of water is converted into steam. The condition of steam at d is known as dry saturated steam. At state c (between the states b and d), the water exists in the form of a mixture of vapour and water and the condition of steam at this state c is known as wet steam. When heat is further added, the temperature will rise along de at constant pressure and the condition of steam after d at any point such as state e, is called superheated steam. Saturated solid line C

Critical point

ase d ph quid + Li olid

Gas or Vapour phase c

a b

S

p

Soli

Pressure

Liquid

e Liquid + Vapour

T=

d

1

2

Solid + Vapour

stan

t

Saturated vapour line

Triple point line Saturated liquid line

Con

3

Volume

Fig. 8.3. Pressure-volume (p-v) diagram for water.

Notes: 1. The condition of the liquid at state a is called sub-cooled liquid, because it exists at a temperature which is less than the saturation temperature corresponding to the pressure. 2. The temperature of the superheated vapour as represented by state e is higher than the saturation temperature corresponding to the pressure. 3. The point C where liquid and vapour phases merge is known as critical point. The temperature corresponding to this critical point is known as critical temperature and the pressure is known as critical pressure. For steam, the critical temperature is 374.15°C and the critical pressure is 221.2 bar. 4. The line 1–2–3 is called a triple point line. Along this line, all the three phases (i.e. solid, liquid and vapour) are in equilibrium. At a pressure below the triple point line, the substance cannot exist in the liquid state, and the substance when heated, transforms from solid to vapour (known as sublimation) by absorbing the latent heat of sublimation from the surroundings. Therefore, the region below the triple point line is solidvapour mixture region. At triple point, the temperature of water is 273 K (0°C) and the pressure is 611.45 N/m2 (4.587 mm of Hg).

8.6 TEMPERATURE-ENTROPY (T-s) DIAGRAM FOR WATER AND STEAM The temperature-entropy (T-s) diagram for water and steam is shown in Fig. 8.4. The vertical ordinate represents the absolute temperature with origin O as absolute zero (i.e. 0°C or 273 K) and abscissa represents the entropy. The advantage of T-s diagram over p-v diagram is that the area under the T-s diagram represents the heat added or rejected during the process while the area under the p-v diagram represents the workdone. The T-s diagram is very useful in the study of thermodynamic properties especially for reversible adiabatic processes. Consider a unit mass of water at 0°C (or 273 K) is heated at constant pressure upto the boiling point (i.e. 100°C). When the heat is added to water from 0°C to 100°C, the entropy of water

Properties of Pure Substances 191 increases with the increase in temperature. By plotting the corresponding values of temperature and entropy, we get a curve OA. This curve OA is known as water line or liquid line. The increase in entropy of water is represented by OA1 (i.e. sf ). The area under the curve OA represents the sensible heat of water (also known as enthalpy of water). e Critical point lin m a e t P y s p4 Dr

374°C Water line

p3

p1 B1

A

Entropy

O

1.0 x=

0.8 x=

ssup

x = 0.6

B

x = 0.4

sfg

sf

A1

C1

B1

A1

p2 B2

0.2

O

B

Super heated region

B3

A2

x=

Tsat

Water region

0

A

A3

x=

Temperature

Tsup

Temperature (T )

C

Entropy (s)

(a)

(b)

Fig. 8.4. Temperature-entropy (T-s) diagram for water and steam.

On further heating of water, the evaporation of water starts from point A and continues upto point B. The evaporation takes place at constant pressure and the heat added during evaporation (i.e. from A to B) is known as latent heat of evaporation (area under the line AB i.e. area ABB1A1). The increase in entropy during evaporation is represented by A1B1 (i.e. sfg). The line AB is evaporation line. When heat is further added after the water is completely evaporated at B, the temperature and entropy of steam increases as shown by the curve BC which is known as superheated line. The area under the curve BC (i.e. area BCC1B1) gives the total heat or enthalpy of superheated steam. The increase in entropy during superheating is represented by B1C1 (i.e. ssup). Now, if the entropy values are plotted corresponding to the absolute temperatures, at different pressures, we get the points A1, A2, A3, ...... etc. and B1, B2, B3 ...... etc. as shown in Fig 8.4 (b). The curve obtained by joining the points A1, A2, A3 .... etc. is known as water line while the curve obtained by joining the points B1, B2, B3 ..... etc. is known as dry steam line. The points where these two curves (i.e. water line and dry steam line) meet is known as critical point (374°C or 647K). Critical point

A

sB Entropy (a) Isothermal process.

0.8 TB

B

x=

x=

0.2 sA

x = 0.6

B

0.8

A

Temperature

TA =TB

x=

Temperature

TA

sA = sB Entropy (b) Isentropic process.

Fig. 8.5. Isothermal and isentropic process on T-s diagram.

192 Engineering Thermodynamics The line AB is divided into 10 parts, each representing the dryness fraction of steam from x = 0 to x = 1.0, as shown in Fig. 8.4(b). The line joining the same values of dryness fraction for different pressures are known as constant dryness fraction lines. An isothermal process on the temperature-entropy (T-s) diagram in represented by a horizontal line, as shown by AB in Fig. 8.5 (a). Since there is no change of temperature during isothermal process, therefore TA = TB. An isentropic process on the temperature-entropy (T-s) diagram is represented by a vertical line, as shown by AB in Fig. 8.5 (b). Since there is no change of enthalpy or total heat during an isentropic process, therefore there is no change of entropy i.e. sA = sB.

8.7 PRESSURE-SPECIFIC VOLUME AND TEMPERATURE (p-v-T ) SURFACE OF A PURE SUBSTANCE When the pressure ( p), specific volume (v) and temperature (T ) are plotted along three mutually perpendicular coordinates, it will result in a p-v-T surface. It represents the equation of state graphically. These surfaces are useful for bringing out the general relationships among the three phases of matter normally under consideration. Fig. 8.6 shows a p-v-T surface of a pure substance such as water. The single phase regions (i.e. solid, liquid and vapour phase), two phase regions (i.e. solidliquid; solid-vapour and liquid-vapour), critical point and triple line are shown in the figure. The saturated liquid line separates the liquid region from the liquid-vapour region where as the saturated vapour line separates the vapour region from the liquid-vapour region.

Critical state c

Solid + Liquid

Liquid

2

Solid

3 1

6

5 Liquid + Vapour

ur po Va

Pressure (p)

4

Saturated liquid line

Constant pressure line

Gas

Saturated vapour line

Triple line a

d

b Solid + Vapour

re

T)

(

tu

p

m

Te

a er

Volume (v)

Fig. 8.6. p-v-T surface for a pure substance (water).

In the two phase region, the lines of constant pressure and constant temperature are same, though the specific volume may change. The triple point actually appears as the triple line on the p-v-T surface, as the pressure and temperature of the triple point are fixed, but the specific volume may vary depending upon the proportion of each phase. The triple line becomes a point on the p-T diagram. Since the freezing temperature normally increases with pressure, the substance contracts on freezing. However, water behaves in the opposite direction. Its freezing temperature decreases with increasing pressure, so it expands on freezing. In Fig. 8.6, 1-2-3-4-5-6 is the constant pressure line.

Properties of Pure Substances 193

8.8 IMPORTANT TERMS AS APPLIED TO STEAM Following are the important terms as applied to steam: 1. Wet steam. When the steam contains moisture or particles of water in suspension, it is said to be wet steam. It means that the evaporation is not complete and the whole latent heat has not been absorbed. 2. Dry saturated steam. When the wet steam is further heated, and it does not contain any particles of water in suspension, then it is known as dry saturated steam. It has received whole of the latent heat and behaves in the same manner as a perfect gas. 3. Superheated steam. When the dry steam is further heated at constant pressure beyond the saturation temperature, it is then called the superheated steam. Since the pressure remains constant, the volume increases when heat is supplied and the volume of one kg of superheated steam is considerably greater than the volume of one kg of saturated steam at the same pressure. 4. Dryness fraction or quality of steam. It is defined as the mass of the dry steam in a certain quantity of steam to the mass of wet steam. It is generally represented by the letter x. If mg is the mass of the dry steam and mf is the mass of water particles in suspension in the steam considered, then mg mg = x = mg + m f mw where mw = Mass of wet steam = mg + mf It may be noted that dryness fraction of dry steam is unity. 5. Sensible heat of water or enthalpy of water. It is defined as the amount of heat required to convert 1 kg of water, at constant pressure, from its freezing point to the temperature of formation of steam (i.e. saturation temperature). It is also referred to as total heat or enthalpy of water or liquid heat. The sensible heat of water is usually expressed by hf . Mathematically, sensible heat of water, hf = Mass of water × Specific heat of water × Rise in temperature = 1 × 4.2 (tsat – 0) = 1 × 4.2 [(tsat + 273) – (0 + 273)] = 4.2 tsat kJ/kg The sensible heat of water (hf ) in kJ/kg may be obtained directly from the steam tables for any given pressure. 6. Latent heat of vaporisation. It is defined as the amount of heat absorbed by 1 kg of water at saturation temperature, for a given pressure, to convert completely into 1 kg of steam at the same temperature. It is usually expressed by hf g. It is also called latent heat of steam. In case of wet steam with dryness fraction x, the heat absorbed during evaporation is x . hfg. 7. Total heat or enthalpy of steam. We know that when the steam is completely dried up, then the total heat present in steam, under normal temperature and pressure conditions, is the heat required to raise the temperature of water from the freezing point to the boiling point or saturation temperature plus the heat required to convert the boiling water into dry steam (i.e. heat required during evaporation). Mathematically, Total heat or enthalpy of steam = Sensible heat + Latent heat The total heat or enthalpy of steam is generally denoted by hg and its value for the dry saturated steam may be read directly from the steam tables. For the wet steam, dry steam and superheated steam may be obtained as follows: 8. Total heat or enthalpy of wet steam. It is the amount of heat required to convert 1 kg of water from its freezing point (i.e. at 0°C) at constant pressure, into wet steam of the given dryness fraction (say x). In other words, the total heat or enthalpy of 1 kg of wet steam (h) is equal to the heat required

194 Engineering Thermodynamics to raise the temperature of 1 kg of water from its freezing point to the boiling point (i.e. hf ) plus the heat required to convert the same water from the boiling point to the wet steam of dryness fraction x (i.e. x × hfg). Mathematically, h = hf + x × hfg 9. Total heat or enthalpy of dry saturated steam. It is the amount of heat required to convert 1 kg of water from its freezing point (i.e. at 0°C) at constant pressure into dry saturated steam. In other words, total heat or enthalpy of 1 kg of dry saturated steam (h) is equal to the heat required to raise the temperature of 1 kg of water from its freezing point to the boiling point (hf ) plus the heat required to convert the same water from the boiling point to the dry saturated steam (i.e. hfg). Mathematically, h = hf + hfg = hg The values of hf , hfg and hg may be read directly from the steam tables at the given pressure. 10. Total heat or enthalpy of superheated steam. It is the amount of heat required to convert 1 kg of water from its freezing point (i.e. 0°C), at constant pressure, into superheated steam. In other words, total heat or enthalpy of superheated steam (hsup) is equal to the heat required to raise the temperature of 1 kg of water from its freezing point to the boiling point (i.e. hf ) plus the heat required to convert the same water from its boiling point to the dry saturated steam (i.e. hfg) plus the heat required to superheat the dry saturated steam upto the desired temperature (tsup). Mathematically, hsup = hf + hfg + cp (tsup – tsat) = hg + cp (tsup – tsat) where cp = Specific heat at constant pressure for superheated steam, and tsat = Saturation temperature at the given constant pressure. Notes: 1. The temperatures tsat and tsup may be taken in °C or in Kelvin (K). 2. (tsup – tsat) is known as degree of superheat. 3. cp (tsup – tsat) is known as heat of superheat.

11. Specific volume and density of steam. The specific volume is defined as the volume of steam per unit mass at a given pressure and temperature. Its unit is m3/kg. The specific volume of the dry saturated steam is expressed by vg, in steam tables. It may be noted that the specific volume of dry saturated steam decreases with the increase in saturation pressure. The density of the steam is defined as the mass per unit of steam i.e. it is reciprocal of the specific volume. Thus, the density of dry saturated steam is 1 rg = (in kg /m3) vg If the steam is wet having a dryness fraction x, then it consists of x kg of dry steam and (1 – x) kg of water in suspension in steam per kg of wet steam. It is represented by vf . Thus, specific volume of 1 kg of wet steam, vw = Volume of dry steam in it + Volume of water in suspension = x vg + (1 – x) vf At low pressure, the value of vw is very small as compared with that of dry saturated steam. Therefore, the term (1 – x) vf may be neglected. \ Specific volume of wet steam, vw = x vg (in m3 /kg) and density of wet steam, 1 1 = rw = vw x vg

Properties of Pure Substances 195 In order to find the specific volume of superheated steam at a given pressure, it is assumed that the superheated steam behaves like a perfect gas and obeys all the gas laws. Since the superheating takes place at constant pressure, therefore applying Charles’ law, we have vsup vsat v = or vsup = sat × Tsup Tsat Tsat Tsup and density of superheated steam,

rsup =

1 vsup

8.9 USE OF STEAM TABLES The various properties of steam at different pressures, such as saturation temperature, sensible heat of water, latent heat of evaporation, total heat (enthalpy) of steam, specific volume and entropy of steam etc. may be obtained during experiments. In actual practice, it is inconvenient and tedious to find relationship between all these properties. The comprehensive data about these properties has been obtained by formulating suitable equations of state in different regions with the help of certain experimental results. The resultant data obtained is put in a tabular form known as steam tables. All these *properties have been tabulated either on pressure basis or temperature basis.

re ta nt lin pre es ss u

re lin

es

Co

B

Cons t temp ant eratu

ns

Co

Specific enthalpy (h) in kJ/kg

Superheated region

n vo stan lum t s e l pec ine ific s

8.10 ENTHALPY-ENTROPY (h-s) DIAGRAM OR MOLLIER CHART FOR WATER AND STEAM

Sa

Co

tura

tion

ns

tan

td Q

P Throttling process

line I s en es A pro tropic sf rac ces s tio ns lin es

ryn

Wet region

Specific entropy (s) kJ/kgK

Fig. 8.7. Enthalpy-entropy (h-s) diagram for water and steam.

An enthalpy-entropy (h-s) diagram (also known as Mollier chart) was introduced in 1904 by Richard Mollier. It is used by engineers for calculating the various properties of steam. The vertical ordinate represents the specific enthalpy (or total heat) in kJ/ kg and the abscissa represents the specific entropy in kJ/kg K. The region below the saturation line represents the wet region while the region above the saturation line represents the superheated condition of steam. The constant

* All these properties and the Mollier chart are given at the end of the book.

196 Engineering Thermodynamics pressure lines are drawn on the diagram through both the wet steam region and superheated steam region. These are straight lines in the former and curved in the latter. The constant dryness fraction lines are plotted in the wet steam region and constant temperatures lines in the superheated region. The constant specific volume lines are drawn in both the wet steam region and superheated steam region. These lines are straight in the wet steam region and curved in the superheated region. Since the entropy remains constant during isentropic process of steam, therefore it is represented by a vertical line AB on the diagram. Similarly, the enthalpy (or total heat) remains constant during throttling process of steam, therefore it is represented by a horizontal line PQ, as shown in Fig. 8.7. The main use of Mollier Chart is used in determining the heat drop during an isentropic expansion or compression of steam. The final condition of the steam may also be obtained from this diagram. Example 8.1. Steam is generated at 9 bar from feed water at 20°C. Find the heat to be supplied to produce 1 kg of dry saturated steam. Solution. Given: Pressure of steam, p = 9 bar Temperature of feed water = 20°C From steam tables, corresponding to 9 bar, we find that hf = 742.6 kJ/ kg ; hfg = 2029.5 kJ/kg We know that total heat of 1 kg of dry saturated steam above 0°C, h = hf + hfg = 742.6 + 2029.5 = 2771.1 kJ/kg Since the feed water is already at 20°C, therefore heat already in water = Mass × Sp. heat of water × Rise in temp. = 1 × 4.2 (20 – 0) = 84 kJ \ Net heat required to produce 1 kg of dry saturated steam = 2771.1 – 84 = 2687.1 kJ/kg Ans. Example 8.2. Find the heat required to produce 2 kg of steam at a pressure of 7 bar and a temperature of 200°C from water at 25°C. Take specific heat of superheated steam as 2.1 kJ/kg K. Solution. Mass of steam, m = 2 kg Pressure of steam, p = 7 bar Temperature of steam, tsup = 200°C Temperature of water tw = 25°C Specific heat of superheated steam, cp = 2.1 kJ/kg K From steam tables, corresponding to a pressure of 7 bar, we find that tsat = 165°C; hf = 697.1 kJ/kg ; hfg = 2064.9 kJ/kg We know that total heat of superheated steam above 0°C, hsup = hf + hfg + cp (tsup – tsat) = 697.1 + 2064.9 + 2.1 (200 – 165) = 2835.5 kJ/kg Sensible heat already present in water per kg = 4.2 (25 – 0) = 105 kJ/kg \ Net heat required = 2835.5 – 105 = 2730.5 kJ/kg and Total heat required = 2 × 2730.5 = 5461 kJ Ans.

Properties of Pure Substances 197 Example 8.3. Determine the condition of steam in the following cases: 1. at a pressure of 10 bar and temperature 200°C; 2. at a pressure of 8 bar and volume 0.22 m3/kg; and 3. at a pressure of 12 bar, if 2700 kJ/kg are required to produce it from water at 0°C. Solution. 1. Condition of steam at a pressure (p) of 10 bar and temperature (t) of 200°C From steam tables, corresponding to a pressure of 10 bar, we find that the saturation temperature (tsat) is 179.9°C. Since the temperature of the given steam is 200°C which is higher than the saturation temperature of 179.9°C, therefore the condition of given steam is superheated. The degree of superheat = 200 – 179.9 = 20.1°C Ans. 2. Condition of steam at a pressure of 8 bar and volume 0.22 m3/kg From steam tables, corresponding to a pressure of 8 bar, the specific volume of dry saturated steam is vg = 0.2403 m3/kg. Since the specific volume of the given steam (0.22 m3/kg) is less than specific volume of dry saturated steam (0.2403 m3/ kg), therefore the condition of given steam is wet. The dryness fraction of steam is given by 0.22 x = = 0.9155 Ans. 0.2403 3. Condition of steam at a pressure of 12 bar and total heat of 2700 kJ/kg From steam tables, corresponding to a pressure of 12 bar, we find that total heat of dry saturated steam is hg = 2782.7 kJ/ kg. Since the total heat of given steam (2700 kJ/ kg) is less than the total heat of dry saturated steam (2787.7 kJ/kg), therefore the condition of given steam is wet. We know that the total heat of given steam h = hf + x hf g 2700 = 798.4 + x × 1984.3 2700 − 798.4 \ x = = 0.958 Ans. 1984.3 Note: From steam tables, corresponding to 12 bar, hf = 798.4 kJ/ kg ; hf g = 1984.3 kJ/kg

Example 8.4. A vessel of 0.3 m3 capacity contains 1.5 kg mixture of water and steam in equilibrium at a pressure of 5 bar. Calculate: (a) the volume and mass of liquid; and (b) the volume and mass of vapour. Solution. Given: Capacity of vessel, v = 0.3 m3 Mass of water and steam, mw = mf + mg = 1.5 kg Pressure, p = 5 bar From steam tables, corresponding to 5 bar, we find that vf = 0.0011 m3/kg ; and vg = 0.3747 m3/kg We know that specific volume of wet steam 0.3 v = vw = = 0.2 m3/kg mw 1.5 Let We also know that

x = Dryness fraction of steam. vw = x vg + (1 – x) vf

198 Engineering Thermodynamics \

0.2 = x × 0.3747 + (1 – x) 0.0011 = 0.3747x + 0.0011 – 0.0011x 0.2 − 0.0011 x = = 0.5324 0.3747 − 0.0011

(a) Volume and mass of liquid It is given that mass of water and steam, mw = mf + mg and mass of steam having dryness fraction x is mg = x × mw \ Mass of liquid (water), mf = mw – mg = mw – x × mw = mw (1 – x) = 1.5 (1 – 0.5324) = 0.7 kg Ans. and volume of liquid (water), = Specific volume of water × Mass of water = vf × mf = 0.0011 × 0.7 = 0.000 77 m3 Ans. (b) Volume and mass of vapour We know that mass of vapour (steam), mg = x × mw = 0.5324 × 1.5 = 0.8 kg Ans. and volume of vapour (steam) = Specific volume of steam × Mass of steam = vg × mg = 0.3747 × 0.8 = 0.3 m3 Ans.

8.11 EXTERNAL WORKDONE DURING EVAPORATION We have already discussed that in the formation of steam from 1 kg of water, a change of state from liquid to vapour takes place. Due to this change, some external work is done against the pressure ( p), and it is included in the latent heat which the steam absorbs during evaporation. Thus, the energy required for the performance of this work is obtained during the latent heat supplied. The amount of work done (or heat absorbed) due to increase in volume during change of water into steam is called external workdone during evaporation. Let p = Pressure on the piston in bar = p × 105 N/m2 = 100 p kN/m2 vw = Volume of water in m3 at pressure p, and vg = Volume of dry saturated steam in m3 at the same pressure p. \ External workdone during evaporation W = Pressure × Change in volume = 100 p (vg – vw) kN-m or kJ Since the volume of water (vw) is very small as compared to volume of dry saturated steam (vg), therefore neglecting the value of vw in the above equation, we have W = 100 p . vg kJ For wet steam having dryness fraction x, external workdone is given by W = 100 p . x vg kJ and for superheated steam, W = 100 p vsup kJ

8.12 INTERNAL ENERGY OF STEAM The internal energy of steam is the actual energy stored in steam. It is equal to the difference between the total heat or enthalpy of steam and the external work done during evaporation. It is

Properties of Pure Substances 199 usually denoted by u. The internal energy of steam for wet steam, dry saturated steam and superheated steam is given as follows; (a) For wet steam, u = h – 100 p . x vg = hf + x hfg – 100 p . x vg kJ/kg (b) for dry saturated steam, u = h – 100 p . vg = hf + hfg – 100 p .vg = hg – 100 pvg kJ/ kg (c) For superheated steam, u = hsup – 100 pvsup = [hg + cp (tsup – tsat )] – 100 pvsup kJ/kg Example 8.5. Calculate the internal energy of 1 kg of steam when its pressure is 10 bar and its dryness fraction is 0.9. Solution. Given: Mass of steam, m = 1 kg Pressure of steam, p = 10 bar Dryness fraction of steam, x = 0.9 From steam tables, corresponding to a pressure of 10 bar, we find that vg = 0.1943 m3/kg ; hf = 762.6 kJ/kg ; hfg = 2013.6 kJ/kg We know that internal energy of wet steam = h – 100 p v = (hf + x hf g) – 100 p (x vg) = (762.6 + 0.9 × 2013.6) – 100 × 10 (0.9 × 0.1943) = 2574.84 – 174.87 = 2399.97 kJ/kg Ans. Example 8.6. Calculate the internal energy of 1 kg of superheated steam at a pressure of 10 bar and a temperature of 280°C. If the steam be expanded at 1.6 bar and dryness 0.8, find the change in internal energy. The specific heat of superheated steam is 2.1 kJ/kg K. Solution. Given: Mass of steam, m = 1 kg Initial pressure of steam, p1 = 10 bar Temperature of steam, tsup = 280°C Final pressure of steam, p2 = 1.6 bar Dryness fraction of steam, x2 = 0.8 Specific heat of superheated steam, cp = 2.1 kJ/kg K Internal energy of superheated steam From steam tables, corresponding to a pressure of 10 bar, we find that tsat = 179.9°C; vg1 = 0.1943 m3/ kg ; hf 1 = 762.6 kJ/kg ; hf g1 = 2013.6 kJ/kg We know that total heat or enthalpy of superheated steam, hsup = hf 1 + hf g1 + cp (tsup – tsat ) = 762.6 + 2013.6 + 2.1 (280 – 179.9) = 2986.4 kJ/kg Now, let us find the specific volume of superheated steam (vsup). We know that vsup vg = Tsup Tsat \

vsup =

vg × Tsup Tsat

=

0.1943 (280 + 273) = 0.2372 m3/kg (179.9 + 273)

We know that internal energy of superheated steam, u1 = hsup – 100 p1 vsup = 2986.4 – 100 × 10 × 0.2372 = 2749.2 kJ/kg Ans.

200 Engineering Thermodynamics Change in internal energy. From steam tables, corresponding to a pressure of 1.6 bar, we find that vg2 = 1.0911m3/ kg ; hf 2 = 475.4 kJ/kg ; hf g 2 = 2220.8 kJ/kg We know that total heat or enthalpy of steam, h2 = hf 2 + x2 hfg2 = 475.4 + 0.8 × 2220.8 = 2252 kJ/kg \ Internal energy, u2 = h2 – 100 p2 x2 vg2 = 2252 – 100 × 1.6 × 0.8 × 1.0911 = 2112.3 kJ/kg and change in internal energy, du = u1 – u2 = 2749.2 – 2112.3 = 636.9 kJ/kg Ans.

8.13 ENTROPY OF WATER Consider 1 kg of water being heated at constant pressure, from the initial temperature (T0) to the saturation temperature (Tsat). The heat absorbed by 1 kg of water during this process is given by dq = Mass × *Specific heat of water × Rise in temperature = 1 × cw (Tsat – T0) = cw dT where dT = Rise in temperature = Tsat – T0 and the change in entropy at an instant when the absolute temperature of water is T, is given by δq cw dT = ds = T T Integrating this expression within the limits T0 and Tsat, we find that the total change of entropy, or

s

Tsat

0

T0

∫ ds =

∫

cw dT T = cw [ log e T ]Tsat 0 T

T T sf = cw loge (Tsat – T0) = cw loge sat = 2.3 cw log sat T 0 T0

Generally the entropy of water is reckoned above the freezing point of water (i.e. 0°C or 273 K). \ Change in entropy of water above 0°C or 273 K is given by

T sf = 2.3 cw log sat 273

8.14 ENTROPY OF EVAPORATION We know that during evaporation of water into dry saturated steam, the temperature and pressure remains constant, and the heat absorbed during evaporation is hf g. \ Change in entropy during evaporation, h fg sf g = Tsat For wet steam of dryness fraction x, the heat absorbed during evaporation will be x × hfg per kg of steam. Thus, change in entropy during evaporation, x × h fg sfg = Tsat

* The specific heat of water may be taken as 4.2 kJ/kg K

Properties of Pure Substances 201

8.15 ENTROPY OF WET AND DRY STEAM The change in entropy of wet steam is equal to the change in entropy of water plus the change in entropy during evaporation of water into wet steam. Mathematically, Change in entropy of wet steam x × h fg = sf + = sf + x × sfg Tsat Similarly, the change in entropy of dry saturated steam is equal to the change in entropy of water plus the change in entropy of water into dry saturated steam. Mathematically, Change in entropy of dry saturated steam, h fg sg = sf + = sf + sf g Tsat

8.16 ENTROPY OF SUPERHEATED STEAM We know that superheating of steam takes place at constant pressure and the temperature rises from saturation temperature (Tsat ) to temperature of superheated steam (Tsup ). Let T be the temperature at any instant between the dry saturated steam and superheated steam and let there is a small rise in temperature (dT ) during this period. Now change in entropy during superheating is given by Heat in superheat (δq ) c ps × dT = ds = T T where cps = Specific heat of superheated steam. Integrating the above expression for the total change in entropy from Tsat to Tsup, we have ssup

∫

sg

\

ds =

Tsup

∫

Tsat

c ps × dT T

T = c ps [ log e T ]Tsup [ s ]sssup g sat

= cps [loge Tsup – loge Tsat]

Tsup Tsup ssup – sg = cps loge = 2.3 cps log Tsat Tsat Tsup ssup = sg + 2.3 cps log Tsat

We have already discussed that change in entropy of dry saturated steam is h fg sg = sf + Tsat \ Change in entropy for converting 1 kg of water from its freezing point (0°C or 273 K) to the temperature of superheated steam is given by h fg Tsup ssup = sf + + 2.3 cps log Tsat Tsat Example 8.7. Find the entropy of one kg of dry saturated steam at a pressure of 5 bar. Solution. Given: Mass of dry saturated steam, m = 1 kg Pressure of steam, p = 5 bar

202 Engineering Thermodynamics From steam tables, corresponding to a pressure of 5 bar, we find that

Tsat = 151.9°C = 151.9 + 273 = 424.9 K; hf g = 2107.4 kJ/kg

We know that entropy of 1 kg of dry saturated steam,

T h fg s = sf + sfg = 2.3 cw log sat + 273 Tsat

424.9 2107.4 + = 2.3 × 4.2 log 273 424.9

... (Q cw for water = 4.2 kJ/kg K)

= 1.856 + 4.96 = 6.816 kJ/kg K Ans. Note: The entropy of one kg of dry saturated steam may be directly seen from the steam tables corresponding to a pressure of 5 bar which is given as sg = sf + sf g = 1.86 + 4.959 = 6.819 kJ/kg K Ans.

Example 8.8. Find the entropy of one kg of superheated steam at 25 bar and a temperature of 290°C. The specific heat of the superheated steam may be taken as 2.1 kJ/kg K. Solution. Given: Mass of steam, m = 1 kg Pressure of steam, p = 25 bar Temperature of steam, Tsup = 290°C = 290 + 273 = 563 K Specific heat of superheated steam, cps = 2.1 kJ/kg K From steam tables, corresponding to a pressure of 25 bar, we find that Tsat = 224°C = 224 + 273 = 497 K ; sg = 6.254 kJ/kg K We know that entropy of 1 kg of superheated steam, Tsup 563 ssup = sg + 2.3 cps log = 6.254 + 2.3 × 2.1 log 497 Tsat = 6.5156 kJ/kg K Ans. Example 8.9. A rigid vessel of volume 0.86 m3 contains 1 kg of steam at a pressure of 2 bar. Evaluate the specific volume, temperature, dryness fraction, enthalpy, internal energy and entropy of steam. Solution. Given: Volume of vessel, v = 0.86 m3 Mass of steam, m = 1 kg Pressure of steam, p = 2 bar From steam tables, corresponding to a pressure of 2 bar, we find that tsat = 120.2°C ; vg = 0.8854 m3/ kg ; hf = 504.7 kJ/ kg ; hf g = 2201.6 kJ/ kg ; sf = 1.53 kJ/kg K ; sfg = 5.597 kJ/ kg K Specific volume of steam We know that specific volume of steam, v 0.86 vg = = = 0.86 m3/kg Ans. m 1

Properties of Pure Substances 203 Temperature of steam We know that temperature of steam = Saturation temperature = 120.2°C Ans. Dryness fraction of steam Since the specific volume of steam (0.86 m3/ kg) is less than the specific volume of dry saturated steam (0.8854 m3/ kg), therefore the steam in the vessel is wet. We know that dryness fraction of steam, 0.86 x = = 0.9713 Ans. 0.8854 Enthalpy of steam We know that enthalpy of steam, h = hf + x hf g = 504.7 + 0.9713 × 2201.6 = 2643.1 kJ/ kg Ans. Internal energy of steam We know that internal energy of steam, u = h – 100 p v = 2643.1 – 100 × 2 × 0.86 = 2471.1 kJ/kg Ans. Entropy of steam We know that entropy of steam, s = sf + x sf g = 1.53 + 0.9713 × 5.597 = 6.966 kJ/kg K Ans. Example 8.10. A vessel of volume 0.05 m3 contains a mixture of saturated water and saturated steam at a temperature of 250°C. The mass of water present is 9 kg. Find the mass, specific volume, enthalpy, internal energy and entropy. Solution. Given: Volume of vessel, v = 0.05 m3 Temperature of saturated water and steam, t = 250°C Mass of water, mf = 9 kg From steam tables, corresponding to a saturation temperature of 250°C, we find that p = 39.776 bar ; vf = 0.00125 m3/kg ; vg = 0.05004 m3/kg ; hf = 1085.8 kJ/kg ; hfg = 1714.6 kJ/kg ; sf = 2.794 kJ/kg K ; sf g = 3.277 kJ/kg K Total mass of water and steam We know that volume of water, vw = Mass of water × Specific volume of water = mf × vf = 9 × 0.00125 = 0.011 25 m3 \ Volume of steam, vs = v – vw = 0.05 – 0.011 25 = 0.038 75 m3 and mass of steam,

mg =

v Volume of steam 0.03875 = s = = 0.774 Specific volume of Steam vg 0.050 04

\ Total mass of water and steam = 9 + 0.774 = 9.774 kg Ans. Specific volume We know that dryness fraction, mg 0.774 = x = = 0.0792 m f + mg 9 + 0.774

204 Engineering Thermodynamics \ Specific volume of wet steam, vw = xvg + (1 – x) vg = 0.0792 × 0.050 04 + (1 – 0.079 2) 0.001 25 = 3.963 × 10–3 + 1.151 × 10–3 = 5.114 × 10–3 m3/ kg Ans. Enthalpy We know that enthalpy of steam, h = hf + xhf g = 1085.8 + 0.0792 × 1714.6 = 1221.6 kJ/kg Ans. Internal energy We know that internal energy, u = h – 100 p . v = 1221.6 – 100 × 39.776 × 5.114 × 10–3 = 1201.26 kJ/ kg Ans. ...(Q v = vw) Entropy We know that entropy of steam, s = sf + x sf g = 2.794 + 0.0792 × 3.277 = 3.0535 kJ/kg K Ans. Example 8.11. A vessel contains wet steam which is 1/3 liquid and 2/3 vapour by volume. The temperature of steam is 151.8°C. Calculate the quality, specific volume, specific enthalpy and specific entropy. Solution. Given: Volume of liquid = 1/3 Volume of vapour = 2/3 Temperature of steam, t = 151.8°C From steam tables, corresponding to a temperature of 151.8°C, we find that vg = 0.377 m3/ kg; hf = 640 kJ/kg; hf g = 2107.5 kJ/ kg ; sf = 1.86 kJ/ kg K ; sf g = 4.96 kJ/ kg K Quality of steam We know that quality or dryness fraction of steam, Volume of vapour 23 x = = = 2/3 = 0.67 Ans. Volume of vapour and liquid 2 3+1 3 Specific volume We know that specific volume of steam, v = x vg = 0.67 × 0.377 = 0.2526 m3/ kg Ans. Specific enthalpy We know that specific enthalpy of steam, h = hf + x sf g = 640 + 0.67 × 2107.5 = 2052 kJ/ kg Ans. Specific entropy We know that specific entropy of steam, s = sf + x sf g = 1.86 + 0.67 × 4.96 = 5.183 kJ/ kg K Ans. Example 8.12. 10 kg of feed water is heated in a boiler at a constant pressure of 15 bar from 14°C. Calculate the enthalpy required and change of entropy when water is converted into (a) Wet steam at 0.95 dryness fraction, and (b) Superheated steam at 300°C. Solution. Given: Mass of feed water, mw = 10 kg Pressure, p = 15 bar

Properties of Pure Substances 205 Temperature of water, tw = 14°C Dryness fraction of steam, x = 0.95 Temperature of superheated steam, tsup = 300°C From steam tables, corresponding to a pressure of 15 bar, we find that tsat = 198.3°C; vg = 0.1317 m3/kg ; hf = 844.6 kJ/kg; hf g = 1945.3 kJ/kg; sf = 2.314 kJ/kg K ; sf g = 4.127 kJ/kg K (a) Enthalpy and change of entropy for wet steam We know that enthalpy of 1 kg of wet steam h = hf + x hf g = 844.6 + 0.95 × 1945.3 = 2692.6 kJ/kg K and enthalpy of 10 kg of wet steam, H1 = 10 × 2692.6 = 26 926 kJ Enthalpy already present in 10 kg of water from 0°C, H2 = 10 × cw (tw – 0) = 10 × 4.187 (14 – 0) = 586.2 kJ \ Total enthalpy required to convert 10 kg of water into wet steam, H = H1 – H2 = 26 926 – 586.2 = 26 339.8 kJ Ans. We know that entropy of 1 kg of wet steam, s = sf + x sf g = 2.314 + 0.95 × 4.127 = 6.2346 kJ/kg K and entropy of 10 kg of wet steam S1 = 10 × 6.2346 = 62.346 kJ/K Entropy of 1 kg of water reckoned from 0°C T 14 + 273 = 2.3 cw log = 2.3 × 4.187 log 273 273 = 0.2091 kJ/kg K ...(Q cw for water = 4.187 kJ/ kg K) and entropy of 10 kg of water, S2 = 10 × 0.2091 = 2.091 kJ/K \ Change in entropy, S = S1 – S2 = 62.346 – 2.091 = 60.255 kJ/K Ans. (b) Enthalpy and change of entropy for superheated steam We know that enthalpy of 1 kg of superheated steam, hsup = hf + hf g + cps (tsup – tsat) = 844.6 + 1945.3 + 2.1 (300 – 198.3) = 3003.47 kJ/ kg and enthalpy of 10 kg of superheated steam, Hsup = 10 × 3003.47 = 30 034.7 kJ/ kg \ Total enthalpy required to convert 10 kg of water into superheated steam = Hsup – H2 = 30 034.7 – 586.2 = 29 448.5 kJ Ans. We know that entropy of 1 kg of superheated steam Tsup ssup = sf + sf g + 2.3 cps log Tsat 300 + 273 = 2.314 + 4.127 + 2.3 × 2.1 log 198.3 + 273 = 6.851 kJ/ kg K

206 Engineering Thermodynamics and entropy of 10 kg of superheated steam \

Ssup = 10 × 6.851 = 68.51 kJ/K Change in entropy = Ssup – S2 = 68.51 – 2.091 = 66.419 kJ/K Ans.

8.17 MEASUREMENT OF DRYNESS FRACTION OR QUALITY OF STEAM In order to estimate the actual state of the wet steam, it is necessary to determine the quality or dryness fraction of steam. Following are the four types of steam calorimeters to determine the dryness fraction of steam: 1. Barrel or tank calorimeter ; 2. Separating calorimeter ; 3. Throttling calorimeter ; and 4. Combined separating and throttling calorimeter. We shall now discuss these calorimeters, in detail, in the following articles:

8.18 BARREL OR TANK CALORIMETER A barrel or tank calorimeter may be used for determining the approximate value of the dryness fraction of steam. It consists of a copper calorimeter placed on the wooden blocks and insulated from all sides as shown in Fig. 8.8. This calorimeter contains a known mass of cold water. It is covered with a wooden lid through which a thermometer passes and dips in the water. The whole assembly is now placed on a platform. A known mass of steam from the main steam pipe is supplied to the calorimeter through a control valve and a sampling tube. This steam on entering the calorimeter condenses and increases the mass and temperature of water. The heat supplied by the condensing steam may be determined by the initial and final temperature of the water in the calorimeter. The pressure at which the steam passes into the calorimeter is indicated by the pressure gauge fitted on the sampling tube. Let p = Pressure of steam in bar, t = Saturation temperature of steam at pressure p (from steam tables), hf g = Latent heat of steam at pressure p (from steam tables), x = Dryness fraction of steam sample, mc = Mass of the calorimeter in kg, cc = Specific heat of the calorimeter, me = Water equivalent of the calorimeter = mc × cc, mw = Mass of cold water in the calorimeter in kg, ms = Mass of steam condensed in water in kg, t1 = Temperature of water in the calorimeter before condensation, t2 = Temperature of water in the calorimeter after condensation, cw = Specific heat of water = 4.2 kJ/kg K Neglecting the heat losses to the surroundings from the system,

Heat lost by steam = Heat gained by water and calorimeter

ms [x hf g + cw (t – t2)] = (mw . cw + mc . cc) (t2 – t1)

Properties of Pure Substances 207 From this expression, the dryness fraction of steam (x) may be obtained. Pressure gauge Control valve

Sampling tube Wooden lid Thermometer Main steam pipe Copper calorimeter Insulation

Water

Steam nozzles Wooden blocks Platform

Fig. 8.8. Barrel or tank calorimeter.

Example 8.13. The following data were observed during an experiment on wet steam by a barrel calorimeter. Mass of empty barrel = 25 kg Mass of barrel and cold water = 135 kg Mass of barrel and hot water = 140 kg Temperature of cold water = 15°C Temperature of hot water = 42°C Steam pressure = 8 bar If the specific heat of the material of the barrel is 0.4 kJ/kg K, estimate the dryness fraction of steam. Solution. Given: Mass of empty barrel, mc = 25 kg Mass of barrel and cold water, mc + mw = 135 kg \ Mass of cold water, mw = 135 – 25 = 110 kg Mass of barrel and hot water, = 140 kg \ Mass of steam condensed, ms = 140 – (mc + mw) = 140 – 135 = 5 kg Temperature of cold water, t1 = 15°C Temperature of hot water, i.e. after condensation, t2 = 42°C Steam pressure, p = 8 bar Specific heat of the material of the barrel, cc = 0.4 kJ/ kg K Let x = Dryness fraction of steam.

208 Engineering Thermodynamics From steam tables, corresponding to a pressure of 8 bar, we find that t = 170.4°C; and hf g = 2046.6 kJ/kg We know that heat lost by steam = ms [x hfg + cw (t – t2)] = 5 [x × 2046.6 + 4.2 (170.4 – 42)] = 10 233 x + 2696.4 and heat gained by water and calorimeter = (mw . cw + mc . cc) (t2 – t1) = (110 × 4.2 + 25 × 0.4) (42 – 15) = 12 744 Equating equations (i) and (ii), we have 10 233 x + 2696.4 = 12744 12 744 − 2696.4 \ x = = 0.98 Ans. 10 233

...(i) ...(ii)

8.19 SEPARATING CALORIMETER A separating calorimeter is quite simple in action and may be used for testing steam of any degree of wetness. Fig. 8.9 shows a schematic diagram of a separating calorimeter. The steam under test (i.e. wet steam) is passed to the calorimeter from the main steam pipe through the control valve and sampling tube, as shown in Fig. 8.9. Steam pressure gauge

Sampling tube Control valve

Stop cock Steam and water separator

Main steam pipe

Inner chamber Perforated cup Graduated water gauge

Outer chamber

Calibrated orifice

Steam outlet

Fig. 8.9. Separating calorimeter.

The steam passes over the perforated cup where the water separates due to the inertia of the droplets. The separated water collects at the bottom of the inner chamber and its amount may be determined by the graduated water gauge. The steam enters into the outer chamber along the sides of the perforated cup and then to the outlet. The amount of steam passing through the outer chamber may determined by condensing it in a weighed quantity of cold water, as it passes from the outlet. The whole calorimeter must be well lagged. Let m = Mass of water removed from the steam in a certain time, M = Mass of dry steam passing in the same time, and x = Dryness fraction of steam taken in the sample.

Properties of Pure Substances 209 \ Dryness fraction of steam,

x =

Mass of condensate M = Mass of steam supplied M + m

8.20 THROTTLING CALORIMETER A throttling calorimeter is a simple apparatus which can be used for determining the dryness fraction of steam. In this calorimeter, the total heat remains constant before and after the throttling process. A schematic diagram of this calorimeter is shown in Fig. 8.10. Main steam pipe

Pressure gauge Thermometer (T1)

Control valve

A Sampling tube

Thermometer (T2)

O B

Manometer

Steam outlet

Fig. 8.10. Throttling calorimeter.

It consists of a separator A into which the steam is admitted through the control valve from the main steam pipe through a sampling tube. The pressure of the steam is measured by the pressure gauge and its temperature is noted by the thermometer T1. This temperature (recorded by thermometer T1) should be same as the saturation temperature corresponding to the pressure of steam in the calorimeter B. Now the steam is passed through a small opening at O, its total heat remaining constant. The temperature and pressure of the steam leaving the calorimeter B is measured by the thermometer T2 and the manometer respectively. The steam after being throttled, is in a superheated state at a lower pressure (say p2) than before throttling (say p1). Since the total heat remains constant, i.e. the total heat before throttling is equal to the total heat after throttling, therefore the dryness fraction of steam before throttling may be obtained as discussed below: Let x = Dryness fraction of steam before throttling, p1 = Pressure of steam before throttling, p2 = Pressure of steam after throttling hf 1 = Sensible heat of water corresponding to pressure p1 (from steam tables), hf g1 = Latent heat of steam corresponding to pressure p1 (from steam tables), hg 2 = Total heat of dry saturated steam corresponding to pressure p2 (from steam tables), tsup = Temperature of superheated steam after throttling, tsat = Saturation temperature corresponding to pressure p2, cp = Specific heat of superheated steam.

210 Engineering Thermodynamics We have discussed above that Total heat before throttling = Total heat after throttling or hf 1 + x hf g1 = hg2 + cp (tsup – tsat) From this expression, the dryness fraction of steam (x) may be obtained. Example 8.14. A throttling calorimeter has steam entering at 10 MPa and leaving out at 0.05 MPa and 100°C. Determine the dryness fraction of steam. Solution. Given: Pressure of steam entering the calorimeter,

p1 = 10 MPa = 10 × 106 Pa = 100 bar

...(Q 1 bar = 105 Pa)

Pressure of steam leaving the calorimeter,

p2 = 0.05 MPa = 0.05 × 106 Pa = 0.5 bar

Temperature of steam leaving the calorimeter,

tsup = 100°C x = Dryness fraction of steam.

Let

From steam tables, corresponding to a pressure of 100 bar, we find that

hf1 = 1408.1 kJ/kg; hf g1 = 1319.7 kJ/kg

and corresponding to a pressure of 0.5 bar,

hg2 = 2646 kJ/ kg; tsat = 81.35°C

We know that Total heat before throttling = Total heat after throttling \

hf1 + x hf g1 = hg2 + cp (tsup – tsat) 1408.1 + x × 1319.7 = 2646 + 2 (100 – 81.35) = 2683.3 x =

...(Taking cp = 2 kJ/kg K)

2683.3 − 1408.1 = 0.966 Ans. 1319.7

8.21 SEPARATING AND THROTTLING CALORIMETER COMBINED The separating calorimeters when used alone gives higher value of the dryness fraction than the actual value. This is due to the fact that the separation of water from the wet steam is not perfect. An accurate estimation of the dryness fraction of steam is done by means of separating and throttling calorimeter combined as shown in Fig. 8.11. In this calorimeter, a sample of wet steam is allowed to pass through the perforated sampling tube into the separating calorimeter. A part of the water is being removed by a separator owing to sudden change of direction of flow. Now, the resulting semi-dry steam flows into a throttling calorimeter through an orifice as shown in the figure. The throttling takes place at constant total heat. The steam coming out of the throttling calorimeter is condensed in the condenser and the mass of the condensate is noted. The amount of water particles separated in the separating calorimeter are also noted. The pressure and temperature before and after the throttling are measured by the pressure gauge and thermometer. Let x1 = Dryness fraction of steam considering the separating calorimeter, and x2 = Dryness fraction of steam entering the throttling calorimeter.

Properties of Pure Substances 211 Now actual dryness fraction of steam, x = x1 × x2 Note: The values of x1 and x2 may be calculated as already discussed in Art. 8.19 and 8.20. Pressure gauge Main steam pipe

Thermometer Orifice

Control valve

Sampling tube Water gauge

Mercury manometer

Water

Separating colorimeter

Throttling calorimeter Steam outlet to condenser

Fig. 8.11. Separating and throttling calorimeter combined.

Example 8.15. In a laboratory experiment, the following observations were recorded to find the dryness fraction of steam by the combined separating and throttling calorimeter: Total quantity of steam passed = 36 kg Water drained from separator = 1.8 kg Steam pressure before throttling = 12 bar Temperature of steam after throttling = 110°C Pressure after throttling = 1.013 bar Specific heat of steam = 2.1 kJ/kg K Determine the dryness fraction of steam before inlet to the calorimeter. Solution. Given: Total quantity of steam passed, M + m = 36 kg Water drained from separator, m = 1.8 kg \ Mass of dry steam, M = 36 – 1.8 = 34.2 kg Steam pressure before throttling, p1 = 12 bar Temperature of steam after throttling, tsup = 110°C Pressure after throttling, p2 = 1.013 bar Specific heat of steam, cp = 2.1 kJ/ kg K We know that dryness fraction for separating calorimeter, M 34.2 = x1 = = 0.95 M +m 36 From steam tables, corresponding to a pressure of 12 bar, we find that hf 1 = 798.4 kJ/kg; and hfg1 = 1984.3 kJ/kg and corresponding to a pressure of 1.013 bar, we find that tsat = 100°C; and hg2 = 2676 kJ/kg Let x2 = Dryness fraction for throttling calorimeter. Since the total heat remains constant during throttling, therefore Total heat before throttling = Total heat after throttling hf1 + x2 × hfg1 = hg2 + cp (tsup – tsat)

212 Engineering Thermodynamics \

798.4 + x2 × 1984.3 = 2676 + 2.1 (110 – 100) = 2697 2697 − 798.4 x2 = = 0.9568 1984.3

We know that dryness fraction of steam before inlet to the calorimeter, x = x1 × x2 = 0.95 × 0.9568 = 0.909 Ans. Example 8.16. During a test to find the dryness fraction of steam, with separating and throttling calorimeters in series, the following readings were obtained: Pressure of steam in the main pipe = 10 bar Manometer reading after throttling = 956 mm of water Barometer reading = 754.91 mm of Hg Temperature of steam after throttling = 111°C Condensate from throttling calorimeter = 0.5 kg Drainage from separating calorimeter = 28.5 g Calculate the dryness fractions of steam entering the calorimeter from the boiler, assuming cp of superheated steam as 2.1 kJ/ kg K. Solution. Given: Pressure of steam in the main pipe, p1 = 10 bar Manometer reading after throttling, 956 133.3 × = *956 mm of water = = 0.0937 bar 13.6 105 Barometer reading = 754.91 mm of Hg = 754.91 ×

133.3 105

= 1.0063 bar

\ Pressure after throttling, p2 = Manometer reading + Barometer reading = 0.0937 + 1.0063 = 1.1 bar Temperature after throttling tsup = 111°C Condensate from throttling calorimeter, M = 0.5 kg Drainage from separating calorimeter, m = 28.5 g = 0.0285 kg Specific heat of superheated steam, cp = 2.1 kJ/kg K Let x = Dryness fraction of steam. We know that dryness fraction for separating calorimeter, M 0.5 = x1 = = 0.946 M + m 0.5 + 0.0285 From steam tables, corresponding to a pressure of 10 bar, we find that hf1 = 762.6 kJ/ kg; and hfg1 = 2013.6 kJ/kg

* 1 mm of Hg = 13.6 mm of water 133.3 = 133.3 N/m2 = bar 105

Properties of Pure Substances 213 and corresponding to a pressure of 1.1 bar, we find that tsat = 102.2°C; and hg2 = 2679.4 kJ/kg Let x2 = Dryness fraction for throttling calorimeter. Since the total heat remains constant during throttling, therefore Total heat before throttling = Total heat after throttling hf1 + x2 hfg1 = hg2 + cp (tsup – tsat) 762.6 + x2 × 2013.6 = 2679.4 + 2.1 (111 – 102.2) = 2697.88 2697.88 − 762.6 \ x2 = = 0.961 2013.6 We know that dryness fraction of steam entering the calorimeter, x = x1 × x2 = 0.946 × 0.961 = 0.909 Ans.

HIGHLIGHTS 1. A pure substance is a single substance which is uniform in chemical composition and does not change even though there may be a change of phase. For engineering purposes, steam is one of the most important pure substance. 2. A pure substance exists in three phases i.e. solid phase (e.g. ice); liquid phase (e.g. water) and gaseous or vapour phase (e.g. steam). 3. The process by which phase changes from solid to liquid (e.g. from ice to water) is known as melting or fusion. The process by which the liquid changes to gaseous state (e.g. water to steam) is called evaporation or boiling. The process of changing the phase from the gaseous state to liquid (e.g. from steam to water) is called condensation. When solid is directly changed to gaseous state, the process is then known as sublimation. 4. When the steam contains moisture or particles of water in suspension, it is then known as wet steam. When it is further heated and contains no particles of water in suspension, it is then known as dry saturated steam. Again, when dry steam is further heated at constant pressure beyond the saturation temperature, it is called superheated steam. 5. The dryness fraction or quality of steam (x) is defined as the ratio of the mass of the dry steam to the mass of the wet steam. Mathematically, mg mg = x = mw mg + m f where mg = Mass of dry steam; and mf = Mass of water particles in suspension in the steam considered. 6. The sensible heat or enthalpy of water is the amount of heat required to convert 1 kg of water at constant pressure from its freezing point to the saturation temperature. It is usually expressed by hf . 7. The latent heat of vaporisation is the heat absorbed by 1 kg of water at saturation temperature for a given pressure, to convert completely into 1 kg of steam at the same temperature. It is expressed by hfg. 8. The total heat or enthalpy of steam is given by h = hf + x hfg ...(For wet steam) = hf + hfg = hg ...(For dry saturated steam) = hf + hfg + cp (tsup – tsat) ...(For superheated steam) where cp = Specific heat of superheated steam, and tsup – tsat = Degree of superheat.

214 Engineering Thermodynamics

i.e.

9. The specific volume of 1 kg of wet steam, vw = x vg + (1 – x) vf = x vg ...(Neglecting vf) 10. The specific volume of superheated steam (vsup) may be obtained by applying Charles’ law, vsup

vsat Tsat Tsup 11. The external workdone during evaporation is given by W = 100 p . x vg kJ ...(For wet steam) = 100 p . vg kJ ...(For dry saturated steam) = 100 p . vsup kJ ...(For superheated steam) where p is the pressure of steam in bar and vg or vsup is in m3/ kg. 12. The internal energy of steam (u) is equal to the difference between the total heat or enthalpy of steam and the external workdone during evaporation. Mathematically, (a) For wet steam u = h – 100 p x vg = hf + x hfg – 100 p x vg kJ/kg (b) For dry saturated steam, u = h – 100 p vg = hf + hfg – 100 p vg = hg – 100 p vg kJ/kg (c) For superheated steam u = hsup – 100 p vsup = [hg + cp (tsup – tsat)] – 100 p vsup kJ/kg 13. Entropy of water above 0°C (or 273 K) is given by

=

T sf = 2.3 cw log sat 273

where cw = Specific heat of water which may be taken as 4.2 kJ/kg K; Tsat = Saturation temperature in K. 14. Change in entropy during evaporation; h fg sfg = ...[For complete evaporation] Tsat =

x h fg Tsat

...[For partial evaporation]

15. Change in entropy of wet steam = sf +

x h fg Tsat

T x h fg = 2.3 cw log sat + 273 Tsat

16. Change in entropy of dry steam,

sg = sf +

h fg Tsat

T h fg = 2.3 cw log sat + 273 Tsat

17. Change in entropy during superheating

Tsup ssup – sg = 2.3 cps log Tsat

Properties of Pure Substances 215 or

Tsup ssup = sg + 2.3 cps log Tsat

18. In order to estimate the actual state of the wet steam, it is necessary to determine the quality or dryness fraction of steam. Following are the four types of steam calorimeters, to determine the dryness fraction of steam: (a) Barrel or tank calorimeter; (b) Separating calorimeter; (c) Throttling calorimeter; and (d) Combined separating and throttling calorimeter.

EXERCISES 1. Calculate the enthalpy of 1 kg of steam at a pressure of 8 bar and a dryness fraction of 0.8.

[Ans. 2358.18 kJ/kg]

2. Determine the specific enthalpy of superheated steam at a pressure of 150 bar and 520°C.

[Ans. 2988.69 kJ/ kg]

3. Calculate the enthalpy of steam at 30 bar, when (a) its dryness fraction is 0.75;

(b) it is dry saturated; and

(c) it is superheated at 400°C.

[Ans. 2353.8 kJ/ kg; 2802.3 kJ/ kg ; 3151.3 kJ/kg]

4. Determine the specific volume of steam at 250°C and a dryness fraction of 0.8. [Ans. 0.04 028 m3/kg] 5. A sample of wet steam exists at 5 bar and possesses dryness fraction of 0.98. With the help of steam tables, determine its temperature, enthalpy and specific volume.

[Ans. 151.9°C; 2705.35 kJ/ kg; 0.3672 m3/ kg]

6. Determine the state of steam i.e. whether it is wet, dry or superheated in the following cases: (a) Pressure 10 bar and specific volume 0.175 m3/kg; and (b) Pressure 15 bar and temperature 220°C.

[Ans. 0.9 ; 21.7°C (Degree of superheat)]

7. Calculate the internal energy of 1 kg of steam at a pressure of 10 bar when the steam is: (a) 0.9 dry; and (b) dry saturated. The volume of water may be neglected. [Ans. 2400 kJ ; 2582 kJ] 8. Using steam tables, determine the volume, enthalpy and internal energy on per kg basis for steam at 12 bar and 0.95 dryness fraction. [Ans. 0.155 m3/ kg; 2683.48 kJ/ kg ; 2497.48 kJ/ kg] 9. Find the internal energy of 1 kg of super heated steam at a pressure of 10 bar and 280°C. If this steam is expanded to a pressure of 5 bar and 0.98 dry, determine the change in internal energy. The specific heat of superheated steam may be taken as 2.1 kJ/kg K. [Ans. 2749.4 kJ/kg; 231.4 kJ/kg] 10. Determine the entropy of 1 kg of wet steam at a pressure of 6 bar and 0.8 dry, reckoned from 0°C. [Ans. 5.7926 kJ/kg K] 11. Steam enters an engine at a pressure of 10 bar and 400°C. It is exhausted at 0.2 bar. The steam at exhaust is 0.9 dry. Find: (a) Decrease in enthalpy; and (b) Change in entropy.

[Ans. (a) 890.4 kJ/kg ; (b) 0.2657 kJ/kg K]

12. Determine the enthalpy, volume, internal energy and entropy of superheated steam at a pressure of 15 bar and a temperature of 220°C. The volume of water may be neglected. The specific heat of superheated steam may be taken as 2.2 kJ/kg K. [Ans. 2837.64 kJ/ kg; 0.1378 m3/ kg; 2630.94 kJ/ kg ; 6.5398 kJ/ kg K]

216 Engineering Thermodynamics 13. A sample of superheated steam exists at 8 bar and a temperature of 280°C. Using steam tables only, find the enthalpy, specific volume and entropy of this sample on one kg basis. Take specific heat of superheated steam as 2.2 kJ/ kg K. [Ans. 3008.62 kJ/ kg ; 0.3 m3/ kg; 7.145 kJ/ kg K] 14. 5 kg of steam is generated at a pressure of 10 bar from feed water at a temperature of 25°C. Calculate, with the help of steam tables, the enthalpy and entropy of steam, when (a) the steam is dry and saturated ; and (b) the steam is superheated upto a temperature of 300°C. Take specific heat for superheated steam as 2.1 kJ/kg K and specific heat for water as 4.187 kJ/ kg K. [Ans. (a) 13357.6 kJ; 31.085 kJ/K; (b) 14618.6 kJ; 33.55 kJ/K] 15. Steam from a small boiler is discharged through a pipe into a barrel calorimeter. After a few minutes, the following observations were recorded Mass of empty barrel = 35 kg Mass of barrel + cold water = 160 kg Mass of barrel + hot water = 164 kg Temperature of cold water = 12°C Temperature of hot water = 28°C Pressure of steam during test = 7.5 bar If the specific heat of the material of barrel is 0.4 kJ/ kg K, calculate the dryness fraction of steam. [Ans. 0.763] 16. In a throttling calorimeter, the pressures before and after throttling are 14 bar and 16 bar respectively. If the temperature after throttling is 120°C; determine the dryness fraction of steam before passing through the calorimeter. Take specific heat for superheated steam as 2.1 kJ/kg K. [Ans. 0.964] 17. The following observations were recorded during an experiment with a combined separating and throttling calorimeter: Mass of water trapped in the separator

= 0.5 kg

Mass of steam condensed after throttling

= 3 kg

Steam pressure before and after throttling = 8 bar and 1.2 bar Temperature of steam after throttling

= 120°C

Specific heat of steam

= 2.1 kJ/kg K

Calculate the dryness fraction of steam in the main.

[Ans. 0.835]

QUESTIONS 1. Define a pure substance. Is iced water a pure substance? 2. Explain the process of steam generation. Show the various stages on p-v and T-s diagrams. 3. Distinguish between saturated liquid and saturated vapour. 4. What is a critical point? Show it on T-s and h-s diagrams. 5. What do you understand by triple point? Explain with the help of sketches. 6. What is the main feature of triple point? State the values of pressure and temperature at the triple point of water. 7. Differentiate between critical point and triple point. 8. What do you mean by the following: (a) Internal latent heat; (b) Internal energy of steam ; and (c) External work of evaporation.

Properties of Pure Substances 217 9. Draw the p-v-T surface of a pure substance and discuss the various regions formed in it. 10. Write a short note on Mollier chart. 11. Define quality or dryness fraction of steam. 12. Name the various methods of finding the dryness fraction of steam. Explain the separating and throttling calorimeter combined.

OBJECTIVE TYPE QUESTIONS 1. Which of the following substance is a pure substance? (a) Oxygen (b) Ammonia (c) Water

(d) all of these

2. The process of changing the phase from the gaseous state to liquid is called (a) evaporation (b) condensation (c) sublimation (d) fusion 3. The sublimation is a process of changing the (a) solid state to a liquid state (b) liquid state to a gaseous state (c) gaseous state to a liquid state (d) solid state directly to gaseous state 4. The temperature at which the water begins to boil, is known as .................... temperature (a) saturation (b) critical 5. The heat supplied to water from 0°C to 100°C, is known as (a) sensible heat of water (b) enthalpy of liquid (c) total heat of water (d) all of these 6. The latent heat of vaporisation at the critical point is (a) less than zero (b) greater than zero (c) equal to zero 7. For steam, (a) the critical temperature is 221.2°C and the critical pressure is 374.15 bar (b) the critical temperature is 374.15°C and the critical pressure is 221.2 bar (c) the critical temperature is 221.2°C and the critical pressure is 221.2 bar (d) the critical temperature is 374.15°C and the critical pressure is 374.15 bar. 8. The point where the saturated liquid line and saturated vapour line meets, is called (a) triple point (b) boiling point (c) critical point (d) ice point 9. The region below the triple point line is (a) solid-vapour region (c) solid-liquid region

(b) liquid-vapour region (d) none of these

10. At triple point, the temperature of water is (a) 100 K (b) 173 K (c) 273 K

(d) 373 K

11. If mg is the mass of dry steam and mf is the mass of water particles in suspension, then the dryness fraction of steam is equal to mf mg mf mg (a) (b) (c) (d) mg mf mg + m f mg + m f

218 Engineering Thermodynamics 12. The amount of heat absorbed by 1 kg of water at saturation temperature for a given pressure, is called (a) sensible heat of water (b) latent heat of vaporisation. (c) enthalpy of steam (d) entropy of steam 13. If tsat is the saturation temperature and tsup is the temperature of superheated steam, then the degree of superheat is given by tsup t (a) tsup – tsat (b) tsup + tsat (c) sat (d) tsup tsat 14. The specific volume of water when heated at 0°C, (a) first increases and then decreases (b) first decreases and then increases (c) increases steadily (d) decreases steadily 15. An accurate estimation of the dryness fraction of steam is done by means of (a) barrel calorimeter (b) separating calorimeter (c) throttling calorimeter (d) combined separating and throttling calorimeter

ANSWERS

1. (d) 6. (c) 11. (d)

2. (b) 7. (b) 12. (b)

3. (d) 8. (c) 13. (a)

4. (a) 9. (a) 14. (b)

5. (d) 10. (c) 15. (d)

9 VAPOUR PROCESSES

9.1 Introduction 9.2 Constant Volume (or Isochoric) Process 9.3 Constant Pressure (or Isobaric) Process 9.4 Constant Temperature (or Isothermal) Process

9.5 Hyperbolic (or pv = C) Process 9.6 Reversible Adiabatic (or Constant Entropy) Process 9.7 Polytropic (or pv n = C) Process 9.8 Throttling (or Constant Enthalpy) Process

9.1 INTRODUCTION Steam is a water vapour and the various processes as discussed for ideal gases are applicable to vapours also, but with different results. The various methods of heating and expanding vapours are as follows: 1. Constant volume (or Isochoric) process ; 2. Constant pressure (or Isobaric) process ; 3. Constant temperature (or Isothermal) process ; 4. Hyperbolic (or pv = C) process ; 5. Reversible adiabatic (or constant entropy) process ; and 6. Polytropic (or pv n = C) process. The above mentioned processes are discussed, in detail, in the following pages.

9.2 CONSTANT VOLUME (OR ISOCHORIC) PROCESS When steam is heated or cooled in a closed vessel, then it is said to be a constant volume process. The constant volume process on p-v, T-s and h-s diagrams is shown in Fig. 9.1. The states 1 and 2 of the steam are shown in the wet region, i.e. quality of steam at states 1 and 2 is wet. Let p1 and p2 = Initial and final pressure of wet steam in bar, x1 and x2 = Initial and final dryness fraction of wet steam. If vg1 and vg2 are the specific volumes of dry saturated steam in m3/ kg (from steam tables) at pressures p1 and p2 respectively, then Initial volume of wet steam, v1 = x1 × vg1 (in m3/ kg) and final volume of wet steam, v2 = x2 × vg2 (in m3/ kg) In case the final condition of steam is dry saturated, then x2 = 1. In that case, v2 = vg2 = x1 × vg1 ...(ii)

220 Engineering Thermodynamics Similarly, if the final condition of steam is superheated, then v2 = vsup and it behaves as an ideal gas. The temperature of the superheated steam (Tsup) may be obtained by applying Charles’ law, vg vsup vsup × T2 i.e. = 2 or Tsup = ...(iii) Tsup vg 2 T2 where T2 and vg2 are the saturation temperature and specific volume of the dry saturated steam at pressure p2 (from steam tables). Superheat region

p2 T2

T1 1 v1 = v2 Volume (a) p-v diagram.

2

Enthalpy

v=C

p1

Superheat region

2

Temperature

Pressure

p2

1 Wet region s1 s2 Entropy (b) T-s diagram.

h2

h1

p1 Saturated curve

2 x2 Wet region

1 x1

s1 s2 Entropy (c) h-s diagram.

Fig. 9.1. Constant volume process.

Since the volume is constant during expansion of steam, therefore v1 = v2 or x1 × vg1 = x2 × vg2 x1 × vg1 \ *x2 = ...(i) vg 2 The other various important relations for the constant volume process are as follows: 1. Workdone We know that for a constant volume process (i.e. for no change in volume), the workdone is zero, i.e. w1 – 2 = pdv = 0 ...(Q dv = 0) 2. Change in internal energy We know that internal energy of steam at state 1, u1 = h1 – 100 p1 v1 = h1 – 100 p1 x1 vg1 (in kJ/kg) and internal energy of steam at state 2, u2 = h2 – 100 p2 v2 (in kJ/kg) = h2 – 100 p2 x2 vg2 ...(for wet steam) = h2 – 100 p2 vg2 ...(for dry saturated steam) = h2 – 100 p2 vsup ...(for superheated steam) \ Change in internal energy, du = u2 – u1 3. Heat transferred We know that heat transferred during state 1 to state 2, q1 – 2 = du + w1 – 2 ... [First Law of Thermodynamics] = du = Change in internal energy ...(Q w1 – 2 = 0) * If x2 is less than x1, then steam is cooled at constant volume.

Vapour Processes 221 Example 9.1. Steam at 15 bar with a dryness fraction of 0.9 and having a volume of 0.03 m3 undergoes non-flow constant volume process to a pressure of 5 bar. Determine: 1. mass of the steam; 2. final condition of the steam; 3. change in enthalpy and internal energy; and 4. heat transferred. Solution. Given: Initial pressure of steam, p1 = 15 bar Initial dryness fraction of steam, x1 = 0.9 Volume of steam, v = 0.03 m3 Final pressure of steam, p2 = 5 bar The process is shown in Fig. 9.2. It is a cooling process. From steam tables, we find that for a pressure of 15 bar, vg1 = 0.132 m3/ kg; hf1 = 844.6 kJ/ kg; h fg1 = 1951.4 kJ/ kg and for a pressure of 5 bar, vg2 = 0.375 m3/kg; hf2 = 640.1 kJ/ kg; hfg2 = 2107.4 kJ/ kg p1

1

2

p2

1 T1

T2

Wet region

2

Superheat region

Enthalpy

Pressure

p1

Temperature

Superheat region

h2

v1 = v2 Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

1

h1

Wet region x2 Entropy

(c) h-s diagram.

1. Mass of the steam We know that initial specific volume of the wet steam, v1 = x1 × vg1 = 0.9 × 0.132 = 0.1188 m3/ kg v 0.03 = \ Mass of the steam, m = = 0.2525 kg Ans. v1 0.1188 2. Final condition of the steam Let x2 = Final dryness fraction of the steam. Since the volume is constant, therefore v1 = v2 or x1 × vg1 = x2 × vg2 x2 =

x1 × vg1 vg 2

Saturation curve

x1

2

Fig. 9.2

\

p2

=

0.9 × 0.132 = 0.317 Ans. 0.375

222 Engineering Thermodynamics 3. Change in enthalpy and internal energy We know that enthalpy of steam at state 1, h1 = hf1 + x1 × hfg1 = 844.6 + 0.9 × 1951.4 = 2601 kJ/ kg and enthalpy of steam at state 2, h2 = hf 2 + x2 × hfg2 = 640.1 + 0.317 × 2107.4 = 1308 kJ/ kg \ Total change in enthalpy = m (h2 – h1) = 0.2525 (1308 – 2601) = – 326.5 kJ/ kg Ans. The negative sign indicates that there is a decrease in enthalpy during cooling process 1-2. We know that initial internal energy of steam at state 1, u1 = h1 – 100 p1 v1 = h1 – 100 p1 x1 vg1 = 2601 – 100 × 15 × 0.9 × 0.132 = 2422.8 kJ/ kg and final internal energy of steam at state 2, u2 = h2 – 100 p2 v2 = h2 – 100 p2 x2 vg2 = 1308 – 100 × 5 × 0.317 × 0.375 = 1248.6 kJ/ kg \ Total change in internal energy, du = m (u2 – u1) = 0.2525 (1248.6 – 2422.8) = – 296.5 kJ Ans. The negative sign indicates that there is a decrease in internal energy during cooling process. 4. Heat transferred We know that heat transferred, q1 – 2 = du + w1 – 2 = du ...(Q w1 – 2 for constant volume = 0) = – 296.5 kJ Ans. The negative sign indicates that the heat is rejected during cooling process. Example 9.2. A pressure cooker contains 1.5 kg of dry saturated steam at 5 bar. Find the quantity of heat which must be rejected so as to reduce the quality to 60% dry. Determine the pressure and temperature of the steam of the new state. From steam tables: At 5 bar: tsat = 151.8°C ; hf = 640.1 kJ/kg ; hfg = 2107.4 kJ/kg ; vg = 0.375 m3/ kg At 2.9 bar: tsat = 132.4°C ; hf = 556.5 kJ/kg ; hfg = 2166.6 kJ/kg ; vg = 0.625 m3/ kg Solution. Given: Mass of steam, ms = 1.5 kg Initial pressure of steam, p1 = 5 bar Final dryness fraction of steam, x2 = 60% = 0.6 It is given that at a pressure of 5 bar, tsat1 = 151.8°C ; hf1 = 640.1 kJ/ kg ; hfg1 = 2107.4 kJ/ kg ; vg1 = 0.375 m3/ kg Let v2 = Final volume of the steam. Since the initial condition of steam is saturated, therefore initial dryness fraction of steam is one (i.e. x1 = 1). Also the volume in the pressure cooker remains constant, therefore v2 = v1 or x2 × vg2 = x1 × vg1 x1 × vg1 1× 0.375 = \ vg2 = = 0.625 m3/ kg x2 0.6

Vapour Processes 223 The constant volume process during cooling from state 1 to state 2 is shown in Fig. 9.3. First of all, let us find the pressure and temperature of the steam of the new state as discussed below:

v=C p2

2

p2 1

T1

Saturation line

2

T2

(a) p-v diagram.

h1

h2

x2 v1 = v2 Volume

Enthalpy

Pressure

1

Temperature

p1 p1

1

Saturation curve

2 x2

s2 s1 Entropy

Entropy

(b) T-s diagram.

(c) h-s diagram.

Fig. 9.3

Pressure and temperature of the steam of the new state From steam tables, corresponding to vg2 = 0.625 m3/ kg, we find that pressure of the steam at the new state, p2 = 2.9 bar Ans. and temperature of the steam, tsat2 = 132.4°C Ans. Quantity of heat It is also given that at a pressure of 2.9 bar, h f 2 = 556.5 kJ/ kg ; hfg2 = 2166.6 kJ/ kg We know that initial internal energy of the steam u1 = h1 – 100 p1 x1 vg1 = hg1 – 100 p1 vg1 = (hf 1 + hfg1) – 100 p1 vg1 ... (Q h1 = hg1; and x1 = 1) = (640.1 + 2107.4) – 100 × 5 × 0.375 = 2747.5 – 187.5 = 2560 kJ/ kg and final internal energy of the steam, u2 = h2 – 100 p2 x2 vg2 = (hf 2 + x2 hfg2) – 100 p2 x2 vg2 = (556.5 + 0.6 × 2166.6) – 100 × 2.9 × 0.6 × 0.625 = 1856.46 – 108.75 = 1747.71 kJ/kg \ Heat transferred, q1 – 2 = u2 – u1 = 1747.71 – 2560 = – 812.29 kJ/kg and total heat transferred = ms × q1 – 2 = 1.5 × – 812.29 = – 1218.4 kJ Ans. The negative sign indicates that the heat is rejected from the steam. Example 9.3. A rigid vessel (2 m3 capacity) holding steam at 20 bar and 300°C, is cooled till the steam becomes just dry and saturated. Determine 1. Mass of the steam in the vessel; 2. Final pressure of the steam; and 3. Amount of the heat transferred from the vessel to the surroundings.

224 Engineering Thermodynamics Solution. Given: Volume of the vessel v = 2 m3 Initial pressure of steam, p1 = 20 bar Initial temperature of steam, tsup = 300°C Since the final condition of steam is dry and saturated, therefore x2 = 1. The cooling process is shown in Fig. 9.4.

1 v=C 2

p2

Saturation line 1

1

h1

tsup p2

300°C p2

p1

Enthalpy

Pressure

p1

Temperature

p1

Saturation curve

h2

2

2 v1 = v2 Volume

Entropy

Entropy

(a) p-v diagram.

(b) T-s diagram.

(c) h-s diagram.

Fig. 9.4

1. Mass of the steam in the vessel From steam tables for superheated steam, corresponding to a pressure of 20 bar and 300°C, we find that specific volume of superheated steam, vsup = 0.1255 m3/ kg \ Mass of the steam in the vessel,

m =

v

=

2 = 15.94 kg Ans. 0.1255

vsup 2. Final pressure of the steam Since the superheated steam is cooled at constant volume till it becomes dry and saturated, therefore vg2 = vsup = 0.1255 m3/ kg From steam tables, corresponding to a specific volume of 0.1255 m3/ kg, we find that final pressure of the steam, p2 = 15.77 bar Ans. 3. Amount of heat transferred From steam tables for superheated steam, corresponding to a pressure of 20 bar and 300°C, enthalpy of superheated steam, hsup1 = 3025 kJ/ kg and enthalpy of saturated steam corresponding to a pressure of 15.77 bar, h2 = 2791.3 kJ/ kg We know that initial internal energy of the superheated steam, u1 = hsup1 – 100 p1 vsup = 3025 – 100 × 20 × 0.1255 = 2774 kJ/ kg

Vapour Processes 225 and final internal energy of the saturated steam, u2 = h2 – 100 p2 v2 = 2791.3 – 100 × 15.77 × 0.1255 = 2593.39 kJ/ kg \ Amount of heat transferred q1 – 2 = u2 – u1 = 2593.39 – 2774 = – 180.61 kJ/ kg Ans. and total amount of heat transferred Q1 – 2 = m × q1 – 2 = 15.94 (– 180.61) = – 2879 kJ Ans. The negative sign indicates that heat is being lost by the steam.

...(Q v2 = vg2 = vsup)

9.3. CONSTANT PRESSURE (OR ISOBARIC) PROCESS We have already discussed in dealing with the formation of steam from water (Art. 8.3) that the heat is supplied at constant pressure when wet steam is converted into dry saturated steam. We also know that the conversion of wet steam into dry saturated steam takes place at constant temperature. Thus a constant pressure process is also known as *constant temperature (or isothermal) process. The generation of steam in steam boilers is an example of constant pressure process. The constant pressure process on p-v, T-s and h-s diagrams is shown in Fig. 9.5. The states 1 and 2 of the steam are shown in the wet region, i.e. the quality of steam at states 1 and 2 is wet. Let x1 and x2 be the initial and final dryness fraction of steam. Since the pressure is constant, therefore the specific volume of the dry saturated steam (vg) in m3/ kg at states 1 and 2 is also constant, i.e. vg1 = vg2 = vg.

p=C

2

v2 Volume (a) p-v diagram.

C

1

Wet region

x2

p=

C p= T

2

x1 v1

Superheat region

Enthalpy

1

Temperature

Pressure

Superheat region

h2

h1

s1 s2 Entropy (b) T-s diagram.

Saturation curve

2 x2

1 x1

Wet region

s1 s2 Entropy (c) h-s diagram.

Fig. 9.5. Constant pressure process.

We know that initial volume of wet steam, v1 = x1 × vg1 = x1 × vg (in m3/ kg) and final volume of wet steam, v2 = x2 × vg2 = x2 × vg (in m3/ kg) In case the final condition of steam is dry saturated, then x2 = 1 and v2 = vg. Similarly, if the final condition of steam is superheated, then v2 = vsup. The various important relations for the constant pressure process are as follows: 1. Workdone If p is the constant pressure in bar, then workdone during the process per kg of steam, w1 – 2 = 100 p (v2 – v1) (in kJ/kg)

* It may be noted that the temperature remains constant only during conversion of wet steam into dry saturated steam. After this, when heat is supplied, the dry saturated steam becomes superheated and during this period, the temperature rises and the pressure remains constant.

226 Engineering Thermodynamics 2. Change in internal energy We know that internal energy of steam at state 1, u1 = h1 – 100 pv1 (in kJ/kg) and internal energy of steam at state 2, u2 = h2 – 100 pv2 (in kJ/kg) = h2 – 100 p x2 vg2 ...(For wet steam) = h2 – 100 p vg ...(For dry saturated steam) = h2 – 100 p vsup ...(For superheated steam) \ Change in internal energy, du = u2 – u1 3. Heat transferred We know that heat transferred during state 1 to state 2, q1 – 2 = du + w1 – 2 ...[First law of Thermodynamics] 3 Example 9.4. A cylinder contains 0.3 m of steam at 5 bar and 300°C. If the steam is cooled at constant pressure until the volume becomes 0.05 m3, determine the final condition of the steam, change in internal energy and the heat absorbed by the steam. Solution. Given: Initial total volume of steam, v1 = 0.3 m3 Pressure of steam, p = 5 bar Temperature of steam, t1 = 300°C Final total volume of steam, v2 = 0.05 m3 The process on the p-v, T-s and h-s diagram is shown in Fig. 9.6. p

2

1

v2

t2

2

2

1 30 0°C

h1

1

t1

Enthalpy

p

Temperature

Pressure

p

h2

2

v1 Volume

Entropy

Entropy

(a) p-v diagram.

(b) T-s diagram.

(c) h-s diagram.

Fig. 9.6

From steam tables of superheated steam, corresponding to a pressure of 5 bar and 300°C, we find that h1 = hsup = 3064.8 kJ/ kg ; vsup = 0.5226 m3/ kg From steam tables, we also find that corresponding to 5 bar, tsat = 151.8°C ; vg2 = 0.375 m3/ kg; hf 2 = 640.1 kJ/ kg ; hf g2 = 2107.4 kJ/ kg We know that mass of the steam, 0.3 m = = 0.574 kg 0.5226

Vapour Processes 227 \ Specific volume of steam at state 2, v 0.05 v2 = 2 = = 0.087 m3/ kg 0.574 m Since this volume (v2′ = 0.087 m3/kg) is less than the volume of saturated steam (vg2 = 0.375 m /kg), therefore the final condition of the steam is wet as shown by point 2 in Fig. 9.6. Final condition of steam Let x2 = Final dryness fraction of steam. We know that v2 = x2 × vg2 v 0.087 \ x2 = 2′ = = 0.232 Ans. vg2 0.375 Change in internal energy We know that initial internal energy of steam at state 1, u1 = h1 – 100 pv1 = hsup – 100 pvsup = 3064.8 – 100 × 5 × 0.5226 = 2803.5 kJ/ kg and final internal energy of steam at state 2, u2 = h2 – 100 pv2 = hf 2 + x2 hfg2 – 100 p x2 vg2 = 640.1 + 0.232 × 2107.4 – 100 × 5 × 0.232 × 0.375 = 1129 – 43.5 = 1085.5 kJ/ kg \ Total change in internal energy, dU = m (u2 – u1) = 0.574 (1085.5 – 2803.5) = – 986 kJ The negative sign indicates that there is a decrease in internal energy. Heat absorbed by the system We know that work transfer from state 1 to state 2 W1 – 2 = 100 p (v2 – v1) = 100 × 5 (0.05 – 0.3) = – 125 kJ \ Heat absorbed by the system, Q1 – 2 = dU + W1 – 2 = – 986 – 125 = – 1111 kJ Ans. The negative sign indicates that the heat is rejected by the system. Example 9.5. Steam is available at 10 bar and 0.9 dry. Determine final dryness fraction of steam when 1. There is a heat loss of 120 kJ from the steam at constant pressure; and 2. The steam temperature decreases to 170°C. Solution. Given: Initial pressure of steam, p1 = 10 bar Initial dryness fraction of steam, x1 = 0.9 1. Final dryness fraction of steam when there is a heat loss of 120 kJ Using Mollier chart, mark point 1 corresponding to pressure of 10 bar and dryness fraction of 0.9, as shown in Fig. 9.7. From the chart, we find that specific enthalpy at point 1, h1 = 2580 kJ/kg Since there is a heat loss of 120 kJ from the steam at constant pressure, therefore mark point 2 on pressure line of 10 bar corresponding to h2 = 2580 – 120 = 2460 kJ/ kg. From the chart, we find that final dryness fraction of steam corresponding to point 2 is Fig. 9.7 x2 = 0.84 Ans. 3

228 Engineering Thermodynamics 2. Final dryness fraction of steam when the steam temperature decreases to 170°C From Mollier chart, we see that a line of 10 bar pressure intersects the saturation curve at A. At this point, the saturation temperature, as read from the chart is 180°C. Now mark point B on the saturation curve such that the temperature at this point is 170°C. From the chart, we find that the pressure line passing through this point B is such that p = 8 bar We also find that the specific volume of saturated steam at 10 bar, vg1 = 0.2 m3/ kg and specific volume of saturated steam at 8 bar vg3 = 0.25 m3/ kg Let x3 = Final dryness fraction of steam. We know that x1 × vg1 = x3 × vg3 0.9 × 0.2 = x3 × 0.25 0.9 × 0.2 \ x3 = = 0.72 Ans. 0.25

9.4 CONSTANT TEMPERATURE (OR ISOTHERMAL) PROCESS The constant temperature (or isothermal) process on p-v, T-s and h-s diagrams is shown in Fig. 9.8. The initial condition of the steam is assumed to the wet, as shown by state 1 in the figure. As the steam expands, the volume increases, but the pressure remains constant. Thus during the conversion of wet steam into dry saturated stream (i.e. during the expansion process 1-A), the pressure remains constant. When the expansion is continued beyond point A, the pressure will fall as shown by A-2 and the final state is superheated steam. p1

p1 A

p2

2

v1

t2 = t1

p2

1 A

x1

2

(a) p-v diagram.

1

A h1

v2 Volume

t2 = t

h2

Enthalpy

1

Temperature

Pressure

p1

1

(b) T-s diagram.

2

Saturation curve

x1

s1 Entropy

p2

s2 Entropy

(c) h-s diagram.

Fig. 9.8. Constant temperature (Isothermal) process.

Note: The constant pressure and isothermal processes are same for wet steam. Similarly, isothermal and hyperbolic processes are identical for superheated steam.

Example 9.6. One kg of steam at a pressure of 100 bar and 0.9 dry is expanded in a reversible isothermal manner till its pressure reduces to 10 bar. Find:1. Final condition of steam; 2. Change of enthalpy; 3. Change in internal energy; 4. Change in entropy; and 5. Heat transfer. Solution. Given: Initial pressure of steam, p1 = 100 bar

Vapour Processes 229 Initial dryness fraction of steam, x1 = 0.9 Final pressure of steam, p2 = 10 bar From steam tables, corresponding to a pressure of 100 bar, we find that tsat1 = 311°C ; vg1 = 0.018 m3/ kg ; h f1 = 1408.1 kJ/ kg ; hfg1 = 1319.7 kJ/kg ; sf1 = 3.361 kJ/ kg K ; sfg1 = 2.259 kJ/ kg K. and corresponding to a pressure of 10 bar, tsat2 = 179.9° C The reversible isothermal process on p-v, T-s and h-s diagram is shown in Fig. 9.9. p1

1

p2

1

2

p2 2

p2

T1 = T2

Enthalpy

Pressure

p1

Temperature

p1

1

t2 = t 1 Saturation curve

x1

2

s1 Volume (a) p-v diagram.

Entropy (b) T-s diagram.

s2 Entropy

(c) h-s diagram.

Fig. 9.9

1. Final condition of steam Since the saturation temperature (tsat1 = 311°C) at state 1 is more than the saturation temperature (tsat2 = 179.9°C) at state 2, therefore the final condition of the steam at state 2 will be superheated having temperature, Tsup = tsat1 = 311°C = 311 + 273 = 584 K 2. Change of enthalpy We know that enthalpy of steam at state 1, h1 = hf1 + x1 × hfg1 = 1408.1 + 0.9 × 1319.7 = 2595.8 kJ/ kg From steam tables for superheated steam, we find that corresponding to 10 bar and 311°C, we find that Enthalpy of steam at state 2, h2 = hsup = 3075.5 kJ/ kg ... (By interpolation) \ Change of enthalpy = h2 – h1 = 3075.5 – 2595.8 = 479.7 kJ/ kg Ans. 3. Change in internal energy First of all, let us find that the specific volume of steam at state 1 and 2. We know that specific volume of steam at state 1, v1 = x1 × vg1 = 0.9 × 0.018 = 0.0162 m3/ kg From steam tables for superheated steam, we find that corresponding to 10 bar and 311°C, the specific volume of steam at state 2, v2 = vsup = 0.2634 m3/ kg ... (By interpolation) We know that initial internal energy of steam at state 1, u1 = h1 – 100 p1 v1 = 2595.8 – 100 × 100 × 0.0162 = 2433.8 kJ/ kg

230 Engineering Thermodynamics and final internal energy of steam at state 2, u2 = h2 – 100 p2 v2 = hsup – 100 p2 vsup = 3075.5 – 100 × 10 × 0.2634 = 2812.1 kJ/ kg \ Change in internal energy, du = u2 – u1 = 2812.1 – 2433.8 = 378.3 kJ/ kg Ans. 4. Change in entropy We know that initial entropy of steam at state 1, s1 = sf1 + x1 × sfg1 = 3.361 + 0.9 × 2.259 = 5.394 kJ/ kg K From steam tables of superheated steam, we find that corresponding to 10 bar and 311°C, the final entropy of steam at state 2, s2 = ssup = 7.164 kJ/ kg K ...(By interpolation) \ Change in entropy, ds = s2 – s1 = 7.164 – 5.394 = 1.77 kJ/ kg K 5. Heat transfer We know that heat transfer, q1 – 2 = T × ds = (311 + 273) 1.77 = 1033.7 kJ/ kg Ans. ...(Q T = Tsup = 311 + 273)

9.5 HYPERBOLIC (OR pv = C) PROCESS

We have already discussed that the superheated steam behaves like a perfect gas, and the process of superheating the steam at constant temperature (i.e. isothermal process) in the superheat region is regarded as hyperbolic process (i.e. pv = Constant). Let one kg of steam having dryness fraction x1 at a pressure p1 (in bar) is expanded hyperbolically till its pressure is p2 (in bar) and dryness fraction x2. We know that Initial volume of steam at state 1, v1 = x1 × vg1 and final volume of steam at state 2, v2 = x2 × vg2 where vg1 and vg2 are the specific volumes of dry saturated steam corresponding to pressure p1 and p2 respectively (from steam tables). Since the expansion is carried out hyperbolically, i.e. according to the law pv = Constant, therefore *p1 v1 = p2 v2 or p1 (x1 × vg1) = p2 (x2 × vg2) p1 ( x1 × vg1 ) \ x2 = ...(i) p2 × vg2 When value of x2 is more than one, then the final condition of steam is superheated. In such a case, v2 = vsup. The temperature of the superheated steam (Tsup) may be obtained by applying Charles’ law to the two conditions of steam, i.e. before and after superheating, i.e. vg vsup = 2 Tsup T2 where T2 is the saturation temperature (in K) corresponding to pressure p2 (from steam tables). * From this expression, if v2 > vg 2, then final condition of steam is superheated.

Vapour Processes 231 The various important relations for the hyperbolic process ( pv = Constant) are as follows: 1. Workdone We know that workdone during hyperbolic expansion from state 1 to state 2, Since p1 v1 = p2 v2

v v w1 – 2 = 100 p1 v1 loge 2 = 2.3× 100 p1 v1 log 2 v 1 v1 v p or 2 = 1 , therefore the above equation may be written as v1 p2 p w1 – 2 = 2.3 × 100 p1 v1 log 1 p2

2. Change in internal energy We know that initial internal energy of steam at state 1, u1 = h1 – 100 p1 v1 and final internal energy of steam at state 2, u2 = h2 – 100 p2 v2 \ Change in internal energy during hyperbolic expansion, du = u2 – u1 = (h2 – 100 p2 v2) – (h1 – 100 p1 v1) = h2 – h1 ...(Q p1 v1 = p2 v2) 3. Heat transferred We know that the heat transferred during state 1 to state 2, q1 – 2 = du + w1 – 2 ...(First Law of Thermodynamics) Example 9.7. Steam at a pressure of 10 bar and 0.9 dry expands to atmospheric pressure hyperbolically. Find: 1. Workdone ; 2. Change in enthalpy ; 3. Change in internal energy ; and 4. Heat absorbed. The specific heat of steam at constant pressure is 2 kJ/ kg K. Solution. Given: Initial pressure of steam, p1 = 10 bar Initial dryness fraction of steam, x1 = 0.9 Final pressure of steam, p2 = Atmospheric pressure = 1.013 bar Specific heat at constant pressure, cp = 2 kJ/ kg K The p-v, T-s and h-s diagrams for the hyperbolic process is shown in Fig. 9.10. From steam tables corresponding to a pressure of 10 bar, we find that T1 = 179.9°C = 179.9 + 273 = 452.9 K ; vg1 = 0.1943 m3/kg ; hf1 = 762.6 kJ/ kg ; hfg1 = 2013.6 kJ/kg and corresponding to a pressure of 1.013 bar, T2 = 100°C = 100 + 273 = 373 K ; vg 2 = 1.673 m3/ kg ; hf 2 = 419.1 kJ/ kg ; hfg2 = 2256.9 kJ/ kg Let v1 and v2 be the initial and final volume of the steam. Since the expansion is hyperbolic, therefore p1 v1 = p2 v2 p1 × x1vg1 10 × 0.9 × 0.1943 pv = or v2 = 1 1 = = 1.726 m3 1.013 p2 p2 Since v2 is more than vg2, therefore the final condition of steam is superheated. In such a case, v2 = vsup = 1.726 m3. The temperature of the superheated steam (Tsup) may be obtained by applying Charles’ law, i.e.

232 Engineering Thermodynamics vsup

Tsup

=

Tsup =

\

vg 2 T2

vsup × T2 vg 2

=

1.726 × 373 = 384.8 K 1.673 p1

1 pv = C 2

p2

v1

v2 Volume

(a) p-v diagram.

T1

h2

1

h1 p2

Tsup 2

2 1

Enthalpy

Pressure

p1

Temperature

p1

p2 Superheat region Saturation curve

T2

Entropy (b) T-s diagram.

Entropy (c) h-s diagram.

Fig. 9.10

1. Workdone We know that workdone during hyperbolic expansion from state 1 to state 2,

v w1 – 2 = 2.3 × 100 p1 v1 log 2 = 230 p1 (x1 vg1) log v1

v2 x1vg 1

1.726 = 230 × 10 (0.9 × 0.1943) log 0.9 × 0.1943 = 402.2 × 0.9943 = 400 kJ/ kg Ans. 2. Change in enthalpy We know that initial enthalpy of steam at state 1 (wet steam), h1 = hf1 + x1 hfg1 = 762.6 + 0.9 × 2013.6 = 2574.8 kJ/ kg and final enthalpy of steam at state 2 (superheated steam), h2 = hsup = hf2 + hfg2 + cp (Tsup – T2) = 419.1 + 2256.9 + 2 (384.8 – 373) = 2699.6 kJ/ kg \ Change in enthalpy, dh = h2 – h1 = 2699.6 – 2574.8 = 124.8 kJ/ kg Ans. 3. Change in internal energy We know that change in internal energy during hyperbolic expansion is equal to change in enthalpy. \ Change in internal energy, du = u2 – u1 = dh = h2 – h1 = 124.8 kJ/ kg Ans. 4. Heat absorbed We know that heat absorbed, q1 – 2 = du + w1 – 2 = 124.8 + 400 = 524.8 kJ/ kg Ans.

Vapour Processes 233

9.6 REVERSIBLE ADIABATIC (OR CONSTANT ENTROPY) PROCESS The *reversible adiabatic process (also known as isentropic process or constant entropy process) on p-v, T-s and h-s diagrams is shown in Fig. 9.11. The states 1 and 2 of the steam are shown in the wet region, i.e. the quality of steam at state 1 and 2 is wet. Let p1 and p2 = Initial and final pressure of steam in bar, x1 and x2 = Initial and final dryness fraction of steam, T1 and T2 = Initial and final temperature of steam corresponding to pressures p1 and p2 respectively, hfg1 and hfg2 = Initial and final enthalpy of evaporation (or latent heat) corresponding to pressures p1 and p2 respectively, sf1 and sf 2 = Initial and final entropy of water corresponding to pressures p1 and p2 respectively, sfg1 and sfg2 = Initial and final entropy of evaporation corresponding to pressures p1 and p2 respectively.

Fig. 9.11. Reversible adiabatic process or Constant entropy process.

We know that initial entropy of steam before expansion i.e. at state 1, x1 × h fg1 s1 = sf1 + = sf1 + x1 × sfg1 ...(i) T1 and final entropy of steam after expansion i.e. at state 2, x2 + h fg2 s2 = sf 2 + = sf 2 + x2 × sfg2 ...(ii) T2 Since the entropy remains constant during reversible adiabatic process (isentropic process), therefore Entropy before expansion (s1) = Entropy after expansion (s2) h fg1 x2 × h fg2 sf 1 + x1 × = sf 2 + T1 T2 or sf 1 + x1 × sfg1 = sf 2 + x2 × sf g2 From these expressions, the final condition of steam (x2) may be obtained. The other various important relations for the reversible adiabatic process are as follows:

* When friction is involved in the process, then it is known as irreversible adiabatic process and there is an increase in entropy.

234 Engineering Thermodynamics 1. Change in internal energy We know that the initial internal energy of the steam before expansion at state 1, u1 = h1 – 100 p1 v1 = h1 – 100 p1 x1 vg1 ...(in kJ/kg) and final internal energy of the steam after expansion at state 2, u2 = h2 – 100 p2 v2 = h2 – 100 p2 x2 vg2 ...(in kJ/kg) \ Change in internal energy, du = u2 – u1 2. Workdone Since there is no heat transfer during the nonflow reversible adiabatic process (i.e. q1–2 = 0), therefore workdone during the process from state 1 to state 2, w1–2 = q1–2 – du ...(First law of thermodynamics) = – du = – (u2 – u1) = u1 – u2 Note: For a steady flow reversible adiabatic process, the workdone during the process is equal to the change in enthalpy, as discussed below: We know that steady flow energy equation, neglecting kinetic energy, is given by h1 + q1 – 2 = h2 + w1 – 2 or w1 – 2 = h1 – h2 ...(Q q1 – 2 = 0)

Example 9.8. Steam at 5 bar having a dryness fraction of 0.9 expands adiabatically and reversibly to a final pressure of 1 bar. Determine the final condition of steam. Solution. Given: Initial pressure of steam, p1 = 5 bar Initial dryness fraction of steam, x1 = 0.9 Final pressure of steam, p2 = 1 bar Let x2 = Final condition of steam. From steam tables, corresponding to a pressure of 5 bar, sf1 = 1.86 kJ/ kg K ; sfg1 = 4.959 kJ/ kg K and corresponding to a pressure of 1 bar, sf 2 = 1.303 kJ/ kg K ; sfg2 = 6.057 kJ/ kg K We know that initial entropy of steam at state 1, s1 = sf1 + x1 × sfg1 = 1.86 + 0.9 × 4.959 = 6.3231 kJ/ kg K and final entropy of steam at state 2, s2 = sf 2 + x2 × sfg2 = 1.303 + x2 × 6.057 kJ/ kg K Since the entropy remains constant during a reversible adiabatic process, therefore s1 = s2 or 6.3231 = 1.303 + x2 × 6.057 6.3231 − 1.303 \ x2 = = 0.829 Ans. 6.057

Example 9.9. One kg of dry saturated steam undergoes an isentropic expansion process from 10 bar to 1 bar. Determine the final condition of steam and the workdone when the expansion takes place 1. in a cylinder fitted with a piston ; and 2. in a turbine.

Vapour Processes 235 Solution. Given: Mass of steam, m = 1 kg Initial pressure of steam, p1 = 10 bar Final pressure of steam, p2 = 1 bar 1. When expansion takes place in a cylinder fitted with a piston (i.e. non-flow process) Since the initial condition of steam is dry saturated, therefore initial dryness fraction of steam, x1 = 1. Let x2 = Final condition of steam. From steam tables, corresponding to a pressure of 10 bar, vg1 = 0.1943 m3/ kg ; hf 1 = 762.6 kJ/ kg ; hfg1 = 2013.6 kJ/ kg ; sf 1 = 2.138 kJ/ kg K ; sfg1 = 4.445 kJ/ kg K and corresponding to a pressure of 1 bar, vg2 = 1.6937 m3/ kg ; hf 2 = 417.5 kJ/ kg ; hfg2 = 2257.9 kJ/ kg ; sf 2 = 1.303 kJ/ kg K ; sfg2 = 6.057 kJ/ kg We know that initial entropy of steam at state 1, s1 = sf 1 + x1 × sfg1 = 2.138 + 1 × 4.445 = 6.583 kJ/ kg K and final entropy of steam at state 2, s2 = sf 2 + x2 + sfg2 = 1.303 + x2 × 6.057 kJ/ kg K Since the entropy remains constant during an isentropic expansion, therefore s1 = s2 or 6.583 = 1.303 + x2 + 6.057 6.583 − 1.303 \ x2 = = 0.872 Ans. 6.057 Workdone We know that initial enthalpy of steam at state 1, h1 = hf1 + x1 × hfg1 = 762.6 + 1 × 2013.6 = 2776.2 kJ/ kg and final enthalpy of steam at state 2, h2 = hf 2 + x2 × hfg2 = 417.5 + 0.872 × 2257.9 = 2386.4 kJ/ kg Initial internal energy of steam at state 1, u1 = h1 – 100 p1 v1= h1 – 100 p1 x1 vg1 = 2776.2 – 100 × 10 × 1 × 0.1943 = 2581.9 kJ/ kg and final internal energy of steam at state 2, u2 = h2 – 100 p2 v2 = h2 – 100 p2 x2 vg2 = 2386.4 – 100 × 1 × 0.872 × 1.6937 = 2238.7 kJ/ kg Since the heat transfer during isentropic process is zero (i.e. q1 – 2 = 0), therefore workdone from state 1 to state 2, w1 – 2 = – du = – (u2 – u1) = u1 – u2 = 2581.9 – 2238.7 = 343.2 kJ/ kg Ans. ...[Q q1 – 2 = du + w1 – 2] 2. When expansion takes place in a turbine When expansion takes place in a turbine (i.e. steady flow process), the final condition of steam will remain same, i.e. x2 = 0.872 Ans. and workdone, w1 – 2 = h1 – h2 = 2776.2 – 2386.4 = 389.8 kJ/ kg Ans.

236 Engineering Thermodynamics

Enthalpy

0.1

bar

70

bar

Example 9.10. Steam expands in a turbine from 70 bar and 500°C to 0.1 bar. If the expansion process is irreversible adiabatic with an isentropic efficiency of 80%; determine the final condition of steam. Also find the work transmitted per kg and the loss of available work due to irreversibilities. Solution. Given: Initial pressure, p1 = 70 bar Initial temperature, T1 = 500°C Final pressure, p2 = 0.1 bar *Isentropic efficiency, h = 80% = 0.8 Final condition of steam Let we solve this Example by using Mollier chart. First of all, mark point 1 corresponding to a pressure of 70 bar and a temperature of 500°C as shown in Fig. 9.12. The vertical line 1-2 shows the isentropic expansion upto a pressure line of 0.1 bar and the curved line 1-2 represents the irreversible adiabatic expansion. From the Mollier chart, we find that 500° h1 C 1 Enthalpy at point 1, h1 = 3410 kJ/ kg Enthalpy at point 2, h2 = 2150 kJ/ kg and dryness fraction of steam at point 2, Saturation curve x2 = 0.82 We know that isentropic efficiency, h2 2 x 2 h1 − h2 Actual heat drop = h = h2 Isentropic heat drop h1 − h2′

3410 − h2 3410 − h2 = 0.8 = 3410 − 2150 1260

2 x2

Entropy Fig. Fig.9.12 9.12

\ h2 = 3410 – 0.8 × 1260 = 2402 kJ/ kg Now draw a horizontal line corresponding to 2402 kJ/ kg, intersecting the pressure line of 0.1 bar at point 2. From the Mollier chart, we find that final condition of steam at point 2, x2 = 0.92 Ans. Work transmitted per kg We know that the work transmitted or actual work, w1 – 2 = h1 – h2 = 3410 – 2402 = 1008 kJ/ kg Ans. Loss of available work due to irreversibilities We know that available work, w1 – 2 = h1 – h2 = 3410 – 2150 = 1260 kJ/ kg \ Loss of available work = (w1 – 2) – (w1 – 2) = 1260 – 1008 = 252 kJ/ kg Ans.

9.7 POLYTROPIC (OR pv n = C) PROCESS The polytropic (or pv n = C) process on p-v, T-s and h-s diagram is shown in Fig. 9.13. The states 1 and 2 of the steam are shown in the wet region, i.e. the quality of steam at the states 1 and 2 is wet. Consider 1 kg of steam being expanded polytropically from state 1 to state 2 in the wet region according to the law pv n = Constant. Let p1 and p2 = Initial and final pressure of steam in bar, x1 and x2 = Initial and final dryness fraction of steam.

* Isentropic efficiency in a steam turbine is defined as the ratio of heat drop to the isentropic heat drop.

Vapour Processes 237 If vg1 and vg2 are the specific volume of dry saturated steam in m / kg (from steam tables) at pressure p1 and p2 respectively, then Initial volume of steam at state 1, v1 = x1 vg1 (in m3/ kg) and final volume of steam at state 2, v2 = x2 vg2 (in m3/ kg) 3

Superheat region

1

2

p2

v1

1

p2

T1

T2

Enthalpy

p1

Temperature

Pressure

p1

2

h1

s1 s2 Entropy

(a) p-v diagram.

(b) T-s diagram.

p2

1

h2

v2 Volume

p1

2 s1 s2 Entropy (c) h-s diagram.

Fig. 9.13. Polytropic (or pv = C) process. n

Since the steam is expanded according to the pv n = Constant, where n is the polytropic index, therefore p1 v1n = p2 v2n or p1 (x1 vg1)n = p2 (x2 vg2)n \

(x2 vg2) = n

p1 ( x1vg1 ) n p2

1

or

p1 ( x1vg ) n n 1 x2 vg2 = p2 1

\

x1vg1 p1 n x2 = vg2 p2

The other various important relations for the polytropic process are as follows: 1. Workdone We know that workdone for a polytropic process from state 1 to state 2, 100 ( p1v2 − p2 v2 ) w1 – 2 = ...(in kJ/kg) n −1 2. Change in internal energy We know that initial internal energy of steam at state 1, u1 = h1 – 100 p1 v1 and final internal energy of steam at state 2, u2 = h2 – 100 p2 v2 \ Change in internal energy, du = u2 – u1

238 Engineering Thermodynamics 3. Heat transferred We know that heat transferred during state 1 to state 2, q1 – 2 = du + w1 – 2 ...[First law of Thermodynamics] Example 9.11. One kg of steam at 5 bar and 0.9 dry is expanded polytropically to 1 bar according to pv 1.1 = Constant. Determine: 1. Final condition of steam ; 2. Workdone ; 3. Change in internal energy ; and 4. Heat transferred. Solution. Given: Mass of steam, m = 1 kg Initial pressure, p1 = 5 bar Initial dryness fraction, x1 = 0.9 Final pressure, p2 = 1 bar From steam tables, corresponding to 5 bar, we find that vg1 = 0.3747 m3/ kg ; hf 1 = 640.1 kJ/ kg ; hfg1 = 2107.4 kJ/ kg and corresponding to 1 bar, we find that vg2 = 1.6937 m3/ kg ; hf 2 = 417.5 kJ/kg ; hfg2 = 2257.9 kJ/ kg 1. Final condition of steam Let x2 = Final dryness fraction of steam. Since the steam expands polytropically according to pv1.1 = Constant, therefore p1 v1n = p2 v2n or p1(x1 vg1) n = p2 (x2 vg2) n 1

\

1

x1vg1 p1 n 0.9 × 0.3747 5 1.1 x2 = = = 0.86 Ans. vg2 p2 1.6937 1

2. Workdone We know that workdone for a polytropic process, 100 ( p1 x1vg1 − p2 x2 vg2 ) 100 ( p1v1 − p2 v2 ) = n −1 n −1 100(5 × 0.9 × 0.3747 − 1× 0.86 × 1.6937) = 1.1 − 1

w1 – 2 =

=

100(1.686 − 1.456) = 230 kJ/ kg Ans. 0.1

3. Change in internal energy We know that initial internal energy of steam, u1 = h1 – 100 p1 v1 = (hf1 + x1 hf g1) – 100 p1 (x1 vg1) = (640.1 + 0.9 × 2107.4) – 100 × 5 (0.9 × 0.3747) = 2536.76 – 168.6 = 2368.16 kJ/ kg and final internal energy of steam, u2 = h2 – 100 p2 v2 = (hf 2 + x2 hfg2) – 100 p2 (x2 vg2) = (417.5 + 0.86 × 2257.9) – 100 × 1 (0.86 × 1.6937) = 2359.3 – 145.7 = 2213.6 kJ/ kg \ Change in internal energy, du = u2 – u1 = 2213.6 – 2368.16 = – 154.56 kJ/ kg Ans. The negative sign indicates that there is a decrease in internal energy.

Vapour Processes 239 4. Heat transferred We know that heat transferred, q1 – 2 = du + w1 – 2 = – 154.56 + 230 = 75.44 kJ/ kg Ans. Example 9.12. 0.8 kg of steam at a pressure of 15 bar and 250°C expands to 1.5 bar. Assuming that the steam expands according to the law pv 1.25 = constant, calculate the final dryness fraction, workdone, heat transferred and change of entropy during the expansion. Solution. Given: Mass of steam, m = 0.8 kg Initial pressure, p1 = 15 bar Initial temperature, T1 = 250°C Final pressure, p2 = 1.5 bar From steam tables for superheated steam, corresponding to a pressure of 15 bar and 250°C, we find that v1 = vsup = 0.152 m3/ kg ; h1 = hsup = 2923.5 kJ/ kg ; and s1 = ssup = 6.71 kJ/ kg K and for saturated steam, corresponding to a pressure of 1.5 bar, we find that vg 2 = 1.159 m3/ kg ; hf 2 = 467.1 kJ/kg ; hfg2 = 2226.3 kJ/kg ; sf 2 = 1.433 kJ/kg K ; sfg2 = 5.79 kJ/ kg K Final dryness fraction of steam Let x2 = Final dryness fraction of steam. Since the steam expands according to the law pv 1.25 = constant, therefore p1 v11.25 = p2 v21.25 \ We know that \

p v2 = v1 1 p2 v2 = x2 vg2 v x2 = 2 = vg2

1

1.25 15 = 0.152 1.5

0.8

= 0.152 × 6.31 = 0.96 m3/ kg

0.96 = 0.828 Ans. 1.159

Workdone We know that workdone during the polytropic process, 100 ( p1v1 − p2 v2 ) 100 (15 × 0.152 − 1.5 × 0.96) = w1 – 2 = n −1 1.25 − 1 100 (2.28 − 1.44) = = 336 kJ/ kg ...(Q n = 1.25) 0.25 \ Total workdone, W1 – 2 = m × w1 – 2 = 0.8 × 336 = 268.8 kJ Ans. Heat transferred We know that initial internal energy of steam, u1 = h1 – 100 p1 v1 = hsup – 100 p1 vsup = 2923.5 – 100 × 15 × 0.152 = 2695.5 kJ/ kg and final internal energy of steam, u2 = h2 – 100 p2 v2 = (hf 2 + x2 hfg2) – 100 p2 v2 = (467.1 + 0.828 × 2226.3) – 100 × 1.5 × 0.96 = 2166.5 kJ/ kg \ Total change in internal energy dU = m × du = m (u2 – u1) = 0.8 (2166.5 – 2695.5) = – 423.2 kJ

240 Engineering Thermodynamics We know that total heat transferred Q1 – 2 = dU + W1 – 2 = – 423.2 + 268.8 = – 154.4 kJ Ans. The negative sign indicates that the heat is rejected by the steam. Change of entropy We know that total change of entropy, dS = m (s2 – s1) = m [(sf 2 + x2 sfg2) – s1] = 0.8 [(1.433 + 0.828 × 5.79) – 6.71] = – 0.3863 kJ/ kg K Ans. The negative sign indicates that there is a decrease in entropy.

9.8 THROTTLING (OR CONSTANT ENTHALPY) PROCESS

Enthalpy

The throttling process or wire drawing process is also known as constant enthalpy or constant total heat process, as shown on the h-s diagram in Fig. 9.14. In this process, the wet steam is caused to flow from a higher pressure to a lower pressure such that the heat transfer (q1 – 2) and work transfer (w1 – 2) are both zero. The enthalpy or total heat of p1 the steam remains constant (i.e. h1 = h2). The throttling process p2 is essentially an adiabatic but irreversible process, causing an increase in entropy. 1 h1 = h 2 2 For a throttling process, the steady flow energy equation may be used at states 1 and 2, which is given by q1 – 2 – w1 – 2 = ( pe2 – pe1) + (ke2 – ke1) + (h2 – h1) Neglecting the changes in potential and kinetic energy, we have s1 s2 Entropy h1 = h2 ...(Q q1 – 2 and w1 – 2 = 0) Fig. 9.14. Throttling process. This shows that during throttling process, the enthalpy remains constant. Example 9.13. Steam at a pressure of 10 bar and 0.9 dryness is throttled to a pressure of 2 bar. Using steam tables only, evaluate the final dryness fraction or degree of superheat. Estimate the change of entropy during this process. Solution. Given: Initial pressure of steam, p1 = 10 bar Initial dryness fraction of steam, x1 = 0.9 Final pressure of steam, p2 = 2 bar Final dryness fraction of steam Let x2 = Final dryness fraction of steam. From steam tables, c orresponding to a pressure of 10 bar, we find that hf 1 = 762.6 kJ/ kg ; hfg1 = 2013.6 kJ/ kg ; sf 1 = 2.138 kJ/ kg K ; sfg1 = 4.445 kJ/ kg K and corresponding to a pressure of 2 bar, we find that hf 2 = 504.7 kJ/ kg ; hfg2 = 2201.6 kJ/ kg ; sf 2 = 1.53 kJ/ kg K ; sf g2 = 5.597 kJ/ kg K We know that the enthalpy of steam before and after the throttling process remains constant, i.e. h1 = h2 or hf 1 + x1 hfg1 = hf 2 + x2 hf g2 762.6 + 0.9 × 2013.6 = 504.7 + x2 × 2201.6 762.6 + 0.9 × 2013.6 − 504.7 \ x2 = = 0.94 Ans. 2201.6

Vapour Processes 241 Change of entropy We know that entropy of steam before throttling, s1 = sf 1 + x1 sfg1 = 2.138 + 0.9 × 4.445 = 6.138 kJ/ kg K and entropy of steam after throttling, s2 = sf 2 + x2 sfg2 = 1.53 + 0.94 × 5.597 = 6.791 kJ/ kg K \ Change of entropy, ds = s2 – s1 = 6.791 – 6.138 = 0.653 kJ/ kg K Ans. Example 9.14. Steam at a pressure of 10 bar and 200°C is throttled to a pressure of 3 bar and then expanded isentropically to a pressure of 0.5 bar. By using Mollier chart, find out the change of entropy and enthalpy during these two processes. Also find the quality of steam at the end of each process. Solution. Given: Initial pressure of steam, p1 = 10 bar Initial temperature, T1 = 200°C Pressure after throttling, p2 = 3 bar Pressure after isentropic process, p3 = 0.5 bar In the h-s diagram, as shown in Fig. 9.15, line 1-2 represents throttling process (constant enthalpy process) and line 2-3 represents isentropic process (constant entropy process). To draw the h-s diagram, first of all, mark point 1 corresponding to a pressure line of 10 bar and a temperature curve of 200°C, as shown in Fig. 9.15. Now draw a horizontal line intersecting the pressure line of 3 bar at point 2. From the Mollier diagram, we find that final condition of steam at point 2 is T2 = 184°C Ans. and h1 = h2 = 2825 kJ/ kg 10 bar 3 bar 1

2

184°C

0.

5

Enthalpy

ba

r

h1 = h2

200°C

h3

3

6.66 Entropy

Saturation curve

x3

7.18

Fig. 9.15

From point 2, draw a vertical line to intersect a pressure line of 0.5 bar at point 3. Now from Mollier chart, we find that s1 = 6.66 kJ/ kg K ; s2 = s3 = 7.18 kJ/ kg K ; and h3 = 2515 kJ/ kg Change of entropy We know that change of entropy = s3 – s1 = 7.18 – 6.66 = 0.52 kJ/ kg K Ans.

242 Engineering Thermodynamics Change of enthalpy We know that change of enthalpy = h1 – h3 = 2825 – 2515 = 310 kJ/ kg Ans. Quality of steam From Mollier Chart, we find that final condition of steam at the end of throttling process at point 2, T2 = 184°C Ans. and final condition of steam at the end of isentropic process (i.e. at point 3), x3 = 0.945 Ans.

HIGHLIGHTS 1. In a constant volume process from state 1 to state 2, workdone per kg of steam, w1–2 = p dv = 0 ...(Q du = 0) 2. In a constant pressure process from state 1 to state 2, workdone per kg of steam, w1–2 = 100 p (v2 – v1) in kJ/ kg where p = Pressure of steam in bar, v1 = Volume of steam at state 1, and v2 = Volume of steam at state 2. 3. The constant pressure and isothermal processes are same for wet steam. Similarly, isothermal and hyperbolic processes are identical for superheated steam. 4. In a hyperbolic (or pv = c) process from state 1 to state 2, the work done per kg of steam,

v w1–2 = 2.3 × 100 p1 v1 log 2 v1

p = 2.3 × 100 p1 v1 log 1 p2

...(Q p1 v1 = p2 v2)

5. In a reversible adiabatic (or constant entropy) process from state 1 to state 2, there is no heat transfer during the non-flow process (i.e. q1 – 2 = 0) and the work done per kg of steam, w1–2 = Change of internal energy from state 1 to state 2 = u1 – u2 6. For a steady flow reversible adiabatic process, the work done per kg of steam, w1–2 = Change in enthalpy from state 1 to state 2 = h1 – h2 n 7. In a polytropic (or pv ) process from state 1 to state 2, the work done per kg of steam, 100 ( p1v1 − p2 v2 ) w1–2 = in kJ/ kg n −1 8. In a throttling process (also known as wire drawing or constant enthalpy or constant total heat process) from state 1 to state 2, the heat transfer (q1 – 2) and work done (w1 – 2) are both zero, i.e. q1–2 = 0 and w1–2 = 0 Thus from the steady flow equation, neglecting potential and kinetic energies, h1 = h2 In other words, in a throttling process, enthalpy remains constant.

EXERCISES 1. Steam at a pressure of 4 bar and a dryness fraction of 0.7 is allowed to expand at a constant volume until the pressure rises to 5.5 bar. Determine: 1. final condition of the steam; 2. change in enthalpy; 3. change in internal energy; and 4. heat transferred. [Ans. 0.945 ; 538.7 kJ/kg ; 490.4 kJ/kg ; 490.4 kJ/kg]

Vapour Processes 243 2. A rigid tank of volume 2 m3, is filled with dry saturated steam at 2 bar. Owing to poor insulation of the tank, the pressure of the steam is found to be 1 bar after sometime. Determine the final state of steam and the amount of energy transferred as heat to the surroundings. [Ans. – 1019.4 kJ/ kg] 3. 0.05 m3 of steam at an initial pressure of 60 bar and an initial temperature of 450°C is cooled at constant volume till the pressure of steam falls to 15 bar. Determine the final quality of steam and the heat transferred. [Ans. 0.342 ; – 1634.8 kJ/ kg] 4. One kg of steam at a pressure of 7 bar and 0.6 dry is heated at constant pressure until it becomes dry and saturated. Find: 1. The increase in volume; 2. Workdone; 3. Heat transfer; and 4. Change in internal energy. [Ans. 0.1091 m3/ kg ; 76.37 kJ/ kg ; 826 kJ/ kg ; 749.63 kJ/ kg] 5. Steam at a pressure of 8 bar and 0.9 dry is heated in a reversible isothermal manner till its pressure reduces to 1 bar. Find: 1. Final condition of steam ; 2. Change in enthalpy ; 3. Change in internal energy ; 4. Change of entropy ; 5. Heat transferred ; and 6. Workdone per kg of steam. [Ans. 170.4°C ; 253.89 kJ/ kg ; 223.67 kJ/ kg ; 1.5054 kJ/ kg K ; 667.5 kJ/ kg ; 443.83 kJ/ kg] 6. 0.05 m3 of steam at 100 bar and 400°C undergoes non-flow hyperbolic expansion process to 40 bar. Determine: 1. final condition of the steam ; 2. workdone during the process ; and 3. heat transferred. [Ans. 350°C ; 458 kJ ; 448.5 kJ] 7. Steam at a pressure of 10 bar and 0.95 dry expands isentropically to a pressure of 4 bar. Find the final condition of the steam by using 1. Steam tables ; and 2. Mollier diagram. [Ans. 0.896] 8. Steam at 20 bar and 250°C is expanded isentropically to 3.5 bar. It is then expanded hyperbolically to 0.6 bar. Using steam tables, determine: 1. Final condition of steam; and 2. Change in specific entropy during hyperbolic process. [Ans. 98°C (superheat); 0.066 kJ/ kg K] 9. Dry and saturated steam at a pressure of 11 bar is supplied to a turbine and is expanded isentropically to a pressure of 1 bar. Calculate: 1. heat supplied ; 2. heat rejected ; and 3. workdone. [Ans. 2362.2 kJ/ kg ; 1955.34 kJ/ kg ; 406.86 kJ/ kg] 10. One kg of steam at a pressure of 1 bar and 0.8 dry is compressed in a cylinder to a pressure of 2 bar. The law of compression is pv1.2 = constant. Find: 1. Final condition of the steam ; 2. Change in enthalpy ; 3. Change in internal energy ; 4. Workdone ; and 5. Heat that passes through the cylinder walls. [Ans. 0.86 ; 174.2 kJ/kg ; 157.4 kJ/kg ; – 83.5 kJ/kg ; 73.9 kJ/kg] 11. One kg of steam at a pressure of 20 bar and 0.96 dry expands to 1 bar. Assuming that the steam expands according to the law pv1.02 = constant, calculate: 1. final dryness fraction ; 2. workdone during expansion ; 3. change in internal energy ; and 4. heat exchange that occurs between the steam and cylinder walls per kg. [Ans. 123°C; 552 kJ/ kg ; 11.65 kJ/ kg ; 563.65 kJ/ kg] 12. Steam initially at 40 bar and 360°C expands to 3.5 bar according to the law pv1.3 = constant. Determine the final condition of the steam ; workdone per kg ; change in internal energy ; and the amount of heat transfer. [Ans. 0.845 ; 390 kJ/ kg ; – 603.42 kJ/ kg ; – 213.42 kJ/ kg] 13. Steam at a pressure of 10 bar is throttled until the pressure is 1 bar and the temperature is 120°C. Determine the quality of steam before it was throttled. [Ans. 0.97] 14. Steam initially at a pressure of 15 bar and 0.95 dryness expands isentropically to 7.5 bar and is then throttled until it is just dry. Using steam tables only, calculate: (a) Change in entropy ; (b) Change in enthalpy ; (c) Change in internal energy, per kg of steam during the entire process. Show the process in a h-s plane. Is the entire process reversible? Justify your statement. [Ans. 1.622 kJ/ kg K ; 87 kJ/kg ; – 51 kJ/ kg]

QUESTIONS 1. Draw p-v, T-s and h-s diagrams for constant volume process. Derive the expression for workdone, heat transfer and change in internal energy. 2. The constant pressure is also known as constant temperature process. Explain the reason. 3. For a vapour undergoing an isothermal process, although the temperature remains constant, the workdone and internal energy is not equal to zero. Explain why.

244 Engineering Thermodynamics 4. Prove that during expansion of steam according to pv = C, the change in internal energy is equal to the change in total heat of steam. 5. State the difference between isothermal and hyperbolic processes. 6. Draw p-v, T-s and h-s diagram for a reversible adiabatic process. Derive the expression of workdone for non-flow process and steady flow process. 7. Write a brief note on polytropic process of steam. 8. Explain throttling process of steam. Show the process on h-s diagram.

OBJECTIVE QUESTIONS 1. In a constant volume process, the heat transferred from state 1 to state 2 is equal to (a) change in enthalpy (b) change in internal energy (c) change in entropy (d) workdone 2. In which of the following process, the workdone from state 1 to state 2 is zero? (a) Constant volume process (b) Constant pressure process (c) Constant temperature process (d) Throttling process 3. The heating of wet steam at a constant temperature till it becomes dry saturated, is similar to (a) constant volume (b) constant pressure (c) constant entropy (d) none of these 4. The isothermal and .................... processes are identical for superheated steam. (a) constant volume (b) constant pressure (c) hyperbolic (d) throttling 5. The process, in which the change in internal energy is equal to the change of enthalpy, is known as (a) hyperbolic process (b) isentropic process (c) polytropic process (d) none of these 6. In a reversible adiabatic process, the workdone from state 1 to state 2, is equal to (a) change in entropy (b) change in enthalpy (c) change in internal energy (d) none of these 7. In an isentropic process, from state 1 to state 2, (a) change in entropy is zero (b) change in enthalpy is zero (c) change in internal energy is zero (d) workdone is zero 8. The reversible adiabatic process on h-s diagram is represented by a (a) horizontal line (b) vertical line (c) curve 9. The throttling process on h-s diagram is represented by a (a) horizontal line (b) vertical line (c) curve 10. The throttling process is also known as (a) wire drawing process (b) constant total heat process (c) constant enthalpy process (d) all of these

ANSWERS

1. (b) 6. (c)

2. (a), (d) 7. (a)

3. (b) 8. (b)

4. (c) 9. (a)

5. (a) 10. (d)

10 AIR STANDARD CYCLES 10.1 Introduction 10.2 Assumptions in Air Standard Cycle 10.3 Important Terms Used in Air Standard Cycles 10.4 Otto Cycle (Constant Volume Cycle) 10.5 Compression Ratio for Maximum Output of an Otto Cycle 10.6 Mean Effective Pressure for Otto Cycle 10.7 Diesel Cycle (Constant Pressure Cycle) 10.8 Mean Effective Pressure for Diesel Cycle 10.9 Dual Cycle 10.10 Mean Effective Pressure for Dual Cycle 10.11 Brayton Cycle

10.12 Difference Between Otto Cycle and Diesel Cycle 10.13 Comparison Between Otto, Diesel and Dual Cycle 10.14 Four Stroke Cycle Petrol Engine 10.15 Two Stroke Cycle Petrol Engine 10.16 Diesel Engine or Compression Ignition Engine 10.17 Four Stroke Cycle Diesel Engine 10.18 Two Stroke Cycle Diesel Engine 10.19 Comparison Between Two Stroke and Four Stroke Cycle Engines

10.1 INTRODUCTION We have already discussed a thermodynamic cycle in Chapter 1 (Art. 1.16) in which a process or processes are performed on a system in such a way that the final state is identical with the initial state. A thermodynamic cycle working with air (or an ideal gas) as a working substance is known as air standard cycle and the thermal efficiency of the air standard cycle is known as air standard efficiency. In most of the gas power cycles, the working substance mainly consists of air. Therefore, for the sake of simplification, thermodynamic analysis of gas power cycles is carried out by devising an idealised cycle known as air standard cycle. These cycles are useful in the study of *internal combustion engines because they represent a limit to which actual cycle may approach and because they are subject to simple mathematical and explanatory treatment. There are many types of cycle used by various types of engines, some require two complete revolutions of the crank for their performance while others may require one complete revolution of the crank. A cycle requiring two complete revolutions of the crank is known as four stroke cycle while a cycle that requires one complete revolutions of the crank for its completion, is called a two stroke cycle.

* As the name implies, the engines in which the combustion of fuel takes place inside the engine cylinder, are called internal combustion engines.

246 Engineering Thermodynamics

10.2 ASSUMPTIONS IN AIR STANDARD CYCLE The following assumptions are made in the air standard cycles: 1. The working substance is taken to be air and it behaves as a perfect gas. 2. The working substance is homogeneous throughout the cycle and does not change in its mass and composition. 3. The engine operates in a closed cycle. 4. All the processes constituting the cycle are reversible, i.e. they take place without any internal friction. 5. The specific heats (i.e. cp and cv) of the working substance remains constant throughout the cycle. 6. The combustion process is replaced by an equivalent heat addition process from an external source. 7. The exhaust process is replaced by an equivalent heat rejection process. 8. No change in potential and kinetic energy takes place during the process.

10.3 IMPORTANT TERMS USED IN AIR STANDARD CYCLES The various important terms, of the vertical piston-cylinder arrangement, as shown in Fig. 10.1 are as follows:

Fig. 10.1. Piston-cylinder arrangement.

1. Bore. It is the diameter (d ) of the cylinder or piston. 2. Stroke. It is the distance moved by the piston in one direction, i.e. either from top dead centre (TDC) to bottom dead centre (BDC) or from BDC to TDC. It is equal to twice the radius of the crank. 3. Top dead centre (TDC). It is the extreme position of the piston near to the head of the vertical cylinder. In case of horizontal cylinder, it is called inner dead centre (IDC). 4. Bottom dead centre (BDC). It is the extreme position of the piston opposite to the head of the vertical cylinder. In case of horizontal cylinder, it is called outer dead centre (ODC). 5. Clearance volume (vc). It is the minimum volume of clearance between the cylinder head and the piston at the top dead centre position. 6. Swept volume or stroke volume (vs). It is the maximum volume swept by the piston in moving from the top dead centre position (TDC) to the bottom dead position (BDC) and vice-versa. If l is the length of stroke and d is the diameter of piston, then Swept volume or stroke volume, π vs = × d 2 × l 4

Air Standard Cycles 247 7. Total volume or full cylinder volume (v). It is the volume occupied by the working fluid when the piston is at the bottom dead centre (BDC). It is equal to the clearance volume (vc) plus the swept volume (vs). Mathematically, total volume, v = vc + vs 8. Clearance ratio (c). It is the ratio of the clearance volume (vc) to the swept volume or stroke volume (vs). Mathematically, clearance ratio, v c = c vs 9. Compression or expansion ratio (r). It is the ratio of total volume (v) to the clearance volume (vc). Mathematically, compression or expansion ratio,

v +v v v 1 = c s = 1+ s = 1+ c vc vc vc 10. Mean effective pressure ( pm). It is the average pressure acting on the piston during the working stroke. It will be able to do the same amount of work, as done by the actual cyclic process. Mathematically, the mean effective pressure is the ratio of workdone per cycle (W) to the swept volume or displacement volume (vs), i.e. W W 4W = = pm = π vs 2 π d 2l × d ×l 4 11. Air standard efficiency. When air is assumed to be the working substance inside the engine cylinder, then the efficiency obtained is called air standard efficiency or ideal efficiency. It is defined as the ratio of workdone to the heat supplied during a cycle. Since workdone is equal to the difference of heat supplied and heat rejected, therefore efficiency of a cycle (or thermal efficiency), Work done Heat supplied − Heat rejected = h = Heat supplied Heat supplied 12. Relative efficiency. The actual efficiency of a cycle is always less than the air standard efficiency of that cycle under ideal conditions. The ratio of actual thermal efficiency to air standard efficiency is known as relative efficiency. Mathematically, relative efficiency,

r =

hR =

Actual thermal efficiency Air standard efficiency

10.4 OTTO CYCLE (CONSTANT VOLUME CYCLE) This cycle was invented by a Frenchman Beau-de-Rochas in 1862. In practical form, this cycle was introduced by a German Engineer Nicholas A. Otto in 1876. This cycle is used for *spark ignition (S.I.) internal combustion engines such as petrol engines. Since the heat is received and rejected at constant volume, therefore this cycle is known as constant volume cycle. An ideal Otto cycle consists of two constant volume and two reversible adiabatic (or isentropic) processes, as shown on p-v and T-s diagrams in Fig. 10.2 (a) and (b) respectively. Let m = Mass of air (in kg) in the engine cylinder at point 1, i.e. when the piston is at outer or bottom dead centre, and p1, v1, T1 = Pressure, volume and temperature of the air respectively at point 1.

* The engines in which the combustion of fuel takes place with the help of a spark plug, are known as spark ignition (S.I.) engines.

248 Engineering Thermodynamics Following are the four processes of an ideal Otto cycle. 1. Process 1-2. This process is a reversible adiabatic (or isentropic) compression process. In this process,the piston moves from the outer or bottom dead centre to the inner or top dead centre. The air is compressed in a reversible adiabatic manner and the temperature of air rises from T1 to T2. In this process, no heat is absorbed or rejected by the air. 3

p2 p4

2 4 Isen. comp. 1

p1

3

T3

Isen. exp.

Temperature

Pressure

p3

l.

o

.v

t ns

Co

T4 T2 T1

4

2 1

ol. t. v

s on

C

vs v2 = v3

v1 = v4 Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 10.2. Otto cycle.

2. Process 2-3. This process is a constant volume heating process. The piston is momentarily at rest at the inner or top dead centre. During the heating process, the heat is absorbed by the air and the temperature of air rises from T2 to T3. We know that heat absorbed by the air during the process 2-3, Q2 – 3 = m cv (T3 – T2) where cv is the specific heat of air at constant volume. 3. Process 3-4. This process is a reversible adiabatic (or isentropic) expansion process. The piston moves from the inner or top dead centre to the outer or bottom dead centre. The air expands in a reversible adiabatic manner and the temperature of air falls from T3 to T4. In this process, no heat is absorbed or rejected by the air, 4. Process 4-1. This process is a constant volume cooling process. The piston is momentarily at rest at the outer or bottom dead centre. During the cooling process, the heat is rejected by the air and the temperature falls from T4 to T1. We know that heat rejected by the air, Q4 – 1 = m cv (T4 – T1) From the above, we see that the air in the cylinder has returned to its original condition, thus completing the cycle. The air standard efficiency of the cycle may be obtained as discussed below: We know that workdone during the cycle, W = Heat absorbed – Heat rejected = m cv (T3 – T2) – m cv (T4 – T1) \ Ideal efficiency or air standard efficiency, m cv (T3 − T2 ) − m cv (T4 − T1 ) Workdone = h = Heat absorbed m cv (T3 − T2 ) = 1 −

T4 − T1 ...(i) T3 − T2

Air Standard Cycles 249 The ratio of compression and expansion (say r) are equal, i.e. v1 v = 4 = r v2 v3 We know that for reversible adiabatic (or isentropic) compression process 1-2,

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

...(ii)

Similarly, for reversible adiabatic (or isentropic) expansion process 3-4,

v T4 = 3 T3 v4

γ −1

1 = r

γ −1

...(iii)

From equations (ii) and (iii), we have TT T1 T = 4 or T4 = 1 3 T2 T2 T3 Substituting this value of T4 in equation (i), we have T1 T3 − T1 T T − T1T2 T2 = 1− 1 3 h = 1 − T3 − T2 T2 (T3 − T2 ) T (T − T ) T 1 = 1 − 1 3 2 = 1 − 1 = 1 − T2 (T3 − T2 ) T2 r

γ −1

...[From equation (ii)]

From the above expression, we see that the air standard efficiency of the Otto cycle depends upon the compression ratio (r) only. It increases with the increase in compression ratio. But, in actual practice, the value of r cannot increase beyond a value of 7 or so. The limitations to the use of higher compression ratios in Otto cycles are due to practical difficulties. In a petrol engine, (which is operated on Otto cycle) in which the cylinder contains an explosive mixture during compression stroke, the maximum compression ratio possible, in practice, is limited by the possibility of preignition and detonation.

10.5 COMPRESSION RATIO FOR MAXIMUM OUTPUT OF AN OTTO CYCLE We have already discussed that the workdone per kg of air of Otto cycle is given by W = cv (T3 – T2) – cv (T4 – T1) = cv [T3 – T2 – T4 + T1] We know that \ We also know that \

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

T2 = T1 r g – 1 v T4 = 3 T3 v4 T4 =

T3

r γ−1

γ −1

1 = r

γ −1

, where r = Ratio of compression.

... (i)

250 Engineering Thermodynamics Substituting the value of T2 and T4 in equation (i), we have

T γ −1 W = cv T3 − T1 r − γ 3−1 + T1 = cv [T3 – T1 r g–1 – T3 r1–g + T1] r This expression is a function of r when the initial temperature T1 and the maximum temperature T3 are fixed. The value of W will be maximum, if the differentiation of W with respect to r is zero, i.e. dW = 0 dr or – T1 (g – 1) (r)g – 2 – T3(1 – g) r – g = 0 – T1 (g – 1) r g – 2 + T3 (g – 1) r – g = 0 T3 (g – 1) r – g = T1 (g – 1) r g – 2

T3 r γ −2 = − γ = r g – 2 + g = r 2g – 2 T1 r

r = T3 T

\

1 2γ −2

1

10.6 MEAN EFFECTIVE PRESSURE FOR OTTO CYCLE We have already discussed that the mean effective pressure ( pm ) is the ratio of workdone to the swept or displacement volume. Mathematically, Workdone (W ) per kg of air Mean effective pressure, pm = Displacement or swepr volume (vs ) = We know that or

cv (T3 − T2 ) − cv (T4 − T1 ) c [(T − T ) − (T4 − T1 ) = v 3 2 ...(i) v1 − v2 v1 − v2

p1 v1 = RT1 ...(General gas equation for one kg of air) c ( γ − 1)T1 RT v1 = 1 = v ...(ii) p1 p1

...(Q R = cp – cv or

For reversible adiabatic (or isentropic) compression process 1-2,

c p − cv R = = g – 1) cv cv

γ

and Substituting the value of

v p p1 v1 = p2 v2 or 2 = 1 ...(iii) p1 v2 g

p1v1 pv p T v = 2 2 or 2 = 2 × 1 T1 T2 p1 T1 v2 p2 from equation (iii), we have p1 γ

\ where

g

v1 v T2 v1 T or 2 = 1 = × v T v T 2 v2 1 2 1

γ −1

= rg – 1

T2 = T1 r g – 1 ...(iv) v r = Ratio of compression = 1 v2

Air Standard Cycles 251 Now considering constant volume process 2-3, where \

p T p p2 = 3 or 3 = 3 = ap T3 T2 p2 T2 p ap = Pressure ratio = 3 p2 T3 = T2 × ap = T1 r g – 1 ap

...[From equation (iv)] ...(v)

Again for the constant volume process 4-1, where \

p4 p T p = 1 or 4 = 4 = ap T4 T1 T1 p1 ap = Pressure ratio = T4 = T1 ap

p4 p1 ...(vi)

Now equation (i) may be written as

pm =

=

=

cv [ (T3 − T2 ) − (T4 − T1 ) ] v1 − v2 cv (T1 r γ −1α p − T1 r γ −1 ) − (T1α p − T1 ) v v1 1 − 2 v1

cv T1[(r γ −1α p − r γ −1 ) − (α p − 1)] cv ( γ − 1) 1 T1 1 − p1 r

=

= where

ap =

p1 r

...[substituting the value v1 from equation (ii)] γ −1

(α p − 1) − (α p − 1) r − 1 ( γ − 1) r

p1r [(α p − 1)(r γ −1 − 1)] ( γ − 1)(r − 1) p3 p = 4 p2 p1

Example 10.1. In an Otto cycle, the temperature at the beginning and end of compression are 43°C and 323°C respectively. Determine the air standard efficiency and compression ratio. Take g = 1.4. Solution. Given: Temperature at the beginning of the cycle,

T1 = 43°C = 43 + 273 = 316 K

252 Engineering Thermodynamics Temperature at the end of compression, T2 = 323°C = 323 + 273 = 596 K Isentropic index, g = 1.4 Compression ratio First of all, let us find the compression ratio (r). We know that for reversible adiabatic (or isentropic) process 1-2 (Refer Fig. 10.2),

v T1 = 2 T2 v1

\

r g – 1 =

γ −1

1 = r

γ −1

T2 596 or r1.4 – 1 = = 1.886 316 T1 1

and r = (1.886) 0.4 = (1.886)2.5 = 4.885 Ans. Air standard efficiency We know that air standard efficiency, γ −1

1.4 −1

1 1 = 1− h = 1 − r 4.885 = 1 – (0.2047)0.4 = 1 – 0.53 = 0.47 or 47% Ans. Example 10.2. An engine of 200 mm bore and 400 mm stroke works on Otto cycle. The clearance volume is 2630 × 103 mm3. The initial pressure and temperature are 1 bar and 50°C. If the maximum pressure is limited to 25 bar, find: 1. The air standard efficiency of the cycle, and 2. The mean effective pressure of the cycle. Assume the ideal conditions. Solution. Given: Bore of cylinder, d = 200 mm Length of stroke, l = 400 mm Clearance volume vc = 2630 × 103 mm3 Initial pressure i.e. pressure before compression, p1 = 1 bar Initial temperature i.e. temperature before compression, T1 = 50°C = 50 + 273 = 323 K Maximum pressure i.e. pressure at the end of constant volume heating process, p3 = 25 bar 1. Air standard efficiency of the cycle We know that swept or stroke volume, π π 2 vs = × d 2 × l = (200) 400 = 12.568 × 106 mm3 4 4

\ Compression ratio,

r =

=

v + vs Total volume = c Clearance volume vc 2630 × 103 + 12.568 × 106 2630 × 103

= 5.78

Air Standard Cycles 253 We know that air standard efficiency

1 h = 1 − r

γ −1

1 = 1− 5.78

1.4 −1

= 1−

1 2.017

= 1 – 0.496 = 0.504 or 50.4% Ans. 2. Mean effective pressure of the cycle We know that for the reversible adiabatic (or isentropic) process 1-2 (Refer Fig. 10.2), p1 v1g = p2 v2g γ

or

v p2 = p1 1 = p1 r g = 1 (5.78)1.4 = 11.66 bar v2

Similarly, for the reversible adiabatic (or isentropic) process 3-4, p3 v3g = p4 v4g γ

or We know that pressure ratio,

v p 25 25 = p4 = p3 3 = γ3 = = 2.144 bar 1.4 v .66 11 r (5.78) 4 ap =

p3 p 2.144 = 4 = = 2.144 p2 p1 1

\ Mean effective pressure of the cycle,

pm =

p1r [(α p − 1)(r γ −1 − 1)] ( γ − 1)(r − 1)

=

1 × 5.78 [(2.144 − 1) (5.781.4 −1 − 1)] (1.4 − 1) (5.78 − 1)

6.727 = 3.518 bar Ans. 1.912 Example 10.3. An engine working on constant volume cycle has a clearance volume of 1 litre and a stroke volume of 6 litres. The suction pressure and temperature are 1 bar and 20°C respectively. The pressure at the end of heat addition is 25 bar. Determine: 1. Pressure and volumes at salient points of the cycle; 2. Thermal efficiency; and 3. Workdone per cycle Take cv1 during heat addition = 0.807 kJ/ kg K ; cv2 during heat rejection = 0.737 kJ/ kg K ; and g = 1.4. Solution. Given: Clearance volume, vc = 1 litre = 1000 cm3 ...( 1 litre = 1000 cm3) 3 Stroke volume, vs = 6 litre = 6000 cm Suction pressure, p1 = 1 bar Suction temperature, T1 = 20°C = 20 + 273 = 293 K Pressure at the end of heat addition, p3 = 25 bar 1. Pressure and volumes at salient points of the cycle Let v1, v2, v3 and v4 = Volumes at points 1, 2, 3 and 4 of the cycle respectively, (Refer Fig. 10.2) p2 and p4 = Pressure at points 2 and 4 respectively. We know that v2 = v3 = vc = 1000 cm3 Ans. =

254 Engineering Thermodynamics and v1 = v4 = vc + vs = 1000 + 6000 = 7000 cm3 Ans. The ratio of compression (v1/v2) and ratio of expansion (v4 /v3) are equal, i.e. v v 7000 r = 1 = 4 = =7 1000 v2 v3 Considering the reversible adiabatic (or isentropic) compression process 1-2, p1 v1g = p2 v2g γ

\

v p2 = p1 1 = p1 r g = 1 × 71.4 = 15.245 bar Ans. v2

Now considering reversible adiabatic (or isentropic) expansion process 3-4, p3 v3g = p4 v4g γ

\

v p 25 p4 = p3 3 = γ3 = 1.4 = 1.64 bar Ans. v r 7 4

2. Thermal efficiency We know that thermal efficiency,

1 h = 1 − r

γ −1

1.4 −1

1 = 1− 7

= 1 – 0.46 = 0.54 or 54% Ans.

3. Workdone per cycle We know that for reversible adiabatic (or isentropic) compression process 1-2, T2 = T1 r g – 1 = 293 (7)1.4 – 1 = 638 K ...[Refer Art. 10.5] For constant volume process 2-3, p3 p = 2 T3 T2 \

T3 =

p3 × T2 25 × 638 = = 1046 K p2 15.245

and for constant volume process 4-1, p4 p = 1 T4 T1 \

T4 =

p4 × T1 1.64 × 293 = = 480.5 K 1 p1

We know that workdone per cycle = Heat absorbed – Heat rejected = cv1 (T3 – T2) – cv2 (T4 – T1) = 0.807 (1046 – 638) – 0.737 (480.5 – 293) = 329.26 – 138.19 = 191.07 kJ/ kg Ans. Example 10.4. In an air standard Otto cycle, the compression ratio is 7 and compression begins at 35°C and 0.1 MPa. The maximum temperature of the cycle is 1100°C. Find : 1. Temperature and pressure at the cardinal points of the cycle ; 2. The heat supplied per kg of air ; 3. The workdone per kg of air ; 4. The cycle efficiency ; and 5. The mean effective pressure of the cycle.

Air Standard Cycles 255 Solution. Given: Compression ratio, r = 7 Temperature at the beginning of compression, T1 = 35°C = 35 + 273 = 308 K Pressure at the beginning of compression, p1 = 0.1 MPa = 0.1 × 106 Pa = 1 bar Maximum temperature of the cycle, T3 = 1100°C = 1100 + 273 = 1373 K p3

3

co

Temperature

Pressure

n.

m p.

2 p2 p4

4

Ise

n.

3

T3

Ise

ex

p.

2 4

T4 T1

1

p1

T2

v1 = v4 Volume (a) p-v diagram.

1

v2 = v 3

Entropy (b) T-s diagram.

Fig. 10.3

The p-v and T-s diagram of the given Otto cycle is shown in Fig. 10.3 1. Temperature and pressure at the cardinal points of the cycle Let T2 and T4 = Temperature at the end of the compression and expansion respectively, i.e. at points 2 and 4, and p2, p3 and p4 = Pressure at points 2, 3 and 4 respectively. We know that for reversible adiabatic (or isentropic) compression process 1-2, \

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

T2 = T1 r g – 1 = 308 (7)1.4 – 1 = 308 × 2.178 = 671 K

...(Q g for air = 1.4)

= 671 – 273 = 398°C Ans. We also know that T1 T = 4 T2 T3 or

T4 =

T1 × T3 308 × 1373 = = 630 K = 630 – 273 = 357°C Ans. T2 671

Considering reversible adiabatic (isentropic) compression volume process 1-2, γ

\

p v p1 v1 = p2 v2 or 2 = 1 = r g p1 v2 g

g

p2 = p1 . r g = 1 × 71.4 = 15.245 bar Ans.

256 Engineering Thermodynamics For constant volume process 2-3, p p2 = 3 T3 T2 T3 1373 = 15.245 × = 31.2 bar Ans. T2 671 Similarly, for constant volume process 4-1, p4 p = 1 T4 T1 \

p3 = p2 ×

\

p4 = p1 ×

T4 630 = 1× = 2.045 bar Ans. T1 308

2. Heat supplied per kg of air We know that heat supplied per kg of air during constant volume process 2-3, Q2 – 3 = cv (T3 – T2) = 0.718 (1373 – 671) = 504 kJ/ kg Ans. ...(Taking cv for air = 0.718 kJ/ kg K) 3. Workdone per kg of air We know that work done per kg of air, W = cv (T3 – T2) – cv (T4 – T1) = cv (T3 – T2 – T4 + T1) = 0.718 (1373 – 671 – 630 + 308) = 380 kJ/ kg Ans. 4. Cycle efficiency We know that cycle efficiency,

1 h = 1 − r

γ −1

1.4 −1

1 = 1− 7

= 1−

1 2.178

= 1 – 0.46 = 0.54 or 54% Ans. 5. Mean effective pressure of the cycle We know that pressure ratio, p p 2.045 ap = 3 = 4 = = 2.045 p2 p1 1 \ Mean effective pressure,

pm =

=

p1r [(α p − 1)(r γ −1 − 1)] ( γ − 1)(r − 1) 1 × 7 [(2.045 − 1) (71.4 −1 − 1)] 7(1.045 × 1.178) = (1.4 − 1) (7 − 1) 0.4 × 6

= 3.59 bar = 359 kN/m2 Ans. Example 10.5. In an air standard Otto cycle, the compression ratio is 7, and the compression begins at 1 bar and 40°C. The heat added is 2500 kJ/kg. Find: (a) work done per kg of air; (b) cycle efficiency; (c) maximum pressure and temperature of the cycle; and (d) mean effective pressure. For air, cv = 0.712 kJ/kgK ; and R = 0.288 kJ/kgK Solution. Given: Compression ratio, r = 7 Pressure at the beginning of compression, p1 = 1 bar Temperature at the beginning of compression, T1 = 40°C = 40 + 273 = 313 K

Air Standard Cycles 257 Heat added, Q2 – 3 = 2500 kJ/ kg Specific heat at constant volume, cv = 0.712 kJ/kgK Gas constant, R = 0.288 kJ/kgK We know that R = cp – cv or cp = R + cv = 0.288 + 0.712 = 1 kJ/kgK ∴ g = cp/cv = 1/0.712 = 1.4 (a) Workdone per kg of air First of all, let us find the temperatures at points 2, 3 and 4 (Refer Fig. 10.2). We know that for reversible adiabatic (or isentropic) process 1-2,

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

1.4 −1

1 = 7

=

1 2.178

\ T2 = T1 × 2.178 = 313 × 2.178 = 681.7 K Heat added during the heating process 2-3 per kg of air, Q2 – 3 = cv (T3 – T2) or 2500 = 0.712 (T3 – 681.7) 2500 \ T3 = + 681.7 = 4193 K 0.712 Now considering the reversible adiabatic (or isentropic) expansion process 4-1, γ −1

γ −1

1.4 −1

v T4 1 1 1 = = = = 3 2.178 T3 r 7 v4 T 4193 \ T4 = 3 = =1925 K 2.178 2.178 We know that workdone per kg of air, W = cv (T3 – T2) – cv (T4 – T1) = cv [T3 – T2 – T4 + T1] = 0.712 [4193 – 681.7 – 1925 + 313] = 1352.3 kJ/ kg Ans. (b) Cycle efficiency We know that cycle efficiency,

1 h = 1 − r

γ −1

1.4 −1

1 = 1− 7

= 1−

1 2.178

= 1 – 0.46 = 0.54 or 54% Ans. (c) Maximum pressure and temperature of the cycle Let p3 and T3 = Maximum pressure and temperature of the cycle. We know that for reversible adiabatic (or isentropic) compression process 1-2, p1 v1g = p2 v2g γ

v p2 v or = 1 = r g ...(Q r = 1 ) v p1 v 2 2 \ p2 = p1 r g = 1(7)1.4 = 15.24 bar Now for the constant volume process 2-3, p3 p = 2 T3 T2

258 Engineering Thermodynamics p2 × T3 15.24 × 4193 = = 93.74 bar Ans. T2 681.7 We have already calculated that maximum temperature of the cycle is T3 = 4193 K Ans. 4. Mean effective pressure We know that pressure ratio, T p 93.74 ap = 3 = 3 = = 6.15 T2 p2 15.24 \

p3 =

\ Mean effective pressure, pm = =

p1r [(α p − 1)(r γ −1 − 1)] ( γ − 1) (r − 1) 1 × 7 [(6.15 − 1) (71.4 −1 − 1) 42.467 = = 17.7 bar Ans. 2.4 (1.4 − 1) (7 − 1)

10.7 DIESEL CYCLE (CONSTANT PRESSURE CYCLE) This cycle was introduced by Dr. Rudolph Diesel in 1893 with an idea to obtain a higher thermal efficiency with a high compression ratio. This cycle is used for *compression ignition (C.I.) internal combustion engines working on diesel oil. Since the heat is supplied at constant pressure, therefore this cycle is also known as constant pressure cycle. An ideal cycle consists of two reversible adiabatic (or isentropic) processes, one constant pressure process and one constant volume process, as shown on p-v and T-s diagrams in Fig. 10.4 (a) and (b) respectively.

Fig. 10.4. Diesel cycle.

m = Mass of air (in kg) in the engine cylinder at point 1, i.e. when the piston is at the outer dead centre, and p1, v1 and T1 = Pressure, volume and temperature of the air respectively at point 1. Following are the four processes of an ideal diesel cycle: 1. Process 1-2. This process is a reversible adiabatic (or isentropic) compression process. In this process, the piston moves from the outer or bottom dead centre to the inner or top dead centre. The air is compressed in a reversible adiabatic (or isentropic) manner and the temperature of air rises from T1 to T2. In this process, no heat is absorbed or rejected by the air. Let

* The engines in which the ignition takes place due to the compression of fuel, are known as compression ignition (C.I.) engines.

Air Standard Cycles 259 2. Process 2-3. This process is a constant pressure heating process. The piston slightly moves from point 2 to point 3, at which the supply of heat is cut-off. Thus, the point 3 is known as cut-off point. During the heating process, the heat is absorbed by the air and the temperature of air rises from T2 to T3. We know that heat absorbed by the air during constant pressure heating process 2-3, Q2 – 3 = m cp (T3 – T2) where cp is the specific heat of air at constant pressure. 3. Process 3-4. This process is a reversible adiabatic (or isentropic) expansion process. In this process, the piston moves from the point of cut-off to the outer or bottom dead centre. The air is expanded in a reversible adiabatic (or isentropic) manner and the temperature of air falls from T3 to T4. In this process, no heat is absorbed or rejected by the air. 4. Process 4-1. This process is a constant volume cooling process. The piston is momentarily at rest at the outer or bottom dead centre. During the cooling process, the heat is rejected by the air and the temperature falls from T4 to T1. We know that heat rejected by the air, Q4 – 1 = m cv (T4 – T1) From the above, we see that the air in the cylinder has returned to its original condition, thus completing the cycle. The air standard efficiency of the cycle may be obtained as discussed below: We know that work done during the cycle, W = Heat absorbed – Heat rejected = m cp (T3 – T2) – m cv (T4 – T1) \ Ideal efficiency or air standard efficiency, mc p (T3 − T2 ) − m cv (T4 − T1 ) Workdone = h = Heat absorbed m c p (T3 − T2 ) = 1 −

cv T4 − T1 1 T4 − T1 = 1− γ T3 − T2 c p T3 − T2

...(Q

cp cv

= g) ...(i)

v Let r = Compression ratio = 1 v2

r = Cut-off ratio =

v3 v2

v4 v v v r = 1 = 1 × 2 = ...(Q v4 = v1) ρ v3 v3 v2 v3 Now considering reversible adiabatic (or isentropic) compression process 1-2. We know that

r1 = Expansion ratio =

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

\ T2 = T1 r g – 1 ...(ii) For the constant pressure heating process 2-3, v v2 = 3 T3 T2 v \ T3 = T2 × 3 = T2 . r = T1 r r g – 1 ...(iii) v2

260 Engineering Thermodynamics Again, considering reversible adiabatic (or isentropic) expansion process 3-4. We know that \

v T4 = 3 T3 v4

γ −1

ρ T4 = T3 r

1 = r1

γ −1

γ −1

ρ = r

γ −1

ρ = T1 r r g – 1 r

γ −1

r ρ

= T1 rg ... r1 = ...(iv)

Now substituting the values of T2, T3 and T4 from equations (ii), (iii) and (iv) respectively in equation (i), we have

h = 1 −

= 1 −

T1 ργ − T1 1 1 ργ − 1 1 − = γ T1 ρ r γ −1 − T1r γ −1 γ r γ −1 (ρ − 1)

1 ργ − 1 r γ −1 γ (ρ − 1)

The effect of compression ratio and cut-off ratio on the efficiency of diesel cycle are as follows: 1. Since the cut-off ratio (r) is always greater than unity, therefore the term within the bracket ργ − 1 i.e. increases with the increase of cut-off ratio. Thus, the efficiency of an ideal diesel γ (ρ − 1) cycle is lower than that of an Otto cycle, for the same compression ratio. 2. The efficiency of an ideal diesel cycle increases with the decrease in cut-off ratio and ργ − 1 approaches maximum (equal to efficiency of Otto cycle) when the term with the bracket i.e. γ (ρ − 1) is unity. In other words, when cut-off ratio, r = 1. 3. The compression ratio for diesel cycle varies from 14 to 18.

10.8 MEAN EFFECTIVE PRESSURE FOR DIESEL CYCLE We know that mean effective pressure,

pm =

=

Workdone (W ) per kg of air Displacement or swept volume (vs ) c p (T3 − T2 ) − cv (T4 − T1 ) v1 − v2

...(i)

The general gas equation for one kg of air is given by p1 v1 = RT1 or \

v1 =

c ( γ − 1)T1 RT1 = v p1 p1

...[Q R = cv (g – 1)]

v 1 r −1 v1 – v2 = v1 1 − 2 = v1 1 − = v1 v1 r r

=

cv ( γ − 1)T1 r − 1 ...(ii) p1 r

Air Standard Cycles 261 We know that for reversible adiabatic (or isentropic) compression process 1-2, γ

and

v p2 = 1 ...(iii) p1 v2 p1v1 pv p T v = 2 2 or 2 = 2 × 1 ...(iv) T1 T2 p1 T1 v2

p1 v1g = p2 v2g or

From equations (iii) and (iv), γ

\

v1 T2 v1 v = T v 2 1 2 v T2 = 1 T1 v2

γ −1

= r g – 1

or T2 = T1 r g – 1 ...(v) Now considering constant pressure process 2-3, v3 v = 2 T3 T2 v \ T3 = T2 × 3 = T2 × r = T1 r g – 1 × r ...[From equation (v)] ...(vi) v2 In the similar way as discussed above, we have for the reversible adiabatic (or isentropic) expansion process 3-4,

v T4 = 3 T3 v4

ρ = r \

γ −1

v v = 3× 2 v2 v4

γ −1

v v = 3× 2 v2 v1

γ −1

...(Q v4 = v1)

γ −1

ρ T4 = T3 r

γ −1

ρ = T1 × rg – 1 × r r

γ −1

= T1 rg

Now the equation (i) may be written as c p (T3 − T2 ) − cv (T4 − T1 ) pm = cv ( γ − 1)T1 r − 1 p1 r =

p1r ( γ − 1)(r − 1)

T3 − T2 T4 − T1 γ − T1 T1

Substituting the values of T2, T3 and T4 from equations (v), (vi) and (vii), we have

pm =

T1r γ −1ρ − T1r γ −1 T1ργ − T1 p1r γ − ( γ − 1)(r − 1) T1 T1

=

p1r [g (r rg – 1 – rg – 1) – ( rg – 1)] ( γ − 1)(r − 1)

=

p1r [g rg – 1 (r – 1) – ( rg – 1)] ( γ − 1)(r − 1)

... (vii)

262 Engineering Thermodynamics Example 10.6. An air standard Diesel cycle has a compression ratio of 14. The pressure at the beginning of the compression stroke is 1 bar and the temperature is 27°C. The maximum temperature of the cycle is 2500°C. Determine the thermal efficiency of the engine. Solution. Given: Compression ratio, r = 14 Pressure at the beginning of the compression stroke, p1 = 1 bar Temperature at the beginning of the compression stroke, T1 = 27°C = 27 + 273 = 300 K Maximum temperature of the cycle, T3 = 2500°C = 2500 + 273 = 2773 K We know that for the reversible adiabatic (or isentropic) compression process 1-2 (Refer Fig. 10.4),

v T1 = 2 T2 v1

γ −1

1 = r

γ −1

\ T2 = T1 r g – 1 = 300 (14)1.4 – 1 = 300 × 2.8737 = 862.11 K Now for the constant pressure heating process 2-3, v3 v T v = 2 or 3 = 3 T3 v2 T2 T2 \ Cut-off ratio,

r =

v3 T 2773 = 3 = = 3.216 v2 T2 862.11

We know that thermal efficiency of the engine,

h = 1 −

= 1 −

(3.216)1.4 − 1 1 ργ − 1 1 = 1− 1.4 −1 r (14) γ (ρ − 1) 1.4 (3.216 − 1) γ −1

1 4.13 = 1 – 0.4635 = 0.5365 or 53.65% Ans. 2.874 3.1

Example 10.7. An ideal Diesel engine has a diameter 150 mm and stroke 200 mm. The clearance volume is 10 per cent of the swept volume. Determine the compression ratio and the air standard efficiency of the engine if the cut-off takes place at 6 per cent of the stroke. Solution. Given: Diameter of cylinder, d = 150 mm = 0.15 m Length of stroke, l = 200 mm = 0.2 m Compression ratio Let r = Compression ratio. We know that the stroke or swept volume, π π vs = × d 2 × l = (0.15)2 0.2 = 3.5 × 10–3 m3 4 4 \ Clearance volume, vc = 10% of swept volume = 0.1 × 3.5 × 10–3 = 0.35 × 10–3 m3 We know that compression ratio,

r =

vc + vs 0.35 × 10−3 + 3.5 × 10−3 = = 11 Ans. vc 0.35 × 10−3

Air Standard Cycles 263 Air standard efficiency It is given that cut-off takes place at 6% of the stroke. Therefore, volume at cut-off, v3 = Clearance volume + Cut off volume = vc + 0.06 vs = 0.35 × 10–3 + 0.06 × 3.5 × 10–3 = 0.56 × 10–3 m3 \

Cut-off ratio =

v3 v 0.56 × 10 −3 = 3 = = 1.6 v2 vc 0.35 × 10 −3

We know that air standard efficiency,

h = 1 −

= 1 −

(1.6)1.4 − 1 1 ργ − 1 1 =1− 1.4 −1 (11) r γ (ρ − 1) 1.4 (1.6 − 1) γ −1

1 1.93 − 1 2.61 1.4 × 0.6

= 1 – 0.424 = 0.576 or 57.6 % Ans. Example 10.8. In an air standard Diesel cycle, the conditions at the beginning of the compression stroke are 300 K and 1 bar. The air is compressed to a pressure of 50 bar and then fuel is injected such that 20 kJ of energy is added per mole of air. Determine the compression ratio, the cut-off ratio and thermal efficiency of the cycle, if cp of air is 3.5 times the gas constant R. Solution. (Refer Fig. 10.4) Given: Temperature at the beginning of the compression stroke, T1 = 300 K Pressure at the beginning of the compression stroke, p1 = 1 bar = 100 kN/m2 Pressure at the end of the compression stroke, p2 = 50 bar = 5000 kN/m2 Heat added at constant pressure, = 20 kJ/mole of air Specific heat at constant pressure for air, cp = 3.5 R = 3.5 × 0.287 = 1.0045 kJ/kg K ... ( R for air = 0.287 kJ/ kg K) Compression ratio v Let r = Compression ratio = 1 v2 We know that p1 v1 = RT1 ...(Considering 1 kg of air) RT 0.287 × 300 \ v1 = 1 = = 0.861 m3 ...( p1 is taken in kN/m2) 100 p1 For reversible adiabatic (or isentropic) compression process 1-2, p1 v1g = p2 v2g \ Compression ratio,

p v1 = 2 v2 p1 r =

1 γ

1

50 1.4 = = 16.352 1

v1 = 16.352 Ans. v2

264 Engineering Thermodynamics Cut-off ratio We know that for the isentropic compression process 1-2, v T2 = 1 T1 v2

γ −1

= rg – 1

\ T2 = T1 rg – 1 = 300 (16.352)1.4 – 1 = 300 × 3.058 = 917.4 K Number of moles in one kg of air Mass of air is gram 1000 = = = 34.48 Molecular mass of air 29 \ Heat added per kg of air Q2 – 3 = Heat added per mole × Number of moles = 20 × 34.48 = 689.6 kJ/ kg We know that heat added per kg of air during the constant pressure process 2-3, Q2 – 3 = cp (T3 – T2) 689.6 = 1.0045 (T3 – 917.4) 689.6 \ T3 = + 917.4 = 1603.9 K 1.0045 We also know that for constant pressure process 2-3, v3 v = 2 T3 T2 \ Cut-off ratio,

r =

v3 T 1603.9 = 3 = = 1.748 Ans. v2 T2 917.4

Thermal efficiency of the cycle We know that thermal efficiency of the cycle, h = 1 −

(1.748)1.4 − 1 1 ργ − 1 1 = 1− 1.4 −1 r (16.352) γ (ρ − 1) 1.4 (1.748 − 1)

= 1 −

1 2.185 − 1 = 1 – 0.37 = 0.63 or 63% Ans. 3.058 1.4 × 0.748

γ −1

Example 10.9. The compression ratio of an ideal air standard Diesel cycle is 15. The heat transfer is 1470 kJ/kg of air. Find the pressure and temperature at the end of each process and determine the cycle efficiency. What is the mean effective pressure of the cycle, if the inlet conditions are 27°C and 1 bar. Solution. (Refer Fig. 10.4) v Given : Compression ratio, r = 1 = 15 v2 Heat transfer, Q2 – 3 = 1470 kJ/ kg Inlet temperature, T1 = 27°C = 27 + 273 = 300 K Inlet pressure, p1 = 1 bar Pressure and temperature at the end of each process Let p2 and T2 = Pressure and temperature at the end of reversible adiabatic (or isentropic) compression process, T3 = Temperature at the end of constant pressure heating process,

Air Standard Cycles 265 p4 and T4 = Pressure and temperature at the end of reversible adiabatic (or isentropic) expansion process. First of all, considering the reversible adiabatic (or isentropic) compression process 1-2, we know that p1 v1g = p2 v2g γ

\

v p2 = p1 1 = p1 r g = 1(15)1.4 = 44.31 bar Ans. v2 γ −1

v T2 = 1 = rg – 1 T1 v2 \ T2 = T1 r g – 1 = 300 (15)1.4 – 1 = 886.3 K Ans. For the constant pressure heating process 2-3, Q2 – 3 = m cp (T3 – T2) 1470 = 1 × 1.005 (T3 – 886.3) ...(Taking cp for air = 1.005 kJ/kg K) 1470 \ T3 = + 886.3 = 2349 K Ans. 1.005 Again for constant pressure process 2-3, v3 v = 2 T3 T2 We also know that

v3 T 2349 = 3 = = 2.65 v2 T2 886.3 Now considering reversible adiabatic (or isentropic) expansion process 3-4, We know that p3 v3g = p4 v4g \ Cut-off ratio,

r =

γ

or

γ

γ

γ

v v v v p4 ρ = 3 = 3 = 3 × 2 = v v v v p3 r 1 4 1 2

...(Q v4 = v1)

1.4

2.65 = = 0.088 15 \ p4 = p3 × 0.088 = 44.31 × 0.088 = 3.91 bar Ans. ...(Q p3 = p2) and \

v T4 = 3 T3 v4

γ −1

ρ T4 = T3 r

v = 3 v1

γ −1

γ −1

v v = 3× 2 v2 v1

γ −1

ρ = r

γ −1

1.4 −1

2.65 = 2349 15

= 1174.2 K Ans. Cycle efficiency We know that cycle efficiency,

h = 1 −

= 1 −

1 1 ργ − 1 = 1− γ −1 (15)1.4 −1 r γ (ρ − 1)

(2.65)1.4 − 1 1.4(2.65 − 1)

1 2.913 2.95 2.31

= 1 – 0.4275 = 0.5725 or 57.25% Ans.

266 Engineering Thermodynamics Mean effective pressure We know that mean effective pressure,

pm =

=

p1r [g rg – 1 ( r – 1) – ( rg – 1)] ( γ − 1)(r − 1) 1 × 15 [1.4 (15)1.4 – 1 (2.65 – 1) – {(2.65)1.4 – 1}] (1.4 − 1)(15 − 1)

= 2.678 [6.824 – 2.913] = 10.47 bar Ans. Example 10.10. Find the air standard efficiencies of the Otto and Diesel cycle on the basis of equal compression ratio of 10 and equal heat rejection of 840 kJ/kg. The suction conditions are 1 bar and 328 K. Solution. v Given: Compression ratio, r = 1 = 10 v2

p3

3

p2

2

p4

Q4 – 1 = 840 kJ/ kg p1 = 1 bar T1 = 328 K

3 Isen. exp.

Isen. comp.

p1

Temperature

Pressure

Heat rejection, Suction pressure, Suction temperature,

4

v2 = v3

4

T4 T2 T1

1

3 3

T3 T3

2

1

v1 = v4 Volume

Entropy

Fig. 10.5. Otto and Diesel cycle.

The p-v and T-s diagram of the Otto and Diesel cycle is shown in Fig. 10.5. The cycle 1-2-3-4 represents the Otto cycle and the cycle 1-2-3-4 represents a Diesel cycle. Considering reversible adiabatic (or isentropic) compression process 1-2 which is common to both the cycles. We know that p1 v1g = p2 v2g γ

v p2 = p1 1 = 1 (10)1.4 = 25.12 bar v2 γ −1

v T2 = 1 = (10)1.4 – 1 = 2.512 T1 v 2 \ T2 = T1 × 2.512 = 328 × 2.512 = 824 K We know that heat rejected during the constant volume process 4-1, Q4 – 1 = m cv (T4 – T1) 840 = 1 × 0.712 (T4 – 328) ...(Taking cv = 0.712 kJ/ kg K) We also know that

Air Standard Cycles 267 840 + 328 = 1508 K 0.712 Now considering reversible adiabatic (or isentropic) expansion process 3-4, \

\

T4 =

v T4 = 3 T3 v4 T3 =

γ −1

v = 2 v1

γ −1

1.4 −1

1 = 10

= 0.398

T4 1508 = = 3789 K 0.398 0.398

We know that heat supplied during the constant volume heating process 2-3, Q2 – 3 = m cv (T3 – T2) = 1 × 0.712 (3789 – 824) = 2111 kJ/kg \ Efficiency of the Otto cycle Heat supplied − Heat rejected 2111 − 840 hotto = = Heat supplied 2111 = 0.602 or 60.2% Ans. Now from the constant volume process 4-1, we have p4 p = 1 T4 T1 \

p4 = p1 ×

T4 1508 = 1× = 4.6 bar T1 328

Considering reversible adiabatic (or isentropic) expansion process 3-4 for Diesel cycle. We know that

T3′ = p3′ p T4 4

γ −1 γ

25.12 = 4.6

1.4 −1 1.4

= (5.46)0.286 = 1.625 ...(Q p3 = p2)

\ T3 = T4 × 1.625 = 1508 × 1.625 = 2450 K We know that heat supplied at constant pressure process 2-3, Q2 – 3 = m cp (T3 – T2) = 1 × 1.005 (2450 – 824) = 1634 kJ/ kg ... (Taking cp = 1.005 kJ/ kg K) \ Efficiency of Diesel cycle, Heat supplied − Heat rejected 1634 − 840 hDiesel = = Heat supplied 1634 = 0.486 or 48.6% Ans.

10.9 DUAL CYCLE This cycle is a combination of Otto and Diesel cycles. Since semi-diesel engines work on this cycle, therefore the dual cycle is also called semi-diesel cycle. An ideal dual cycle consists of two reversible adiabatic (or isentropic) processes, two constant volume processes and one constant pressure process, as shown on p-v and T-s diagrams in Fig. 10.6 (a) and (b) respectively. Let m = Mass of air (in kg) in the engine cylinder at point 1, i.e. when the piston is at the outer dead centre, and p1, v1 and T1 = Pressure, volume and temperature of the air respectively at point 1.

268 Engineering Thermodynamics Following are the five processes of an ideal dual cycle: 1. Process 1-2. This process is a reversible adiabatic (or isentropic) compression process. In this process, the piston moves from the outer or bottom dead centre to the inner or top dead centre. The air is compressed in a reversible adiabatic (or isentropic) manner and the temperature of air rises from T1 to T2. In this process, no heat is absorbed or rejected by the air.

p2

2

p5

4

Temperature

3

Pressure

p 3 = p4

Isen. exp. Isen. comp.

p1 v2 = v3

v4

T4

5

T3 T5 T2

1

T1

4 3 2

C v= v=

p=

C 5

C

1

v1 = v5

Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 10.6. Dual cycle.

2. Process 2-3. This process is a constant volume heating process. The piston is momentarily at rest at the inner or top dead centre. During the heating process, the heat is absorbed by the air and the temperature of air rises from T2 to T3. We know that heat absorbed by the air during constant volume heating process 2-3, Q2 – 3 = m cv (T3 – T2) where cv is the specific heat of air at constant volume. 3. Process 3-4. This process is a constant pressure heating process. The piston slightly moves from the inner or top dead centre towards outer or bottom dead centre (i.e. from point 3 to point 4). The supply of heat is cut-off at point 4. During the heating process, the heat is absorbed by the air at constant pressure and the temperature of air rises from T3 to T4. We know that heat absorbed by the air during constant pressure heating process 3-4, Q3 – 4 = m cp (T4 – T3) where cp is the specific heat of air at constant pressure. 4. Process 4-5. This process is a reversible adiabatic (or isentropic) expansion process. In this process, the piston moves from the cut-off point 4 to the outer or bottom dead centre. The air is expanded in a reversible adiabatic (or isentropic) manner and the temperature of air falls from T4 to T5. In this process, no heat is absorbed or rejected by the air. 5. Process 5-1. This process is a constant volume cooling process. The piston is momentarily at rest at the outer or bottom dead centre. During the cooling process, the heat is rejected by the air and the temperature falls from T5 to T1. We know that heat rejected by the air, Q5 – 1 = m cv (T5 – T1) From the above, we see that the air in the cylinder has returned to its original condition, thus completing the cycle. The air standard efficiency of the cycle may be obtained as discussed below: We know that work done during the cycle W = Heat absorbed – Heat rejected = (Q2 – 3 + Q3 – 4) – Q5 – 1 = [m cv (T3 – T2) + m cp (T4 – T3)] – m cv (T5 – T1)

Air Standard Cycles 269 \ Ideal efficiency or air standard efficiency,

h =

Work done Heat absorbed

m cv (T3 − T2 ) + m c p (T4 − T3 ) − m cv (T5 − T1 ) = m cv (T3 − T2 ) + m c p (T4 − T3 ) = 1 −

T5 − T1 ...(i) (T3 − T2 ) + γ (T4 − T3 )

r =

v1 = Compression ratio ; v2

ap =

p3 = Pressure ratio ; and p2

r =

v4 v4 = = Cut-off ratio v3 v2

Let

...(Q v3 = v2)

Now considering the reversible adiabatic (or isentropic) compression process 1-2. We know that

v T2 = 1 T1 v2

γ −1

= rg – 1

\ T2 = T1 r g – 1 ...(ii) For a constant volume heating process 2-3, we know that p3 p = 2 T3 T2 p3 × T2 = ap T2 = ap T1 r g – 1 ...[From equation (ii)] ...(iii) p2 For constant pressure heating process 3-4, we know that v v4 = 3 T3 T4

or

T3 =

\

T4 =

v4 × T3 = r T3 = r . ap T1 r g – 1 ...[From equation (iii)] ...(iv) v3

Considering reversible adiabatic (or isentropic) expansion process 4-5,

v T5 = 4 T4 v5

γ −1

v v = 4 × 2 v2 v5

γ −1

v v = 4 × 2 v2 v1

γ −1

ρ = r

\

ρ T5 = T4 r

γ −1

ρ = r ap T1 r g – 1 × r

= rg ap T1

Now substituting the values T2, T3, T4 and T5 is equation (i), we have h = 1 −

...(Q v5 = v1)

γ −1

γ −1

...[From equation (iv)]

(ργ α p T1 − T1 ) (α p T1 r γ −1 − T1 r γ −1 ) + γ ( ρα p T1 r γ −1 − α p T1 r γ −1 )

270 Engineering Thermodynamics = 1 − = 1 −

T1 ( ργ α p − 1) T1 (α p r γ −1 − r γ −1 ) + γ T1 α p r γ −1 (ρ − 1) (α p ργ − 1) r γ −1[(α p − 1) + γ α p (ρ − 1)]

10.10 MEAN EFFECTIVE PRESSURE FOR DUAL CYCLE We know that mean effective pressure, Workdone (W ) per kg of air pm = ...(i) Displacement or swept volume (vs ) =

cv (T3 − T2 ) + c p (T4 − T3 ) − cv (T5 − T1 ) v1 − v2

We also know that swept volume, v 1 r − 1 vs = v1 – v2 = v1 1 − 2 = v1 1 − = v1 ...(ii) v r r 1

The general gas equation for one kg of air is given by RT1 cv ( γ − 1) T1 = p1 v1 = R T1 or v1 = ...[Q R = cp – cv = cv (g – 1)] p1 p1 Substituting the value of v1 in equation (ii),

vs = v1 – v2 =

cv ( γ − 1) T1 r − 1 p1 r

\ Mean effective pressure,

pm =

=

cv (T3 − T2 ) + c p (T4 − T3 ) − cv (T5 − T1 ) cv ( γ − 1) T1 r − 1 p1 r T3 − T2 γ (T4 − T3 ) T5 − T1 p1r + − ...(iii) ( γ − 1) (r − 1) T1 T1 T1

From Art. 10.9, we have γ −1

α p T1 r − T1 r T3 − T2 = T1 T1

ρ α p T1 r T4 − T3 = T1

and

γ −1

γ −1

= r g – 1 (ap – 1)

− α p T1 r γ −1

T1

= rg – 1 ap ( r – 1)

ργ α p T1 − T1 T5 − T1 = = rg ap – 1 T1 T1

Now equation (iii) becomes,

pm =

p1r [rg – 1 (ap – 1) + g rg – 1 ap (r – 1) – (r g ap – 1)] ( γ − 1) (r − 1)

Air Standard Cycles 271 Example 10.11. A high speed oil engine working on a dual combustion cycle has a pressure of 1 bar and 50°C before compression. The air is then compressed isentropically to 1/15th of its original volume. The maximum pressure is twice the pressure at the end of isentropic compression. If the cut-off ratio is 2, determine the temperature at the end of each process and the ideal efficiency of the cycle. Take g = 1.4. Solution. (Refer Fig. 10.6) Given: Pressure of air before compression, p1 = 1 bar Temperature of air before compression, T1 = 50°C = 50 + 273 = 323 K Volume of air after isentropic compression, 1 v2 = v1 15

Maximum pressure,

v1 = 15 v2 p3 = 2p2

\ Pressure ratio,

ap =

\ Compression ratio,

r =

p3 =2 p2

v4 v4 = =2 v3 v2 We know that ideal efficiency of the cycle, Cut-off ratio,

r =

h = 1 −

= 1 − = 1 −

(α p .ργ − 1) r γ −1 (α p − 1) + γ α p (ρ − 1) (2 × 21.4 − 1)

(15)1.4 −1 [ (2 − 1) + 1.4 × 2 (2 − 1) ] 4.278 = 1 – 0.381 = 0.619 or 61.9% Ans. 2.954 (1 + 2.8)

Note: The ideal efficiency of the cycle may also be obtained by determining T2, T3, T4 and T5 and then applying equation (i) as discussed in Art. 10.9.

10.11 BRAYTON CYCLE The Brayton cycle is a constant pressure cycle used in gas turbine power plant. This cycle is also known as Joule’s cycle. An ideal cycle consists of two reversible adiabatic (or isentropic) processes and two constant pressure processes, as shown on p-v and T-s diagram in Fig. 10.7 (a) and (b) respectively. Let m = Mass of air (in kg) in the engine cylinder at point 1, and p1, v1 and T1 = Pressure, volume and temperature of air respectively at point 1. Following are the four processes of an ideal Brayton cycle: 1. Process 1-2. This process is a reversible adiabatic (or isentropic) compression process. In this process, the air is compressed in a reversible adiabatic (isentropic) manner and the temperature rises from T1 to T2. In this process, no heat is absorbed or rejected by the air.

272 Engineering Thermodynamics 2. Process 2-3. This process is a constant pressure heating process. During the heating process, the heat is absorbed by the air and the temperature of air increases from T2 to T3. We know that heat absorbed by the air during constant pressure process 2-3, Q2 – 3 = m cp (T3 – T2) where cp is the specific heat of air at constant pressure.

Fig. 10.7. Brayton cycle.

3. Process 3-4. This process is a reversible adiabatic (or isentropic) expansion process. In this process, the air is expanded in a reversible adiabatic (or isentropic) manner and the temperature of air falls from T3 to T4. In this process, no heat is absorbed or rejected by the air. 4. Process 4-1. This process is a constant pressure cooling process. During the cooling process, the heat is rejected by the air and the temperature falls from T4 to T1. We know that heat rejected by the air, Q4 – 1 = m cp (T4 – T1) From the above, we see that the air in the cylinder has returned to its original condition, thus completing the cycle. The air standard efficiency of the cycle may be obtained as discussed below: We know that workdone by the air, W = Heat absorbed – Heat rejected = m cp (T3 – T2) – m cp (T4 – T1) \ Ideal efficiency or air standard efficiency, m c p (T3 − T2 ) − m c p (T4 − T1 ) Workdone h = = Heat absorbed m c p (T3 − T2 ) T4 − T1 ...(i) T3 − T2 Now considering reversible adiabatic (or isentropic) compression process 1-2. We know that = 1 −

γ −1 T2 p = v1 = 2 v T1 p1 2

γ −1 γ

...(ii)

and for reversible adiabatic (or isentropic) expansion process 3-4,

v T3 = 4 T4 v3

γ −1

p = 3 p4

γ −1 γ

p = 2 p1

γ −1 γ

...(iii) ...( p3 = p2 and p4 = p1 )

Air Standard Cycles 273 From equations (ii) and (iii), T p T2 = 3 = 2 T4 p1 T1 Now equation (i) may be written as

γ −1 γ

= (rp )

γ −1 γ

or

T T4 = 3 ...(iv) T2 T1

T T1 4 − 1 T1 = 1 − T1 h = 1 − T2 T T2 3 − 1 T2 1 = 1 − γ −1 (rp )

...[From equation (iv)]

...[From equation (ii)]

γ

where rp = Compression ratio or pressure ratio =

p2 p1

Example 10.12. A gas turbine operates on Brayton cycle which takes in air at 1 bar and 15°C. The air is compressed to 5 bar and heated in a combustion chamber to 800°C. The hot air expands in the turbine to 1 bar. Find: 1. Power developed; and 2. Thermal efficiency of the cycle. Take g = 1.4; and cp = 1.005 kJ/kg K. Solution. (Refer Fig. 10.7) Given: Initial pressure of air, p1 = 1 bar Initial temperature of air, T1 = 15°C = 15 + 273 = 288 K Pressure at the end of compression, p2 = 5 bar Temperature at which air is heated in combustion chamber, T3 = 800°C = 800 + 273 = 1073 K Pressure to which air is expanded, p4 = p1 = 1 bar Isentropic index, g = 1.4 Specific heat at constant pressure, cp = 1.005 kJ/ kg K Considering the isentropic compression process 1-2, we have

p T2 = 2 T1 p1

γ −1 γ

5 = 1

1.4 − 1 1.4

= (5)0.286 = 1.584

\ T2 = T1 × 1.584 = 288 × 1.584 = 456.2 K Similarly, for the isentropic expansion process 3-4, we have \

p T4 = 4 T3 p3 T4 =

γ −1 γ

1 = 5

1.4 − 1 1.4

1 = 5

T3 1073 = = 677.4 K 1.584 1.584

0.286

=

1 1.584

274 Engineering Thermodynamics We know that heat absorbed by the air during constant pressure heating process 2-3, Q2 – 3 = m cp (T3 – T2) = 1 × 1.005 (1073 – 456.2) = 620 kJ/ kg and heat rejected by the air during constant pressure cooling process 4-1, Q4 – 1 = m cp (T4 – T1) = 1 × 1.005 (677.4 – 288) = 391 kJ/ kg 1. Power developed We know that workdone or power developed = Heat absorbed – Heat rejected = 620 – 391 = 229 kJ/kg Ans. 2. Thermal efficiency of the cycle We know that thermal efficiency of the cycle, Heat absorbed − Heat rejected 620 − 391 h = = Heat absorbed 620 = 0.369 or 36.9% Ans. Example 10.13. The pressure ratio and maximum temperature of a Brayton cycle are 5 : 1 and 923 K respectively. The air enters the compressor at 1 bar and 298 K. Calculate for 1 kg of air flow, the compressor work, turbine work and the efficiency of the cycle. Solution. (Refer Fig. 10.7) p Given: Pressure ratio, rp = 2 = 5 p1 Maximum temperature, T3 = 923 K Pressure of air entering the compressor, p1 = 1 bar Temperature of air entering the compressor, T1 = 298 K Compressor work Considering the reversible adiabatic (or isentropic) compression process 1-2. We know that

p T2 = 2 T1 p1

γ −1 γ

= (rp )

γ −1 γ

=

1.4 − 1 (5) 1.4

= (5)0.286 = 1.584

\ T2 = T1 × 1.584 = 298 × 1.584 = 472 K We know that compressor work per kg of air flow w1 – 2 = m cp (T2 – T1) = 1 × 1 (472 – 298) = 174 kJ/ kg ...(Taking cp for air = 1 kJ/kg K) Turbine work Considering the reversible adiabatic (or isentropic) expansion process 3-4, we know that

p T3 = 3 T4 p4

γ −1 γ

p = 2 p1

γ −1 γ

= (rp )

γ −1 γ

= (5)

1.4 − 1 1.4

= 1.584

T3 923 = = 582.7 K 1.584 1.584 We know that turbine work per kg of air flow, w3 – 4 = m cp (T3 – T4) = 1 × 1 (923 – 582.7) = 340.3 kJ/ kg Ans. \

T4 =

Air Standard Cycles 275 Efficiency of the cycle We know that net work done during the cycle w = Turbine work – Compressor work = 340.3 – 174 = 166.3 kJ/kg and heat supplied, q = m cp (T3 – T2) = 1 × 1 (923 – 472) = 451 kJ/kg \ Efficiency of the cycle, w 166.3 h = = 0.3687 or 36.87% Ans. q 451

10.12 DIFFERENCE BETWEEN OTTO CYCLE AND DIESEL CYCLE

The following are the main differences between an Otto cycle and a Diesel cycle: 1. In Otto cycle, both the supply of heat and rejection of heat takes place at constant volume. In Diesel cycle, the heat is supplied at constant pressure and the heat is rejected at constant volume. 2. The air standard efficiency of an Otto cycle depends only on the compression ratio whereas the air standard efficiency of Diesel cycle depends upon the compression ratio and cut-off ratio both. 3. The air standard efficiency of Otto cycle is higher than Diesel cycle.

10.13 COMPARISON BETWEEN OTTO, DIESEL AND DUAL CYCLE The comparison between Otto, Diesel and dual cycle for different parameters are as follows: 1. For equal compression ratio and heat input The p-v and T-s diagram for the three cycles is shown in Fig. 10.8. The Otto cycle is represented by 1-2-3-4; Diesel cycle by 1-2-3-4 and the dual cycle is shown by 1-2-2-3-4.All the three cycles start at the same initial pressure, volume and temperature at point 1. The air is compressed in a reversible adiabatic (or isentropic) manner from state 1 to state 2. The heat is supplied under varying conditions for different cycle. Since the same amount of heat is supplied to each cycle, therefore the cycle which rejects the least amount of heat will be more efficient. We know that from T-s diagram, the heat rejected by the Otto cycle, Diesel cycle and dual cycle is given by the area a-1-4-b, a-1-4b and a-1-4-b respectively. The thermal efficiency of the cycle is given by 3

2

p=C

3

3

3 3

p=C

4 4 4 1

Temperature

Pressure

2

2

v= 2

(a) p-v diagram.

3

C

v=C

4 4

4

1

a Volume

C

p=

b b Entropy

b

(b) T-s diagram.

Fig.10.8. Otto, Diesel and dual cycles for equal compression ratio and heat input.

hth = 1 −

Heat rejected (QR ) QR =1− Heat supplied (QS ) Constant

276 Engineering Thermodynamics From the above, we see the Otto cycle rejects least amount of heat (area a-1-4-b). Thus, Otto cycle has the maximum efficiency. Thus the order of efficiencies for the three cycles is as follows: (hth)Otto > hth (Dual) > hth (Diesel) 2. For same maximum pressure and temperature and same heat rejection 3 3

Temperature

Pressure

2

2 2

4

3 2 2 2

p=c

3

v=c

v=c

4

1

1

a

b

Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 10.9. Otto, Diesel and dual cycles for same maximum pressure and temperature.

The p-v and T-s diagram for the three cycles is shown in Fig. 10.9. The Otto cycle is represented by 1-2-3-4; Diesel cycle by 1-2-3-4 and the dual cycle is shown by 1-2-3 -3 -4. All the three cycles have the same maximum pressure ( p3 ) and temperature (T3). Since all the cycles reject equal amount of heat (area a-1-4-b), therefore thermal efficiency

hth = 1 −

Heat rejected (QR ) Constant =1− Heat supplied (QS ) Heat supplied (QS )

The cycle which has higher addition, will be more efficient. We see that the Diesel cycle is more efficient than Otto cycle. Thus (hth)Diesel > (hth)Dual > (hth)Otto

10.14 FOUR STROKE CYCLE PETROL ENGINE Spark plug Exhaust Inlet valve valve

(a) Suction or charging

(b) Compression stroke. stroke.

(c) Expansion or working stroke.

(d) Exhaust stroke.

Fig. 10.10. Four stroke cycle petrol engine.

In 1876, Doctor Otto, a German Engineer produced an engine working on the cycle known as Otto cycle or four stroke cycle. It requires four strokes of the piston to complete the cycle of operations in the cylinder. The action of a four stroke cycle petrol engine (i.e. spark ignition engine) using a fuel-air mixture (known as charge) is discussed below. It may be noted that the petrol engines uses a carburettor which prepare a mixture for the engine.

Air Standard Cycles 277 1. Suction stroke. This is also known as charging stroke. The piston moves downwards from the top dead centre (TDC) position creating a vacuum inside the cylinder, as shown in Fig. 10.10 (a). During this stroke, the inlet (or intake) valve is kept open so that the mixture of air and fuel from the carburettor enters the cylinder. The pressure inside the cylinder may theoretically be assumed to remain atmospheric. At the end of the stroke, when the piston reaches the bottom dead centre (BDC) position, the inlet valve is closed. Thus the cylinder is filled up with fresh charge. 2. Compression stroke. During the upward movement of the piston from the bottom to the top dead centre, both the valves are closed as shown in Fig. 10.10(b). The charge is compressed isentropically to some relatively high pressure and temperature. This completes one revolution of the crank shaft. Shortly before the end of compression stroke (i.e. before the top dead centre is reached), the compressed charge is ignited by means of a *spark plug. The combustion of the charge is theoretically assumed to be instantaneous so that it may be taken to take place at constant volume. 3. Expansion or working stroke. The high pressure set up by the combustion of charge, pushes the piston downwards, as shown in Fig. 10.10(c). The thrust on the piston is created causing the rotation of the crankshaft. This stroke is, therefore, known as expansion or working stroke. In this case, both the valves are closed. This is the only part of the cycle in which positive or useful work is done. 4. Exhaust stroke. When the piston is at the bottom dead centre position, the exhaust valve opens and the inlet valve is closed. Now the upward movement of piston expels out the products of combustion remaining in the cylinder through the exhaust valve into the atmosphere, as shown in Fig. 10.10(d).

10.15 TWO STROKE CYCLE PETROL ENGINE A two stroke cycle petrol engine was first successfully devised by Sir Duglad Clerk in 1880. In this cycle, all the four operations i.e. suction, compression, expansion and exhaust takes place during two strokes of the piston. Hence, in a two stroke cycle, there is one working stroke after every revolution of the crankshaft. A two stroke engine has ports instead of valves. The working of a two stroke cycle engine is shown in Fig. 10.11. The crank case of the engine is designed as a compressor in which air or mixture of air and fuel is compressed. It is then forced into the engine cylinder. Following are the four stages of the two stroke cycle petrol engine: 1. First stage. The piston is moving downwards from the top dead centre (TDC) towards bottom dead centre (BDC). When the piston is in the position as shown in Fig. 10.11 (a), the space above it contains the expanded gases after ignition. These expanded gases leave the cylinder through the uncovered exhaust port having done work on the piston. 2. Second stage. After a small fraction of the crank revolution, a transfer port is also uncovered by the piston as shown in Fig. 10.11 (b), and connects the cylinder with the crank case which contains slightly compressed charge of air mixed with fuel. This charge is transferred to the upper part of the cylinder through the transfer port. The piston is so shaped (i.e. crowned) that the fresh charge of fuel and air will move to the top of the cylinder and push out the remaining exhaust gases through the exhaust port. This process of removing the exhaust gases completely from the upper part of the cylinder is known as scavenging. The piston is now at the bottom dead centre (BDC). 3. Third stage. As the piston starts moving upwards from the bottom dead centre (BDC) position, the inlet port closes first, allowing a short period for the fresh charge to continue the scavenging process. It may be noted that during the upstroke of the piston, the compression of the charge begins with both the transfer port and exhaust port closed, as shown in Fig. 10.11 (c).

* Since the fuel is ignited by means of a spark plug, therefore the petrol engines are known as spark ignition (or S.I.) engines.

278 Engineering Thermodynamics Spark plug

Cylinder Crown

Transfer port

Exhaust port Inlet port Piston

Compression of mixture Connecting rod Crank case

Crank

(a)

(b)

Compression of mixture begins

Compression of mixture completed

Mixture

Expansion of mixture (c)

Mixture being taken in (d)

Fig. 10.11. Two stroke cycle petrol engine.

4. Fourth stage. The upward movement of the piston during compression stroke lowers the pressure (below atmospheric pressure) in the crankcase so that the fresh charge of air mixed with fuel vapour is drawn into the crankcase through the inlet port which is uncovered by the piston, as shown in Fig. 10.11 (d). The charge is compressed during the compression stroke and a little before the piston reaches top dead centre (TDC), the charge is ignited with the help of a spark plug. The hot high pressure pushes the piston downwards with full force and expansion of the burnt gases takes place. The cycle is thus completed and repeated again.

10.16 DIESEL ENGINE OR COMPRESSION IGNITION ENGINE The Diesel engine (or compression ignition engine) was invented in 1893 by a German Engineer, Doctor Rudolf Diesel, with a view to obtain a practicable engine of high thermal efficiency. The first engine was built in 1897 by M/S Mirrless, Watson and Co. of Glasgow.

Air Standard Cycles 279 This engine differs from the petrol engine because in this case, the mixture is ignited by means of a fuel injection valve by giving a fine fuel spray in place of sparking plug in petrol engines. Hence there are three valves i.e. inlet valve, exhaust valve and fuel injection value, as shown in Fig. 10.12. The three valves have the following functions: 1. Inlet valve. It admits air into the cylinder from the engine. 2. Exhaust valve. It allows the burnt gases to escape through a silencer to the atmosphere. 3. Fuel injection valve. It has to perform the following functions: (a) It supply the fuel at the correct time. (b) It breaks up the fuel into fine spray. (c) It distributes the fuel to all parts of the combustion space. (d) It delivers the correct amount of fuel. The essential features of a diesel engine are as follows: (a) The oil fuel is ignited by compression alone. (b) The fuel injection is by means of compressed air. (c) The combustion takes place with no rise in pressure (i.e. at constant pressure) during injection. (d) The combustion commences as soon as injection begins. The diesel engines are designed to work either on four stroke cycle or two stroke cycle as discussed in the following pages.

10.17 FOUR STROKE CYCLE DIESEL ENGINE The four strokes i.e. suction, compression, expansion and exhaust operations occuring in a Diesel engine are as follows: Inlet valve

Fuel injection valve Exhaust valve Exhaust

(a) Suction or Charging stroke.

(b) Compression stroke.

(c) Expansion or Working stroke.

Fig. 10.12. Four stroke cycle diesel engine.

(d) Exhaust stroke.

280 Engineering Thermodynamics 1. Suction stroke. The suction starts with the piston at the top dead centre. During this stroke, the inlet valve is open and the exhaust valve and fuel injection valve are closed as shown in Fig. 10.12 (a). The downward movement of the piston from top dead centre (TDC) towards bottom dead centre (BDC) causes suction in the cylinder, which draws in a fresh charge of air from atmosphere at atmospheric pressure. It may be noted that during suction stroke, the pressure remains constant with increase in volume. 2. Compression stroke. The upward movement of the piston from bottom dead centre (BDC) towards top dead centre (TDC), causes the compression of air as all the valves, i.e. inlet, exhaust and fuel injection valve are closed, as shown in Fig. 10.12 (b). The air is compressed to a high pressure and temperature (about 35 bar and 800°C). Since the compression ratio is very high (15 to 20), therefore the volume is decreased. 3. Expansion or working stroke. When the piston reaches at the end of compression stroke, the inlet and exhaust valves are closed whereas the fuel injection valve remains open during a small part of the expansion stroke, as shown in Fig. 10.12 (c). The fuel is admitted through a fuel valve in the form of a fine spray. The heat produced by the high compression of air raises the temperature of air sufficiently to ignite the fuel as soon as it is injected into the cylinder, and the combustion continues till the fuel valve remains open. The injection of fuel is so regulated that theoretically the combustion is assumed to take place at constant pressure. Now, the hot high pressure products of combustion pushes the piston downwards to the bottom dead centre and hence doing the work. 4. Exhaust stroke. When the piston reaches near the bottom dead centre (BDC), the exhaust value opens, whereas the inlet and fuel valve remain closed, as shown in Fig. 10.12 (d). The hot gases are now discharged into the atmosphere at constant volume, in order to make room for fresh air to enter into the cylinder. The exhaust valve closes after the piston reaches the top dead centre (TDC). The cycle is thus completed.

10.18 TWO STROKE CYCLE DIESEL ENGINE The working of a two stroke cycle Diesel engine is shown in Fig. 10.13. Following are the four stages of the two stroke cycle Diesel engine: 1. First stage. The piston moves downwards from the top dead centre (TDC) towards bottom dead centre (BDC) during the expansion or power stroke. It first uncovers the exhaust port as shown in Fig. 10.13 (a) and the cylinder pressure drops to atmospheric pressure as the combustion products leave the cylinder through the exhaust port. 2. Second stage. After a small fraction of the crank revolution, a transfer port is also uncovered by the piston, as shown in Fig. 10.13 (b). The slightly compressed air in the crankcase enters the engine cylinder. The piston is so shaped (i.e. crowned) that the fresh air will move up to the top of the cylinder and push out the remaining exhaust gases through the exhaust port. This process of removing the exhaust gases completely from the upper part of the cylinder is known as scavenging. It also prevents the fresh air flowing directly to the exhaust port and being lost. The piston is now at the bottom dead centre (BDC). 3. Third stage. As the piston starts moving upwards from the bottom dead centre (BDC) position, the inlet port closes first. The upward movement of the piston compresses the air, with both the transfer port and exhaust port closed as shown in Fig. 10.13 (c). 4. Fourth stage. The upward movement of the piston during compression stroke lowers the pressure in the crankcase so that the fresh air is drawn into the crankcase through the inlet port which is uncovered by the piston, as shown in Fig. 10.13 (d). A little before the piston reaches the top dead centre (TDC), the fuel is forced through the open fuel injection valve under pressure in the form of a very fine spray into the engine cylinder; which gets ignited due to high temperature of the compressed air. It may be noted that rate of fuel injection is such as to maintain the pressure of gases

Air Standard Cycles 281 approximately constant during combustion. Now the hot high pressure drive the piston downwards with full force and expansion of gases takes place. The cycle is thus completed and repeated again. Fuel valve

Cylinder Transfer port

Expansion of products

Exhaust port Piston

Burnt gases out

Inlet port Connecting rod

Air compressed in crank case Crank (a)

(b)

Air in

Air

(c )

(d)

Fig. 10.13. Two stroke cycle diesel engine.

10.19 COMPARISON BETWEEN TWO STROKE AND FOUR STROKE CYCLE ENGINES

The comparison between two stroke and four stroke cycle engines are as follows: 1. The two stroke engine gives one working or power stroke for each revolution of the crank. In a four stroke cycle engine, there is only one working or power stroke for every two revolutions of the crank. 2. The power produced by a two stroke cycle engine is almost double than that of a four stroke cycle engine. 3. A two stroke cycle engine requires a lighter flywheel as it gives a uniform torque on the crankshaft. A four stroke cycle engine requires a larger flywheel due to greater fluctuation of speed.

282 Engineering Thermodynamics

4. For the same power, a two stroke cycle engine requires less space. 5. A two stroke cycle engine has inlet, exhaust and transfer ports, whereas a four stroke cycle engine has two complicated valves for inlet and outlet which are operated by a cam and lever arrangement. 6. The thermal efficiency of a four stroke cycle engine is higher than that of a two stroke cycle engine. This is due to the fact that the compression ratio of a four stroke cycle engine is higher than that of a two stroke cycle engine. 7. A two stroke engine is used in light vehicles such as motor cycle, scooter and marine engines. A four stroke engine is used for small and medium, stationary power engines and heavy vehicles. 8. The two stroke engines are much easier to start than four stroke engines. 9. The initial cost of a two stroke engine is considerably less than a four stroke engine. 10. The consumption of lubricating oil in two stroke engines is more than that of a four stroke engine because of greater wear and tear.

HIGHLIGHTS 1. A thermodynamic cycle working with air (or an ideal gas) as a working substance is known as air standard cycle and the thermal efficiency of the air standard cycle is known as air standard efficiency. 2. A cycle requiring two complete revolutions of the crank is known as four stroke cycle, while a cycle that requires one complete revolution of the crank for its completion is known as two stroke cycle. 3. The air standard efficiency of an ideal Otto cycle (constant volume cycle) is given by

h = 1 −

T1 1 =1− T2 r

γ −1

where T1 and T2 = Temperature at the beginning and at the end of reversible adiabatic (or isentropic) compression respectively, and v1 v2 4. The compression ratio (r) for maximum output of an Otto cycle is given by

r = Ratio of compression =

T r = 3 T

1 2γ − 2

1

where T1 = Initial temperature, and T3 = Maximum temperature. 5. The mean effective pressure ( pm ) for an Otto cycle is given by where

pm =

p1r [(α p − 1) (r γ −1 − 1)] ( γ − 1) (r − 1)

p1 = Initial pressure, and p p ap = Pressure ratio = 3 = 4 p2 p1

Air Standard Cycles 283 6. The air standard efficiency of an ideal Diesel cycle (constant pressure cycle) is given by where

h = 1 −

1 ργ − 1 r γ (ρ − 1) γ −1

r = Compression ratio =

v1 , and v2

v3 v2

r = Cut-off ratio =

7. The efficiency of an ideal Diesel cycle is lower than that of an Otto cycle, for the same compression ratio. 8. The efficiency of an ideal Diesel cycle approaches maximum when the cut-off ratio (r) is unity. 9. The mean effective pressure ( pm ) for Diesel cycle is given by p1 r pm = [ g r g – 1 ( r – 1) – ( rg – 1)] ( γ − 1) (r − 1) where p1 = Initial pressure. 10. The air standard efficiency of a dual cycle is given by where

h = 1 −

(α p ργ − 1) r γ −1[(α p − 1) + γ α p (ρ − 1)]

r = Compression ratio =

ap = Pressure ratio =

r = Cut-off ratio =

v1 , v2

p3 , and p2 v4 v4 = ...(Q v3 = v2) v3 v2

11. The mean effective pressure ( pm) for the dual cycle is given by p1r pm = [r g – 1 (ap – 1) + g rg – 1 ap (r – 1) – (rg ap – 1)] ( γ − 1) (r − 1) 12. The air standard efficiency of an ideal Brayton cycle is given by

where

h = 1 −

T1 =1− T2

1 (rp )

γ −1 γ

rp = Pressure ratio or compression ratio =

p2 . p1

13. In a four stroke cycle engine, the working cycle is completed in four strokes of the piston or two revolutions of the cranckshaft. 14. In a two stroke cycle engine, the working cycle is completed in two strokes of the piston or one revolution of the crankshaft. 15. The process of removing products of combustion (burnt gases) completely from the cylinder of the engine, is known as scavenging.

284 Engineering Thermodynamics

EXERCISES 1. An engine working on the Otto cycle, has a cylinder diameter of 150 mm and a stroke of 225 mm. The clearance volume is 1250 cm3. Find the air standard efficiency of this engine. Take g = 1.4. [Ans. 43.6%] 2. An engine working on ideal Otto cycle has temperature and pressure at the beginning of isentropic compression as 25°C and 1.5 bar respectively. If the thermal efficiency of the engine is 48% and g = 1.4, find the compression ratio. Also find the temperature and pressure at the end of compression. [Ans. 5.13; 14.8 bar, 300°C] 3. An engine working on a Otto cycle is supplied with air at 100 kPa and 35°C. The compression ratio is 8. The heat supplied is 2100 kJ/kg. Calculate the maximum pressure and temperature of the cycle. Also find the cycle efficiency and mean effective pressure of the cycle. [Ans. 94.35 bar; 3632.4 K ; 56.5% ; 1535 kPa] 4. A single cylinder petrol engine has a swept volume of 500 cm3 and a clearance volume of 50 cm3. If the pressure and temperature at the beginning of compression are 100 kN/m2 and 25°C and maximum temperature of the cycle is 1055°C. Calculate (a) Air standard efficiency ; (b) Pressure and temperature at each characteristic point, and (c) Mean effective pressure of the cycle. [Ans. 61.7% ; 504.8°C, 235.8°C, 2870.45 kN/m2; 4901 kN/m2; 170.74 kN/m2; 313.2 kN/m2] 5. In an Otto cycle, air at 1 bar and 17°C is compressed isentropically until the pressure is 15 bar. The heat is added at constant volume until the pressure rises to 40 bar. Calculate the air standard efficiency and the mean effective pressure for the cycle. Take cv = 0.717 kJ/ kgK; Ru = 8.314 kJ/ kg mole K. [Ans. 53.86% ; 5.7 bar] 6. In a Diesel engine, the compression ratio is 13 and the cut-off of fuel takes place at 8% of the stroke. Find the air standard efficiency of the engine. Take g for air as 1.4. [Ans. 58.25%] 7. A Diesel engine has a swept volume of 9.33 litres and the clearance volume is 8% of the stroke. The initial pressure and temperature in the cycle is 1 bar and 27°C. The compression ratio is 15. If the cut-off takes place at 8% of the stroke, calculate: 1. cycle efficiency ; and 2. heat supplied per cycle. [Ans. 60.37% ; 886.2 kJ/ kg] 8. An ideal Diesel cycle operates on 1 kg of standard air with initial pressure of 1 bar and a temperature of 35°C. The pressure at the end of compression is 33 bar and the cut-off takes place at 6% of the stroke. Determine: 1. the compression ratio; 2. the percentage clearance; 3. the heat supplied; and 4. the heat rejected. Take g = 1.4 and cp = 1 kJ/ kg K. [Ans. 12.14; 8.97% of stroke volume; 559 kJ/ kg ; 230 kJ/ kg] 9. A Diesel engine has a bore of 250 mm and a stroke of 400 mm. The cut-off takes place at 5 per cent of the stroke. Estimate: (a) air-standard efficiency ; 2. mean effective pressure, if the clearance volume and pressure at the end of suction stroke are 1.2 litres and 1 bar respectively. [Ans. 63.5% ; 6.04 bar] 10. In an air standard Diesel cycle, the compression ratio is 16. The temperature and pressure at the beginning of isentropic compression is 15°C and 1 bar respectively. The heat is added until the temperature at the end of constant pressure process is 1480°C. Calculate: (a) Cut-off ratio; (b) Heat supplied per kg; (c) Cycle efficiency; and (d) Mean effective pressure. [Ans. 2; 880.36 kJ/ kg ; 61.3% ; 6.95 bar] 11. The following data pertains to a compression ignition engine working on air standard Diesel cycle: Cylinder bore = 150 mm ; Stroke length = 250 mm; Clearance volume = 400 cm3. Calculate the air standard efficiency of the engine if fuel injection takes place at constant pressure for 5% of the stroke. How this efficiency will be affected if the fuel supply continues upto 8% of the stroke. [Ans. Efficiency decreases by 3.28%] 12. Two engines are to operate on Otto and Diesel cycles, with the following data: Maximum temperature = 1227°C ; Exhaust temperature = 427°C ; Ambient condition 1 bar and 27°C. Find the compression ratios; maximum pressures and efficiencies. [Ans. Otto cycle : 6.72 ; 33.6 bar ; 53.3%, Diesel cycle: 12.3 ; 33.6 bar ; 58.1%]

Air Standard Cycles 285 13. An air standard dual cycle has a compression ratio of 10. The pressure and temperature at the beginning of compression are 1 bar and 27°C. The maximum pressure is 40 bar and the maximum temperature is 1400°C. Determine: (a) the temperature at the end of constant volume heat addition; (b) cut-off ratio ; (c) workdone per kg of air; and (d) the cycle efficiency. Assume for air, cp = 1.004 kJ/kg K; and cv = 0.717 kJ/ kg K [Ans. 1200 K; 1.394 ; 465 kJ/kg; 58.5%] 14. An engine working on a Brayton cycle has the initial and final pressures as 1 bar and 3 bar respectively. The temperature of air before compression is 25°C and it is heated to 650°C. Determine: 1. power developed per kg of air; and 2. efficiency of the engine. Take cp = 1 kJ/ kg K ; and cv = 0.715 kJ/ kg K [Ans. 135 kJ ; 25.9%] 15. A gas turbine plant is working on the Brayton cycle between temperatures 27°C and 800°C. (a) Find the pressure ratio at which the cycle efficiency approaches the Carnot cycle efficiency. (b) Find the pressure ratio at which the workdone per kg of air is maximum. (c) Compare the efficiency at this pressure ratio with the Carnot efficiency for the given temperatures. [Ans. 86.1; 9.3 ; 0.653]

QUESTIONS 1. (a) Define air standard cycles. (b) List the assumptions made in the analysis of air standard cycles. 2. Define the following terms used in air standard cycles: (a) Clearance volume ; (b) Swept volume ; (c) Stroke length ; (d) Compression ratio ; (e) Air standard efficiency ; and ( f ) Relative efficiency. 3. What is an Otto cycle? Show that the efficiency of the Otto cycle depends only on the compression ratio. 4. Show that for the Otto cycle, the air standard efficiency can be expressed as 1 h = 1 − γ −1 r where r is the compression ratio. 5. Explain the limitations of compression ratio in actual engine using Otto cycle. 6. Show that the compression ratio for maximum output of an Otto cycle is

T r = 3 T

1 2γ − 2

1

where T1 is the initial temperature and T3 is the maximum temperature. 7. Derive an expression for the mean effective pressure of the Otto cycle. 8. Describe the ideal cycle for a Diesel engine and derive the air standard efficiency of Diesel cycle with the help of neat diagram. 9. What is the effect of compression ratio and cut-off ratio over the efficiency of a Diesel engine? 10. Show that the efficiency of Diesel cycle is less than that of Otto cycle for the same compression ratio. 11. Derive an expression for the air standard efficiency of the dual cycle with the help of p-v and T-s diagrams. 12. Compare the Otto, Diesel and Dual cycles for the same compression ratio and same heat input. 13. Show that for the same maximum pressure and temperature of the cycle and the same heat rejection hDiesel > hDual > hOtto

286 Engineering Thermodynamics 14. Derive the expression of optimum pressure ratio for maximum net work output in an ideal Brayton cycle. What is the corresponding cycle efficiency? 15. Describe with the aid of neat sketches the working principle of a four stroke cycle petrol engine. 16. Discuss the working of a two stroke cycle petrol engine with the help of neat sketches. 17. What do you understand by scavenging? 18. Explain with neat sketches, the working of four stroke Diesel engine. 19. Describe the working of two stroke cycle Diesel engine with the help of neat sketches. 20. Distinguish between four stroke and two stroke engines.

OBJECTIVE TYPE QUESTIONS 1. An Otto cycle is also known as (a) constant volume cycle (b) constant pressure cycle (c) constant temperature cycle (d) none of these 2. An Otto cycle consists of (a) one constant volume, one constant pressure and two reversible adiabatic processes (b) two constant pressure and two isothermal processes (c) two constant volume and two reversible adiabatic processes (d) two constant pressure and two reversible adiabatic processes 3. The air standard efficiency of an otto cycle is given 1 (a) 1 – r g – 1 (b) 1 − r

γ −1

1 (c) 1 + r g – 1 (d) 1 + r

γ −1

where r = Compression ratio ; and g = Ratio of specific heats. 4. The compression ratio is the ratio of (a) swept volume to clearance volume (b) swept volume to total volume (c) total volume to swept volume (d) total volume to clearance volume 5. A Diesel cycle is also known as (a) constant volume cycle (b) constant pressure cycle (c) constant temperature cycle (d) none of these 6. A Diesel cycle consists of (a) one constant volume, one constant pressure and two reversible adiabatic processes (b) two constant pressure and two isothermal processes (c) two constant volume and two reversible adiabatic processes (d) two constant pressure and two reversible adiabatic processes 7. The efficiency of Diesel cycle, for the same compression ratio, is .................... Otto cycle. (a) equal to (b) greater than (c) less than 8. The efficiency of Diesel cycle increases with the (a) decrease in cut-off ratio (b) increase in cut-off ratio (c) constant cut-off ratio (d) none of these 9. The efficiency of Diesel cycle approaches to Otto cycle efficiency when (a) cut-off ratio is decreased (b) cut-off ratio is increased (c) cut-off ratio is zero (d) cut-off ratio is constant

Air Standard Cycles 287 10. An Otto cycle efficiency is higher than Diesel cycle efficiency for the same compression ratio and heat input because in Otto cycle (a) maximum temperature is higher (b) heat rejection is lower (c) combustion is at constant volume (d) expansion and compression are isentropic 11. The dual combustion cycle consists of (a) one constant pressure, two constant volume and two isentropic processes (b) one constant volume, two constant pressure and two isothermal processes (c) one isothermal, two constant volume and two isentropic processes (d) none of the above 12. For the same compression ratio, the efficiency of dual cycle is (a) less than Diesel cycle (b) greater than Diesel cycle (c) less than Diesel cycle and greater than Otto cycle (d) greater than Diesel cycle and less than Otto cycle 13. For the same maximum pressure and temperature (a) Otto cycle is more efficient than Diesel cycle (b) Diesel cycle is more efficient than Otto cycle (c) Dual cycle is more efficient than Otto and Diesel cycles (d) Dual cycle is less efficient than Otto and Diesel cycles 14. A cycle which consists of two reversible adiabatic processes and two constant pressure processes is known as (a) Otto cycle (b) Diesel cycle (c) Dual cycle (d) Brayton cycle 15. In a four stroke cycle engine, the working cycle is completed in (a) one revolution of the crankshaft (b) two revolutions of the crankshaft (c) three revolutions of the crankshaft (d) four revolutions of the crankshaft 16. The petrol engines are also known as .................... engines. (a) spark ignition (b) compression ignition 17. In a two stroke cycle engine, the working cycle is completed in .................... revolution of the crankshaft. (a) one (b) two (c) three (d) four 18. The diesel engines are also known as .................... engines. (a) spark ignition (b) compression ignition 19. In diesel engines, the ignition takes place due to the heat produced in the engine cylinder at the end of (a) suction stroke (b) exhaust stroke (c) compression stroke (d) expansion stroke 20. A diesel engine, during suction stroke, draws (a) air only (b) diesel only (c) a mixture of diesel and air

ANSWERS

1. (a) 6. (a) 11. (a) 16. (a)

2. (c) 7. (c) 12. (d) 17. (a)

3. (b) 8. (a) 13. (b) 18. (b)

4. (d) 9. (c) 14. (d) 19. (c)

5. (b) 10. (b) 15. (b) 20. (a)

11 MIXTURE OF GASES

11.1 Introduction 11.2 Mass Fraction 11.3 Mole Fraction 11.4 Volume Fraction 11.5 Dalton’s Law of Partial Pressures

11.6 Amagat’s Leduc Law 11.7 Properties of Gas Mixtures 11.8 Change of Entropy During Mixing Process 11.9 Volumetric Analysis and Gravimetric Analysis

11.1 INTRODUCTION In engineering applications, we often come across thermodynamic systems which are mixture of gases. If the mixture of gases remains homogeneous in composition throughout a process, and do not react chemically with one another, then it may be considered as a single pure substance. Following are the examples of various mixtures: 1. Dry air. It is a mixture of several gases like nitrogen, oxygen, carbon-dioxide etc. 2. Ordinary atmospheric air. It is a mixture of dry air and water vapour. 3. Flue gases. These consist of a mixture of various gases like carbon-dioxide, sulphur dioxide, nitrogen, water vapour etc. In general, the mixture of pure substances may be divided into the following two groups: (a) A mixture involving only gases considered as ideal gases; and (b) A mixture involving ideal gases and vapours. The thermodynamic properties of mixtures may be determined experimentally and tabulated or related by general thermodynamic relations in the similar way as for single pure substance. However, the thermodynamic behaviour of a mixture of gases and vapour depends upon the individual properties of its constituent gases. Thus, it may not be practical to tabulate the properties of all the mixtures except for the cases of constant chemical composition such as dry air. Hence, it is important and necessary to deduce the properties of any mixture from the properties of its constituents.

11.2 MASS FRACTION A homogeneous mixture of two or more ideal gases is called an ideal gas mixture. Let us consider a container initially divided into three separate compartments of volumes v1, v2 and v3, as shown in Fig. 11.1. Assume that these three compartments are filled with different ideal gases at the same temperature (T ) and pressure ( p). The total mass (m) of the mixture will be the sum of the masses of individual constituents, i.e. m = m1 + m2 + m3

1

2

3

p v1 T m1

p v2 T m2

p v3 T m3

Fig. 11.1. Mass fraction.

Mixture of Gases 289 Dividing throughout by the total mass (m), we have m m m 1 = 1 + 2 + 3 ...(i) m m m The mass fraction of the constituent may be defined as the ratio of the mass of the constituent to the total mass of the mixture. It is usually denoted by y. m \ Mass fraction of gas 1, y1 = 1 m Mass fraction of gas 2,

y2 =

m2 m

m3 m Now the equation (i) may be written as y1 + y2 + y3 = 1 This shows that the sum of mass fractions is unity.

and mass fraction of gas 3,

y3 =

Note: The mass of each gas may be calculated from the gas equation, pv = mRT. p v3 p v1 p v2 \ m1 = ; m2 = ; and m3 = , R3 T R2 T R1 T where R1, R2 and R3 are the gas constants in kJ / kg K and depends upon the type of gas.

11.3 MOLE FRACTION The mole fraction of the constituent may be defined as the ratio of the number of moles of a constituent to the total number of moles in a mixture. It is usually denoted by x. Let n1, n2 and n3 be the number of moles of each constituent. \ Total number of moles in the mixture, n = n1 + n2 + n3 Dividing throughout by the total number of moles (n), we have n n n 1 = 1 + 2 + 3 = x1 + x2 + x3 n n n where x1, x2 and x3 are the mole fraction of each constituent. The above equation shows that the sum of mole fractions is unity. When the analysis of a gas is made on the basis of moles, it is called molar analysis. Notes: 1. The number of moles of each gas can be calculated from the gas equation, pv = n Ru T. p v3 p v1 p v2 \ n1 = ; n2 = ; and n3 = Ru T Ru T Ru T where Ru is the universal gas constant in kJ/ kg mole K and is same for all the gases. Its value is 8.314 kJ/kg mole K. 2. The mass of any constituent (say i) of the molecular mass (M) is related to the mole content as mi = ni Mi where mi = Mass of the component i, ni = Number of moles of the component i, and Mi = Molecular mass of the ith component.

11.4 VOLUME FRACTION The volume fraction of the constituent may be defined as the ratio of the volume occupied by a constituent theoretically assumed to be separated at the pressure and temperature of the mixture to the total volume of the mixture. It is usually denoted by z. Let v1, v2 and v3 be the volume occupied by each constituent.

290 Engineering Thermodynamics \ Total volume of the mixture, v = v1 + v2 + v3 Dividing throughout by the total volume (v), we have v v v 1 = 1 + 2 + 3 = z1 + z2 + z3 v v v where z1, z2 and z3 are the volume fraction of each constituent. The above equation shows that the sum of volume fractions is unity. The volume of each gas can be calculated from the gas equation, pv = n Ru T. n R T n R T n R T \ v1 = 1 u ; v2 = 2 u ; and v3 = 3 u p p p and for the gas mixture,

v =

n Ru T p

The volume fraction of each constituent is given by n R T v n p = 1 = x1 z1 = 1 = 1 u × v p n Ru T n v n Similarly, z2 = 2 = 2 = x2 v n v n and z3 = 3 = 3 = x3 v n This shows that for an ideal gas mixture, the volume fraction is equal to the mole fraction.

11.5 DALTON’S LAW OF PARTIAL PRESSURES The mixture of ideal gases obeys a law known as Dalton’s law of partial pressures. According to this law, each gas of a mixture occupying a certain volume behaves independently of others. The pressure exerted by each gas is, therefore, called a partial pressure and the total pressure exerted by the mixture is equal to the sum of the partial pressures exerted by its individual constituent of the gases measured alone at the temperature and volume of the mixture. For example, if p1, p2, p3 etc. are the partial pressures of the individual gases contained in the mixture, then the total pressure ( p) of the gas mixture is p = p1 + p2 + p3 + ... 1 p1 T v m1

2 +

p2 T v m2

3 +

p3 T v m3

Mixture T = Constant

T, v p = p1+p2+p3

Fig. 11.2. Dalton’s law of partial pressures.

This law is represented schematically in Fig. 11.2. The three ideal gases 1, 2 and 3 occupying volume v, when mixed result in a mixture of volume v. The final pressure is the sum of the partial pressures of the individual constituent of the gases. Let n1, n2 and n3 be the number of moles of the gases 1, 2 and 3. Since there is no chemical action, therefore, the mixture is in a state of equilibrium with the equation of state, p1 v = n1 Ru T ...(i)

Mixture of Gases 291 p2 v = n2 Ru T ...(ii) and p3 v = n3 Ru T ...(iii) Now for the ideal gas mixture, pv = n Ru T = (n1 + n2 + n3) Ru T ...(iv) where n = n1 + n2 + n3 = Total number of moles in the mixture. From the above equations (i), (ii), (iii) and (iv), we have n p n 1 = 1 = x1 or p1 = p × 1 = p . x1 p n n

n p2 n2 = = x2 or p2 = p × 2 = p . x2 p n n

and

p3 n3 n = = x3 or p3 = p × 3 = p . x3 n p n

The ratio concerned.

p p1 p2 , ; and 3 may be defined as the partial pressure ratio of the constituent p p p

11.6 AMAGATS LEDUC LAW This law states that the volume of a mixture of ideal gases is equal to the sum of the partial volumes or separted volumes of the constituent gases, when each exists at the temperature and pressure of the mixture. Mathematically, v = v1 + v2 + v3 This law is represented schematically in Fig. 11.3. 1 p, v1 T, n1

2 +

Mixture

3

p, v2 T, n2

+

p, v3 T, n3

T = Constant

p, T v = v1+v2 + v3 n = n1 + n 2 + n 3

Fig. 11.3. Amagats Leduc law.

Applying gas equation to each gas, we have

pv1 = n1 Ru T or n1 =

pv1 ...(i) Ru T

pv2 = n2 Ru T or n2 =

pv2 ...(ii) Ru T

p v3 = n3 Ru T or n3 =

pv3 ...(iii) Ru T

and

n = n1 + n2 + n3 pv3 pv1 pv2 + + = Ru T Ru T Ru T We know that

=

p (v + v2 + v3) ...(iv) Ru T 1

292 Engineering Thermodynamics Now for the gas mixture, the ideal gas equation may be written as pv pv = n Ru T or n = ...(v) Ru T From equations (iv) and (v), we get v = v1 + v2 + v3 This is Amagats Leduc law. When the gas equation for the individual constituents is divided by the gas equation of the entire gas mixture, then pv1 n1 Ru T v n = or 1 = 1 pv n Ru T v n Similarly

v n v2 n2 = ; and 3 = 3 v n v n

This shows that the ratio of the partial volume of any constituent to the total volume of the mixture (i.e. volume fraction) is equal to the mole fraction of that constituent.

11.7 PROPERTIES OF GAS MIXTURES The various properties of the gas mixtures are as follows: 1. Specific volume and density. The specific volume (vs) of each constituent is the volume of the mixture (v) divided by the mass of the constituent (m). Mathematically, v v v vs1 = ;v = ;v = and so on m1 s2 m2 s3 m3 where vs1, vs2 and vs3 are the specific volume of each constituent having mass m1, m2 and m3 respectively. The specific volume of the mixture is given by v v vs = = m m1 + m2 + m3 + ... or

m + m2 + m3 m m m 1 = 1 = 1 + 2 + 3 + ... vs v v v v

1 1 1 + + + ... vs1 vs 2 vs 3 Since density is the reciprocal of specific volume, therefore density of the mixture, r = r1 + r2 + r3 + ... 2. Internal energy. The internal energy of a mixture of gases is equal to the sum of the internal energies of the individual components, each taken at the temperature and volume of the mixture (i.e. sum of the partial energies). Mathematically, u = u1 + u2 + u3 + ... The total internal energy of each component on mass basis is given by mu = m1 u1 + m2 u2 = m3 u3 + ... \ Specific internal energy of the gas mixture, m u + m2 u2 + m3 u3 + ... m m m u = 1 1 = 1 u1 + 2 u2 + 3 u3 + ... m m m m

=

= y1 u1 + y2 u2 + y3 u3 + ... (in kJ/ kg)

Mixture of Gases 293 Note: The total internal energy on the mole basis is given by nu = n1 u1 + n2 u2 + n3 u3 + ... \ Molar specific internal energy, n n n u = 1 u1 + 2 u2 + 3 u3 + ... n n n = x1 u1 + x2 u2 + x3 u3 + ... (in kg / kg mole)

3. Enthalpy. The enthalpy of a mixture of gases is equal to the sum of the enthalpies of the individual components, each taken at the temperature and volume of the mixture (i.e. sum of partial enthalpies). Mathematically, h = h1 + h2 + h3 + ... In the similar way as discussed above, the total enthalpy of each component on a mass basis is given by mh = m1 h1 + m2 h2 + m3 h3 + ... \ Specific enthalpy of the gas mixture, m h + m2 h2 + m3 h3 + ... h = 1 1 m m m m = 1 h1 + 2 h2 + 3 h3 + ... m m m = y1 h1 + y2 h2 + y3 h3 + ... (in kJ / kg)

Note: Total enthalpy of each component on mole basis is given by n h = n1 h1 + n1 h2 + n3 h3 + ... \ Molar specific enthalpy of the gas mixture, h = x1 h1 + x2 h2 + x3 h3 + ... (in kJ / kg mole)

4. Gas constant. The gas constant for the mixture (R) may be obtained from the gas equation, mRT pv = mRT or p = . Now for the gases 1, 2 and 3, at the same temperature (T) and volume (v), v m R T m1 R1 T m R T ; p2 = 2 2 ; and p3 = 3 3 v v v According to Dalton’s law of partial pressures, p = p1 + p2 + p3

or \

p1 =

m RT m R T m R T mRT = 1 1 + 2 2 + 3 3 v v v v mR = m1 R1 + m2 R2 + m3 R3 m m m R = 1 R1 + 2 R2 + 3 R3 m m m

= y1 R1 + y2 R2 + y3 R3

Notes: 1. The molecular mass of the mixture (Mmix) may be obtained as discussed below: We know that gas constant, Universal gas constant ( Ru ) R = Molecular mass Now for the gases 1, 2 and 3, Ru R R R = y1 × u + y2 × u + y3 × u M mix M3 M1 M2

294 Engineering Thermodynamics or \

y 1 y y = 1 + 2 + 3 M mix M1 M 2 M 3 Mmix =

1 y1 / M 1 + y2 / M 2 + y3 / M 3

where M1, M2 and M3 are the molecular mass of gas 1, 2 and 3 respectively. 2. The molecular mass of the mixture (Mmix) is also given by the following equation, i.e.

Mmix =

Total mass of the mixture (m) Total number of moles in the mixture (n))

5. Specific heat at constant volume. We have already discussed that the total internal energy of a mixture of gases on mass basis is given by m u = m1 u1 + m2 u2 + m3 u3 ...(i) We know that internal energy of an ideal gas per kg is u = cv dT where cv = Specific heat of the gas at constant volume. Now equation (i) may be written as m (cv )mix dT = m1 cv1 dT + m2 cv2 dT + m3 cv3 dT m m m or (cv )mix = 1 cv1 + 2 cv 2 + 3 cv3 m m m = y1 cv1 + y2 cv2 + y3 cv3 where y1, y2 and y3 is the mass fraction. In the similar way as discussed above, we may obtain the molar specific heat at constant volume (cvm) as n (cvm)mix dT = n1 cvm1 dT + n2 cvm2 dT + n3 cvm3 dT n n1 n cv m1 + 2 cv m 2 + 3 cv m3 n n n = x1 cv m1 + x2 cv m2 + x3 cv m3 6. Specific heat at constant pressure. The total enthalpy of a mixture of gases on mass basis is given by m h = m1 h1 + m2 h2 + m3 h3 ...(i) We know that enthalpy of an ideal gas per kg is h = cp dT where cp = Specific heat of the gas at constant pressure. Now equation (i) may be written as m (cp)mix dT = m1 cp1 dT + m2 cp2 dT + m3 cp3 dT m m m or (cp)mix = 1 c p1 + 2 c p 2 + 3 c p 3 m m m = y1 cp1 + y2 cp2 + y3 cp3 where y1, y2 and y3 is the mass fraction. In the similar way as discussed above, we may obtain the molar specific heat at constant pressure (cp m) as n (cpm)mix dT = n1 cpm1 dT + n2 cpm2 dT + n3 cpm3 dT \

n n1 n c pm1 + 2 c pm 2 + 3 c pm3 n n n = x1 cpm1 + x2 cpm2 + x3 cpm3

\

(cvm)mix =

(cpm)mix =

Mixture of Gases 295 Note: We have already discussed in chapter 3 (Art. 3.14) that molar specific heat where

cm = c . M

M = Molecular mass.

\ Molar specific heat at constant volume,

cvm = cv . M

and molar specific heat at constant pressure,

cpm = cp . M

The difference between the two molar specific heats (cpm – cvm) is equal to universal gas constant (Ru), i.e.

cpm – cvm = Ru = R . M

6. Entropy. According to Gibb’s theorem, the total entropy of a mixture of ideal gases is the sum of the entropies of the constituent gases when each gas is at the same temperature and volume as that of the gas mixture. The total entropy of each component on the mass basis is given by m s = m1 s1 + m2 s2 + m3 s3 and entropy per unit mass is m m m s = 1 s1 + 2 s2 + 3 s3 = y1 s1 + y2 s2 + y3 s3 m m m Similarly total entropy of each component on the mole basis is given by n s = n1 s1 + n2 s2 + n3 s3 and molar specific entropy, n n n s = 1 s1 + 2 s2 + 3 s3 = x1 s1 + x2 s2 + x3 s3 n n n

11.8 CHANGE OF ENTROPY DURING MIXING PROCESS The mixing of gases occurs in many engineering processes and systems. Let us consider two gases 1 and 2 separated from one another by a suitable partition in an insulated vessel. After the partition is removed, the gases diffuse into one another at the same temperature and pressure. It may be noted that whenever there is a mixing of ideal gases, there is a certain increase in entropy. In the mixing process, each gas may be considered to undergo a free expansion process from its initial pressure, with no work done. Thus, the mixing process is an irreversible process. We know that during the irreversible process, there is an increase in entropy of the system according to the principle of increase of entropy. Let the gases 1 and 2 mix adiabatically and isothermally. We know that the change in entropy of gas 1, on mole basis, is given by

p p ds 1 = – n1 Ru loge 1 = – 2.3 n1 Ru log 1 p p

Similarly, change in entropy of gas 2,

p p ds 2 = – n2 Ru loge 2 = – 2.3 n2 Ru log 2 p p

\ Total change of entropy of the mixture, ds = ds1 + ds2

p p = – 2.3 n1 Ru log 1 – 2.3 n2 Ru log 2 p p

296 Engineering Thermodynamics

p p = – 2.3 Ru n1 log 1 + n2 log 2 p p

In the similar way, the total change of entropy, on mass basis, is given by

p p ds = – 2.3 m1 R1 log 1 + m2 R2 log 2 p p m ... Ru = R × M , and n = . M

Example 11.1. A rigid vessel contains 0.5 kg of carbon monoxide and 2 kg of nitrogen at a temperature of 300 K and a pressure of 8 MPa. Calculate the volume of the vessel using ideal gas equation of state. Solution. Given: Mass of carbon monoxide, mCO = 0.5 kg Mass of nitrogen, mN = 2 kg 2 Temperature, T = 300 K Pressure, p = 8 MPa = 8 × 106 N / m2 = 8 × 103 kN / m2 Volume of the vessel Let v = Volume of the vessel in m3. First of all, let us find the number of moles in CO and N2. We know that number of moles of carbon monoxide, m Mass of CO = CO nCO = Molecular mass of CO M CO 0.5 = = 0.017 86 28 and number of moles of nitrogen, mN 2 2 = nN = = 0.071 43 2 M N 2 28

...(Q MCO = 28)

...(Q MN = 28) 2

\ Total number of moles in the mixture, n = nCO + nN = 0.017 86 + 0.071 43 = 0.089 29 2 Now using the ideal gas equation, pv = n Ru T n R T 0.089 29 × 8.314 × 300 \ v = u = = 0.0278 m3 Ans. p 8 × 103 ...(Q Ru = 8.314 kJ / kg mole K) Example 11.2. A gas mixture consists of 0.5 kg of carbon monoxide and 1 kg of carbon dioxide. Determine: 1. Mass fraction of each component ; 2. Mole fraction of each component; 3. Average molar mass ; and gas constant of the mixture. Solution. Given: Mass of carbon monoxide, mCO = 0.5 kg Mass of carbon dioxide, mCO = 1 kg 2 \ Total mass of the mixture, m = mCO + mCO = 0.5 + 1 = 1.5 kg 2

Mixture of Gases 297 1. Mass fraction of each component We know that mass fraction of carbon monoxide, m 0.5 1 = Ans. yCO = CO = m 1.5 3 and mass fraction of carbon dioxide, mCO2 1 2 = = Ans. yCO = 2 m 1.5 3 2. Mole fraction of each component First of all, let us find the number of moles of carbon monoxide (CO) and carbon dioxide (CO2). We know that number of moles of carbon monoxide, m 0.5 nCO = CO = = 0.017 86 M CO 28 ...[Q Molecular mass of CO (MCO ) = 28] and number of moles of carbon dioxide, mCO2 1 = nCO = = 0.022 73 ... (Q MCO2 = 44) 2 M CO2 44 \ Total number of moles of the mixture, n = nCO + nCO = 0.017 86 + 0.022 73 = 0.040 59 2 We know that mole fraction of carbon monoxide, n 0.01786 xCO = CO = = 0.44 Ans. n 0.04059 and mole fraction of carbon dioxide, nCO2 0.022 73 = xCO = = 0.56 Ans. 2 n 0.04059 3. Average molar mass and gas constant of the mixture We know that average molar mass of the mixture Total mass of the mixture (m) *Mm = Total number of moles of the mixture (n) = and gas constant of the mixture,

R =

1.5 = 36.955 kg / mole Ans. 0.04059 Ru 8.314 = = 0.225 kJ / kg K Ans. M m 36.955

...(Q Universal gas constant (Ru) is same for all gases and its value is 8.314 kJ / kg mole K)

* The molar mass of the mixture (Mm) may also be obtained as follows: Mm = xCO × Molar mass of CO + xCO × Molar mass of CO2

mCO2 m = xCO × CO + xCO × 2 nCO nCO2

= 0.44 ×

2

0.5 1 + 0.56 × = 12.318 + 24.637 = 36.955 kg/mole 0.01786 0.02273

298 Engineering Thermodynamics Note: The gas constant may also be obtained as follows: We know that

R =

=

mCO2 m1 Ru m2 Ru m R Ru × + × × = CO × u + m M1 m M 2 m M CO m M CO2 0.5 8.314 1 8.314 × + × = 0.1 + 0.126 = 0.226 kJ / kg K 1.5 28 1.5 44

Example 11.3. A mixture of 1 kg of oxygen and 2 kg of nitrogen occupies 1.2 m3 volume at temperature 300 K. Assuming perfect gas behaviour, determine the following parameters for the gas mixture. 1. Specific volume ; 2. Gas constant ; 3. Pressure ; and 4. Molecular mass. Take Ru = 8.314 kJ/kg mole K. Solution. Given: Mass of oxygen, mO = 1 kg 2 Mass of nitrogen, mN = 2 kg 2 Volume of mixture, v = 1.2 m3 Temperature of mixture, T = 300 K 1. Specific volume We know that total mass of the mixture, m = mO + mN = 1 + 2 = 3 kg 2 2 v 1.2 = \ Specific volume of the mixture = = 0.4 m3/ kg Ans. m 3 2. Gas constant We know that gas constant of the mixture, mO mN 2 R R 1 8.314 2 8.314 × u = × + × Rmix = 2 × u + 3 32 3 28 m M O2 m M N2 = 0.0866 + 0.198 = 0.2846 kJ / kg K Ans. 3. Pressure Let p = Pressure of the mixture. We know that pv = mRT mRT 3 × 0.2846 × 300 = \ p = = 213.45 kN /m2 Ans. v 1.2 ...(Q R = Rmix) 4. Molecular mass We know that mass fraction of oxygen, mO 1 yO = 2 = = 0.333 2 m 3 and mass fraction of nitrogen, mN 2 2 = = 0.67 yN = 2 m 3 We know that molecular mass of the mixture, 1 1 Mmix = = yO2 / M O2 + yN 2 / M N 2 0.333/ 32 + 0.67 / 28 =

1 1 = = 29.07 Ans. 0.0104 + 0.024 0.0344

Mixture of Gases 299 Note: The molecular mass of the mixture may also be obtained as discussed below: We know that number of moles of oxygen,

nO = 2

and number of moles of nitrogen,

nN = 2

mO2

=

1 = 0.03125 32

mN 2

=

2 = 0.071 43 28

M O2

M N2

\ Total number of moles in the mixture, n = nO + nN = 0.031 25 + 0.071 43 = 0.102 68 2 2 We know that molecular mass of the mixture, Total mass of the mixture (m) Mmix = Total number of moles in the mixture (n) =

3 = 29.2 Ans. 0.102 68

Example 11.4. A vessel having 0.115 m3 volume is filled with a mixture of 1.5 kg of carbon dioxide and 1 kg of nitrogen at 25°C. Determine: (a) pressure of mixture; (b) gas constant of the mixture; (c) mole fraction of each constituent; and (d) molecular mass of the mixture. Solution. v = 0.115 m3

Given: Volume of vessel,

Mass of carbon dioxide, mCO = 1.5 kg Mass of nitrogen,

2

mN = 1 kg 2

T = 25°C = 25 + 273 = 298 K

Temperature,

(a) Pressure of the mixture Let p = Pressure of the mixture. First of all, let us find the number of moles in CO2 and N2. We know that number of moles of carbon dioxide,

nCO = 2

mCO2 Mass of CO 2 = Molecular mass of CO 2 M CO2

1.5 = = 0.034 44 and number of moles of nitrogen,

nN = 2

mN 2 M N2

=

1 = 0.0357 28

...(Q MCO = 44) 2

...(Q MN = 28)

\ Total number of moles in the mixture, n = nCO + nN = 0.034 + 0.0357 = 0.0697 Now using the ideal gas equation, \

2

2

pv = n Ru T p =

n Ru T 0.0697 × 8.314 × 298 = = 1501.6 N/m2 Ans. v 0.115

2

300 Engineering Thermodynamics (b) Gas constant of the mixture We know that gas constant of the mixture, mCO2 mN 2 Ru R × + × u Rmix = m M CO2 m M N2 1.5 8.314 1 8.314 × + × = 0.1134 + 0.1188 2.5 44 2.5 28 = 0.2322 kJ/ kg K Ans. ...(Q m = mCO2 + mN 2 = 1.5 + 1 = 2.5 kg) (c) Mole fraction of each constituent We know that mole fraction of carbon dioxide, nCO2 0.034 = xCO = = 0.4878 Ans. 2 n 0.0697 and mole fraction of nitrogen, nN 0.0357 xN = 2 = = 0.5122 Ans. 2 n 0.0697

=

(d) Molecular mass of the mixture We know that mass fraction of carbon dioxide, mCO2 1.5 = yCO = = 0.6 2 m 2.5 and mass fraction of nitrogen,

yN = 2

mN 2 m

=

1 = 0.4 2.5

We know that molecular mass of the mixture,

Mmix =

=

1 1 = yCO2 / M CO2 + yN 2 / M N 2 0.6 / 44 + 0.4 / 28 1 1 = = 35.8 Ans. 0.0279 0.0136 + 0.0143

Note: The molecular mass of the mixture may also be obtained as follows: Total mass of the mixture m 0.25 = = Mmix = = 35.8 Ans. Total number of moles n 0.0697

Example 11.5. A mixture consisting of 1 kg of helium (He) and 2.5 kg of nitrogen (N2 ) at 25°C and 1 bar, is compressed in a reversible adiabatic process to 7 bar. Calculate: 1. The final partial pressure of the constituents; 2. The final temperature and change in internal energy of the mixture during the process. Take molecular mass of He = 4 kg ; cv for N2 = 0.743 kJ/ kg K ; cv for He = 3.14 kJ / K ; cp for N2 = 1.04 kJ / kg K ; cp for He = 5.23 kJ / kg K. Solution. Given: Mass of helium, mHe = 1 kg Mass of nitrogen, mN = 2.5 kg 2 Initial temperature of mixture, T1 = 25°C = 25 + 273 = 298 K

Mixture of Gases 301 Initial pressure of mixture, p1 = 1 bar Pressure of mixture after compression, p2 = 7 bar Molecular mass of helium, MHe = 4 kg Specific heat at constant volume for nitrogen, (cv )N = 0.743 kJ / kg K 2 Specific heat at constant volume for helium, (cv )He = 3.14 kJ / kg K Specific heat at constant pressure for nitrogen, (cp )N = 1.04 kJ / kg K 2 Specific heat at constant pressure for helium, (cp )He = 5.23 kJ / kg K 1. Final partial pressure of the constituents First of all, let us find the mole fraction (x) for He and N2. We know that number of moles of helium, m Mass of helium 1 = He = = 0.25 nHe = Molecular mass of helium M He 4 Similarly, number of moles of nitrogen, mN 2 2.5 = nN = = 0.09 2 M N2 28

...(Q MN = 28) 2

\ Total number of moles in the mixture, n = nHe + nN = 0.25 + 0.09 = 0.34 2 We know that mole fraction of helium, n 0.25 xHe = He = = 0.7353 n 0.34 nN 0.09 and mole fraction of nitrogen, xN = 2 = = 0.2647 2 n 0.34 Since the partial pressure ratio is equal to their mole fraction, therefore partial pressure of helium, pHe = p2 × xHe = 7 × 0.7353 = 5.1471 bar Ans. and partial pressure of nitrogen, pN = p2 × xN = 7 × 0.2647 = 1.8529 bar Ans. 2 2 2. Final temperature and change in internal energy of the mixture Let T2 = Final temperature of the mixture. We know that mass fraction of helium, m 1 yHe = He = = 0.2857 ...(Q m = mHe + mN2 = 1 + 2.5 = 3.5 kg) m 3.5 mN 2 2.5 = and mass fraction of nitrogen, yN = = 0.7143 2 m 3.5

302 Engineering Thermodynamics Specific heat at constant pressure for the mixture, cp = yHe (cp)He + yN 2 (cp)N 2 = 0.2857 × 5.23 + 0.7143 × 1.04 = 1.4942 + 0.7429 = 2.2371 kJ / kg K Similarly, specific heat at constant volume for the mixture, cv = yHe (cv )He + yN (cv )N 2 2 = 0.2857 × 3.14 + 0.7143 × 0.743 = 0.8971 + 0.5307 = 1.4278 kJ/ kg K \ Ratio of specific heats for the mixture, c p 2.2371 = g = = 1.5668 cv 1.4278 We know that for reversible adiabatic process,

T2 p2 = T1 p1

γ −1 γ

7 = 1

1.5668 − 1 1.5668

= (7)0.3617 = 2.02

\ T2 = T1 × 2.02 = 298 × 2.02 = 602 K Ans. and change in internal energy, dU = m cv (T2 – T1) = 3.5 × 1.4278 (602 – 298) = 1519.18 kJ Ans. Example 11.6. 0.3 m3 of helium at 20 bar and 30°C is mixed with 0.7 m3 of oxygen at 5 bar and 5°C by opening the valve between two tanks. Calculate the heat transfer, if the final temperature of the mixture is 25°C. Solution. Given: Volume of helium, vHe = 0.3 m3 Pressure of helium, pHe = 20 bar = 20 × 102 kN / m2 Temperature of helium, THe = 30°C = 30 + 273 = 303 K Volume of oxygen, vO = 0.7 m3 2 Pressure of oxygen, pO = 5 bar = 5 × 102 kN / m2 2 Temperature of oxygen, TO = 5°C = 5 + 273 = 278 K 2 Temperature of the mixture, Tmix = 25°C = 25 + 273 = 298 K First of all, let us find the mass and specific heat at constant volume for helium (He) and oxygen (O2). We know that gas constant for helium, Ru 8.314 RHe = = Molecular mass of helium ( M He ) 4 = 2.0785 kJ / kg K R 8.314 and gas constant for oxygen, RO = u = = 0.26 kJ / kg K 2 M O2 32 From the characteristic equation of gas, pv = mRT, we have Mass of helium,

mHe =

pHe . vHe 20 × 102 × 0.3 = = 0.9527 kg RHe THe 2.0785 × 303

Mixture of Gases 303 and mass of oxygen,

mO = 2

pO2 vO2 RO2 TO2

=

5 × 102 × 0.7 = 4.842 kg 0.26 × 278

\ Total mass of the mixture, m = mHe + mO = 0.9527 + 4.842 = 5.7947 kg 2 Specific heat at constant volume for helium, RHe 2.0785 = (cv) He = = 3.15 kJ / kg K γ He − 1 1.66 − 1 Specific heat at constant volume for oxygen, RO2 0.26 = (cv)O = = 0.65 kJ / kg K 2 γ O2 − 1 1.4 − 1

...(Q gHe = 1.66)

...(Q gO = 1.4) 2

and specific heat at constant volume for the gas mixture, mHe (cv ) He + mO2 (cv )O2 0.9527 × 3.15 + 4.842 × 0.65 (cv)mix = = 5.7947 m 3 + 3.1473 = = 1.061 kJ / kg K 5.7947 We know that initial internal energy before mixing, U1 = mHe (cv)He THe + mO (cv)O TO 2 2 2 = 0.9527 × 3.15 × 303 + 4.842 × 0.65 × 278 = 909.3 + 874.95 = 1784.25 kJ and final internal energy after mixing, U2 = m (cv )mix . Tmix = 5.7947 × 1.061 × 298 = 1832.15 kJ \ Heat transfer, Q = U2 – U1 = 1832.15 – 1784.25 = 47.9 kJ Ans. Example 11.7. Find the increase in entropy when 2 kg of oxygen at 60°C are mixed with 6 kg of nitrogen at the same temperature. The initial pressure of each constituent is 103 kPa and is same as that of the mixture. Solution. Given: Mass of oxygen, mO = 2 kg 2 Temperature of oxygen, TO = 60°C = 60 + 273 = 333 K 2 Mass of nitrogen, mN = 6 kg 2 Initial pressure of each constituent, pO = pN = 103 kPa 2 2 We know that number of moles of oxygen, Mass of oxygen (mO2 ) 2 nO = = = 0.0625 2 32 Molecular mass of oxygen ( M O2 ) and number of moles of nitrogen, mN 2 6 = nN = = 0.2143 2 M N 2 28 Total number of moles in the mixture, n = nO + nN = 0.0625 + 0.2143 = 0.2768 2 2 We know that mole fraction of oxygen, nO pO 0.0625 xO = 2 = 2 = = 0.2258 2 n p 0.2768

304 Engineering Thermodynamics and mole fraction of nitrogen,

xN = 2

nN 2 n

=

pN 2 p

=

0.2143 = 0.7742 0.2768

We know that increase in entropy,

pO pN ds = – 2.3 nO Ru log 2 – 2.3 nN Ru log 2 2 2 p p

= – 2.3 × 0.0625 × 8.314 log (0.2258) – 2.3 × 0.2143 × 8.314 log (0.7742) = 0.7724 + 0.4555 = 1.2279 kJ / kg Ans. Example 11.8. Nitrogen and hydrogen are mixed in a steady flow adiabatic process in the ratio of 3 kg of hydrogen and one kg of nitrogen. The hydrogen enters at 1.5 bar and 35°C and the nitrogen at 1.5 bar and 260°C. The pressure after mixing is 1.45 bar. Determine: (a) the final temperature of the mixture; and (b) the net change in entropy of mixture. The specific heat at constant pressure for nitrogen and hydrogen may be taken as 1.04 kJ / kg K and 14.4 kJ / kg K respectively. The gas constant for nitrogen is 0.297 kJ / kg K and for hydrogen it is 4.15 kJ / kg K. 1 H2 Solution. Given: Mass of hydrogen mH = 3 kg 2 Mass of nitrogen, mN = 1 kg 2 2 Pressure of hydrogen, (p1 )H = 1.5 bar 2 Mixture Temperature of hydrogen, TH = 35°C 2 2 = 35 + 273 = 308 K Pressure of nitrogen, ( p1 )N = 1.5 bar 2 Temperature of nitrogen, TN = 260°C 2 N2 = 260 + 273 = 533 K 1 Fig. 11.4 Pressure after mixing p = 1.45 bar Specific heat at constant pressure for nitrogen, (cp)N = 1.04 kJ / kg K 2 Specific heat at constant pressure for hydrogen, (cp)H = 14.4 kJ / kg K 2 Gas constant for nitrogen, RN = 0.297 kJ / kg K 2 Gas constant for hydrogen, RH = 4.15 kJ / kg K 2 The process is shown in Fig. 11.4. (a) Final temperature of the mixture Let T = Final temperature of the mixture. For a steady flow adiabatic process, Q = 0 ; and W = 0 Thus the initial enthalpy of the individual gases is equal to the final enthalpy of the mixture. We know that initial enthalpy of hydrogen and nitrogen = mH (cp)H TH + mN2 (cp)N TN 2

2

2

2

2

= 3 × 14.4 × 308 + 1 × 1.04 × 533 = 13 305.6 + 554.4 = 13 860 kJ and final enthalpy of the mixture = mH (cp)H T + mN (cp)N T 2

2

2

2

...(i)

= 3 × 14.4 T + 1 × 1.04 T = 44.24 T ...(ii)

Mixture of Gases 305 Equating equations (i) and (ii), 44.24 T = 13 860 13860 \ T = = 313.3 K Ans. 44.24 (b) Net change in entropy of the mixture First of all, let us find the number of moles of each constituent (i.e. hydrogen and nitrogen). We know that number of moles of hydrogen, mH 2 3 = = 1.5 ...[Q Molecular mass of hydrogen, MH = 2] nH = 2 2 M H2 2 and number of moles of nitrogen, mN 2 1 = nN2 = = 0.0357 M N 2 28

...(Q MN = 28) 2

\ Total number of moles in the mixture, n = nH + nN = 1.5 + 0.0357 = 1.5357 2 2 Now mole fraction of hydrogen, nH 1.5 xH = 2 = = 0.977 2 n 1.5357 nN 2

0.0357 = = 0.023 n 1.5357 \ Partial pressure of hydrogen after mixing, ( p2)H = p × xH = 1.45 × 0.977 = 1.417 bar 2 2 and partial pressure of nitrogen, ( p2)N = p × xN = 1.45 × 0.023 = 0.0334 bar 2 2 We know that change of entropy for hydrogen, and mole fraction of nitrogen,

xN = 2

T *(ds)H = 2.3 mH (c p ) H 2 log TH 2 2 2

( p1 ) H 2 + RH 2 log ( p2 ) H 2

313.3 1.5 = 2.3 × 3 14.4 log + 4.15 log 308 1.417 = 6.9 [0.1067 + 0.103] = 1.447 kJ / K and change of entropy for nitrogen,

T (ds)N = 2.3 mN (c p ) N 2 log TN 2 2 2

( p1 ) N 2 + RN 2 log ( p2 ) N 2

313.3 1.5 = 2.3 × 1 1.04 log + 0.297 log 533 0.0334 = 2.3 (– 0.323 + 0.491) = 0.386 kJ / kg \ Change of entropy of the mixture (ds)mix = (ds)H + (ds)N = 1.447 + 0.3086 = 1.833 kJ /K Ans. 2

* Refer Chapter 7

2

306 Engineering Thermodynamics

11.9 VOLUMETRIC ANALYSIS AND GRAVIMETRIC ANALYSIS The mixture of gases may be analysed either on the volumetric basis or on mass basis (i.e. gravimetric basis). Such analysis are called volumetric analysis and gravimetric analysis respectively. In order to convert the volumetric analysis of a gas mixture into mass or gravimetric analysis, first of all multiply the volume of each constituent to its own molecular mass to obtain the mass of that constituent. Add up these masses to obtain the total mass of the mixture. Now divide the mass of each constituent to the total mass of the mixture which gives the mass fraction of each constituent. The procedure is best explained in the following example. Example 11.9. A mixture of gases has the following volumetric composition: CO2 = 12%; O2 = 4% ; N2 = 82%; CO = 2% Calculate: 1. Gravimetric composition; 2. Molecular mass of the mixture; and 3. Gas constant for the mixture. Solution. Given: Volume of carbon dioxide (CO2) in 1 m3 of the mixture = 12% = 0.12 Volume of oxygen (O2) = 4% = 0.04 Volume of nitrogen (N2) = 82% = 0.82 Volume of carbon monoxide (CO) = 2% = 0.02 1. Gravimetric composition We have already discussed in Art. 11.4 that the volume fraction (z) is equal to the mole fraction (x). We also know that *mass of the constituent per mole of the mixture = x × M, where M is the molecular mass of the constituent. Using these relations, the results may be tabulated as follows: Constituent % by volume **kg mole per Molecular Mass of the or mole of mixture mass (M) constituent % by mole or mole fraction per mole of or volume the mixture fraction (x) m=x×M (a) (b) (c) (d)

% Mass (d ) = × 100 Σ (d )

CO2

12

0.12

44

5.28

5.28 × 100 = 17.55 30.08

O2

4

0.04

32

1.28

1.28 × 100 = 4.26 30.08

N2

82

0.82

28

22.96

22.96 × 100 = 76.33 30.08

CO

2

0.02

28

0.56

0.56 × 100 = 1.86 30.08

Total

100

1.00

S(d) = 30.08

* We know that mole fraction of the constituent (say xa) Number of moles of the constituent (say na ) = Total number of moles in the mixture (n)

100

...[For constituent a]

a nd mass of the constituent = Number of moles of the constituent × Molecular mass of the constituent \ Mass of the constituent per mole of the mixture (i.e. when n = 1) = Mole fraction of the constituent (x) × Molecular mass of the constituent (M) ** We may also say that number of moles of the constituent per mole of the mixture i.e. mole fraction (x).

Mixture of Gases 307 2. Molecular mass of the mixture We know that molecular mass of the mixture = Sm = S (d) = 30.08 Ans. 3. Gas constant for the mixture R 8.314 R = u = = 0.2764 kJ / kg K Ans. Σ m 30.08 Example 11.10. A mixture of gases at a pressure of 4 bar has a temperature of 150°C. A sample is analysed and volumetric analysis is found to be CO2 = 14%; O2 = 5%; and N2 = 81% Determine the gravimetric analysis and partial pressure of the gases in the mixture. If 2.3 kg of mixture is cooled at constant pressure to 15°C; find the final volume. Solution. Given: Pressure of mixture of gases, p = 4 bar = 400 kN/m2 Temperature of mixture, T = 150°C Volume of carbon dioxide (CO2) = 14% = 0.14 Volume of oxygen (O2) = 5% = 0.05 Volume of nitrogen (N2) = 81% = 0.81 Gravimetric analysis As discussed in the previous example, the results of the gravimetric analysis are given in following table. Constituent % by volume or % by mole

(a)

Volume fraction Molecular Mass of the or mole fraction mass (M) constituent (x) per mole of the mixture m=x×M (b) (c) (d)

% Mass (d ) = × 100 Σ (d )

CO2

14

0.14

44

6.16

6.16 × 100 = 20.2 30.44

O2

5

0.05

32

1.6

1.6 × 100 = 5.3 30.44

N2

81

0.81

28

22.68

22.68 × 100 = 74.5 30.44

Total

100

1.00

S(d) = 30.44

Partial pressure of the gases in the mixture We know that volume fraction is equal to the mole fraction. \ Partial pressure of carbon dioxide, pCO = xCO × p = 0.14 × 4 = 0.56 bar Ans. 2 2 Partial pressure of oxygen, pO = xO × p = 0.05 × 4 = 0.20 bar Ans. 2

2

100

308 Engineering Thermodynamics and partial pressure of nitrogen, pN = xN × p = 0.81 × 4 = 3.24 bar Ans. 2 2 Final volume Given: Mass of the mixture, mmix = 2.3 kg Temperature, T = 15°C = 15 + 273 = 288 K We know that number of moles in the mixture, Mass of the mixture (mmix ) n = Total mass of the constituents per mole of the mixture (Σ d ) 2.3 = = 0.0756 30.44 Using the gas equation, we have pv = n Ru T \

v =

n Ru T 0.0756 × 8.314 × 288 = = 0.453 m3 Ans. p 400

HIGHLIGHTS 1. When the mixture of gases remains homogeneous in composition throughout a process and do not react chemically with one another, then it may be considered as a single pure substance. The dry air, ordinary atmospheric air and flue gases are few examples of various mixtures. 2. The mass fraction of the constituent may be defined as the ratio of the mass of the constituent to the total mass of the mixture. It is usually denoted by y. 3. The mole fraction of the constituent may be defined as the ratio of the number of moles of the constituent to the total number of moles in a mixture. It is usually denoted by x. 4. The mass of any constituent of the molecular mass (M) is related to the mole content as follows: m = n × M where m = Mass of the constituent, n = Number of moles of the constituent, and M = Molecular mass of the constituent. 5. The volume fraction of the constituent may be defined as the ratio of the volume occupied by a constituent theoretically assumed to be separated at the pressure and temperature of the mixture to the total volume of the mixture. It is usually denoted by z. 6. For an ideal gas mixture, the volume fraction is equal to the mole fraction. 7. According to Dalton’s law of partial pressures, the total pressure exerted by the mixture is equal to the sum of the partial pressures exerted by its individual constituent of the gases measured alone at the temperature and volume of the mixture. Mathematically, p = p1 + p2 + p3 + ... 8. The ratio of the partial pressure of any constituent to the total pressure of the mixture is called partial pressure ratio of that constituent. 9. According to Amagat’s Leduc law, the volume of a mixture of ideal gases is equal to the sum of partial volumes or separated volumes of the constituent gases, when each exists at the temperature and pressure of the mixture. Mathematically, v = v1 + v2 + v3 + ...

Mixture of Gases 309 10. The ratio of the partial volume of any constituent to the total volume of the mixture (i.e. volume fraction) is equal to the mole fraction of that constituent. 11. The gas constant for the mixture (R) is given by R = y1 R1 + y2 R2 + y3 R3 + ... where R1, R2, R3 = Gas constants for the individual constituents; and y1, y2, y3 = Mass fraction for the individual constituents. 12. The molecular mass of the mixture (Mmix) is given by 1 Mmix = y1 / M1 + y2 / M 2 + y3 / M 3 + ... where M1, M2 and M3 are the molecular masses for the individual constituent. The molecular mass of the mixture may also be obtained by the following relation, i.e. Total mass of the mixture (m) Mmix = Total number of moles in the mixture (n) 13. The specific heat at constant volume for the mixture of gases (cv)mix is given by m m m (cv)mix = 1 cv1 + 2 cv 2 + 3 cv3 + ... m m m = y1 cv1 + y2 cv2 + y3 cv3 + ... where y1, y2 and y3 are the mass fractions of individual constituents. In the similar way, molar specific heat at constant volume for the mixture of gases is given by n n n (cvm)mix = 1 cv m1 + 2 cv m 2 + 3 cv m3 n n n = x1 cvm1 + x2 cvm2 + x3 cvm3 + ... ...[Q cvm = cv × M] 14. The specific heat at constant pressure for the mixture of gases (cp)mix is given by (cp)mix = y1 cp1 + y2 cp2 + y3 cp3 + ... and (cpm)mix = x1 cpm1 + x2 cpm2 + x3 cpm3 + ... ...[Q cpm = cp × M] 15. Total change of entropy of the mixture of two gases is given by

p p ds = – 2.3 Ru n1 log 1 + n2 log 2 p p

...(On mole basis)

p p = – 2.3 m1 R1 log 1 + m2 R2 log 2 p p

...(On mass basis)

EXERCISES 1. Air is treated as a mixture of ideal gases which has 0.23 kg of oxygen and 0.77 kg of nitrogen. Find the gas constant and molecular mass of air. [Ans. 0.288 kJ / kg K; 28.82] 2. A rigid tank contains 2 kg of nitrogen at 2 MPa. A sufficient quantity of oxygen is added to increase the pressure to 2.7 MPa while the temperature remains constant at 38°C. Determine the mass of oxygen added. [Ans. 0.8 kg] 3. A closed vessel has a capacity of 0.5 m3. It contains 0.23 kg of oxygen and 0.77 kg of nitrogen. Some quantity of carbon dioxide is forced into the vessel so that the temperature remains at 20°C and the pressure rises to 1.5 MPa. Find the total masses of oxygen, nitrogen and carbon dioxide in the cylinder. [Ans. mO = 1.36 kg; mN = 4.55 kg ; mCO = 4.515 kg] 2

2

2

310 Engineering Thermodynamics 4. A mixture of two gases in the proportion of 40% and 60% at 20°C occupies a vessel of 5 m3 capacity. If the value of gas constant for the gases is 287 J / kg K and 294 J/ kg K respectively, find the partial pressure, total pressure and the mean value of gas constant for the mixture. The total mass of mixture is 1.5 kg. [Ans. 10.1 kPa, 15.5 kPa ; 25.6 kPa ; 291.2 J / kg K] 3 5. A vessel having 1.8 m of volume is filled with a mixture of 3 kg of nitrogen and 5 kg of carbon dioxide at 20°C. Determine: (a) pressure of the mixture; (b) gas constant of the mixture ; (c) mole fraction of each constituent; and (d) molecular mass of the mixture. [Ans. 298.8 kPa; 0.229 kJ / kg K; 0.107, 0.114; 36.2] 6. A mixture of 6 kg of oxygen and 9 kg of nitrogen has a pressure of 3 bar and a temperature of 20°C. Calculate the following for the mixture: (a) Mole fraction of each component; (b) Average molecular mass; (c) Specific gas constant; (d) Volume and density; (e) Partial pressure and partial volumes. [Ans. 0.3684, 0.6315; 29.5; 0.282 kJ / kg K; 4.13 m3, 3.632 kg / m3; 110.6 kPa, 189.4 kPa, 1.52 m3, 2.61 m3] 7. A cylinder containing 1 kg of carbon monoxide at 1.7 bar and 38°C is connected through a valve to a second cylinder containing 1 kg of nitrogen at 1.03 bar and 15°C. The valve is opened and adiabatic mixing takes place. Find the change of entropy. Take specific heat at constant pressure for both carbon monoxide and nitrogen as 1.045 kJ / kg K. The gas constant for both carbon monoxide and nitrogen may be taken as 0.3 kJ / kg K. [Ans. 0.4251 kJ / K] 8. Two kg of carbon dioxide at 38°C and 1.4 bar are mixed with 5 kg of nitrogen at 150°C and 1.03 bar to form a mixture at a final pressure of 70 kPa. The process occurs adiabatically in a steady flow apparatus. Calculate the final temperature of the mixture and the change in entropy. Take specific heat at constant pressure for carbon dioxide and nitrogen as 0.85 kJ / kg K and 1.04 kJ / kg K respectively. [Ans. 395.4 K; 1.829 kJ / K] 9. A mixture of gases has the following volumetric composition: CO2 = 14% ; CO = 1% ; O2 = 5%; and N2 = 80% Find: (a) Gravimetric composition ; (b) Molecular mass of the mixture ; and (c) Gas constant of the mixture [Ans. CO2 = 21.24%, CO = 0.96%, O2 = 0.55%, N2 = 77.25%; M = 29; R = 0.2867 kJ / kg K] 10. The volumetric composition of a sample of gaseous fuel is H2 = 48% ; CH4 = 22% ; C2H4 = 3% ; CO2 = 4% ; CO = 16% ; and N2 = 7%. Calculate: (a) Molecular mass of the mixture, and (b) specific gas constant of the mixture.

[Ans. 13.58; 0.612 kJ/ kg K]

QUESTIONS 1. Define the following terms as applied to mixture of gases: (a) Mass fraction; (b) Mole fraction; and (c) Volume fraction. 2. Show that for an ideal gas mixture, the volume fraction is equal to mole fraction. 3. Write an expression between the mass, molecular mass and number of moles of the constituent. 4. Explain Dalton’s law of partial pressures. 5. What do you understand by partial pressure ratio? 6. What is Amagat-Leduc law? 7. Derive expression for specific volume and density of mixture of gases. 8. Write a short note on (a) internal energy of a gas mixture; and (b) enthalpy of a gas mixture. 9. Write the expression for entropy changes for mixture of gases undergoing a process.

Mixture of Gases 311

OBJECTIVE TYPE QUESTIONS 1. The mass of any constituent (say m) of the gas having molecular mass (M) is related to the mole content (n) as (a) m = n/M (b) m = n × M (c) m = n + M (d) m = n – M 2. The volume fraction for an ideal gas mixture is .................... mole fraction. (a) equal to (b) less than (c) greater than 3. The partial pressure of a constituent of an ideal gas is (a) equal to the mole fraction (b) inversely proportional to the mole fraction (c) directly proportional to the mole fraction (d) none of these 4. The volume of a mixture of ideal gases is equal to the sum of the partial volume of the constituent gases when each exists at the same temperature and pressure of the mixture. This law is known as (a) Joule’s law (b) Charles’ law (c) Dalton’s law (d) Amagats Leduc law 5. If cvm is the molar specific heat at constant volume and cpm is the molar specific heat at constant pressure, then universal gas constant (Ru) is given by (a) Ru = cvm + cpm (b) Ru = cpm – cvm (c) Ru = cpm / cvm (d) Ru = cpm × cvm

ANSWERS

1. (b)

2. (a)

3. (c)

4. (d)

5. (b)

12 VAPOUR POWER CYCLES 12.1 Introduction 12.2 Carnot Vapour Cycle 12.3 Limitations of Carnot Vapour Cycle 12.4 Rankine Cycle 12.5 Difference Between Rankine Cycle and Carnot Cycle for a Vapour 12.6 Rankine Cycle Using Superheated Steam 12.7 Modified Rankine Cycle 12.8 Methods of Improving the Efficiency of Rankine Cycle

12.9 Reheat Cycle 12.10 Regenerative Cycle 12.11 Regenerative Cycle with an Open Feed Water Heater 12.12 Regenerative Cycle with a Closed Feed Water Heater 12.13 Comparison Between Open and Closed Feed Water Heaters 12.14 Regenerative Cycle with Two Feed Water Heaters.

12.1 INTRODUCTION A heat engine is a device which while operating in a *cycle, converts heat energy supplied to it, into mechanical energy. The cycle which continuously converts heat into work is called a power cycle. In a power cycle, the working fluid repeatedly performs a succession of processes. If the working fluid is alternately vaporised and condensed, then the cycle is termed as vapour power cycle. 2

Steam engine or Turbine Work out Generator

Boiler 3 Heat supplied

1 Cooling water in Heat rejected

Work in

Feed Pump

4

Condenser

Fig. 12.1. Simple steam power plant.

* We have already discussed that if the processes in the cycle are performed in the reverse order, then the cycle becomes a reversed cycle and the device operating on the reversed cycle is called a refrigerator or heat pump which is also a heat engine operating in the reversed order.

Vapour Power Cycles 313 In vapour power cycles, the working fluid undergoes a change in phase as it passes through the equipment. For example, in a steam power plant, as shown in Fig. 12.1, the boiler, turbine, condenser and feed pump constitute the essential elements of a heat engine, in which the working fluid is water which is alternately vaporised and condensed.

12.2 CARNOT VAPOUR CYCLE We have already discussed Carnot cycle in Chapter 6 (Art. 6.6). The Carnot cycle is a reversible cycle and has the highest efficiency between the two given temperature limits and it forms the basis of comparison for the study of power cycles. Such a cycle consists of two isothermal processes and two reversible adiabatic processes as discussed below: Saturated liquid line 1

2

T1 = T2

Pressure

Temperature

p 1 = p2

p 3 = p4

3

4 v1

v4 v2 volume

v3

2

1

T3 = T4

Saturated dry line

4

3

4 s1 = s4

3 s2 = s3 Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.2. Carnot vapour cycle.

1. Isothermal expansion. This process is represented a horizontal line 1-2 on the p-v and T-s diagrams as shown in Fig. 12.2. During this process, water is heated in a boiler and converted into steam in the dry saturated form as represented by point 2. The temperature and pressure during the conversion process of water into steam remains constant, i.e. T1 = T2 and p1 = p2. The heat absorbed by the water during the isothermal process (i.e. during the conversion process) is given by q1–2 = T1 (s2 – s1) 2. Reversible adiabatic expansion. This process is represented by the line 2-3. During this process, the dry saturated steam at state 2 enters the steam engine or turbine where it is expanded in a reversible adiabatic manner to state 3. The temperature and pressure during this process falls to T3 and p3 respectively. We have already discussed that during the reversible adiabatic process, no heat is supplied or rejected. Therefore, there is no change of entropy (i.e. s2 = s3). It may be noted that the steam at state 3 is wet steam. 3. Isothermal compression. This process is represented by the line 3-4. During this process, the wet steam at state 3 enters the condenser where it is condensed isothermally. The heat is rejected to the cooling medium (i.e. water) at a constant temperature and pressure, i.e. T3 = T4 and p3 = p4. The heat rejected during the isothermal compression is given by q3–4 = T3 (s3 – s4) = T3 (s2 – s1) ...(Q s3 = s2 and s4 = s1)

Note: The steam at state 4 is also a wet steam.

4. Reversible adiabatic compression. This process is represented by the line 4-1. During this process, the wet steam at state 4 enters the feed pump where it is compressed in a reversible adiabatic manner to state 1. The temperature and pressure rises to T1 and p1 respectively. We know that during reversible adiabatic process, no heat is supplied or rejected. Therefore there is no change of entropy (i.e. s4 = s1). This completes the cycle.

314 Engineering Thermodynamics We know that efficiency of the Carnot cycle Workdone during the cycle hCarnot = Heat supplied =

Heat supplied − Heat rejected q1− 2 − q3 − 4 = Heat supplied q1− 2

=

T1 ( s2 − s1 ) − T3 ( s2 − s1 ) T1 − T3 T = =1− 3 T1 ( s2 − s1 ) T1 T1

We see that for the highest efficiency of the Carnot cycle, the heat supplied should be at the highest possible temperature and rejected as the lowest possible temperature.

12.3 LIMITATIONS OF CARNOT VAPOUR CYCLE Though the efficiency of the Carnot cycle is the maximum for the given values of T1 and T3, yet it is not considered as a theoretical cycle for the steam power plant due to the following practical difficulties: 1. The isothermal (constant temperature) processes 1-2 and 3-4 can only be achieved when the piston moves very slowly to allow the heat transfer to take place at constant temperature. Similarly, the reversible adiabatic processes 2-3 and 4-1 can only be achieved when the piston moves very fast so that the heat transfer is negligible due to the very short time available. From the p-v diagram of the Carnot cycle, we see that the isothermal expansion process 1-2 and reversible adiabatic process 2-3 takes place in the same stroke (i.e. outward stroke) which means that for the part of the stroke, the piston moves very slowly and for the remaining part of the stroke, the piston moves very fast. Similar is the case for isothermal compression process (3-4) and reversible adiabatic compression process (4-1), i.e. during the inward stroke of the piston. This variation of speed in the same stroke is not possible. 2. It is difficult to control the quality of condensate coming out of the condenser (state 4). In other words, the outlet condition of wet steam from the condenser can not be controlled. 3. The wet steam (state 3) from the outlet of the turbine contains water droplets at high velocity which can pit and corrode the turbine blades. 4. It is difficult to compress the wet vapour (state 4) in the reversible adiabatic manner to the saturated state 1. 5. The size of the feed pump or compression has to be very large to handle the mixture of water and vapour corresponding to state 4. 6. The work ratio is low, i.e. the ratio of the net output to the gross output is low. 7. The Carnot vapour cycle does not permit superheating of steam. Example 12.1. In a Carnot cycle, heat is supplied at 350°C and is rejected at 25°C. The working fluid is water which while receiving heat, evaporates from liquid at 350°C to steam at 350°C. From the steam tables, the entropy change for this process is 1.4378 kJ / kg K. If the cycle operates on a stationary mass of 1 kg of water, find the heat supplied, workdone and heat rejected per cycle. What is the pressure during heat reception? Show the cycle with temperature-entropy diagram. Solution. Given: Temperature at which heat is supplied, T1 = T2 = 350°C = 350 + 273 = 623 K

Vapour Power Cycles 315

Temperature

Temperature at which heat is rejected, T3 = T4 = 25°C = 25 + 273 = 298 K Entropy change for the process, s2 – s1 = s3 – s4 = 1.4378 kJ/ kg K Iso. exp. Mass of water, mw = 1 kg 2 T1 = T2 1 The temperature-entropy (T-s) diagram for the cycle is shown in Fig. 12.3. Isen. Isen. comp. exp. Heat supplied per cycle We know that heat supplied per cycle per kg, T3 = T4 4 Iso. comp. 3 q1–2 = (s2 – s1) T1 = 1.4378 × 623 s1 = s4 s2 = s3 = 895.75 kJ/ kg Ans. Entropy Workdone per cycle Fig. 12.3 We know that workdone per cycle per kg, w = (s2 – s1) (T1 – T3) = 1.4378 (623 – 298) = 467.28 kJ / kg Ans. Heat rejected per cycle We know that heat rejected q3–4 = (s3 – s4) T3 = 1.4378 × 298 = 428.47 kJ / kg Ans. Note: Heat rejected is also given by q3–4 = q1–2 – w = 895.75 – 467.28 = 428.47 kJ/ kg

Pressure during heat reception The pressure of water during heat reception corresponding to 300°C, from steam tables, is 165.35 bar. Ans.

12.4 RANKINE CYCLE Steam engine or Turbine Steam Work output 2 Throttle valve

3

Condenser

Boiler

Cooling water Heat rejected

1

Heat supplied

4 Extraction pump Feed pump

Saturated water

Hotwell

Fig. 12.4. Schematic diagram of a plant working on Rankine cycle.

The Rankine cycle is a modification of the Carnot cycle in which the condensation process (i.e. the process 3-4 as shown in Fig. 12.5) is extended into the liquid phase. The schematic arrangement of a steam engine or a turbine plant is shown in Fig. 12.4. The cycle is represented by 1-2-3-4 on the p-v and T-s diagrams as shown in Fig. 12.5 (a) and (b) respectively.

316 Engineering Thermodynamics Consider 1 kg of saturated water at pressure p1 and temperature T1, as represented by point 1 in Fig. 12.5. 1

2 2

Pressure

Temperature

p 1 = p2

p 3 = p4

4

3

3

1

T1 = T2

T3 = T4

2

4 3

2

3

volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.5. Rankine cycle.

The cycle consists of the following four processes: (a) Process 1-2 (Isothermal expansion process). The saturated water at point 1 is converted into *dry saturated steam (point 2) in a steam boiler at constant pressure p1 = p2 and constant temperature T1 = T2. The heat absorbed in converting water into dry saturated steam is latent heat of vaporization (hf g1 = hf g2) corresponding to pressure p1 = p2. (b) Process 2-3 (Reversible adiabatic or isentropic expansion process). The dry saturated steam at point 2 expands isentropically along the curve 2-3 in a steam engine or a steam turbine. The steam at point 3 is a wet steam having dryness fraction (say x3). Let the pressure and temperature of this wet steam is p3 and T3 respectively. Since no heat is supplied or rejected during isentropic process, therefore there is no change of entropy. (c) Process 3-4 (Isothermal compression process). The wet steam (at point 3) from the engine or turbine is now condensed isothermally in a condenser. The heat is rejected at constant pressure ( p3) and constant temperature (T3) until the whole steam is converted into saturated water at point 4. The pressure and temperature at point 4 is p4 = p3 and T4 = T3 respectively. The heat rejected by the steam during the process 3-4 is the latent heat equal to x3 hfg3. (d) Process 4-1 (Sensible heating). The extraction pump and the feed pump increases the pressure of water (at point 4) from p4 to p1. The feedwater at temperature T4 is heated in a boiler (sensible heating) at constant volume from temperature T4 to the saturation temperature (T1) corresponding to pressure p1, before being converted into steam along 1-2. The heat absorbed by the water during the sensible heating ( process 4-1) is given by (hf 1 – hf 4) or (hf 2 – hf 3), where hf 1 = hf 2 is the sensible heat of water corresponding to pressure p1 = p2 and hf 4 = hf 3 is the sensible heat of water corresponding to pressure p4 = p3. From the above, we see that the heat absorbed during the cycle = Heat absorbed during the process 1-2 in converting water into dry saturated steam + Heat absorbed during the sensible heating process 4-1 = hf g2 + (hf 2 – hf 3) = hf 2 + hf g2 – hf 3 = h2 – hf 3 ...(Q For dry steam, h2 = hf 2 + hfg2) We know that heat rejected during the cycle = h3 – hf 3 = hf 3 + x3 hf g3 – hf 3 = x3 hf g3

* The steam may be in the wet, dry saturated or superheated condition. In this case, steam is converted into dry saturated steam.

Vapour Power Cycles 317 \ Workdone during the cycle = Heat absorbed – Heat rejected = h2 – hf 3 – x3 hf g 3 = h2 – (hf 3 + x3 hf g3) = h2 – h3 and efficiency of the cycle or Rankine efficiency, h − h3 Workdone = 2 hR = Heat absorbed h2 − h f 3 Notes: 1. In case the saturated water at point 1 is not converted into dry saturated steam, but it is wet as shown by point 2, then the Rankine cycle becomes 1-2-3-4. In this case, the heat absorbed in converting the water into wet steam is x2 hf g2 , where x2 is the dryness fraction at point 2. Thus, the heat absorbed during the cycle will be = h2 – hf 3 = hf 2 + x2 hf g2 – hf 3 2. If m is the mass of steam (in kg / h) used during the cycle, then total workdone or power output m (h2 − h3 ) = in kJ / s or kW 3600

12.5 DIFFERENCE BETWEEN RANKINE CYCLE AND CARNOT CYCLE FOR A VAPOUR The following are the differences between the Rankine cycle and the Carnot cycle: 1. In Rankine cycle, a pump is used to pressurise the vapour, whereas in the Carnot cycle, the wet vapour is compressed, using a compressor. Thus, the work of compression in Carnot cycle is much larger than the Rankine cycle. This means that the work ratio of Carnot cycle is low. 2. In Rankine cycle, only a part of heat is supplied isothermally at constant higher temperature. In Carnot cycle, all the heat is supplied isothermally. Thus, the efficiency of Rankine cycle is lower than that of a Carnot cycle, working between the same maximum and minimum temperatures. 3. In Rankine cycle, steam is completely condensed during heat rejection, which is practicable. In Carnot cycle, the steam is still wet at the end of heat rejection. In other words, it is difficult to control the condensation process at precisely the correct dryness fraction (at state 4 in Fig. 12.5) so that the state 1 is exactly obtained after the isentropic compression (process 4-1). This makes Carnot cycle for vapour impracticable. 4. In Rankine cycle, superheating of steam is permissible. The Carnot cycle does not permit superheating. Example 12.2. A simple Rankine cycle steam power plant operates between the temperature limits of 260°C and 95°C. Steam is supplied to the turbine in a condition of complete saturation and the expansion in the turbine is isentropic. Draw the ideal Rankine cycle and Carnot cycle using steam and capable of working between the same temperature limits in a T-s or h-s diagram. Estimate and compare the efficiency of the two cycles while working between the given conditions. Solution. Given: Maximum temperature, T1 = T2 = 260°C = 260 + 273 = 533 K Minimum temperature, T3 = T4 = 95°C = 95 + 273 = 368 K Efficiency of the Rankine cycle The T-s and h-s diagrams for the Rankine cycle are shown in Fig. 12.6 (a) and (b) respectively.

318 Engineering Thermodynamics First of all, let us find the dryness fraction of steam at point 3 (x3) after the isentropic expansion process 2-3. From steam tables, corresponding to a maximum temperature of 260°C, we find that h2 = hg2 = 2796.4 kJ / kg ; s2 = sg2 = 6.001 kJ / kg K and corresponding to a minimum temperature of 95°C, we find that hf 3 = 398 kJ/ kg; hf g3 = 2270.2 kJ/ kg ; sf 3 = 1.25 kJ/ kg K; sf g3 = 6.167 kJ/ kgK

Fig. 12.6

We know that for isentropic expansion process 2-3, Entropy at point 2 = Entropy at point 3 or s2 = sf 3 + x3 × sf g3 6.001 = 1.25 + x3 × 6.167 6.001 − 1.25 \ x = = 0.77 6.167 We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hfg3 = 398 + 0.77 × 2270.2 = 2146 kJ/ kg \ Efficiency of the Rankine Cycle, h − h3 2796.4 − 2146 650.4 = = hR = 2 = 0.2712 or 27.12% Ans. h2 − h f 3 2796.4 − 398 2398.4 Efficiency of the Carnot cycle We know that efficiency of the Carnot cycle, T − T3 533 − 368 = h = 1 = 0.3095 or 30.95% Ans. T1 533 Example 12.3. In a Rankine cycle, the maximum pressure of steam supplied is 6 bar. The dryness fraction is 0.9. The exhaust pressure is 0.7 bar. Find the theoretical workdone and Rankine efficiency. Solution. Given: Maximum pressure of steam, p2 = 6 bar Dryness fraction of steam, x2 = 0.9 Exhaust pressure, p3 = 0.7 bar The Rankine cycle with wet steam having dry fraction x2 = 0.9 and isentropic expansion 2-3, is shown on T-s and h-s diagrams in Fig. 12.7 (a) and (b) respectively

Vapour Power Cycles 319 First of all, let us find the dryness fraction of steam (x3) after the isentropic expansion at point 3. From steam tables, corresponding to a pressure of 6 bar, we find that hf 2 = 670.4 kJ/ kg ; hf g2 = 2085.1 kJ/ kg ; sf 2 = 1.931 kJ/ kg K; sf g2 = 4.827 kJ/ kg K and corresponding to a pressure of 0.7 bar, we find that hf 3 = 376.8 kJ/ kg; hf g3 = 2283.3 kJ/ kg; sf 3 = 1.192 kJ/ kg K; sf g3 = 6.288 kJ/ kg K p2 2

h2

Temperature

1

Enthalpy

x2

2 x2

p3 Saturation curve

h3

3

x3

3

4

x3 s2 = s3

s2 = s3

Entropy

Entropy

(a) T-s diagram.

(b) h-s diagram.

Fig. 12.7

Now for the isentropic expansion process 2-3, Entropy before expansion (s2 ) = Entropy after expansion (s3 ) sf 2 + x2 sf g2 = sf 3 + x3 sf g3 1.931 + 0.9 × 4.827 = 1.192 + x3 × 6.288 6.2753 = 1.192 + 6.288 x3 6.2753 − 1.192 = 0.808 6.288 We know that enthalpy of steam at point 2, h2 = hf 2 + x2 hf g2 = 670.4 + 0.9 × 2085.1 = 2547 kJ/ kg and enthalpy of steam at point 3, h3 = hf3 + x3 hfg3 = 376.8 + 0.808 × 2283.3 = 2221.7 kJ/ kg Theoretical workdone We know that theoretical workdone = h2 – h3 = 2547 – 2221.7 = 325.3 kJ/ kg Ans. Rankine efficiency We know that Rankine efficiency, h − h3 2547 − 2221.7 325.3 = = hR = 2 = 0.15 or 15% Ans. h2 − h f 3 2547 − 376.8 2170.2 \

x3 =

Example 12.4. A turbine working on a Rankine cycle is supplied with dry saturated steam at 25 bar and exhaust takes place at 0.2 bar. For a steam flow rate of 10 kg/s, determine: (a) Quality of steam at the end of expansion; (b) Turbine shaft work; (c) Power required to drive the pump; (d) Work ratio; (e) Rankine efficiency; and ( f ) Heat flow in the condenser. Solution. Given: Pressure at which steam is supplied p1 = p2 = 25 bar

320 Engineering Thermodynamics

Temperature

Exhaust pressure, p3 = p4 = 0.2 bar Steam flow rate, m = 10 kg/s 2 1 The T-s diagram for the Rankine cycle is shown in Fig. 12.8. From steam tables, corresponding to pressure of 25 bar, we find that h2 = hg2 = 2801 kJ/ kg; s2 = sg2 = 6.253 kJ/ kg K 4 and corresponding to pressure of 0.2 bar, we find that 3 hf 3 = 251.5 kJ/ kg ; hf g3 = 2358.4 kJ/kg; sf 3 = 0.832 kJ/ kg K ; sf g3 = 7.077 kJ/ kg K Entropy (a) Quality of steam at the end of expansion Fig. 12.8 Let x3 = Quality of steam or Dryness fraction of steam at the end of expansion (i.e. at point 3). We know that for isentropic expansion process 2-3, Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sf g3 or 6.253 = 0.832 + x3 × 7.077 6.253 − 0.832 \ x3 = = 0.766 Ans. 7.077 (b) Turbine shaft work We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 251.5 + 0.766 × 2358.4 = 2058 kJ/ kg \ Turbine shaft work = m (h2 – h3) = 10(2801 – 2058) = 7430 kJ/s or kW Ans. ...(Q 1 kJ/ s = 1 kW) (c) Power required to drive the pump From steam tables, corresponding to a pressure of 0.2 bar, we find that specific volume of water, vf = 0.001 017 m3 / kg \ Power required (or work required) to drive the pump = m vf ( p1 – p4) 100 = 10 × 0.001 017 (25 – 0.2) 100 = 25.22 kJ / s or kW Ans. ...( p1 and p4 are taken in kPa) (d) Work ratio We know that work ratio Net work required Turbine work − Pump work = = Gross work done Turbine work = (e) Rankine efficiency We know that Rankine efficiency,

hR =

7430 − 25.22 = 0.9966 Ans. 7430

h2 − h3 2801 − 2058 743 = = = 0.2914 or 29.14% Ans. h2 − h f 3 2801 − 251.5 2549.5

Vapour Power Cycles 321 ( f) Heat flow in the condenser We know that heat flow in the condenser = m (h3 – hf 4) = 10 (2058 – 251.5) = 18 065 kJ/ s or kW Ans. ...(Q hf 4 = hf 3) Example 12.5. Dry and saturated steam at 15 bar is supplied to a steam engine. The exhaust takes place at 1.1 bar. Calculate: (a) Rankine efficiency; (b) Steam consumption per kWh, if the efficiency ratio is 0.65; and (c) Carnot cycle efficiency for the given pressure limits, using steam as the working fluid. Solution. Given: Pressure at which steam is supplied, p2 = 15 bar Exhaust pressure, p3 = 1.1 bar Efficiency ratio = 0.65 The T-s diagram of the Rankine cycle is shown in Fig. 12.9. From steam tables, corresponding to a pressure of 15 bar, we find that tsat 2 = t2 = 198.3°C; h2 = hg2 = 2790 kJ/ kg; s2 = sg2 = 6.441 kJ/ kg K

and corresponding to a pressure of 1.1 bar, we find that tsat3 = t3 = 102.3°C; hf 3 = 428.8 kJ / kg;

hf g3 = 2250.8 kJ / kg; sf 3 = 1.333 kJ / kg;

sf g3 = 5.995 kJ/ kg

First of all, let us find the dryness fraction of steam (x3 ) after the isentropic expansion process 2-3. We know that for isentropic expansion process,

Temperature

T1 = T2

T3 = T4

2

1

4

3

Entropy

Fig. 12.9 Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sf g3 or 6.441 = 1.333 + x3 × 5.995 6.441 − 1.333 \ x3 = = 0.852 5.995 We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 428.8 + 0.852 × 2250.8 = 2346.5 kJ / kg (a) Rankine efficiency We know that heat supplied during the cycle = h2 – hf 3 = 2790 – 428.8 = 2361.2 kJ / kg and workdone during the isentropic expansion = h2 – h3 = 2790 – 2346.5 = 443.5 kJ / kg Since the efficiency ratio is 0.65, therefore actual or useful workdone = 0.65 × 443.5 = 288.3 kJ/ kg We know that Rankine efficiency, Actual workdone 288.3 = hR = = 0.122 or 12.2% Ans. Heat supplied 2361.2

322 Engineering Thermodynamics (b) Steam consumption per kWh We know that steam consumption per kWh Heat equivalent to 1 kWh 3600 = = Power or work developed in kW 288.3 = 12.5 kg / kWh Ans. (c) Carnot cycle efficiency We know that Carnot cycle efficiency, T − T3 471.3 − 375.3 = hCarnot = 2 = 0.2034 or 20.34% Ans. T2 471.3 ...(Q T2 = 198.3 + 273 = 471.3 K; and T 3 = 102.3 + 273 = 375.3 K)

12.6 RANKINE CYCLE USING SUPERHEATED STEAM The p-v and T-s diagram of the Rankine cycle using superheated steam is shown in Fig. 12.10 (a) and (b) respectively. In this case, the water is first evaporated into dry saturated steam (as shown by operation 1-2) and then it is superheated (as shown by operation 2-2).

Fig. 12.10. Rankine cycle using superheated steam.

Now the superheated steam is expanded isentropically as shown by the process 2-3 and finally it is condensed as shown by the process 3-4. In this case, the enthalpy of superheated steam is given by hsup = h2 = hg2 + cp (T2 – T2) where T2 = Tsup = Temperature of superheated steam. The value of enthalpy of superheated steam may be directly seen from the steam tables for superheated steam corresponding to a given pressure and temperature. Following are the advantages of using superheated steam: 1. The process of superheating the steam leads to a higher temperature of steam at the inlet to the engine or turbine without increasing the maximum pressure in the cycle. 2. It delivers more work as represented by the area 2-2-3-3. 3. There is also an increase in the heat input as represented by the area 2-2-a-a-2. 4. Since the average temperature at which the heat is added increases, therefore overall effect is an increase in efficiency of the cycle. 5. The superheating of steam increases the dryness fraction of steam at the end of isentropic expansion in the turbine (as shown by point 3). In other words, the quality of steam leaving the turbine increases and thus the moisture content of the steam at the turbine exit is reduced.

Vapour Power Cycles 323 Example 12.6. A steam turbine receives steam at 15 bar and 350°C and exhausts to the condenser at 0.06 bar. Determine the thermal efficiency of the ideal Rankine cycle operating between these two limits. Solution. Given: Pressure at which turbine receives steam, p2 = 15 bar Temperature of steam, T2 = Tsup = 350°C Exhaust pressure, p3 = 0.06 bar The Rankine cycle with superheated steam is shown in Fig. 12.10. From steam tables of superheated steam corresponding to a pressure of 15 bar and 350°C, we find that h2 = hsup = 3148.7 kJ/ kg ; and s2 = ssup = 7.104 kJ/ kg K and from steam tables of dry saturated steam corresponding to a pressure of 0.06 bar, we find that hf 3 = 151.5 kJ/ kg; hfg3 = 2416 kJ/ kg; sf 3 = 0.521 kJ/ kg K; sfg3 = 7.81 kJ/ kg K Let x3 = Dryness fraction of steam leaving the turbine or entering the condenser. We know that for isentropic expansion process 2-3, Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sf g3 or 7.104 = 0.521 + x3 × 7.81 7.104 − 0.521 \ x3 = = 0.843 7.81 and enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 151.5 + 0.843 × 2416 = 2188.2 kJ/ kg We know that the thermal efficiency the ideal Rankine cycle, h − h3 3148.7 − 2188.2 960.5 = = hR = 2 = 0.32 or 32% Ans. h2 − h f 3 3148.7 − 151.5 2997.2 Example 12.7. Steam at 50 bar and 400°C expands in a Rankine cycle to 0.34 bar. For a mass flow rate of 150 kg /s of steam, determine (a) the power developed; ( b) the thermal efficiency; and (c) specific steam consumption. Solution. Given: Pressure at which steam is supplied, p1 = p2 = 50 bar Temperature of steam, T2 = Tsup = 400°C Exhaust pressure, p3 = p4 = 0.34 bar Mass flow rate of steam, m = 150 kg/ s The p-v and T-s diagram of the Rankine cycle is shown in Fig. 12.11. From steam tables of superheated steam, corresponding to a pressure of 50 bar and a temperature of 400°C, we find that h2 = ssup = 3198.3 kJ/ kg ; s2 = ssup = 6.651 kJ/ kg K and from steam tables of dry saturated steam, corresponding to a pressure of 0.34 bar, we find that hf 3 = 301.5 kJ/ kg ; hf g3 = 2329 kJ/ kg; sf 3 = 0.98 kJ/ kg K; sf g3 = 6.747 kJ/ kg K

324 Engineering Thermodynamics Let x3 be the dryness fraction of steam after the isentropic expansion i.e. at point 3. We know that for the isentropic expansion process 2-3,

T2 = Tsup 2

0.34

T1

Temperature

1

Pressure

50

3

4

2

T3 = T4

2

1

4

3

Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.11

Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sfg3 or 6.651 = 0.98 + x3 × 6.747 \

x3 =

6.651 − 0.98 = 0.84 6.747

and enthalpy of steam at point 3, h3 = hf 3 + x3 hfg3 = 301.5 + 0.84 × 2329 = 2257.86 kJ/ kg (a) Power developed We know that workdone or power developed during the cycle = m (h2 – h3) = 150 (3198.3 – 2257.86) = 141 066 kJ/s or kW Ans. ...(Q 1 kJ /s = 1 kW) (b) Thermal efficiency We know that thermal efficiency of the cycle

hR =

h2 − h3 3198.3 − 2257.86 940.44 = = h2 − h f 3 3198.3 − 301.5 2896.8

= 0.3246 or 32.46% Ans. (c) Specific steam consumption We know that specific steam consumption Heat equivalent to 1 kWh 3600 3600 = = = Power developed in kW h2 − h3 3198.3 − 2257.86 = 3.828 kg / kWh Note: Specific steam consumption is defined as the mass rate of steam required for unit power. Mathematically, Mass of steam flow in kg/ h Specific steam consumption = ... (in kg / kWh) Power in kW =

150 × 3600 = 3.828 kg / kWh Ans. 141 066

Vapour Power Cycles 325

Temperature

Example 12.8. In an Ideal Rankine cycle, steam at 150 bar and 500°C enters a turbine and the condenser operate at 10 kPa. If the power output of the cycle is 100 MW, determine: (a) the mass flow of steam; (b) the thermal efficiency of the cycle ; and (c) the mass flow rate of cooling water through the condenser, if the water enters the condensers at 30°C and leaves at 40°C. Solution. Given: Pressure of steam at which it enters the turbine, p2 = 150 bar Temperature of steam, T2 = Tsup = 500°C Exhaust pressure, p3 = 10 kPa = 0.1 bar Power output, P = 100 MW = 100 × 103 kW The Rankine cycle with superheated steam is shown in Fig. 12.12. From steam tables of superheated steam, corresponding to a pressure of 150 bar and 500°C, we 2 T1 = Tsup find that h2 = hsup = 3310.6 kJ/ kg 2 1 s2 = ssup = 6.349 kJ/ kg K and from steam tables of dry saturated steam corresponding to a pressure of 0.1 bar, T3 = T4 hf 3 = 191.8 kJ/ kg; hfg3 = 2392.9 kJ/ kg 4 3 sf 3 = 0.649 kJ/ kg K; sfg3 = 7.502 kJ/ kg K First of all, let us find the dryness fraction of steam (x3) after the isentropic expansion process 2-3 in the Entropy turbine. We know that for isentropic expansion process. Fig. 12.12 Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sfg3 or 6.349 = 0.649 + x3 × 7.502 6.349 − 0.649 \ x3 = = 0.76 7.502 We know that enthalpy of steam at point 3, h3 = hf3 + x3 hfg3 = 191.8 + 0.76 × 2392.9 = 2010.4 kJ/ kg (a) Mass flow of steam Let m = Mass flow of steam in kg /s. We know that the workdone or power developed (P), 100 × 103 = m (h2 – h3) = m (3310.6 – 2010.4) = 1300.2 m \

m =

100 × 103 = 76.9 kg/s Ans. 1300.2

(b) Thermal efficiency of the cycle We know that thermal efficiency of the cycle, h − h3 3310.6 − 2010.4 1300.2 = = hR = 2 h2 − h f 3 3310.6 − 191.8 3118.8 = 0.417 or 41.7% Ans.

326 Engineering Thermodynamics

Temperature

(c) Mass flow rate of cooling water through the condenser Let mw = Mass flow rate of cooling water through the condenser in kg/s, TWE = Temperature of water entering the condenser = 30°C = 30 + 273 = 303 K ...(Given) TWL = Temperature of water leaving the condenser = 40°C = 40 + 273 = 313 K ...(Given) Now for the energy balance of the condenser, Heat given out by the steam = Heat taken in by the water m (h2 – hf 3) = mw cw (TWL – TWE) 76.9 (3310.6 – 191.8) = mw × 4.187 (313 – 303) 239 836 = 41.87 mw 239 836 \ mw = = 5728 kg / s Ans. 41.87 Example 12.9. A steam turbine receives superheated steam at a pressure of 17 bar and having a degree of superheat of 110°C. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically. Calculate: (a) Heat supplied; (b) Heat rejected; (c) Net workdone; and (d) Thermal efficiency. Solution. Given: Pressure at which steam is received by the turbine, p2 = 17 bar Degree of superheat, T2 – T2 = 110°C Exhaust pressure, p3 = p4 = 0.07 bar 2 T2 = Tsup The T-s diagram is shown in Fig. 12.13. T2 1 2 First of all, let us find the temperature of the superheated steam at point 2 (i.e. T2 = Tsup ). From steam tables, corresponding to a pressure of 17 bar, we find that saturation temperature of steam at point 2, T3 = T4 T2 = 204.3°C 4 3 \ T2 = Tsup = 110 + T2 = 110 + 204.3 = 314.3°C Entropy Now from steam tables of superheated steam, Fig. 12.13 corresponding to a pressure of 17 bar and a temperature of 314.3°C, we find that h2 = h sup = 3064.3 kJ / K ; s2 = ssup = 6.909 kJ / kg K and from steam tables of dry saturated steam, corresponding to a pressure of 0.07 bar, we find that hf 3 = 163.4 kJ/ kg; hfg3 = 2409.2 kJ/ kg; sf 3 = 0.559 kJ/ kg K; sf g3 = 7.718 kJ/ kg K Let x3 be the dryness fraction of steam after the isentropic expansion process 2-3. We know that for isentropic process, Entropy before expansion at point 2 (s2) = Entropy after expansion at point 3 (s3) = sf 3 + x3 sf g3 6.909 = 0.559 + x3 + 7.718 6.909 − 0.559 \ x3 = = 0.823 7.718

Vapour Power Cycles 327 We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 163.4 + 0.823 × 2409.2 = 2146.2 kJ/ kg (a) Heat supplied We know that heat supplied = h2 – hf 3 = 3064.3 – 163.4 = 2900.9 kJ/ kg Ans. (b) Heat rejected We know that *heat rejected = h3 – hf 3 = 2146.2 – 163.4 = 1982.8 kJ/ kg Ans. (c) Net workdone We know that **net workdone = h2 – h3 = 3064.3 – 2146.2 = 918.1 kJ/ kg (d) Thermal efficiency Net workdone 918.1 = We know that thermal efficiency = Heat supplied 2900.9 = 0.3165 or 31.65% Ans.

12.7 MODIFIED RANKINE CYCLE We have already discussed that the area of the pressure-volume ( p-v) diagram of the Rankine cycle represents the workdone. The work obtained near the tail end of the p-v diagram is very small, as shown in Fig. 12.14. In fact, it is not even sufficient to overcome the work lost in friction in the tail end part of the stroke. Therefore, in modified Rankine cycle, the isentropic expansion is terminated at some point 3 (before the expansion is complete) by opening the exhaust port. This causes a sudden pressure drop at constant volume due to the steam communicating with the outside atmosphere. The line 3-4 represents this process. By doing so, we considerably reduce the stroke length and hence the size of the cylinder, without any appreciable change in the workdone, as shown by 3-3-4. The cycle 1-2-3-4-5-1 is now known as modified Rankine cycle. This cycle is used in steam engines whereas steam turbines work on Rankine cycle. From the p-v diagram, as shown in Fig. 12.14 (a), the workdone during the cycle, per kg of steam is given by w = Area 1-2-3-4-5 = Area 1-2-b-a + ***Area 2-3-c-b – Area 4-c-a-5 = 100 p2 v2 + (u2 – u3) – 100 p4 v4 ...(i) ...(The pressure is taken in bar and volume in m3) We know that internal energy at point 2,

u2 = h2 – 100 p2 v2

and internal energy at point 3, u3 = h3 – 100 p3 v3 Substituting the values of u2 and u3 in equation (i), we have

w = 100 p2 v2 + [(h2 – 100 p2 v2) – (h3 – 100 p3 v3)] – 100 p4 v4

= h2 – h3 + 100 p3 v3 – 100 p4 v3 ...(Q v4 = v3) = h2 – h3 + 100 ( p3 – p4) v3 ...(ii) * ** ***

Heat rejected is also given by x3 hfg3 = 0.823 × 2409.2 = 1982.8 kJ / kg Net workdone is also given by (Heat supplied – Heat rejected). \ Net workdone = 2900.9 – 1982.8 = 918.1 kJ / kg Since the heat transfer during the isentropic process is zero, therefore the workdone during isentropic expansion (Area 2-3-c-b) is equal to change in internal energy (i.e. u2 – u3).

328 Engineering Thermodynamics We know that heat supplied during the cycle = h2 – hf 4

Work lost

1

T1 = T2

2

3

3

p3 p 4 = p5

Temperature

2

1

Pressure

p1 = p2

5 a

v2

T4 = T5

3

4 c v4

b

5

4

Work lost

3

Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.14. Modified Rankine cycle.

\ Efficiency of the modified Rankine cycle, (h − h3 ) + 100 ( p3 − p4 ) v3 Workdone hMR = = 2 h2 − h f 4 Heat supplied Example 12.10. Steam at 5.6 bar and 0.9 dry is supplied to a steam engine, where in the cutoff occurs at one-half stroke, and the release pressure is 2.5 bar. The steam is released from the engine at constant volume to a back pressure of 1 bar. Neglecting clearance volume, find the modified Rankine efficiency of the engine. Solution. Given: Pressure at which steam is supplied, p2 = p1 = 5.6 bar Dryness fraction of steam, x2 = 0.9 Volume at cut-off, v2 = v3 / 2 Release pressure, p3 = 2.5 bar Back pressure, p4 = p5 = 1 bar The p-v and T-s diagram for the modified Rankine cycle is shown in Fig. 12.15 (a) and (b) respectively.

2

1

Temperature

1

Pressure

5.6

2

3

3

2.5

4

1 5 v2

4

5

v3 = v4

s2 = s3

Volume

Entropy

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.15

From steam tables, corresponding to a pressure of 5.6 bar, we find that vg2 = 0.3367 m3 /kg; hf2 = 658.8 kJ/ kg; hfg2 = 2093.7 kJ/ kg; sf 2 = 1.904 kJ/ kgK; sfg2 = 4.877 kJ/ kg K

Vapour Power Cycles 329 Corresponding to a pressure of 2.5 bar, we find that

vg3 = 0.7184 m3 / kg; hf 3 = 535.4 kJ/ kg; hfg3 = 2181 kJ/ kg; sf 3 = 1.607 kJ/ kg K;

sf g3 = 5.445 kJ/ kg K

and corresponding to a pressure of 1 bar,

hf 4 = 417.5 kJ/kg

First of all, let us find the dryness fraction of steam after the isentropic expansion process 2-3 (i.e. at point 3). Let x3 be the dryness fraction at point 3. We know that for isentropic process 2-3,

Entropy before expansion (s2)

= Entropy after expansion (s3)

sf 2 + x2 sfg3 = sf 3 + x3 sf g3 1.904 + 0.9 × 4.877 = 1.607 + x3 × 5.445

6.2933 = 1.607 + 5.445 x3 6.2933 − 1.607 \ x3 = = 0.86 5.445 We know that enthalpy of steam at point 2, h2 = hf2 + x2 hfg2 = 658.8 + 0.9 × 2093.7 = 2543.13 kJ/kg and enthalpy of steam at point 3, h3 = hf 3 + x3 hfg3 = 535.4 + 0.86 × 2181 = 2411 kJ/kg Volume of steam at point 2, v2 = x2 vg2 = 0.9 × 0.3367 = 0.303 m3 /kg and volume of steam at point 3, v3 = 2v2 = 2 × 0.303 = 0.606 m3 / kg ...(Q It is given that v2 = v3 / 2) \ Modified Rankine efficiency, (h − h3 ) + 100 ( p3 − p4 ) v3 hMR = 2 h2 − h f 4

=

(2543.13 − 2411) + 100 (2.5 − 1) 0.606 132.13 + 90.9 = 2543.13 − 417.5 2125.63

= 0.105 or 10.5% Ans. Example 12.11. A steam engine takes dry steam at 20 bar and exhausts at 1.2 bar. The pressure at the release is 3 bar. Find: 1. The theoretical loss of work per kg of steam due to incomplete expansion; and 2. The loss in Rankine efficiency due to restricted expansion of steam. Solution. Given: Pressure of steam supplied to the engine, p2 = p1 = 20 bar Exhaust pressure, p4 = p5 = 1.2 bar Release pressure, p3 = 3 bar The p-v and T-s diagram is shown in Fig. 12.16 (a) and (b) respectively. In the figure, the modified Rankine cycle is represented by 1-2-3-4-5-1 while the Rankine cycle is represented by 1-2-3-5-1.

330 Engineering Thermodynamics 1. Theoretical loss of work per kg of steam due to incomplete expansion From steam tables, corresponding to a pressure of 20 bar, we find that v2 = vg2 = 0.0995 m3 / kg ; h2= hg2 = 2797.3 kJ/ kg ; s2 = sg2 = 6.337 kJ/ kg K Corresponding to a pressure of 3 bar, we find that vg3 = 0.6055 m3 / kg; hf 3 = 561.4 kJ / kg ; hf g3 = 2163.2 kJ / kg; sf 3 = 1.672 kJ / kg K; sfg3 = 5.319 kJ / kg K and corresponding to a pressure of 1.2 bar, hf 3 = 439.4 kJ/ kg; hf g3 = 2244.1 kJ/ kg; sf 3 = 1.361 kJ/ kg K; sf g3 = 5.937 kJ/ kg K

Fig. 12.16

First of all, let us consider the modified Rankine cycle 1-2-3-4-5-1. Let x3 be the dryness fraction of steam after the isentropic expansion process 2-3. We know that for isentropic process, Entropy before expansion (s2) = Entropy after expansion (s3) = sf 3 + x3 sfg3 \ 6.337 = 1.672 + x3 × 5.319 6.337 − 1.672 or x3 = = 0.877 5.319 We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hfg3 = 561.4 + 0.877 × 2163.2 = 2458.5 kJ / kg and volume of steam at point 3, v3 = x3 × vg3 = 0.877 × 0.6055 = 0.531 m3 / kg We know that work obtained in modified Rankine cycle, wMR = h2 – h3 + 100 ( p3 – p4) v3 = 2797.3 – 2458.5 + 100 (3 – 1.2) 0.531 = 434.38 kJ/ kg ...(i) Now let us consider the Rankine cycle 1-2-3-5-1. Let x3 be the dryness fraction of steam after isentropic expansion process 2-3. We know that for isentropic process, Entropy before expansion (s2) = Entropy after expansion (s3) = sf 3 + x3 sfg3 6.337 = 1.361 + x3 × 5.937 6.337 − 1.361 or x3 = = 0.838 5.937 \ Enthalpy of steam at point 3 h3 = hf 3 + x3 hfg3 = 439.4 + 0.838 × 2244.1 = 2320 kJ / kg We know that workdone obtained in Rankine cycle, wR = h2 – h3 = 2797.3 – 2320 = 477.3 kJ / kg

Vapour Power Cycles 331 \ Theoretical loss of work due to incomplete expansion = wR – wMR = 477.3 – 434.38 = 42.92 kJ / kg Ans. 2. Loss of Rankine efficiency due to restricted expansion of steam We know that heat supplied during the modified Rankine cycle, q = h2 – hf 4 = h2 – hf 3 ...(Q hf 4 = hf 3) = Heat supplied during the Rankine cycle = 2797.3 – 439.4 = 2357.9 kJ / kg \ Modified Rankine efficiency, Workdone ( wMR ) 434.38 = hMR = = 0.184 Heat supplied (q ) 2357.9 and Rankine efficiency,

hR =

wR 477.3 = = 0.202 q 2357.9

\ Percentage loss in Rankine efficiency 0.202 − 0.184 = = 0.089 or 8.9% Ans. 0.202 Note: Since the heat supplied during the modified Rankine cycle is equal to the heat supplied during the Rankine cycle, therefore Percentage loss in Rankine efficiency = Percentage loss of work w − wMR 42.92 = = R = 0.09 or 9% Ans. wR 477.3

Example 12.12. The cylinder of a steam engine is 300 mm in diameter and piston stroke is 580 mm. The steam at admission is at 10 bar and 300°C. It expands isentropically to 0.7 bar and is then reduced at constant volume to condenser at 0.28 bar. Determine: 1. The modified Rankine efficiency. 2. The new stroke if the same amount of steam from original condition is expanded isentropically to condenser pressure. 3. The new Rankine efficiency. 4. The workdone in kilo-joules by the extraction of boiler feed pump per kg of water returned to the boiler. Solution. Given: Diameter of engine cylinder, D = 300 mm = 0.3 m Length of piston stroke, L = 580 mm = 0.58 m Admission pressure of steam, p2 = p1 = 10 bar Temperature of steam, T2 = Tsup = 300°C Pressure after isentropic expansion, p3 = 0.7 bar Release pressure, p4 = p5 = 0.28 bar We know that volume of the engine cylinder, π π v = × D2 L = (0.3)2 0.58 = 0.041 m3 4 4

332 Engineering Thermodynamics The p-v and T-s diagram for the modified Rankine cycle is shown in Fig. 12.17 (a) and (b) respectively. 2

10

2

3

0.7

0.28 5

3

4 v3 = v4 L

1

Temperature

Pressure

T2 = Tsup 1

3 4

5

v3

2

3

Entropy

L Volume

(a) p-v diagram.

(b) T-s diagram.

Fig. 12.17

From steam tables for superheated steam, corresponding to a pressure of 10 bar and 300°C, we find that h2 = hsup = 3052.1 kJ / kg; s2 = ssup = 7.125 kJ / kg K Now from steam tables for dry saturated steam, corresponding to a pressure of 0.7 bar, we find that vg3 = 2.3647 m3 / kg; hf 3 = 376.8 kJ / kg; hfg3 = 2283.3 kJ / kg; sf 3 = 1.192 kJ/ kg K; sf g3 = 6.288 kJ / kg K and corresponding to a pressure of 0.28 bar, we find that hf 4 = 282.7 kJ / kg 1. Modified Rankine efficiency First of all, let us find the dryness fraction of steam after isentropic expansion process 2-3 (i.e. at point 3). Let x3 be the dryness fraction at point 3. We know that for isentropic process 2-3. Entropy before expansion (s2) = Entropy after expansion (s3) = sf 3 + x3 sf g3 or 7.125 = 1.192 + x3 × 6.288 7.125 − 1.192 \ x3 = = 0.9435 6.288 We know that volume of steam at point 3, v3 = x3 × vg3 = 0.9435 × 2.3647 = 2.231 m3 / kg and enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 376.8 + 0.9435 × 2283.3 = 2531.1 kJ / kg \ Modified Rankine efficiency, (h − h3 ) + 100 ( p3 − p4 ) v3 hMR = 2 h2 − h f 4 =

(3052.1 − 2531.1) + 100 (0.7 − 0.28) 2.231 3052.1 − 282.7

=

521 + 93.7 = 0.222 or 22.2% Ans. 2769.4

Vapour Power Cycles 333 2. New stroke length It is given that the same amount of steam from the original condition (i.e. at 10 bar and 300°C) is expanded isentropically to the condenser pressure of p3 = 0.28 bar (i.e. point 3). Let x3 be the dryness fraction of steam at point 3. Now from steam tables, corresponding to a pressure of 0.28 bar, we find that vg3 = 5.5778 m3 / kg; hf 3 = 282.7 kJ / kg; hf g3 = 2340 kJ/kg; sf 3 = 0.925 kJ / kg K; sfg3 = 6.868 kJ / kg K We know that for the isentropic process 2-3, Entropy before expansion (s2) = Entropy after expansion (s3) = sf 3 + x3 sfg3 or 7.125 = 0.925 + x3 × 6.868 7.125 − 0.925 \ x3 = = 0.9027 6.868 We know that mass of steam at point 3,

m3 =

Cylinder volume (v) 0.041 = = 0.018 38 kg Volume of steam at point 3 (v3 ) 2.231

\ Volume of steam at point 3 v3 = m3 x3 vg3 = 0.018 38 × 0.9027 × 5.5778 = 0.0925 m3 Let L be the new stroke length. Therefore the new stroke volume at point 3, π v3 = × D2 L 4 \

0.0925 = L =

π (0.3)2 L = 0.0707 L 4 0.0925 = 1.308 m Ans. 0.0707

3. New Rankine efficiency We know that enthalpy of steam at point 3, h3 = hf 3 + x3 hf g3 = 282.7 + 0.9027 × 2340 = 2395 kJ / kg \ New Rankine efficiency h − h3′ 3052.1 − 2395 657.1 = = hR = 2 h2 − h f 3′ 3052.1 − 282.7 2769.4 = 0.2373 or 23.73% Ans. 4. Workdone The work is done by the extraction and boiler feed pumps in raising the pressure from 0.28 bar to 10 bar. From steam tables, corresponding to a pressure of 0.28 bar, we find that volume of water at point 5, v5 = 0.001021 m3 / kg \ Workdone by the extraction and boiler feed pumps = 100 ( p1 – p5) v5 = 100 (10 – 0.28) 0.001 021 = 0.9924 kJ / kg Ans.

334 Engineering Thermodynamics

12.8 METHODS OF IMPROVING THE EFFICIENCY OF RANKINE CYCLE

It is a known fact that most of the electric power in the world is produced by steam power plants. Thus, a small increase in the thermal efficiency will lead to a large saving in the fuel requirement. Therefore, every effort is made to improve the efficiency of the cycle on which the steam power plants operate. Following are the ways of improving the thermal efficiency of the vapour power cycle: 1. By decreasing the average temperature at which the heat is rejected from the working fluid in the condenser. It may be noted that lowering of the operating pressure of the condenser automatically lowers the temperature of steam and thus the temperature at which heat is rejected. Though in this case, the thermal efficiency increases but the moisture content of the steam leaving the turbine also increases, which is highly undesirable in turbines because it decreases the turbine efficiency and erodes the turbine blades. 2. By increasing the average temperature at which heat is added to the steam. It is done by superheating the steam to high temperature. It may be noted that when the steam is superheated to the higher temperature, the moisture content of the steam leaving the turbine decreases. Hence the moisture problem in the turbine is reduced. In other words, when the steam is superheated, the quality of steam leaving the turbine increases and hence the efficiency of the cycle increases. 3. Another way of increasing the average temperature during heat addition process is to increase the operating pressure of the boiler. This, in turn, raises the average temperature at which heat is added to the steam and thus raises the thermal efficiency of the cycle. Though the increasing of boiler pressure increases the thermal efficiency of the Rankine cycle, but it also increases the moisture content of the steam to unacceptable levels. In order to overcome this effect, the simple ideal Rankine cycle is modified with a *reheat process, i.e. the steam in the turbine is expanded in two stages and reheated it in between. The reheating of steam is a practical solution to the excessive moisture problem in turbines and it is frequently used in modern steam power plants. In short, the reheating of steam has the following advantages in addition to the increase in efficiency of the turbine. 1. The quality of steam (i.e. the dryness fraction of steam) leaving the turbine increases. Thus the erosion of turbine blades reduces. 2. The workdone through the turbine increases. 3. Since there is a reduction in specific steam consumption, therefore, the amount of water required in the condenser is reduced.

12.9 REHEAT CYCLE The schematic diagram of a power plant working on reheat cycle is shown in Fig. 12.18 and the enthalpy-entropy (h-s) diagram is shown in Fig. 12.19. The cycle differs from the simple Rankine cycle in the fact that the expansion process takes place in two stages. In the first stage, the steam is expanded isentropically in a high pressure (H.P.) turbine from an initial pressure p1 to intermediate pressure p2, as shown by the vertical line 1-2 in Fig. 12.19. This steam is sent back to the boiler where it is reheated in a reheater at constant pressure p2, usually to the inlet temperature of the first (high pressure) turbine stage. In the second stage, the steam expands isentropically through a low pressure (L.P.) turbine, to the condenser pressure p4, as shown by the vertical line 3-4, as shown in Fig. 12.19. If h1, h2, h3 and h4 are the enthalpies or total heat of steam at points 1, 2, 3 and 4 respectively, then the total heat supplied during a reheat cycle is equal to the sum of the total heat at point 1 and the heat supplied during reheating process 2-3.

* The ideal reheat Rankine cycle (simply known as reheat cycle) is discussed in Art.12.9.

Vapour Power Cycles 335 \ Total heat supplied, Q = h1 + [(h3 – h2) – hf 4] and workdone by the turbine WT = Total heat drop = (h1 – h2) + (h3 – h4) 1

Turbine 1

L.P.

H.P.

Boiler Super heater

2

3

Reheater

4

3

Condenser

Pump

Fig. 12.18. Power plant working on reheat cycle. Constant temperature line

3 h3 1

Re

he

Enthalpy

ati

ng

h1

p1 h2

2 p2

h4

4 p3

Sa tu cu ratio rve n

2 Entropy

Fig. 12.19. Enthalpy-entropy (h-s) diagram of reheat cycle.

We know that efficiency of the reheat cycle (h1 − h2 ) + (h3 − h4 ) Workdone (WT ) h = = Total heat supplied (Q) h1 + (h3 − h2 ) − h f 4 Note: In case the steam is not reheated, then the expansion of steam through the turbine will be along the vertical line 1-2 as shown in Fig. 12.19 and the efficiency will be h − h2′ h = 1 , which is same as for Rankine cycle. h1 − h f 2′

Example 12.13. Steam at a pressure of 15 bar and 250°C is expanded through a turbine at first to a pressure of 4 bar. It is then reheated at a constant pressure to the initial temperature of 250°C

336 Engineering Thermodynamics

Enthalpy

and is finally expanded to 0.1 bar. Using Mollier chart, estimate the workdone per kg of steam flowing through the turbine and the amount of heat supplied during the process of reheat. Compare the workoutput when the expansion is direct from 15 bar to 0.1 bar without any reheat. Assume all expansion processes to be isentropic. Solution. Given: Inlet pressure, p1 = 15 bar Inlet temperature, T1 = 250°C Intermediate pressure, p2 = 4 bar 250°C 3 Final pressure p3 = 0.1 bar h3 1 The process of reheating is represented on h-s h1 diagram, as shown in Fig. 12.20. The point 1 represents the initial condition of steam at entrance to the turbine p1 corresponding to a pressure of 15 bar and a temperature Saturation h2 of 250°C. curve 2 p2 The vertical line 1-2 represents the isentropic h4 expansion process in the first stage upto a pressure of 4 x4 p2 = 4 bar. The process 2-3 represents the reheating h2 of steam at a constant pressure p2 and upto the initial p3 2 temperature of 250°C. The vertical line 3-4 represents further isentropic expansion in the second stage, upto the Entropy pressure p3 = 0.1 bar. Now the point 4 represents the final Fig. 12.20 condition of steam at exhaust. From the Mollier chart, we find that h1 = 2915 kJ / kg; h2 = 2655 kJ / kg; h3 = 2960 kJ / kg; h4 = 2340 kJ / kg; and x4 = 0.87 Workdone per kg of steam We know that workdone per kg of steam = (h1 – h2) + (h3 – h4) = (2915 – 2655) + (2960 – 2340) = 260 + 620 = 880 kJ / kg Ans. Heat supplied during the process of reheat From steam tables, corresponding to a pressure of 0.1 bar, we find that sensible heat of water at point 4, hf 4 = hf 2 = 191.8 kJ / kg We know that heat supplied during the process of reheat = Heat supplied during the process 2-3 = (h3 – h2) – hf 4 = (2960 – 2655) – 191.8 = 113.2 kJ / kg Workoutput when the expansion is direct The direct isentropic expansion of steam from pressure 15 bar to 0.1 bar is shown by the vertical line 1-2 in Fig. 12.20. From the Mollier chart, we find that enthalpy at 2, h2 = 2120 kJ/kg \ Workoutput = Total heat drop between 1-2 = h1 – h2 = 2915 – 2120 = 795 kJ/ kg Ans. Example 12.14. In a 15 MW steam power plant operating on ideal reheat cycle, steam enters the H.P. turbine at 15 MPa and 600°C. The condenser is maintained at a pressure of 10 kPa. If the moisture content at the exit of the L.P. turbine is 10.4%, determine; (a) reheat pressure; (b) thermal efficiency; and (c) specific steam consumption.

Vapour Power Cycles 337

600°C

g

h3 h1

1

tin

bar

Assume steam to be reheated to the initial temperature. Solution. Given: Power produced by the plant = 15 MW = 15 × 106 W = 15000 kW = 15 000 kJ / s Pressure of steam entering H.P. turbine, p1 = 15 MPa = 150 bar = 150 × 102 kPa Temperature of steam, T1 = 600°C Condenser pressure, p3 = 10 kPa = 0.1 3

Enthalpy

Re h

ea

Since the moisture content at the exit of the L.P. turbine is 10.4% or 0.104, therefore dryness fraction h2 of steam at exit of the L.P. turbine is given by p1 2 x4 = 1 – 0.104 = 0.896 Sa tur p cu ation 2 The process of reheat cycle is represented on rve x4 h4 the h-s diagram, as shown in Fig. 12.21. The point 1 =0 4 .89 represents the initial condition of steam correspond6 ing to a pressure of p1 = 150 bar and temperature T1 = p3 600°C. Now, mark point 4 corresponding to a pressure p3 = 0.1 bar and dryness fraction, x4 = 0.896. From Entropy this point 4, draw a vertical line upwards meeting the Fig. 12.21 line of 600°C at point 3. Draw a vertical line through point 1 intersecting the constant pressure line at point 2. From the Mollier chart, we find that p2 = 40 bar From the Mollier chart, we find that h1 = 3580 kJ / kg; h2 = 3140 kJ / kg; h3 = 3680 kJ / kg; h4 = 2340 kJ / kg From steam tables, corresponding to a condenser pressure of 0.1 bar, we find that hf 4 = 191.8 kJ / kg (a) Reheat pressure From the Mollier chart, we find that the reheat pressure (during process 2-3) is p2 = 40 bar Ans. (b) Thermal efficiency We know that thermal efficiency,

h =

=

(h1 − h2 ) + (h3 − h4 ) h1 + (h3 − h2 ) − h f 4 (3580 − 3140) + (3680 − 2340) 440 + 1340 = 3580 + [ (3680 − 3140) − 191.8] 3580 + 348.2

= 0.453 or 45.3% Ans. (c) Specific steam consumption We know that workdone by the turbine = (h1 – h2) + (h3 – h4)

338 Engineering Thermodynamics

= (3580 – 3140) + (3680 – 2340)

= 440 + 1340 = 1780 kJ / kg \ Steam required =

Power developed by the plant in kJ / s 15 000 = Workdone in kJ / kg 1780

= 8.427 kg / s and specific steam consumption Steam required in kg / h 8.427 × 3600 = = Power developed in kW 15 000 = 2.0225 kg / kWh Ans.

12.10 REGENERATIVE CYCLE We have already discussed that in the Rankine cycle, the condensate which is at a fairly low temperature is pumped to the boiler. Same is done in the reheat cycles. Thus, there is irreversible mixing of the cold condensate with the hot boiler water. This results in the loss of cycle efficiency. Therefore, in order to overcome this shortcoming, it is necessary to raise the temperature of the liquid leaving the pump (feed water), before it enters the boiler. One method of doing this is by bleeding or extracting small amount of steam from the turbine, at certain points during its expansion, and using this steam for heating the feed water to the saturation temperature corresponding to the boiler pressure before it enters the boiler. This method of feed heating is called regenerative feed heating and the cycle is called regenerative cycle. It may be noted that the device where the feed water is heated is known as a regenerator or feed water heater. The regeneration (or feed water heating) in addition to improving the cycle efficiency, it also provides a convenient means of deaerating the feed water (i.e. removing the air that leaks in at the condenser) to prevent corrosion in the boiler. Due to these advantages, regeneration is used in all modern steam power plants. A feed water heater is essentially a heat exchanger where the heat is transferred from the steam to the feed water either by mixing the two fluid streams (open feed water heaters) or without mixing them (closed feed water heaters). The regeneration with both types of feedwater heaters is discussed in the following pages:

12.11 REGENERATIVE CYCLE WITH AN OPEN FEED WATER HEATER The schematic diagram of a steam power plant with one open feed water heater is shown in Fig. 12.22 (a) and T-s diagram of the cycle is shown in Fig. 12.22 (b). This cycle is also known as single stage regenerative cycle. Consider one kg of steam enters the turbine at a boiler pressure (say p1) as represented by point 1. It expands in the turbine isentropically to an intermediate pressure (say p2). Let a small quantity of wet steam (say m kg) is extracted at state 3 and passed on to feed water heater, while the remaining steam (1 – m) kg continues to expand isentropically upto point 3. This steam is now condensed in the condenser and leaves it as a saturated liquid at state 4 at the condenser pressure (say p3). The condensate from the condenser is pumped into the feed water heater, where it mixes up with the steam extracted from the turbine. The fraction of steam extracted is such that the mixture leaves the feed water heater as a saturated liquid at state 5. A second pump raises the pressure of water to

Vapour Power Cycles 339 the boiler pressure (state 6). The cycle is completed by heating the water in the boiler to the turbine inlet state 1. 1 kg steam 1

Open feed water heater Boiler

2 m

(1 – m) 3

6 1 kg

5

Temperature

Turbine

6 5

(1 – m)

1

1 kg

m kg 2 (1 – m) kg 3

4

4 Condenser Feed pump

Entropy

(a) Schematic steam power plant with one open feed water heater.

(b) T-s diagram.

Fig. 12.22. Regenerative cycle with an open feed water heater.

Enthalpy

The cycle on the h-s diagram is represented as shown in Fig. 12.23. Let h1, h2 and h3 be the enthalpies at states 1 h1 1, 2 and 3 respectively. Since in the feed water heater, the bled steam m kg from the turbine is mixed up with the Satu feed water (1 – m) kg from the condenser, therefore for ratio n cu the energy balance of the heater, rve Heat lost by bled steam h2 = Heat gained by feed water 2 or m (h2 – hf 2) = (1 – m) (hf 2 – hf 3) ...(i) where hf 2 and hf 3 are the sensible heat or enthalpy of h3 3 water (from steam tables) corresponding to pressure p2 and p3 respectively. Now workdone by the turbine per kg of feed water Entropy between states 1 and 2, Fig. 12.23 w1 = h1 – h2 and workdone by the turbine per kg of feed water between states 2 and 3 w2 = (1 – m) (h2 – h3) \ Total workdone by the turbine per kg of feed water = w1 + w2 = (h1 – h2) + (1 – m) (h2 – h3) and total heat supplied per kg of feed water = h1 – hf 2 We know that efficiency of the regenerative cycle Total workdone (h1 − h2 ) + (1 − m) (h2 − h3 ) = h = h1 − h f 2 Heat supplied It may be noted that when there is no regenerative feed heating (i.e. when m = 1), then the cycle becomes the Rankine cycle and the efficiency will be h − h3 hR = 1 h1 − h f 3

340 Engineering Thermodynamics

12.12 REGENERATIVE CYCLE WITH A CLOSED FEED WATER HEATER The schematic diagram of a steam power plant with one closed feed water heater is shown in Fig. 12.24 (a) and T-s diagram of the cycle is shown in Fig. 12.24 (b). In this case, heat is transferred from the extracted steam to the feed water at different pressures without any mixing takes place. In an ideal closed feed water heater, the feed water is heated to the exit temperature of the extracted steam which leaves the heater as a saturated liquid below the exit temperature of the extracted steam because a temperature difference of atleast a few degrees is required for any effective heat transfer to take place. The condensed steam is either pumped to the feed water line or delivered to another heater or to the condenser through a device called trap, which allows the liquid to be throttled to a lower pressure region but traps the vapour. The enthalpy of steam remains constant during the throttling process. 1 Boiler Turbine

2 3

6

1

Temperature

Closed feed water heater

6 2

5

4 Mixing chamber

3

5 Feed pump

4

Condenser

(a) Schematic steam power plant with one closed feed water heater.

Entropy (b) T-s diagram.

Fig. 12.24. Regenerative cycle with one closed feed water heater.

12.13 COMPARISON BETWEEN OPEN AND CLOSED FEED WATER HEATERS The comparison between open and closed feed water heaters are as follows: 1. The open feed water heaters are simple and inexpensive whereas the closed feed water heaters are more complex because of internal piping network, thus they are more expensive. 2. The open feed water heaters have good heat transfer characteristic, whereas the heat transfer in closed feed water heater is less effective as the two streams are not allowed to be in direct contact. 3. In an open feed water heater, a separate pump is required for each heater to handle the feed water. However, closed feed water heaters do not require a separate pump for each heater because the extracted steam and feed water are at different pressures. In order to obtain best of these systems, most steam power plants use a combination of open and closed feed water heaters. Example 12.15. A Rankine cycle using steam operates with inlet condition of dry saturated steam at 10 bar and condenser pressure of 0.1 bar. Regenerative heating with a mixing type of heater is done at 2 bar. Determine the improvement in efficiency over the simple cycle. Also find the reduction in work per kg of steam circulated. Solution. Given: Inlet pressure of steam,

Vapour Power Cycles 341

Enthalpy

p1 = 10 bar Condenser pressure, p3 = 0.1 bar 1 Pressure at which regenerative heating is done, h1 p1 p2 = 2 bar Sa tu The h-s diagram of the cycle is shown in Fig. 12.25. cu rati h2 rv on From Mollier chart, we find that p2 2 e *h1 = 2780 kJ / kg; h2 = 2495 kJ / kg ; h3 = 2085 kJ / kg 0 h3 .9 p3 3 From steam tables, corresponding to a pressure of 2 bar, 0.7 we find that 9 hf 2 = 504.7 kJ / kg and corresponding to a pressure of 0.1 bar, we find that Entropy hf 3 = 191.8 kJ / kg Fig. 12.25 Improvement in efficiency over the simple cycle First of all, let us find the efficiency of the regenerative cycle. Let m be the mass of steam bled in kg, for regenerative heating. We know that for the energy balance of the feed water heater, Heat lost by bled steam = Heat gained by feed water m (h2 – hf 2) = (1 – m) (hf 2 – hf 3) m (2495 – 504.7) = (1 – m) (504.7 – 191.8) = (1 – m) 312.9 1990.3 m = 312.9 – 312.9 m 312.9 or m = = 0.136 kg 1990.3 + 312.9 \ Total workdone in the regenerative cycle = (h1 – h2) + (1 – m) (h2 – h3) = (2780 – 2495) + (1 – 0.136) (2495 – 2085) = 285 + 354 = 639 kJ / kg * The values of h1, h2 and h3 may be obtained from the steam tables as follows: From steam tables, corresponding to a pressure of 10 bar, we find that h1 = hg1 = 2776.2 kJ / kg; s1 = sg1 = 6.583 kJ / kg K Corresponding to a pressure of 2 bar, we find that hf 2 = 504.7 kJ / kg; hf g2 = 2201.6 kJ / kg; sf 2 = 1.53 kJ / kg K; sfg2 = 5.597 kJ / kg K and corresponding to a pressure of 0.1 bar, we find that hf 3 = 191.8 kJ / kg; hfg3 = 2392.9 kJ / kg; sf 3 = 0.649 kJ / kg K; sfg3 = 7.502 kJ / kg K Now let us find the dryness fraction of steam at point 2 (i.e. x2) and at point 3 (i.e. x3) We know that entropy at point 1 (s1) = Entropy at point 2 (s2) = sf 2 + x2 sf g2

6.583 = 1.53 + x2 × 5.597 or x2 =

6.583 − 1.53 = 0.9028 5.597

\ Enthalpy at point 2, h2 = hf 2 + x2 hf g2 = 504.7 + 0.9028 × 2201.6 = 2492.3 kJ / kg Similarly and

s1 = s3 = sf 3 + x3 sfg3 6.583 = 0.649 + x3 × 7.502 or x3 =

6.583 − 0.649 = 0.791 7.502

h3 = hf 3 + x3 hfg3 = 191.8 + 0.791 × 2392.9 = 2084.56 kJ / kg

342 Engineering Thermodynamics and total heat supplied per kg of water = h1 – hf 2 = 2780 – 504.7 = 2275.3 kJ / kg \ Efficiency of regenerative cycle Total workdone 639 = = = 0.281 or 28.1% Total heat supplied 2275.3 We know that workdone for simple Rankine cycle = h1 – h3 = 2780 – 2085 = 695 kJ / kg and heat supplied during Rankine cycle = h1 – hf 3 = 2780 – 191.8 = 2588.2 kJ / kg Workdone 695 = \ Efficiency of the Rankine cycle = = 0.2685 or 26.85% Heat supplied 2588.2

Enthalpy

We see that improvement in efficiency over simple Rankine cycle 0.281 − 0.2685 = = 0.0465 or 4.65% Ans. 0.2685 Reduction in work From above, we see that reduction in work per kg of steam for the regenerative cycle = 695 – 639 = 56 kJ / kg Ans. Example 12.16. In a single heater regenerative cycle, the steam enters the turbine at 30 bar and 400°C and the exhaust pressure is 0.1 bar. The feed water heater is a direct contact type which operates at 4 bar. Find the efficiency, the steam rate of the cycle and the mean temperature of heat addition. Neglect pump work. Solution. 1 400°C h1 Given: Pressure at which steam enters the turbine, 2 h2 p1 = 30 bar Sa tur p1 Temperature of steam, T1 = 400°C cur ation ve p2 Exhaust pressure, p3 = 0.1 bar h3 p3 3 Pressure at the feed water heater, 0.8 38 p2 = 4 bar The h-s diagram of the regenerative cycle is shown in Fig. 12.26. From the Mollier chart, we find that Entropy h1 = 3230 kJ / kg ; h2 = 2740 kJ / kg ; Fig. 12.26 h3 = 2195 kJ / kg ; x3 = 0.838 From steam tables, corresponding to a pressure of 4 bar, we find that hf 2 = 604.7 kJ / kg and corresponding to a pressure of 0.1 bar, we find that hf 3 = 191.8 kJ / kg Efficiency of the cycle First of all, let us find the steam bled from the turbine per kg of steam supplied. Let m be the mass of steam bled in kg. We know that for the energy balance of the feed water heater, Heat lost by bled steam = Heat gained by feed water m (h2 – hf 2) = (1 – m) (hf 2 – hf 3) m (2740 – 604.7) = (1 – m) (604.7 – 191.8)

Vapour Power Cycles 343 or \

2135.3 m = 412.9 – 412.9 m 412.9 m = = 0.162 kg 2135.3 + 412.9

and total workdone by the turbine per kg of feed water = (h1 – h2) + (1 – m) (h2 – h3) = (3230 – 2740) + (1 – 0.162) (2740 – 2195) = 490 + 456.7 = 946.7 kJ / kg We know that total heat supplied per kg of water = h1 – hf 2 = 3230 – 604.7 = 2625.3 kJ / kg \ Efficiency of the cycle, Total work done 946.7 = h = = 0.36 or 36% Ans. Total heat supplied 2625.3 Steam rate We know that steam rate in kg / kWh =

Heat equivalent to 1 kWh 3600 = Total work or power developed in kJ / kg 946.7

= 3.8 kg / kWh Ans. Mean temperature of heat addition From steam tables of superheated steam, corresponding to a pressure of 30 bar and a temperature of 400°C, we find that s1 = ssup = 6.925 kJ / kg K and from steam tables of dry saturated steam, corresponding to a pressure of 0.1 bar, Tsat = T3 = 45.83°C = 45.83 + 273 = 318.83 K; sf 3 = 0.6448 kJ / kg K Let T = Mean temperature of heat addition. We know that workdone = Change in temperature × Change in entropy = (T – T3 ) (s1 – sf 3) or 946.7 = (T – 318.83) (6.925 – 0.6448) = 6.2802 T – 2002.3 946.7 + 2002.3 \ T = = 469.6 K 6.2802 = 469.6 – 273 = 196.6°C Ans.

12.14 REGENERATIVE CYCLE WITH TWO FEED WATER HEATERS A schematic diagram of a regenerative cycle with two bleeding points at 2 and 3 is shown in Fig. 12.27. The T-s and h-s diagrams of the cycle is shown in Fig. 12.28 (a) and (b) respectively. Consider one kg of steam enters the turbine at a boiler pressure (say p1) as represented by point 1. It expands in the turbine isentropically to pressure p2 corresponding to point 2. At this point, a small quantity of steam say m1 kg is extracted and passed on to the feed water heater (No. I) for feed heating purposes. The rest of the steam i.e. (1 – m1) kg continues to expand isentropically to point 3. Again at point 3, a quantity of steam say m2 is extracted and passed to the feed water heater (No. II) for feed heating, as shown in Fig. 12.27. The remaining steam i.e. (1 – m1 – m2) kg of steam expands further isentropically upto point 4. The steam is now condensed in the condenser and leaves it as a saturated liquid at point 5 at the condenser pressure.

344 Engineering Thermodynamics 1 kg 1

h1

4

Turbine

h4 h2

2

h3

Boiler

(1 – m1 – m2)

3

Condenser m1 kg

m2 kg

8 5 hf 4

F.W.P. 1 kg F.W.P.

I

hf 2

hf 3 6 Feed water heaters

F.W.P.

II (1 – m1 – m2)

Fig. 12.27. Schematic diagram of regenerative cycle with two feed water heaters. 1 h1

1 1 kg

1 kg

6 5

2 (1 – m1) kg

m2 kg

3 (1 – m1 – m2) kg

(1 – m1 – m2) kg

Enthalpy

m1 kg

2

h2 p2

3

h3 p3 h4 p4

4

4

n tio ra tu rve Sa cu

Temperature

8 7

p1

Entropy

Entropy

(a) T-s diagram.

(b) h-s diagram.

Fig. 12.28

It is now pumped at the same pressure to feed water heater (No. II) and then pumped to feedwater heater (no. I) where it mixes up with the steam extracted from the turbine and finally pumped to the boiler. In the similar way as discussed for an open feed water (Art.12.11), we have for the heat balance of feed water heater No. I, Heat lost by steam bled at point 2 = Heat gained by feed water m1 (h2 – hf 2) = (1 – m1) (hf 2 – hf 3) hf 2 − hf 3 or m1 = h2 − h f 3 Similarly for the heat balance of feed water heater No. II, Heat lost by steam bled at point 3 = Heat gained by feed water m2 (h3 – hf 3) = (1 – m1 – m2) (hf 3 – hf 4)

Vapour Power Cycles 345 or

m2 =

(1 − m1 ) (h f 3 − h f 4 ) h3 − h f 4

We know that total workdone in the turbine per kg of feed water = Workdone in the turbine between 1 and 2 + Workdone in the turbine between 2 and 3 + Workdone in the turbine between 3 and 4 = (h1 – h2) + (1 – m1) (h2 – h3) + (1 – m1 – m2) (h3 – h4) and total heat supplied per kg of feed water = h1 – hf 2 \

Efficiency of the plant =

Total workdone Total heat supplied

Enthalpy

Example 12.17. In a steam turbine plant, the steam enters the turbine at 40 bar and 350°C. It exhausts to a condenser at 0.1 bar. Steam is extracted at 8 bar and 2 bar for the purpose of heating the feed water in two open feed water heaters. The feed water leaves each heater at the temperature of the condensing steam. The appropriate pumps are used for the water leaving the condenser and the two feed water heaters. Calculate the net workdone per kg of steam and thermal efficiency of the cycle. Solution. Given: Pressure of steam entering the turbine, 1 p1 = 40 bar h1 350° C Temperature of steam, T1 = 350°C Condenser pressure, p4 = 0.1 bar h2 Sa Pressure of steam bled at point 2, 2 tu x2 ar ra b tio p2 = 8 bar 40 n cu r a rv Pressure of steam bled at point 3, h3 e 8b x3 3 p3 = 2 bar ar The h-s diagram for the plant, as shown in Fig. 2b 12.29, is drawn as discussed below: h4 4 x4 bar First of all, mark point 1 corresponding to a 0.1 pressure of 40 bar and a temperature of 350°C. Now draw a vertical line 1-4 upto a pressure line of 0.1 bar, intersecting the lines of 8 bar and 2 bar at points Entropy 2 and 3 respectively, as shown in Fig. 12.29. Fig. 12.29 From the Mollier diagram, we find that h1 = 3090 kJ / kg; h2 = 2760 kJ/ kg ; x2 = 0.988 h3 = 2500 kJ/ kg; x3 = 0.907; h4 = 2090 kJ / kg; and x4 = 0.792 From steam tables, corresponding to a pressure of 8 bar, 2 bar and 0.1 bar respectively, we find that hf 2 = 720.9 kJ / kg; hf 3 = 504.7 kJ/ kg; hf 4 = 191.8 kJ/kg We know that mass of steam bled at point 2, h f 2 − h f 3 720.9 − 504.7 216.2 = = m1 = = 0.096 kg h2 − h f 3 2760 − 504.7 2255.3

346 Engineering Thermodynamics Similarly, mass of steam bled at point 2,

m2 =

(1 − m1 ) (h f 3 − h f 4 ) h3 − h f 4

=

(1 − 0.096) (504.7 − 191.8) 282.86 = 2500 − 191.8 2308.2

= 0.123 kg Net workdone per kg of steam We know that net workdone per kg of steam = (h1 – h2) + (1 – m1) (h2 – h3) + (1 – m1 – m2) (h3 – h4) = (3090 – 2760) + (1 – 0.096) (2760 – 2500) + (1 – 0.096 – 0.123) (2500 – 2090) = 330 + 235 + 320 = 885 kJ / kg Ans. Thermal efficiency of the cycle We know that total heat supplied per kg of feed water = h1 – hf 2 = 3090 – 720.9 = 2369.1 kJ / kg \ Thermal efficiency of the cycle,

h =

Total workdone 885 = = 0.3736 or 37.36% Ans. Total heat supplied 2369.1

HIGHLIGHTS 1. The efficiency the Carnot vapour cycle is given by T hCarnot = 1 – 3 T1 where T1 and T3 = Maximum and minimum temperature in the cycle. 2. In order to obtain the highest efficiency of the Carnot cycle, the heat supplied should be at the highest temperature and rejected at the lowest possible temperature. 3. Though the efficiency of the Carnot cycle is the maximum, yet it is not considered as a theoretical cycle for the steam power plant. 4. A steam engine or a steam turbine plant works on the Rankine cycle which is a modification of the Carnot cycle. 5. The Rankine cycle, as compared to Carnot cycle, has high work ratio. 6. In Rankine cycle using superheated steam, the dryness fraction of steam after isentropic expansion increases. In this case, the workdone increases but the specific steam consumption decreases. 7. In a modified Rankine cycle, the stroke length is reduced and hence the size of the cylinder is reduced. Thus, the modified Rankine cycle is used in steam engines while the steam turbines work on Rankine cycle. 8. The efficiency of the Rankine cycle may be improved by the following methods: (a) By decreasing the average temperature at which heat is rejected from the working fluid in the condenser. This is done by lowering the operating pressure of the condenser.

Vapour Power Cycles 347 (b) By increasing the average temperature at which heat is added to the steam. This is done by increasing the operating pressure of the boiler. Though this method increases the thermal efficiency of the Rankine cycle, but it also increases the moisture contents of the steam to unacceptable levels. In order to overcome this effect, the steam in the turbine is expanded in two stages and reheated in between them. 9. The reheating of steam has the following advantages in addition to the increase in efficiency of the turbine: (a) The dryness fraction of steam leaving the turbine increases. Thus erosion of turbine blades reduces. (b) The workdone through the turbine increases. (c) There is a reduction in specific steam consumption and thus the amount of water required in the condenser is reduced. 10. The method of extracting (or bleeding) a small amount of steam from the turbine, at certain points during its expansion, and using this steam for heating the feed water to the saturation temperature corresponding to the boiler pressure before it enters the boiler, is called regenerative feed heating.

EXERCISES 1. A steam engine uses steam at 10 bar and 0.9 dry and exhausts at 1.1 bar. Determine the dryness fraction at the end of isentropic expansion and the efficiency of the Rankine cycle. [Ans. 0.8; 16.1%] 2. Dry and saturated steam at a pressure of 11 bar is supplied to a turbine and is expanded isentropically to a pressure of 1 bar. Calculate: (a) The heat supplied; (b) The heat rejected; and (c) Theoretical thermal efficiency. [Ans. 2362 kJ / kg; 1955.3 kJ / kg; 17.2%] 3. A steam power plant works between 30 bar and 40 kPa. The steam supplied to the turbine is dry saturated. Calculate the cycle efficiency and steam consumption, if the cycle used is Rankine cycle. [Ans. 27.56%; 5.256 kg / kWh] 4. Compare the values of Rankine cycle and Carnot cycle efficiencies, if the maximum and minimum temperatures are 400°C and 40°C. The steam in case of Rankine cycle is supplied at 20 bar. [Ans. 33.4%; 53.5%] 5. In an ideal Rankine cycle, the steam condition at the turbine inlet is 20 bar and 350°C. The condenser pressure is 0.08 bar. Determine the cycle efficiency. If the steam flow rate is 2000 kg / h, what is the power output in kW? [Ans. 32.4%, 533.6 kW] 6. A steam power plant operates between a boiler pressure of 4 MPa and 300°C and a condenser pressure of 50 kPa. Determine the thermal efficiency of the cycle, the work ratio and the specific steam flow rate, assuming the cycle to be a simple ideal Rankine cycle. [Ans. 28.76% ; 0.9946; 4.774 kg / kWh] 7. A steam turbine receives steam at a pressure of 20 bar and superheated by 87.6°C. The exhaust pressure is 0.07 bar and expansion of steam takes place isentropically. Using steam tables, calculate the following: (a) Heat supplied assuming that heat pump supplies water to the boiler at 20 bar; (b) Heat rejected; (c) Net workdone; (d) Workdone by the turbine; (e) Thermal efficiency; and ( f) Theoretical steam consumption. If the actual steam consumption is 5 kg / kWh, what is the efficiency ratio of the turbine? [Ans. 2861.6 kJ / kg ; 1934.4 kJ / kg ; 920.2 kJ / kg ; 922.2 kJ / kg ; 31.14% ; 3.9 kg / kWh ; 0.78]

348 Engineering Thermodynamics 8. A steam engine uses dry saturated steam at a pressure of 10 bar and the back pressure is 0.7 bar. The pressure at release is 3.8 bar. Assuming the pressure drop to take place at constant volume, find the efficiency of the modified Rankine cycle. Neglect clearance. [Ans. 13.27%] 9. Steam at 17 bar and 250°C is supplied to a steam engine where it is expanded isentropically to a release pressure of 3.5 bar. The steam is released from the engine at constant volume into the condenser where it is condensed at a pressure of 0.5 bar and the condensate is pumped back to the boiler. The steam flow rate is 1200 kg / h. Neglecting pump work, determine with the help of steam tables, (a) the power output of the engine in kW; and (b) the efficiency of the cycle. [Ans. 153.44 kW; 17.88%] 10. Neglecting pump work, calculate heat supplied, workdone and thermal efficiency of a steam power plant in which steam is supplied at 40 bar and 300°C. The condenser pressure is 0.2 bar. The isentropic efficiency of the turbine is 90%. Calculate the quantity of steam required in kg per hour for the plant to develop 800 kW. Draw the cycle in a h-s plane. [Ans. 2718.5 kJ/kg; 778.5 kJ/kg; 28.64%; 3699.4 kg/h] 11. Find the ideal cycle efficiency and steam consumption of a reheat cycle operating between pressures of 30 bar and 0.04 bar with initial temperature of 450°C. Assume the first expansion to be carried out to the point where steam is dry saturated and the steam is reheated to the original temperature of 450°C. The pump work may be neglected. [Ans. 38.4% ; 2.416 kg / kWh] 12. In a reheat steam power cycle, steam at 500°C expands in a high pressure turbine till it is saturated vapour. It is reheated at constant pressure to 500°C and then expands to a low pressure turbine to 40°C. If the moisture content at the turbine exhaust is limited to 15%, assuming all ideal processes, calculate reheat pressure, steam pressure at inlet to the high pressure turbine and cycle efficiency. [Ans. 40.4 bar; 225 bar; 44.14%] 13. In a regenerative cycle with one feed water heater, dry saturated steam is supplied from the boiler at a pressure of 30 bar. It is condensed in the condenser at a pressure of 1 bar. The steam is bled from the turbine at a pressure of 5 bar and enters the open feed water heater. Determine the amount of bled steam per kg of steam supplied and the efficiency of the cycle. Find the improvement in efficiency due to regeneration in percentage, over the simple Rankine cycle. [Ans. 0.1074 kg ; 24.48% ; 5.06%] 14. A steam power plant is operating on the ideal regenerative cycle with one feed water heater. Steam enters the turbine at 30 bar and 400°C and is condensed in the condenser at a pressure of 0.1 bar. Some quantity of steam leaves the turbine at a pressure of 6 bar and enters the open feed water heater. Determine the fraction of the steam extracted from the turbine and thermal efficiency of the cycle. [Ans. 0.1814 kg; 36.1%] 15. A steam power plant operates on an ideal regenerative cycle utilising steam as the working fluid and two open feed water heaters. Steam enters the turbine at a pressure of 50 bar and a temperature of 400°C. It exhausts to the condenser at a pressure of 0.1 bar. Steam is bled to the two open feed water heaters at a pressure of 5 bar and 0.5 bar. The feed water is heated in each heater to the saturation temperature of the bled steam. The condensate is also pumped at this temperature into the feed line immediately after the heater. Find: (a) the masses of the steam bled in the turbine per kg of steam entering the turbine; (b) the net workdone per kg of steam; and (c) the thermal efficiency of the cycle. [Ans. 0.128 kg, 0.061 kg; 999.23 kJ / kg; 39.11%]

QUESTIONS 1. Give reasons why Carnot cycle can not be considered as a theoretical cycle for steam power plant even though its efficiency is maximum. 2. Draw the Rankine cycle on T-s diagram using dry saturated steam and obtain an expression for the Rankine cycle efficiency.

Vapour Power Cycles 349 3. Explain why the Rankine cycle rather than the Carnot cycle is used as a standard of reference for the performance of steam power plants. 4. Describe briefly the Rankine cycle using superheated steam. 5. Why is modified Rankine cycle adopted for reciprocating steam engines? Derive an expression for the efficiency of the modified Rankine cycle. 6. What are the methods employed to increase the efficiency of the Rankine cycle? 7. Draw the layout of a steam turbine plant using intermediate reheating of steam. Show the corresponding thermodynamic cycle on h-s diagram. Briefly discuss the advantages of reheating. 8. Explain briefly the advantages and limitations of using regenerative cycle in the modification to Rankine cycle. 9. Derive an expression for the efficiency of a regenerative cycle with one open feed water heater. 10. Derive an expression for the efficiency of a regenerative cycle with two open feed water heaters.

OBJECTIVE TYPE QUESTIONS 1. In a Carnot vapour cycle, (a) the outlet condition of wet steam from the condenser cannot be controlled (b) the work ratio is low (c) the thermal efficiency is greatly affected by the temperature at which heat is supplied to the working fluid (d) all of the above 2. A steam engine works on (a) Carnot cycle (b) Rankine cycle (c) Joule cycle (d) Otto cycle 3. The efficiency of Rankine cycle is .................... the Carnot cycle (a) equal to (b) greater than (c) less than 4. In a Rankine cycle with superheated steam, (a) the workdone increases (b) the dryness fraction of steam after isentropic expansion increases (c) the specific steam consumption decreases (d) all of the above 5. The modified Rankine cycle is adopted in the operation of (a) steam engines (b) steam turbines (c) gas turbines (d) none of these 6. The efficiency of the Rankine cycle can be increased by (a) decreasing the maximum pressure of the cycle (b) increasing the temperature at which heat is rejected (c) increasing the exhaust pressure of the condenser (d) lowering the exhaust pressure 7. In a steam power plant, the effect of lowering the exhaust pressure is to .................... the thermal efficiency and also the moisture content of the steam leaving the turbine. (a) decrease (b) increase

350 Engineering Thermodynamics 8. The effect of superheating the steam at the turbine inlet without increase in maximum pressure is (a) to decrease both thermal efficiency and quality of steam (b) to decrease thermal efficiency but to increase quality of steam (c) to increase thermal efficiency but to decrease quality of steam (d) to increase both thermal efficiency and quality of steam 9. The reheating of steam in a turbine (a) increases the workdone through the turbine (b) reduces erosion of the turbine blades (c) increases the thermal efficiency of the turbine (d) all of the above 10. The process of extracting or bleeding of a small amount of steam from the turbine at certain points during its expansion, and using this steam for heating the feed water to the saturation temperature corresponding to the boiler before it enters the boiler, is known as (a) reheating (b) bleeding (c) regenerative feed heating (d) none of these

ANSWERS

1. (d) 6. (d)

2. (b) 7. (b)

3. (c) 8. (d)

4. (d) 9. (d)

5. (a) 10. (c)

PROPERTIES OF STEAM

Properties of Saturated Water and Steam 353 Table 1: Properties of Saturated Water and Steam (Pressure based) Pressure in bar

Saturation temperature in °C

(p)

(t )

0.00611

0.000

Specific volume in m3/kg Water (vf )

Steam (vg )

Specific enthalpy in kJ/kg

Water (hf )

Specific entropy in kJ / kg K

Evaporation (hfg )

Steam (hg )

Water (sf )

Evaporation (sfg )

Steam (sg )

0.001 000

206.16

0.0

2501.6

2501.6

0.000

9.158

9.158

0.02

17.51

0.001 001

67.012

73.5

2460.2

2533.7

0.261

8.464

8.725

0.04

28.98

0.001 004

34.803

121.4

2433.1

2554.5

0.423

8.053

8.476

0.06

36.18

0.001 006

23.741

151.5

2416.0

2567.5

0.521

7.810

8.331

0.08

41.54

0.001 008

18.104

173.9

2403.2

2577.1

0.593

7.637

8.230

0.10

45.83

0.001 010

14.674

191.8

2392.9

2584.8

0.649

7.502

8.151

0.12

49.45

0.001 012

12.361

206.9

2384.2

2591.2

0.696

7.391

8.087

0.14

52.58

0.001 013

10.693

220.0

2376.7

2596.7

0.737

7.297

8.033

0.16

55.34

0.001 015

9.4324

231.6

2370.0

2601.6

0.772

7.215

7.987

0.18

57.83

0.001 016

8.4446

242.0

2363.9

2605.9

0.804

7.142

7.946

0.20

60.09

0.001 017

7.6492

251.5

2358.4

2609.9

0.832

7.077

7.909

0.25

64.99

0.001 020

6.2040

272.0

2346.4

2618.4

0.893

6.939

7.832

0.30

69.13

0.001 022

5.229

289.3

2336.1

2625.4

0.944

6.825

7.769

0.35

72.71

0.001 025

4.5255

304.3

2327.2

2631.5

0.988

6.729

7.717

0.40

75.89

0.001 027

3.9932

317.7

2319.2

2636.9

1.026

6.645

7.671

0.45

78.74

0.001 028

3.5761

329.6

2312.0

2641.6

1.060

6.570

7.630

0.50

81.35

0.001 030

3.2401

340.6

2305.4

2646.0

1.091

6.504

7.595

0.60

85.95

0.001 033

2.7317

359.9

2293.6

2653.5

1.146

6.387

7.533

0.70

89.96

0.001 036

2.3647

376.8

2283.3

2660.1

1.192

6.288

7.480

0.80

93.51

0.001 039

2.0869

391.7

2274.0

2665.7

1.233

6.202

7.435

0.90

96.71

0.001 041

1.8691

405.2

2265.6

2670.9

1.270

6.126

7.395

1.00

99.63

0.001 043

1.6937

417.5

2257.9

2675.4

1.303

6.057

7.360

1.013

100.00

0.001 044

1.6730

419.1

2256.9

2676.0

1.307

6.048

7.355

1.20

104.8

0.001 048

1.4281

439.4

2244.1

2683.4

1.361

5.937

7.298

1.40

109.3

0.001 051

1.2363

458.4

2231.9

2690.3

1.411

5.836

7.247

1.60

113.3

0.001 055

1.0911

475.4

2220.8

2696.2

1.455

5.747

7.202

1.80

116.9

0.001 058

0.9772

490.7

2210.8

2701.5

1.494

5.668

7.162

2.00

120.2

0.001 061

0.8854

504.7

2201.6

2706.3

1.530

5.597

7.127

2.50

127.4

0.001 068

0.7184

535.4

2181.0

2716.4

1.607

5.445

7.052

3.00

133.5

0.001 074

0.6055

561.4

2163.2

2724.7

1.672

5.319

6.991

3.50

138.9

0.001 079

0.5240

584.3

2147.3

2731.6

1.727

5.212

6.939

4.0

143.6

0.001 084

0.4622

604.7

2132.9

2737.6

1.776

5.118

6.894

4.5

147.9

0.001 089

0.4137

623.2

2119.7

2742.9

1.820

5.034

6.855

5.0

151.9

0.001 093

0.3747

640.1

2107.4

2747.5

1.860

4.959

6.819

6.0

158.8

0.001 101

0.3155

670.4

2085.1

2755.5

1.931

4.827

6.758

7.0

165.0

0.001 108

0.2727

697.1

2064.9

2762.0

1.992

4.713

6.705

8.0

170.4

0.001 115

0.2403

720.9

2046.6

2767.5

2.046

4.614

6.660

9.0

175.4

0.001 121

0.2148

742.6

2029.5

2772.1

2.094

4.525

6.619

10.0

179.9

0.001 127

0.1943

762.6

2013.6

2776.2

2.138

4.445

6.583

354 Engineering Thermodynamics p

t

vf

vg

hf

hf g

hg

sf

sf g

sg

12.0

188.0

0.001 139

0.1632

798.4

1984.3

2782.7

2.216

4.303

6.519

14.0

195.0

0.001 149

0.1407

830.1

1957.7

2787.8

2.284

4.181

6.465

16.0

201.4

0.001 159

0.1237

858.5

1933.2

2791.7

2.344

4.074

6.418

18.0

207.1

0.001 168

0.1103

884.5

1910.3

2794.8

2.398

3.978

6.375

20.0

212.4

0.001 177

0.0995

908.6

1888.7

2797.3

2.447

3.890

6.337

22.0

217.2

0.001 185

0.0907

930.9

1868.2

2799.1

2.492

3.809

6.301

24.0

221.8

0.001 193

0.0832

951.9

1848.5

2800.4

2.534

3.735

6.269

26.0

226.0

0.001 201

0.0768

971.7

1829.7

2801.4

2.574

3.665

6.239

28.0

230.0

0.001 209

0.0714

990.5

1811.5

2802.0

2.611

3.600

6.211

30.0

233.8

0.001 216

0.0666

1008.3

1794.0

2802.3

2.646

3.538

6.184

32.0

237.4

0.001 224

0.0624

1025.4

1776.9

2802.3

2.679

3.480

6.159

34.0

240.9

0.001 231

0.0587

1041.8

1760.3

2802.1

2.710

3.424

6.134

36.0

244.2

0.001 238

0.0554

1057.5

1744.2

2801.7

2.740

3.371

6.111

38.0

247.3

0.001 245

0.0524

1072.7

1728.4

2801.1

2.769

3.321

6.090

40.0

250.3

0.001 252

0.0497

1087.4

1712.9

2800.3

2.797

3.272

6.069

42.0

253.2

0.001 259

0.0473

1101.6

1697.8

2799.4

2.823

3.225

6.048

44.0

256.1

0.001 266

0.0451

1115.4

1682.9

2798.3

2.849

3.180

6.029

46.0

258.8

0.001 273

0.0430

1128.8

1668.3

2797.1

2.874

3.136

6.010

48.0

261.4

0.001 279

0.0411

1141.8

1653.9

2795.7

2.897

3.094

5.991

50.0

263.9

0.001 286

0.0394

1154.5

1639.7

2794.2

2.921

3.053

5.974

52.0

266.4

0.001 293

0.0378

1166.9

1625.7

2792.6

2.943

3.013

5.956

54.0

268.8

0.001 300

0.0363

1179.0

1611.8

2790.8

2.965

2.974

5.939

56.0

271.1

0.001 306

0.0349

1190.8

1598.2

2789.0

2.986

2.936

5.923

58.0

273.4

0.001 312

0.0336

1202.4

1584.6

2787.0

3.007

2.899

5.906

60.0

275.6

0.001 319

0.0324

1213.7

1571.3

2785.0

3.027

2.863

5.891

62.0

277.7

0.001 325

0.0313

1224.9

1558.0

2782.9

3.047

2.828

5.875

64.0

279.8

0.001 332

0.0302

1235.8

1544.8

2780.6

3.066

2.794

5.860

66.0

281.9

0.001 338

0.0292

1246.5

1531.8

2778.3

3.085

2.760

5.845

68.0

283.9

0.001 345

0.0283

1257.1

1518.8

2775.9

3.104

2.727

5.831

70.0

285.8

0.001 351

0.0274

1267.5

1506.0

2773.4

3.122

2.694

5.816

72.0

287.7

0.001 358

0.0265

1277.7

1493.2

2770.9

3.140

2.662

5.802

74.0

289.6

0.001 365

0.0257

1287.8

1480.5

2768.2

3.157

2.631

5.788

76.0

291.4

0.001 371

0.0249

1297.7

1467.8

2765.5

3.174

2.600

5.774

78.0

293.2

0.001 378

0.0242

1307.5

1455.3

2762.7

3.191

2.569

5.760

80.0

295.0

0.001 384

0.0235

1317.2

1442.7

2759.9

3.208

2.539

5.747

82.0

296.7

0.001 391

0.0228

1326.7

1430.3

2757.0

3.224

2.510

5.734

84.0

298.4

0.001 398

0.0222

1336.2

1417.8

2754.0

3.240

2.481

5.721

86.0

300.1

0.001 404

0.0216

1345.4

1405.5

2750.9

3.256

2.452

5.708

88.0

301.7

0.001 411

0.0210

1354.7

1393.1

2747.8

3.271

2.423

5.694

90.0

303.3

0.001 418

0.0205

1363.8

1380.8

2744.6

3.287

2.395

5.682

92.0

304.9

0.001 425

0.0200

1372.8

1368.5

2741.3

3.302

2.367

5.669

94.0

306.5

0.001 432

0.0195

1381.7

1356.3

2738.0

3.317

2.340

5.657

96.0

308.0

0.001 439

0.0190

1390.6

1344.1

2734.7

3.332

2.313

5.645

98.0

309.5

0.001 446

0.0185

1399.4

1331.9

2731.3

3.346

2.286

5.632

Properties of Saturated Water and Steam 355 p

t

vf

vg

hf

hf g

hg

sf

sf g

sg

100.0

311.0

0.001 453

0.0180

1408.1

1319.7

2727.8

3.361

2.259

5.620

110.0

318.0

0.001 489

0.0160

1450.6

1258.8

2709.4

3.431

2.129

5.560

120.0

324.6

0.001 527

0.0143

1491.7

1197.5

2698.2

3.497

2.003

5.500

130.0

330.8

0.001 567

0.0128

1531.9

1135.1

2667.0

3.561

1.880

5.441

140.0

336.6

0.001 611

0.0115

1571.5

1070.9

2642.4

3.624

1.756

5.380

150.0

342.1

0.001 658

0.0103

1610.9

1004.2

2615.1

3.686

1.632

5.318

160.0

347.3

0.001 710

0.0093

1650.4

934.5

2584.9

3.747

1.506

5.253

170.0

352.3

0.001 770

0.0084

1691.6

860.0

2551.6

3.811

1.375

5.186

180.0

357.0

0.001 840

0.0075

1734.8

779.0

2513.8

3.877

1.236

5.113

190.0

361.4

0.001 926

0.0067

1778.7

691.8

2470.5

3.943

1.090

5.033

200.0

365.7

0.002 037

0.0059

1826.6

591.6

2418.2

4.015

0.926

4.941

210.0

369.8

0.002 202

0.0050

1886.3

461.2

2347.5

4.105

0.717

4.822

220.0

373.7

0.002 668

0.0037

2010.3

186.3

2196.6

4.293

0.288

4.581

221.2

374.15

0.003 170

0.0032

2107.4

000.0

2107.4

4.443

0.000

4.443

356 Engineering Thermodynamics Table 2: Properties of Saturated Water and Steam (Temperature based) Saturation temperature in °C

Pressure in bar

(t )

(p) 0

0.006 11

1

0.006 57

2

0.007 06

3

0.007 58

4

0.008 13

5

0.008 72

6

0.009 35

7

0.010 01

8

0.010 72

9

0.011 47

10

0.012 27

11

0.013 12

12

0.014 01

13

0.014 97

14

0.015 97

15

0.017 04

16

0.018 17

17

0.019 36

18

0.020 62

19

Specific volume

Water (vf )

Steam (vg )

Specific enthalpy in kJ/kg

Water (hf )

Specific entropy in kJ / kg K

Evaporation (hfg )

Steam (hg )

Water (sf )

Evaporation (sfg )

Steam (sg )

206.16

0.0

2501.6

2501.6

0.000

9.158

9.158

192.61

4.2

2499.2

2503.4

0.015

9.116

9.131

179.92

8.4

2496.8

2505.2

0.031

9.074

9.105

168.17

12.6

2494.5

2507.1

0.046

9.033

9.079

157.27

16.8

2492.1

2508.9

0.061

8.992

9.053

147.16

21.0

2489.7

2510.7

0.076

8.951

9.027

137.78

25.2

2487.4

2512.6

0.091

8.910

9.001

129.06

29.4

2485.0

2514.4

0.106

8.870

8.976

120.97

33.6

2482.6

2516.2

0.121

8.830

8.951

113.44

37.8

2480.3

2518.1

0.136

8.790

8.926

106.43

42.0

2477.9

2519.9

0.151

8.751

8.902

99.909

46.2

2475.5

2521.7

0.166

8.712

8.878

93.835

50.4

2473.2

2523.6

0.181

8.673

8.854

88.176

54.6

2470.8

2525.4

0.195

8.635

8.830

82.900

58.8

2468.5

2527.2

0.210

8.596

8.806

77.978

62.9

2466.1

2529.1

0.224

8.558

8.782

73.384

67.1

2463.8

2530.9

0.239

8.520

8.759

69.095

71.3

2461.4

2532.7

0.253

8.483

8.736

65.087

75.5

2459.0

2534.5

0.268

8.446

8.714

0.021 96

0.001 000

61.341

79.7

2456.7

2536.4

0.282

8.409

8.691

20

0.023 37

0.001 002

57.838

83.9

2454.3

2538.2

0.296

8.372

8.668

22

0.026 42

0.001 002

51.492

92.2

2449.6

2541.8

0.325

8.299

8.624

24

0.029 82

0.001 003

45.926

100.6

2444.9

2545.5

0.353

8.228

8.581

26

0.033 60

0.001 003

41.034

108.9

2440.2

2549.1

0.381

8.157

8.538

28

0.037 78

0.001 004

36.728

117.3

2435.4

2552.7

0.409

8.087

8.496

30

0.042 42

0.001 004

32.929

125.7

2430.7

2556.4

0.437

8.018

8.455

32

0.047 53

0.001 005

29.572

134.1

2425.9

2560.0

0.464

7.950

8.414

34

0.053 18

0.001 006

26.601

142.4

2421.2

2563.6

0.491

7.883

8.374

36

0.059 40

0.001 006

23.967

150.7

2416.4

2567.1

0.518

7.816

8.334

38

0.066 24

0.001 007

21.627

159.1

2411.7

2570.8

0.545

7.751

8.296

40

0.073 75

0.001 008

19.546

167.5

2406.9

2574.4

0.572

7.686

8.258

42

0.081 99

0.001 009

17.692

175.8

2402.1

2577.9

0.599

7.622

8.221

44

0.091 00

0.001 009

16.036

184.2

2397.3

2581.5

0.625

7.559

8.184

46

0.1009

0.001 010

14.557

192.6

2392.5

2585.1

0.651

7.497

8.148

48

0.1116

0.001 011

13.233

200.9

2387.7

2588.6

0.678

7.435

8.113

50

0.1234

0.001 012

12.046

209.3

2382.9

2592.2

0.704

7.374

8.078

52

0.1361

0.001 013

10.890

217.6

2378.1

2595.7

0.729

7.314

8.043

54

0.1500

0.001 014

10.022

226.0

2373.2

2599.2

0.755

7.254

8.009

56

0.1651

0.001 015

234.4

2368.4

2602.8

0.780

7.196

7.976

9.1587

Properties of Saturated Water and Steam 357 t

p

vf

vg

hf

hf g

hg

sf

sf g

sg

58

0.1815

0.001 016

8.3808

242.7

2363.5

2606.2

0.806

7.137

7.943

60

0.1992

0.001 017

7.6785

251.1

2358.6

2609.7

0.831

7.080

7.911

62

0.2184

0.001 018

7.0437

259.5

2353.7

2613.2

0.856

7.023

7.879

64

0.2391

0.001 019

6.4690

267.8

2348.8

2616.6

0.881

6.967

7.848

66

0.2615

0.001 020

5.9482

276.2

2343.9

2620.1

0.906

6.911

7.817

68

0.2856

0.001 022

5.4756

284.6

2338.9

2623.5

0.930

6.856

7.786

70

0.3116

0.001 023

5.0463

293.0

2334.0

2627.0

0.955

6.802

7.757

72

0.3396

0.001 024

4.6557

301.4

2329.0

2630.3

0.979

6.748

7.727

74

0.3696

0.001 025

4.3000

309.7

2324.0

2633.7

1.003

6.695

7.698

76

0.4019

0.001 027

3.9757

318.2

2318.9

2637.1

1.027

6.642

7.669

78

0.4365

0.001 028

3.6796

326.5

2313.9

2640.4

1.051

6.590

7.641

80

0.4736

0.001 029

3.4091

334.9

2308.8

2643.7

1.075

6.538

7.613

82

0.5133

0.001 031

3.1616

343.3

2303.8

2647.1

1.099

6.487

7.586

84

0.5557

0.001 032

2.9350

351.7

2298.6

2650.4

1.123

6.436

7.559

86

0.6011

0.001 033

2.7272

360.1

2293.5

2653.6

1.146

6.386

7.532

88

0.6495

0.001 035

2.5365

368.5

2288.4

2656.9

1.169

6.337

7.506

90

0.7011

0.001 036

2.3613

376.9

2283.2

2660.1

1.193

6.287

7.480

92

0.7561

0.001 038

2.2002

385.4

2278.0

2663.4

1.216

6.239

7.454

94

0.8146

0.001 039

2.0519

393.8

2272.8

2666.6

1.239

6.190

7.429

96

0.8796

0.001 041

1.9153

402.2

2267.5

2669.7

1.261

6.143

7.404

98

0.9430

0.001 042

1.7893

410.6

2262.2

2672.8

1.285

6.095

7.380

100

1.013

0.001 044

1.6730

419.1

2256.9

2676.0

1.307

6.048

7.355

110

1.433

0.001 052

1.2099

461.3

2230.0

2691.3

1.419

5.820

7.239

120

1.985

0.001 061

0.891 52

503.7

2202.3

2706.0

1.528

5.601

7.129

130

2.701

0.001 070

0.668 14

546.3

2173.6

2719.9

1.634

5.392

7.026

140

3.614

0.001 080

0.508 50

589.1

2144.0

2733.1

1.739

5.189

6.928

150

4.760

0.001 091

0.392 45

632.1

2113.2

2745.3

1.842

4.994

6.836

160

6.181

0.001 102

0.306 76

675.5

2081.3

2756.7

1.943

4.805

6.748

170

7.920

0.001 114

0.242 55

719.2

2047.9

2767.1

2.042

4.621

6.663

180

10.027

0.001 128

0.193 80

763.1

2013.2

2776.3

2.139

4.443

6.582

190

12.551

0.001 142

0.156 32

807.5

1976.7

2784.2

2.236

4.268

6.504

200

15.549

0.001 156

0.127 16

852.3

1938.6

2790.9

2.331

4.097

6.428

210

19.077

0.001 172

0.104 24

897.7

1898.5

2796.2

2.425

3.929

6.354

220

23.198

0.001 190

0.086 04

943.7

1856.2

2799.9

2.518

3.764

6.282

230

27.976

0.001 209

0.071 45

990.3

1811.7

2802.0

2.610

3.601

6.211

240

33.478

0.001 229

0.059 65

1037.6

1764.6

2802.2

2.702

3.439

6.141

250

39.776

0.001 251

0.050 04

1085.8

1714.7

2800.5

2.794

3.277

6.071

260

46.943

0.001 276

0.042 13

1134.9

1661.5

2796.4

2.885

3.116

6.001

270

55.058

0.001 303

0.035 59

1185.2

1604.6

2789.8

2.976

2.954

5.930

280

64.202

0.001 332

0.030 13

1236.8

1543.6

2780.4

3.068

2.790

5.858

290

74.461

0.001 366

0.025 54

1290.0

1477.6

2767.6

3.161

2.624

5.785

300

85.927

0.001 404

0.021 65

1345.1

1406.0

2751.1

3.255

2.453

5.708

310

98.700

0.001 448

0.018 33

1402.4

1327.6

2730.0

3.351

2.277

5.628

320

112.89

0.001 500

0.015 48

1462.6

1241.1

2703.7

3.450

2.092

5.542

358 Engineering Thermodynamics p

t

vf

vg

hf

hf g

hg

sf

sf g

sg

330

128.63

0.001 562

0.012 99

1526.5

1143.6

2670.1

3.553

1.896

5.449

340

146.05

0.001 639

0.010 78

1595.5

1030.7

2626.2

3.662

1.681

5.343

350

165.35

0.001 741

0.008 80

1671.9

895.7

2567.6

3.780

1.438

5.218

360

186.75

0.001 896

0.006 94

1764.2

721.3

2485.5

3.921

1.139

5.060

370

210.54

0.002 214

0.004 97

1890.2

452.6

2342.8

4.111

0.704

4.814

374.15

221.20

0.003 170

0.003 17

2107.4

0.0

2107.4

4.443

0.000

4.443

Saturation temperature (in °C) t

6.983

32.9

45.8

54.0

60.1

69.1

Pressure (in bar) p

0.01

0.05

0.1

0.15

0.2

0.3

8.688

13.06

8.502

9.748

8.368

6.493

17.20 2688 8.447

11.51 2688 8.261

8.585 2686 8.126

5.714 2685 7.936

v h s

v h s

v h s

v h s

8.179

2781

2782

2783

2783

19.51

9.008

2784

39.04

2688

34.42

v

9.751

8.768

9.512

s

2784

195.3

s

2689

150

h

172.2

v

100

h

Property of superheated steam

8.396

2879

7.268

8.584

2879

10.91

8.718

2880

14.61

8.903

2880

21.23

9.223

2880

43.66

9.966

2880

218.4

200

8.593

2977

8.040

8.781

2977

12.07

8.915

2977

16.16

9.100

2977

24.14

9.420

2978

48.28

10.163

2978

241.4

250

8.774

3076

8.811

8.962

3076

13.22

9.096

3077

17.71

9.281

3077

26.45

9.601

3077

52.90

10.344

3077

264.5

300

8.943

3177

9.581

9.130

3177

14.37

9.264

3177

19.25

9.449

3177

28.76

9.769

3177

57.51

10.512

3178

287.6

350

9.101

3279

10.35

9.288

3279

15.53

9.422

3280

20.80

9.607

3280

31.06

9.927

3280

62.13

10.670

3280

310.7

400

Temperature of superheated steam in °C

9.391

3489

11.89

9.578

3489

17.84

9.712

3489

23.89

9.897

3489

35.68

10.217

3489

71.36

10.960

3489

356.8

500

9.654

3706

13.43

9.842

3706

20.15

9.975

3706

26.98

10.160

3706

40.30

10.480

3706

80.59

11.223

3706

403.0

600

Specific volume (v ) in m3 /kg; Specific enthalpy (h) in kJ/ kg; Specific entropy(s) in kJ/kg K

Table 3: Properties of Superheated Steam

9.897

3929

14.70

10.084

3929

22.45

10.217

3929

30.07

10.402

3929

44.91

10.722

3929

89.82

11.465

3929

449.1

700

10.121

4159

16.51

10.309

4159

24.76

10.442

4159

33.16

10.628

4159

49.53

10.947

4159

99.06

11.690

4159

495.3

800

Properties of Superheated Steam 359

t

75.9

81.3

99.6

120.2

133.5

143.6

151.8

p

0.4

0.5

1.0

2.0

3.0

4.0

5.0

8.045

3.890

7.801

3.420 2682 7.695

1.696

s

v h s

v

7.078

0.4710

6.930

–

– – –

– – –

– – –

– – –

v h s

v h s

v h s

v h s

2777

s

–

–

2753

2762

0.9602

7.280

2770

0.9602

7.614

2676 7.360

h

1.937

7.940

2780

2781

2684

4.866

4.279

v

150

h

100

7.060

2827

0.4252

7.172

2862

0.5345

7.312

2866

0.7166

7.507

2871

1.081

7.834

2876

2.173

8.158

2878

4.356

8.263

2879

5.448

200

7.271

2962

0.4745

7.380

2965

0.5953

7.517

2968

0.7965

7.708

2971

1.199

8.033

2976

2.406

8.355

2976

4.821

8.460

2977

6.028

250

7.460

3065

0.5226

7.566

3067

0.6550

7.702

3071

0.8754

7.892

3072

1.316

8.215

3075

2.639

8.537

3076

5.284

8.641

3076

6.607

300

7.633

3168

0.5701

7.738

3170

0.7140

7.873

3174

0.9536

8.062

3174

1.433

8.384

3176

2.871

8.705

3177

5.747

8.810

3177

7.185

350

7.793

3272

0.6172

7.898

3274

0.7725

8.032

3277

1.031

8.221

3277

1.549

8.543

3278

3.103

8.864

3279

6.209

8.9v68

3279

7.763

500

8.087

3484

0.7108

8.191

3485

0.8893

8.324

3487

1.187

8.513

3487

1.781

8.834

3488

3.565

9.154

3489

7.134

9.258

3489

8.918

500

8.351

3702

0.8040

8.455

3703

1.005

8.590

3704

1.341

8.776

3704

2.013

9.097

3705

4.028

9.417

3705

8.058

9.522

3706

10.07

600

8.595

3926

0.8969

8.698

3927

1.121

8.831

3928

1.496

9.020

3928

2.224

9.939

3928

4.490

9.660

3929

8.981

9.764

3929

11.23

700

8.820

4157

0.9896

8.924

4157

1.1237

9.057

4158

1.650

9.244

4158

2.475

9.565

4159

4.952

9.885

4159

9.905

9.990

4159

12.38

800

360 Engineering Thermodynamics

t

158.8

165.0

170.4

175.4

179.9

212.4

233.8

p

6.0

7.0

8.0

9.0

10.0

20

30

–

–

–

–

–

–

–

–

–

–

–

– – –

– – –

– – –

– – –

– – –

– – –

– – –

v h s

v h s

v h s

v h s

v h s

v h s

v h s –

–

–

–

–

–

–

–

–

–

150

100

–

–

–

–

–

–

6.695

2829

0.2061

6.753

2835

0.2305

6.817

2840

0.2610

6.888

2846

0.3001

6.968

2851

0.3522

200

6.289

2858

0.070 61

6.547

2904

0.1115

6.926

2944

0.2328

6.980

2948

0.2597

7.040

2951

0.2933

7.106

2955

0.3364

7.182

2958

0.3940

250

6.541

2995

0.081 15

6.788

3025

0.1255

7.124

3052

0.2580

7.176

3055

0.2874

7.233

3057

0.3242

7.298

3060

0.3714

7.373

3062

0.4344

300

6.744

3117

0.090 51

6.957

3138

0.1386

7.301

3158

0.2825

7.352

3160

0.3144

7.409

3162

0.3544

7.473

3164

0.4058

7.546

3166

0.4743

350

6.921

3231

0.099 28

7.126

3248

0.1511

7.464

3264

0.3065

7.515

3266

0.3410

7.571

3267

0.3842

7.634

3269

0.4397

7.707

3270

0.5136

400

7.233

3456

0.1161

7.431

3467

0.1756

7.761

3478

0.3540

7.811

3480

0.3937

7.866

3481

0.4432

7.929

3482

0.5069

8.001

3483

0.5920

500

7.507

3682

0.1324

7.701

3690

0.1995

8.028

3698

0.4010

8.077

3699

0.4458

8.132

3699

0.5018

8.195

3700

0.5737

8.267

3701

0.6697

600

7.756

3911

0.1483

7.948

3917

0.2232

8.272

3923

0.4477

8.321

3924

0.4976

8.376

3924

0.5600

8.438

3925

0.6402

8.510

3925

0.7471

700

7.985

4146

0.1641

8.176

4150

0.2466

8.500

4155

0.4943

8.548

4155

0.5493

8.603

4156

0.6181

8.665

4156

0.7065

8.736

4157

0.8245

800

Properties of Superheated Steam 361

t

250.3

263.9

275.6

285.6

295.0

303.3

311.0

p

40

50

60

70

80

90

100

–

–

–

–

–

–

–

– – –

– – –

– – –

– – –

– – –

– – –

– – –

v h s

v h s

v h s

v h s

v h s

v h s

v h s –

–

–

–

–

–

–

–

–

–

–

–

–

–

150

100

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

200

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

250

–

–

–

–

–

–

5.793

2787

0.024 23

5.934

2841

0.029 45

6.071

2887

0.036 16

6.212

2927

0.045 32

6.364

2963

0.058 84

300

5.947

2926

0.022 41

6.039

2959

0.025 78

6.133

2990

0.029 94

6.231

3018

0.035 22

6.336

3045

0.042 20

6.451

3070

0.051 92

6.584

3094

0.066 42

350

6.213

3097

0.026 39

6.286

3118

0.029 91

6.364

3139

0.034 28

6.448

3158

0.039 89

6.541

3177

0.047 34

6.646

3196

0.057 76

6.769

3214

0.073 34

400

6.596

3373

0.03275

6.657

3385

0.03673

6.723

3398

0.04170

6.796

3410

0.04808

6.879

3421

0.05659

6.975

3433

0.06850

7.089

3445

0.086 35

500

6.902

3624

0.038 31

6.958

3633

0.042 79

7.019

3641

0.048 39

7.088

3649

0.055 59

7.166

3657

0.065 19

7.251

3666

0.078 62

7.368

3674

0.098 78

600

7.166

3868

0.043 53

7.220

3874

0.048 52

7.279

3881

0.054 76

7.345

3887

0.062 78

7.345

3893

0.073 47

7.510

3900

0.088 44

7.618

3905

0.1109

700

7.406

4113

0.048 56

7.458

4118

0.054 06

7.516

4122

0.060 94

7.581

4127

0.069 78

7.581

4132

0.081 57

7.743

4136

0.098 07

7.849

4141

0.12 28

800

362 Engineering Thermodynamics

t

318

324.6

330.8

336.6

342.1

347.3

352.3

p

110

120

130

140

150

160

170

–

–

–

–

–

–

–

– – –

– – –

– – –

– – –

– – –

– – –

– – –

v h s

v h s

v h s

v h s

v h s

v h s

v h s –

–

–

–

–

–

–

–

–

–

–

–

–

–

150

100

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

200

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

250

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

300

–

–

–

5.311

2621

0.009 76

5.443

2693

0.011 46

5.562

2754

0.013 21

5.666

2805

0.015 10

5.764

2850

0.017 21

5.857

2890

0.019 61

350

5.760

2922

0.013 03

5.824

2951

0.014 28

5.883

2977

0.015 66

5.951

3006

0.017 23

6.016

3031

0.019 02

6.081

3055

0.021 08

6.148

3078

0.023 51

400

6.264

3284

0.017 97

6.305

3297

0.019 29

6.345

3309

0.020 78

6.394

3324

0.022 51

6.441

3337

0.024 49

6.491

3350

0.026 79

6.543

3362

0.029 50

500

6.603

3562

0.021 72

6.639

3571

0.023 20

6.677

3581

0.024 87

6.716

3589

0.026 80

6.758

3597

0.029 02

6.802

3608

0.031 60

6.850

3614

0.034 66

600

6.886

3823

0.025 07

6.919

3829

0.026 72

6.954

3837

0.028 57

6.991

3842

0.030 72

7.030

3848

0.033 19

7.072

3854

0.036 07

7.117

3860

0.039 47

700

7.137

4079

0.028 21

7.168

4084

0.030 03

7.201

4090

0.032 07

7.237

4093

0.034 14

7.274

4098

0.037 16

7.315

4103

0.040 33

7.358

4107

0.044 08

800

Properties of Superheated Steam 363

t

357.0

361.4

365.7

369.8

374.15

p

180

190

200

210

221.2

–

–

–

–

–

–

–

– – –

– – –

– – –

– – –

s

v h s

v h s

v h s

v h s –

–

–

–

–

–

–

–

–

h

–

–

150

v

100

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

200

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

250

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

300

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

350

5.398

2733

0.008 16

5.486

2781

0.009 07

5.556

2819

0.009 95

5.628

2857

0.010 89

5.695

2890

0.011 91

400

6.064

3208

0.013 03

6.108

3227

0.013 91

6.142

3239

0.014 77

6.184

3255

0.015 73

6.223

3270

0.016 79

500

6.441

3518

0.016 22

6.474

3526

0.017 20

6.505

3537

0.018 15

6.536

3545

0.019 22

6.569

3553

0.020 40

600

6.739

3792

0.018 95

6.768

3798

0.020 04

6.796

3806

0.021 10

6.824

3810

0.022 29

6.854

3817

0.023 60

700

6.988

4057

0.021 35

7.025

4061

0.022 67

7.051

4067

0.023 83

7.078

4070

0.025 15

7.107

4075

0.026 60

800

364 Engineering Thermodynamics

INDEX

Index ♦ 367

A Absolute pressure, 11 Adiabatic process, 61 — system, 4 Amagat’s Leduc law, 291 Applications of Thermodynamics, 1 Assumptions in air standard cycle, 246 — steady flow process, 90 Atmospheric pressure, 11 Available energy, 162 Availability, 170 — in a steady flow process, 170 — of a closed system, 170 Avogadro’s hypothesis, 29

B Barrel calorimeter, 206 Boyle’s law, 25 Brayton cycle, 271

C Carnot cycle, 120 — efficiency of, 123 — practical difficulties of, 124 — theorem, 124 — vapour cycle, 313 Change of entropy — for ideal gases, 145 — during constant volume (isochoric) process, 148 — during constant pressure (isobaric) process, 150 — during constant temperature (isothermal) process, 152 — during mixing process, 295 — during reversible adiabatic (isentropic) process, 154 — during polytropic ( pvn = C) process, 156 — during thermodynamic processes, 148 Charles’ law, 26 Characteristic equation of a gas, 29 Classical thermodynamics, 2 Classification of thermodynamic processes, 49 Clausius inequality, 141 Closed system, 3 Combination of Boyle’s and Charles’ law, 28

Comparison of heat and work, 9 — between Otto, Diesel and Dual cycle, 275 — between two stroke and four stroke cycle engines, 281 — between open and closed feed water heaters, 340 Compression ignition engine, 278 — ratio for maximum output of an Otto cycle, 249 Concept of continuum, 4 — temperature, 19 Constant enthalpy process, 240 — entropy process, 233 — pressure cycle, 258 — volume cycle, 247 — temperature process, 58, 228 — pressure process, 54, 225 — volume process, 52, 219 Corollaries of Carnot’s theorem, 125

D Dalton’s law of partial pressures, 290 Diesel cycle, 258 — engine, 278 — mean effective pressure for, 269 Density, 10 Deviation of real gas from ideal gas, 38 Difference between Otto cycle and Diesel cycle, 275 — Rankine cycle and Carnot cycle, 316 — two specific heats, 34 Different forms of stored energy, 9 Dual cycle, 267

E Efficiency of Carnot cycle, 123 Energy, 9 Engineering applications of steady flow energy equation, 98 Enthalpy of a gas, 34 — entropy (h-s) diagram for water and steam, 195 Entropy of evaporation, 200 — superheated steam, 201 — water, 200 — wet and dry steam, 201 Equality of temperature, 19 Equivalence of Kelvin-Planck and Clausius statement, 117

368 ♦Engineering Thermodynamics Extensive property, 5 External workdone during evaporation, 198

F First law of thermodynamics for a closed or non-flow system undergoing a cycle, 45 — as applied to steady flow process, 91 — undergoing a change of state, 46 — limitations of, 47 Flow energy, 91 — system, 3 Force, 3 Free expansion process, 83 Four stroke cycle petrol engine, 276 — Diesel engine, 279

G Gas, 1 Gauge pressure, 11 General gas equation, 29 Gibbs function, 173 Graphical representation of formation of steam, 189

H Heat, 6 Helmholtz function, 171 Hetrogeneous system, 4 Homogeneous system, 4 Hyperbolic process, 230

I Ideal gas, 25 Important terms as applied to steam, 193 — used in air standard cycle, 246 Intensive property, 5 Internal energy, 10 — of steam, 198 Irreversible process, 6 — non-flow process, 83 Irreversibility, 173 Isentropic process, 61 Isobaric process, 54 Isolated system, 4 Isometric or isochoric process, 52 Isothermal process, 58

J Joule’s law, 28 Joule Thomson porous plug experiment, 97

K Kinetic energy, 10 Kelvin-Planck statement, 116

L Law of conservation of energy, 10 — degradation of energy, 116 Laws of Thermodynamics, 13 Limitations of Carnot vapour cycle, 314

M Macroscopic approach, 2 Mass, 2 — fraction, 288 Measurement of dryness fraction of steam, 206 — of pressure, 12 — temperature, 19 Mechanical work, 7 Mean effective pressure for diesel cycle, 260 — dual cycle, 270 — Otto cycle, 250 Methods of improving the efficiency of Rankine cycle, 334 Microscopic approach, 2 Modified Rankine cycle, 327 Molar specific heats of a gas, 35 Mole fraction, 289 Mollier chart for water and steam, 195

N Non-flow system, 3 Normal temperature and pressure, 13

O Otto cycle, 247

P Path function, 8 Perfect gas, 25 Perpetual motion machine of first kind (PMM-I), 47 — of second kind (PMM-II), 118 Phase, 4 Phases of water, 188

Index ♦ 369 Point function, 8 Polytropic process, 68, 236 Potential energy, 10 Power, 9 Practical difficulties of Carnot cycle, 124 Pressure, 11 — measurement of, 12 Pressure-specific volume-temperature ( p-v-T) surface of a pure substance, 192 Pressure-volume ( p-v) diagram for water and steam, 190 Principle of increase of entropy, 144 Process of formation of steam, 189 Properties of gas mixtures, 292 Pure substance, 2

Q Quasi-static process, 6

R Rankine cycle, 315 — using superheated steam, 322 Ratio of specific heats, 35 Regnault’s law, 34 Regenerative cycle, 338 — with an open feed water heater, 338 — with a closed water heater, 340 — with two feed water heaters, 343 Reheat cycle, 334 Relation between pressure, volume and temperature during an adiabatic change, 64 Reversible process, 6 —adiabatic process, 233

S Scope of Thermodynamics, 1 Separating calorimeter, 208 — and throttling calorimeter, 210 Specific heat, 7 — of a gas, 33 — volume, 10

Standard temperature and pressure, 13 Statements of Second Law of Thermodynamics, 116 Statistical thermodynamics, 2 Steady flow energy equation applied to various processes, 95

T Tank calorimeter, 206 Temperature-entropy (T-S) diagram, 141 — measuring scales, 21 Thermodynamic cycle, 5 — equilibrium, 5 — process, 5 — property, 5 — state and path, 4 — system, 3 — work, 7 Throttling calorimeter, 209 — process, 97, 240 Types of non-flow processes, 51 — thermodynamic system, 3 Two stroke cycle Diesel engine, 280 — petrol engine, 277

U Unavailable energy, 162 Universal gas constant, 29 Use of steam tables, 195

V Vapour, 1 Volumetric analysis and gravimetric analysis, 306 — fraction, 289

W Weight, 2 Workdone for a closed or non-flow process, 49 — steady flow process, 94 Working substance, 2