Engineering Thermodynamics [1 ed.] 9781783320509, 9781842658437

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ENGINEERING THERMODYNAMICS

ENGINEERING THERMODYNAMICS

S. Murugan

α Alpha Science International Ltd. Oxford, U.K.

Engineering Thermodynamics 394 pgs.  |  174 figs.  |  28 tbls.

S. Murugan Associate Professor Department of Mechanical Engineering National Institute of Technology Rourkela Copyright © 2014 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K. www.alphasci.com All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher. ISBN 978-1-84265-843-7 E-ISBN 978-1-78332-050-9 Printed in India

PREFACE I am pleased to introduce the first edition of the textbook titled “Engineering Thermodynamics” to the students who have opted for “Engineering Thermodynamics” in their B.Tech Degree or AMIE as a part of their course. Thermodynamics is one of the important subjects in the courses of B.E/B.Tech Mechanical Engineering, Electrical and Electronics Engineering, Chemical Engineering, and Instrumentation Engineering, or in AMIE. The subject “Thermodynamics” requires special attention by any student as it involves both theory and analysis. Many students have an aversion to learn the subject as it has many concepts that make it a little complex. The subject is a foundation for their future, but, every student who learns it faces difficulty in understanding it thoroughly and deeply. Keeping this in mind, this book is written to make the students understand the subject easily. Although “Thermodynamics” is taught in Physics and Chemistry at the School level, the subject “Engineering Thermodynamics” is quite different for engineering students, as they apply the concepts of Thermodynamics in various applications, such as calculating engine efficiency, and heat or work values in thermal systems, such as boilers, turbines, compressors etc. The subject also deals with energy interactions in various systems, such as refrigeration and air conditioning systems. After interaction and discussion with many readers of this subject, the author developed an interest to write this book. In each and every chapter, the necessary illustrations and examples are given, in areas where one may find some difficulty in understanding the theory. Solved and unsolved problems are given in each and every chapter, so that the readers will be able to practise different kinds of problems. This will enable the readers to solve similar kind of problems in their real time applications. Besides these problems, objective type and theory questions are also included at the end of each chapter. This will make the readers review and practise the subject with more confidence. This book is also useful to the readers of different categories who appear for competitive or aptitude examinations. This book is written in an unconventional and a “Student friendly” style, so as to make any one be more confident in the subject. The contents of the book cover the syllabus framed by many of the Universities. The author believes that the book will definitely help the readers immensely. Any criticism or comments or suggestions are welcome for the improvement of the book.

S. Murugan

ACKNOWLEDGEMENTS I would like to express my sincere thanks to Mr. N. K. Mehra, Publisher and Managing Director, Narosa Publishing House Pvt. Ltd., New Delhi, who readily accepted to publish this book. My special thanks to all the people involved in preparation of this textbook. I would also like to register my whole hearted thanks to the Editor and reviewers for their valuable comments to improve the layout of the book. It is worth noting that without the students’ feedback it is impossible to get an appreciation of the book. Every comment by the students who read and gave their suggestions, made me to take a lot of time to improve the content of the book, according to their need. It is quite difficult to succeed in any task, if there is no encouragement from one’s family members. It’s my pleasure in thanking my parents, wife Mrs. S. Asha and daughter Mrs. M. Geetha who gave me full support in completing this task. S. Murugan

CONTENTS Preface Acknowledgements

...v ...vii

1.  UNITS AND MEASUREMENT

1.1

1.1 Introduction to Thermodynamics 1.2 Units 1.2.1 Base Units 1.2.2 Derived Units 1.2.3 Supplementary Units 1.3 Other Systems of Units 1.4 Definition and Measurement of Properties 1.4.1 Mass and Weight 1.4.2 Force 1.4.3 Pressure 1.4.4 Temperature 1.4.5 Energy 1.4.6 Heat 1.4.7 Work

2.  INTRODUCTION TO IDEAL GASES

...1.1 ...1.1 ...1.1 ...1.2 ...1.2 ...1.3 ...1.3 ...1.3 ...1.3 ...1.4 ...1.8 ...1.12 ...1.13 ...1.14

2.2

2.1 Introduction ...2.1 2.2 Gas Laws ...2.1 2.2.1 Boyle’s Law ...2.1 2.2.2 Charles’ Law ...2.2 2.2.3 Gay Lussac Law ...2.2 2.3 Equation of State ...2.2 2.4 Avogadro’s Hypothesis ...2.3 — 2.5 Universal Gas Constant (R) ...2.3 2.6 Internal Energy of Ideal Gas ...2.4

x 2.7 2.8 2.9 2.10

2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20

Contents

Enthalpy of an Ideal Gas ...2.4 Joule’s Law ...2.5 Specific Heat Capacity ...2.5 Specific Heats of a Gas ...2.6 2.10.1 Heating of a Gas at Constant Pressure ...2.6 2.10.2 Heating of a Gas at Constant Volume ...2.6 Relation between Cp, and Cv ...2.7 Molar Specific Heats of Gases ...2.7 Properties of Non-Reactive Mixtures ...2.7 Gibbs-Dalton’s Law ...2.8 Amagat-Leduc Law of Partial Volume ...2.9 Molecular Weight of a Mixture of Ideal Gases ...2.10 Graham’s Law of Diffusion ...2.11 Relationship between Volumetric (Molar), and Gravimetric Analysis (Mass Analysis) ...2.11 Conversion From Gravimetric or Molar (Volumetric) Fraction ...2.12 Equations Relating Mass Fraction with Extensive Properties of the Mixture of Gases ...2.13

3.  THERMODYNAMIC SYSTEM AND FIRST LAW

3.1

3.1 Law of Conservation of Energy 3.2 Thermodynamic System 3.2.1 Types of Thermodynamic Systems 3.3 Terminologies 3.4 Thermodynamic Properties 3.4.1 Intensive Property 3.4.2 Extensive Property 3.5 State 3.5.1 Change of State 3.6 Thermodynamic Process 3.7 Path 3.8 Thermodynamic Cycle 3.9 Equilibrium 3.9.1 Mechanical Equilibrium 3.9.2 Chemical Equilibrium 3.9.3 Thermal Equilibrium 3.9.4 Thermodynamic Equilibrium 3.9.5 Zeroth Law of Thermodynamics 3.10 Reversible Process

...3.1 ...3.1 ...3.1 ...3.3 ...3.3 ...3.3 ...3.3 ...3.3 ...3.4 ...3.4 ...3.4 ...3.4 ...3.4 ...3.5 ...3.5 ...3.5 ...3.5 ...3.5 ...3.5

Contents

xi

3.11 Work Transfer ...3.6 3.11.1 Sign Convention of Work Transfer ...3.6 3.11.2 Work a Path Function ...3.6 3.12 Heat Transfer ...3.7 3.12.1 Sign Convention of Heat Transfer ...3.7 3.13 First Law of Thermodynamics ...3.7 3.13.1 Perpetual Motion Machine of the First Kind (PMM–I) ...3.8 3.13.2 Various Thermodynamic Processes ...3.9 3.14 Flow Processes or Open Systems ...3.18 3.14.1 Steady Flow Energy Equation (SFEE) ...3.18 3.14.2 Applications of Steady Flow Processes ...3.20 3.15 Heat Absorbing Systems ...3.20 3.15.1 Boiler ...3.20 3.15.2 Evaporator ...3.21 3.16 Heat Rejection Systems ...3.22 3.16.1 Condenser ...3.22 3.17 Work Developing Systems ...3.22 3.17.1 Turbine ...3.23 3.18 Work Absorbing Systems ...3.24 3.18.1 Rotary Compressor ...3.24 3.18.2 Reciprocating Compressor ...3.24 3.19 Other Systems ...3.25 3.19.1 Steam Nozzle ...3.25 3.20 Determination of Work Transfer in a Steady Flow Process ...3.26

4.  SECOND LAW OF THERMODYNAMICS

4.1

4.1 Limitations to the First Law of Thermodynamics 4.2 Heat Engine, Refrigerator and Heat Pump 4.2.1 Heat Engine 4.2.2 Refrigerator 4.2.3 Heat Pump 4.3 Statements of the Second Law of Thermodynamics 4.3.1 Kelvin Planck Statement 4.3.2 Clausius Statement 4.4 Violation of the Kelvin-Planck and Clausius Statements 4.5 Reversibility 4.5.1 Irreversible Process 4.6 Carton Cycle 4.7 Carnot Theorem

...4.1 ...4.2 ...4.2 ...4.2 ...4.3 ...4.4 ...4.4 ...4.4 ...4.5 ...4.7 ...4.7 ...4.8 ...4.9

xii 4.8 Thermodynamic Temperature Scale 4.9 Entropy of Perfect Gases 4.9.1 Entropy 4.9.2 Clausius Inequality 4.9.3 Entropy - Property of a System 4.9.4 Change in Entropy of a Perfect Gas in Various Thermodynamic Processes

Contents

...4.11 ...4.13 ...4.13 ...4.14 ...4.16 ...4.18

5.  AVAILABILITY AND IRREVERSIBILITY

5.1

5.1 Introduction 5.2 Available Energy 5.2.1 Energy Interaction with a Constant Temperature Energy Source 5.2.2 Heat Transfer in a Heat Exchange Process 5.3 Availability in Thermodynamic Systems 5.3.1 Reversible Work in Steady Flow Process 5.3.2 Unsteady Open or Closed System 5.4 Dead State

...5.1 ...5.2 ...5.2 ...5.3 ...5.4 ...5.4 ...5.5 ...5.6

6.  PROPERTIES OF PURE SUBSTANCE

6.1

6.1 Pure Substance ...6.1 6.1.1 Existence of Pure Substance ...6.1 6.1.2 Terminology ...6.2 6.2 Steps of Steam Formation ...6.2 6.3 Pressure-Volume-Temperature Diagram ...6.3 6.3.1 T-v Diagram of Water–Steam ...6.4 6.4 Saturation Pressure and Temperature ...6.4 6.5 p-V Diagram for Water ...6.5 6.6 Property Diagrams of Water ...6.7 6.7 Steam Tables ...6.7 6.8 Thermodynamic Properties of Pure Substances (Water and Steam) ...6.9 6.8.1 Pressure ...6.9 6.8.2 Specific Volume ...6.9 6.8.3 Density of Steam ...6.10 6.8.4 Enthalpy of Water and Steam ...6.10 6.8.5 Specific Internal Energy ...6.11 6.8.6 Entropy of Steam ...6.12 6.9 Determination of the Dryness Fraction ...6.14 6.9.1 Separating Calorimeter ...6.14

Contents

xiii

6.9.2 Throttling Calorimeter ...6.15 6.10 Non-Flow Processes of Pure Substances ...6.17 6.10.1 Constant Volume Process ...6.18 6.10.2 Constant Pressure Process ...6.19 6.10.3 Constant Temperature Process ...6.20 6.10.4 Hyperbolic Process ...6.21 6.10.5 Reversible Adiabatic Process (Isentropic Process or Constant Entropy Process) ...6.21 6.10.6 Polytropic Process ...6.22 6.11 Work Done and Internal Energy and Heat Transfer Calculations ...6.23 6.11.1 Constant Volume Process ...6.23 6.11.2 Constant Pressure Process ...6.24 6.11.3 Constant Temperature Process ...6.24 6.11.4 Hyperbolic Process ...6.25 6.11.5 Reversible Adiabatic Process (Isentropic Process) ...6.26 6.11.6 Polytropic Process ...6.26 6.11.7 Throttling Process ...6.27 6.12 Mollier Diagram ...6.28

7.  REAL GAS MIXTURE

7.1

7.1 Introduction ...7.1 7.2 Real Gases ...7.1 7.3 Forces of Attraction ...7.1 7.3.1 Van der Waal’s Force of Attraction ...7.2 7.3.2 Effect of the Actual Volume Occupied by the Molecules ...7.2 7.3.3 Effect of the Size of Molecules ...7.2 7.3.4 Effect of the Mass of Molecule ...7.3 7.4 Van der Waal’s Equation ...7.3 7.4.1 Correction for the size of the Molecules ...7.5 7.4.2 Correction on Pressure ...7.5 7.4.3 Determination of Critical Pressure (pc), Critical Volume (v—c ) and Critical Temperature (Tc) ...7.6 ­ 7.4.4 Demerits of Van der Waal’s Equation ...7.8 7.4.5 Redlich-Kwang Equation ...7.8 7.4.6 Virial Equation ...7.9 7.4.7 Dieterici Equation ...7.9 7.4.8 Beattie Bridge Man Equation ...7.9 7.4.9 Berthelot Equation ...7.10 7.4.10 Mcleod Equation ...7.11 7.4.11 Wohl’s Equation ...7.12

xiv 7.5 Compressibility Factor 7.5.1 Reduced Properties

8.  THERMODYNAMIC RELATIONS

Contents

...7.12 ...7.13

8.1

8.1 Introduction ...8.1 8.2 Overview of Mathematical Relations ...8.1 8.2.1 Cyclic Relation ...8.2 8.2.2 Gibbs Free Energy (or) Gibbs Function ...8.3 8.2.3 Helmotz Free Energy (or) Helmotz Function ...8.3 8.3 Maxwell’s Relations ...8.4 8.4 Tds Equations ...8.5 8.5 Useful Relations between Partial Derivatives ...8.7 8.6 The Clausius Equations ...8.8 8.7 Equation for Internal Energy and Enthalpy ...8.10 8.8 Equation for Enthalpy ...8.11 8.9 Joule-Thomson Coefficient ...8.12 8.10 Thermodynamic Relations using Cp and Cv ...8.15 8.11 Isothermal Compressibility ...8.18 8.12 Coefficient of Linear Expansion and Volume Expansion ...8.18 8.13 Isothermal Compressibility and Adiabatic Compressibility ...8.19

9. PSYCHROMETRY

9.1

9.1 Introduction 9.2 Terms and Definition 9.2.1 Dry Air 9.2.2 Dry Bulb Temperature 9.2.3 Wet Bulb Temperature 9.2.4 Dew Point Temperature 9.2.5 Specific Humidity or Humidity Ratio 9.2.6 Relative Humidity 9.2.7 Degree of Saturation 9.2.8 Carrier Equation for Partial Pressure of Water Vapour 9.3 Psychrometer 9.3.1 Sling Psychrometer 9.4 Specific Enthalpy of Moist Air 9.5 Adiabatic Saturation 9.6 Psychrometric Chart 9.7 Psychrometric Processes

...9.1 ...9.1 ...9.1 ...9.2 ...9.2 ...9.2 ...9.3 ...9.4 ...9.4 ...9.5 ...9.5 ...9.5 ...9.6 ...9.7 ...9.8 ...9.9

Contents

xv

9.7.1 Sensible Cooling 9.7.2 Sensible Heating 9.7.3 Humidification 9.7.4 Dehumidification 9.7.5 Cooling and Dehumidification 9.7.6 Heating and Humidification 9.8 Sensible Heat Factor 9.9 Mixing of Air Streams

...9.10 ...9.11 ...9.12 ...9.12 ...9.13 ...9.14 ...9.15 ...9.16

10.  FUELS AND COMBUSTION

10.1

10.1 Introduction 10.2 Classification of Fuels 10.2.1 Solid Fuels 10.2.2 Fuel Analysis 10.2.3 Liquid Fuels 10.2.4 Gaseous Fuels 10.3 Some Important Fuel Properties 10.3.1 Density 10.3.2 Viscosity 10.3.3 Calorific Value or Heating Value of Fuels 10.3.4 Flash Point and Fire Point 10.3.5 Pour Point 10.3.6 Cloud Point 10.3.7 Ignition Temperature 10.4 Determination of the Heating Value of a Fuel 10.4.1 Lower Calorific Value 10.4.2 Heating Value of Gaseous Fuels 10.4.3 Experimental Method for Determination of the Heating Value 10.5 Basic Combustion Equations 10.6 Air Fuel Ratios for Combustion 10.6.1 Stoichiometric Air Fuel Ratio 10.6.2 Excess Air 10.6.3 Determination of Stoichiometric or Theoretical Air Fuel Ratio 10.7 Conversion of Fuel Constituents Given in Volume to Mass Percentage 10.8 Excess Air Supplied Per kg of Fuel 10.9 Quantity of Carbon in Flue Gas

...10.1 ...10.1 ...10.1 ...10.2 ...10.3 ...10.6 ...10.9 ...10.9 ...10.9 ...10.9 ...10.10 ...10.10 ...10.10 ...10.10 ...10.11 ...10.11 ...10.11 ...10.12 ...10.14 ...10.15 ...10.15 ...10.15 ...10.15 ...10.17 ...10.18 ...10.18

xvi 10.10 Quantity of Flue Gas Per kg of Fuel Burnt 10.11 Flue Gas Analysis 10.11.1 Construction 10.11.2 Method of Analysis 10.11.3 Advantages 10.11.4 Limitations 10.12 Flame Temperature 10.13 Internal Energy of Combustion 10.14 Enthalpy of Combustion 10.15 Enthalpy of Formation

11.  AIR STANDARD CYCLES 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Why Air Standard Cycles? A Thermodynamic Cycle Air Standard Cycle Assumptions The Carnot Cycle The Stirling Cycle The Ericsson Cycle Important Air Standard Cycles 11.7.1 The Otto Cycle 11.7.2 The Diesel Cycle 11.7.3 The Dual Cycle 11.7.4 The Lenoir Cycle 11.7.5 The Atkinson Cycle 11.7.6 The Brayton Cycle 11.8 Comparison between Otto, Diesel and Brayton Cycles 11.9 Comparison of Air Standard Efficiencies 11.9.1 Same Compression Ratio and Heat Supplied 11.9.2 Same Maximum Pressure and Heat Supplied 11.9.3 Same Maximum Temperature and Heat Rejected 11.9.4 Same Compression Ratio and Heat Rejected 11.9.5 Same Maximum Pressure and Output Index

Contents

...10.19 ...10.19 ...10.20 ...10.20 ...10.20 ...10.20 ...10.20 ...10.22 ...10.22 ...10.23

11.1 ...11.1 ...11.1 ...11.2 ...11.2 ...11.5 ...11.6 ...11.6 ...11.7 ...11.12 ...11.17 ...11.20 ...11.22 ...11.25 ...11.28 ...11.29 ...11.29 ...11.30 ...11.30 ...11.31 ...11.31 ...I.1

CHAPTER

1 1.1

UNITS AND MEASUREMENT

INTRODUCTION TO THERMODYNAMICS

The science, which deals with the analysis of various machines by quantity, which involves the transfer of energy into useful work, is called thermodynamics. Many energy conversion devices require the transfer of energy into work. Thermodynamics is applied in various thermal equipments like steam turbines, boilers, condensers, cooling towers, heat exchangers, reciprocating engines, refrigerators, air conditioners, heat pumps etc. It is also used in internal combustion engines, turbo jets and rockets etc. In all these cases, the design of the thermal equipments essentially requires an in-depth knowledge of thermodynamics. Hence, the study of engineering thermodynamics is very important in the field of engineering.

1.2 UNITS In thermodynamics, each and every parameter is measured and the final quantity is expressed in a value or number followed by a unit. Over many years, a large number of standards have been defined for physical measurements and many systems of units have been derived. There was an attempt to simplify the language of science by the adoption of a system of units. Such a system is called the SI unit system, commonly known as the International System of Units (abbreviated SI from the French le System International units), and is used universally. This system of units was the outcome of a resolution of the 9th general conference of weights and measures (CGPM) in 1960, and from that, the establishment of a complete set of rules for units of measurement, was made. The SI unit system has three divisions: 1. Base units 2. Derived units 3. Supplementary units.

1.2.1 Base Units The base unit is the first division of SI units, in which seven kinds of fundamental quantities are measured. They are given in Table 1.1.

1.2

Engineering Thermodynamics

(i) Meter (m): It is defined as 1650 763.73 of the wavelength in vacuum of the orange-red light emitted by 86 36K in the transition of 2p10 to 5 d5. (ii) Kilogram (kg): It is defined as the mass of a platinum-iridium cylinder kept at the International Bureau of weights and measures at serves near Paris, France. Originally it is equal to the mass of a cubical decimeter of water at its maximum density. (iii) Second (s): It is the time spent by 9 192 631 770 cycles of radiation from hyperfine transition in Cesium, when unperturbed by the external fields. (iv) Ampere (A): It is the current, which is maintained in each of two parallel wires of infinite length placed in vacuum, 1 meter apart, so as to produce a force of 2 × 10–7 newtons per meter length. (v) Kelvin (K): It is the fraction of 1/273.16 of the thermodynamic temperature of the triple point of water. (vi) Candela (cd): It is the luminous intensity in a given direction due to a source which emits monochromatic radiation of frequency of 540 × 1012 Hz and whose radiant intensity in that direction is 1/683 watt per steradian. (vii) Mole (mol): It is defined as the amount of substance of a system, which contains as many elementary entities as there are atoms in 0.012 kg of the carbon isotope 6C12. Table 1.1 Basic Measurements with Units Measurable quantity Length Mass Time

Units Meter Kilograms Seconds

Symbol m kg s

Electric current Temperature Luminous intensity Amount of substance

Ampere Kelvin Candela Mole

A K Cd mol

1.2.2 Derived Units The derived units are obtained from the base units. For example, the unit of density is kg/m3, which is obtained from the base units of mass (kg) and length (m).

1.2.3 Supplementary Units Two supplementary units are defined in SI units. 1. Radian 2. Steradian. These two units are meant for plane and solid angles respectively. They are given in Table 1.2. (a) Radian (rad): It is the plane angle between two radii of a circle, which cut on the circumference of an arc equal in length to the radius.

1.3

Units and Measurement

(b) Steradian (sr): It is the solid angle, which has its vertex in the centre of a sphere, and cuts off an area of the surface of the sphere, equal to that of a square, with sides of equal length to the radius of the same sphere. Table: 1.2 Angular Measurements with Units

1.3

Measurable quantity

Units

Symbol

Plane angle

Radian

rad

Solid angle

Steradian

sr

OTHER SYSTEMS OF UNITS

Some other systems of units are still in use. They are (i) CGS (ii) FPS (iii) MKS.

1.4

DEFINITION AND MEASUREMENT OF PROPERTIES

1.4.1 Mass and Weight The amount of matter contained by a body is known as the mass. The basic unit of mass is the kilogram. Weight is the force exerted by gravity on the mass of a body or substance. The weight of the body varies from place to place, because the value of the constant of proportionality varies from place to place. As per Newton’s second law of motion, the weight w = 1/gc . mg Where,

gc = Constant of proportionality (gc = 1)



m = Mass



g = Acceleration due to gravity.

1.4.2 Force Force is that which produces or tends to produce a change in the state of rest or of uniform motion, of a body. Newton’s second law of motion states that the applied force or impressed force is directly proportional to the rate of change of momentum. Suppose,

m = Mass of the body

u = Initial velocity of the body v = Final velocity of the body a = Constant acceleration t = Time required to change the velocity from u to v

Momentum = Mass × velocity

1.4

Engineering Thermodynamics

Change of momentum = mv – mu ...(1.1) The rate of change in momentum = (mv – mu)/t ...(1.2) = m(v – u)/t ...(1.3) = ma where,

a = (v – u)/t

Therefore, Force

F = Mass × acceleration.

...(1.4)

1.4.3 Pressure The normal force exerted per unit area in the surface of a body is termed as pressure. The various units used for pressure in the SI system of units are given below. For example, the force “F” is applied on a piston in a cylinder as shown in Fig. 1.1. The area of the piston is “A”. The pressure exerted normal to the plane surface of the piston is given by Pressure (P) = Force/Area of the piston Pressure “p” Piston Cylinder

Fig. 1.1 Pressure exerted on a piston

There are various units followed for pressure. They are given below

1 bar = 105 N/m2



1 Pa = 1 N/m2

1 Atmospheric pressure = 1.013 × 105 N/m2 Atmospheric pressure = 760 mm of Hg

1 mm of Hg = (1013 × 102)/760 = 133.3 N/m2 1 N/m2 = 760/(1013 × 102 ) = 7.5 × 10–3 mm of Hg

1 ata (atmospheric technical absolute) = 736 mm of Hg.

1.5

Units and Measurement

1.4.3.1  Elements of Pressure There are three elements in the measurement of pressure: (i) Atmospheric pressure (ii) Gauge pressure (iii) Absolute pressure (i) Atmospheric pressure (Patm): It is defined as the pressure produced by a column of mercury 760 mm high, and is denoted as atm. The value of the atmospheric pressure is 1.013 bar. (ii) Gauge pressure (Pg): The value read by the pressure gauge is called gauge pressure. Generally, all the pressure gauges read the difference between the absolute pressure and atmospheric pressure only. (iii) Absolute pressure: The actual pressure of the system is known as absolute pressure. Generally, thermodynamic relations are concerned with absolute pressure. The relation between the gauge, atmospheric and absolute pressures is shown in Fig. 1.2(a). The relation between the vacuum, atmospheric and absolute pressures is shown in Fig. 1.2(b).

Gauge pressure Absolute pressure

Vacuum Atmospheric pressure Absolute pressure

(a)

(b)

Fig. 1.2 (a) Graphical representation of gauge, atmospheric and absolute pressures (b) Graphical representation of vacuum, atmospheric and absolute pressures

The relation between above three pressures is given below

pabs = patm + pg

...(1.5)

* The above relation is applicable only when the gauge pressure is positive. When the pressure is below atmospheric, then the gauge pressure will be negative; the pressure then measured is called vacuum.

pabs = patm – pvacuum ...(1.6)

1.6

Engineering Thermodynamics

l Normal

temperature and pressure (NTP): Normal temperature and pressure refers to the conditions of atmospheric pressure at 760 mm of Hg and 20°C.

l Standard

temperature and pressure (STP): STP refers to the temperature and pressure of any gas in standard atmospheric pressure are taken at 760 mm of Hg and 0°C.

1.4.3.2  Measurement of Pressure Various instruments are commonly used for measuring pressure. A few of them are: (i) Piezometer tube (ii) Manometer (iii) Mechanical gauge (i) The Piezometer tube: It is the simplest device used for measuring the intensity of pressure in liquids. Figure 1.3 shows a piezometer tube. The bottom end of a glass tube is connected to the liquid in the tank, whose pressure is to be measured. The top of the tube is kept open to the atmosphere. The liquid rises in the tube till the equilibrium is attained between the pressure exerted by the liquid column and that in the container. The height of the liquid column from the point, at which the pressure of the liquid in the tank is to be measured, gives the pressure of the liquid at the point in terms of the head of the liquid. The piezometer cannot be used for measuring vacuum and high pressures. Open to the atmosphere

h

Fig. 1.3 Piezometer tube

(ii) Manometer: The manometer is a commonly used instrument used to measure the pressure. One example is a U-tube manometer. The U-tube manometer may be open or closed type. In open type, both the ends of the limbs are open and exposed to the atmosphere. In a closed type, one end of the limb is closed while the other limb is exposed to the atmosphere. Generally water or mercury is used as a fluid in the manometer. Fig. 1.4(a) shows the closed type U-tube manometer which is used to measure the gauge pressure.

1.7

Units and Measurement

h1

h2

h1

h2

Fig. 1.4 (a) Gauge pressure indicating Fig. 1.4 (b) Gauge pressure indicating positive pressure negative pressure

One end of the manometer is connected to the tank, in which the pressure of the fluid is to be measured and the other end is open to the atmosphere. Let “Z” is the difference between the level of water or mercury in the two limbs, “ρ” is the density of water or mercury and “g” is gravitational constant, then the gauge pressure is as given below Gauge pressure pg = Z ρg 

[Z = h2 – h1]

...(1.7)

The level of water or mercury in the two limbs are different while the measurement of vacuum is shown in Fig. 1.4 (b). (iii) Mechanical gauge: The mechanical gauges are extensively used in many commercial applications. The working of mechanical gauges is based on the elastic deformation. The bourdon tube or bourdon gauge is one of the most commonly used mechanical gauges used to measure tyre pressure in cycle shops, pressure in compressors etc. Fig. 1.5(a) shows the photograph of the sectional view of the bourdon gauge. The parts of the bourdon gauge are shown in Fig. 1.5(b). The bourdon tube has an elliptical cross sectional tube which is connected to the fulcrum through a linkage. A pointer is connected to the fulcrum which will indicate the pressure on the graduated scale fixed on the tube. When a fluid,whose pressure has to be measured enters into the bourdon tube, an elastic deformation occurs and actuates the linkage to move the pointer and in turn the pointer indicates the pressure.

Scale Fulcrum wheel Bourdon tube Fulcrum Indicator Gas in

Fig. 1.5 (a) Photograph of bourdon gauge (b) Components of bourdon gauge

1.8

Engineering Thermodynamics

1.4.4 Temperature The property that determines the hotness or level of heat intensity of a body is called temperature. It is directly proportional to the level of intensity of heat. If the temperature of the body is high then the body is hot. Similarly, a low temperature of the body indicates that the body is cool. It is an intensive property of a system. A common instrument used to measure the temperature is the “Thermometer”. The following scales are used to measure the temperature. (a) Centigrade scale (b) Fahrenheit scale (c) Kelvin scale (d) Rankine scale (a) Centigrade or Celsius scale: The point at which water freezes under atmospheric pressure is taken as the reference point 0°C on the measurement scale, and the temperature at which water boils at atmospheric pressure, is marked 100°C. The difference between these two values is divided into hundred units on the scale. This is also called the Celsius scale. (b) Fahrenheit scale: The point at which water freezes under atmospheric pressure is taken as the reference point 32°F on the measurement scale and the temperature at which water boils is marked 212°F. The difference between these two values is divided into a hundred and eighty equal units on the scale. (c) Kelvin scale: When the reference point is taken as the triple point of water for measuring the temperature, then it is known as the Kelvin temperature scale. The temperature of the triple point of water is 273.16°C. (d) Rankine scale: The absolute Fahrenheit scale is called as Rankine scale. The relationship between the temperature scales and the Rankine scale are given below

°F = 1.8°C + 32

...(1.8)



°C = (°F – 32) × 5/9

...(1.9)



K = °C + 273.15

...(1.10)



R = °F + 459.67.

...(1.11)

1.4.4.1  Absolute Temperature The zero readings of the Celsius scale and the Fahrenheit scale are chosen arbitrarily. This will help when the changes of temperature in a process are known. But when the equations relating to the fundamental laws are written, the value of temperature may be used in the equation. In such cases the value of

1.9

Units and Measurement

temperature whose reference point is true zero or absolute zero will be considered. The lowest temperature possible to be attained using any thermodynamic process is called absolute temperature. At this temperature, the entropy of the substance is minimum. The temperature measured from this zero is called the absolute temperature. The values of the absolute temperatures taken in the Celsius scale and Fahrenheit scale are –273.15°C and –459.67°F respectively.

1.4.4.2  Measurement of Temperature The most important property considered in thermodynamics, is the temperature. Heat cannot be measured directly by any device. It can be only determined with the help of an expression, in which temperature is involved. Since the temperature is directly measurable and proportional to heat energy, it is convenient to measure the temperature only to quantify heat. Similarly, enthalpy and internal energy are also determined with the help of the known temperature. A few instruments widely used to measure temperature are: (i) Gas filled thermometer (ii) Electrical resistance thermometer (iii) Thermocouple (iv) Liquid-in-glass thermometer. (i) Gas Filled Thermometer: The gas filled thermometer specifically operates at constant volume, with the gas expanding or contracting according to the change in temperature. The gas filled thermometer fundamentally works according to the Charles’ law. As per the law, the given mass of an ideal gas at constant volume produces an absolute pressure in direct proportion to the absolute temperature of gas under ideal conditions. Figure 1.6 shows a gas filled thermometer which operates at constant volume. PO

Capillary tube h Gas bulb

Limb A

Mercury manometer

Fig. 1.6 Gas filled thermometer

Limb B

1.10

Engineering Thermodynamics

The gas filled in a gas bulb which is in contact with a capillary tube. A mercury manometer is connected to the capillary tube. The other end of the manometer is open to the atmosphere. The mercury level in the manometer limbs can be adjusted, so that the mercury level in limb A will touch the top of the capillary. When the thermometer is kept on a body, the gas in the bulb expands as it is heated up, so that it pushes the mercury down. The limb B is adjusted, so that the mercury again reaches the top of the capillary, and the pressure in the bulb can be estimated by using the expression.

P = Patm + rmercury gh ...(1.12)

where,

P = Pressure measured in bulb

Patm = Atmospheric pressure

rmercury = Density of mercury



g = 9.81 m/s2



h = Difference between the levels of mercury in two limbs.

Let the initial and final pressure measured by the thermometer be p1 and p2 respectively. Then, use the gas equation p1v = RT1 and p2v = RT2. The change in the temperature of the body whose temperature is to be measured, is (T2 – T1) = v(p2 – p1)/R

...(1.13)

or,

...(1.14)

DT = v Dp/R

Some gas filled thermometers operate at constant pressure, in which the mercury levels are to be adjusted, to maintain the difference between two levels of mercury (h), and the volume unit mass of gas “v” (small) will vary corresponding to the temperature of a body or substance. Then the change in temperature of body is given as

DT = D v p/R

...(1.15)

(ii) Electrical Resistance Thermometer: An electrical resistance thermometer can be used for the measurement of temperature. The measurement of the change of resistance of a metal due to temperature change, is done by a Wheatstone bridge. An ordinary Wheatstone bridge is not used for the measurement of changes in resistance. A resistance thermometer is shown in Fig. 1.7, in which platinum is used as a metal wire. In the platinum used electrical resistance thermometer, the accuracy of the measurement is very high. where,

R = Rm(1 + aT + bT2) (1.16) Rm = Resistance of platinum metal wire when it is kept in melting ice.

T = Temperature of a body which is to be measured.

a, b = Constants.

Resistance thermometers are used for the calibration of thermometers.

1.11

Units and Measurement Galvanometer

R1

R2

R3

R

Fig. 1.7 Electrical resistance thermometer

(iii) Thermocouple: It is the simplest and most widely used device for temperature measurement. The thermocouple works on the principle of Seebeck effect. When two dissimilar metals at different temperatures are joined together, then an emf is generated in the circuit (Fig. 1.8). The emf generated is dependent on the temperature difference between the hot and cold junctions. The function is given by,

E = aq + bq2 ...(1.17)

where,

E = emf generated in the circuit.



a and b = Constants.



q = Temperature difference between the hot and cold junctions.

The values of a and b depend upon the metals used. The value of a is in the range of 40-50 mV/°C, and the value of b is in the range of 0.04-0.4 mV/°C. For every 100°C of temperature difference 5 mV of emf will be generated in the circuit. Cold junction (reference junction) Metal wore A (Iron) Copper wire

emf measuring instrument

Metal wore B (Constantan)

Fig. 1.8 Thermocouple

A few thermocouple metal pairs that are commonly used for temperature measurements are Iron – Constantan, Copper – Constantan, Chromel – Alumel, Platinum/Platinum rhodium etc.

1.12

Engineering Thermodynamics

(iv) Pyrometers: These are used to measure temperatures in the range of 1200°C to 3500°C. The pyrometer receives all the radiation from a definite area of a hot body, and reproduces it on sensitive temperature transducers like thermopile, bolometer, etc.

1.4.5 ENERGY The capacity to do work is called energy. Energy may be in two forms: (a) Stored energy, and (b) Transition energy. (i) Stored energy: Energy possessed in a system is called stored energy. A few forms of stored energy are kinetic energy, potential energy, internal energy, etc. (ii) Transition energy: When the energy possessed by the system crosses the boundaries of the system, it is called transition energy. These energies are found in the following forms; n Kinetic energy: It is the energy possessed by a body or a substance when it is exactly in motion. If a fluid of mass “m” is moving with a velocity V m/sec, then the kinetic energy possessed by it during its motion is given as: 1 mV2 kg × m2/s2 2 1 = mV2 N-m (1 N = kg × m/s2 ) ...(1.18) 2 where, m = mass of the body in kg



Kinetic energy (KE) =



V = Velocity of the body.

n Potential energy: The energy possessed by a body or substance due to its rise above the earth’s surface is known as potential energy. If a fluid of mass “m” is raised above the earth’s surface to a height of “Z”, then the potential energy possessed by it is given as: Potential energy (P.E) = mgZ.

...(1.19)

n Flow energy: The energy possessed by a body or a substance during the transfer of mass across the boundaries of a system, is called as flow energy. If the pressure of a fluid is “p” and the volume is “V”, and it is assumed to cross over its boundaries against its volume, then it is given by,

Flow energy = Pressure exerted by the fluid × Area of fluid covered × Length of the fluid film



Flow energy = p × A × l

...(1.20)

= p × V (N/m2) × m3 = p × V N-m

  where,

V = Volume occupied by the fluid in

...(1.21) m3.

n Internal energy: It is an amount of energy possessed by a body or a system due to its molecular arrangement and their motion. It is denoted by “U”.

1.13

Units and Measurement

n Total energy: All engineering processes carry out the conversion of one form of energy to another form of energy. The sum of all the above-three types of energy is termed as total energy, and is denoted by “E”. It may be expressed as an equation given below

E = Potential energy + Kinetic energy

+ Internal energy

E = PE + KE + U

...(1.22)

For a stationary system, PE = 0 (as effect of gravity is negligible)

KE = 0

Then the equation becomes E = U. Energy is the thermodynamic property of a system. The energy is a point function. Its value depends on the end states only, and not on the path by which the change of state occurs in a process.

1.4.5.1  Law of Conservation of Energy The law of the conservation of energy states that the energy can neither be created nor be destroyed. One form of energy can be converted into another form. The conversion may be from electrical energy to mechanical energy, heat energy to mechanical energy, chemical energy to heat energy, and so on.

1.4.6 Heat It is a form of energy in transit, which occurs due to the temperature difference between two points in a substance or a body. Heat can be transferred across the boundary, only when a system at a higher-level temperature passes to a system at a lower level temperature. It can never happen in the reverse direction by nature. This condition will be discussed in the Second law of thermodynamics. Heat is measured in Joules or Kilo Joules. Heat can only be sensed due to the temperature difference. Heat can be transferred by three modes which are (a) Conduction (b) Convection and (c) Radiation. (a) Conduction: It is the transfer of heat within one or more bodies by physical contact, but in the absence of fluid motion. An example of conduction is a metal rod heated up at one end by an external source of heat. Heat is transferred from the source end to the other end of the rod by the movement of molecules from one end to the other end. (b) Convection: It is the transfer of heat within a fluid or gas, or between two media, by the motion of molecules from one region to another. Water heated up in a vessel is an example of convection heat transfer. (c) Radiation: It is the transfer of heat from a hot body to a cold one movement of particles in space or between the gas media.

1.14

Engineering Thermodynamics

Heat may be available in two forms in a substance. They are: (i) Sensible heat, and (ii) Latent heat (i) Sensible heat: The amount of heat contained in a unit mass of any substance by its temperature difference is called the sensible heat of the substance. The unit of the sensible heat in the SI unit is J/kg or kJ/kg. (ii) Latent heat: It is defined as the amount of heat added or rejected, to obtain the phase transformation of any substance. The unit of the latent heat in the SI unit is J/kg or kJ/kg. The three types of latent heat are: (a) Latent heat of melting or fusion (b) Latent heat of vaporization or evaporation (c) Latent heat of sublimation (a) Latent heat of melting or fusion: The amount of heat required per unit mass of a substance at melting point (constant) and at 1 atmosphere pressure, to convert it from the solid to the liquid state is called the latent heat of melting or fusion. (b) Latent heat of vaporization or evaporation: The amount of heat required to vaporize a unit mass of liquid at its boiling point (constant) and under 1 atmosphere pressure, is called the latent heat of vaporization. (c) Latent heat of sublimation: It is the amount of heat released for unit mass of substance to bring the gas to solid.

1.4.7 Work Consider 1 Newton of force is applied on a body so that it moves from point A to point B, which are 1 m apart form each other, as shown in Fig. 1.9. The amount of work done may be determined by multiplying the force and the distance travelled by the body. The unit of the work done is given in N-m or J. It is mathematically derived in mechanics.

Body F

A

Body

B

Fig. 1.9 Work done by force an a body

1.15

Units and Measurement



Work done = Force acted on the body × distance travelled by the body



W = F × l N-m.

There are a few forms of work transfer considered in thermodynamics. They are given below: (a) Displacement work (b) Shaft work (c) Electrical work (d) Paddle wheel work (e) Flow work (f) Work done of stretching (g) Work done in changing the area of a surface film (h) Magnetization of a paramagnetic solid (i) Free expansion (a) Displacement work: Let us consider a thermodynamic system, that consists of a cylinder and piston assembly, as shown in Fig. 1.10. The system is in thermodynamic equilibrium. The initial pressure and volume of the gas filled in the cylinder are p1, V1 respectively. When the gas is allowed to expand the piston moves a certain distance “x”. The final pressure and volume are p2 and V2 respectively. If the area of the cylinder is “A”, then the amount of work done by the gas in a small displacement is Displacement Work = Force × Distance travelled P

Fig. 1.10 Displacement work



dW = p × A × dx ...(1.23)



dW = F × dx      [As F = p × A]



dW = p dV ...(1.24)



W = ∫p dV ...(1.25)

(b) Shaft work: Consider a solid shaft which is rotated by an electrical motor, as shown in Fig. 1.11. There is an amount of work transfer as the rotation

1.16

Engineering Thermodynamics

of the shaft can raise a weight by a pulley action. The torque produced in the shaft is “T” m-N, and the rotational speed is “N” rpm then the shaft work obtained is

W = Angular velocity of the shaft × Torque produced in the shaft



W = v × t ...(1.26)



W = 2p N t watts

...(1.27)

Electric motor

Fig. 1.11 Shaft work

(c) Electrical work: If a current (I) flows through a resistor, with potential voltage V, then the rate at which the work done will be equal to the product of the potential voltage and the flow of current. This is shown in Fig. 1.12.

W = VI

Current (I) Fig. 1.12 Electrical work

(d) Paddle wheel work: Let us consider a fluid stored in a cylinder, as shown in Fig. 1.13. A stirring device is used for stirring the fluid inside the cylinder. At the end of the stirring rod a weight is attached. As the weight is lowered the stirring device stirs the fluid in the cylinder. This work is called Paddle wheel work.

System

Load (W)

Fig. 1.13 Paddle wheel work

The amount of work transferred into the fluid is

W = ∫t dw ...(1.28)

where,

t = Torque transmitted by the shaft



w = The rotating angle of the shaft.

1.17

Units and Measurement

(e) Flow work: When a fluid flows across the boundary of a system, the energy transferred by a device like a pump, to make the fluid flow through the required space, is termed as the flow energy. This is applicable in a flow process or in an open system. If the pressure and volume of the fluid are p and V respectively, then the flow energy through a small boundary is

W = pV

...(1.29)

(f) Work done in changing the area of a surface film: Due to the surface tension on the fluid, the surface of the liquid is reduced or shrunk. If the change in the surface area of the fluid is dA and the surface tension is s then the work done on the surface of the liquid due to the surface tension is given as:

W = – ∫ s dA ...(1.30)

The negative sign refers to the work done on the fluid. An example of this type of work done is expansion of ink drop on a surface of ink drop on a surface. Figure 1.14 shows the work done on a surface film. Initial surface area of film Final surface area of film

Fluid

Fig. 1.14 Work done in a surface film

(g) Magnetization of paramagnetic solid: The work done on a unit volume of a magnetic material through which the magnetic field is uniform, is given as

W = –∫ HdI ...(1.31)

where,

H = Field strength



dI = Component of magnetization.

(h) Work done on a stretching wire: A wire is stretched with some pull as shown in Fig. 1.15. When we stretch a wire there is a small work done on the wire. l

Fig. 1.15 Stretching a wire



W = –A E L

(∈22 − ∈12 ) ...(1.32) 2

1.18

Engineering Thermodynamics



A = Cross sectional area of the wire



E = Modulus of elasticity



L = Length of the wire



∈ = Strain

As the work is done on the wire to stretch it, the negative sign is placed in the value of the work transfer. Table 1.3 Various forms of Work Transfer Form of work

Force

Type of displacement

Work

Compression or expansion work

Pressure

Change in volume

pdV (flow process) –Vdp (non flow process) 2pNt

Shaft work

Torque

Rotation

Electrical work

Electric field

Polarization

EI

Flow work

Pressure

Volume

pV

Magnetic work

Magnetic field

Magnetic moment

–H.dI

Surface tension

Surface tension

Peripheral area

–sdA

Paddle wheel work

Torque

Change in volume

–tdV

Work in a stretching wire

Stress

Strain

–EALe2/2

Work transfer is a path function, and its value depends on the path by which the change of state occurs in a process, but not on the end states of the process.

Objective Type Questions 1. The barometric pressure at the base of a hill is 750 mm of Hg, and at the top 600 mm of Hg. If the average air density is 1 kg/m3, the height of the hill is approximately (a) 2000 m

(b) 3000 m

(c) 4000 m

(d) 5000 m

2. A new temperature scale °ρ in which the boiling and freezing of point of water at one atmosphere are 100°C and 300°C respectively. Correlate the scale with Centigrade scale. The reading of 0°ρ on the centigrade scale is (a) 0°C

(b) 50°C

(c) 100°C

(d) 150°C

3. Pressure reaches a value of absolute zero (a) At a temperature of –273 K (b) Under vacuum condition (c) At the earths centre (d) When a molecule momentum of a system become zero

1.19

Units and Measurement

4. Work done on the system in a free expansion process is (a) Positive

(b) Negative

(c) Zero

(d) Maximum

5. Which of the following is correct? (a) Absolute pressure = gauge pressure + atmospheric pressure (b) Gauge pressure = absolute pressure + atmospheric pressure (c) Atmospheric pressure = absolute pressure + gauge pressure (d) Absolute pressure = gauge pressure – atmospheric pressure 6. The temperature at which the entropy becomes zero is called (a) Absolute scale of temperature (b) Triple point temperature (c) Absolute zero temperature (d) None of these 7. The absolute zero pressure will be (a) When the molecular momentum of the system becomes zero (b) At sea level (c) At a temperature of 273 K (d) Under vacuum condition (e) At the centre of the earth 8. Absolute zero temperature is taken as (a) –273°C

(b) 273°C

(c) 237°C

(d) –373°C

Theory Questions 1. Define: (i) Force (ii) Pressure. 2. What are the elements of pressure? 3. What are the different measurements used to measure the pressure? Explain with sketches. 4. Name the different scales followed. 5. What is meant by absolute temperature? 6. With the help of a neat sketch, explain the principle of gas filled thermometer. 7. With the help of a neat sketch, explain the principle of an electrical resistance thermometer. 8. What is a pyrometer? 9. What are the different forms of energy? Define each of them. 10. What is the law of conservation of energy? 11. Define heat. What are the three modes of heat transfer?

1.20

Engineering Thermodynamics

12. Define Conduction, Convection and Radiation. 13. What is meant by sensible heat and latent heat? 14. Define the latent heat of fusion. 15. Define the latent of sublimation. 16. Define the latent of evaporation. 17. Define work. 18. Explain with sketches the different forms of work.

Unsolved Problems 1. A pressure gauge fitted on a vessel measures a gauge pressure of 25 bar. The atmospheric pressure is 1.03 bar. If the value of g is 9.81 m/s2, determine the absolute pressure in the vessel. [Ans: 26.03 bar] 2. A tank is filled with 15 litres of ideal gas at 760 mm pressure. The tank is connected to an evacuated vessel that has a capacity of 10 litres. Find the resultant pressure. [Ans: 1.5 bar] 3. A mercury manometer is used to measure the pressure drop of water flowing in a pipe line. If the difference of mercury levels in the two limbs of the manometer shows 10 cm, determine the pressure drop. Assume the density of mercury is 13.6 × 103 kg/m3 and 1 × 103 kg/m3. [Ans: 0.12 bar] 4. An ideal gas is filled in an insulated and rigid cylinder. A paddle wheel is inserted into the cylinder, and is rotated by an electric motor for 2 minutes. The power of the motor is 50 W, if the pressure of the gas is 10 bar and the temperature is 27°C. Calculate the final pressure and temperature of the gas. [Ans: 19 bar, 315°C] 5. A gas is filled in a piston cylinder assembly. The piston can move freely inside the cylinder. The piston is subjected to 1 bar of atmospheric pressure. A paddle wheel is placed in the cylinder volume, and rotated continuously till the volume of the gas is doubled. The specific heat of gas at constant pressure is 0.29 kJ/mol K and the room temperature is 27°C. Calculate the work done by the paddle wheel. [Ans: 8.73 kJ] 6. A capillary tube is closed at one end, and the other end is open and has a mercury pellet of length 19 cm. If it is held vertically with the open end downwards, the length of the air columns is 20 cm. If the atmospheric pressure is 760 mm of Hg, find the length of the air column when it is kept vertically with its opened end-up?

CHAPTER

2

INTRODUCTION TO IDEAL GASES

2.1 INTRODUCTION There are two types of gases based on their nature (i) ideal gases and (ii) real gases. A gas, which obeys the gas laws and the general gas equation pV = mRT, for all values of temperature and pressure, is called an ideal or perfect gas. A gas, which does not obey the gas equation or gas law, is called real gas. There are no attractive or repulsive forces between two molecules and occupying no space between them. Ideal gases can be completely evaporated from there liquid state. The values of p, V and T calculated, using the gas equation, differ from the observed values. It is observed from the study of various gases, that all of them exhibit ideal behavior only at high temperature and low pressure.

2.2

GAS LAWS

2.2.1 Boyle’s Law In 1662, Robert Boyle formulated the Boyle’s Law. It states “The absolute pressure of the given quantity of gas is inversely proportional to its volume when the temperature of the gas is held constant”. Suppose the pressure and volume of the gas are denoted by p and V then, Fig. 2.1 shows the change of volume of a gas as the pressure changes between two points 1 and 2. 1

p 2 V

Fig. 2.1 Change of volume with pressure

or, So,

p ∝ 1/V      (when T = Constant) pV = Constant p1V1 = p2V2

...(2.1) ...(2.2)

2.2

Engineering Thermodynamics

2.2.2 Charles’ Law In 1787, Jacques Charles formulated the Charles’ law. It states “The volume of the given quantity of a gas is directly proportional to its absolute temperature when the absolute pressure is held constant”. V ∝ T      (when p = Constant)

V/T = Constant

...(2.3)

Between two states 1 and 2, the equation may be given as V1 V = 2 ...(2.4) T1 T2



Gay Lussac confirmed Charles’ law in 1802 and found that the temperature of gas influences the volume of gas.

2.2.3 Gay Lussac Law It states “The absolute pressure of the given quantity of gas is directly proportional to the absolute temperature of it when the volume is held constant”. p ∝ T     (when V = Constant)

i.e.

p/T = Constant

or,

...(2.5)

Between two states 1 and 2, it may be given as

p1 p = 2 T2 T1



      (when

V = Constant)

...(2.6)

1 of the 273 volume of the gas at 0°C. The volume at temperature t°C is given by For each degree rise in temperature, the volume will increase by

V0 t Vt = V0 + V= ( t + 273 ) . ...(2.7) 0 273 273



2.3

EQUATION OF STATE

An equation which relates the pressure, volume and temperature of a gas is called the equation of state or characteristic equation of the gas. This is obtained by combining Boyle’s and Charle’s law. At constant volume or,

p ∝ T p/T = C

...(2.8)

At constant temperature or,

p ∝ 1/V pV = C

...(2.9)

2.3

Introduction to Ideal Gases

Combining the above two equations when both V and T vary, We get

pV ∝ T



pV = mRT ...(2.10)

For unit mass

pV = RT

...(2.10 (a))

Where R is called the characteristic gas constant. It is different for different gases. For air,

2.4

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

AVOGADRO’S HYPOTHESIS

According to Avogadro’s hypothesis, equal volumes of gases, under identical conditions of temperature and pressure, contain the same number of molecules. The masses of a given volume of different gases are proportional to their molecular weights. Let us consider two gases A and B, having different molecular weights MA and MB respectively.

MA ρA = ...(2.11) ρB MB

or,

VB M = A ...(2.12) VA MB

Otherwise

MAVA = MBVB

This shows that the product of molecular weight and volume of all the MV  gases are the same at a given temperature. The product   is known as  m  the molar volume of the gas, Consider the ideal gas equation (2.10(a))

pV = mRT

Multiply both the sides by the molecular weight of an ideal gas and devide by mass of gas occupying that volume. M pV mR T M = ...(2.13) m m pVmol = MRT pV MR = mol ...(2.14) T For many gases; MARA = MBRB = MCRC ...... = MURU.

2.5

...(2.15)

UNIVERSAL GAS CONSTANT ( R )

One mole of a gas refers to a quantity of gas, whose mass is numerically equal to the molecular weight of that gas. Thus, 1 g mole of oxygen refers to 32 g of

2.4

Engineering Thermodynamics

oxygen and 1 kg mole of oxygen refers to 32 kg of oxygen. To convert the mass of a gas into the molecular mass of that gas, m Number of moles n = M where, Thus

m = Mass and M = Molecular weight 64 g of O2 =

64 = 2 g mole of Oxygen 32

4 kg mole of N2 = 4 × 28 = 112 kg of Nirogen.

Avogadro’s law of states that 1 g mole of all gases at a pressure of 760 mm Hg and temperature of 0°C is equal to 22.4 litres. At

p = 760 mm Hg = 1.013 bar = 1.013 × 105 N/m2

t = 0°C i.e., T = 273.15 K

V = molar specific volume = volume of 1 mole of gas

= 22.4 litres/g mole

p V = R T

where R is the universal gas constant and applies to the mole of the gas.

R =

1.013 × 10 5 × 22.4 × 10–3 =8314.3 N-m/g mol K 273.15

= 8314.3 J/g mol K = 8.3143 kJ/g mol K. As per Avogadro’s hypothesis the number of molecules in one mole of any gas is the same and is known as Avogadro’s number (N). The value of N is 6.023 × 1026 molecules/kgmol. The Boltzman constant (k) is the ratio of the universal gas a constant ( R ) and Avogadro’s number (N). k = R /N k = 13.8 × 10–24 J/K.

2.6

INTERNAL ENERGY OF IDEAL GAS

The internal energy (U) of an ideal gas is the energy stored in the gas. It is the product of the mass of the gas, the specific heat capacity of the gas at constant volume and the temperature. It is expressed as follows:

2.7

U = mCvT kJ.

...(2.16)

ENTHALPY OF AN IDEAL GAS

The total enthalpy or heat content (H) of an ideal gas is given by the following expression:

2.5

Introduction to Ideal Gases



H = U + pV ...(2.17)

where,

U = Internal energy of the ideal gas, kJ



p = Pressure of the gas, N/m2



V = Volume of the gas, m3.

For an ideal gas

pV = mRT

Therefore,

H = U + mRT- ...(2.18)



H = mCvT + mRT ...(2.19)



H = mT(Cv + R)



H = mT(Cp)

or mCpT

...(2.20)

Since the first term “U”, the internal energy is dependent on the temperature and the second term is also temperature dependent, the enthalpy of an ideal gas is a function of temperature. or,

H = f (T) ...(2.21)

It is to be noted that the internal energy of an ideal gas depends on the specific heat capacity of the gas at constant volume and temperature, whereas the enthalpy of the gas depends on the specific heat capacity of the gas at constant pressure and the temperature of the gas.

2.8

JOULE’S LAW

Joule’s law states that the internal energy of an ideal gas depends only on its absolute temperature. If the temperature of the ideal gas changes from T1 to T2, the change in the internal energy remains the same, irrespective of the pressure and volume changes.

2.9

SPECIFIC HEAT CAPACITY

It is the quantity of heat required to raise the temperature of a unit mass of any substance by one degree. It is generally denoted by C. Let the mass of the substance be “m” and the specific heat capacity “C” and the temperature of the substance is raised from T1 to T2. Then the heat required to raise the temperature may be determined as follows:

Q = mC(T2 – T1)kJ ...(2.22)



C = Q/[m × (T2 – T1)] ...(2.23)

The values of specific heat for heating and cooling the gas may be taken in two ways: (a) Specific heat at constant pressure (Cp) and (b) Specific heat at constant volume (Cv). It is to be noted that the Cp value is always greater than the Cv.

2.6

Engineering Thermodynamics

2.10 SPECIFIC HEATS OF A GAS The specific heat of a substance is defined as the quantity of heat required to raise the temperature of one kg of that substance by 1°C. While solids and liquids have only one specific heat, the specific heat of a gas depends on the nature of the process and the following two specific heats are commonly used as the specific heats for other processes (a) Specific heat at constant pressure (Cp) (b) Specific heat at constant volume (Cv) The unit of specific heat is kJ/kg K or J/kgK.

2.10.1 Heating of a Gas at Constant Pressure Consider a gas contained in a cylinder. Let a frictionless piston move in the cylinder. Suppose the mass of the gas contained in the cylinder is ‘m’ kg. The gas is heated up at constant pressure and the temperature of the gas increase from T1 to T2. Due to the heating of the gas, the volume increases from V1 to V2. The piston moves and does work

Constant volume gas heating

P

2

1

V

(a)

(b)

Fig. 2.2 (a) Heating of gas at constant pressure (b) Change of volume with pressure

Total heat supplied to the gas

Q = mCp(T2 – T1) kJ q = Cp(T2 – T1) kJ/kg.

...(2.24) ...(2.25)

2.10.2 Heating of a Gas at Constant Volume Let the heat be supplied keeping the volume constant. As the volume remains constant, there is no work transfer. The temperature increases from T1 to T2 and the specific heat of the gas at constant volume is Cv,

Q = mCv(T2 – T1) kJ

...(2.26)

The entire heat supplied goes to increase the internal energy of the gas.

Q = DU = U2 – U1 = mCv(T2 – T1)kJ ...(2.27)

The internal energy is a property of the gas and its change, therefore, a function of temperature only.

2.7

Introduction to Ideal Gases

The solids and liquids are heated at constant pressure. The gases are heated at constant pressure or constant volume.

2.11 RELATION BETWEEN Cp AND CV For a constant pressure process,

Q = mCp (T2 – T1) ...(2.28)



Q = (U2 – U1) + p(V2 – V1) (from first law of thermodynamics)

or,

mCp (T2 – T1) = (U2 – U1) + p(V2 – V1) ...(2.29)

But the change in internal energy (U2 – U1) = mCv (T2 – T1)

...(2.30)

Subtituing the Eqn. (2.25) in Eqn. (2.24) we get;

mCp (T2 – T1) = mCv (T2 – T1) + p(V2 – V1)

= mCv (T2 – T1) + mR(T2 – T1) ...(2.31) Dividing the above equation by m(T2 – T1) in both sides we get;

Cp = Cv + R

...(2.32)

or, Cp – Cv = R

...(2.33)

Defining the ratio of two specific heats Cp and Cv as g Cp/Cv = g

...(2.34)

Cp – Cv = R

Solving Cv =

R

( g – 1)

...(2.35)

Substitute Cv value in Eqn. (2.33) and simplify the equation to get Cp

 R  Cp =  g .  g −1

2.12 MOLAR SPECIFIC HEATS OF GASES When a gas is heated up, the molecular weight of the gas is considered. This is because, the mass of the gas is equal to its molecular weight. Based on this, a molar specific heat is used for the calculations. The molar specific heat is defined as the amount of heat required to raise the temperature through one degree for one gram of substance. The molar specific heats of gases are taken at constant pressure and constant volume. They are denoted as Cp and Cv respectively.

2.13 PROPERTIES OF NON-REACTIVE MIXTURES Many thermodynamic calculations involve a homogeneous mixture of different gases. Such is the case of air, which is a mixture of nitrogen and oxygen, with a

2.8

Engineering Thermodynamics

small trace of argon and other gases present. A homogeneous mixture of gases can be regarded as a single substance if the constituents do not react chemically with each other. Suppose ideal gases A, B, C...i are mixed together. Let ma, mb, mc...mi be the mass of the components A, B, C...i and na, nb, nc and ni be the moles of the components respectively. Total mass of the mixture m = ma + mb + mc + md + ...... mi

...(2.36)

A description of the mixture may be given by its gravimetric analysis; this is the mass fraction of each constituent. For instance, the mass fraction of gas i in the mixture is gi =

mi mi = ...(2.37) ma + mb + ...mi ∑ mi i

If Mi stands for the molecular weight of the ith component then the number of moles of that component is given by mi ...(2.38) Mi Mole fraction of the ith component of a mixture, is defined as the ratio of the number of moles of that component to the total number of moles of the mixture.

ni =



xi = mole fraction of ith component =

where,

ni n n = total number of moles = na + nb +.....ni ...(2.39)

2.14 GIBBS-DALTON’S LAW The thermodynamic properties of a mixture of perfect gases may be calculated from the properties of the individual gases by means of Gibbs-Dalton Law. This law is in two parts; the first part is concerned with the intensive property – pressure and the second with the extensive properties, internal energy, enthalpy and entropy. Dalton’s law of partial pressures states that the pressure of a mixture of perfect gases is equal to the sum of the partial pressures that each component would exert if it alone occupied if it alone occupied the volume of the mixture at the temperature of the mixture. Consider a mixture of ideal gases, A, B, C and D. Let pa, pb, pc and pd be their partial pressures. If the gases are ideal, the perfect gas equation gives

pa = na R T/Va,

pb = nb R T/V,....., pi = ni RT ...(2.40)

p = nRT/V p a na pb nb pi ni = = xa , = = xb , = = x. ...(2.41) p n p n p n

2.9

Introduction to Ideal Gases

xa + xb + xc + ..... +xi = 1 ...(2.42) pa p p + i + .... i = 1 or p + p + p + ..... p = p. ...(2.43) a b c i p p p



2.15 AMAGAT-LEDUC LAW OF PARTIAL VOLUME The partial volume of a component of a mixture of ideal gases is the volume would be occupied by that component at the pressure and temperature of the mixture. Figure 2.3 shows the ideal gas mixture which has different gases A, B, C, D...i. Denoting the partial volume by VA, VB, VC and VD

= pVa n= n= ni RT a RT, pVb b RT, pVi

...(2.44)

pV = nRT ...(2.45) TA

T VA A

B

C

D

A nA

TB VB

TC VC

B nB

(a)

C nC

TD VD D nD

(b) Fig. 2.3 Ideal gas mixture

Va na Vb nb Vi ni = = xi ...(2.46) V= n= x A , V= n= xB , V n Already it is shown that na + nb + .... ni = 1  

...(2.47)

Va Vb Vi + = ..... = 1 ...(2.48) V V V

Va + Vb +....Vi = V.

...(2.49)

The total volume is thus equal to the sum of the partial volumes. Volumetric analysis of the ith component is defined as the ratio of the partial volume of the ith component to the total volume of the mixture. Vi x= mole fraction yi = = i V

...(2.50)

The internal energy, enthalpy and entropy of a mixture of ideal gases are respectively equal to the sum of the internal energies, enthalpies and entropies that each component would have, if it alone occupied the volume of the mixture at the temperature of the given mixture. Mathematically this may be stated as

U = Ua + Ub + Uc + .... = SUi ...(2.51)

2.10

Engineering Thermodynamics



H = Ha + Hb + Hc + .... = SHi ...(2.52)



S = Sa + Sb + Sc + .... = SSi. ...(2.53)

2.16 MOLECULAR WEIGHT OF A MIXTURE OF IDEAL GASES Consider the ith component of the mixture which has a mass mi and molecular weight Mi. If the number of moles ni, then

mi = ni Mi

If the mole fraction of the then

ith component

...(2.54) is xi, the total number of moles ni,



ni = xi n ...(2.55)



mi = xin Mi

...(2.56)

The total mass of the mixture is m = ma + mb + .... mi

...(2.57)

= xan ma + xb n mb + .... xi n mi

...(2.58)

= nSxiMi ...(2.59) Therefore, the molecular weight of the mixture is given by

M =

m nSxi M i = = Sxi M i ...(2.60) n n

Consider a mixture of gases A, B, C,....I whose masses are ma, mb, .... mi and partial pressures are pa, pb,....pi respectively. If p and V are the pressure and volume of the mixture, we have

p = pa + pb + ..... pi

...(2.61)



m = ma + mb + ..... mi

...(2.62)

We know that,

pV = mRT



paV = ma Ra T



pbV = mbRb T



piV = mi Ri T

= (ma Ra + mb Rb + ..... mi Ri)T ...(2.63) = mRT

m m  ma  R a + b R b + .... i R i  ...(2.64) R =  m m m  

= (ga Ra + gb Rb +..... gi Ri) =

∑ gi R i ...(2.65) i

Where gi is the gravimetric or mass fraction of ith component.

2.11

Introduction to Ideal Gases

2.17 GRAHAM’S LAW OF DIFFUSION Consider a gas contained in a vessel. Assume that the vessel is porous, so that the gas tends to escape from the vessel through the pores. This is called diffusion and the rate of diffusion depends on the gas. Graham’s law of diffusion states that the rate of diffusion, r of a gas is inversely proportional to r proportional to M, where r the density and M is the molecular weight. r =

For two gases A and B,

M ...(2.66) r

ra M a = rb M b

M =

=

rb ...(2.67) ra

m ma + mb + .....mi = ...(2.68) n n na ma + nb mb + ..... ni mi n

na nb n Ma M b + .... i M i = = n n n = xa Ma + xb Mb + ....xi Mi

...(2.69) ...(2.70)

= Sxi Mi ...(2.71) ni Where xi is the molar fraction of the ith component = . n

2.18 RELATIONSHIP BETWEEN VOLUMETRIC(MOLAR) AND GRAVIMETRIC ANALYSIS (MASS ANALYSIS) (a) Conversion from volumetric to gravimetric analysis: As shown earlier the volumetric analysis of a mixture of ideal gases is same as the molar analysis. Volumetric analysis of the ith component yi = ni . n Mass of the ith component = mi = xi Mi = yi Mi

Vi = molar analyzing input V

ith = xi =

...(2.72)

Total mass of the mixture m = xa ma + xb mb + .... xi mi ...(2.73) = ya ma+ ybmb + .... yi mi = Gravimetric fraction of the

ith

∑ yi M ...(2.74) i

yM m component = i = i i ...(2.75) m Sy i M i

Molecular weight of the mixture =

Sy i M i Sy i

. ...(2.76)

2.12

Engineering Thermodynamics

2.19 CONVERSION FROM GRAVIMETRIC TO MOLAR (VOLUMETRIC) FRACTION Let ma, mb, mc ..... + mi be the mass components A, B .....i of a mixture of gases. Total mass of the mixture m = ma + mb + mc ..... + mi

...(2.77)

Mass fraction of gravimetric fraction of A = ga = ma/m ...(2.78)

B = gb = mb/m ...(2.79)

Where Moles of

A = na = ma/Ma



B = nb = mb/Mb



I = ni = mi/Mi.

Total number of moles of the mixture n = na + nb + ..... ni =

ma m m + ..... i = ∑ Mi ...(2.80) Ma Mi i

ma Ma = volume fraction of A ...(2.81) m ∑ Mi i mi ni Mi = I = x= ...(2.82) i mi n ∑M i

na = Mole fraction of A = x= a n

Mole fraction of

Molecular weight of the mixture =

Ni =

m m a + mb + .... + mi ...(2.83) = mi n ∑M i

Ratio of individual mass fraction to mass of each constituent Sum of all ratios of mass fraction to mass constituent

The values of volume and mass fraction for a few gases are given in Table. 2.1. Table 2.1 Volume and Mass Fraction for a Few Gases Gas

Molecular weight, m

Value of n

Mass fraction, n/m

Fraction volume

Argon

40

0.4

0.1

0

Carbon Dioxide

44

0.18

0

0

Hydrogen

2

0.0

0

0

Nitrogen

28

22

0.79

0.8

Oxygen

32

6.7

0.21

0.2

2.13

Introduction to Ideal Gases

2.20 EQUATIONS RELATING MASS FRACTION WITH EXTENSIVE PROPERTIES OF THE MIXTURE OF GASES Let us consider the internal energy of gases be U, enthalpy be H and entropy be S. Total internal energy of the mixture is given U = mu = m 1u1 + m2u2 + .... miui =

∑ mi ui ...(2.84) i

m1Cv1 T + m2 Cv2 T + .... miCvi T = ∑ mi C vi T ...(2.85) i

The specific heat at constant volume Cv of the mixture would be given by mCv = m1 Cv1 + m2Cv2+ ..... + miCvi ...(2.86)



Cv =

m m1 m C v 1 + 2 C v 2 + ..... + i C vi ...(2.87) m m m

= g1Cv1 + g2Cv2 + ..... giCvi =

∑ giC vi ...(2.88) i

Where gi is the mass fraction of the ith component. The enthalpy of the mixture is given by

H = mh = m1h1+ m2h2+....+ mihi ...(2.89)



mCp T = m1Cp1 T + m2Cp2 T +.....+ miCpi T ...(2.90)

This gives the specific heat at constant pressure as Cp = g1Cp1 + g2Cp2 +.....+ giCpi =

∑ giC pi ...(2.91)

The above equations enable the internal energy and enthalpy of a mixture of ideal gases to be determined, in terms of the specific heats of the constituents. The entropy change may be calculated as follows: ds = C P

 dT   dp dp  dT ...(2.92) = −R −  ∑ gi R i   ∑ gi C Pi  t p  i  T  i  p

  T   p S2 – S1 =  ∑ gi C pi  ln 2 −  ∑ gi R i  ln 2 . ...(2.93)  i  T1  i  p1

Solved Problems Problem 2.1 The density of a gas at NTP is 1.3 kg/m3. Calculate the gas constant and molecular weight of the gas. Find the specific volume of the gas at 1300 kN/m2 and 308 K. Solution NTP means 0°C and 1 atm.

pV = mRT

2.14

Engineering Thermodynamics



ρ =

p as m = ρ when v = 1 RT



R =

p 1.013 × 10 5 = = 267.9 J/kgK rT 1.3 × 273



R 8314.3 M = = = 31.03 kg R 267.91

when m = 1 then v = vs, where Vs is the specific volume of g as



RT 267.91 × 308 = = 0.0365 m 2 kg. vs = 5 P 13 × 10



Problem 2.2 A balloon is considered as a sphere of 18 m in diameter. The surrounding air is at 16°C and barometer reads 75 cm of Hg. Determine the load the balloon can lift if it is filled with H2 at 21°C and atmospheric pressure. Solution V (volume of balloon) =

3

4 3 4 9 πR = × π   = 3053.6 m 3 3 3 1

R 8314.3 = = 4157.15 J/kgK. M 2 Pressure of H2 in balloon = Atmospheric pressure

R (for H2) =

= Mass of H2 in balloon =

75 × 1.013 × 105 = 99967.1 N/m2 76 pH2 VH2 RT

=

99967.1 × 3053.6 = 250.6 kg 4157.15 × ( 21 + 273 )

Volume of air displaced by balloon = Volume of balloon = 3053.6 m3 Mass of air displaced by balloon 75 × 1.013 × 10 5 × 3053.6 pa va 76 = = = 3693.1 kg 287 × (16 + 278) RaTa The load lifted by the balloon = Buoyant force. = (ma – mH2)g = (3693.1 – 250.6) × 9.81 = 33.771 kN. Problem 2.3 Determine the volume of a tank which has to store 2 kg mole of any gas at 293 K and 3 bar absolute pressure. What is the mass of (i) Hydrogen and (ii) Nitrogen that could be confined in this tank? Solution Using the universal gas equation

—

pVM = RT

—

where VM is the molar volume and R is universal gas constant.

2.15

Introduction to Ideal Gases

Vm = or,

RT 8314 × 293 = = 8.12 m3/kg mol. p 300 × 10 3

Volume of tank = 2 × 8.12 m3 = 16.24 m3

Mass of H2 which can be stored at the same pressure and temperature. = 2 × mH2 = 2 × 2 = 4 kg

Similarly the mass of N2 which can be stored at the same pressure and temperature = 2 × mN = 2 × 28 = 56 kg. 2

Problem 2.4 The quality of steel produced can be improved by providing oxygen instead of air for the reduction process. A steel company manufactures 10000 ton of steel per day. The oxygen supplied is 20 m3 per ton of steel produced. The volume of oxygen supplied is measured at 1 bar absolute pressure and 27°C. Determine the mass of oxygen required per day in the steel plant. Solution The mass of oxygen can be calculated by using the characteristic equation as

pV = mRT V = 104 × 20 = 2 × 105 m3/day

R 8314.3 = 260 J/kgK. = m 32 Substituting the values in the above equation

R =



m =

pV 1 × 10 5 × 20 × 10000 = RT 260 × ( 27 + 273 )

= 256,410 kg/day = 256.41 tons/day. Problem 2.5 Vessel A of 0.5 m3 capacity contains nitrogen at a pressure of 10 bar and 30°C. Vessel B contains 1 kg nitrogen at 5 bar – ab and 20°C. The vessels A and B are connected to each other through a value. After opening the connecting value and establishing equilibrium, the final temperature was measured and found to be 30°C. Determine the final pressure after mixing. Neglect the volume of the pipe connecting the vessels. A 3 Va = 0.5 m Ta = 30°C pa = 10bar

Valve

B Tb = 20°C pb = 5bar mb = 1kg

Fig. p. 2.1 Vessel filled with nitrogen

Solution The arrangement of the system is shown in Fig. p. 2.1.

2.16

Engineering Thermodynamics

RN = 2

R 8314.3 = = 297 J/kgK mN 2 28

The mass of nitrogen in vessel A is given by

ma =

pa Va 10 × 10 5 × 0.5 = = 5.56 kg. RTa 297 × 303

The volume of vessel B can be calculated as Vb =

mb RTb 1 × 297 × 293 = = 0.174 m3. pb 5 × 10 5

Total mass after mixing m = ma + mb = 5.56 + 1 = 6.56 kg Total volume after mixing V = Va + Vb = 0.5 + 0.174 = 0.674 m3 Applying the characteristic gas equation to the system

pV = mRT p =

mRT 6.56 × 297 × 303 = V 0.674

= 0.876 × 106 N/m2 or 8.76 bar.

Objective Type Questions 1. Boyle’s law relates the pressure and volume of ideal gas (a) When the temperature is constant (b) When the entropy is constant (c) When the enthalpy is constant (d) When the internal energy is constant 2. Charles’ law relates the volume and temperature of ideal gas (a) When the pressure is constant (b) When the entropy is constant (c) When the enthalpy is constant (d) When the internal energy is constant 3. Characteristic equation for gas (a) pV = mRT (b) p/V = mRT (c) pV = mR/T (d) pV2 = mRT 4. The gas constant R is the ratio of (a) The universal gas constant and molecular mass of the gas (b) The pressure and volume of the gas (c) The pressure and temperature of the gas (d) The volume and temperature of the gas

2.17

Introduction to Ideal Gases

5. The gas constant is the difference between (a) Pressure and volume

(b) Pressure and temperature

(c) Volume and temperature

(d) Cp and Cv

6. The adiabatic index is the ratio of (a) Pressure and volume

(b) Pressure and temperature

(c) Cp/Cv

(d) Volume and temperature

7. The value of the universal gas constant (a) 8.314 kJ/kgmolK

(b) 7.314 kJ/kgmolK

(c) 8.134 kJ/kgmolK

(d) 8.413 kJ/kgmolK

8. The volume of 1 kg of air at 0°C and 1.013 bar is (a) 0.664 m3

(b) 0.774 m3

(c) 0.674 m3ρ

(d) 0.647 m3

9. The value of the gas constant for air is (a) 287 J/kgK

(b) 297 J/kgK

(c) 279 J/kgK

(d) 278 J/kgK

10. The volume of gas at NTP (0°C and 1.013 bar) whose mass is equal to kgmol is (a) 22.14 m3/kgmol

(b) 21.12 m3/kgmol

(c) 22.4146 m3/kgmol

(d) 22.31 m3/kgmol.

Theory Questions 1. Define an ideal gas. 2. State Boyle’s law. 3. State Charles’ law. 4. Prove that the characteristic gas equation is given by pV = mRT. 5. What are the types of the specfic heat of gas? 6. Prove Cp – Cv = R. 7. State Avogadro’s hypothesis. 8. Define the Boltzman constant. 9. What is internal energy? What is the SI unit for it? 10. What is meant by enthalpy? 11. Using Amagat’s model, derive the expression for the calculation of entropy change during the mixing of two non reacting ideal gases. 12. State the Avogadro’s hypothesis and its significance. 13. What is an equation of state?

2.18

Engineering Thermodynamics

Unsolved Problems 1. 2 kg of a gas undergoes a non-flow reversible constant volume process from an initial pressure of 500 kN/m2 and temperature of 333K to a final pressure of 1600 kN/m2. Determine the heat added, change in enthalpy and entropy. 2. If one kg of O2 at 1.013 bar and 15°C has a specific volume of 0.7 m3 and the ratio of specific heats = 1.395, determine the values of the specific heats. 3. A system contains two vessels A and B with a connecting valve. Vessel A contains 4.29 m3 of N2 at 15.5 bar and 43.3°C. Vessel B contains 0.91 kg of N2 at 5.6 bar and 15.55°C. After connecting the vessels, the valve is opened; the resulting equilibrium temperature is 32.2°C. Find the final pressure. 4. An air pump is employed to extract air from a receiver of 2 m3 capacity which contains air at atmospheric pressure. If the pump draws air at a uniform rate of 7.5 litres/sec and assuming the temperature of the receiver remains constant, find the time taken by the pump for the pressure in the receiver to become 0.22 bar. Take the atmospheric pressure as 1.013 bar. 5. A sphere 2 m in diameter contains O2 at 21 bar and 27°C. How many drums of 15 cm diameter and 0.61 m long, which are initially devoid of any gas can be filled from this container to a pressure of 3.5 bar and 18°C. Assume that the temperature of O2 left in the sphere remains at 27°C. 6. A perfectly insulated system contains 0.0564 m3 of H2 at 23.9°C and 5.6 bar and it receives an input of paddle work at a constant pressure until the temperature is 65.5°C. Determine (a) the heat exchange (b) ∆u (c) Work input (d) Net work (e) ∆h and (f) ∆s. For H2 Cv = 9.52 and Cp = 13.68 kJ/kgK. 7. Determine the molecular volume of any perfect gas at 600 N/m2 and 30°C. The Universal gas constant may be taken as 8314 J/kg mole-K.

CHAPTER

3 3.1

THERMODYNAMIC SYSTEM AND FIRST LAW

LAW OF CONSERVATION OF ENERGY

The law of the conservation of energy states that energy can neither be created nor be destroyed. One form of energy can be converted into another form. The conversion of energy may be from electrical energy to mechanical energy, heat energy to mechanical energy, chemical energy to heat energy and so on.

3.2

THERMODYNAMIC SYSTEM

A thermodynamic system is defined as any space or any matter or group of matter within a prescribed boundary, considered for a thermodynamic analysis and the space outside the boundary of the system is called the surroundings. Both the system and the surroundings will together form the universe. The system, boundary and surroundings are shown in Fig. 3.1 (a). Boundary

Surroundings System Boundary

System Thermodynamic System

(a)

(b)

Fig. 3.1 (a) Illustration of the system, boundary and surroundings (b) Gas cylinder

A gas filled in a gas cylinder is an example of the thermodynamic system as shown in Fig. 3.1 (b). The gas filled in the tank is assumed as the system and the surface of the tank is the boundary. The space outside the tank forms the universe. The boundary of the system may either be fixed or movable.

3.2.1 Types of Thermodynamic Systems Thermodynamic systems are generally classified into three as given on next page:

3.2

Engineering Thermodynamics



Thermodynamic Systems |  ↓ ↓ ↓ losed System       Open System C Isolated System

3.2.1.1  Closed System A system that has a fixed mass in it and allows the transfer of energy only into the system and out of it, is called a closed system. For example, a gas is filled in a cylinder and heat is supplied to the gas from an external source; then, the system is said to be closed (Fig. 3.2). The mass of the gas is fixed. As the piston expands due to the rise in the temperature of gas, the work is done. Heat and work, flow across the boundary of the system. Cylinder Piston in initial position Piston in final position

Fig. 3.2 Closed system

3.2.1.2  Open System A system that allows mass and energy transfer into and out of the system is referred to as an open system, which is illustrated in Fig. 3.3. A good example of an open system is a centrifugal compressor as shown in Fig. 3.3. In a compressor, both the mass and energy flow across the system. Unlike the closed system, the mass is not fixed. Therefore, for thermodynamic calculations, the control volume is defined in open systems. Control volume

Mass in

Mass out

Fig. 3.3 Open system

3.2.1.3  Isolated System A system, which does not have a transfer of either mass or energy into and out of the system, is referred to as an isolated system. A good example of an isolated

Thermodynamic System and First Law

3.3

system is a thermo flask. In a thermo flask neither mass nor energy flows into the system or out of it.

3.3 TERMINOLOGIES There are a few important terms used in thermodynamics as given below: 1. Thermodynamic property 2. State 3. Change of state 4. Process 5. Cycle

3.4

THERMODYNAMIC PROPERTIES

Measurable physical quantities in units are called as the thermodynamic properties of a system. Example: Pressure, volume, temperature, density etc.

Thermodynamic property is classified into two types: 1. Intensive property 2. Extensive property

3.4.1 Intensive Property If the thermodynamic property of a system is independent of its mass, then it is called as an intensive property. Examples of intensive property are temperature and pressure.

3.4.2 Extensive Property If the thermodynamic property of a system depends on its mass, then it is called as extensive property. Examples of extensive properties are volume and energy. Suppose a mixture of gas A and gas B is filled in a tank. The volume of the tank is total volume. The volume occupied by the gas A is a specific volume. In this case, the volume of the gas depends on the mass of the gas. The specific extensive property of a system is considered to be the intensive property of the system. If the total internal energy of the system U is divided by the mass m, then we get the specific internal energy, which is the intensive property. U u = ...(3.1) m

3.5 STATE When the properties of a system are measured by its properties in a sufficient condition, then the system is said to be in a definite state. A gas filled in a

3.4

Engineering Thermodynamics

cylinder is bound to have a few described properties like pressure, temperature, volume etc. With such properties, if the system is said to exist, then it remains in a particular state. A graphical form with the help of some thermodynamic properties can describe the state of a system. For example, the p-V diagram in which the initial and final states are described by (p1, v1) and (p2, v2).

3.5.1 Change of State If there is a change in the property values of a system in a certain condition, then the system is said to have undergone a change of state. This is shown in Fig. 3.4. 2 p 1 v

Fig. 3.4 Change of state

3.6

THERMODYNAMIC PROCESS

If a thermodynamic system undergoes a change of state between two equilibrium states and the initial and final values of the properties are definite then the system is said to undergo a thermodynamic process.

3.7 PATH The way in which the final state is reached from the initial state, i.e., the number of changes of states in a process is called a path. When a system has its chemical composition and physical structure, then it is said that it exists in a phase. If a system exists in a single phase then the system is a homogeneous system. A system, which exists in more than one phase then it is called heterogeneous system.

3.8

THERMODYNAMIC CYCLE

If there are a number of thermodynamic processes carried out and the final state is brought back to the initial state, so that the initial conditions of the system are restored then the system is said to undergo a thermodynamic cycle.

3.9 EQUILIBRIUM We know that the thermodynamic system exists in a state. If there is no considerable change in any of its important properties, then the system is said to be in an equilibrium state. There are three conditions of equilibrium. They are

3.5

Thermodynamic System and First Law

(a) Mechanical equilibrium (b) Thermal equilibrium (c) Chemical equilibrium

3.9.1 Mechanical Equilibrium The state in which there is no unbalanced force between the system and the surroundings, is known as mechanical equilibrium.

3.9.2 Chemical Equilibrium If there is no chemical reaction or transfer of matter from one part of a solution to another, then the system is said to be in chemical equilibrium.

3.9.3 Thermal Equilibrium A system which satisfies the mechanical and chemical equilibrium and has no change in any of its properties, when it is separated from its surroundings by a diathermic wall, then the system is said to be in thermal equilibrium.

3.9.4 Thermodynamic Equilibrium A thermodynamic system that satisfies all the three equilibrium conditions said to be in the thermodynamic equilibrium.

3.9.5 Zeroth Law of Thermodynamics When a system A is in thermal equilibrium with a system B and system B is also in thermal equilibrium with system C, individually then the system A and B are in thermal equilibrium with each other.

3.10 REVERSIBLE PROCESS It is defined as a process, which proceeds in such a manner that the system apparently in equilibrium state at all times. It is also called a quasi static process. It is shown in Fig. 3.5. 2 p

1 Equilibrium states v

Fig. 3.5 Reversible process

3.6

Engineering Thermodynamics

3.11 WORK TRANSFER 3.11.1  Sign Convention of Work Transfer In thermodynamics, work transfer is an important parameter. The work done by the thermodynamic system is considered as positive work transfer, whereas the work done on the system is considered to be negative work transfer. This is illustrated in Fig. 3.6. Boundary

Work done on the system (Negative work value) System Surroundings

Work done on the system (Negative work value)

Fig. 3.6 Sign convention of work transfer

Examples: Positive work transfer: Shaft output from an engine. Negative work transfer: Work given to an air compressor by an electrical motor.

3.11.2  Work a Path Function Let us assume that a gas is filled in a cylinder. The initial state of the gas is A. The gas occupies a certain volume “V1” m3 and a pressure of “p1” bar. Let the gas change its state, such that it reaches the final state B. The final volume and pressure of the gas at state B are “V2” m3 and “p2” bar respectively. The change of state may happen along the paths A-1-B or A-2-B or A-3-B. It is known that the area under p-V represents the work done or work transfer in a system. This is shown in Fig. 3.7. B 3 p

2 1 A V

Fig. 3.7 Work A path function

It is clear from the figure that the work done or work transfer in the three processes A-1-B or A-2-B or A-3-B is quite different. Therefore, the work done

3.7

Thermodynamic System and First Law

or work transfer in a thermodynamic system does not depend only on the end states, but, also on the path by which the system reaches the final state from the initial state.

3.12 HEAT TRANSFER 3.12.1  Sign Convention of Heat Transfer Heat transfer is an important parameter in thermodynamic calculations, like work transfer. The heat that enters the system is considered as positive heat transfer. The heat that leaves out of the system is considered as negative heat transfer. This is illustrated in Fig. 3.8. Boundary Heat leaves out of the system (Negative heat value) System Heat transfer into the system (Positive heat value)

Surroundings

Fig. 3.8 Sign convention of heat transfer

Examples: Positive heat transfer: Heat supplied to a vessel. Negative heat transfer: Heat rejected from a room to the atmosphere. Heat transfer is a path function and its value depends on the path by which the change of state occurs in a process and not on the end states of the process.

3.13 FIRST LAW OF THERMODYNAMICS Thermodynamics deals only with various laws. The analysis of energy transfer in the thermodynamic process is derived from the first law of thermodynamics. The law is concerned with the transfer of energy and work. The first law of thermodynamics is formulated on the basis of the law of conservation of energy. The law states “When a thermodynamic system undergoes a cyclic process, there is some the algebraic sum of heat transfer is equal to the algebraic sum of work transfer”.

∫∮dQ = ∮dW

[Qalgebraic] = [Walgebraic]

...(3.2) ...(3.3)

It is a fact that heat and work are mutually convertible. But, when a system undergoes a change of state in a process, There is some amount of heat transfer in the system, in which a part of the heat energy will be converted into work energy and the remaining will be stored as internal energy in the system. In other

3.8

Engineering Thermodynamics

words, the net difference between the heat and work in a change of state of a system, will be stored in the system itself. Neglecting the kinetic and potential energy, the energy stored in the system is the internal energy of the system.

Q – W = ∆E

...(3.4)

or,

Q – W = ∆U

...(3.5)

Let us consider a system as shown in Fig. 3.9. Qb

Qa

System

Wa

Boundary Qb

Fig. 3.9 System with heat and work flow

The quantities of heat and work transfer of the system are given in the Fig. 3.9. Applying the first law of thermodynamics to a system has a change of state

(Qa – Qb) = (Wa – Wb) + ∆E

...(3.6)

or,

(Qa – Qb) = (Wa – Wb) + ∆U

...(3.7)

(Neglecting kinetic, potential and flow energies)

3.13.1  Perpetual Motion Machine of the First Kind (PMM–I) A device, which produces work without any heat input, is called a perpetual motion machine of first kind. It violates the first law of thermodynamics. In order to develop a certain amount of work, there must be a supply of heat to the device. Perpetual machine of motion machine of first kind (PMM - I) No heat PMM - I Work output

Fig. 3.10 Perpetual motion machine of first kind (PMM-I)

The following sign conventions are to be followed in the energy transfer calculations:

3.9

Thermodynamic System and First Law

• Heat flow into the system

= Positive

• Heat flow out of the system = Negative • Work done by the system

= Positive

• Work done on the system

= Negative

3.13.2  Various Thermodynamic Processes In thermal applications, the thermodynamic processes are carried out in two ways. They are: (i) Non-flow process and (ii) Flow process. The mass is fixed in a non-flow process. But, in the flow process the mass is unfixed. Again, the non-flow process is divided into two types as, (i) Reversible non-flow process and (ii) Irreversible non-flow process. The types are shown in Fig. 3.11. Non-Flow Process

Reversible non-flow process

Irreversible non-flow process

Fig. 3.11 Types of non-flow process

The flow processes are carried out in two types of thermal systems. They are; (i) Steady flow system in which the flow of fluid and energy are invariant with time and (ii) Unsteady flow system in which the flow of fluid and energy are variant with time, across the control volume. This is shown in Fig. 3.12. Flow Process

Steady flow process

Unsteady flow process

Fig. 3.12 Types of flow process

Now, let us consider the non-flow process. It is classified into a few types as shown in Fig. 3.13.

3.10

Engineering Thermodynamics

Reversible Non-Flow Processes Constant Volume Isochoric Process

Constant Pressure Isobaric Process

Constant Hyperbolic Temperature Process Isothermal Process

Reversible Adiabatic Isentropic Process

Polytropic Process

Fig. 3.13 Reversible non-flow processes

3.13.2.1  Constant Volume Process Assume that a gas of mass “m “is heated up in a cylinder. The initial temperature is T1 and final temperature is T2. The initial and final pressures of the gas in the system are p1 and p2 respectively. 1. P-V-T relationship: We know that the general gas equation is pV = mRT



...(3.8)

As the volume is constant throughout the process then the gas equation becomes p = Constant T p1 p2 p = = T1 T2 T



2 p 1 V

Fig. 3.14 Constant volume process q Work

transfer during the process



dW = p dV

...(3.9)

Integrating the Eqn. (3.9) between two limits 1 and 2 2

W1–2 = ∫ pdV 1

q Change



W1–2 = 0 (As there is no change in volume dV = 0) ...(3.10) in internal energy DU = mCv(T2 – T1) kJ

...(3.11)

3.11

Thermodynamic System and First Law q Heat

transfer during the process

As per first law of thermodynamics

Q = W1–2 + DU



Q = 0 + m Cv(T2 – T1)



Q = m Cv (T2 – T1)kJ

q Change



...(3.12) ...(3.13)

in enthalpy DH = m cp(T2 – T1)kJ

• There is no change in volume. Therefore, no work transfer occurs in the process. • The pressure and temperature increase when the gas is heated up. During the cooling of the gas, the pressure and temperature will decrease. • The amount of heat transfer during the process is equal to the change in internal energy.

3.13.2.2  Constant Pressure Process Assume that a gas of mass "m" is heated in a cylinder at constant pressure. The initial temperature is T1 and the final temperature is T2. The initial and final volumes of the gas in the system are V1 and V2 respectively. 1. p-V-T relationship: We know that the general gas equation is

pV = m RT

As the volume is constant throughout the process, the gas equation becomes

V = Constant T V1 V2 V = = T T1 T2

p

1

2 Work transfer W1–2 V

Fig. 3.15 Constant pressure process q Work

transfer during the process 2

W1–2 = ∫ pdV 1

3.12

Engineering Thermodynamics

W1–2 = p(V2 – V1) kJ

...(3.14)

Change in internal energy DU = m Cv(T2 – T1) kJ



...(3.15)

Heat transfer during the process: As per first law of thermodynamics

Q = W1–2 + DU



Q = p(V2 – V1) + mCv(T2 – T1)



Q = m R(T2 – T1) + mCv(T2 – T1)

[As pV = mRT]

...(3.16) ...(3.17)



Q = m(T2 – T1) [R + Cv] [since R + Cv = Cp] ...(3.18)



Q = mCp(T2 – T1)kJ

...(3.19)

DH = mCp (T2 – T1)kJ

...(3.20)

q Change

in enthalpy



• There is no change in the pressure during the process. Work transfer depends mainly on the change in volume. • The heat transfer in the process depends on the temperature at the beginning and end of the process. The heat transfer during the process is equal to the change in enthalpy.

3.13.2.3  Constant Temperature Process Assume that a gas of mass “m “is heated in a cylinder. The initial temperature is T1 and final temperature is T2. The initial and final pressures of the gas in the system are p1 and p2 respectively. 1. p-V-T relationship: We know that the general gas equation is

pV = mRT

As the temperature is constant during the process, the gas equation becomes

pV = Constant 1 p 2 V

Fig. 3.16 Constant temperature process

Hence the process is hyperbolic. The p-V diagram is shown in Fig. 3.16.

Thermodynamic System and First Law q Work

3.13

transfer during the process

dW = p dV From the p-V-T relationship p1 V1 p2 V2 = p = ...(3.21) V V Substituting the work done in the equation and integrating between the two limits 2

W1–2 = ∫ pdV 1

W1–2 =

2

∫1

 p1 V1    dV  V 

V  = p1V1 ln  2  ...(3.22)  V1  q Change in internal energy DU = mCv(T2 – T1)kJ ....(3.23) DU = 0 kJ [As there is no change in temperature T2 = T1] q Heat transfer during the process As per first law of thermodynamics Q = W1–2 + DU • The pressure and volume are inversely proportional to each other. When the pressure increases in the compression, the volume decreases. The pressure of a gas decreases in the expansion and hence volume increases. • Since there is no change in the internal energy of the system, as the temperature remains constant, the heat transfer during the process is fully converted to work transfer. V  Q = W1–2 = p1V1 ln  2  + 0 ...(3.24)  V1  V  = p1V1 ln  2  ...(3.25)  V1  q Change



in enthalpy DH = mCp(T2 – T1) kJ ...(3.26) DH = 0 kJ [As there is no change in temperature T2 = T1]

3.13.2.4 Reversible Adiabatic Process Assume that a gas of mass “m” is heated in a cylinder, which is insulated so that there will be no effect of the outside temperature. The initial temperature

3.14

Engineering Thermodynamics

is T1 and the final temperature is T2. The initial and final pressures of the gas in the system are p1 and p2 respectively. The initial and final volumes of the gas in the system are V1 and V2 respectively. The adiabatic index is g. No heat transfer will occur during the process. 1. p–V–T relationship: We know that the general gas equation is

pV = mRT

As the volume is constant throughout the process, the gas equation becomes

pVg = Constant



pVg = p1V1g = p2V2g 1 Pv = Constant P 2 W1–2



V

Fig. 3.17 Adiabatic process q Work



transfer during the process dW = p dV

From the p-V-T relationship

p =

p 1 V1g

=

p 2 V2g

...(3.27) Vg Vg Substituting the work done in the equation and integrating between the two limits, 2

W1–2 = ∫ p d V 1

W1–2 = ∫

2

1

(

g  p1 V1  g  V

)  dV

...(3.28)

 

( −g+ 1)  gV = p1 V1    –g + 1 

=

p1 V1g

(

...(3.29)

 V2( −g + 1) − V1−( g + 1)   ( –g + 1) 

(

 V2( −g+ 1) − V1−( g+ 1) g = p2 V2   ( –g + 1) 

)   

)   

...(3.30)

3.15

Thermodynamic System and First Law

Multiplying with inside terms and rearranging the terms, we will get

(

 p2 V2g × V2(– g+ 1) – p1 V1g × V1–( g+ 1) =   ( –g + 1)   ( p2 V2 – p1 V1 )  =    ( –g + 1) 

)   

...(3.31) ...(3.32)

 ( p2 V2 – p1 V1 )  =   ( 1 – g )  

...(3.33)

 ( p2 V2 – p1 V1 )  =   ( 1 – g )  

...(3.34)

 ( p1 V1 – p2 V2 )  W1–2 =   kJ ( g – 1)  

...(3.35)

q Change

in internal energy DU = m Cv(T2 – T1)kJ q Heat transfer during the process As per first law of thermodynamics Q = W1–2 + DU Q = 0 q Change in enthalpy DH = mCp(T2 – T1) kJ

...(3.36)

...(3.37) ...(3.38) ...(3.39)

3.13.2.5  Polytropic Process Assume that a gas of mass “m” is heated polytropically in a cylinder. The initial temperature is T1 and the final temperature is T2. The initial and final pressures of the gas in the system are p1 and p2 respectively. The initial and final volumes of the gas in the system are V1 and V2 respectively. The polytropic index is n. 1. p–V–T relationship: We know that the general gas equation is

pV = m RT

As the volume is constant throughout the process, then the gas equation becomes 1 p

n

pV = Constant 2 V

Fig. 3.18 Polytropic process

3.16

Engineering Thermodynamics



pVn = Constant



pVn = p1V1n = p2V2n

q Work



transfer during the process dW = p dV

From the p-V-T relationship

p =

( p1 V1n ) = ( p2 V2n )

...(3.40) Vn Vn Substituting in the work done equation and Integrating between two limits 2

W1–2 = ∫ p dV 1



W1–2 =

2

∫1

(

n  p1 V1  n  V

)  dV ...(3.41)  

 V( −n + 1)  = p1 V1n   – n + 1   

 ( − n + 1) – V1( − n + 1) n V = p1 V1  –n + 1 

  

...(3.42)

 ( − n + 1) – V1( − n + 1) n V2 = p2 V2  –n + 1 

  

...(3.43)

Multiplying with inside terms and rearranging the terms, we will get

 p2 V2n × V2 ( – n + 1) – p1 V1n × V1( – n + 1) =  – n+1 

   

 ( p2 V2 – p1 V1 )  =    ( –n + 1 ) 

...(3.44)

 ( p2 V2 – p1 V1 )  =   ( 1 – n )  

...(3.45)

 ( p2 V2 – p1 V1 )  =   ...(3.46) ( 1 – n )  

 ( p1 V1 – p2 V2 )  =   ...(3.47) ( n – 1)   (T1 – T2 ) = mR ( n – 1)

3.17

Thermodynamic System and First Law q Change

in internal energy DU = mCv(T2 – T1) kJ

q Heat

...(3.48)

transfer during the process

As per first law of thermodynamics

Q = W1–2 + DU

...(3.49)

(T – T2 ) = mR 1 + m Cv(T2 – T1) kJ ( n − 1)

...(3.50)

( T1 – T2 ) + mC (T – T ) kJ ...(3.51) = m C p – C v v 2 1 ( n – 1)

(

)

(

)

Cp – Cv – C v ...(3.52) = m ( T1 – T2 ) ( n – 1)

 ( g – 1)  – 1 C v ...(3.53) = m ( T1 – T2 )   ( n – 1)   ( g – 1) – ( n – 1)  = m ( T1 – T2 )   C v ...(3.54) ( n – 1)    ( g – n)  C v  R = C p – C v ; ...(3.55) = m ( T1 – T2 )   ( n – 1)  = C v ( g – 1) Substituting CV =

R in Eqn. (3.55), we get ( v – 1)

Q = m ( T1 – T2 )



= m ( T1 – T2 ) =

R ( g – n) C v ( n – 1)

...(3.56)

R ( g – n) ( g – 1)

W1− 2 ( g – n ) ( g – 1)

...(3.57) ...(3.58)

Change in enthalpy l It

DH = m Cp(T2 – T1) kJ.

...(3.59)

is important to note that the sign conventions for heat should be carefully followed in the case of heating or expansion and cooling or compression of gases. The expansion of gases is considered to be positive work and the cooling of gas is considered as negative work. Similarly, the expansion or heating of gas is considered as positive heat transfer, whereas the cooling or compression of gas is negative heat transfer. Table 3.1 gives the sign convention used in heat and work transfer calculations.

3.18

Engineering Thermodynamics Table 3.1 Sign Conventions for Heat and Work Transfer

Heating or Expansion

Cooling or Compression

Work Transfer

Positive

Negative

Heat Transfer

Positive

Negative

3.14 FLOW PROCESSES OR OPEN SYSTEMS In open flow systems, there are both mass and energy transfer into the system or out of the system. An example, is an air compressor that is commercially available in cycle shops or fuel filling stations, which is used for the inflation of air.

3.14.1  Steady Flow Energy Equation (SFEE) The open systems or flow processes may be of time dependent or independent. Generally, the time dependent flow process is referred to as the steady flow and the other is the unsteady flow process. Let us consider an open system which is shown in Fig. 3.19. Working fluid in

Control volume

Qin

Work W1-2

Z1

Z2 Working fluid out

Fig. 3.19

The following factors are taken into account.

m1 = Mass flow rate of the fluid at inlet (kg/sec)

m2 = Mass flow rate of the fluid at outlet (kg/sec) A1 = Area at inlet (m2) A2 = Area at outlet (m2)

p1 = Pressure of the fluid at inlet (KN/m2)



p2 = Pressure of the fluid at outlet (KN/m2)

3.19

Thermodynamic System and First Law



v1 = specific volume of the fluid at inlet (m3/kg)



v2 = specific volume of the fluid at outlet (m3/kg)

V1 = Velocity of the fluid at inlet (m/s) V2 = Velocity of the fluid at outlet (m/s)

U1 = Internal energy of the fluid at inlet (kJ)



U2 = Internal energy of the fluid at outlet (kJ)

Z1 = Elevation of inlet from the ground level (kJ) Z2 = Elevation of outlet from the ground level (kJ)

W = Work transfer during the process (kJ)



Q = Heat transfer during the process (kJ)

A continuity equation is given as, the rate of flow of a substance considering mass, entering the control volume of a steady flow system is equal to the rate of flow of a substance leaving the system. m =



A 1 V1 A 2 V2 = v1 v2

...(3.60)

As per the energy balance at both sides of the inlet and the outlet. The sum of all energies at the inlet = Sum of all energies at the outlet Heat transfer + [Kinetic energy + Potential energy + Flow energy + Internal energy]inlet = Work transfer + [Kinetic energy + Potential energy + Flow energy + Internal energy]outlet         V2 V2 Q +  m1 1 + m1 gZ 1 + m1 p1 v1 + m1 u1   = W +  m2 2 + m2 gZ 2 + m2 p2 v2 + m2 u2   2 2        

V2 V 2     Q − W =  m2 2 − m1 1   + {[ m2 gZ 2 + m1 gZ 1 ] + [ p2 v2 ] m2 − [ p1 v1 ] m1 + [ u2 – u1 ]} 2 2    



...(3.61)

{

}

Q −= W m  V22 − V22  + mg [ Z 2 – Z1 ] + [ p2 v2 − p1 v1 ] + [ u2 + u1 ] \(m1 = m2 = m) ...(3.62) Q – W = rKE + rPE + [(p2V2 + U2) – (p1V1 – U1)]

...(3.63)

Q – W = rKE + rPE + [H2 – H1] ...(3.64) Q – W = r KE + rPE + rH kJ

...(3.65)

Since the mass flow rate of the substance is taken in kg/s, the energies are obtained in kJ/s.

3.20

Engineering Thermodynamics

3.14.2  Applications of Steady Flow Processes Most of the thermal equipments are considered as open systems or steady flow processes. The applications of steady flow processes, according to their use are given below Applications of steady flow processes with examples

A. Heat absorption systems 1. Boiler

B. Heat rejection systems

C. Work developing systems

D.Work absorbing systems

1. Condenser

1. Water turbine 2. Steam /gas turbine

1. Reciprocating air compressor 2. Rotary air compressor

3. IC Engine

3. Centrifugal pump

2. Heat exchanger

E.Other systems

Steam Nozzle

3.15 HEAT ABSORBING SYSTEMS In these systems heat energy is absorbed or gained by the system so that the temperature and enthalpy of the system will increase. The amount of heat absorbed or gained by the system will be the difference in the enthalpy at the outlet and inlet of the process.

3.15.1 Boiler The boiler is a device or an equipment in which steam is generated by adding heat to water. The boiler is also called as steam generator. Figure 3.19(a) shows the flow diagram of a steam boiler and Fig. 3.19(b) shows the sectional image of a steam boiler. The boiler is assumed to be a steady flow heat absorbing system. The assumptions made for the boiler are given below 1. No work transfer (W = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible The steady flow energy equation is

Q – W = r KE + rPE + rH kJ



Q = rH

...(3.66)



Q = [H2 – H1] kJ

...(3.67)

3.21

Thermodynamic System and First Law Flue gas out Steam out

Flue gas out

Steam out

Water in

Water in Heat in Q1–2

(a)

(b)

Fig. 3.19 (a) Flow diagram of steam boiler (b) Sectional image of steam boiler

3.15.2 Evaporator An evaporator is a container in which fluid is raised to vapour with the absorption of heat. An example is evaporator coil in a refrigerator. The enthalpy of the fluid increases. Figure 3.20(a) illustrates the arrangement of an evaporator and Fig. 3.20(b) shows the image of an evaporator. The following assumptions are taken for the evaporator. 1. No work transfer (W = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible The steady flow energy equation is

Q – W = rKE + rPE + rH kJ



Q = rH



Q = [H2 – H1] kJ

Fluid in

EVAPORATOR

...(3.68)

Fluid out

Heat in Q1–2

(a)

(b)

Fig. 3.20 (a) Flow diagram of an evaporator (b) Image of an evaporator

3.22

Engineering Thermodynamics

3.16 HEAT REJECTION SYSTEMS In these systems, the heat energy is rejected or lost in the process so that the temperature and enthalpy of the system will decrease. The amount of heat rejected or lost by the system will be the difference in the enthalpy, at the inlet and outlet of the system.

3.16.1 Condenser The image of a steam condenser is shown in Fig. 3.21. A condenser is a device which receives hot fluid from a turbine or refrigerator to condense it, and maintain the pressure of the fluid at the exit of the condenser below that of the atmosphere. In the condenser the following assumptions are noted 1. No work transfer (W = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible The flow diagram and image of the steam condenser is shown in Fig. 3.22. The steady flow energy equation is given below:

Q – W = r KE + rPE + rH kJ



Q = rH



Q = [H2 – H1]



–Q = [H2 – H1]

since heat is taken out of the system  ...(3.69)

Q = [H1 – H2 ]

Water outlet

To ejector vacuum system

Steam Baffle

Flanged cover plate

Tubesheet

Flanges

Baffle

Baffle

Water inlet

Tubesheet Condensate

Fig. 3.21 The image of a surface condenser

3.17 WORK DEVELOPING SYSTEMS In these systems, work is developed or produced in the process, so that the work output will be achieved in the system. The amount of work developed

3.23

Thermodynamic System and First Law Hot fluid

Coolant in

CONDENSER

Coolant out

Condensate

Fig. 3.22 Flow diagram and image of a steam condenser

or produced by the system will be the difference in the enthalpy at the inlet and outlet of the system.

3.17.1 Turbine A turbine is a device, which utilizes the heat contained by steam or gas and performs work. By losing the enthalpy of steam or gas, work is produced by the turbine. The flow diagram of a steam turbine is shown in Fig. 3.23(a) and the sectional view of the steam turbine is shown in Fig. 3.23(b). The following assumptions are considered for deriving the equation for a turbine 1. No heat transfer (Q = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible The steady flow energy equation is

Q – W = r KE + rPE + rH kJ –W = rH –W = [H2 – H1] W = [H1 – H2] W = [H1 – H2] ...(3.70) Steam in

Steam or gas in

Turbine work W1–2

Steam or gas out Steam out

Fig. 3.23 (a) Flow diagram of a steam turbine (b) Sectional view of a steam turbine

3.24

Engineering Thermodynamics

3.18 WORK ABSORBING SYSTEMS In these systems, work is given or absorbed in the process, so that work will be given to the system. The amount of work absorbed or given to the system will be the difference in the enthalpy at the inlet and outlet of the process.

3.18.1  Rotary Compressor Figure 3.24(a) shows the flow diagram of the rotary compressor and Fig. 3.24(b) shows the image of a rotary compressor. The assumptions made for the rotary compressor are given below To condenser Air out

Discharge port

Rotor vane

Discharge reed

Rotor slot

Compressor work W1–2

Air in

Suction port

(a)

Oil

Cylinder Rotor

(b)

Fig. 3.24 (a) Flow diagram of the rotary compressor (b) Image of the rotary compressor

1. No heat transfer (Q = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible The steady flow energy equation is

Q – W = r KE + rPE + rH kJ



–W = rH



–W = [H2 – H1]                 



W = [H1 – H2] ...(3.71)

3.18.2  Reciprocating Compressor The line diagram and photograph of a reciprocating air compressor are shown in Fig. 3.25(a) and Fig. 3.25(b) respectively. The assumptions made for the reciprocating air compressor are as follows 1. Change in kinetic energy is negligible 2. Change in potential energy is negligible

3.25

Thermodynamic System and First Law

The steady flow energy equation is

Q – W = r KE + rPE + rH kJ



Q – W = rH



Q – W = [H2 – H1 ]



Q = W + [H2 – H1] ...(3.72) Air out

Air in

Air in Piston Cylinder

Air out through discharge port

(a)

(b)

Fig. 3.25 (a) Line diagram of the reciprocating compressor (b) Photograph of the reciprocating compressor

3.19 OTHER SYSTEMS 3.19.1  Steam Nozzle Let us take example of a domestic pressure cooker. At the top of the lid a nozzle is fitted, through which the a steam escapes when the pressure inside the cooker shoots up. Steam nozzles are generally used in steam turbines, in which the velocity of steam is increased and the pressure reduces when the steam passes through a small cross section. The increased velocity of the steam is allowed to strike over the steam turbine blades, to rotate the turbine by which mechanical work is obtained. The entire steam nozzle is assumed to be fully insulated. 1

2

Fluid out

Fluid in

1

2

 

(a)

     

(b)

Fig. 3.26 (a) A steam nozzle (b) A steam nozzle

3.26

Engineering Thermodynamics

The steady flow energy equation is

Q – W = r KE + rPE + rH kJ



rH = m[V22 – V12] ...(3.73) H2 – H1 = m[V22 – V12] ...(3.74)

[V22 – V12] = [2g(H2 – H1)]/m ...(3.75) V22 = V12 + [2g(H2 – H1)]/m ...(3.76) V22 = V12 +[2g(h2 – h1)]

...(3.77)

V22 = V12 +

...(3.78)

[2 gr h]

Therefore, the velocity of the steam at the exit of the nozzle, 2 V2 =  V1 + 2g∆ h  .

...(3.79)

3.20 DETERMINATION OF WORK TRANSFER IN A STEADY FLOW PROCESS The work done in a non-flow process is given as w = p dV. The work transfers in a steady flow process differs from the non-flow process. According to the first law of thermodynamics, the heat transfer in a change of state is given as q = du + pdV



...(3.80)

The steady flow energy equation is that Q – W = du + d(p V) + d(KE) + d(PE)



...(3.81)

= du + p dV + Vdp + d(KE) + d(PE) q – w = dq + Vdp + d(KE) + d(PE)



dq = Vdp + d(KE) + d(PE)

...(3.82) ...(3.83)

1

2

p Vdp V

Fig. 3.27 Work done in a steady flow process

Neglecting the kinetic and potential energies at initial and final conditions, we get

dW = –Vdp

Integrating between two states 1 and 2

...(3.84)

3.27

Thermodynamic System and First Law 2

2

         ∫ dW = 0 – ∫1 Vdp

...(3.85)

1

2

W1-2 = – ∫ Vdp

...(3.86)

1

Applying the above equation for various processes, we get various work transfer equations as given in Table 3.2. Table 3.2 Work Transfer for Various Processes in Steady Flow Thermodynamic

Expansion work

process

Compression work

W1–2 = –∫Vdp

Constant volume Constant Pressure

W1–2 = –∫Vdp

–V(p2 – p1)

V(p1 – p2)

0

0 V  p1V1 log  2   V1 

 V2    V1 

p1V1 log 

Constant temperature Reversible adiabatic Process

 υ ( p1 V1 − p2 V2 )  ( υ – 1)

 υ ( p1 V1 − p2 V2 )  –  ( υ – 1)

Polytropic process

n ( p1 V1 − p2 V2 )  ( n – 1)

n ( p1 V1 − p2 V2 )  – ( n – 1)

Summary: The first law of thermodynamics is applied to the following; (A) Thermodynamic cycle ∫dQ = ∫dW (B) During a change of state in a process Q = W + DU Table 3.3 gives the p-V-T relationship, work transfer, change in internal energy, heat transfer, change in enthalpy for different thermodynaic processes. It also gives the p-V and p-T diagram of different thermodynamic processes of non-flow processes. Table 3.3 Different Thermodynamic Processes and their Property Relationships Type of process Constant volume process (Isochoric) Constant pressure process (Isobaric)

p-V-T relationship p/T = C

Work transfer

Change in internal energy

Heat transfer

Change p-V diagram in enthalpy

0

mCv DT

m Cv DT

m Cp DT

p/T = p1/T1

V/T = C

V/T = V1/T1 = V2/T2

2

1 p

p

= p2/T2

p-T diagram

2

1 V

T

p(V2 – V1)

m Cv DT

m Cp DT

m Cp DT

1

1 p

2 V

p

2 T

[Table Cont...]

3.28

Engineering Thermodynamics pV = C pV = p1V1 = pV2

Constant Temperature process (Isothermal)

2.3 p1V1 log10 (V2/V1)

0

2.3 p1V1 log10 (V2/V1)

0

1

1 p

2

p

2

V

Hyperbolic process

pV = C pV = p1V1 = pV2

2.3 log10 V1)

p1V1 (V2/

0

2.3 p1V1 m Cv DT log10 (V2/V1)

T

1 p

2 V

pVg = C pVg = p1Vg1 = pVg2

Reversible adiabatic process

[p2V2 – p1V1] /(g – 1)

0

m Cv DT

m Cp DT

1

1 p

p

2

2 V

Polytropic process

pVn = C pVn = p1V1n = pV2n

[p2V2 – p1V1] /(n – 1)

T

m Cv DT [p2V2 – p1V1] / m Cp DT Varies as pern the value (n – 1) + mCv of “n” (T2 – T1)

Solved Problems Problem 3.1 In a chemical processing industry various thermodynamic systems were involved, where a particular thermodynamic system undergoes a thermodynamic cycle. Four processes are involved in the cycle. The values of work transfer, heat transfer and internal energy in each process are tabulated in Table p. 3.1. Find the rate of work. Table p. 3.1 Work, Change Internal Energy and Heat Transfer in Thermodynamic Processes Thermodynamic process

Work transfer kJ/min

Change in internal energy kJ/min

Heat transfer kJ/min

A-B

250

—

300

B-C

–800

400

—

C-D

–600

D-A

—

Solution Process A – B QA – B = WA – B + DUA – B

300 = 250 + DU



DUA – B = 50 kJ/min

Process B – C QB – C = WB – C + DUB – C

QB – C = – 800 + 400

QB – C = – 400 kJ/min Process C – D QC – D = WC – D + DUC – D

200 = –600 + DUC – D

200 100

50

3.29

Thermodynamic System and First Law

DU = 800 kJ/min



Process D – A QD–A = WD – A + DUD – A

50 = W + 100



W = –50 kJ/min Thermodynamic cycle ∫dQ = ∫dW



Q algebraic = Walgebraic

QA–B + QB–C + QC–D + QD–A = WA–B + WB–C + WC–D + WD–A

300 – 400 + 200 + 50 = 250 + 400 – 600 + 100.

Problem 3.2 A metal drum with a diameter of 1.1 m is filled with 1 kg mole of nitrogen at 100°C. The gas is cooled to 60°C at constant volume. Determine the work transfer, change in internal energy and change in enthalpy and heat transfer. The ratio of the specific heats is 1.4 and 1 kg of mole nitrogen is 28 kg. Assume that the length of the cylinder is equal to diameter of the cylinder. Given

m = 1 kg – mole = 28 kg;

Solution Constant volume

V = 1.04 m2

T1 = 100 + 273 = 373 K; T2 = 60 + 273 = 333 K;

g = 1.4. V T1

Constant volume V = 1.04 m3

T V

Fig. p. 3.1



Cylinder diameter = 1.1 m

π 2 dL 4 π = × 1.12 × 1.1 4 π = × 1.33 4



Volume of cylinder =

V = 1.04 m3 R R = u M =

8.314 28

3.30

Engineering Thermodynamics

= 0.297 kJ/kg K. Cv = R/(g – 1) = 0.297/(1.4 – 1) = 0.74 kJ/kg K. Cp = Cv × g = 0.74 × 1.4 = 1.036 kJ/kg K. 1. Work transfer during the process 2

W1–2 = ∫ p dV 1

W1–2 = 0 (As there is no change in volume dV = 0) kJ. 2. Change in internal energy

DU = m cv(T2 – T1) kJ



DU = 1 × 0.74 × (333 – 373)



DU = 1 × 0.74 × (–40)



DU = –21.904 kJ/kg

3. Heat transfer during the process As per first law of thermodynamics

Q = DU



Q = –21.904 kJ/kg

4. Change in enthalpy

DH = m cp(T2 – T1) kJ



DH = 1 × 1.036 × (333-373)



DH = 1 × 1.036 × (–40)



DH = –41.44 kJ/mole.

Problem 3.3 In a heating system, 0.5 kg of a perfect gas is heated from 100° to 300°C at a constant pressure of 2.8 bar. Determine the work transfer, change in internal energy, change in enthalpy and heat transfer. Take the gas constant R = 0.28 kJ/kg K. Cp = 1 kJ/kg K and Cv = 0.72 kJ/kg K Given

m = 0.5 kg;



p = 2.8 bar = 2.8 × 105 N/m2;

T1 = 100 + 273 = 373 K; T2 = 300 + 273 = 573 K; Solution Work transfer during the process 2

W 1-2 = ∫ pdV 1

3.31

Thermodynamic System and First Law T1 p

T2 T

Fig. p. 3.2

= p(V2 – V1) kJ = mR (T2 – T1)

= 0.5 × 0.28 × (573 – 373) W1–2 = 28 kJ. Change in internal energy

DU = m Cv(T2 – T1) kJ

= 0.5 × 0.72 × (200) = 72 kJ. Heat transfer during the process As per first law of thermodynamics

Q = W1-2 + DU

= 28 + 72 = 100 kJ. Change in enthalpy

DH = m cp(T2 – T1) kJ DH = m cp(T2 – T1)



DH = 0.5 × 1 × 200



DH = 100 kJ.

Problem 3.4 In a closed cylinder a perfect gas is heated in a reversible isothermal process from 1 bar and 40°C to 10 bar. Find the work done per kg of gas. Take R = 287 J/kg K. Given

T = 40 + 273 = 313 K;



m = 1 kg;



p1 = 1 bar = 1 × 105 N/m2;



p2 = 10 bar = 10 × 105 N/m2; T2 P T1 V

Fig. p. 3.3

3.32

Engineering Thermodynamics

Solution Work done, W = 2.3 m R T log10(v2/v1) = 2.3 m R T log10(p1/p2) = 2.3 × 1 × 287 × log10 (1/10) = –206.61 kJ/kg. Problem 3.5 A sample of 1 kg of air at a pressure of 7 bar and a temperature of 363 K undergoes a reversible polytropic process, which may be represented as pV1.1 = constant. The pressure at the end of the process is 1.4 bar. Evaluate (a) the final specific volume and temperature. (b) the work done and the heat transfer during the process. Assume R = 287 J/kg K and g = 1.4. Given T1 = 363 K;

m = 1 kg;



p1 = 7 bar = 7 × 105 N/m2;



g = 1.4;



R = 287 J/kg K;



p2 = 1.4 bar = 1.4 × 105 N/m2;

Solution 1.1 pv = C

p

V

Fig. p. 3.4



n = 1.1.



p1V1 = mRT1.

V1 = mRT1/p1 = 0.15 m3/kg .

 p  T1  1  = T2  p2 

T2 =

 p  1   p2 

( ( n–1) /n )    

T1 ( ( n–1) /n )       

= 313.5 K.

T1 v  =  2  T2  v1 

( n −1)

v2 = 0.649 m3/kg.

3.33

Thermodynamic System and First Law

Work done,

W =

( p1 v1 − p2 v2 ) ( g – 1)

= 141.4 kJ/kg. Heat transfer,

 ( g – n )  q =  × W  ( g – 1 ) 

= 106.05 kJ/kg. Problem 3.6 A steam turbine in a steam power plant receives steam at 40 bar. The specific enthalpy and specific volume of steam at the inlet of the turbine are 3214 kJ/kg and 0.07 m3/kg, respectively. The steam after expanding in the turbine has the pressure at the exit of 35 bar. The enthalpy and volume of steam are 3202.6 kJ/kg and 0.084 m3/kg, respectively. If the steam is carried in a 0.2 m diameter pipe the heat loss in the pipeline is 8.5 kJ/kg. Calculate the mass flow rate of steam. p1 = 40 bar;

Given

h1 = 3214 kJ/kg;



v1 = 0.073 m3/kg;



p2 = 35 bar;



h2 = 3202.6 kJ/kg;



v2 = 0.084 m3/kg;



Q = 8.5 kJ/kg

Solution 1 Turbine 2

Fig. p. 3.5

( ) 

 h2 + V22  h + V12  m m 1  + gz1 + Q1 – 2 =   2 2   

 

+ gz2 + w1–2

 V22   V2  h1 +  1  + Q1–2 = h2 +    2   2  [potential energy, gz1 = gz2 = 0; w1-2 = 0 ]

 V2   V2  3214 × 1000 +  1  – 8.5 × 1000 = 3202.6 × 1000 +  2  .  2   2 

( ) ( )  = 2.9 × 1000

 V22 – V12   2 

 

...(1)

3.34

Engineering Thermodynamics

From the continuity equation,

A 2 V2 A 1 V1 = v2 v1



V2 V1 = 0.084 0.073

V1 = 0.86 V2

...(2)

On substituting Eqn.(2) in Eqn.(1), we get

V22 = 22273.42

V2 = 149.24 m/s.



A 2 V2 v2  π × 0.2 2 × 149.24   4 ( )  m = 0.08 m =

= 55.79 kg/s. Problem 3.7 Air enters a compressor at a velocity of 60 m/s, pressure of 100 kPa, temperature of 40°C and leaves the compressor at a velocity of 90 m/s, pressure of 500 kPa, temperature of 120°C. Consider the system as adiabatic. Find the power of the motor for a mass rate of flow of 40 kg/min. Write the assumptions made. Given

V1 = 60 m/s;

p1 = 100 kPa = 100 × 1000 N/m2;

T1 = 313 K;

V2 = 90 m/s;

p2 = 500 kPa = 500 × 1000 N/m2;

T2 = 393 K;

m = 40 kg/min = 0.66 kg/s.

Solution

Compressor 1

Fig. p. 3.6

Since the system is adiabatic, g = 1.4; Using continuity equation,

A 1 V1 = m v1

2

3.35

Thermodynamic System and First Law

For unit area, V1 = m v1



60 0.66 = 90.9 m3/kg. v1 =



For the same area,

V2 V1 = v2 v1 v2 =

90  60     90.9 

= 136.35 m3/kg. Work done during the adiabatic compression, w = mR



T2 – T1 ( g – 1)

= 0.66 × 0.287 ×

(393 – 313) ( 1.4 – 1)

= 37.88 kW. Power required to run the compressor = Work done by the compressor. Therefore,

w = –37.88 kW. (Negative sign indicates the work done on the system).

Problem 3.8 In a forging industry, the atmospheric air drawn into a compressor, is compressed from 100 kPa and at 22­°C to a pressure of 1 MPa while being cooled at the rate of 16 kJ/kg by circulating water through the compressor casing. The volume flow rate of the air inlet condition is 150 m3/ min and the power input to the compressor is 500 kW. Determine (a) The mass flow rate. (b) Temperature of the air exit. Neglect the datum head. Given

p1 = 100 kPa;



p2 = 1 Mpa;



q = 16 kJ/s;



v1 = 150 m3/min = 2.5 m3/s;



w = –500 kW;

T1 = 22 + 273 = 295 K. Solution For the reciprocating compressor,

q = (h2 – h1) + w

h2 – h1 = q – w

3.36

Engineering Thermodynamics



rh = 16 – (–500) = –516 kJ. = m Cp(T2 – T1)

516 = 1 × 1.005 × (T2 – 295)

T2 = 806 K. = 533°C

Mass flow rate = Volume flow rate × Density

= 2.5 × 1.2 = 3 kg/s. Problem 3.9 In a steam power plant the steam enters a highly insulated steam turbine at 10 Mpa and 500°C at the rate of 3 kg/s and leaves at 50 kPa. If the power output of the turbine is 2 MW, determine the temperature of the steam at the exit the of turbine. Neglect the kinetic energy changes. Given P1 = 10 Mpa = 10 × 106 N/m2; T1 = 500 + 273 = 773 K;

m = 3 kg/sec;



p2 = 50 kPa = 50 × 103 N/m2;



w = 2M = 2 × 103 kW. 1 Turbine 2

Fig. p. 3.7

For the turbine,

w = h1 – h2 h1 – h2 = m Cp(T1 – T2) 2000 = 3 × 2.1 × (773 – T2)

T2 = 455 K = 182°C. Problem 3.10 A process industry uses a blower unit to supply air. The blower unit draws 1 kg/s of air at 20°C and consumes a power of 15 kW. The inlet and outlet velocities of air are 100 m/s and 150 m/s respectively. Find the exit of air temperature, assuming adiabatic conditions. Take the Cp of air as 1.005 kJ/kg K. Given T1 = 20 + 273 = 293 K;

w = 15 kW;

V1 = 100 m/s;

3.37

Thermodynamic System and First Law

Solution 1

Blower unit

2

Fig. p. 3.8

V2 = 150 m/s;   V12  V2  m  h1 +  + gz1 + Q = m  h2 + 2  + gz2 + w) 2  2    Since potential energy (qg = qg) and heat (Q) becomes 0,

w =

=

( h1 – h2 ) + ( V12 ) – ( V22 ) 2

( ) ( )

mC p ( T1 – T2 ) + V12 – V22 2 3

1 × 1.005 × 10 × (293 – T2 ) + ((100 × 100) – (150 × 150)) 2 T2 = 301.7 K (or) 28.7°C



–15 × 103 =

Since work is done on the blower work, it is considered negative. Problem 3.11 An oil industry uses a heat exchanger in which oil is cooled from 90°C to 30°C. Water drawn from a small pond is used to cool the oil in the heat exchanger. The inlet temperature of the water is 25°C and the outlet temperature is 70°C. The enthalpies of the oil and water at the inlet and outlet conditions are given below: Water hinlet = 593.64 kJ/kg;

houtlet = 699.77 kJ/kg;

Oil hoilin = 748.19 kJ/kg;

hoilout = 605.43 kJ/kg;



Calculate the mass flow rate of water required to cool the oil for cooling 2.78 kg/s of oil? Given Tw1 = 25 + 273 = 298 K; Tw2 = 90 + 273 = 363 K; Toil1 = 70 + 273 = 343 K; Toil2 = 30 + 273 = 303 K;

moil = 2.78 kg/s.



hw1 = 593.64 kJ/kg.



hw2 = 699.77kJ/kg.



hoil1 = 748.19 kJ/kg.



hoil2 = 605.43 kJ/kg.

3.38

Engineering Thermodynamics

Solution hw = hw2 – hw1. = 106.13 kJ/kg. hoil = hoil1 – hoil2



= 142.76 kJ/kg. = moil × Cpoil × (To1 – To2) Cpoil = 1.18 kJ/kg K. Water in Oil in Heat exchanger Oil out Water out

Fig. p. 3.9

Now, writing the energy balance for water and oil,

hoil = hwater

moil × Cpoil × (To1 – To2) = mw × Cpw(Tw2 – Tw1) Therefore,

mw = 1.44 kg/s.

Objective Type Questions 1. Which one is not the extensive property of a thermodynamic system? (a) Mass

(b) Total volume

(c) Total internal energy

(d) Density

2. One bar in SI unit is equal to (a) 1 × 105 N/m2

(b) 1 × 104 N/m2

(c) 1 × 103 N/m2

(d) 1 N/m2

3. Unit of power is (a) Watt

(b) Joule

(c) Joule-metre

(d) None of the above

3.39

Thermodynamic System and First Law

4. The law that states that heat and work are mutually convertible is known as (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) None of these 5. A definite area or space where some thermodynamic process takes place is known as (a) Thermodynamic system

(b) Thermodynamic process

(c) Thermodynamic cycle

(d) Thermodynamic law

6. An open system is one in which (a) Heat and work cross the boundary of the system, but the mass does not (b) The mass crosses the boundary of the system but the heat and work do not (c) Both heat and work as well as mass cross the boundary of the system (d) Neither the heat and work nor the mass cross the boundary of the system 7. An isolated system (a) Is a specified region where the transfer of energy and/or mass take place (b) Is a region of constant mass and only energy is allowed to cross the boundaries (c) Cannot transfer either energy or mass to or from the surroundings (d) Is one in which the mass within the system is not necessarily constant (e) None of these 8. In an extensive property of a thermodynamic system (a) Extensive heat is transferred (b) Extensive energy is utilised (e) None of these

(c) Extensive work is done (d) All of the above

9. Which of the following is an intensive property of a thermodynamic system? (a) Volume

(b) Temperature (c) Mass

(d) Energy

10. Which of the following is an extensive property of a thermodynamic system (a) Pressure

(b) Volume

(c) Temperature (d) Density

11. When two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. This statement is called (a) Zeroth law of thermodynamic

(c) First law of thermodynamic

(b) Second law of thermodynamic

(d) Kelvin Planck’s law

12. In a thermodynamic process, when the heat and work interact the internal energy of the system increased by 30 kJ. The amount of heat and work transfer is respectively.

3.40

Engineering Thermodynamics

(a) –50 kJ and –80 kJ

(b) 50 kJ and 80 kJ

(c) –50 kJ and 80 kJ

(d) 50 kJ and –80 kJ

13. Which one of the following can be considered as the property of a system? (a) ∫ p dV (c)

 dT   pdV   +  T   V  

∫ 

(b)

∫ V dp



 dT   Vdp   (d) ∫  +   T   T  

14. The internal energy of a certain system is given by a formula E = 25 + 0.25 t kJ, where t is the temperature. If the system executes a work which is 0.75 kNm, then the heat transfer is (a) –1.0 kJ

(b) –0.5 kJ

(c) 0.5 kJ

(d) 1 kJ

15. Air with an enthalpy of 100 kJ/kg, is compressed by an air compressor to a pressure and temperature at which the enthalpy becomes 200 kJ/kg. The loss of heat is 40 kJ/kg from the compressor as the air passes through it. Neglecting kinetic and potential energies the power required for an air mass flow of 0.5 kg/s is (a) 30 kW

(b) 50 kW

(c) 70 kW

(d) 90 kW

16. Which of the following relationship is valid only for reversible processes undergone by a closed system of a simple compressible substance which neglects kinetic and potential energy? (a) δQ = dU + δW

(b) TdS = dU + pdV

(c) TdS = dU + δW

(d) δQ = dU + pdV

17. A tank containing air is tilted by a paddle wheel, the work output of the paddle wheel is 9000 kJ and the heat transferred to the surrounding of the tank is 3000 kJ. The external work done by the system is (a) 0

(b) 3000 kJ

(c) 6000 kJ

(d) 9000 kJ

p   18. A gas expands from pressure p1 to p2  p2 = 1  . If the process of expansion 10   is isothermal, the volume at the of end expansion is 0.55 m3. If the process of expansion is adiabatic the volume at the end of expansion is closer to (a) 0.45 m3

(b) 0.55 m3

(c) 0.65 m3

(d) 0.28 m3

19. A steam boiler is a (a) Heat absorbing system

(b) Heat rejection system

(c) Work absorbing system

(d) Work developing system

20. A reciprocating compressor is a (a) Heat absorbing system

(b) Heat rejection system

(c) Work absorbing system

(d) Work developing system

Thermodynamic System and First Law

3.41

Theory Questions 1. Define a system; draw a sketch and indicate on it, the system, boundary and surroundings. 2. Define thermodynamic systems and their surroundings with examples. 3. Prove that work is a path function. 4. Distinguish between closed, open and isolated systems. 5. Define a thermodynamic system. Name various types of systems. 6. Explain why in thermodynamic calculations, it is necessary to use absolute temperature. 7. What do you mean by transfer of heat? What are its positive and negative directions? 8. What is meant by thermodynamic equilibrium? 9. State the Zeroth law of thermodynamics. 10. How does the homogeneous system differ from the heterogeneous one? 11. Explain the following terms (i) state (ii) process (iii) thermodynamic cycle. 12. Distinguish between intensive properties and extensive properties by giving examples. 13. What do you mean by a quasi static process? 14. Define reversible work. 15. Define the first law of thermodynamics. 16. What is the convection for positive and negative work? 17. What are the corollaries to the first law of thermodynamics? 18. Prove that for an isolated system, there is no change in the internal energy. 19. State the first law for a closed system undergoing a change of state. 20. What are the point function and path function? Give examples for each. 21. State the differences between open and closed system. 22. What is the thermodynamics definition of “work”? 23. Distinguish between heat and work transfer. 24. What do you understand by flow systems? Classify them with an example. 25. An insulated rigid vessel is divided into two parts by a membrane. One part of the vessel contains air at 10 MPa and the other part is fully evacuated. The membrane ruptures and the air fills the entire vessel. Is there any heat and/or work transfer during this process? Justify your answer. 26. Define internal energy and prove that it is a property of a system. 27. State the first law of the thermodynamics as referred to closed system undergoing a cyclic change.

3.42

Engineering Thermodynamics

28. State the first law of the thermodynamics and prove that Q = W + ∆U for a non-flow process. 29. What do you mean by the perpetual motion machine of the first kind (PMMI)? 30. Deduce an expression for the work done by a gas in a system during the reversible polytropic process. 31. Explain clearly the difference between the non-flow and steady flow processes. 32. Derive the steady flow energy equations for the following cases. (a) Reciprocating air compressor (b) Steam boiler (c) Steam nozzle (d) Steam turbine 33. Mathematically state the steady flow energy equation and apply it to a condenser. 34. Define the term “Steam nozzle”. What are the types of nozzles? 35. State the relationship between the velocity of steam and heat drop during any part of a steam nozzle. 36. Explain why nozzles are made convergent and divergent. 37. Derive the general equation for a steady flow system and apply the equation to a nozzle and derive an equation for the velocity at the exit. 38. Derive the expressions for the work and heat transfer during the isothermal process. 39. Write down the steady flow energy equation and simplify it for the isentropic flow through the nozzle. 40. Define the terms thermodynamic equilibrium, properties, cycle and work done.



Unsolved Problems

1. A system follows a particular path A from state 1 to state 2. The system receives 220 kJ of heat and develops 120 kJ of work in the process. The same system follows in path B from state 2 to state 1 in which 150 kJ of work is done on the system. Calculate the heat transfer in path B. 2. 1 kg of water at 28°C and 1 bar is continuously heated in a vessel till it is continuously evaporated. Calculate the heat and work transfer if the following data are considered. Specific volume of water = 0.001 m3/kg Specific volume of vapour = 1.694 m3/kg

3.43

Thermodynamic System and First Law

Enthalpy of vapour = 2675.47 kJ/kg Enthalpy of water = 125.6 kJ/kg 3. A rigid insulated tank of 3 m3 volume is divided into two compartments; one compartment of volume 1 m3 contains an ideal gas at 1 bar and 27°C, while the second compartment of volume 2 m3 contains the same gas at 10 bar and 727°C. If the partition between the two is ruptured, determine the final state (Temperature and pressure) of gas. 4. A nozzle is a device for increasing the velocity of a steadily flowing steam. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the velocity is 60 m/s at the discharge end, the enthalpy is 2762 kJ/kg. the nozzle is horizontal and there is negligible heat loss from it. (A1 = 0.1 m2, v1 = 0.187 m3/kg, v2 = 0.498 m3/kg). Determine the velocity of steam at the exit of the nozzle. 5. A slow chemical reaction takes place in a fluid at the constant pressure of 0.1 MPa. The fluid is surrounded by a perfect heat insulator during the reaction which begins at state 1 and ends at state 2. The insulation is then removed and there is 105 kJ of heat flow to the surroundings as the fluid goes to state 3. The following data are observed for the fluid at states 1, 2, 3. State

1

2

3

V(m3)

0.03

0.3

0.06

t(°C)

20

270

20

For the fluid system, calculate E2 and E3, if E1 = 0. 6. A turbine operates under steady flow conditions, receiving steam at the following state pressure of 1.2 MPa, temperature of 461 K, enthalpy 2785 kJ/kg, velocity 33.33 m3/s and elevation 3 m. The steam leaves the turbine at the state pressure of 20 kPa, enthalpy 2512 kJ/kg, velocity 100 m/s and elevation 0 m. Heat is lost to the surroundings at the rate of 0.29 kJ/s. If the rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the turbine in kW? 7. A gas undergoes a thermodynamic cycle consisting of the following processes Process 1-2 constant pressure p = 1.4 bar, V1 = 0.028 m3, W1-2 = 10.5 kJ. Process 2-3 compression with pV = constant U3 = U2. Process 3-1 constant volume, U1 – U3 = –26.4 kJ.

There are no significant changes in internal energy for the process 1-2. 8. The heat capacity at a constant pressure of a certain system is a function of temperature only and may be expressed as

 41.87  2.093 +   J / °C  t + 100 

3.44

Engineering Thermodynamics

where it is the temperature of the system in °C. The system is heated while it is maintained at a pressure of 1.013 bar, until its volume increases from 2000 cm3 to 2400 cm3 and its temperature increases from 273 K to 373 K. Calculate the work done in the process. 9. A perfect gas flows through a turbine at a rate of 3 kg/s. The inlet and exit velocity are 30 and 140 m/s respectively. The initial and final temperatures are 600°C and 200°C respectively. Heat loss is 45 kJ/s. calculate the power developed. Assume Cp = 1.1 kJ/kg. 10. A centrifugal air compressor operating at a steady state has an air intake of 1.2 kg/min. The inlet and exit conditions are as follows Properties

p(kPa)

T°C

u(kJ/kg)

v(m3/kg)

Inlet

100

0

195.14

0.784

Exit

300

55

220.95

0.465

γ – n Prove that Q = mC v  ( T2 – T1 ) for a polytropic process of index n.  1 – n  11. A system is taken through a series of processes, as a result of which it is restored to the initial state. The work and heat interactions for some of the processes are measured and they are given below Process

W(kJ)

Q(kJ)

u(kJ)

1-2

100

100

–—

2-3

–—

–150

200

3-4

–250

–—

–—

4-1

300

–—

80

12. (i) A quantity of gas has a volume of 0.14 m3, pressure 1.5 bar and temperature 100°C. If the volume at the end of the constant pressure process is 0.112 m3, find the final temperature, work done, change in internal energy and the heat energy given off. Take Cp = 1.005 kJ/kgK, R = 285 J/kgK. 13. A mass of air is initially at 700 kPa and 260°C and occupies 0.028 m3. The air is expanded at a constant pressure to 0.084 m3. A polytropic process with n = 1.5 is then carried out, followed by a constant temperature process. All the processes are reversible. (i) Sketch the cycle in p-V and T-s diagrams. (ii) Find the heat energy received and heat energy rejected in the cycle (iii) Find the efficiency of the cycle. 14. Air, at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s, and its temperature is raised to 800°C. It then enters a turbine at the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle, where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s, calculate:

Thermodynamic System and First Law

3.45

(i) The rate of heat transfer to the air in the heat exchanger. (ii) The power output from the turbine assuming no heat loss and, (iii) The velocity at the exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h = Cpt, where Cp is 1.005 kJ/kg K and “t” is the temperature. 15. Air undergoes a cyclic process in a cylinder and piston arrangement. First the atmospheric air at 1 bar and 27°C is compressed adiabatically to 10 bar, then expanded isothermally up to initial pressure, then brought to the initial conditions under constant pressure, find out (i) Change in internal energy (ii) Change in enthalpy (iii) Heat transfer (iv) Work transfer for each process and also for the cycle. 16. 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 0.2 m3 and 1.2 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 is in m3/kg. Calculate the net heat transfer and the maximum internal energy of the gas attained during expansion. 17. A room for four persons has two fans, each consuming 0.18 kW power and three 100 W lamps. Ventilated air at the rate of 80 kg/h enters with an enthalpy of 84 kJ/kg and leaves with an enthalpy of 59 kJ/kg. If each person puts out heat at the rate of 630 kJ/h, determine the rate at which heat is to be removed by a room cooler, so that a steady state is maintained in the room. 18. In a steam generator, compressed liquid water at 10 MPa and 30°C enters a 30 mm diameter tube at the rate of 3 litres/s. Steam at 9 MPa and 400°C exists in the tube. Find the rate of heat transfer to the water. 19. In an air compressor, air flows steadily at the rate of 0.5 kg/s. At the entry to the compressor, the air has a pressure of 105 kPa and specific volume of 0.86 m3/kg and at the exit of the compressor, the corresponding values are 705 kPa and 0.16 m3/kg. Neglect the kinetic and potential energy changes. The internal energy of the air leaking through the compressor is 95 kJ/kg greater than that of the air entering. The cooling water in the compressor absorbs 60 kJ/s of heat from the air. Find the power required to drive the compressor. 20. A piston and cylinder machine contains a fluid system, which passes through a complete cycle of four processes. During a cycle, the sum of all the heat transfers is –170 kJ. The system completes 100 cycles per min. Complete the following table showing the method for each item and complete the net rate of work output in kW.

3.46

Engineering Thermodynamics Process

Q(kJ/min)

W(kJ/min)

∆E(kJ/min)

a-b

0

2170

—

b-c

21000

0

—

c-d

–2100

—

–36000

d-a

—

—

—

21. Air flows steadily at the rate of 0.5 kg/s through an air compressor, entering at 7 m/s velocity, 100 kPa pressure and 0.95 m3/kg volume and leaving at 5 m/s, 700 kPa and 0.19 m3/kg. The internal energy of the air leaving is 90 kJ/kg greater than that of the air entering. Cooling water from the compressor jackets absorbs the heat from the air at the rate of 58 kW. (i) Compute the rate of shaft work input to the air in kW (ii) Find the ratio of the inlet pipe diameter to that of the outlet pipe diameter. 22. A closed system consists of air which is initially at 1.5 bar and 67°C. The volume doubles as the system undergoes a process according to the law pV1.2 = C. Find the work done, heat transfer and the change in entropy during this process. For air R = 0.287 kJ/kg K and g = 1.4. 23. A perfect gas is expanded in a non-flow process according to pVn = C, from the initial state point 1. Sketch on a common p-V diagram for all the five expansion processes when n = 0, 1 between 1 and g, g and ∞. Name all the processes and write the expressions for the work done in each case. 24. A cylinder contains a gas at 500 kPa occupying 0.2 m3. The gas is expanded to 100 kPa. During expansion, the volume was inversely proportional to the pressure. Find the volume at the end of the expansion. Calculate the work done. 25. Air of mass 0.5 kg is compressed reversibly and adiabatically from an initial state at 80 kPa, 60°C to 0.4 MPa and then expanded at a constant pressure to the original volume. Sketch the process on the p-V plane and determine the heat and work transfer. For air, assume R = 0.287 kJ/kgK and Cv = 0.713 kJ/kgK. 26. Air flows steadily at the rate of 0.5 kg/s through an air compressor entering at 7 m/s velocity, 100 kPa pressure and 0.95 m3/kg specific volume and leaving at 5 m/s, 700 kPa and, 0.19 m3/kg. The internal energy of air leaving is 90 kJ/kg greater than that of the air entering. Cooling water in the compressor jackets absorbs the heat at the rate of 58 kW. Calculate the rate of shaft work input to the compressor. 27. A steam turbine receives steam at a pressure of 1 MPa, 300°C. The steam leaves the turbine at a pressure of 15 kPa. The work output of the turbine was measured and found to be 600 kJ/kg of steam flowing through the turbine. Determine the efficiency of the turbine. 28. Air expands from a pressure of 10 bar, 1 m3 and 327°C to a pressure of 1 bar. As per the law of expansion pV1.3 = constant. Find the heat and work transfer, the entropy and temperature change in the process.

Thermodynamic System and First Law

3.47

29. Air flows at a rate of 0.75 kg/s through an air compressor entering at 5 m/s velocity, 1 bar an 0.95 m3/kg specific volume. Air leaves the compressor at 10 m/s, 10 bar with a specific volume of 0.19 m3/kg. The internal energy of the air leaving is 90 kJ/kg more than that of the air entering. The heat transfer from the compressor to the atmosphere is 50 kW. Find the shaft work. 30. In an isentropic flow through a nozzle, air flows at the rate of 600 kg/h. At the inlet to the nozzle, the pressure is 2 MPa and temperature is 127°C. The exit pressure is 0.5 MPa. The initial air velocity is 300 m/s. Determine (i) The exit velocity of air and (ii) The inlet and exit cross-sectional areas of the nozzle. 31. A centrifugal pump delivers 2750 kg of water per minute from an initial pressure of 0.8 bar absolute to a final pressure of 2.8 bar absolute. The suction is 2 m below and the delivery is 5 m above the centre of the pump. If the suction and delivery pipes are 15 cm and 10 cm in diameter respectively, make the calculations of the power required to run the pump. 32. Air in a closed stationary system expands in a reversible adiabatic process from 0.5 MPa, 15°C to 0.2 MPa. Find the final temperature and the change in enthalpy, the heat transferred and the work done per kg of air. 33. A mass of air is initially at 260°C and 700 kPa and occupies 0.028 m3. The air is expanded at a constant pressure to 0.084 m3. A polytropic process with n = 1.5 is then carried out, followed by a constant temperature process. All the processes are reversible. (i) Sketch the cycle in p-V and T-s diagrams. (ii) Find the heat energy received and heat energy rejected in the cycle (iii) Find the efficiency of the cycle. 34. Air at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a turbine at the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle, where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s, calculate: (i) The rate of heat transfer to the air in the heat exchanger. (ii) The power output from the turbine assuming no heat loss and, (iii) The velocity at the exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h = Cpt, where Cp is 1.005 kJ/kg K and “t” is the temperature.

CHAPTER

4 4.1

SECOND LAW OF THERMODYNAMICS

LIMITATIONS TO THE FIRST LAW OF THERMODYNAMICS

We know that the first law of thermodynamics is derived on the basis of the law of conservation of energy. The first law of thermodynamics states that when a thermodynamic system undergoes a cyclic process the algebraic sum of heat transfer is equal to the algebraic sum of work transfer.

∫ dQ = ∫ dW

...(4.1)

It is a fact that heat and work are mutually convertible. But, when a system undergoes a change of state in a process and an amount of heat is given to the system, then a part of the heat energy will be converted into work and the remaining is stored as internal energy into the system itself. There are a few limitations to the first law of thermodynamics as given below: l The direction of the process by which the transformation takes place is not

clearly mentioned.

l Though

heat and work are mutually convertible, work may be fully converted into heat. But, heat may not be fully converted into work, due to some heat rejected.

l The law does not explain the feasibility of the process, the duration of the

process (How long will the process take? etc.)

These limitations are the basis to form the second law of thermodynamics. The following terms are used in the study of the second law of thermodynamics. 1. Thermal Reservoir 2. Source 3. Sink 1. Thermal Reservoir: It is a body in which a definite quantity of heat may be extracted or stored without any change in the temperature. 2. Source: It is a body in which a definite quantity of heat is supplied to a heat engine. The source is assumed to be a high temperature thermal reservoir. 3. Sink: It is a body in which a definite quantity of heat is rejected from heat engine. The sink is assumed to be a low temperature thermal reservoir.

4.2

Engineering Thermodynamics

4.2

HEAT ENGINE, REFRIGERATOR AND HEAT PUMP

4.2.1 Heat Engine The diagram of a heat engine is shown in Fig. 4.1(a)

l A cyclically operating device whose objective is to receive heat from a heat



l After producing useful work, a part of the heat will be rejected to the low



l Assume



l The

source at a high temperature, and to produce an amount of useful work, is called a heat engine. temperature reservoir, called a sink.

that the source is at Ts temperature and supplies Qs amount of heat to the heat engine. The engine rejects Qr amount of heat to the sink. efficiency of the heat engine is h =

Work output Qs – Qr = Heat input Qs

                                h = 1 –

Qr Qs

It is to be noted that the efficiency of the heat engine will always be less than 100%.

4.2.2 Refrigerator The diagram of a refrigerator is shown in Fig. 4.1(b).

l A cyclically operating device whose objective is to maintain the temperature



l Assume



l The

amount of heat rejected to the atmosphere = Qatm = Qr + Wr



l The

coefficient of performance COP =



l COP

=

Qr Wr



l COP

=

Qr Q atm – Q r



l The



l One

of a body below its atmospheric temperature or surrounding temperature by extracting heat from the body and rejecting it to the atmosphere with the help of an external work, is called a refrigerator. that the body is at Tr temperature and Qr amount of heat is extracted from it. With the aid of an external work, this heat is rejected to the atmosphere. Desired effect Work input

rate at which the heat can be removed is called the capacity of refrigerator. It is expressed in Ton of Refrigeration (TR). ton of refrigeration is defined as the quantity of heat removed to freeze one ton of water at 0°C to ice at 0°C in 24 hours. The value of one ton refrigeration is 3.5 kW.

4.3

Second Law of Thermodynamics

Let

m = 1 ton = 1000 kg.

Latent heat of ice = 305 kJ/kg. Ton of Refrigeration =

mh 1000 × 305 = t 24 × 3600

= 3.5 kW.

4.2.3 Heat Pump The diagram of a heat pump is shown in Fig. 4.1(c).

l A

cyclically operating device whose objective is to continuously extract heat from a body at low temperature and to supply a certain quantity of heat to a body at high temperature with the help of an external work, is called a heat pump. Source Ts

Hot body Th

Atmosphere Tatm

Qs

Qh W = Qh – Qatm

W = Qs – Qr E

R

Qh W = Qatm – Qh

Qr

Qr



HP

Sink Tr

Body or refrigerated space Tr

Atmosphere Tatm

(a)

(b)

(c)

Fig. 4.1 (a) Heat Engine (b) Refrigerator (c) Heat pump



l In

the heat pump, the body which is at high temperature is considered as a room space and the body which is at low temperature is assumed to be the atmosphere.



l Assume



l Amount



l The

coefficient of performance (COP)Pump =



l The

coefficient of performance (COP)Pump =

that the body is at Th temperature, and Qh amount of heat is extracted from it. With the aid of an external work (heat pump) this heat is pumped to the hot body (room space). of heat pumped to the room space = Qpump = Wp + Qr Desired effect . Work input Q pump Q pump – Q r

.

4.4

Engineering Thermodynamics

4.3

STATEMENTS OF THE SECOND LAW OF THERMODYNAMICS

There are two statements followed in the second law of thermodynamics. They are 1. The Kelvin planck statement, and 2. The clausius statement.

4.3.1 Kelvin Planck Statement It is impossible to construct a device operating in a cyclic process, which is to produce no other effect than the extraction of heat from a single reservoir and perform an equivalent amount of work. This implies that it is impossible for a device to convert heat in to work with a single reservoir. In other words, there must be two reservoirs, namely the source and the sink, which have to be operated at different temperatures (one is a source and the other is a sink). This is illustrated in Fig. 4.2(a). Reservoir 1 Thermal reservoir Ts

Source Ts

(Impossible) E

E No Qr

Sink Tr Reservoir 2

(a)

(b)

Fig. 4.2 (a) Illustration of Kelvin Planck statement (b) PMM-II

PMM-II (Perpetual motion machine of its second kind): It is a device that receives heat as energy from a single reservoir and converts it into an equivalent amount of work. Figure 4.2(b) shows the PMM-II. Since, there is no heat rejection to a sink, the efficiency of this will become 100%. It is practically impossible.

hmax =

Qs Qs

...(4.2)

hmax = 100 %  (Impossible)

4.3.2 Clausius Statement It is impossible to construct a device operating in a cyclic process that is to transfer heat from a body at low temperature to a body at high temperature without an aid of an external work. It means that it is only possible for a device

4.5

Second Law of Thermodynamics

that will transfer any amount of heat from a body at low temperature to a body at a high temperature when there is some external work done on the device.

4.4

VIOLATION OF THE KELVIN-PLANCK AND CLAUSIUS STATEMENTS

Though the Kelvin-Planck and Clausius statements appear to be different these two statements are interlinked. These two statements are meant equal to be each other. It is to be proved that both the statements are equivalent with each other. Body (high temp.) External (W)

Body (high temp.)

P

P

Body (low temp.)

Body (low temp.)

(a)

(b)

Fig. 4.3 (a) Illustration of Clausius statement (b) Impossibility of Clausius statement

Two cases are considered to prove the equivalence of the two statements which are explained below:

CASE – I Left Side of the Diagram Assume that, a heat pump is operating between two reservoirs; a source and a sink at temperatures Ts and Tr. The amount of heat transferred to the body at higher temperature is Qr, and the heat extracted from the body at low temperature is also Qr. It shows that without an external aid of work, Qr quantity of heat is transferred from the body at low temperature to a body at higher temperature. This violates the clausius statement. Source at Ts Qs

Qs

Qh = Qs P

E Qr Sink at Tr

P Qr

W = Qs – Qr E

Qr

Qr

Sink at Tr

Fig. 4.4 (a) Heat pump and heat engine with source and sink (b) Heat pump and heat engine only with sink

4.6

Engineering Thermodynamics

Right Side of the Diagram A heat engine is considered on the right hand side of the diagram, which operates between the same two reservoirs A and B. The hot source is at Ts temperature and the sink is at Tr temperature. If Qs amount of heat is supplied to the engine, and Qr amount of heat is rejected to the sink, then the work done by the heat engine is Qs – Qr. This satisfies the Kelvin Planck statement. But, as shown in Fig. 4.4(b), when we consider both the heat pump and the heat engine together, then it shows that the heat engine operates with a single reservoir (sink), and receives the heat Qs from the heat pump directly, and produces the same amount of work Qs – Qr which is not possible. This is because the heat rejected to the sink in the heat pump is Qr. This violates the Kelvin-Planck statement.

CASE –II Let us consider a perpetual motion machine of the first kind which produces a network in a cycle. This is shown in Fig. 4.5(a). The machine is operated with only a single reservoir. Then, it violates the Kelvin-Planck statement. On the other hand, referring to Fig. 4.5(b), the heat pump should extract Qatm amount of heat from the sink at Tatm temperature, and to transfer it to the hot body at Th temperature with the external aid of mechanical work W. But, taking HE and HP constituting together a cycle, Qs amount of heat is supplied as it is, and there is no heat rejection (Qr = 0) to the sink leads. Since Qr = 0, there will be no extraction of heat Qr, but the sole heat Qs is transferred to the source at Ts. This leads to a violation of Clausius statement. Source at Ts = Th Qs = Qh E

Source at Ts = Th

Qs = Qh

Qh + Qatm

W

P

Qr = 0

E Qr = 0

Sink at Tr = Qatm (a)

Qh

No heat rejection

P Qr

Qr

No heat extraction

Sink at Tr (b)

Fig. 4.5 (a) Heat pump operated by a PMM-I (b) Heat pump operated by PMM-I provided and with out taking heat form sink

Using the second law of thermodynamics, we shall determine the possible efficiency of an engine. The possibility of occurrence of the process can also be determined.

4.7

Second Law of Thermodynamics

4.5 REVERSIBILITY When a process is exactly restored with all its elements and surroundings to the initial state of the process, then it is known as a reversible process. The conditions for reversibility are: l The

process must pass through a continuous series of states and through the same states the process must return to its original state.

l There l The

must be no evidence of the events that occurred in the process.

system must be in thermodynamic equilibrium.

l There

must be no dissipative effects.

A few examples of a reversible process are given below: (i) Expansion of spring: Suppose a spring is stretched; then, it expands. If the spring is released, then it returns to its original state. Thus, after the compression or expansion of the process, the spring returns to its original state. (ii) Electrolysis: Assume that the electrolysis of water is carried out, as shown in Fig. 4.6. As the current is supplied, the hydrogen and oxygen molecules reach the negative and positive electrodes respectively. The process is continued even if the supply of current is stopped. The molecules may recombine and form water.

Positive

Negative

Electrolysis

Fig. 4.6 Electrolysis of water

(iii) Any mechanical process carriedout in a conservative force field is also considered as a reversible process.

4.5.1 Irreversible Process Any process, that does not satisfy the conditions of reversibility is called an irreversible process. Examples of irreversibility are; (a) Diffusion, (b) Free expansion and (c) Combustion.

4.8

Engineering Thermodynamics

4.6

CARTON CYCLE

The reversible engine was invented by Sadi Carnot. As per the Kelvin-Planck statement there must be two thermal reservoirs, one is to supply heat to the engine and second is to receive the lost heat from the engine. So there are two isothermal processes are taken place. There are two reversible adiabatic processes by which work transfer taken place. Let us assume that a reversible engine is operated between two reservoirs, which are at two different temperatures. Source T1

Piston and cylinder assembly Sink T2

Fig. 4.7 Illustration of Carnot cycle 3 4

p

3

4

2

1

T

2 1

S

V

(a)

(b)

Fig. 4.8 (a) p-V diagram (b) T-S diagram of Carnot cycle

For better understanding let us start from 3-4 in which heat is supplied.

Process 3-4 The piston and cylinder assembly is brought in contact with a heat source which is at T1. The source supplies Q1 amount of heat to the air in the piston and cylinder assembly at constant volume. Heat supplied

Qs = Q3–4 = T3(S4 – S3) kJ

...(4.3)

Process 4-1 Now, the piston and cylinder assembly is brought in contact with a diathermic wall so there is no heat transfer from the cylinder or into the cylinder. The heat transfer is zero. The hot air in the cylinder expands reversibly and adiabatically, so that an expansion work is obtained.

4.9

Second Law of Thermodynamics

Process 1-2 The piston cylinder assembly is now brought in contact with a low temperature sink which is at T1. Q2 amount of heat is rejected from the cylinder to the sink. Heat rejection

Qr = Q1-2 = T1(S1 – S2) kJ

...(4.4)

Process 2-3 Now, the piston and cylinder assembly is brought in contact with a diathermic wall, so that no heat interaction takes place between the cylinder and the surroundings. The air is compressed by the movement of piston towards left direction. The work is done on the air to compress it. Heat supplied

Qs = Q3-4 = T3(S4 – S3)

Heat rejection

Qr = Q1–2 = T1(S1 – S2)

Carnot efficiency hCarnot =

heat supplied − heat rejected heat supplied

=

T3 (S 4 – S 3 ) – T1 (S 1 – S 2 ) T3 ( S 4 – S 3 )

...(4.5)



( T3 – T1 )( S 4 – S 3 ) ( as S 4 =S 1 , S 2 =S 3 ) T3 ( S 4 – S 3 )

...(4.6)

T3 – T1 T3

...(4.7)

hCarnot =

=

= 1 –

4.7

T1 T3

...(4.8)

CARNOT THEOREM

The Carnot theorem states that no heat engine working between given temperature T1 and T2 can have efficiency greater than that of reversible engine working between the same temperatures. This can be proved by showing that assumption to the contrary (of the statement of the theorem) leads to violation of the second law. Source T1

A

A Q2 Sink T1

Fig. 4.9 Reversible engine connected to irreversible engine

4.10

Engineering Thermodynamics

Let as consider a heat engine A as shown in Fig. 4.9 operating on cycle, receives heat Q1 from the source at T1 performs work W1 and rejects heat Q to the sink at T2. Then

W1 = Q1 – Q ηa =

...(4.9a)

W1 Q 1 – Q Q = =1– Q1 Q1 Q1

Let as consider a heat engine B has the same cycle duration, operating between the same source and sink. The amount of heat rejected to the sink is Q1. Then

W = Q1 – Q2



hb =

...(4.10)

Q W Q1 – Q 2 = =1– 2 Q1 Q1 Q1

Let as assume that it is possible that

ηb > ηa



  Q2  Q  1 –  >  1 –  Q1  Q1   

...(4.11a)

or

Q < Q2

...(4.11b)

and

W < W1

...(4.11c)

As B is a reversible, we can construct a device B1 which works in opposite direction to that of B. B1 takes in one cycle heat Q2 from the second reservoir, needs work W to be done on it and rejects heat to the first reservoir. Consider that the device obtained by combining A and B1 in such way that out of the work W1 delivered by W is fed to B1. Then the sole effect of the combined device is to be absorbed is (Q2 – Q) amount of heat, which is positive quantity by virtue, from the engine reservoir at T2 and do work (W1 – W),

(Q2 – Q) = (W1 – W)

Without effecting or using any other heat exchanger, this violates the Planck’s statement of the second law. Hence the assumption ha > hb is absurd.

ηa > ηb

...(4.11d)

A corollary of this theorem is that all reversible heat engines working between temperature T1 and T2 have the same efficiency. If R1 and R2 are such engines and R2 drive R1, backwards (R1 is reversible) We get

ηr1  ηr2

...(4.11e)

If R2 drives R1 backwards We get

ηr2  ηr1

Both the results can hold only if ηr1  ηr2

...(4.11f)

4.11

Second Law of Thermodynamics

Therefore a reversible engine, the efficiency does not depend on the working substance, but only of temperature of the source and the sink. or

4.8

hr = f(T1, T2)

...(4.11g)

THERMODYNAMIC TEMPERATURE SCALE

It is understood that efficiency of a heat engine which operates in a cycle between two thermal reservoirs is given below

h =

= =

Work output Heat supplied W Q

( Qs – Qr )

...(4.12)

...(4.13)

 Qr  = 1 –   .  Qs 

...(4.14)

Qs

As per the second law of thermodynamics, the two thermal reservoirs must be at different temperatures for any engine, operating in a cycle. Also, the efficiencies of all reversible engines operating at the same temperatures are equal. Therefore, the efficiency of a reversible engine is dependent on the operating temperatures. If the efficiency of a reversible engine is mentioned against the function of operating temperatures, then or,

h = Ψ(Ts, Tr)

...(4.15)

  = Ψ(Ts, Tr) 

...(4.16)

Qs = ∅(Ts, Tr) Qr

...(4.17)

Q 1– r  Qs

Now, let us consider a reversible engine A operating between two sources at Ts1 and a sink at temperature Tr1. Between the same two reservoirs, two reversible engines B and C are kept in series as shown in Fig. 4.10. Now, for engine A the efficiency is given by

ha = Y1(Ts1, Tr1)

Qs1 = ∅(Ts1, Tr1) Qr 1

...(4.18) ...(4.19)

Now, the engine receives Qs2 amount of heat from source at Ts, and rejects heat Qr2 at an intermediate temperature of Ti. The rejected heat, which is received by engine C at Ti is rejected to sink at Tr.

4.12

Engineering Thermodynamics Source Ts

Qs1

Eb Qri

Wa = Qs1 – Qr1 Ea

Qsi = Qri Ec

Qr1

Qr1 Sink Tr

Fig. 4.10 Thermodynamic temperature scale

Let us consider a series of reversible engines as shown in Fig. 4.10. Qs1 = ∅(Ts1, Ti) Q ri Q ri Q = si = ∅ ( Ti , Tr 1 ) And for engine C   Qr 1 Qr 1 So, for engine B

...(4.20) ...(4.21)



Q si = ∅(Ti, Tr1) Qr 1

...(4.22)



Qs1 Q ri Qs1 = Q si Q si Qr 1

...(4.23)



T  Qs1 ∅  s1  T Q ri =  i  Q si  Ti  ∅  Qr 1  Tr 1 

...(4.24)



 Qs1    = ∅(Ts, Tr)  Q si 

...(4.25)

Q  The ratio  s 1  is a function of Ts, Tr only.  Q si   Q  θ ( Ts ) Therefore, it can be written as  s 1  =  Q si  θ ( Tr )  Q s 1   Q s 1  Ts 1         =  =   Q si   Q ri  Tri

...(4.26) ...(4.27)

4.13

Second Law of Thermodynamics

where, Ts1 = Source temperature, Tri = Intermediate sink temperature or subsequent engine source temperature. For two reservoirs between T and Ttw the ratio is given as

Q = Ttw Qtw

where, suffix ‘tw’ stands for triple point of water. Q T So, = ...(4.28) Qtw 273.16  Q  T = 273.16 ×    Qtw  Between two reservoirs T1 and T2, it can be given as, or,

...(4.29)

Q1 T = 1 . ...(4.30) Q2 T2



Since all the reversible engines are in series, the sink of the first engine is the source of the second engine. If the work output of all the reversible engines are the same, then

W = h1 Q1 = h2 Q2 = …. = hn–1 Qn–1 = hn Qn ...(4.31)

But, as per the efficiency expression related with the temperatures of source and sink  ( T1 – T2 )   ( T2 – T3 )   ( Tn −1 – Tn )  .... =    W =   Q=   Q=  Q n ...(4.32) 1 2 T1 T2 Tn–1       As 

Qn Q1 Q 2 = = .... = T1 T2 Tn

   W = (T1 – T2 ) = (T2 – T3) = .... = (Tn–1 – Tn ).

...(4.34)

The above expressions are useful to calibrate the temperature scale. This scale is called the thermodynamic temperature scale. The intervals may be small or large, based on the requirement of the scale. Using this temperature scale we can calibrate thermometers.

4.9

ENTROPY OF PERFECT GASES

4.9.1 Entropy Entropy is an important thermodynamic property, which is very useful in thermodynamic calculations. In fact, instead of defining the term entropy, it is easier to define the change in entropy, as it is dependent on the amount of heat transfer in a process. It is the ratio of the amount of heat transfer during a process to its absolute temperature.

4.14

Engineering Thermodynamics

Consider a reversible process in which the amount of heat transfer in a small segment in the process is dQ and the absolute temperature is T, then the change in entropy in the small segment is given by S

T

S

Fig. 4.11 T-s diagram

dQ T Between two states in a process,

dS =

         ∫ dS = ∫

dQ T

...(4.35)

...(4.36)

The unit for the entropy is kJ/K and the specific entropy is given in kJ/kg K.

4.9.2 Clausius Inequality Let us consider a reversible engine, absorbing heat Q1 from the source at T1 and rejecting heat Q2 to the sink at T2. Q1 Q 2 – = 0 T1 T2



...(4.37)

dQ = 0 ...(4.38) T Let us also consider an irreversible engine working between the same temperatures, absorbing heat Q1 at T1, because its efficiency will be less than that of the reversible one. The work done by it will be less than the corresponding work done by the reversible engine.

∫



Now, if Q2 amount of heat is rejected by the irreversible engine, then

(Q1 – Q′2) < (Q1 – Q2)

dQ  Q 1 Q′2  – >  T  T1 T2  Since the efficiency of the irreversible engine is or       

∫

 Q′2   T2  1 –  < 1 –  Q T1  1                  Let us consider any irreversible cycle as shown in Fig. 4.12

...(4.39) ...(4.40)

4.15

Second Law of Thermodynamics

A1

B1

C1

T

A2

B2

C2

s

Fig. 4.12 T-s diagram

Consider three adiabatic processes pass through the diagram at the points A1, A2, B1, B2 and C1, C2 as shown in Fig. 4.12. This adiabatic are formed by particle of isotherms A1B1, A2B2, B3C3 and B4C4 such that QA1B1 = QAB

...(4.42)

Then for any cycle say A1 B1 A2 B2

 δQ   ∫  = A1B1A2B2 = 0  T 

...(4.43)

(it is a Carnot cycle) When the adiabatics are drawn close to one another, the temperature along AB as well along A1B1 may be approximated by that at their points of intersection, so that Q Q           ...(4.44) =   T AB  T1 A1B1 Also QBB1 = 0 = QB′B2 = QA′A = QAA1 ...(4.45) (as they are portion of adiabatic)  δQ  = 0 ...(4.46)  ∫   T ABB′A′ Also, when a large number of infinitesimal cycles, they will all together, equivalent to the original cycle and as for each one of these infinitesimal cycle Hence,

δQ = 0, T

...(4.47)

 δQ   ∫  original cycle = 0  T 

...(4.48)

we shall have

∫

δQ will be negative for at T least some of the infinitesimal cycles and zero for others. If the original cycle is irreversible, the value

∫

4.16

Engineering Thermodynamics

Hence, in general dQ ≤ 0 T Therefore, for the irreversible cycle         ∫

       

∫

...(4.49)

dQ < 0 T

...(4.50)

This is known as Clausius inequality, the equality sign holds good if the cycle is reversible. The Clausius inequality is derived by using Carnot’s theorem which is a statement of the second law.

4.9.3 Entropy - Property of a System Any thermodynamic process is defined by its starting and ending of states. States are defined with definite values of the properties of a system. Let us consider entropy of a system which has unit mass and we have to prove that it is a property of the system. B

P

T

Q

R

A s

Fig. 4.13 T-s diagram showing different processes followed in a system

Now, assume that there are two end states, as shown in Fig. 4.13. There are two cycles constituted beginning from state A and reaching state B, through two processes. They are in a reversible manner; then (i) Process A-B-A via path P and Q (ii) Process A-B-A via path P and R B  dq  =    A ∫    T  APBQA

A

∫

B

 dq  T   APBRA

...(4.51)

We should have to prove that any path can be followed between two end states in thermodynamic cycle. But, their end values should not have any difference when the cycle follows different paths.

4.17

Second Law of Thermodynamics

Process A-B-A ( via Path P and Q)  dq 

∫  T APBQA = A ∫



B  dq 

+ T   APB

B

∫

A

 dq  ...(4.52) T   BQA

Similarly, Process A-B-A (via Path P and R)

 dq         ∫  T APBRA =

∫

A

B  dq 

+ T   APB

A

B

∫

 dq  ...(4.53) T   BRA

As per the clausius theorem for any reversible process  dq 

∫  T  = 0

     

A

∫

B  dq 

+ T   APB

B

...(4.54)

∫

A

 dq  = 0 T   BQA

...(4.55)

Via P, Q Similarly, B  dq  +       A ∫    T  APB

B

∫

A

 dq  = 0 ...(4.56) T   BRA

Via P, R Comparing Eqns. (4.54) and (4.55) A  dq  =      B ∫  T    BQA

B

∫

A

 dq  ...(4.57) T   BRA

BQA and BRA indicate two reversible paths. In the above equation, it is clear that the value of dq/T is some irrespective of path followed. The above expression is an independent path joining two states A and B. Though the paths of the process are different, the values of states A and B are the same in the two paths. Since

B

B

∫

A

 dq  = ds ...(4.58) T   BQA

A ∫ [dq /T ] depends only on end states A and B, it is proved that the

entropy is a point function and property of the system.

Graphical representation of entropy change: The change in entropy may be obtained in the graphical form also (i.e. temperature entropy diagram). Let us take the reversible process AB shown in Fig. 4.14.

4.18

Engineering Thermodynamics

As per definition ds =



For reversible adiabitic

dq T

ds = 0 because there is no change in entropy

Q

T

B

A

s

Fig. 4.14 Heat transfer represented in T-s diagram



δq = T ds

...(4.59)

b



qrev = T a∮ ds

...(4.60)

For m kg,

qrev = T(Sb – Sa).

...(4.61)

4.9.4 Change in Entropy of a Perfect Gas in Various Thermodynamic Processes We know that there are a few thermodynamic processes by which a system may undergo a process. Now, The change in the entropy of a perfect gas can be determined in each of the process.

4.9.4.1 Constant Volume Process Let us assume that p1 and p2 are the initial and final pressure of the gas respectively. T1 and T2 are the initial and final temperature of the gas. The mass of the gas is “m“. The gas is heated at a constant volume. Then for a small change in the temperature the amount of heat transferred during the constant volume process is given as

dq = mCv dT

...(4.62)

Divide the Eqn. (4.62) by temperature T and integrate within two limits dq



∫T

        

∫ dS = m C v ∫

= m Cv

∫

dT T dT T

...(4.62a)

4.19

Second Law of Thermodynamics

We get

(S2 – S1) = m Cv ln

T2 T1

...(4.63)

∆S = m Cv ln

T2 T1

...(4.64)



2

T

1 s

Fig. 4.15 T-s diagram for constant volume process

4.9.4.2 Constant Pressure Process Let us assume that v1 and v2 are the initial and final volumes of the gas respectively. T1 and T2 are the initial and final temperature of the gas. The mass of the gas is “m“. The gas is heated at a constant volume. Then for a small change in the temperature the amount of heat transferred during the constant pressure process is given as dq = m Cp dT



...(4.65)

Divide the Eqn. (4.65) by temperature T and integrate within two limits          ∫

dq dT = m Cp ∫ T T

...(4.65a)

dT T

...(4.65b)

          ∫ dS = m C p ∫ We get

(S2 – S1) = m Cp ln

T2 T1

...(4.65c)

ΔS = m Cp ln

T2 T1

...(4.66)

2 T 1 s

Fig. 4.16 T-s diagram for constant pressure process

4.20

Engineering Thermodynamics

4.9.4.3 Constant Temperature Process Let us assume that p1 and p2 are the initial and final pressures of the gas respectively. T1 and T2 are the initial and final temperatures of the gas. The mass of the gas is “m”. The gas is heated at a constant temperature. Then, for a small change in the temperature the amount of heat transferred during the constant pressure process is given as 1

T

2

s

Fig. 4.17 T-s diagram for constant temperature



Q = 2.3 m R T log

V2 V1

...(4.67)

Then, as per definition

Change in entropy =



S2 – S1 =



Heat transferred Absolute temperature

r  2.3 mRT log  2  T  r1 

V ∆S = 2.3 m R log  2  V1

  . 

...(4.67a)

4.9.4.4 Reversible Adiabatic Process or Isentropic Process In the case of a reversible adiabatic process, the system is considered to be highly insulated. It means that there will be no heat transferred from or out of the gas. 2 T 1 s

Fig. 4.19 T-s diagram for isoentropic process

So,

Q = 0

...(4.68)

ΔS = 0

...(4.69)

In this process the entropy of the gas remains the same throughout the process.

4.21

Second Law of Thermodynamics

4.9.4.5 Polytropic Process Let us assume that p1 and p2 are the initial and final pressures of the gas respectively. T1 and T2 are the initial and final temperatures of the gas. The specific volume of the gas at the initial and final conditions are v1 and v2 respectively. The mass of the gas is “m”. The gas is heated at polytropically. Then for a small change in the temperature the amount of heat transferred during the polytropic process is given as

dW dQ = ( γ – n ) × γ – 1 ( )

...(4.70)



dQ = ( γ – n ) ×

pdV ( γ – 1)

...(4.71)

Divide the Eqn. (4.71) by T           But

dQ = T

{( γ – n ) × pdV}

...(4.72)

( γ – 1) T

p mR = T V

...(4.73)

2 T

1 s

Fig. 4.19 T-s diagram for polytropic process

So Eqn. (4.72) becomes

dS =

{(

γ – n) × m R

( γ – 1)

dV V

Integrating within the two limits of 1 and 2

dS =

          ∫ dS =

{(

∫

γ – n ) × mR

{(

( γ – 1)

dV V

γ – n) × m R

( γ – 1)

}

}

dV V





}

...(4.74)

...(4.75)



 V    2.3 ( γ – n ) × mR ln  2    V1              S = ( γ – 1)

...(4.76)

...(4.77)

4.22

Engineering Thermodynamics

Solved Problems Problems 4.1 A cyclic heat engine operates between a source temperature of 600°C and a sink temperature of 30°C. What is the least rate of heat rejection per kW net output of the engine? Given

Ts = 600 + 273 = 873 K

Tr = 30 + 273 = 303 K Solution T   303  Maximum efficiency, hmax = hrev = 1 −  s  = 1−  = 0.652  873   T1 

hmax = 65.2% Source Ts = 873 Qs Wnet = Qs – Qr

Sink Tr = 303 k

Fig. p. 4.1

Qs =

Wnet . hmax

=

1 0.652

Heat supplied,

= 1.533 kW Heat rejection,

Qr = Q1 – Wnet.

= 1.533 – 1 = 0.533 kW. Problem 4.2 A domestic food freezer maintains a temperature of –10°C. The ambient air temperature is 30°C. If heat is rejected from the freezer at the continuous rate of 1.75 kJ/s, what is the least power necessary to pump this heat out continuously? Given Freezer temperature, Tr = –10 + 273 = 263 K. Ambient air temperature, TAtm = 30 + 273 = 303 K. For the minimum power requirement,

QAtm . TAtm

4.23

Second Law of Thermodynamics

Solution Atmosphere TAtm = 303 QAtm

W=?

R QRef

Freezer TRef = 258

Fig. p. 4.2



QAtm =

= Therefore,

Q Ref TAtm TRef 1.75 × 303 = 2.050 kJ/s. 263

W = QAtm – Qref = 2.05 – 1.75

= 0.30 kJ/s = 0.30 kW. Problem 4.3 An engine works between the temperature limits of 1600 K and 350 K. What can be the maximum thermal efficiency of this engine? Given Ts = 1600 K; Tr = 350 K. Solution

T  hmax = 1 −  r   Ts 

 350  = 1 −  0.781 =  1600  hmax = 78.1% Source Ts = 1775 K

E

Sink Tr = 375

Fig. p. 4.3

4.24

Engineering Thermodynamics

Problem 4.4 A reversible engine is supplied with heat from two constant temperature sources at 900 K and 600 K and it rejects heat to a constant temperature sink at 300 K. The engine develops work equivalent to 90 kJ/s and rejects heat at the rate of 56 kJ/s. Estimate: (i) The heat supplied by each source. (ii) The thermal efficiency of the engine. Given Ts1 = 900 K; Ts2 = 600 K;  T2 = T4 = 300 K;

W = 90 kJ/s; Q2 + Q4 = 56 kJ/s. Qs1 = heat supplied by the first source. Qs2 = heat supplied by the second source.

Solution

h1 (first source) =

W1 Qs1

 Qr 1  = 1 −    Qs1 

 Tr  = 1 –    Ts 1 

 300  = 1 –    900  = 0.67 = 67%

Work obtained from the first source, W1 = Qs1 – Qr1 = 0.67 Qs1 The heat rejected to the sink,

Qr1 = Qs1 – W1

= Qs1 – 0.67 Qs1 = 0.33 Q1. Source 1 Ts1 = 900 K

Source 2 Ts2 = 600 K W1 + W2 = W

Qs1

Qs1

Sink Tr1 = Tr2 = 300 K

Fig. p. 4.4

4.25

Second Law of Thermodynamics

Similarly, the efficiency of the engine when the heat is supplied from the second source,

h2 =

W2 Qs2

 Qr 2   = 1 −   Qs 2   Tr 2  = 1 –    Ts 2   300  = 1 –    600  = 0.5 = 50% Work obtained by the engine from the second source, W2 = Qs2 – Qr2 = 0.5 Qs2 Heat rejected to the sink,

Qr2 = Qs2 – W2

= Qs2 – 0.5 Qs2 = 0.5 Q3. W.k.t. total work obtained from the engine (W), 90 W1 + W2 = 0.67 Qs1 + 0.5 Qs2 ...(1) Total heat rejected to the sink,

Qr1 + Qr2 = 56 = 0.33 Qs1 + 0.5 Qs2 ...(2)

By solving the equations we get,

Qs1 = 100 kJ/s

and

Qs2 = 46 kJ/s.

Thermal efficiency of the engine,

hmax =

=

W

( Qs1 + Qs 2 ) W Q ( s1 + Qs 2 )

= 0.616 = 61.6%. Problem 4.5 A cold storage is to be maintained at –15°C, while the surroundings are at 35°C. The heat leakage from the surroundings into the cold storage is estimated to be 35 kW. The actual COP of the refrigeration plant is one third of an ideal plant working between the same temperatures. Find the power required to drive the plant.

4.26

Engineering Thermodynamics

Given

TRef = –15 + 273 = 258 K;

TAtm = 25 + 273 = 298 K;

Solution

QRef = 35 kW.

COPactual =



1 × COPideal 3 TRef

COPactual =

( TAtm – TRef )



258 (298 − 258)

=

= 6.475. Therefore, COPactual =

1 2.158 × 6.475 = 3

Also

W =

=

Q2 COPactual 29 2.158

= 13.438 kW. Atmosphere TAtm = 308 K

W

QAtm = 41.60 kW R QRef = 29 kW Cold storage Tr = 268 K

Fig. p. 4.5

Problem 4.6 Two reversible heat engines A and B are arranged in a series. Engine A rejects heat directly to B. Engine A receives 200 kJ at a temperature of 421°C from a hot source, while the engine B is in communication with a cold sink at a temperature of 4.4°C. If the work output of A is twice that of B, find (a) The intermediate temperature and efficiency of each engine (b) The heat rejected to the cold sink. Given Ts1 = 421 + 273 = 694 K.   

Ts2 = intermediate temperature which is to be determined.

4.27

Second Law of Thermodynamics

Solution Ts3 = 4.4 + 273 = 277.4 K. Wa = 2Wb. For the maximum power output, Ts1 – Ts2 = 2 × (Ts2 – Ts3)

694 – Ts2 = 2 × (Ts2 – 277.4)

Ts2 = 416.3 K = 143.3°C.

 Ts 2  ha = 1 –    Ts 1 

 416.3  = 1 –    694  = 0.4 = 40%

T  hb = 1 –  s 3   Ts 2 

 277.4  = 1 –    416.3  = 0.33 = 33%

ha =

W1 Qs1 Source Ts1 = 694 K Qs1 WA HE A Qs2 WB

HE B

Qr Sink Ts4 = 694 K

Fig. p. 4.6

W1 = 0.4 × 200 = 80 kJ.

Q2 = Qs1 – W1

= 200 – 80 = 120 kJ.

4.28

Engineering Thermodynamics

Similarly,

W2 = hb × Qs2

= 39.6 kJ.

Qs3 = Qs1 – W2

= 80.4 kJ. Problem 4.7 A reversible heat engine receives heat from a reservoir at a temperature of 700°C and rejects heat at a temperature of T2. A second reversible engine receives the heat rejected by the first engine and rejects the heat to a sink at the temperature of 37°C. Calculate the temperature T2 for (a) an equal efficiency for the both engines, (b) an equal work output of both the engines. Given: Ts1 = 700 + 273 = 973 K; Tr = 37 + 273 = 310 K. Solution

 Ts 2 ha = 1 –   Ts 1

  Ts 2  =  ...(1)   973 



 Tr hb = 1 –   Ts 2

  310  = 1 −   ...(2)   Ts 2 

At equal efficiency,

         

ha = hb  310  T  1 –  s 2 = 1 −    973   Ts 2 

Ts2 = 549 K = 276°C.

T   549  ha = 1 –  s 2  = 1 –    973   973 

= 0.435 = 43.5% = hb. Source Ts1 = 973 K Qs1 Engine A Engine B Qr Sink Tr = 310 K

Fig. p. 4.7

4.29

Second Law of Thermodynamics

For equal work output, Wa = Wb

Ts1 – Ts2 = Ts2 – Tr

973 – Ts2 = Ts2 – 310

Ts2 = 641.5 K

= 368.5°C. Problem 4.8 In a space heating, 3500 kJ/min of heat is required by a heat pump. The work required to operate the heat pump is 8 kJ/h. Determine the COP and the heat absorbed from outside. Given Solution

Qp = 3500 kJ/min; W = 480 kJ/min. W = Qr – Qs Qp Qs = Qp – W

= 3500 – 480 = 3020 kJ/min.  QP  COP(Heat pump) = 1 –    QAtm   200000  = 1 –   = 7.14.  172000  Space

W Qs Atmosphere

Fig. p. 4.8

Problem 4.9 A vessel of 2.5 m3 capacity contains 1 kg mole of nitrogen at 100°C. If the gas is cooled to 30°C, calculate the change in the specific entropy. The ratio of the specific heats is 1.4 and 1 kg mole nitrogen is 28 kg. Given

V = 2.5 m3;



M = 1 kg mole = 28 kg;

T1 = 100 + 273 = 373 K; T2 = 30 + 273 = 303 K;

g = 1.4.

4.30

Engineering Thermodynamics

Solution

R =

=

R M 8.314 28

= 0.297 kJ/kg K. Cv =

R ( g – 1)

=

0.297 14 ( – 1)

= 0.74 kJ/kg K. Cp = Cv × g = 0.74 × 1.4 = 1.036 kJ/kg K.

T  ΔS = 2.3 m Cv log  2   T1   

= – 0.1536 kJ/kg K. 1 T 2 S

Fig. p. 4.9

Problem 4.10 0.5 kg of a perfect gas is heated from 100 to 300°C at a constant pressure of 2.8 bar. It is then cooled to 100°C at constant volume. Find the overall change in the entropy. Consider Cp = 1 kJ/kg K and Cv = 0.72 kJ/kg K. Given m = 0.5 kg; p = 2.8 bar = 2.8 × 105 N/m2; T1 = 100 + 273 = 373 K; T2 = 300 + 273 = 573 K; T3 = 100 + 273 = 373 K. Solution Let us find ΔS1:

T  (S2 – S1) = 2.3 m Cp log  2   T1   573  = 2.3 × 0.5 × 1 × log    373  = 0.214 kJ/K.

4.31

Second Law of Thermodynamics

At constant volume, Δ S2,

 T3  (S3 – S2) = 2.3 m Cv log    T2 

 373  = 2.3 × 0.5 × 1 × log    573  = – 0.154 kJ/K. Overall change in entropy,

ΔS = S3 – S1

= ΔS1 + ΔS2

= 0.214 – 0.154 = 0.06 kJ/K. 2 T 1 S

Fig. p. 4.10

Problem 4.11 A certain quantity of a perfect gas is heated in a reversible isothermal process from 1 bar at 40°C to 10 bar. Find the work done and entropy per kg of gas. Take R = 287 J/kg K. Given T = 40 + 273 = 313 K; m = 1 kg; p1 = 1 bar = 1 × 105 N/m2; p2 = 10 bar = 10 × 105 N/m2. Solution Work done,

V  W = 2.3 m R T log  2   V1 

p  = 2.3 m R T log  1   p2   1  = 2.3 × 1 × 287 × log    10  = – 206.61 kJ/kg. Change in entropy,

p  ΔS = 2.3 m R log  1   p2 

= 2.3 × 1 × 287 × log  1   10  = – 0.6601 kJ/kg K.

4.32

Engineering Thermodynamics

Other way, Since,

W1–2 = Q1–2 Heat supplied Absolute temperature

ΔS =

=

–206.61 313

= –0.6601 kJ/kg K. T

2

1

S

Fig. p. 4.11

Problem 4.12 1 kg of air at a pressure of 7 bar and a temperature of 363 K undergoes a reversible polytropic process which may be represented by pV1.1 = constant. The final pressure is 1.4 bar. Evaluate (a) the final specific volume, temperature and increase in the entropy. (b) the work done and the heat transfer during the process. Assume R = 287 J/kg K and g = 1.4. Given

T1 = 363 K;  m = 1 kg;

p1 = 7 bar = 7 × 105 N/m2;  g = 1.4;

R = 287 J/kg K;  p2 = 1.4 bar = 1.4 × 105 N/m2;

n = 1.1. Solution

p1v1 = mRT1. v1 =

mRT1 p1

= 0.15 m3/kg.

 p T1 1 =  T2   p2 

T2 =

 p  1   p2 

  

( n–1) /n 

  

T1   

( n–1) /n 

= 313.5 K. v  T1 =  1 T  v2            1

( n −1)

  

4.33

Second Law of Thermodynamics



v2 = 0.649 m3/kg.  ( g − n )   p1  ΔS = m Cv   ln    n   p2 

= 0.314 kJ/K. Work done,

w =

( p1 v1 – p2 v2 ) ( g – 1)

= 141.4 kJ/kg. Heat transfer,

q =

( g – n) ×w ( g – 1)

= 106.05 kJ/kg. 2 T 1 S

Fig. p. 4.12

Problem 4.13 An ideal gas of molecular mass 30 and specific heat ratio of 1.38 is compressed according to the law of pV1.25 = constant. The process has a change from a pressure of 1 bar and 15°C to a pressure of 16 bar. Calculate the temperature at the end of compression, the heat received or rejected, the work done by the gas during the process and the change in the entropy. Assume that the mass of the gas is 1 kg. Use only the calculated values of Cp and Cv. Given T1 = 15 + 273 = 288 K;  m = 1 kg; p1 = 1 bar = 1 × 105 N/m2; m = 30;  p2 = 16 bar = 16 × 105 N/m2. Solution          

T1 ( n–1) /n  = ( p1 – p2 )   T2

T2 = 501.4 K.

R =

=

Ru m

8314 30

= 277 J/kg K. Cv =

R

( g – 1)

4.34

Engineering Thermodynamics

=

277 ( 1.4 – 1)

= 0.73 kJ/kg K. Cp = Cv × g

= 1.007 kJ/kg K. we know that Tvg–1 = constant

v1 = 0.8 m3. T1 ( n –1)  = ( v1 – v2 )   T2

          v2 = 0.08 m3/kg. Work done,

w =

( p1 v2 – p1 v1 ) ( n – 1)

= 236.8 kJ. Heat transfer

 ( g – n)  Q =   ×w  ( g – 1 ) 

= 81 kJ.  ( g – n)   p1  Change in entropy, Δs =   × ln    p2   n  = –0.21 kJ/K. 2 T 1 S

Fig. p. 4.13

Problem 4.14 A gas at 95°C and 1 bar is compressed with an index of compression 1.3, the volume compression ratio being 6:1. The maximum pressure is 25 bar. Assuming that the ratio of specific heats as 1.38 and Cv = 0.754 kJ/kg K, find the change in entropy during the compression stroke which takes place at (i) a polytropic process (ii) a constant volume process. Solution Given

p1 = 1 bar = 1 × 105 N/m2; 



p2 = 25 bar = 25 × 105 N/m2;



n = 1.3; g = 1.38; Cv = 0.754 kJ/kgK;

v           1 = 6 ; T1 = 95 + 273 = 368 K. v2

4.35

Second Law of Thermodynamics

 p ( n–1) /n  T1  =  1  T2   p2               T2 = 773.4 K. V  Change in entropy, ΔS = m Cv(g – n) ln  2   V1  = – 0.108 kJ/kg K. Change in entropy at constant volume,

T  (Δs) v = m Cv ln  1   T2 

= 0.6 kJ/K.

Objective Type Questions 1. A heat engine working on the Carnot cycle receives heat at the rate of 40 kW from a source at 1200 K and rejects it to a sink at 300 K. The heat rejected is (a) 30 kW

(b) 20 kW

(c) 10 kW

(d) 5 kW

2. An electric power generating station produces 400 MW. If the coal releases 36 × 108 KJ/h of energy, the rate of rejection of heat from the power plant is (a) 200 MW

(b) 400 MW

(c) 600 MW

(d) 800 MW

3. The efficiency of a Carnot engine is given as 0.75. If the cycle direction is reversed, what will be the value of C.O.P. for the Carnot refrigerator? (a) 0.27

(b) 0.33

(c) 1.27

(d) 2.33

4. An inventor claims that the heat engine has the following specifications: Power developed = 50 kW; fuel burned per hour = 3 kg; Heating value of fuel = 75,000 kJ/kg; Temperature limits = 627°C and 27°C; Cost of fuel = ` 30/kg; value of power = ` 5/kWh. The performance of his engine is (a) Possible

(b) Not possible (c) Economical (d) Uneconomical

5. During which of the following processes does heat rejection take place in a Carnot vapour cycle? (a) Constant volume

(b) Constant pressure

(c) Constant temperature

(d) Constant entropy

6. Which one of the following changes/sets of changes in the source and sink temperature (T1 and T2 respectively) of a reversible engine will result in the maximum improvement in efficiency? (a) T1 + ΔT

(b) T2 – ΔT

(c) (T1 + ΔT) and (T2 – ΔT)

4.36

Engineering Thermodynamics

7. Heat transfer takes place according to the (a) Zeroth law of Thermodynamics (b) First law of Thermodynamics (c) Second law of Thermodynamics (d) Third law of Thermodynamics 8. A heat pump operating on the Carnot cycle pumps heat from a reservoir at 300 K to a reservoir at 600 K. The coefficient of performance is (a) 1.5

(b) 0.5

(c) 2

(d) 1

9. The heat absorbed and rejected during a polytropic process is equal to 1/2

 ( g – n )  (a)    ( g – 1 ) 

 ( g – n )  × work done (b)   × work done  ( n – 1 )  2

 ( g – n )   ( g – n )  (c)   × work done (d)   × work done  ( g – 1 )   ( g – 1 )  10. Clausius inequality is stated as δQ δQ (a) ∫ δQ < 0 (b) 0 (c) ∫ δQ = ∫ T > 0 (d) ∫ T ≤0 11. An electric motor of 5 kW is subjected to a braking test for 1 Hour. The heat generated by the frictional forces in the process is transferred to the surroundings at 20°C. The resulting entropy change will be (a) 22.1 kJ/K

(b) 30.2 kJ/K

(c) 61.4 kJ/K

(d) 82.1 kJ/K

12. If a system undergoes an irreversible adiabatic process, then (symbols have usual meanings)  δQ  (a) ∫   = 0 and ΔS > 0  T  (c)

 δQ   > 0 and ΔS = 0 T 

∫ 

 δQ  (b) ∫   = 0 and ΔS = 0  T   δQ  (d) ∫   < 0 and ΔS < 0  T 

13. 1600 kJ of energy is transferred from a heat reservoir at 800 K to another heat reservoir at 400 K. The amount of entropy generated during the process would be (a) 6 kJ/K

(b) 4 kJ/K

(c) 2 kJ/K

(d) Zero

14. A cyclic heat engine receives 600 kJ of heat from a 1000 K source and dQ  rejects 450 kJ to a 300 K sink. The quantity Φ   and the efficiency of  T  the engine are respectively. (a) 2.1 kJ/K and 70%

(b) –0.9 kJ/K and 25%

(c) 0.9 kJ/K and 70%

(d) –2.1 kJ/K and 25%

15. Which one of the following statements applicable to a perfect gas will also be true for an irreversible process? (Symbols have the usual meanings). (a) dQ = dU + pdV

(b) dQ = T dS

(c) TdS = dU + pdV

(d) None of the above

4.37

Second Law of Thermodynamics

 δQ  16. When a system undergoes a process such that ∫   = 0 and ΔS > 0, the  T  process is (a) Irreversible adiabatic (b) Reversible adiabatic (c) Isothermal (d) Isobaric 17. A system of 100 kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/kg–K to 0.4 kJ/kg–K. At the same time, the entropy of the surroundings decreases from 80kJ/K to 75kJ/K. The process is (a) Reversible and Isothermal (c) Reversible

(b) Irreversible (d) Impossible

18. Considering the relationship Tds = dU + pdV between the entropy, internal energy, pressure and temperature, which of the following statement is correct? (a) It is applicable only for a reversible process (b) For an irreversible process, TdS > dU + pdV (c) It is valid only for an ideal gas (d) It is equivalent to the Ist law, for a reversible process 19. An industrial heat pump works between the temperatures of 27°C and –13°C. The rates of heat supply and heat rejection are 750 W and 1000 W respectively. The COP of the heat pump is (a) 7.5

(b) 4.0

(c) 6.5

(d) 3.0

20. The COP of the heat pump is (a) Greater than the COP of the heat engine by unity (b) Lesser than the COP of the heat engine by unity (c) Equal to the COP of the heat engine (d) Cannot be ascertained

Theory Questions 1. Give the following statements of the second law of thermodynamics. (i) Clausius statement (ii) Kelvin-Planck statement. 2. What is a thermal energy reservoir? 3. What do you mean by the perpetual motion machine of second kind (PMM-II)? 4. Describe a heat engine, a refrigerator and a heat pump. 5. What do you mean by ‘Clausius inequality’? 6. State the Carnot Theorem. Describe the working principle of a Carnot cycle. Why is it not practically possible?

4.38

Engineering Thermodynamics

7. Derive an expression for the efficiency of a reversible heat engine. 8. What do you mean by the term ‘Entropy’? 9. What do you mean by a reversible process? 10. What is an irreversible process? Give some examples of irreversible processes. 11. What do you mean by Entropy Change, Entropy Generation and Entropy Transfer? 12. Enunciate the Clausius’ statement of second law of thermodynamics. 13. Mention any four factors which render processes irreversible. 14. What is the importance of a temperature-entropy diagram? 15. State the Kelvin-Planck statement of second law of thermodynamics. 16. What is the process involved in a Carnot cycle. Sketch the same in p-V and T-s diagram. 17. What is a perpetual motion machine of the second kind? 18. A reversible heat engine operates between a source at 800°C and a sink at 30°C. What is the least rate of heat rejection per kW network output of the engine? 19. State the second law of thermodynamics. Also write its physical significance. 20. What are the limitations of the first law of thermodynamics? 21. What is meant by a ton of refrigeration? 22. Define the Carnot cycle with p-V and T-s diagrams. 23. Establish the inequality of Clausius and express entropy change in irreversible process. 24. Prove that the Kelvin-Planck and Clausius statements stating second law of thermodynamics are equivalent. 25. State and prove the Clausius inequality and hence deduce that the property entropy exists. 26. Deduce the expression for the efficiency of a Carnot engine. 27. A reversible engine works between three thermal reservoirs A, B and C. The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperatures TA and TB, respectively, and rejects heat to the thermal reservoir C, kept at temperature TC. The efficiency of the engine is a times the efficiency of the reversible engine, which works between the two reservoirs A and C. Prove that TA T = ( 2a – 1) + 2 ( 1 – a ) A TB TB

Second Law of Thermodynamics

4.39

28. A Carnot engine receives energy from a reservoir at Tres through a  is proportional to the heat exchangers where the heat transfer rate Q H  = K(T – T ). It rejects heat temperature difference and is given by Q H res H at a given low temperature TL. To design the engine for maximum power, show that the high heat temperature TH should be selected as TH = Tres TL . 29. In order to check the validity of the second law, m1 kg of the fluid at a temperature of T1 is entropically and adiabatically mixed with m2 kg of the fluid at temperature T2(T1 > T2). Determine the change in the entropy of the universe, and find an expression for the same for an equal mass of fluid. Also prove that change in entropy is necessarily positive and is given T + T2 by log 1 . 2 T1 T2 30. Derive the following relation for the entropy change of an ideal gas p  v  = ∆S C p ln  2  + C v ln  2  v  1  p1  31. Write expressions for the change of entropy of a perfect gas during: (a) Constant volume process in terms of pressure and temperature (b) Constant pressure process in terms of volume and temperature

Unsolved Problems 1. Prove that no heat engine can be more efficient than a reversible Cornot heat engine, when both of them are operating between the same thermal reservoirs. 2. Two Carnot refrigerators A and B operate in a series. The refrigerator A absorbs energy at the rate of 1 kJ/s from a body at 300K and rejects energy as heat to a body at T. The refrigerator B absorbs the same quantity of energy which is rejected by the refrigerator A from the body at T, and rejects energy as heat to a body at 1000K. If both the refrigeratiors have the same COP, calculate (i) The temperature T of the body (ii) The COP of the refrigerators and (iii) The rate at which energy is rejected as heat to the body at 1000K. (Ans: 547.72 K, 1.211, 3.333 kJ/s) 3. A reversible heat engine receives heat from a high temperature reservoir at T1 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 it to a cold reservoir at 280K. Make calculations for temperature T1: (i) for equal thermal efficiencies of the two engines, (ii) For the two engines to deliver the same amount of work. (Ans: 2285.7 K, 1320 K)

4.40

Engineering Thermodynamics

4. An inventor claims to have designed a heat engine, which absorbs 260 kJ of energy as heat from a reservoir at 52°C and delivers it as 72 kJ work. He also states that the engine rejects 100 kJ and 88 kJ of energy to the reservoirs at 27°C and 2°C, respectively. State with justification,whether his claim is acceptable or not. (Ans: Not acceptable) 5. A Carnot engine operates between two reservoirs at the temperature of T1°K and T2°K. The work output of the engine is 0.6 times that of the heat rejected. Given that the difference in temperature between the source and sink is 200°C, calculate the source temperature, sink temperature and thermal efficiency                     (Ans: 533.3 K, 333.33 K, 37.5 %) 6. One kg of air is heated from 27°C to 127°C from a heat source at 127°C. Find the entropy change of air, heat source and of the universe. What will be the change of entropy of the universe, if the air is heated firstfrom a heat source at 77°C and then from another heat source at 127°C.? Comment on the results. (Ans: 0.28912 kJ/kg°C, –0.25125 kJ/kg°C, 0.03787 kJ/kg°C) 7. Water, initially a saturated liquid at 100°C, is contained within a pistoncylinder assembly. Water undergoes a process, during which the piston moves freely in the cylinder, and no heat transfer takes place with the surroundings, but the state is changed to the corresponding vapour state. If the change of state is brought about by the action of a paddle wheel, determine the network per unit mass in kJ/kg, and the amount of entropy produced per unit mass. (Steam tables can be referred) (Ans: 2087.6 kJ/kg, 6.048 kJ/kg K) 8. A cyclic heat engine operates between the source temperature of 1000°C and a sink temperature of 40°C. Find the least rate of heat rejection per kW net output of the engine? (Ans: 0.326 kJ/s) 9. A Carnot heat engine draws heat from a reservoir at temperature T1 and rejects the heat to another reservoir at temperature T3. The Carnot forward cycle engine drives a Carnot reverse cycle engine or Carnot refrigerator which absorbs the heat from the reservoir at a temperature T2 and rejects the heat to the reservoir at a temperature T3. The higher temperature T1 is equal to 600 K and the lower temperature T2 is equal to 300 K, determine the (i) Temperature T3 such that the heat supplied to the engine Q1 is equal to the heat absorbed by the refrigerator Q2. (ii) The efficiencyof the Carnot engine and COP of Carnot refrigerator. (Ans: 400 K) 10. A heat engine receives heat from a source at 1200 K at the rate of 500 kJ/s, and rejects the waste heat to a medium at 300 K. The power output of the heat engine is 180 kW. Determine the reversible power and irreversible rate for this process. (Ans: 375 kJ/s (kW), 195 kW) 11. Carbon steel balls (r = 7833 kg/m3) and Cp = 0.465 kJ/kg°C, 8 mm in diameter, are annealed by heating them firstto 900°C in a furnace, and then allowing them to cool slowly to 100°C in ambient air at 35°C. If 2500 balls are to be annealed per hour, determine the (i) rate of heat transfer from

Second Law of Thermodynamics

4.41

the balls to the air and (ii) rate of entropy generation due to heat loss from the balls to the air. (Ans: 1953 kJ/hour, 3.544 kJ/°C per hour) 12. A domestic food freezer maintains a temperature of –15°C. The ambient air is at 30°C. If heat leaks into the freezer at a continuous rate of 1.75 kJ/s, what is the least power necessary to pump this heat out continuously? 13. Three identical bodies of constant heat capacities are at temperatures 300 K and 100 K. If no work or heat is supplied from outside, then what is the highest temperature to which any one of these bodies can be raised by the operation of a heat engine or refrigerators? 14. A vessel of volume 0.04 m3 contains a mixture of saturated water and saturated steam at a temperature of 250°C. The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and the internal energy. 15. One kg of air at 310 K is heated at constant pressure, by bringing it in contact with a hot reservoir at 1150 K. Find the entropy change of air, the hot reservoir and universe, assuming Cp of air to be 1.005 kJ/kg K.

16. Two reversible heat engines A and B are arranged in a series. ‘A’ rejects heat directly to ‘B’. Engine A receives 200 kJ at 427°C from a heat source, and rejects sit to a sink at 7°C. The work output of A is twice that of B. Find the intermediate temperatures between A and B, efficiency of each engine, and the quantity of heat rejected to the sink. Sketch the arrangement and label all the results. 17. Two reversible heat engines A and B are arranged in a series with A rejecting heat directly to B through an intermediate reservoir. Engine A receives 200 kJ of heat from a reservoir at 421°C and engine B is in thermal communication with a sink at 4.4°C. If the work output of A is twice that of B, find (i) the intermediate temperature between A and B, (ii) the efficiency of each engine and (iii) the heat rejected to the cold sink. 18. Air expands from a pressure of 10 bar at 427°C to a pressure of 1 bar, according to the law of expansion pV 1.25 = constant. Find the final temperature and specific volume. Also calculate the heat transfer and the entropy change. 19. A reversible engine works between a source at 987°C and two sinks one at 127°C and another at 27°C. The energy rejected is the same at both the sinks. If the heat input rate is 100 kW, compute the heat rejection rate to each sink and the engine efficiency. 20. A heat engine operating between two reservoirs at 300 K and 100 K is used to drive a heat pump, which extracts heat from the reservoir at 300 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, determine the rate of heat rejection from the heat pump if the rate of supply of heat to the engine is 50 kW.

4.42

Engineering Thermodynamics

21. A reversible power cycle is used to drive a heat pump cycle. The power cycle takes in Q1 heat unit at T1, and rejects Q2 at T2. The heat pump extracts Q4 amount of heat energy from the sink at T4, and discharges Q3 at T3. Develop Q an expression for the ratio 3 in terms of the four temperatures. What Q1 Q must be the relationship of the temperatures T1, T2, T3 and T4 for 3 to Q1 exceed the value of unity?

CHAPTER

5

AVAILABILITY AND IRREVERSIBILITY

5.1 INTRODUCTION All the thermal applications may be of heat absorption or heat rejection systems, and work absorption or work developing systems. With example starting from small compressors to power plants, the thermal equipment involved in them are influenced by various parameters. The most important factors are availability and irreversibility. They define the quality of energy available in the thermodynamic process of a system. In principle, the thermal efficiency of the device is the function of input and output. Therefore, it is essential to measure the energy losses in the system. In fact, such measurements are useful to find the energy wasted in a particular thermodynamic process. A knowledge of the available energy and waste in energy helps to improve the performance of a thermodynamic application. The basics of the second law of thermodynamics reveal that a complete conversion of work to heat is possible, but the conversion of heat into work is not possible. The energy, which is available for conversion, is known as the available energy. The energy, which is not available for conversion is called unavailable energy. Thermal systems which exist above their absolute temperature possess energy. From the second law of thermodynamics point of view, it is proven that part of the heat energy is converted into work from the supplied heat. The second law identifies the directions of the flow of available and unavailable energy. The unavailable energy is also termed as irreversibility. In all thermal systems, the determination of the available energy and irreversibility is very important. In work developing systems like turbines, internal combustion engines etc., the efficiency can be improved by knowing the irreversibility. The determination of the availability helps us to know how much work can be performed by the system, during the change of one state to another. This chapter helps us to find the maximum work obtainable in a process. We know to calculate the maximum work which could be produced in a process. The availability of a system gives information about the amount of work that could be done or achieved by a system. It also reveals the amount of work obtainable in a system, which could perform a work during a change of state.

5.2

Engineering Thermodynamics

5.2

AVAILABLE ENERGY

There are two grades of energy, (1) high grade and (2) low grade energy. Examples of high grade energy are mechanical work, electrical work, hydraulic power, wind power, tidal power and kinetic energy. Heat derived from combustion of fuel energy from fusion and fission and solar energy are examples of low grade energy. Low grade energy can be used to get high grade energy source, but complete conversion of low grade to high grade energy is impossible. The part of low grade energy available for conversion into high grade energy is referred as useful work or available energy.

5.2.1 Energy Interaction with a Constant Temperature Energy Source Consider a heat engine which is shown in Fig. 5.1. Q1 amount of heat supplied from a reservoir (source) at T1 and Q0 amount of heat is rejected to the low temperature reservoir which is at T0 Source Q1 E

Available energy T

W = Q1 – Q0

Unavailable energy

Sink, Q0

Fig. 5.1 Heat engine

We know that ,

s

Fig. 5.2 Available and unavailable energy

W = Q1 – Q0 …(5.1)

Q Q Also for reversible engine, = 1 = 0 T0 T1

…(5.2)

 Q  W = Q1  1 − 0  …(5.3) Q1    T = Q1  1 − 0 T1 

  …(5.4) 

 T  In cyclic reversible engine, dW = dQ1  1 − 0  T1   This is represented in T–s diagram in Fig. 5.2.

…(5.5)

5.3

Availability and Irreversibility

The area below the surrounding temperature T2 is unavailable energy which is T2(S2 – S1) T2 = T0 when sink temperature is equal to atmospheric temperature.

5.2.2 Heat Transfer in a Heat Exchange Process Consider a heat received through a heat exchange process carried out at constant pressure and rejects heat to the sink as shown in Fig. 5.2. Let Wrev = (Q1 – Q2) – Q0 = (Q1 – Q2) – T0∆S

...(5.6) …(5.7)

Heat transfer through a finite temperature difference. Consider a heat engine operating between temperatures T1 and T0 shown in Fig. 5.11. We know that,

W = Q1 – Q0

…(5.8)



 Q W = Q1  1 − 0 Q1 

  

…(5.9)

 T = Q1  1 − 0 T1 

  

…(5.10)

 T  = T1∆S  1 − 0  T1  

…(5.11)

= (T1 – T0) ∆S

…(5.12)

Let us now assume that Q1 amount of heat is transferred through a finite temperature difference from the source at T1 to the engine absorbing heat T1′ lower then T1. Engine receives heat Q1 at T1′ and rejects Q2′. The engine reversibly operates between two temperatures T1 and T0 At T1 temperature, Q1 = T1′ ∆S         

…(5.13)

∆S′ > ∆S as T1 > T1′ Q2′ = T0 ∆S and Q2′ = T0 ∆S′

Since

∆S′ > ∆S

Therefore,

Q2′ > Q2



W′ = Q1 – Q′2

…(5.14)

…(5.15)

= T1 ∆S′ – T0 ∆S′

…(5.16)



…(5.17)

W = Q1 – Q2 = T1 ∆S – T0 ∆S

5.4

Engineering Thermodynamics

Since

∆S′ > ∆S and Q2′ > Q2



Q2′ > Q = W′ < W

Subtracting Eqn. (5.17) from Eqn. (5.12),

W – W′ = T0 (∆S′ – ∆S)

…(5.18)

Therefore, whenever there is a heat transfer through finite temperature difference, then there will be a reduction in available energy. The decrease in the available energy is the product of the lower feasible temperature of the heat reduction and additional entropy change in the system while receiving heat transfer from the same source. It is to be noted that there will be greater heat rejection Q2′ and greater unavailable energy when the temperature is greater during a heat flow through a finite temperature. Energy degrades each time and therefore, the second law is also sometimes referred to as the law of degradation of energy.

5.3

AVAILABILITY IN THERMODYNAMIC SYSTEMS

The term availability of a given system is defined as a maximum useful work that can be obtained in a process in which the system comes to equilibrium with its surroundings. It depend upon the state of system and surroundings. Suppose a system only exchange heat with surroundings. Then, the availability is the maximum possible useful work, that can be obtained from the system since that the system attains a dead state. The dead state is referred a system which is in thermal and mechanical equilibrium with its surroundings.

5.3.1 Reversible Work in Steady Flow Process In a steady flow process, dm1 = dm2 = dm

…(5.19)

Then, the steady flow energy equation is : H1 +

1 1 (mv12) + mgz1 + Q = H2 + (mv22) + mgz2 + W 2 2

…(5.20)

The subscript 1 and 2 refers the inlet outlet conditions At the outlet system be in equilibrium with the surrounding at pressure p0 and temperature T0. If kinetic and potential energy are negligible then

W = (H – H0) + Q

…(5.21)

If the term Q is greater, then the useful work will be larger or when Q is maximum then W is maximum. Suppose S1 and S0 are entropies of the system at the inlet and outlet of the system then,

(∆S)system = S1 – S0

…(5.22)

5.5

Availability and Irreversibility

Q (∆S)surr. = –   T0



  

…(5.23)

From the entropy principle, (S1 – S0) – (Q/T0) > 0

…(5.24)

or,

Q ≤ T0 (S1 – S0)

…(5.25)

The useful work,

W ≤ (H – H0) + T0 (S1 – S0)

…(5.26)



W ≤ (H – T0 S1) + (H0 – T0 S0)

…(5.27)

Hence,

W ≤ B – B0

…(5.28)

Where

B = H – T0 S

…(5.29)

B is called availability function for steady flow and is also called Keenam function. Reversible work in a non-flow process We know that in non-flow process, dm1 = dm2 = 0



…(5.30)

From the first law of thermodynamics

Qsys – W = U1 – U2

…(5.31)

From the second law of thermodynamics for the reversible process

(∆S)universe = (∆S)system + (∆S)surroundings

Where Where,

…(5.32)

(∆S)system = (S2 – S1) Q sys Tsys

=

Q surr. = (∆S)surroundings Tsurr.

…(5.33)

(Q)system = T0(S2 – S1)

Substituting the above in Eqn. 5.31 T0(S2 – S1) – W = U1 – U2

…(5.34)



…(5.35)

W = (U2 – U1) – (S1 – S2)

For a closed system, when undergoing change in volume, work done against the pressure

(W)atm. = p0(V2 – V1)

…(5.36)



(W)max. useful = (W)max – (W)min

…(5.37)

= [(U1 – U2) – T0(S1 – S2)] – p0(V2 – V1)

…(5.38)

= [(U1 – U2) + p0(V1 – V2)] – T0(S1 – S2)

…(5.39)

5.3.2 Unsteady Open or Closed System Consider a system whose initial volume is V1 and final volume is V2. The entire work w of the system cannot be delivered since a certain quantity of the work will be spent in pushing out to the surroundings whose conditions are p0 and T0.

5.6

Engineering Thermodynamics

The actual work delivered by the system less the work performed on the surrounding is called as the actual work done on the atmosphere is p0(V2 – V1) (W)useful = (W)act – p0(V2 – V1)



(W)max. useful = (W)max – p0(V2 – V1)



(W)max = U – TS

= [ (U1–U2)+ p0(V1 – V2)] – T0(S1 – S2)

…(5.40) …(5.41) …(5.42) …(5.43)

(W) max useful = (U1– p0V1 – T0 S1) – (U2 – p0V2 – T0 S2) …(5.44)



Let, (U1 – p0V1 – T0 S1) = Ф1

…(5.45)

(U2 – p0V2 – T0 S2) = Ф2

…(5.46)

Then,

(W)max useful = Ф1 – Ф2

…(5.47)

The work is obtained as a part from a decrease in the internal energy of the system and a part from the heat withdrawn from the environment.

5.4

DEAD STATE

It is a state of a system when the system and surroundings are in equilibrium with each other and there is no work performed within the system. At the dead state, the fixed quantity of matter is assumed to be sealed in an envelope which is resistant to mass flow at rest relative to the surrounding and internally equilibrium at the temperature and pressure of the surrounding. At the dead state, both the system and surrounding possesses energy, but the value of useful energy is zero, because there is neither possibility of a spontaneous change within the system or the surrounding, nor can there be an interaction between them. Energy is said to be degraded each time as it flows through a finite temperature difference.

Solved Problems Problem 5.1 In a food industry, one kg of steam is stored in a tank. It expands in such a way, that its pressure increases from 500 to 2000 kPa. The temperature at initial and final conditions are 150°C and 130°C. The pressure and temperature of the atmosphere is 130 kPa and 25°C. The system receives 50 kJ of heat at 110°C. Determine the maximum work during the expansion. Given data

P1 = 500 kPa = 500 N/m2

T1 = 150°C = 523 K

P2 = 2000 kPa = 1000 N/m2

T2 = 130°C = 503 K Ta = 25°C = 298 K

Tr = 110°C = 383 K

Qa = 50 kJ

5.7

Availability and Irreversibility

Solution For closed system exchanges heat with atmosphere and a reservoir from Eqn. (5.25)

 T  Wmax = [(Ee - Ta Se) – (Eb – TaSb)] – QR  1 – a  Tr   Property values of air are obtained from steam tables at pressures p1 = 500 kPa = 500 N/m2 = 500 × 10–5 bar = 5 × 10–3 bar p2 = 2000 kPa = 2000 N/m2 = 2000 × 10–5 bar



= 2 × 10–2 bar. Problem 5.2 A compressor is used in a forging industry. Atmospheric air is drawn in to the compressor with an initial pressure and temperature of 10 bar and 20°C respectively. The machine delivers the compressed air at a pressure of 15 bar. The mass flow rate of air is 3 kg/s. The power input to the compressor is 400 kW. During compression, the heat removed from the air to the atmosphere is 40 kJ/kg. Calculate the change in entropy of the universe and irreversibility of compression. Given data Initial pressure

p1 = 10 bar = 100 kN/m2

Initial temperature

T1 = 20°C = 293 K

Mass of air

m = 3 kg/m3

Power input

W = 400 kW

Pressure

p2 = 5 bar = 500 kN/m2

Final temperature

T2 = ?

Heat transferred

q = 40 kJ/kg

Solution As per SFEE for compressor

W = (H2 – H1) + Q



W – Q = (H2 – H1)



W – Q = mCp(T2 – T1)



m Cp(T2 – T1) = 400 – 3 × 40



m Cp(T2 – T1) = 400 – 120 = 280 kJ/S



(T2 – T1) =

280 + T1 mC p



T2 =

280 + T1 mC p

5.8

Engineering Thermodynamics

 280  =   + 119.4  3 × 1.005 

T2 = 412.40 K

From the property relation

T ΔS = dH – vdP dH vdP – T T



ΔS =



dS = mCp



dT dP – mR T P

(S2 – S1) = mC p ln

T2 P – mR ln 2 T1 P1

ΔS = 3 × 1.005 × 2.3 log 10

41240 1194

= 3 × 0.287 × 2.3 × log 10

15 × 10 2 10 × 10 2

= 3.3842 kJ/K. Problem 5.3 A steel plant in which a heat exchanger was used for the heat removal from a hot fluid, has the following data Mass flow rate of water = 30 kg/min Inlet temperature of water = 50°C Outlet temperature of water = 105°C Inlet temperature of hot gases = 200°C Mass flow rate of hot gases = 70 kg/min Change in entropy of universe. Assume Cp of water = 4.187 kJ/kg K, CPgas = 1 kJ/kg K Solution Heat lost by hot gas = heat gained by water

mg Cpg (tgi – tgo) = mω Cpω (Tωo – Tωi)

70 30 × 1 × (473 – t go ) = × 4.187 × (232 – 378) 60 60    

Tgo = 473 +

30 × 4.187 × 55 70



Tg = 473 –

30 × 4.187 × 55 70

= 571.69 K

5.9

Availability and Irreversibility

(ΔS)water = mC pω ln



=

Tω0 Tω1

30 378 × 4.187 × ln 60 323

= 0.3291 kJ/K (S)gas = mC pg ln



=

Tg out Tg in

Tg out 70 × 1 × ln 60 473

70 571.69 × 1 × ln 60 473 = 0.48265 kJ/K =

(ΔS)universe = (ΔS)water + (ΔS)universe

= 0.3291 + 0.4826 = 0.8117 kJ/K. Problem 5.4 A compressor was used to compress one kg of gas polytropically. The initial pressure of the gas was 0.1 bar. After compression, the temperature and pressure of the gas reached 350 K and 1 bar respectively. Calculate the change is entropy if the polytropic process follows pV1.25 = C. Assume that CV = 0.7 kJ/kg K, R = 0.287 kJ/kg K. Given data Mass of air = 1 kg; Final temperature

T2 = 350 K;

Initial pressure

p1 = 0.1 bar;

Final pressure

p2 = 1 bar;

CV = 0.7 kJ/kg K,

R = 0.287 kJ/kg K.

Solution We know that for a polytropic process n−1

T2  p2  n =  T1  p1             T2 T1 = ( n −1) n p2 p1

p  T1 = T2 ×  1   p2 

( n −1)

n

5.10

Engineering Thermodynamics

T1 = 350 ×  1   0.1 



( 1.25 − 1)

1.25

= 350 × (0.1)(0.25/1.25) = 350 × 0.6 = 220.8 K. For a steady flow process the change in entropy is given as

DS =

2 ∫ dH T

–

2 ∫ VdH T

= mC p

dp dT – mC p T p

= mC p

p T2 – mC p 2 T p1

= 1 × 1.005 × 2.3 × log

350 2208

1 0.1 = 1.5930 – 10.05 = 1 × 1.005 ×

= –8.457 kJ/kg K Negative sign indicates that there is a decrease in entropy. Problem 5.5 Two kg of water in a closed vessel is heated up from 5°C to 100°C by a heating coil. What is the entropy change in water, external source and the universe at 60°C. Given data: m = 2 kg Initial temperature

Tb = 5°C

Final temperature

Te = 100°C

Temperature of the universe T = 60°C Solution It is understood that water is heated up in a finite temperature difference. The change in entropy between water and source will be positive, because the entropy of water increases, and that of the source decreases.

(ΔS)water = mc ln

tb ta

ta = Initial temperature of water 5°C (275°K) tb = Final temperature of water 600°C (333 K)

(ΔS)water = 2 × 4.187 × ln

333 278

5.11

Availability and Irreversibility

= 2 × 4.187 × 0.805 = 1.5116 kJ/K. Heat absorbed by system from the source = Q = mc ΔT

Q = 2 × 4.187 × (333 – 278)

= 2 × 4.187 × 55 = 460.57 kJ. Q T It should be noted that the negative sign indicates that the heat is supplied from the source to the water. Change in entropy of source (ΔS)source = –



(ΔS)source = –

460.57 333

= – 1.383 kJ/K

(ΔS)universe = (ΔS)system + (ΔS)source

= 1.5116 – 1.383 = 0.1286 kJ/K.

Problem 5.6 A metal treatment industry heats up 500 kg of steel at 900°C for metal treatment. After treatment the hot steel has to be cooled to 150 K. The ambient temperature is 30°C and the specific heat capacity of the steel is 0.5 kJ/kg; calculate the available energy, and the unavailable energy in the steel. Given data Initial temperature of steel

T1 = 900  C = 1173 K

Final temperature

T2 = 150  C = 423 K

Specific heat capacity of steel C5 = 0.5 kJ/kgK Ambient temperature

Ta = 30°C = 303 K

Solution Change in entropy ΔS = mCp ln

T2 T1 423 1173



ΔS = 500 × 0.5 ln



ΔS = – 254.98 kJ/K

Available energy = Q – To ΔS

Q = m Cps(T1 – T2)

= 500 × 0.5 × (1173 – 423) = 187500 kJ or,

Q = 187.5 MJ To ΔS = 303 (–254.98) – 77258.94 kJ

5.12

Engineering Thermodynamics

Available energy = Q – To ΔS = 187.5 – 303 × (–254.98) = + 187.5 + 772.58 = 264.75 MJ Unavailable energy = To ΔS = 303 × (–254.98) = –77258.94 kJ or, = –77.6 MJ Problem 5.7 A sugar industry uses a steam generator for producing steam. The steam generator leaves the exhaust at an initial temperature of 800°C. The ambient temperature is 30°C. The gas is cooled by air in a heat exchange. The inlet and exit temperatures of the air are 25°C and 200°C respectively. The pressure of the air at the inlet and exit conditions are, 1 bar and 2 bars respectively. Assume that the mass of the gas is 200 kg, and the specific heat of gas is 2 kJ/kg K. Calculate the maximum work which can be obtained (i) from the gas (ii) from the air (iii) loss in the available energy Given data Inlet temperature of air = 25°C = 278 K Exit temperature of as = 300°C = 573 K Inlet pressure of air = 1 bar Exit pressure of air = 2 bar Exit temperature of gas = 800°C = 1073 K Ambient temperature = 30°C = 303 K Solution I. Maximum work obtained from gas = Qgas – To (ΔS)gas We know that

Qgas = m Cp ΔT

= 200 × 2 × (1073 – 573) = 200000 kJ or,

Qgas = 200 MJ (ΔS)gas = mCg ln

Tg out Tg in

573 1073 = –250.93 kJ/K = 200 × 2 ln

5.13

Availability and Irreversibility

So, maximum work obtained from gas = Qgas – To (ΔS)gas = 200000 – 303 × (–250.93) = 200000 + 76031.79 = 276031.79 kJ or, = 276.03 MJ. II. Maximum work obtainable from air = Qair – To (ΔS)air. Since we have to know the mass of air, an energy balance equation is put.

Heat lost by gas = Heat gained by air

mgCp gas(Tgas in – Tgas out) = mair Cp air (Tair out – Tair in)

Qgas = mair Cp air (Tair out – Tair in)

Therefore,

mair =

(

Qgas

C p air × Tair out – Tair in

        mair =

276031.79 1.004 × (200 – 25)

=

276.03 × 103 1.004 × 75

)

= 1571.04 kg of air

(ΔS)air = mair Cp air ln

Tair out Tair in

= 1571.04 × 1.004 × ln

573 323

= 904.17 kJ/K. Maximum work obtainable from air = Qgas – To ΔS = 326830.71 – 303 × 974.17 = 326830.71 – 273964.9 = 52865.81 kJ or, = 52.86 MJ. Loss in available energy = Maximum work obtainable by gas – Maximum work obtained from hot air = 326.83 MJ – 52.86 MJ = 273.97 MJ. Problem 5.8 A casting industry uses a furnace, in which two gases A and B are released. The mass of gas A and B are the same and 500 kg. The exit temperatures of gas A and B are 900°C and 500°C respectively. The ambient temperature is 27°C. The specific heat of the gas is 2.0 kJ/kgK. Both gases are sent to a mixer, where they are mixed. Can we use the mixer as the source of a

5.14

Engineering Thermodynamics

heat engine? Comment on this. The temperature of gases is reduced to 40°C at the exit of the heat engine. Calculate the available energy if (i) The two gases are used separately as sources in the heat engine. (ii) The two gases are mixed together and used in the heat engine. Given data Mass of gas

mA = 500 kg

Exit temperature of gas A Ta out = 900°C = 1173 K Ambient temperature = 27°C = 300 K Exit temperature of gas B Tb out = 500°C = 773 K Mass of gas

mB = 500 kg.

Solution (i) Available energy obtainable by gas A = QA – To ΔSA

QA = mA Cp A (TA out – TA in)

= 500 × 2 × (900 – 40) = 860000 kJ

(ΔS)A = mA Cp A ln

= 500 × 2

TA out TA in

( 900 + 273 ) ( 40 + 273 )

= + 1321.1 kJ/K. Therefore, available energy obtained by gas A = QA – To ΔSA = 860000 – 300 × (+1321.1) = 860000 – 396334.99 = 463665.01 kJ or, = 463.66 MJ. (ii) Available energy obtained from gas B = QB – To ΔSB

QB = mBCpB (TB in – TB out)

= 500 × 2 × (500 – 40) = 460000 kJ

(ΔS)B = mBCpB ln

TB out TB in

= 500 × 2 × ln

( 500 + 273 ) ( 40 + 273 )

5.15

Availability and Irreversibility

773 313 = 909.075 kJ/K. = 1000 × ln

Available energy of gas B = QB – To ΔSB

= 460000 – 300 × 909.075 = 460000 – 271222.75 = 188777.2 kJ = 188.77 MJ. Total available energy from gas A and B = Available energy from gas A

+ Available energy from gas B

= 463.66 + 188.77 = 652.43 MJ. (ii) Temperature of gases A and B after mixing =

TA in TB in

2 900 + 500 = 2 = 700°C = 973 K Available energy of mixed gases A and B = Qmix – To (ΔS)mix

Qmix = mmix Cp mix (Tmix in – Tmix out)

= (500 + 500) × 2 × (700 – 40) = 1280000 kJ = 1280 MJ

(ΔS)mix = mmix Cp mix ln

Tmix out Tmix in

= (500 + 500) × 2 × ln

973 313

= 2268.36 kJ/K. Therefore, available energy from mixed gases A and B = Qmix – ToΔSmix = 1280000 – 300 × 2268.36 = 1280000 – 680508 = 599492 kJ or, = 599.4 MJ. Comment: From the above solution, the available energy obtained from the mixed gases A and B is lesser than the sum of available energy obtained in gases A and B.

5.16

Engineering Thermodynamics

Problem 5.9 A pneumatic machine uses air, which is compressed by a compressor. The atmospheric air is initially at 10 bar and 20°C. After compression, the compressor delivers the compressed air at a pressure of 15 bars. The mass flow rate of air is 3 kg/s. The power input to the unit is 400 kW. During compression, the heat removed from the air to the atmosphere is 40 kJ/ kg. Calculate the change in the entropy of the universe and the irreversibility of compression. Given data p1 = 10 bar = 100 kN/m2

Initial pressure

Initial temperature Tb = 20°C = 293 K Mass of air

m = 3 kg/m3

Power input

W = 00 kW

Pressure

p2 = 5 bars = 500 kN/m2

Final temperature Heat transferred

T2 = ?

q = 40 kJ/kg.

Solution As per SFEE for compressor

W = (H2 – H1) + Q



W – Q = (H2 – H1)



W – Q = mCp(T2 – T1)



mCp(T2 – T1) = 400 – 3 × 40



mCp(T2 – T1) = 400 – 120

= 280 kg/s

(T2 – T1) = 280/m Cp

T2 = (280/m Cp) + T1 = (280/m Cp) + 119.4 T2 = 412.40 K. From the property relation TDS = dH – vdp

 dH   vdp  DS =  –   T   T 

 dp   dT   – mR   T   p             p2   T2  (S2 – S1) = mC p ln   – mR ln   Π  p1 

∫ ds = mC p ln 



 412.40  DS = 3 × 1.005 × 2.3 log10    119.4 

Availability and Irreversibility

5.17

 (15 × 10 2 )  = 3 × 0.287 × 2.3 × log10  2   (10 × 10 )  = 3.3842 kJ/K. Problem 5.10 Air is stored in a tank whose volume is 0.8 m3. Assume that the pressure and temperature of the atmosphere are 1 bar and 25°C respectively. Calculate the maximum possible work that could be done, when the air enters the tank to attain equilibrium with the atmosphere. Given data

To = 25°C = 298 K



po = 1 bar = 1 × 102 kN/m2

Solution An open system, which exchanges heat only with the atmosphere, is given by      δWmax = – d(E – ToS) + (e1 + p1v1 – ToS1) + (e2 + p2v2 – ToS2)m2 Integrating the above equation we get,      Wmax = (E1 – ToS1) – (e2 – ToS2) + (eo + povo – ToSo) (m2 – m1) Since the atmosphere air enters the tank, the third term in the expression is denoted with subscripts (i.e., eo + povo – ToSo). It should be noted that the mass of atmosphere changes only from the initial state (m1) to the final state (m2). Now the condition satisfies the problem as;

(E1 – ToS1) = 0



m1 = 0

Therefore,

Wmax = (E2 – ToS2) + m2(e2 + povo – ToSo)

Since the gas brought to equilibrium with atmosphere doing some work done,

p2 = po,  T2 = To

Therefore,

e2 = eo  v2 = 0

So,

Wmax = poVo × m2

= poVo = poVtank = 1 × 102 × 0.8 = 80 kJ The maximum work which could be done is 80 kJ. Problem 5.11 A steady flow apparatus is one through which steam enters at 800 kPa, and 200°C, with negligible velocity and leaves at the exit. The final conditions of the steam at the exit are 120 kPa and 120°C respectively, with a velocity of 150 m/s. Heat is exchanged only with the atmosphere, which is at 20°C. If the mass flow rate is 7200 kg/h, calculate the maximum possible power output. Figure p. 5.1 shows the inlet and outlet conditions of the steam in the steady flow apparatus.

5.18

Engineering Thermodynamics

Given data

p1 = 800 kPa p2 = 120 kPa



= 800 kN/m2 ; = 120 kN/m2 ;

T1 = 200°C

= 473 K ;

To = 20°C

= 293 K;

T2 = 200°C

= 393K

m = 7200 kg/h = 2 kg/s



Solution  u1 + p1 v1 + v12   u2 + p2 v2 + v22 – Wmax =     2 + gz1 – To s1   2 + gz2 – To s2 Neglecting initial velocity and change in potential energy We get Wmax =

( h1 – To s1 ) – ( h2 – v22 ) 2 – To s1

  

 u + P2 v2 + v22  – 2 m  2 + gz – T s  2 0 2  

 v2  or, Wmax = m[(h1 – h2) – To(s1 – s2)] –  2  2×m   p1

m, p 2 , T2

T1

To

Fig. p. 5.1 Steady flow apparatus

From the steam tables, the specific entropy values are obtained for the initial and final conditions corresponding to the given pressures p1 and p2 Corresponding to pressure p1 = 800 kPa = 800 kN/m2 = 8 bar,

h2 = 2769 kJ/kg;  s2 = 606663 kJ/kg;

Corresponding to pressure p1 = 120 kPa = 120 KN/m2 = 1.2 bar,

h2 = 2683 kJ/kg;  s2 = 7.298 kJ/kg;

Substituting the values in the equation for Wmax We get,

 v2  Wmax = m[(h1 – h2) – To(s1 – s2)] –  2  2 × m

Wmax = 2[(2769 – 2683) – 293 (6.663 – 7.290)] 2  −  150 × 10–3 ×  2 

(

)

= 543.97 kJ/s

The maximum possible power output is Wmax = 271.94 kJ/s

5.19

Availability and Irreversibility

Problem 5.12 A bus station uses a reciprocating air compressor for inflating the air. The air is compressed from 1.1 bar and 20°C to 9 bar and 40°C. The compressed air is cooled only by the atmosphere. Neglecting kinetic energy, and potential energy. Calculate the least possible work required per kg of air. The inlet and outlet conditions of the compressor are shown in Fig. p. 5.2. Given data

T1 = 20°C = 293 K;



T2 = 40°C = 313 K



p1 = 1 bar = 1 × 102 kN/m2;



p2 = 9 bar = 9 × 102 kN/m2

Solution For a steady flow process,

ΔK = ΔPE = 0;

Wmax = (h1 – Tos1) – (h2 – Tos2) Wmax = (h1 – h2) – (s1 – s2) To

or,

Air T1 p1 Compressor p2 T2

Fig. p. 5.2

To calculate (h1 – h2); By definition, (h1 – h2) = Cp(T2 – T1) or,

(h1 – h2) = 0.72 (293 – 313)

= 14.4 kJ/kg To calculate (s1 – s2); By definition

(s2 – s1) =

1

∫

2

( dh – υdp ) T

For open system dh 1 υdp – T 2∫ T 1 T 1 υdp = 2 ∫ C p – 2 ∫ T T

∫

1

or,

(s2 – s1) =



 p1  T   (s1 – s2) =  C p ln 1  –  R ln  T2   p2  

2

5.20

Engineering Thermodynamics



  293     1  (s1 – s2) = 0.7 ln    – 0.287 ln    313    9    

or,

  293     1  (s1 – s2) = 0.7 × 2.3 log 10    – 0.287 × 2.3 log 10     313     9  

= 0.58373 kJ/kg K wmax = (h1 – h2) – To(s1 – s2) Therefore,

wmax = 14.4 – 293(0.58373)

wmax = –156.62 kJ/kg The least possible work = 156.62 kJ/kg.

Problem 5.13 A steam generator supplies steam to a steam turbine at 1 MPa and 350°C, with a negligible velocity. At the exit of the steam turbine, the steam is found to be 0.4 MPa, 130 m/s and 0.9 dry. The atmospheric pressure is 1.013 bar. The plant rejects heat to a water pond at 18°C. Calculate (i) The power output, (ii) The maximum power at end states (iii) The maximum power which could be obtained from the exhaust steam. The inlet and outlet conditions of the steam turbine are represented in Fig. p. 5.3. Given data

T1 = 350°C = 623 K;

p2 = 0.4 MPa = 40 × 102 kN/m2; p1 = 1 MPa = 103 kN/m2;

V2 = 130 m/s;

X2 = 0.92;

Po = 1.03 bar T1 p1 Turbine T2 p2

Fig. p. 5.3

Solution The property values of h and s at the initial and final conditions of steam are obtained from the steam tables corresponding to their pressure. At pressure

p1 = 10 bar;



h1 = 2778 kJ/kg;



s1 = 6.586 kJ/kg

At pressure

p2 = 4 bar;

5.21

Availability and Irreversibility



h2 = 2739 kJ/kg



s2 = 6.891 kJ/kg K.

During the dead state, water is assumed to be a compressed liquid. At 1.013 bar, the h and s values of water are the same as the values of the saturated liquid at the same temperature. So,

ho = 2676 kJ/kg



so = 7.344 kJ/kgK.

(a) Power output of the turbine using steady flow energy equation. 2 W = h1 – h2 – v2 2 = (2778 – 2739) – (130 × 10–3)2



= 38.991 kJ/s (b) Maximum work at end states:   v2 Wmax = (h1 – ToS1) –  h2 + 2 – To S2    2      0.132  = (2778 – 291 × 6.586) –  2739 +  – 291 × 6.891 2     [ 2676 – 291 × 7.344] = 733.727 – 538.83 = 194.83 kJ/s

Objective Type Questions 1. The availability of a system (a) Does not depend upon the conditions of the system. (b) Does not depend upon the conditions of the surroundings (c) Depends upon the conditions of the system only (d) Depends upon the conditions of the system and the surroundings. 2. Whenever a certain quantity of heat is transferred from system (a) Available energy increases

(b) Available energy decreases

(c) Unavailable energy increases

(d) Unavailable energy decreases

3. Irreversibility is zero in the case of (a) Adiabatic process

(b) Isothermal process

(c) Non-flow process

(d) Reversible process

5.22

Engineering Thermodynamics

4. Consider the following statements, which statement is/are true (a) Availability is generally conserved (b) Availability can either be negative or positive (c) Availability is the maximum theoretical work obtainable (d) Availability can be destroyed in irreversibility 5. A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during the process is to be used as a source of energy. The ambient temperature is 303 K and the specific heat is 0.5 kJ/kg K. The available energy of the billet is (a) 490.44 MJ

(b) 10.35 MJ

(c) 30.95 MJ

(c) 0.10 MJ

6. One kg of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is (a) Equal to the entropy change of the reservoir (b) Equal to the entropy change of water (c) Equal to zero (d) Always positive 7. Consider the two processes given below: (a) A heat source at 1200 K loses 2500 kJ of heat to a sink at 800 K (b) A heat source at 800 K loses 2000 kJ of heat to a sink at 500 K 8. Which of the following statements is correct? (a) Process (i) is more irreversible than process (ii) (b) Process (ii) is more irreversible than process (i) (c) The irreversibility of processes (i) and (ii) is equal (d) Both the processes are reversible.

Theory Questions 1. What is the availability? 2. What do you understand by the term “dead state”. 3. What is irreversibility, and what are its causes? 4. What is meant by available energy? 5. What is the difference between the maximum work and useful work? 6. Does the availability of a system depend upon the conditions of the system and surroundings? 7. What is the second law efficiency? 8. What do you understand by availability? Deduce the availability function for open and closed cycles.

5.23

Availability and Irreversibility

Unsolved Problems 1. The mass of air is 0.2 kg, and has a temperature of 573 K; it is heated reversibly at constant pressure till its temperature reaches 2066 K. Find the available and unavailable energies of the heat added, taking T0 = 303 K and Cp = 1.0047 kJ/kgK. 2. In a chemical plant, 10 kmol/s of a particular gas is cooled from 700 K to 320 K in a heat exchanger, by using air as a coolant. The air enters the heat exchanger at 300 K and leaves at 330 K. The ambient conditions are 298K and 0.1 MPa. Assume Cp(gas) = 3.5 Cp air, R = 8.314 kJ/kgK. Calculate the following: (i) The change in the availability of air. (ii) The change in the availability of the gas. (iii) The loss in the available energy associated with the energy transfer from the gas to the air. 3. If 1 kg of superheated steam at 2.5 MPa and 523 K, is allowed to undergo a process in which the steam reduces to dry saturated steam at 0.1 MPa, calculate the maximum useful work. The ambient conditions are 298 K and 0.1 MPa. The superheated steam at 2.5 MPa and 523 K is v = 0.087 m3/kg; u = 2662.6 kJ/kg; s = kJ/kgK The saturated steam at 0.1 MPa vg = 1.694 m3/kg ug = 2506.1 kJ/kg; sg = 7.3594 kJ/kgK 4. Derive an expression for the change in availability for a closed system, when the (i) System exchanges heat only to the atmosphere. (ii) System exchanges heat to the atmosphere and a heat reservoir. 5. An inlet condition of an adiabatic compressor is p1 = 1 bar, T1 = 288 K, and the outlet condition is p2 = 5.5 bar, the mass is flowing at the rate of 0.01 kg/s, and the efficiency of the compressor is 75%. After leaving the compressor, the air is cooled to 313 K in an after cooler. Calculate; (a) The power which is required for driving the compressor (b) The rate of irreversibility for the overall loss (compressor and cooler). 6. In a particular chemical industry, the exit temperature of the products leaving the reactor is 1273 K, while the ambient temperature is 298 K, and the Cp of products are equal 1.0 kJ/kgK. Its is decided to convert the energy possessed by vapor into work. Therefore, the hot products are cooled in an air cooled heat exchanger. The inlet and exit conditions are Tin = 298 K, Pin = 50 kPa, Tex = 573K, Pex = 10 kPa, if the production rate of the reactor is 83 kg/min. So, calculate

5.24

Engineering Thermodynamics

(a) The maximum work obtained if the reactor products could be directly employed in a heat engine. (b) The maximum work that can be obtained from the hot air. (c) The loss in the available energy due to the energy transfer process. 7. Air enters a rotary compressor at a pressure of 110 kPa and a temperature of 294K; then it is compressed adiabatically up to pressure 660 kPa and temperature 523 K. Calculate the entropy production and irreversibility for the unit mass flow rate. Neglect the K.E changes. 8. Suppose one kilo mole of an ideal gas is available at 1 MPa and at ambient temperature of 300 K. If a gas is allowed to expand to 0.1 MPa while in contact with the ambient atmosphere, and the final temperature is same as the ambient temperature, calculate the maximum work that can be obtained from the gas. 9. 500 kJ of heat is removed from a constant temperature heat reservoir maintained at 835 K. The heat is received by a system at a constant temperature of 720 K. The temperature of the surroundings is 298 K. The lowest available temperature is 280 K. Illustrate the problem by the T-s diagram and calculate the net loss of available energy as a result of this irreversible heat transfer. 10. A lump of 1000 kg steel at 1200 K is to be cooled to 400 K, if it is desired to use steel as a source of energy. Calculate the available energy and unavailable energy, ambient temperature is 300 K and the specific heat capacity of steel is 0.5 kJ/kgK. 11. An ideal gas is flowing through an insulated pipe at the rate of 3 kg/sec. There is a 10% pressure drop from the inlet to the exit of that pipe. What is the rate of energy loss, because of the pressure drop due to friction? Take R = 0 – 287 kJ/kgK, and the ambient temperature as 300 K. 12. One kg of ice at –5°C is exposed to the atmosphere, which is at 20°C. The ice melts and comes into thermal equilibrium with the atmosphere. (i) Determine the entropy increase of the universe. (ii) What is the minimum amount of work necessary to convert the water back to ice at –5°C? Assume Cp for ice as 2.093 kJ/kgK, and the latent heat of fusion of ice as 333.3 kJ/kgK. 13. Two kg of air at 500 kPa and 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, 5°C for this process. Determine: (i) The maximum work (ii) The change in availability, and (iii) The irreversibility

Availability and Irreversibility

5.25

For air, take Cv = 0.718 kJ/kgK, u = CvT, where Cv = constant and pV = mRT, where p is pressure in kPa, V is volume in m3, m is mass in kg, R is a constant equal to 0.287 kJ/kg K and T is the temperature in K. 14. A gas flows through a pipe at the rate of 2 kg/s. Because of inadequate insulation, the gas temperature decreases from 800 to 790°C between two sections in the pipe. Neglecting the pressure losses, calculate the irreversibility rate (or rate of energy degradation) due to this heat loss. Take To = 300 k and a constant Cp = 1.1 kJ/kg K. For the same temperature drop of 10°C, when the gas cools from 80°C to 70°C due to heat loss, what is the rate of energy degradation? Take the same values of To and Cp. What inference can you draw from this example? 15. One kg of air is contained in a piston cylinder assembly at 10 bar pressure and 500 K temperature. The piston moves outwards and the air expands to 2 bar pressure and 350 K temperature. Determine the maximum work obtainable. Assume the environmental conditions to be 1 bar and 290 K. Also calculate for the availability in the initial and final states.

CHAPTER

6 6.1

PROPERTIES OF PURE SUBSTANCE

PURE SUBSTANCE

A substance, which has a fixed chemical composition throughout, is termed as a pure substance. It may exist in more than one phase. Examples of pure substances are water, nitrogen, helium and carbon dioxide. The pure substance may be a single chemical element or a mixture of various elements. It must be noted that a mixture of oil and water does not form a pure substance, because oil is not miscible with water. A mixture of two or more phases is a pure substance till the chemical composition of all the phases is the same. e.g., a mixture of ice and water. Steam is the most important working fluid in many industrial applications. It is widely used in power plants, sugar, pulp and chemical industries. Steam is the main source for indirect heating in such applications. The processes involved in such industries are analyzed with the help of the properties of the working fluid. So, it is better to study the properties of a pure substance, for an understanding of the thermal applications in which it is used.

6.1.1 Existence of Pure Substance A pure substance exists in three forms. They are (i) Solid (ii) Liquid and (iii) Vapour or Gas. Water is a very good example of pure substance; when it exists in a solid form, then it is known as ice. When the latent heat of freezing is removed from water at its freezing point then it becomes ice. It starts to become a liquid, if the latent heat of fusion or melting is added to the ice. When water possesses only sensible heat, then it is referred to as saturated water. If the latent heat of evaporation is added to the saturated water (fully boiled) then it becomes vapour, known as steam. If the vapour contains a small amount of water and exists in two phases (vapour-water) then the mixture is referred to as wet steam, and water free vapour is called dry steam. Hence, the fraction of dry steam present in the total mass of steam is important. This is because, in many industrial applications, dry steam is required rather wet steam, as wet steam has a tendency to corrode the equipment and reduce its performance and life.

6.2

Engineering Thermodynamics

6.1.2 Terminology 6.1.2.1 Dryness Fraction The fraction of water vapour present in the steam is called as the dryness fraction. It is denoted as “x”. If the dry steam exists in the vapour phase with a significant amount of heat contained in it, then the steam is called super heated steam. The dryness fraction of steam is the ratio of the mass of dry steam to the total mass of steam formed with water. In other words, it is the ratio of a specific volume of dry steam to the specific volume of steam formed. x =



=

Mass of actual dry steam Mass of steam present with water particles mg mg + m f

where, mf = mass of water vapour suspended per kg of steam

mg = mass of dry steam per kg of steam

The phenomenon of the formation of superheated steam from ice is described in the subsequent section.

6.2

STEPS OF STEAM FORMATION

A representation of the steps involved in steam formation from ice, is illustrated in Fig. 6.1. The variation of temperature with specific enthalpy during phase transformation is also shown in the figure. Let us consider a unit mass of ice, which is at say –5°C and at 1 atm pressure, which is assumed to be unchanged as shown in Fig. 6.1. The arrangement maintains a constant pressure over the ice. The following stages are involved in the formation of steam. 1. Line 1-2 represents the increase in the temperature of the ice from –5°C to 0°C. The volume of ice increases with increasing temperature. At 0°C the ice starts to melt. 2. Line 2-3 indicates the melting of ice into water at a constant temperature of 0°C. During this period the latent heat of fusion is absorbed by the ice. 3. Line 3-4 shows that the period at which the conversion of water into steam takes place. By the addition of sensible heat, water rises in temperature to its boiling point (100°C), at which the pressure is 1.013 bar (i.e., Normal atmospheric pressure). This is shown in point 4. When water reaches the boiling point, it starts evaporating. 4. Line 4-5 represents evaporation. During this period, the latent heat of evaporation of water is added. The temperature remains unchanged, but water converts fully into steam, by the addition of the latent heat of evaporation. There is a large increase in the volume. Note that the dry steam behaves exactly like a perfect gas at high temperatures and low pressures.

6.3

Properties of Pure Substance Superheated saturated 6 steam

T Saturated water with steam

Ice with saturated Saturated water water

4

Saturated steam

Latent heat of evaporation

Superheating

5

Boiling point

Ice 2

3 Latent heat of fusion

Melting point 1

Fig. 6.1 T-h Diagram of ice in to superheated steam

5. On further heating, the temperature of the dry steam rises. The process of obtaining steam at constant pressures is called superheating. 6. The rise in temperature during the superheating of steam is called the degree of superheat.

Degree of superheat = (Tsup – Tsat)

It must be noted that the specific heat of superheated steam is 2.1 kJ/kg K.

6.3

PRESSURE-VOLUME-TEMPERATURE DIAGRAM

Let us consider the sequence of processes happening in a phase transformation. From Fig. 6.2, it is seen that there is a mere rise in the melting point. There is a drop in the boiling point and increase in the change in volume, which is found in evaporation. The two influencing temperatures of water become equal, when the pressure comes down to 0.006112 bar. This temperature is called the triple point of water. This has been internationally accepted, and found to be the reference point for the ideal gas temperature scale. The three phases solid (ice), liquid (water), steam (gas) co-exist in thermal equilibrium. This is shown as the triple point line in Fig. 6.2. Above the atmospheric pressure, are curves similar the of curve at atmospheric pressure. When the change in volume becomes zero at a particular pressure then the horizontal position of the curve

6.4

Engineering Thermodynamics

will drastically reduce to a point. The pressure, volume and temperature at this point are referred to as critical pressure, critical volume and critical temperature. Their values are pc = 221.2 bar, Tc = 647.3 K, vc = 0.00317 m3/kg

Recalling the phase rule for a system at equilibrium, it relates:

P = number of phases that can co-exist,



C = number of components making up the phases, and



F = degrees of freedom.

Where these three variables are related in the equation

P + F = C + 2

Applying the phase rule to ice, water and steam

P = 3 – Water, ice, steam



F = 0 – No degrees of freedom,

It should be noted that all three-phases co-exist at equilibrium, as per the phase rule. At the triple point of water it has only one pressure and temperature.

6.3.1 T-v Diagram of Water-Steam The T-v diagram (Liquid – vapour two-phase diagrams) is shown in Fig. 6.2. p > pc

TC = 647.3°K

Temperature

374.15°C

Critical point p TC TC = 647.3

Liquid region

b

Supe

rheat

Liquid vapour mixture or wet vapour

ed re

gion

1.01325 bar G3

d

Triple point line

0.06112 bar

373.15°K

c

C

f

273.16°K g

E

0.0317 m3/kg

F2

Fig. 6.3 Pressure-specific volume of steam

6.5

p-V DIAGRAM FOR WATER

Figure 6.3 represents the p-V diagram for saturated liquid and saturated vapour at various temperatures; for e.g., the curve a-b-c-d. The curve a-b represents the unsaturated liquid. At point b the unsaturated liquid becomes saturated liquid. The curve b-c represents a combined liquid and vapour of a two-phase mixture or wet vapour. If many curves are brought to represent the saturated liquid and saturated vapour then they exactly form two curves. (i) The saturated liquid line, and (ii) The saturated vapour curve. These two curves intersect at a critical point. The saturated liquid line separates the liquid phase and the liquid-vapour two-phase, whereas the saturated vapour line splits the liquid vapour twophase and the vapour phase. The vapour phase is shown on the right side of the saturated vapour line. It is common to say that superheated vapour is always dry and, the temperature of superheated vapour is greater than the saturation temperature. The degree of superheat is the difference between the superheat

6.6

Engineering Thermodynamics

temperature of a substance, and the saturation temperature corresponding to the actual pressure. The term, compressed liquid is used, when a liquid in a state lies on the left of the saturated liquid line. In such a case, the liquid pressure, is greater than the saturation pressure corresponding to the actual temperature of liquid. When the state of the substance is to be determined, the pressure and temperature, are sufficient in the case of a saturated liquid. But, for wet vapour, the pressure and temperature are not sufficient, because, for the corresponding pressure and temperature the state of the substance may be marked on any point of the line b-c. Therefore, it is important to give the pressure and volume or temperature volume data, to determine the state of wet vapour. A three dimensional model, which gives a clear picture of the phase transformation, is known as p-v-T surface. This is shown in Fig. 6.4.

TC

Pressure

Compressed liquid

Solid

Su vap perhe our ate

d

We tripl t vapo e p o ur int Sol

id-v

Spe c

ific

r ba

Gas

Iso

Line

apo ur

volu me

e

ur

at

r pe

m Te

Fig. 6.4 p-V-T surface

Figure 6.5 shows the difference between sublimation, melting and boiling with respect to pressure and temperature Fusion curve (water)

Normal substance

Critical point

Pressure P (bar)

Boiling

0.006112 bar

Triple point Sublimation

273.1 273.16

373.1 647.3 Temperature (T)°K

Fig. 6.5 Difference between sublimation, melting and boiling

6.7

Properties of Pure Substance

6.6

PROPERTY DIAGRAMS OF WATER

A few property diagrams of water are used in various thermodynamic processes. They are (i) the pressure volume (p-V) diagram (ii) the temperature entropy T-s diagram (iii) the enthalpy entropy (h-s) diagram and (iv) the pressure enthalpy (p-h) diagram. All the diagrams show the change of state from a sub-cooled saturated liquid to a superheated corresponding to a constant pressure taken above the triple point line. p-V diagram: In the p-V diagram of the saturated liquid and vapour, their specific volumes are denoted by vf and vg respectively. To determine the specific volume during the evaporation in the two phase region, first the dryness fraction is considered. The p-V diagram is shown in Fig. 6.6(a). T-s diagram: The temperature entropy diagram is shown in Fig. 6.6(b). Ssup denotes the specific entropy of superheated vapour. Vapour Saturated liquid region

Critical point of water superheat region Critical point

p

Two phase region

 f Evaporation

T

g sup

s1

Superheat Two phase region region evaporation sg sf sfg Saturated liquid

Saturated liquid curve Saturated vapour curve Triple point line Specific volume (v)

(a)

ssup

Saturated vapour line Triple point line Specific entropy (s)

(b)

Fig. 6.6 (a) p-V diagram (b) T-s diagram

6.7

STEAM TABLES

Steam tables are most important for any thermodynamic calculations, where, properties of steam are used. There are four main steam tables used. (i) Saturated water and steam (T) tables (ii) Saturated water and steam (P) tables (iii) Superheated steam tables (iv) Supercritical steam tables. Table 6.1 lists the values of saturation pressure, specific volume of saturated water(vf),

6.8

Engineering Thermodynamics

and vapour (vg) corresponding to saturation temperature. The table includes the values of the specific enthalpy of saturated water (hf), evaporation (hfg), and vapour (hg), and also, the specific entropy of saturated water (sf), evaporation(sfg) and vapour (sg). Table 6.2 gives the above values corresponding to the saturation pressure. Superheated steam Table 6.3 provides the specific volume, specific enthalpy, specific entropy corresponding to saturation pressure and superheated temperature. Table 6.4 provides the specific volume, specific enthalpy, and specific entropy corresponding to the saturation pressure and super critical temperature. Interpolation is followed to obtain the property values at an intermediate temperature or pressure, which is not in the tables. For example, the given temperature is 41°C. If the property values are given only for 40°C and 42°C in the table, then the interpolation method is used to get the values at 41°C. Table 6.1 Saturated Water and Steam (T) Table T(°C)

Ps(bar)

Vg (m3/kg)

hf

0.01

0.006112

206.1

0

hfg

hg

sf

2500.8 2500.8

sfg

0

sg

9.155 9.155

T(°C) 0.01

Table 6.2 Saturated Water and Steam (P) Table Ps(bar)

T(°C)

vg(m3/kg)

hf

hfg

0.010

7.0

129.2

29

2485

hg

sf

sfg

sg

2514 0.106 8.868 8.974

Ps(bar) 0.010

Table 6.3 Superheated Steam Table T°C→ 50

100

150

200

250

300

350

400

v

149.1

172.2

193.5

218.4

241.4

264.5

287.6

310.7

h

2595

2689

2784

2880

2978

3077

3178

3280

s

9.241

9.512

9.751

9.966

Ps (bar)

0.01

10.163 10.344 10.512 10.670

Table 6.4 Supercritical Steam Table Ps (bar)

0.01

T°C→ 350

375

400

425

450

500

600

700

v

0.143

0.153

0.168

0.190

0.224

0.334

0.540

0.693

h

1572

1709

1862

2030

2227

2641

3200

3579

s

3.525

3.742

3.971

4.218

4.494

5.047

5.731

6.144

Properties of Pure Substance

6.8

6.9

THERMODYNAMIC PROPERTIES OF PURE SUBSTANCES (WATER AND STEAM)

There are a few thermodynamic properties which are used in the steam calculation. 1. Pressure 2. Specific volume 3. Specific internal energy 4. Specific enthalpy 5. Specific entropy.

6.8.1 Pressure The pressure of saturated water, saturated steam and superheated steam are obtained from the table values corresponding to the temperature.

6.8.2 Specific Volume The specific volume of water and steam increases with increasing temperature. For low pressures, the specific volume of water is very small, compared to the volume of steam. It is well understood that the change in volume occurs mainly during evaporation. There is no change in volume from water to dry steam, until the pressure and temperature reach the critical point. In many cases, steam is used below the critical pressure.

6.8.2.1 Specific Volume of Saturated Water (vf ) The specific volume of saturated liquid can be obtained for the given pressure or temperature.

6.8.2.2 Specific Volume of Wet Steam The specific volume of wet steam may be determined, either by using the specific volume of saturated steam (vg ) or the specific volume of saturated water (vf ), if the dryness x is given as follows (i) Saturated liquid (ii) Vapour. Let us consider a unit mass of wet steam with a dryness fraction x. In the unit mass of steam x kg is contained by dry steam and (1 – x)kg is carried by the suspended water particles. Therefore,

vwet = xvg + (1 – x) vf

The specific volume of wet steam = vwet = xvg m3/kg The dryness fraction of steam can be obtained from the values of vwet and vg.

6.10

Engineering Thermodynamics

6.8.2.3 Specific Volume of Dry Steam Corresponding to the given pressure or saturation temperature, the value vg can be obtained from the steam tables.

6.8.2.4 Specific Volume of Superheated Steam The specific volume of superheated steam can also be obtained from the superheat tables, for the corresponding pressure (or) saturation/superheated temperature. The specific volume of superheated steam can be determined as follows. The superheating of steam takes place at a constant pressure process. According to the constant pressure process calculations v = Constant           T Between saturated and superheated condition, vg T           sat or,

=

vsup Tsup

 vg  vsup = Tsup *    Tsat 

6.8.3 Density of Steam We know that the density of a substance is the reciprocal of the specific volume of the substance. If the specific volume of the given steam is v, then the density (ρ) is given below

Density = ρ = 1/v kg/m3.

6.8.4 Enthalpy of Water and Steam Saturated liquid or any quality of steam has its enthalpy value. From the previous definitions, it is also called the “total heat content” of the substance. According to the first law of thermodynamics h = u + pv where,

h = specific enthalpy in kJ/kg

u = specific internal energy in kJ/kg

p = pressure in bar (or N/m2)



v = specific volume in m3/kg

Applying the above equation to water and any quality of steam, we get the following.

Properties of Pure Substance

6.11

6.8.4.1 Saturated Water For a unit mass of saturated water, the specific enthalpy of water is denoted as hf kJ/kg, where the subscript f denotes the condition of the saturated water. Its value is obtained from the steam tables.

6.8.4.2 Wet Steam The specific enthalpy of wet steam is given as hg = hf + xhfg kJ/kg



Since wet steam has water particles in it, the dryness fraction x is taken into account.

6.8.4.3 Dry Steam The value of the dryness fraction for dry saturated steam is 1. Therefore, the specific enthalpy is

hg = hf + xhfg kJ/kg.

6.8.4.4 Superheated Steam The specific enthalpy of superheated steam can be determined by adding the specific enthalpy of dry saturated steam after dry steam is attained. hsup = hg + Cpsup (Tsup – Ts) where, hsup = Specific enthalpy of superheated steam kJ/kg hg = Specific enthalpy of dry saturated steam kJ/kg

Cpsup = Specific heat capacity of superheated steam kJ/kg k



Tsup = Temperature of superheated steam K



Tsat = Temperature of saturated steam K

Also,

hsup = [(hf + hfg) + Cpsup (Tsup – Ts)].

The values of specific enthalpy for saturated water and different forms of steam may be obtained from the steam tables. Using the enthalpy values, the internal energy of saturated water and steam can be evaluated using the equation. In some steam tables, the value of internal energy is also provided corresponding to the pressure and temperature.

u = h – pv.

6.8.5 Specific Internal Energy Saturated water

uf = hf – pf vf kJ/kg

Wet steam = ug = hg – pg xvg = (hf + xhfg) – pg xvg kJ/kg

6.12

Engineering Thermodynamics

For steam as x = 1, the equation becomes = ug = (hf + hfg) – pg vg kJ/kg ug = hg – pgvg kJ/kg

Superheated steam

usup = hsup – pg vsup

= [(hg + Cpsup (Tsup – Ts)) – pgvsup].

6.8.6 Entropy of Steam In any thermodynamic process, heat and work are involved since the p-V and T-s diagrams represent the work and heat transfer in the process. It is necessary to know how to determine the entropy of a process. Steam of different qualities is widely used in various thermal applications. This section deals with entropy calculations for various qualities of steam and saturated liquid. Like enthalpy, entropy is determined for saturated water, wet steam, dry steam and superheated steam.

6.8.6.1 Specific Entropy of Liquid We know that the entropy change of a substance is given by δq T Let us consider the unit mass of a saturated liquid, which is to be raised to vapour at constant pressure heating. ds =



δq T δq = Amount of heat transferred in unit mass of liquid. ds =

Then Where,

T = Absolute temperature of liquid.

Now, the amount of heat transferred δq under a constant pressure is δq = Cp dT ...(6.2)



Substituting Eqn. (6.2) in Eqn. (6.1) ds =

C p dT T

Integrating between initial and final states 1 and 2 2

        

2

∫1 ds = ∫1 C p

dT T

We get (s2 – s1) = Cp [ln T2 – ln T1] = Cp ln

T2 . ...(6.3) T1

6.13

Properties of Pure Substance

It is important that when the initial temperature of water is taken at the triple point (273.16 K), the entropy at the initial condition is zero; substituting s1 = 0 and T1 = 273.16 K in Eqn. (6 3) we get

s2 = Cp ln

T2 273.16

[The final state is at the saturation temperature of water Tf ] Therefore,

sf = Cp ln

T2 . 273.16

...(6.4)

6.8.6.2 Specific Entropy of Evaporation During the evaporation, the specific entropy of a unit mass of liquid is determined as follows. We know that the heat contained by a substance during its evaporation is equal to the specific entropy of evaporation. q = hfg kJ/kg

Therefore, or,

dq T q s = T

ds =

sfg =

h fg T

kJ/kg k.

6.8.6.3 Specific Entropy of Steam We know that the formation of steam begins only when the saturated liquid water reaches the boiling point of 100°C. Further addition of heat to the saturated water leads to the formation of wet steam and then, dry saturated steam. The specific entropy of wet steam is equal to the sum of the specific entropy of the saturated liquid and the specific entropy of evaporation,

sg = sf + x . sfg

...(6.6)

Substituting Eqns. (6.4) and (6.5) in Eqn. (6.6), we get;

sg = Cp ln

Tf 273.16

+x

h fg Ts



...(6.7)

For dry saturated steam, the dryness fraction value is equal to 1 Therefore, Eqn. (6.7) becomes

sg = C p ln

Tf 273.16

+

h fg Ts



...(6.8)

6.8.6.4 Superheated Steam After reaching the dry saturated condition, the steam is further heated at constant pressure to attain the superheated state. The value of the specific heat capacity for steam is taken during the superheated state.

6.14

Engineering Thermodynamics

Let,

Cp sup = Specific heat of superheated of steam = 2.1 kJ/kg K.

The entropy in the superheated region is obtained by integrating between two limits Tsup and Tsat. ds = C p sup ∫

We get

sup

sat

      ∫

sup

sat

ds = C p sup ∫

sup

sat

     (ssup – s g ) = C p sup ln

Therefore,

6.9

dT T

dT T

Tsup Ts

...(6.9)

Tsup   ssup = S sat + C p sup ln . Ts  

...(6.10)

...(6.11)

DETERMINATION OF THE DRYNESS FRACTION

The value of the dryness fraction of steam is most important in many cases, as it influences the major properties like, the specific volume, specific enthalpy, specific internal energy etc. There are two methods used to determine the dryness fraction of steam. (i) The separating calorimeter (ii) The throttling calorimeter

6.9.1 Separating Calorimeter Figure 6.7 (a) shows the schematic diagram of a separating calorimeter. The main components of a separating calorimeter are 1. Calorimeter tank 2. Pressure gauge 3. Perforated tray 4. Water drain valve The water vapour is separated from the wet steam by means of a centrifugal action taking place, when the wet steam flows around, so that the dry steam and water are collected separately, and their masses are found out. Wet steam enters at the top of the calorimeter. Near the entrance to the calorimeter tank a perforated tray is suspended. A water drain valve is provided at the bottom of the calorimeter. A glass gauge with calibrated readings is fitted on the side of the calorimeter tank. The exit for dry steam is provided at the top of the tank. When steam enters through the entry and strikes the perforated

6.15

Properties of Pure Substance

tray, its direction of flow changes. As the density of water is higher than the dry saturated steam, water falls due to the centrifugal action. Dry steam in the form of vapour leaves through the exit of the calorimeter tank. Dry steam will go to a condenser where it will be condensed. There are three factors, mainly considered in order to obtain better results in the determination of dryness fraction. (i) Prevention of condensation taking place inside the calorimeter. For this, sufficient warm up must be provided for measurement. (ii) Mixing of steam with water must be carefully avoided. Otherwise the mass of water will slightly increase than the actual. (iii) Proper thermal insulation must be provided to avoid the influence of atmospheric air on the condensation of steam. Pressure gauge

vV Vv vV

I Steam pipe Sampling tube

Pressure gauge

Perforated box

Steam exit

Fig. 6.7(a) Schematic of the separating calorimeter

If M is the mass of dry steam condensed and m is the mass of water removed from the wet steam in the calorimeter, Then the dryness fraction is, x=

Mass of dry steam M . = Total mass of steam carrying dry steam M+m

6.9.2 Throttling Calorimeter The schematic diagram of a throttling calorimeter is shown in Fig. 6.7(a). The main components of the throttling calorimeter are (i) Throttling chamber (ii) Sampling tube

6.16

Engineering Thermodynamics

(iii) Pressure gauge (iv) U-Tube manometer (v) Thermometer. Pressure gauge

Throttle valve Steam pipe

Sampling tube Thermometer

Manometer Steam exit

Fig. 6.7(b) Throttling calorimeter

Steam from a steam pipe is sent through a sampling tube having holes in it. The steam then goes to a throttle chamber through a stop valve and a pressure gauge. At the entry of the throttling chamber there is a throttle orifice, which allows the steam into the throttling chamber. The pressure and temperature of the steam after throttling are measured by the U-tube water manometer and the thermometer provided. The throttled steam is then sent to a condenser where it is condensed. The mass of the condensate is then determined. The stop valve is used to close the steam flow to the throttling chamber. Partial closure of the stop valve will lead to a slight deviation in the actual measurements of the temperature and pressure. The following points should be considered in the determination of the dryness fraction of steam, using the throttling calorimeter. (i) When the temperature of steam after throttling is above 100°C, it is assumed that steam is being superheated. (ii) The pressure of steam after throttling should be nearly equal to the atmospheric pressure. (iii) The pressure of steam before throttling is measured only when the flow of steam becomes steady.

6.17

Properties of Pure Substance

Table 6.5 Various Properties of Saturated Water, Steam and Superheated Steam Property

Saturated water Evaporation

Specific volume m3/kg

vf

–

Specific Enthalpy kJ/kg

hf

x hfg

Specific Internal energy

uf

–

Wet steam

Dry steam Superheated steam

xvg

vg

hg = hf + xhfg

hg = hf + hfg

vsup =

vg Tg

× Tsup

hsup = hg + Cpsup (Tsup – Ts)

ug = hg – pxvg ug = hg – pvg usup = hsup – pvsup ssup = sg +

Specific entropy

T sf = ln s 273

External work

W=100 p(vg – vf) or w = 100 pvg

sfg =

hfg T

–

sg = sf + xsfg

sg = sf + xsfg

wg = 100 pxvg

wg = 100 pxvg

In

Tsnp Ts

Wsup = 100 p vsup

6.10 NON-FLOW PROCESSES OF PURE SUBSTANCES The following non-flow processes are carried out for pure substances. They are (i) Constant volume process (Isochoric) (ii) Constant pressure process (Isobaric) (iii) Constant temperature process (Isothermal) (iv) Hyperbolic process (v) Reversible adiabatic process (Isentropic) (vi) Polytropic process In this section, the above processes applied to closed systems, containing pure substances (saturated liquid and vapour) are studied. According to the first law of thermodynamics, for a kg of pure substance,

dq = du + δw

It should be noted, that in the equation the work done (w) in the non-flow processes and flow processes are given below For non-flow process δw = pdv Flow process

δw = –vdp

The gas equation of state is not valid in the case of vapours. Consider the unit mass of a working substance, with a dryness fraction ‘x’. Initially, the wet steam is heated at a constant volume. The pressure and temperature increase. p1 and p2 = Pressure of substance (bar) at the initial and final condition respectively.

6.18

Engineering Thermodynamics

h1 and h2 = Specific enthalpy of the substance at the initial and final condition respectively in (kJ/kg) u1 and u2 = Specific internal energies of the substance at the initial and final conditions respectively. The value of the dryness fraction x2 can be determined for the non-flow processes, by using the fundamental p-V relationship.

6.10.1  Constant Volume Process In a constant volume process, the volume of the substance is constant throughout the process. p2 2

2

p2 p1

p1

p

T

1

1

v

s

(a) p-V diagram

(b) T-s diagram

Fig. 6.8 Constant volume process

Initial volume of wet steam v1 = x1vg



1

Where vg = specific volume of wet steam m3/kg at pressure p1 1

Final volume of steam Where Therefore,

v2 = x2 vg

2

vg2 = specific volume of steam at final pressure p2 v1 = v2

v1 = v2 x1 v g 1 Therefore,       x2 = vg 2 or, per unit mass

If

x2 < 1 final condition of steam is wet

x2 = 1 final condition of steam is dry saturated

x2 > 1 final condition of steam is superheated steam

The final temperature of steam at superheated condition is calculated as follows

Let, the temperatures of steam at the saturated and superheated condition s at pressure p2 be Tg2 and Tsup respectively.

6.19

Properties of Pure Substance

Between saturated and superheated condition. vg1 = T           g 2

vsup v2 = Tsup Tsup

v Tsup = Tg 2 × 2 vg 2           As volume is content

δw = 0



δq = du q = u2 – u1



q = (h2 – p2v2) – (h1 – p1v1).



...(6.12)

6.10.2  Constant Pressure Process Constant pressure heating is shown in the p-V and T-s diagrams in Fig. 6.9(a) and (b) respectively. The generation of steam in boilers is a typical example of constant pressure heating. Therefore, V1 = V2 m3 or, per unit mass v1 = v2. Considering the dryness fraction of steam, x1vg1 = x2vg2. The values of the specific volume at the initial and final conditions may be determined if x1 and x2 values are known. The final condition of steam is therefore (i) Wet steam when x2 < 1 (point 2’) (ii) Dry steam x2 = 1 (point 2) (iii) Superheated steam when x2 > 1 (point 2’’) x1vg1 = x2vg2 x1 v g 1 x2 = vg 2                 

2 1

1

2 2

p

1

2

2

v

(a) p-V diagram

2

T

S (b) T-s diagram

Fig. 6.9 Constant pressure process

6.20

Engineering Thermodynamics

The temperature of superheated steam can be determined, using v1 Tsup = Tg 2 × v g2



2

w = ∫ p dv = p(v2 – v1)

Work done

1

Applying the first law of thermodynamics, we get; q = w + (u2 – u1)



...(6.13 a)

= p(v2 – v1) + (u2 – u1) = (h2 – h1).

...(6.13 b)

6.10.3  Constant Temperature Process In the wet region the constant temperature process will also be at constant pressure. Once the steam reaches the saturation point, during the constant temperature process, the pressure will change. This is shown in the p-V and T-s diagrams in Fig. 6.10. Until the steam reaches saturation, the heating of the wet steam at constant temperature process is similar to the constant pressure process. Therefore, the value of x2 may be determined by equating the initial and final values of the specific volume of steam. p2 p

p 1 < p2

p1

1 1

2

2

T

2

2

p 2 < p2

v

s

(a) p-V diagram

(b) T-s diagram

Fig. 6.10 Constant temperature process

In the wet region

w = p(v2 – v1)

...(6.14 a)



q = (h2 – h1)

...(6.14 b)



v1 = x1 vg1,



v2 = x2 vg2.

...(6.15)

6.21

Properties of Pure Substance

6.10.4  Hyperbolic Process This process follows the law pv = Constant. Hyperbolic process is also an isothermal process in the superheated region as the steam behaves like a gas in the superheated region. The T-s diagram is shown in Fig. 6.11. or,

p1 x1vg1 = p2x2vg2 x2 =



p1 x1 v g 1 p2 v g 2 p1

p2

1

T

2

s Fig. 6.11 Hyperbolic process



Tsup = Tg2 ×

v2 vg 2

Work done during hyperbolic process w =

2

∫1 p dv

v  = pv ln  2   v1  q = w + (u2 – u1)

v  q = pv ln  2  + (u2 – u1)  v1 



v  q = pv ln  2  + (h2 – p2v2) – (h1 – p1v1). ...(6.16 b)  v1 

...(6.16 a)

6.10.5 Reversible Adiabatic Process (Isentropic Process or Constant Entropy Process) In a reversible adiabatic process the entropy remains constant. Figure 6.12 shows the representation of isentropic process in p-V and T-s diagram. So, the value of the entropy at initial condition is equal to the value of the entropy at final condition. Therefore,

s1 = s2

6.22

Engineering Thermodynamics

1

1

p T

2

2

v

s

(a) p-V diagram

(b) T-s diagram

Fig. 6.12 Reversible adiabatic process

or,

sf1 + x1sfg1 = sf 2 + x2sfg2 s f 1 + x1

h fg 1 T1

= s f 2 + x2

h fg 2 T2

where s f, h fg, T are specific entropy (kJ/kg K) specific enthalpy of vaporization (kJ/kg), and temperature of steam (K) respectively. The subscript 1 and 2 indicate the initial and final condition of steam corresponding to pressures p1 and p2 respectively.

q = w + (u2 – u1)

As

q = 0



w = (u1 – u2)

= (h1 – p1u1) – (h2 – p2u2) ...(6.17)

6.10.6  Polytropic Process The polytropic process follows pvn = Constant Per unit mass pvn = p1vn1 = p2vn2 1

p 2

v

Fig. 6.13 Polytropic process

Considering the quality of steam

p1 (x1vg1)n = p2(x2vg2)n

6.23

Properties of Pure Substance



x2 =

x1 v g 1  p11/n   . v g 2  p2 

1. If x2 > 1, the steam is superheated steam at the end of the process.

The specific volume of the superheated steam may be determined as follows. p1(x1vg1)n = p2(vsup)n p  vsup = x1 v g 1  1   p2 

1/n

.

The final temperature of the superheated steam may be determined in the usual method, using Charles’ law. 2. If x2 > 1, then the final condition of the steam is superheated steam.

6.11 WORK DONE AND INTERNAL ENERGY AND HEAT TRANSFER CALCULATIONS 6.11.1 Constant Volume Process 1. Work done per kg of steam Since the volume is constant, the change in volume is zero. Therefore, the work done in the process, w1–2 = Pressure of steam × Change in volume w1–2 = 0.



...(6.19)

2. Change in internal energy The change in the internal energy in the constant volume process is given by

Δu = u2 – u1, kJ/kg

Δu = (h2 – 100 p2v2) – (h1 – 100 p1v1)

...(6.20 a)

For calculations it is necessary to find out the quality of the steam for both initial and final conditions. For example, The initial condition of the steam is wet, and the final condition was found to be dry. Then,

Δu = u2 – u1

conversion [1 bar = 100 kN/m2]

Δu = (h2 – 100 p2v2) – (h1 – 100 p1v1)

= [(h2 – 100 p2vg2) – (hf1 – x1hg1) – (100 p1 x1vg1)]

...(6.20b)

where, x2 = 1.

Suppose the final condition of steam is superheated then

Δu = (h2 – 100 p2v2) – (h1 – 100 p1v1)

= [(hsup – 100 p2vsup) – (hg – 100 p1x1vg1)] = [(hg + Cpsup(tsup – ts) – (100 p2vsup) – (hfg + x1hfg1) – (100 p1x1vg1)]

...(6.20c)

6.24

Engineering Thermodynamics

where

vsup Tsup

vsat . Tsat

=

3. Heat transferred (Heat absorbed or rejected) = q1–2 = w + Δu kJ/kg

q = 0 + Δu



q = u2 – u1 kJ/kg.

...(6.21)

6.11.2  Constant Pressure Process Figures 6.14 (a) and (b), show the p-V and T-s diagram respectively. According to the first law q = w + Δu q = pdv + Δu

1 p

1

2

T 2

v

s

(a) p-V diagram

(b) T-s diagram

Fig. 6.14 At constant pressure process

= p(v2 – v1) + (u2 – u1) If the initial condition of the steam is wet

q = (u2 – u1) + p(v2 – v1)

= (u2 + pv2) – (u1 + p1v1) = (h2 – h1).

...(6.22)

6.11.3  Constant Temperature Process When the unit steam reaches the saturated state the process carried out under constant temperature is the same as constant pressure process. Once the steam reaches the saturated condition, the steam behaves like an ideal gas. Heat transfer during the process;

q = TΔs.

6.25

Properties of Pure Substance

It should be noted that the specific volume of dry steam remains constant corresponding to the pressure. This is because of the values obtained for specific volume in the steam table, for the initial and final conditions are the same (as pressure remains constant), and only the vg value is applicable.

6.11.4  Hyperbolic Process An isothermal process in the superheated region is called a hyperbolic process. The steam in the superheated region behaves like an ideal gas; therefore, it follows pv = C. In the hyperbolic process, the work done is determined similar to the non-flow process of a perfect gas. 2

As we know w1–2 = ∫ pdv



1

2  p1 v1  w1–2 = ∫1   dv  v  2  gv  = p1 v1 ∫1    v 

= p1v1 ln(v) v = p1 v1 ln  2  v1 when the pressure is measured in bar

  

...(6.23)

v  w1–2 = 2.3 × 100 × p1v1 ln  2  kJ/kg  v1  Change in the internal energy

The change in the internal energy is similar to the isothermal process

Δu = u2 – u1

= (h2 – p2v2 ) – (h1 – p1v1)

...(6.24 a)

= (h2 – h1) (As p2v2 = p1v1) kJ/kg

...(6.24 b)

Heat transferred during the process According to the first law of thermodynamics, the heat transferred during the process per kg of substance q1–2 = w1–2 + Δu v  = p1v1 ln  2  + (h2 – h1) kJ/kg  v1  p  = p2v2 ln  2  + (h2 – h1) kJ/kg  p1 

...(6.25)

6.26

Engineering Thermodynamics

6.11.5   Reversible Adiabatic Process (Isentropic Process) As discussed in the earlier chapters, a process in which no heat transfer takes place between the system and surroundings, and vice versa, is known as an adiabatic process. The entropy of the process remains constant. Therefore,

s1 = s2

Suppose the initial and final conditions are wet. Then, sf1 + x1sfg1 = sf 2 + x2 sfg2

...(6.26)

Change in internal energy (Δu)

Δu = u2 – u1 kJ/kg

u1 = Initial internal energy of wet steam kJ/kg u2 = Final internal energy of wet steam kJ/kg u1 = h1 – 100 p1v1 = [(hf1 + x1hfg1) – (100 p1 x2vg1)]

...(6.27a)

u2 = h2 – 100 p2v2 = [(hf 2 + x2hfg2) – (100 p2 x2 vg2)]

...(6.27b)

The values of h2 and v2 vary with the final conditions of steam (i.e., dry saturated or superheated steam). Heat transfer in the process: Since the process is reversible adiabatic there will be no heat transfer in the process. Therefore,

q1–2 = 0

Work done during the process: According to the first law

q1–2 = W1–2 + Δu

W1–2 = –Δu (as q1–2 = 0).

...(6.28)

6.11.6  Polytropic Process In a polytropic process, the steam follows the law pvn = C. Between two states 1 and 2

pvn = p1v1n = p2v2n p =

(i) Change in internal energy We know that

p1 v1n vn

Δu = u2 – u1

Suppose the initial condition of steam is wet, then

u1 = h1 – 100 p1v1 kJ/kg

6.27

Properties of Pure Substance

Where,

v1 = x1vg1 and h1 = hf1 + xhfg1

If the final condition of steam is dry saturated, then

u2 = h2 – 100 p2v2

Where,

v2 = vg2 m3/kg



h2 = hfg2 + hg2

Therefore, Δu = u2 – u1 = [(hf2 + hfg2 – 100 p2vg2) – (hf1 + xhfg1 – 100 x1 p1 vg1)] ...(6.29) (ii) Work done in the polytropic process Work done in the polytropic process is given by 2

w1–2 = ∫ pdv



1

n 2  p1 v1 = ∫1   vn

=

  dv 

p1 v1 – p2 v2 . n– 1

...(6.30)

Suppose the pressure is given in bar, then  p1 v1 – p2 v2 w1–2 = 100  n– 1 



 . 

The v1 and v2 values depend on the quality of steam. (iii) Heat transfer in the process We know that

q1–2 = w1–2 + Δu

 p1 v1 – p2 v2  =   + ( u2 – u1 ) n– 1  

...(6.31)

6.11.7  Throttling Process Throttling is a process, in which a restriction is placed in the flow of a fluid. It is applicable to a low speed fluid flow. In a pipe line, fluid enters at the inlet and leaves the exit; between the inlet and the outlet there is a valve which, partially restricts the flow by closing it. The following points are to be noted;

l The

change in the kinetic energy and potential energy between the inlet and the outlet is negligible.



l Work



l Change

transfer is negligible in the internal energy is negligible.

6.28

Engineering Thermodynamics



l During

the throttling, the dry steam will be superheated. If the initial condition before throttling is wet, then final condition will be dry.



l Throttling



l Applying the above conditions is steady flow energy equation, we get for

is irreversible adiabatic steady flow process.

a unit mass

u u



h1 = h2 kJ/kg

If the initial condition of steam is known, and the final condition of the steam has to be determined, then, hf1+ x1hfg1 = hf2 + x2hfg2

 h f 1 + x1 h fg 1 – h f 2 Therefore,      x2 =   h fg 2 

 .  

...(6.32)

6.12 MOLLIER DIAGRAM The Molllier diagram is shown in Fig. 6.15. It is a graphical representation, in which all the property values are put together, so as to enhance the thermodynamic calculations easier. The thermodynamic properties, such as specific enthalpy, and entropy can be determined with reference to a third thermodynamic property. It is useful in solving the reversible adiabatic and steady flow processes. The diagram constitutes the saturation line. Above the saturation line is the superheated region and below the saturation line wet region falls. The wet region has a constant temperature line with increasing pressure. In the Mollier diagram, we find the following lines 1. Constant dryness fraction lines 2. Constant volume lines 3. Constant pressure lines 4. Constant temperature lines 5. Constant entropy lines 6. Constant enthalpy lines In a certain process, 1 kg of steam initially has a dryness fraction of 0.8 and enthalpy of 2500 kJ/kg. The final enthalpy is 2300 kJ/kg. Find the change in entropy during the process, using the Mollier diagram. In the Mollier diagram, below the saturation line we find a line having x = 0.8. Now, mark a point 1 on the vertical scale, in which the specific enthalpy value is 2500 kJ/kg. Draw a horizontal line from the point, to meet this constant dryness fraction line at A. From A, erect a vertical line to meet the x-axis. So that the value of entropy is determined. Similarly, for the final enthalpy of 2300 kJ/kg, point 2 is marked on the y-axis. A horizontal line is drawn from point 2 to meet the constant dryness fraction line at B. From B a vertical line is drawn to find the value of entropy (s2) on the x-axis.

6.29

Properties of Pure Substance

If the specific enthalpy is taken as a y-axis, and the specific entropy is in x-axis, it is useful in solving the reversible adiabatic process and steady flow processes. Constant pressure line

Constant temperature line

Vapour region

h

Saturation line

Wet region 2500

x= 0.8

Constant specific volume line

Constant dryness line fraction

Specific entropy(s)

Fig. 6.15 Mollier diagram



s1 = 5.84 kJ/kg K



s2 = 6.64 kJ/kg

Therefore, change in entropy

Δs = s2 – s1

= 6.64 – 5.84 = 0.8 kJ/kg K.

Solved Problems Problem 6.1 Get the value of the specific volume vg at (i) the given saturation temperature of 150°C (ii) for a given pressure of 2 bar, referring to the steam tables.

6.30

Engineering Thermodynamics

Solution (i) Referring to the steam tables, obtain the value of vg corresponding to temperature 150°C, as given in Table p. 6.1. Table p. 6.1 Saturation temperature ts

Pressure (bar)

Specific volume vg(m3/kg)

150°C

4.76

0.392

vg = 0.392



m3/kg

(ii) Referring to the steam tables, obtain the value of vg corresponding to a pressure of 2 bar, as given in Table p. 6.1(a) Table p. 6.1(a) Pressure (bar)

Saturation temperature (°C)

Specific volume vg(m3/kg)

2 bar

120.2

0.8856

vg = 0.8856 m3/kg



Problem 6.2 The steam is heated at a constant volume in a container at a pressure of 3.5 bar and dryness fraction 0.7 so that the final pressure is 4.5 bar. Find the quality of steam at the final condition (dryness fraction), and the heat transferred by unit mass of steam.

2 p

T

2

1

1 s

v

(a)

(b)

Fig. p. 6.1 (a) p-V diagram (b) T-s diagram in constant volume process

Given data Initial pressure = 3.5 bar Final pressure = 4.5 bar Initial dryness fraction x1 = 0.8 bar At pressure p1 - 3.5 bar the property values are obtained from the steam tables.

T1 = 138.9°C



hf1 = 584 kJ/kg

6.31

Properties of Pure Substance

hfg1 = 2549 kJ/kg

vg1 = 0.5241 m3/kg

Solution At pressure p2 = 4.5 bar the property values are obtained from the steam tables.

T2 = 147.9°C



hf2 = 623 kJ/kg



hfg2 = 2558 kJ/kg

vg2 = 0.4139 m3/kg q = w + (u2 – u1)

w = ∫p dv = 0 (as v = constant )



q = (u2 – u1).

(i) Final condition of steam We know that at constant volume process v1 or,

x1vg1 = x2vg2 x2 =

=

x1 v g 1 vg 2

0.7 × 0.05241 0.4139

x2 = 0.886

Since x2 < 1, the final condition of steam is wet. (ii) Change in the internal energy we know that change in the internal energy u = u2 – u1, kJ/kg

(u2 – u1) = (hg2 – 1000 p2v2) – (h1 – 100 p1v1)

hg2 = hf2 + x2hfg = 623 + 0.886 × 2558 = 288.38 hg1 = hf + x2hfg1 = 5484 + 0.7 × 2549 = 2889.38 kJ/kg u2 = hg2 – 100 p2v2 = 2889.38 – 100 × 4.5 × 0.4139 = 2889.38 – 100 × 4.5 × 0.4139 = 2703.125 kJ/kg

u1 = hg1 – 100 p1v1

kJ/kg

6.32

Engineering Thermodynamics

= 2368.38 – 100 × 3.5 × 0.5241 = 2368.3 – 183.43 = 2184.87 kJ/kg. Therefore, change in internal energy u = u2 – u1 = 2703.125 – 2148.87 = 518.26 kJ/kg. Heat transferred per kg of steam The heat transfer during the constant volume process is equal to the amount of change in the internal energy. q1–2 = u q1–2 = 518.26 kJ/kg.



Problem 6.3 Superheated steam at a pressure of 8 bar and temperature of 230°C is cooled in a drum. The final pressure of the steam is 4 bar. Find the final quality of the steam, change in internal energy, work done, heat transferred in the process and mass of steam. Assume that the volume of the drum is 0.15 m3. Given data

p1 = 8 bar



p2 = 4 bar

V1 = 0.15 m3

T1 = 230°C = 503 K 1 1 p

T

2

2 v

(a)

s

(b)

Fig. p. 6.2 (a) p-V diagram (b) T-s diagram in constant volume process

Solution Corresponding to the initial pressure p1 = 8 bar, the proper values are obtained from the steam tables.

t2 = 170.84°C = 443.84 K



vg1 = 0.2403 m3



hf1 = 721 kJ/kg

hfg1 = 2048 kJ/kg T1 > Tsat steam is superheated initially. Corresponding to the final pressure p2 = 4 bar, the property values are obtained from the steam tables.

6.33

Properties of Pure Substance



Ts = 143.6°C = 416.6 K



vg2 = 0.4623 m3/kg



hf2 = 605 kJ/kg



hfg2 = 2134 kJ/kg

Since the steam is in a superheated condition, it is necessary to find its value. vsup T           sup

=

vg1 Ts 1

vsup = v g 1 ×

=

Tsup T1

0.2403 × 503 443.84



v1 = vsup = 0.272 m3/kg



v2 = v1 = 0.272 m3/kg



vg2 = 0.4623 m3/kg

Since v2 < vg2, the steam is wet at the end of the process.

x2 =

=

v2 vg1 0.2723 0.4623

Change in internal energy u1 = hsup – 100 p1vsup u2 = h2 – 100 p2v2 hsup = hg1 + Cpsup(Tsup – T1) = (721 + 2048) + 2.1(503 – 443.84) = 2893.236 kJ/kg

h2 = (hf2 + x2hfg2)

= (605 + 0.589 × 2134) = 1861.926 kJ/kg

u1 = 2893.266 – 100 × 8 × 0.2723

= 2893.266 – 217.84 = 2675.42 kJ/kg

u2 = h2 – 1000 p2v2

= 1861.926 – 100 × 4 × 0.589 × 0.4623

6.34

Engineering Thermodynamics

= 1861.926 – 108.0 = 1753.0 kJ/kg Δu = u2 – u1



= 1753.0 – 1861.926 = –108.91 kJ/kg. The negative value indicates, that there is loss in the internal energy due to cooling. (ii) Work done W1–2 = 0 (constant volume process) (iii) Heat transferred

Heat transferred per kg of steam q1-2 = w + Δu (as v = Constant, w1–2 = 0)



q1–2 = Δu (as w1–2 = 0).



Therefore, the heat transferred per kg of steam is q1–2 = –108.91; that means 108.91 kJ/kg of heat is rejected from superheated steam so that the cooling in done. (iv) Mass of steam We know that the volume of the drum V = 0.15 m3 The initial volume of steam Vsup = 0.2723 m3/kg

Mass of steam =

=

Volume of drum Initial volume steam V Vsup

0.15 0.2723 = 0.55 kg. =

Problem 6.4 Water is supplied and heated in a boiler, so that steam is generated at constant pressure. The temperature of the water is 40°C, after generation of the steam the final condition of the steam is dry saturated. The pressure of steam generated is 15 bar. Given

T1 = 40 + 273 = 313 K p = 15 bar

Solution Corresponding to T1 = 40°C, the property values of saturated liquid are obtained from the tables.

hf1 = 167.5 kJ/kg.

Final condition of steam Similarly, for a pressure 15 bar, the property values are obtained from the tables.

6.35

Properties of Pure Substance



Ts = 198.3°C



vg2 = 0.1317 m3/kg



hf 2 = 8446 kJ/kg



hfg2 = 1945.3 kJ/kg



h2 = 2790 kJ/kg



x2 = 1.

Therefore, The heat absorbed by the system during the process = q = h2 – h1 q = (hf 2 + hfg2) – (hf1)

Per kg of steam

= 2790 – 167.5 = 2622.5 kJ/kg. Problem 6.5 Steam at a pressure of 8 bar, a with a dryness fraction of 0.85 expands in a cylinder. The final pressure of the steam is 2 bar. The expansion takes place reversibly and at constant temperature; then, calculate (i) the change in the internal energy, (ii) heat supplied per kg of steam, (iii) The work done.

1

2

1 T

p

2

v

s

Fig. p. 6.3 (a) p-V diagram (b) T-s diagram in constant temperature process

Given data

p1 = 8 bar = 8 × 105 N/m2



x1 = 0.85



p2 = 2 bar.

Solution We have to find the quantity of steam at the final condition. The property values of steam at p1 = 8 bar are obtained from the steam tables. vg1 = 0.2403 m3/kg hg1 = 721 kJ/kg hfg1 = 2048 kJ/kg Similarly, the property values of steam at p2 = 2 bar are obtained from the steam tables. ug2 = 0.8856 m3/kg

Ts2 = 120.2°C

ht2 = 505 kJ/kg

6.36

Engineering Thermodynamics

hfg2 = 2530 kJ/kg

For the processes p1 = 8 bar, the saturation temperature is 170.4°C.

T2 = T1 = 170.4°C.

T2 > Ts2. Therefore the final condition of steam is superheated. (i) Determination of vsup As we know that = v2         

v= sup

vs × Tsup

=

Ts

0.8856 + 443.4 393.2

vsup = 0.9986 m3/kg

The value of vsup can also be obtained from the steam tables and from the superheated tables. From steam tables for p2 = 2 bar, We get

T2 = 170.4°C



hsup = (By interpolation)



vsup = 1 m3/kg.

Change in internal energy

Δu = u2 – u1



Δu = (hsup – p2vsup) – [(hf1 + xhfg1) – p1 x1vg1)]



vsup = vs ×

Tsup Ts

Substituting the values of the properties we get

Δu = u2 – u1 u2 = hsup – 100 p2vsup

hsup at 170.4°C = 2770 kJ/kg Similarly, hsup at 200°C = 2876 kJ/kg. For 50°C rise in temperature, the increase in enthalpy is = 2876 – 2770 kJ/kg = 106 kJ/kg For 1°C the increase in enthalpy =

106 kJ/kg 50

For 20.4°C rise in temperature = 2.12 kJ/kg The increase in enthalpy = 2.12 × 20.4 = 43.248 kJ/kg.

Properties of Pure Substance

Therefore, at 170.4°C, the enthalpy of superheated steam (hsup) is

hsup = 2813.2 kJ/kg



h1 = hf1 + x1 hfg1

= 721 + 0.85 × 2048 = 2461.8 kJ/kg

u2 = hsup – 100 p2 vsup

= 2813.2 – 100 × 2 × 0.9986 = 2813.2 – 199.72 = 2613.48 kJ/kg Similarly,

u1 = h1 – 100 P1x1vg1

= 2461.8 – 100 × 8 × 0.85 × 0.2403 = 2461.8 – 163.4 = 2134.9 kJ/kg Change in internal energy = Δu = u2 – u1 = 2613.48 – 2134.9 = 478.48 kJ/kg Heat Transfer during the process q = TΔs Where,

Δs = s2 – s1.

At

p2 = 2 bar, T2 = 150°C



s2 = ssup = ssup at 150°C

= 7.280 kJ/kg Similarly, ssup at 200°C = 7.507 kJ/kg Therefore, ssup 170.4°C = [(7.507 – 7.250) 20.4] + 7.280 = 7.3726 kJ/kg. Corresponding to 8 bar

sf1 = 2.046 kJ/kg



sfg1 = 4.617 kJ/kg



s1 = sf1 + x1sfg1

= 2.046 + 0.85 × 4.617 = 2.046 + 3.92445 = ssup – s1 = 5.97 kJ/kg

Δs = s2 – s1

= 7.3726 – 5.97 = 1.403 kJ/kg K. Heat transferred during the process = TΔs

6.37

6.38

Engineering Thermodynamics

= 443.4 × 1.402 = 621.64 kJ/kg Work done during the process = w = q – Δu = 621.64 – 478.48 = 143.166 kJ/kg. Problem 6.6 One kg of dry saturated steam at a pressure of 10 bar expands hyperbolically so that the final pressure reaches 0.8 bar. Determine (i) the final condition of steam (ii) change in the internal energy, (iii) heat transfer during the process and (iv) work done by steam. 1 1 T

p

2

2 v

s

Fig. p. 6.4 (a) p-V diagram (b) T-s diagram in hyperbolic process

Given data

p1 = 10 bar = 10 × 105 N/m2 = 103 kN/m2 p2 = 0.8 bar = 0.8 × 105 N/m2

= 0.8 × 102 kN/m2

x1 = 1(dry saturated).

Solution The property values are taken from the steam tables corresponding to p1 = 10 bar

Ts1 = 179.9°C = 452.9 K

vg1 = 0.1944 m3/kg Similarly, for

hf1 = 763 kJ/kg

fg1 = 2015 kJ/kg p2 = 0.8 bar

vg2 = 2.2.087 m3/kg

Ts2 = 93.5°C = 366.5 K

Since during hyperbolic process pv = C p1v1 = p2v2

p1x1vg1 = p2x2vg2 p1vg1 = p2x2vg2 p1vg1 = p2x2vg2 x2 =

( p1 vg1 ) ( p2 v g 2 )

6.39

Properties of Pure Substance

=

(10 × 105 × 0.1944) (0.8 × 105 × 2.087)

= 1.164 x2 > 1. Therefore, the final condition of steam is superheated Again

p1x1vg1 = p2vsup vsup = p1vg1 ×

=

x1 p2

(10 × 105 × 1 × 0.1944) (0.8 × 105)

= 2.43 m3/kg. We know that vsup × T2



Tsup =



Tsup =

or,

Tsup = 153.73°C

vg 2 2.43 × 3.665 426.73 K 2.087

Change in the enthalpy

h1 = hf1 + x1hfg1

Since

x1 = 1



h1 = hf1 + hfg1



h1 = 763 + 2015

= 2778 kJ /kg

h2 = hsup = hg2 + Cpsup (Tsup – Ts2)

= 26654 + 2.1 (153.73 – 93.5) = 2791.4 kJ/kg

Δh = h2 – h1

= 2791.4 – 2778 = 13.483 kJ/kg

v  w = p1 v1 ln  2   v1 

 10  = 10 × 100 × 0.1944 ln    0.8  = 491 kJ/kg Change in the internal energy As

p1v1 = p2v2

6.40

Engineering Thermodynamics

q = w + Δh



= 491 + 13.483 kJ/kg = 504.483 kJ/kg. Problem 6.7 A steam turbine receives oky steam at a pressure of 20 bar and 300°C. The exhaust pressure is 0.007 bar, during the expansion of steam. Determine (i) the heat transfer (ii) the work done during the process. Given data

p1 = 20 bar Tsup = 300°C = 573K p2 = 0.07 bar

Solution Corresponding to p1 = 20 bar, the following values are obtained from superheated steam tables.

s1 = 6.768 kJ/kg



Ts = 212.4°C



vsup = 0.1255 m3/kg



h1 = 3025 kJ/kg.

Corresponding to p2 = 0.07 bar, the following values are obtained from the steam tables

Ts = 39°C



sf2 = 0.559 kJ/kg K



hf = 163 kJ /kg



vg2 = 20.53 m3/kg



sfg2 = 7.715 kJ/kg K



hg = 2572 kJ/kg.

Final condition of steam Since the process is isentropic, the specific entropy is constant throughout the process.

s2 = s1



sf 2 + x2 sfg2 = s1



0.559 + x2 + 7.715 = 6.768



7.715 x2 = 6.209



x2 = 0.8047

The final condition of steam is wet. Change in the internal energy

u1 = hsup – 100 p1vsup

= 3025 – 100 × 20 × 0.1255 = 3025 – 257 = 2774 kJ/kg

6.41

Properties of Pure Substance

u2 = h2 – 100 p2x2vg2



= (hf2 + x1hfg2) – 100 p2x2vg2 = (163 + 1 × 2572) – 100 × 0.07 × 0.805 × 20.53 = 2735 – 113.6 = 2621.4 kJ/kg Since the process is isentropic, the heat transfer is zero. 1 1 p 2

2

T

s

v

Fig. p. 6.5 (a) p-V diagram (b) T-s diagram



Δu = u2 – u1

= 2621.4 – 2774 kJ/kg = – 152.6 kJ/kg Work done According to the finest law q = w + Δu               w = q – Δu              w = 0 – (– 152.6) = 152.6 kJ/kg. During the expansion of steam in the turbine, the work obtained from it, is 697.11 kJ/kg. Problem 6.8 In a polytropic process, the steam initially dry saturated and at a pressure of 12 bar is allowed to expand, so that the pressure at the end of the process is 2 bar. Assume that the process is reversible and the value of the polytropic index is 1.25. Determine (i) the temperature at the end of the process (ii) the work done (iii) heat transfer and (iv) change in the entropy. Given data

p1 = 12 bar



n = 1.25



p2 = 2 bar

Solution Final quality of steam Corresponding to p1 = 12 bar, the property values are obtained from the stream tables.

vg1 = 0.1632 m3/kg

6.42

Engineering Thermodynamics

hf1 = 505 kJ/kg sf1 = 2.216 kJ/kg K



hfg1 = 2202 kJ/kg sfg1 = 4.307 kJ/kg K



p2 = 2 bar, we find

Similarly for

vg2 = 0.8856



hf2 = 798 kJ/kg sf12 = 1.530 kJ/kg



hfg2 = 1986 kJ/kg sfg2 = 5.597 kJ/kg.



We know that in a polytropic process  x1 v g 1   p1 1 /n  x2 =   v g 2   p2   



Assuming that the initial steam is dry saturated, we get  1 × 0.1632   12  x2 =  ×   0.08856   2 



1/1.25

= 0.1842 × 4.192 = 0.7725.

1 p

1

T

2

2 V

S

Fig. p. 6.6 (a) p-V diagram (b) T-s diagram in polytropic process

The final condition of steam is wet. Work done      w =



=

100 × ( p1 v1 – p2 v2 ) (n – 1) 100 × ( p1 v g 1 – p2 v g 2 ) (n – 1)



6.43

Properties of Pure Substance

= 100(12 × 0.1632 – 2 × 0.7725 × 0.8856) =

100(1.9584 – 1.3682) 0.25

w = 236.05 kJ/kg. Change in the internal energy

u2 = h2 – 100 p2v2

= (hf2 + x2hfg2) – 100 P2x2v2 = (505 + 0.7725 × 2202) – (100 × 2 × 0.7725 × 0.8856) = 2206.04 – 136.82 = 2206.04 kJ/kg

u1 = h1 – 100 P1v1

= hg1 – 100 P1vg1 = (hg1 – 1986) – 100 × 12 × 0.1632 = 2784 – 195.84 = 2588.16 kJ/kg

Δu = u2 – u1

= – 382.12 kJ/kg. Heat transferred

q = w + Δu

= 236.05 – 382.12 = – 146.07 kJ/kg. 1

T

2

Fig. p. 6.7 T-s diagram

Change in entropy

s1 = sf1 + x1sfg1



s1 = 2.216 + 1 × 4.307

= 6.523 kJ/kg

6.44

Engineering Thermodynamics

s2 = sf2 + x2sfg2



= 1.530 + 0.7725 × 5.597 = 5.8536 kJ/kg K Change in entropy

Δs = s2 – s1



Δs = 5.8536 – 6.523



Δs = –0.6694 kJ/kg K.

Problem 6.9 One kg of steam at 10 bar with 0.85 dry ness is throttled to a pressure of 2 bar. Determine the final quality of steam. Given data

p1 = 10 bar

x1 = 0.85 p2 = 2 bar Solution Corresponding to p1 = 10 bar, the enthalpy values are obtained from the steam tables. hf1 = 763 kJ/kg hfg1 = 2065 kJ/kg

Similarly at

p2 = 2 bar

we find that

hf 2 = 511 kJ/kg



hfg2 = 2198 kJ/kg

Therefore,

x2 =



x2 =

( h f 1 + x1 h fg 1 – h f 1 ) h fg 2 (763 + 0.82 × 2014) – 511 2198

= 1.68 since x2 > 1, the final quality of steam is superheated. Problem 6.10 In a certain process 1 kg of steam initially has a dryness fraction of 0.85 and enthalpy 2500 kJ/kg. The final enthalpy is 2300 kJ/kg. Find the change in entropy during the process using the Mollier diagram. Solution In the Mollier diagram below the saturation line we find a line having x = 0.85. This line is the constant dryness fraction line having x = 0.9. Now, mark a point 1 on the vertical scale in which the specific enthalpy value is 2200 kJ/kg. Draw a horizontal line from the point to meet this constant dryness fraction line at A. From A, erect a vertical line that meets the x-axis, so, that the value of entropy is determined. Similarly, for the final enthalpy of 2300 kJ/kg, point 2 is marked on the y-axis. A horizontal line is fawn from point 2 to meet the constant dryness fraction line at B. From B a vertical line is erected to find the value of entropy (S2) in x-axis.

6.45

Properties of Pure Substance



s1 = 5.84 kJ/kg K



s2 = 6.64 kJ/kg

Therefore the change in entropy is Δs = s2 – s1 = 6.64 – 5.84 = 0.8 kJ/kg K.

Objective Type Questions 1. The dryness fraction of steam depends exactly on (a) The mass of dry steam present in the total mass of steam (b) The mass of wet steam present in total mass of steam (c) entropy of steam (d) none of these 2. When saturated liquid at 40°C is throttled to –20°C, the quality of steam at the exit is (a) 0.1869

(b) 0.212

(c) 0.231

(d) 0.788

3. For the dryness fraction of wet steam (a) x < 1

(b) x > 1

(c) x = 1

(d) x = 0

(c) x = 1

(d) x = 0

(c) x = 1

(d) x = 0

4. Dry steam has a dryness fraction (a) x < 1

(b) x > 1

5. Saturated liquid has a dryness fraction (a) x < 1

(b) x > 1

6. For the dryness fraction of superheated steam (a) x < 1

(b) x > 1

(c) x = 1

(d) x = 0

7. The critical temperature of steam is (a) Tc = 647.3 K

(b) Tc = 657.3 K (c) Tc = 627.3 K (d) Tc = 617.3 K

8. The critical pressure of steam is (a) pc = 231.2 bar

(b) pc = 211.2 bar

(c) pc = 221.2 bar

(d) pc = 201.2 bar

9. The critical volume of steam is (a) Vc = 0.00317 m3/kg

(b) Vc = 0.00307 m3/kg

(c) Vc = 0.00307 m3/kg

(d) Vc = 0.0027 m3/kg

10. The specific volume of wet steam is (a) xvg m3/kg

(b) x – vg m3/kg

(c) x + vg m3/kg

(d) x/vg m3/kg

6.46

Engineering Thermodynamics

11. The specific volume of dry steam is (a) x/vg m3/kg

(b) vg m3/kg

(c) x + vg m3/kg

(d) None of (a), (b) and (c)

12. The specific enthalpy of wet steam is (a) hf + xhfg

(b) hf – xhfg

(c) hfg – xhfg

(d) hf /xhfg

(b) sf – xsfg

(c) sfg – shfg

(d) sf /xsfg

Ts h fg

(c) Ts – hfg

(d) Ts + hfg

T (c)  sup  Ts

 T (d)  s  Tsup 

13. Specific entropy of wet steam is (a) sf + xsfg 14. sfg is equal to (a)

h fg Ts



(b)

15. Degree of superheat is given by (a) (Tsup – Ts)

(b) (Tsup + Ts)

  

   

16. Water has a critical specific volume of 0.0003155 m3/kg. A closed and rigid steel tank with a volume of 0.025 m3/kg, contains a mixture of water and steam at 0.1 Mpa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level of the mixture in the tank (a) Will fall (b) Will rise (c) Will remain the same (d) Either rise or fall depending on the amount of heat added.

Theory Questions 1. What is a pure substance? Give same examples. 2. Define wet and dry steam. 3. What do you mean by the degree of superheat? 4. How dose evaporation differ from boiling? 5. Draw a skeleton of the p-V diagram for water and show an isotherm passing from the compressed liquid state to the superheated vapour state through the vaporization process. 6. Determine whether water at the following states is a compressed liquid, a superheated vapour or a mixture of saturated water steam. (a) 18 MPa, 0.003 m3/kg

(b) 130°C, 200 kPa.

7. Draw the p-v-T surface for water, and also indicate its salient features.

6.47

Properties of Pure Substance

8. Find the change in the entropy of 1 kg of ice which is heated from –5°C to 0°C. It melts into water at 0°C. CPice = 2.093 kJ/kg K. The pressure during heating is maintained at 1 atm. The latent heat of fusion of ice = 334.96 kJ/kg. 9. Define the dryness fraction of steam. 10. What is the volume occupied by 5 kg of dry saturated steam at 10 bar? 11. A system contains air in the form of a liquid-vapour mixture in equilibrium. Can this mixture be treated as a pure substance? Justify your answer. 12. Explain with a neat sketch, the construction of the Mollier diagram and give its use in thermodynamic process representation. 13. Differentiate clearly between wet steam, dry steam, saturated steam and superheated steam. 14. Name all the lines available in the Mollier chart. 15. Derive the expression for the work done and the heat transfer of steam in a constant pressure process. Draw the p-V and T-s diagram. 16. Derive the expression for the work done and the heat transfer of steam in a constant volume process. Draw the p-V and T-s diagram. 17. Derive the expression for the work done and the heat transfer of steam in a constant temperature process. Draw the p-V and T-s diagram. 18. Derive the expression for the work done and the heat transfer of steam in a reversible adiabatic process. Draw the p-V and T-s diagram. 19. Derive the expression for the work done and the heat transfer of steam in a polytropic process. Draw the p-V and T-s diagram. 20. With a neat skecth explain, the method of determining the dryness fraction using the seperating calorimeter. 21. How do you determine the dryness fraction of steam, using the throttling calorimeter.

Unsolved Problems 1. A vessel of volume 0.04 m3 contains a mixture of saturated water and saturated steam at a temperature of 250°C. The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume, enthalpy, entropy and the internal energy. 2. Three kg of dry saturated steam at 10 bar expands to 2 bar according to an isothermal or isentropic process in a piston-cylinder arrangement. Compute the work done, heat transfer and change in entropy in each process. 3. One kg of steam is contained in an elastic balloon of spherical shape which supports an internal pressure proportional to its diameter. The initial

6.48

Engineering Thermodynamics

condition of the steam is saturated vapour at 110°C. Heat is transferred to the steam until its pressure reaches 200 kPa. Determine: (i) The final temperature (ii) The heat transferred. Take the Cp of steam = 2.25 kJ/kgK 4. Determine the volume change when 1 kg of saturated water is completely vaporized at a pressure of (1) 1 kPa (2) 100 kPa (3) 10,000 kPa. 5. It is desired to produce 1 kg of steam at 6 bar with the following qualities: (1) 90% dry (2) 100% dry (3) 250°C. The heat content of the water is 100 kJ/ kg. Find, using the steam tables, the quantity of heat required in each of the above cases. Take the Cp of superheated steam as 2.3 kJ/kg K. 6. A rigid vessel of volume 5 m3 contains 0.05 m3 of saturated liquid water and the rest of the volume as saturated vapour at 0.1 MPa. The heat is transferred until the vessel is filled with the superheated vapour. Determine the heat transferred, and the work done during the process. 7. One kg of steam initially dry saturated at 1.1 MPa, expands in a cylinder following the law pv1.3 = C. The pressure at the end of expansion is 0.1 MPa. Determine (i) the final volume (ii) the final dryness fraction (iii) the work done (iv) the change in the internal energy (v) the heat transferred.

CHAPTER

7

Real Gas Mixture

7.1 INTRODUCTION As already discussed in Chapter 2, there are two types of gases based on nature; (i) ideal gases and (ii) real gases. A gas, which obeys the gas laws and the general gas equation pV = mRT, for all values of temperature and pressure, is called an ideal or a perfect gas. A gas, which does not obey the gas equation or gas law, is called a real gas. There is no attractive or repulsive forces between two molecules, occupying no space. Ideal gases can be completely evaporated from the liquid state. The values of p, V and T calculated using the gas equation, differs from the observed values. The study of various gases reveals that all gases exhibit ideal behavior, only at high temperature and low pressure. At high temperature and low pressure the density of gases are low. Due to large intermolecular distance, attractive and repulsive forces are neglected, and low density satisfies the space occupying factor.

7.2

REAL GASES

At high pressure and low temperatures, there is a deviation in the behaviour of real gases from the behavior of the ideal gases. The causes for deviation of real gas from ideal gas behavior are: (a) Forces of attraction (b) Effect of the actual volume occupied by the molecule (c) Effect of the size of molecules (d) Effect of the mass of the molecule The dutch physican J.D Van der Waals was the first scientist to introduce the correction factor in the gas equation for real gases.

7.3

FORCES OF ATTRACTION

The first important reason for the deviation of real gases from the ideal gases is that certain forces of attraction exist between the molecules of the real gases. This force is called Van der Waal’s force of attraction.

7.2

Engineering Thermodynamics

7.3.1 Van der Waal’s Force of Attraction There are two factors affecting the Van der Waal’s force of attraction. (a) The effect of temperature and (b) The effect of pressure.

7.3.1.1  Effect of Temperature on Van der Waal’s Force The real gas deviates from the ideal gas due to Van der Waal’s force which has significant effect at low temperatures. The molecules move at low velocities at low temperatures. Therefore, the force of attraction between molecules cannot be overcome. At high temperatures, the molecules will be moving with high velocities. So, the molecules will oppose Van der Waal’s forces effectively and because of this, the deviation of the gas is the minimum.

7.3.1.2  Effect of Pressure An increase of pressure at constant temperature will affect the deviation of real gases from ideal gases considerably. This is because of the distance between the molecules decreases. As a result of this, the average distance between the molecules is small. Due to this, the force of attraction increases, so that the deviation of the behavior is more pronounced. At low pressure, intermolecular distance is more, as a result the force attraction is very less, and so the deviation of the behavior is also less.

7.3.2 Effect of the Actual Volume Occupied by the Molecules At high pressures, the total volume occupied by the gas is relatively very small. But, the fraction of the total volume actually occupied by the molecules themselves is appreciable. Therefore, the gases behave like real gases at high pressures. At low pressures, the total volume occupied by the gas is considerable, and the molecules keep themselves apart. So, the fraction of the total volume occupied by the gas molecules is small. Thus, the gases tend to behaves as ideal gases at low pressure.

7.3.3 Effect of the Size of Molecules The size of the molecules of the gas is a major factor at a given temperature and pressure. The attractive forces depend on the electron cloud of the molecules. The attractive forces are significant between large sized molecules. The attractive force is less between small sized molecules. Hence, at a given temperature and pressure, the real gases consisting of large molecules deviate from the ideal gases. Examples of gases which deviate the maximum from ideal gases are Carbon dioxide and Sulphur dioxide. The deviation is less in the case of Hydrogen and Helium.

7.3

Real Gas Mixtures

7.3.4 Effect of the Mass of Molecule It is obvious that larger molecules have a greater molar mass. The molecules of greater mass in a gas will move at slower velocities. So, the force of attraction is appreciable and hence, they deviate from an ideal gas. But the molecules of lower mass in a gas will move faster. Therefore the force of attraction is negligible. The gas deviates slightly from the ideal gas. The deviation of behavior from the ideal gas for certain real gases which is obtained experimentally in Fig. 7.1. N2 H2 CO2

pV

p

Fig. 7.1 Deviation of behavior from ideal gas of certain real gases

Figure 7.2. depicts the values of pV against pressure p at different temperatures. The value of pV for an ideal gas remains constant for all the values of pressure at a given temperature. 290 K

pV

298 K 323 K 343 K

p

Fig. 7.2 Variation of pV with p at different temperatures

This will give a horizontal line in the graph. For Carbon dioxide, the pV value decreases with an increase in the pressure at first and reaches a minimum value. Then the value pV continues to increase with the increase in pressure. For Hydrogen and Helium, the pV and pressure both increase. It is seen that as the temperature is raised, the depression in the curve becomes less. For Nitrogen, the depression is the maximum at 290 K, and the least at 343 K.

7.4

VAN DER WAAL’S EQUATION

From early definitions we understand that the real gases differ slightly from the behavior of ideal gas at high pressures and at low temperatures. The corrected equation of state for real gases was deduced by J.O. Van der Waal. There are two factors considered for the correction. They are:

7.4

Engineering Thermodynamics

(i) Size of molecules of the real gas. (ii) Force of attraction between the molecules of the real gas. B

A

Fig. 7.3 Real gas in a vessel

It is assumed that the force of attraction influences each molecule of a real gas. This force of attraction between molecules is called cohesion. This affects the movement of molecules when they are closer. Consider a real gas, which is contained in a vessel as shown in Fig. 7.3. Molecule A is inside the vessel, and molecule B is near the wall of the vessel. The neighbour molecules in all directions attract molecule A. Molecule B is attracted inside the wall, obviously there is no force of attraction on molecule B from the otherside of the wall. The inward force on molecule B will be the minimum at a distance equal to the radius of a molecule, and the maximum when, the molecule is about to strike the wall. The velocity during the striking of the wall is less than that of the molecule striking the wall without the intermolecular forces. The velocity of the molecule is reduced due to the intermolecular forces, which decreases the pressure on the wall. The decrease in the pressure depends on (a) Number of molecules present in the unit volume. (b) Number of molecules striking per unit area of walls of a vessel in a second. But, these factors are proportional to the density of the gas. Therefore, the decrease in pressure ∝ (density of gas)2

p ∝ (ρ)2 ..(7.1)



1 p ∝   V



p ∝

2

1 V2

a ...(7.2) V2 where a = Proportionality constant. or,

p =

7.5

Real Gas Mixtures

7.4.1 Correction for the Size of the Molecules Let us assume that the volume of a vessel is V. The molecules move freely inside the vessel. The size of the molecules is considerably small so that the effective volume available is the actual volume. The molecules are generally considered as spherical objects. Suppose the affected volume of the total molecules is “b”, then the effective volume is less when the molecules are in motion and less when they are at rest.

7.4.2 Correction on Pressure The actual pressure exerted by the gas molecules = Measured pressure (p) decreases the pressure of the molecules on the wall. a ...(7.3) v2 Comprising the two correction on pressure and volume, the gas equation become pm = p +



a    pm + 2  ( v – b ) = R T ...(7.4) v  



It may be written as p =



a RT – 2 . ( v – b) v

...(7.5)

Table 7.1 Values of a, and b for Various Gases for Van der Waal’s Equation

a (kNm4/kg-mol2)

b (m3/kg-mol)

Air

135.33

0.0362

CO2

362.85

0.0423

He

3417.6

0.0228

H2O(vapour)

551.13

0.03

Hg (vapour)

2031.9

0.065

O2

139.2

0.031

Gas

Equation (7.4) can be expanded as pv 3 – ( pb + RT ) v 2 + av – ab = 0

...(7.6)

The above equation is in cubic form of volume, and therefore the Van der Waal’s equation of state has three roots of each pressure.

7.6

Engineering Thermodynamics

7.4.3 Determination of Critical Pressure (pc), Critical Volume (v—c ) and Critical Temperature (Tc) ­

Figure 7.4 shows the different isotherms(constant temperature curves) for Van der Waals' gas on pressure volume (p-V) diagram. Tc pc

Isotherm

p

T2

T1

T3 T4 v

Fig. 7.4 Isotherms and critical pressure and temperature on p-V diagram

The critical point is the point at which all the three phases meet in the p-V diagram. It is seen from the figure, that only one value of specific volume can be obtained at a temperature, equal to or above the critical temperature. Also, three values of specific volumes can be obtained below the critical temperature. Let us take the Eqn. (7.5)

 RT   a  pe =   –  2  ...(7.7) v – b v 

The values of the constants “a” and “b” for different gases are different as the constants vary with properties of gas. Therefore, the values of “a” and “b” should be obtained empirically, for a particular region of pressure and temperature. At the critical point, the equation is written as

 RTc   a   –  2  ...(7.8)  vc – b   vc 

pc = 

Differentiating the above equation with respect to volume

 –RT  ∂ pc  c   =   ( v – b )2  ∂ vc   c

  2a   +   ...(7.9)   vc3  

It is seen from Fig. 7.4, that at temperatures below the critical temperature, the Van der Waal isotherms deviate much as the volume increases. An isotherm indicates those equilibrium states, at which the temperature is constant. Therefore, the slope of the isothermal curve in the p-V diagram is given by Eqn.(7.9).

7.7

Real Gas Mixtures

∂p  At critical point the slope is equal to  c  ; whose, value is zero.  ∂ vc T

It can also be said to be a point of inflection, as the isotherm at T3 approaches a concave shape upward to the left of this point, and concave downward to the right of this point. Therefore,

 ∂ pc   ∂ vc

  = 0 

..(7.10)

 ∂2 pc  2  ∂ vc

  = 0 

...(7.11)

And also

Substituting Eqn. (7.10) in Eqn. (7.9)  RT c   vc – b 2 



(

)

  2a   =  3  ...(7.12)   vc  

Again differentiating the Eqn. (7.10), we get

 ∂ 2 pc  2  ∂ vc

 2R T  c  =  3   ( vc – b )



 ∂ 2 pc  2  ∂ vc

  = 0 

...(7.14)

 2R T   6 a  c   –   = 0  ( vc – b )3   vc4 

...(7.15)





  6a   –  4  ...(7.13)   vc  

 2R T  c   =  6 a  ...(7.16)  v4   ( vc – b )3   c 

From Eqns. (7.12) and (7.15) We get

v c = 3 b ...(7.17)

Substituting v c = (3 b) in Eqn. (7.16), we get

 2R T   6a  c     ...(7.18) =  ( 3b – b )3   ( 3 b ) 4 

7.8

Engineering Thermodynamics



 6a  2R Tc  =   4 3   8 b   81 b

  ...(7.19) 

 6a  Tc =   ...(7.20)  276 R  Substituting the values vc and Tc in Eqn. (7.8) and we get

 a pc =  2  27 b

  ...(7.21) 



v c = 3 b ...(7.22)

 8a  Tc =   ...(7.23)  27 bR  Comparing the values of pc, vc and Tc from the above equations, we get

 RTc  8   =   = 2.667 ...(7.24) 3  pc vc 

The ratio RTc is called critical coefficient. pc vc

7.4.4 Demerits of Van der Waal’s Equation There is much deviation between the values of critical co-efficient obtained from the Van der waals' equation and that from experimentation. The proportional constant a and co volume b vary with temperature. Experimentally the value of vc is found to be 2 b.

7.4.5 Redlich-Kwang Equation An improved version of Van der Waal’s equation is formed by Redlich-Kwang.

 RT    a T  ...(7.25) p =     ( v – b )   ( v – b ) R

where,

 0.4275 R 2 Tc2.5   ...(7.26) a =  pc  



 0.0867 RTc  b =   ...(7.27) pc  

7.9

Real Gas Mixtures

7.4.6 Virial Equation —

The product of pressure and critical volume (v c) is given in a series expansion. It is given by where,

B C D pvc = 1 + 2 + 3 + 4 + ....... ...(7.28) RT v v v B, C, D = Second, third and fourth virial coefficients.

7.4.7 Dieterici Equation The Dieterici equation of state is given in the exponential form.  a  pc( v c – b) = exp   RT  RT v 

...(7.29)

Where the values of a and b are

 4R 2 T 2 a =  2 2 c  e p c 



 RT b =  2 c  e pc

  ...(7.30) 

  . ...(7.31) 

7.4.8 Beattie Bridge Man Equation The Beattie Bridge man equation is given by where,

 RT  A p =  2  ( I – ε ) ( v + B )  2  ...(7.32) v  v 



a  A = Ao 1 –  ...(7.33) v 



b  B = Bo 1 –  v 

...(7.34)



 c  ε = 1 –  . v T 3  

...(7.35)

There are five constants A, B, a, b and c which have to be determined for each gas. The values of the constants Ao, Bo, a, b and c for various gases are given in Table 7.2.

7.10

Engineering Thermodynamics Table. 7.2 Values of constants Ao, Bo, a, b and c

Gas

Ao x 103 N-m2/ kgmol2

a m3/ kgmol

Bo m3/ kgmol

b m3/ kgmol

c x 104 m3 – K3/ kgmol

O2

150.68

0.025

0.0461

0.1042

4.8

H2

19.9

–0.005

0.029

–0.044

0.05

N2

135.8

0.026

0.054

–0.007

4.2

CO2

507

0.071

0.1074

0.0723

66

NH3

241.2

0.17

0.0341

0.191

176.9

Air

131.5

0.0193

0.0461

–0.001

4.34

He

2.158

0.0598

0.0140

0.0

48

7.4.9 Berthelot Equation

p =

RT a − ...(7.36) v − b T vc 2

At critical points

pc =

RTc a ...(7.37) − v − b Tc vc 2



 ∂ pc   ∂ vc

  –RTc   2 a  ...(7.38) +  =  2 3   (vc − b )   Tc vc 

 ∂ pc    = 0  ∂ vc 

...(7.39)

 RT   2a  c So,      ...(7.40)  =  2 3  ( vc – b )   Tc vc  Again differentiating Eqn. (7.37) with respect to volume, we get

 ∂2 p  =  2c   ∂vc 

 2RT   6 a  c  = –  0 ...(7.41)  ( v – b )3   Tc vc4   c 

 ∂ 2 pc   2  = 0  ∂vc 

 2RTc        ( v – b )3  c

  6a =   T v 4  c c 

...(7.42)   ...(7.43) 

7.11

Real Gas Mixtures

From Eqns. (7.40) and (7.43) we get vc = 3 b ...(7.44)



b =

or,

vc ...(7.45) 3

Substituting in Eqn. (7.43), we get  

   

   2 RT  6a   c   =  4  ...(7.46)  ( 3 vc – b )3   Tc ( 3b )  

 2 RTc  3  8b

 6a    =  4  ...(7.47)  Tc ( 3b )    9 R Tc2 a =   8

Substituting b =

  × (3 b). 

...(7.48)

vc in Eqn. (7.47), we get 3 b =



RTc . ...(7.49) 8 pc

Therefore, the Berthelot equation becomes

 9 R Tc2 ( 3 b )   v – vc  RTc  pc +   =  8 Tc vc2   3  

  9 RTc2 p +   c   8 

 3b   2   Tc vc

...(7.50)

   v – vc  R Tc ...(7.51)     =    3 



 R Tc b =   8 pc

  ...(7.52) 



 9 R Tc2 ( 3b )   27 R 2 Tc3 a =   =  8    64 p c

  . ...(7.53)  

7.4.10  Mcleod Equation The Mcleod equation is formulated with four constants a, b’, B and c. The equation is given below a   RT  p + 2  ( v − b ') = v      

7.12

Engineering Thermodynamics

where

b’ = A – Bp + Cp2



a   p =  p + 2  v  

for normal substances b′ = v 2 .

7.4.11  Wohl’s Equation Wohl proposed a three constant equation of state. The equation is as follows;   c   RT   a  +  3    −   v – b   ( v – b) v   v  2 2   R Tc     pc

where

 96 a =  2  15

and

 R 3 Tc3 c =  256   3   15   pc2

7.5

  RTc   ; b =     pc 

  . 

COMPRESSIBILITY FACTOR

In order to show the difference of deviations for real gases from the ideal gas behavior a factor called the compressibility factor (Z) is used

Z =

pv v = ...(7.54) R T vi

where v = volume occupied by one mole of the gas at a given temperature; the compressibility factor Z is the function of temperature and pressure of the real gases. The amount by which the actual value varies from a value unity gives a measure of the deviation from the ideal behavior. Suppose   

( pv )ideal

= R T ...(7.55)

Then

Z =

pv ...(7.56) RT



Z =

pv Actual volume = . ...(7.57) Ideal volume ( pv )ideal

Figure 7.5 shows the variation of the compressibility factor with reduced pressure(pr) for different temperatures. It is apparent from the figure that as the reduced pressure tends to zero, the compressibility factor tends to value 1 at all temperatures. When the value of temperature tends to infinity, then the compressibility factor value becomes zero.

7.13

Real Gas Mixtures

T = 5 TC T = 3 TC Z

T = 2 TC

T = 2.5 TC

T = 1.4 TC

T = 1.2 TC T = 1.0 TC

T = 0 TC

Reduced Pressure (Pr) Fig. 7.5 Compressibility chart Tr = 1.7

Tr = 1.6

Tr = 1.5

pv– Z =  RT

Tr = 1.2 Tr = 1.10 Tr = 1.0

Reduced pressure (Pr)

Fig. 7.6 Generalized compressibility chart

Therefore, the compressibility factor is the ratio of the product of pressure and volume of a real gas to the product of pressure and volume of an ideal gas. It is also an index of the deviation of real gas from ideal behavior. The above Fig. 7.6, clearly indicates that gases like CO2, N2, CH4 have the value of Z decreasing initially and increasing finally for an ideal gas Z = 1, at all temperatures and pressures.

7.5.1 Reduced Properties The advantages of the compressibility chart are (i) Charts can be produced for low ranges of values. (ii) The deviation of a real gas to a perfect gas can be easily found out.

7.14

Engineering Thermodynamics

But, at the same time, a major disadvantage of the compressibility chart is that successive experimentation is required to determine the unknown properties. To avoid this, the properties must be quoted as fractions of the critical values of the concerned gases. For example;

pR =

p ...(7.58) pc



vR =

v vc

TR =

T Tc



...(7.59)



...(7.60)

These pR, vR and TR are shown as reduced properties. Suppose, the determined pR, vR and TR values are equal for two gases, whose area at the same pc and Tc. This is known as the law of corresponding states; if it become true then it is possible for us to show the variation of a graphical from for all the gases.

Z with pR, TR and vR in Zc

We know that

 pv  Z =   ...(7.61)  RT 

Substituting the values for p, V and T from the reduced fractions we get

 p × pR × vc × vR Z =  c R × Tc TR 

But Zc = Therefore, If

pc vc Tc



  ...(7.62)  ...(7.63)

p v  Z = Zc  R R  ...(7.64)  TR  vR = f(pR, TR), then

Z = f(pR, TR) for all gases. Zc

Unfortunately, such a set of curves will differ for all gases. It cannot be said that Z → 1 as pR → 0 for all the gases.

7.15

Real Gas Mixtures

.

Solved Problems Problem 7.1 Determine the specific volume of methane at 120°C and 60 bar. Compare the results with those obtained by the ideal gas equation. Given T = 120°C and p = 60 bar = 46.39 × 105 N/m2 Solution Get the values of Tc, pc and molecular mass (M) for methane. Tc = 190.7 K

pc = 45.8 atm = 46.39 × 105 N/m2



M = 16.043

—  Ru  ­R =   M —  8314  R =    16.043  —  8314  R =    16.043  — R = 518.23 kJ/kg K

 273 + 120  Tr =   Tc    273 + 120  Tr =    190.7 

Tr = 1.536  p  pr =    pc   60  pr =    46.39  pr = 1.29

From the compressibility chart obtain the value for "Z" for Tr and pr. We know that

Z = 0.92 pV = ZRT —

Substitute the values of Z, p, R and T and get the value for v.   v = 0.00953 m3/kg Now, the ideal gas equation is pV = mRT

 RT   v =   p 

Calculating the value for v,

7.16

Engineering Thermodynamics



 RT  v =    p 



v = 0.01037 m3/kg

Both the values obtained with the compressibility chart and ideal gas equation are Problem 7.2 Calculate the pressure exerted by the unit mass of CO2 at 100°C, if the specific volume is 1 m3/kmol. Use the Van der Waals' equation of state. Also, compare the results with these of the ideal gas equation. Given Specific volume = 1 m3/kmol Temperature

T = 100°C = 373 K

Solution (i) Specific volume V = 1 × 44 = 44 m3/kg.



Take the constant values a and b from the tables for CO2, and substituting them in the Van der Waals' equation. We know that Van der Waals' equation is



a   RT  p + 2  ( v – b) = v    RT   a  p =  – 2  v –b v 



a = 362.85 kNm4/kmol4



b = 0.0423 m3/kmol



R = 189 J/kgK



p = 1603.746 – 187.422



p = 1416.32 N/m2.

(ii) Ideal gas equation pV = mRT or,

 mRT  p =    V 

Therefore,

 mRT  p =    V 

= 1602.2 N/m2. Problem 7.3 One mole of n-butane at 200°C occupies 0.93 m3. Estimate the pressure using (a) the Redlich-Kwang equation (b) the ideal gas equation. Given

T = 200°C



V = 0.93 m3.

7.17

Real Gas Mixtures

Solution Get the values of the critical constant for pressure, temperature and molecular mass from the referred tables.

pc = 37.96 bar,

Tc = 152°C —  Ru  R =   M



R = 143J/kg K.

(a) Redlich-Kwang equation

 R T  a T  p =   ...(7.25)   v – b  ( v – b ) v 

where,

 0.4275 R 2 Tc2.5 a =  pc 



 0.0867 R Tc  –3 b =  =  1.388 × 10 ...(7.27)  p c  

  = 8575.39 ...(7.26) 

Substitute the values for a, b in Eqn. (7.23) to get the pressure.

p = 72838.817 – 18.782



p = 72.82 bar.

(c) Ideal gas equation or,

pV = mRT p = mRT/V.

Objective Type Questions 1. Nitrogen at an initial pressure of 10 bar, 1 m3 and 300 K is expanded isothermally to a final volume of 2 m3. The p-V-T relation is a   RT  p + 2  ( v – b) = v          where a > 0. The final pressure will be (a) Slightly less than 5 bar (b) Slightly more than 5 bar (c) Will be exactly 5 bar (d) Cannot be determined. 2. Real gases (a) Always obey the gas law and general equations (b) Sometimes they obey the gas law and general equation (c) Do not obey the gas law and general equation (d) Some time obey and sometimes do not obey.

7.18

Engineering Thermodynamics

3. The real gases deviate from the ideal gases because (a) Of their force of attraction (b) Of their pressures (c) Of their enthalpy (d) None of these 4. The real gas deviates from the ideal gas, due to Van der Waals' force which is mostly found to be at (a) Low temperatures (b) High temperatures (c) Equal temperatures (d) Equal masses. 5. An increase of pressure at constant temperature will affect the deviation of real gases from ideal gases, due to (a) Closeness of the molecules (c) Closeness of the temperature

(b) Closeness of the enthalpies (d) None of these

6. The attractive forces are high between (a) Large sized molecules (b) Small sized molecules (c) Medium sized molecules (d) None of these 7. The molecules of greater mass in a gas will move at (a) Slow speed (b) Medium speed (c) High speed (d) Depending on other factors 8. Van der Waals' equation is (a)  p + a  ( v – b ) = RT v2  

b (b)  p + 2  ( v – b ) = RT v  

b b   (c)  R + 2  ( v – b ) = PT (d)  R + 2  PT = ( v – b ) v  v    9. The force of attraction between molecules is called (a) Cohesion (b) Collision (c) Repulsion (d) None of these

Theory Questions 1. Write the four major causes for the deviation of real gases from ideal gases. 2. Discuss the different parameters that affect Van der Waal’s force of attraction, 3. Draw the generalized compressibility chart. 4. Discuss in detail the causes of the deviation of real gases from ideal gases. 5. Explain the construction, and the uses of generalized compressibility chart. 6. Sketch the compressibility chart and indicate its uses. 7. Determine the change in the internal energy and change in the entropy, when a gas obeys Van der Waals’ equation.

7.19

Real Gas Mixtures

8. Derive Van der Waals' equation. 9. Write the (i) Redlich-Kwang equation (ii) Virial equation (iii) Dieterici equation.(iv) Beattie Bridgeman equation. 10. Derive the Berthelot equation.

Unsolved Problems 1. (i) Determine the change of internal energy, enthalpy and entropy when a gas obeys Van der Waals' equation. (ii) Prove that the difference in the specific heat capacity is equal to

Cp – Cv = R and Cp – Cv = (TVB2)/KT.

2. A mixture of 2 kg of oxygen (MW = 32 kg/mol) and 2 kg argon (MW = 40 kg/mol) is present in an insulated piston/cylinder arrangement at 100 kPa, 300 K. The piston now compresses the mixture to half of its initial volume. Find the final pressure, temperature and piston work. Assume Cv for oxygen and argon as 0.6618 kJ/kg K and 0.3122 kJ/kg K respectively. 3. Calculate the pressure of steam at a temperature of 500ºC and a density of 24 kg/m3 using: (i) The ideal-gas equation (ii) The Van der Waals’ equation (iii) The Redlich-Kwang equation (iv) The compressibility factor and (v) The steam table. 4. A mixture of ideal gases consists of 3 kg of N2 and 5 kg of CO2 at a pressure of 300 kPa and at 20°C. Find (i) The mole fraction of each constituent, (ii) Equivalent molecular wt. of the mixture (iii) Equivalent gas constant of the mixture (iv) The partial pressures and partial volumes and (v) The volume and density of the mixture, and Cp and Cv values of the mixture. Assume the value of

Cp Cv

for CO2 = 1.286 and for N2 = 1.4.

CHAPTER

8

THERMODYNAMIC RELATIONS

8.1 INTRODUCTION Generally,thermodynamic properties such as pressure(p), volume(V), temperature(T), internal energy(u), specific enthalpy(h), specific entropy(s), Gibb's function(g) and Helmotz function(a) are used to specify the thermodynamic state of a system in which only p, V and T are directly measurable. The laws of thermodynamics along with the methods of the differential calculus enable to derive a number of useful general thermodynamic relations. It will be useful to introduce different combinations of some of the properties with p, V and T that will help to determine the other properties. The rate of change of one property with respect to the other can be obtained where the reaming properties are kept constant. The methods of partial differentiation play an important role in forming the general thermodynamic relations.

8.2

OVERVIEW OF MATHEMATICAL RELATIONS

Let us consider a thermodynamic property that is a function of x and y. The thermodynamic property is a variable.

Z = f(x, y) ...(8.1)

The partial a differentiation of the above equation is written as

 ∂z   ∂z  dz =   dx +   dy ...(8.2)  ∂x  y  ∂y x

where

 ∂z  A =   and B =  ∂x  y

 ∂z    dy  ∂y x

Then Eqn. (8.1) becomes;

dz = Adx + Bdy ...(8.3)

The partial derivatives of A with respect to “x” and B with respect to “y” can be given as

8.2

Engineering Thermodynamics



 ∂2 z   ∂A     =   ∂x x  ∂x ∂y x



 ∂B   ∂2 z    =   ...(8.5)  ∂y  y  ∂y ∂x  y

...(8.4)

Comparing Eqns. (8.4) and (8.5), we get  ∂B   ∂A   =   ...(8.6)          ∂ x  x  ∂y  y  ∂2 z   ∂2 z  = as           .  ∂x∂y x  ∂y∂x  y Note that when x, y and z are point functions, then the differentials are exact differentials. An example of this is explained below. We know that, the work done in a system is a function of two independent variables, the pressure and volume. but,

dW = d(pV) ...(8.7) W ≠ f(p.V)

...(8.8)

Differentiating the Eqn. (8.7) with respect to p and V we get

 ∂w   ∂w   dp +  dW =   dV ...(8.9)  ∂V  p  ∂p  v

We know that the work done in a constant volume is zero, and hence, the first term of Eqn. (8.9) is zero. Therefore, Eqn. (8.9) becomes

 ∂w  dW =   dV . ...(8.10)  ∂V  p

8.2.1 Cyclic Relation Consider the Eqn. (8.1) as “z” depends on two independent variables x and y The Eqn. (8.1) can be written as

x = f (z, y) ...(8.11)

Differentiating the above equation with respect to z and y, we get

 ∂x   ∂x  dx =   dy +   dz ...(8.12) ∂ y  ∂z  y  z

Substituting the expressions for dx in equation, we get

8.3

Thermodynamic Relations

  ∂z   ∂x   ∂z    ∂x   ∂z    dz =      +    dy +     dz ...(8.13)  ∂z y  ∂x  y   ∂x y  ∂y  z  ∂y x 

Or,

  

 ∂x   ∂z  dz =     dz +  ∂z y  ∂x  y

  ∂z   ∂z   ∂z        +    dy   ∂x y  ∂y  z  ∂y x 

 ∂x ∂z    dz 1 −      =  ∂z  y  ∂x  y  

 ∂z   ∂x   ∂z       +    dy ...(8.14)  ∂x  y  ∂y  z  ∂y x 

Here, y and x are the independent variables. “z” can be varied keeping “y“ as constant. In that case y = 0 and z # 0. In order to get this condition, the terms in the brackets must be equal to zero for any value assigned to y and z. Therefore, we shall write it as,  ∂x   ∂z      = 1  ∂z  y  ∂x  y



...(8.15)

Or,      

 ∂x   ∂x     ∂z  =  ∂z  ...(8.16)  y  y

And, ∂z      ∂x  y

 ∂x   ∂z  0 ...(8.17)   +  =  ∂y  z  ∂y x

By bringing the right side equation, we get  ∂z     ∂x  y

 ∂x   ∂z    =   = –1.  ∂y  z  ∂y x

...(8.18)

8.2.2 Gibbs Free Energy (or) Gibbs Function It is defined as the difference between the total enthalpy, and the product of the temperature and total entropy of a thermodynamic system.

G = H – TS

...(8.19)

Or, in specific terms g = h – Ts.

...(8.20)

The above equation is considered as a state function, and is related to the flow availability.

8.2.3 Helmotz Free Energy (or) Helmotz Function It is the difference between the total internal energy and the product of temperature and total entropy of a thermodynamic system.

8.4

Engineering Thermodynamics



F = U – TS

...(8.21)

Or, in specific

f = u – Ts.

...(8.22)

8.3

MAXWELL’S RELATIONS

According to the first law of thermodynamics dq = du + pdv ...(8.23)



From the entropy calculation we know that dq = Tds



...(8.24)

Substituting Eqn. (8.24) in Eqn. (8.23) Tds = du + pdv ...(8.25) du = Tds – pdv ...(8.26)

Or,

u = f (T, p)



...(8.27)

If Eqn. (8.26) is in the form of dz = Adx + Bdy

Or,

Then Eqn. (8.26) can be written as

 ∂p   ∂T  du =   dv –   dv ...(8.28)  ∂v S  ∂s  v

Where,

A = T  and  B = (–p)

Therefore,

 ∂T   –∂v    =   ...(8.29)  ∂v s  ∂s  v

Again considering the first law dq = dh – vdp ...(8.30)

Or,

Tds = dh – vdp ...(8.31)

Or,

dh = Tds + vdp ...(8.32)

In the above fashion, the terms in Eqn. (8.32) can be written as  ∂T   ∂v  dh =   ds +   dp ...(8.33) ∂ p  ∂s  p  s



 ∂T   ∂v    =    ∂s  p  ∂p s

...(8.34)

We know that the Helmotz function that

f = u – Ts

In differential form

df = du – (Tds + sdT)

...(8.35)

8.5

Thermodynamic Relations

Or,

df = du – Tds – sdT ...(8.36)

Substituting Eqn. (8.26) for (du), in Eqn. (8.36), we get

df = du – Tds – sdT



df = –pdv – sdT

...(8.37)

 ∂p   −∂s  df =   dT +  −  dv  ∂v T  ∂T  v

...(8.38)

With respect to

Comparing Eqn. (8.36) with the equation dz = Adx + Bdy

 ∂p   ∂s    =  ∂T  ...(8.39)  v  ∂v T

Consider Eqn. (8.22) for Gibb’s function g = h – Ts



It can be written in differential form as

dg = dh – d(Ts) ...(8.40)

Or,

dg = dh – (Tds + sdT)

Or, dg = dh – Tds – sdT

...(8.41) ...(8.42)

Substituting Eqn.(8.32) for (dh), in Eqn. (8.42) and simplifying we get,

dg = dh – Tds – sdT



dg = vdp – sdT.

...(8.43)

Again it can be written with respect to temperature as

 ∂s   ∂v  dg = –   dT +   dp ...(8.44) p ∂  ∂T  p  T

Comparing Eqn. (8.44), with Adx + Bdy, we get

 ∂s   ∂v    =   . ...(8.45)  ∂T T  ∂T  p

Equations (8.29), (8.34), (8.39) and (8.45) are known as Maxwell’s relations.

8.4 Tds EQUATIONS Let us assume that entropy s is a function of T and v

s = f (T, v) ...(8.46)

Partially differentiating we get

 ∂s   ∂s  ds =   dT +   dv ...(8.47)  ∂T v  ∂v T

8.6

Engineering Thermodynamics

Multiplying the above equation by T  ∂s   ∂s  Tds = T   dT + T   dv ...(8.48)  ∂T v  ∂v T Substituting Eqn. (8.48) in Eqn. (8.26), we get

  ∂s    ∂s  du = T   dT +  T   – p  dv ...(8.49)  ∂T v   ∂v T  We know that

u = f (T, v) ...(8.50)

Writing the partial differentiation of the above equation, we get

 ∂u   ∂u  du =   dT +   dv ...(8.51)  ∂T v  ∂v T

Comparing Eqn. (8.49) with Eqn. (8.51) We get,

 ∂u   ∂s    = T    ∂T  v  ∂T  v

 ∂u   ∂s  = T –p And,       =      ∂v T  ∂u T

...(8.52) ...(8.53)

By definition  ∂s   ...(8.54)  ∂T  v

cv = T 

We know from Maxwell’s Eqn. (8.39) that        

 ∂p   ∂s    =  v ∂  T  ∂T  v

Then Eqn. (8.48) becomes  ∂p  Tds = c v dT + T   dv ...(8.55)  ∂T  v Suppose

s = f (T, p)

 ∂s   ∂s  ds =   dT +   dp ∂ T  p  ∂p T

...(8.56)

Multiplying Eqn. (8.56) by T, we get  ∂s   ∂s  Tds = T   dT + T   dp ...(8.57)  ∂T  p  ∂p T By definition

∂s  cp = T    ∂T  p

8.7

Thermodynamic Relations

Consider Eqn. (8.54) and rearranging the terms, we get  ∂s  ∂v            = –   ...(8.58)  ∂T  p  ∂p T Substituting the above terms in Eqn. (8.57), we get  ∂v  Tds = c v dT – T   dp . ...(8.59)  ∂T  p Equations (8.55) and (8.59)are known as Tds equations.

8.5

USEFUL RELATIONS BETWEEN PARTIAL DERIVATIVES

Equating, (8.55) and (8.59),  ∂p   ∂v  Tds = c v dT + T   dp ...(8.60)  dv = c p dT – T   ∂T  p  ∂T  v ∂v   ∂p  T  T   ∂ T p  ∂ T  v dv + dp dT = Cp – Cv Cp – Cv

(



)

(

)

...(8.61)

T = f(v, p) ...(8.62)

 ∂T   ∂T   dp dT =   dv +   ∂v  p  ∂p 

...(8.63)

Comparing Eqns. (8.61) and (8.63), we get  ∂p  T  ∂ T   ∂T            =  ∂υ  p Cp – Cv

(

)



 ∂p  T  T ∂  ∂T  v            =   ∂v  p Cp − Cv

(

)

...(8.64)

...(8.65)

Equations (8.64) and (8.65) yield identical expression for (Cp – Cv)

 ∂v   ∂p  T          C p − C v =    ...(8.66)  ∂T  p  ∂T  v

 ∂p   ∂T   ∂v  But,     = −1    ∂T  v  ∂v  p  ∂p  T  ∂p   ∂v   ∂p               = −  ∂T  p  ∂v T  ∂T υ Substituting in Eqn. (8.66),

...(8.67)

8.8

Engineering Thermodynamics 2

 ∂v   ∂p  −T         C p − C v =     ∂T  p  ∂v T

...(8.68)

Equation (8.68) is an important one in thermodynamics. It shows that Cp 2 ∂v  can never be less than Cv since   is always positive. Thus the expression  ∂T  p Cp – Cv cannot be negative and Cp is always greater than Cv. The only exception is Cp = Cv, when water has the maximum density at 4°C. Equation 8.68 also shows that Cp → Cv as T → 0. This implies that at absolute zero temperature the two heat capacities are equal and g = 1. Ratio of heat capacities =



Cp Cv

∂s  T   Cp  ∂T  p = = g = ∂s  Cv T    ∂T  v

 ∂s     ∂T  p  ∂s     ∂T  v

...(8.69)

 ∂s   ∂T   ∂p  But,       = − 1  ∂T  p  ∂p s  ∂s T  ∂s   ∂T   ∂v        = −1  ∂T  v  ∂v s  ∂s T

\

 ∂T   ∂v       ∂v s  ∂s T g =  ∂T   ∂p       ∂p s  ∂s T

...(8.70)



 ∂p     ∂v s g = .  ∂p     ∂v T

...(8.71)

Since Cp is always greater than Cv, the slope of an isentropic process curve at any given state in the p–V diagram is always greater than that of the isothermal at the same point.

8.6

THE CLAUSIUS EQUATIONS

Consider two phases 1 and 2 of a pure substance in a system is equilibrium with each other. The system is closed so that the mass remains constant. If at in equilibrium, a small amount of phase 1 is transferred to phase 2, at constant temperature, the free energy change dg accompanying the transfer will be zero. If one mole of substance is transferred, maintaining equilibrium then

8.9

Thermodynamic Relations



∆g = g1 – g2 = 0

...(8.72)

Or,

g1 = g2.

...(8.73)

The pressure is a function of temperature only during the change of phase and it is not dependent on volume. Thus, the molar Gibbs free energy (chemical potential) for a single component system is the same for different phases in equilibrium. Now, suppose the pressure and temperature of the system are changed by very small amount of dp and dT respectively, so that the system is still in equilibrium. This change in parameters will result in a change in the gibbs energy in each phase. Let the changed values be g1 + dg1 and g2 + dg2 in the two phases respectively. Since the two phases are still in equilibrium, therefore,

g1 + dg1 = g2 + dg2 ...(8.74)

Substituting Eqn. (8.73) in Eqn. (8.74), we get dg1 = dg2



...(8.75)

dg = vdp – sdT

...(8.76)

Hence,

dg1 = v1dp – s1dT

...(8.77)



dg2 = v2dp – s2dT

...(8.78)

From Eqn. (8.43)

So,

v1dp – s1 dT = v2dp – s2dT ...(8.79)  dp   s2 – s1   ...(8.80)  =  dT   v2 – v1 

 dp   ∆s  or,           . ...(8.81) =  dT   ∆v 

∆s and ∆v are the differences of the molar entropies and molar volumes of dp the two phases respectively. is the rate of change of vapour pressure with dT temperature. If ∆h is the enthalpy change for the reversible transfer of a mole of substance from phase 1 to phase 2 at temperature T then

∆s =

Eqn. (8.81) then becomes  dp  =    dT 

∆h T

...(8.82)

   ∆h  ∆h =    . ...(8.83)  T ( v2 − v1 )   T∆v 

Equation (8.83) is known as the clapeyron equation. It is applicable to all phase transition of pure substance. If ∆h and ∆v are assumed to be constant, then 2  ∆h  2  dT            ∫1 dp =     ...(8.84)  ∆v  ∫1  T 

8.10

Engineering Thermodynamics

It gives

(p2 – p1) =

∆h  T2  ln   . ...(8.85) ∆v  T1 

This equation is preferred when the temperature interval is appreciable. The clapeyron equation may also be derived from the Maxwell’s relation.  ∂p   ∂s            =   ...(8.86)  ∂T  v  ∂v T

The vapour pressure of a substance does not depend on the volume, so, dp  dp  . Further phase transitions occur at constant   may be replaced by dT  dT  v ds  ds  temperature. Hence   may be written as   which is the rate of change  dv   dv T of entropy with volume. If ∆s is the change is the molar entropy accompanying ds ∆s a change in the molar volume ∆v, then = . dv ∆v  ∆p   ∆s  So,         . ...(8.87)  =  ∆T  v  ∆v T

8.7

EQUATION FOR INTERNAL ENERGY AND ENTHALPY

As per the first law of thermodynamics, when a system undergoes an infinitesimal reversible process between two equilibrium states, then consider Eqn. (8.26).

du = Tds – dw

From the Tds equations take Eqn. (8.55)

 ∂p  ds = Cv dT + T   dv  ∂T  v

Substitute the above equation in Eqn. (8.26) If

   ∂p  du = C v d T + T   dv  T – pdv ...(8.88) T ∂  v   u = f (T, v) ...(8.89)  ∂u    ∂u  du =   dT +   dv  ...(8.90)  ∂v T   ∂T  v

Comparing Eqns. (8.88) and (8.90),

  ∂p    ∂u    = T   – p  ...(8.91)  ∂v T   ∂T  v 

8.11

Thermodynamic Relations

Equation (8.91) is an important relation in thermodynamics. It helps to determine the change in internal energy, with respect to the change in volume at constant temperature from the equation of state. Conversely, if internal energy “u“ is known, a differential equation can be written, whose solution will give the equation of state. Consider Eqn. (8.25): Tds = du + pdv u = f (p, T) and v = f (p, T)

If then we can write,

 ∂u   ∂u  du =   dT +   dT  ∂T  p  ∂p  v

...(8.92. a)

 ∂v   ∂v  dv =   dp +   dT  ∂T  v  ∂p T

...(8.92. b)

and

Substituting du and dv in Eqn. (8.42) we get  

 ∂u    ∂v    ∂u   ∂v  Tds =   dp +   dT  + p   dp +   d T  ...(8.93)  ∂T  p   ∂T  p  ∂p  T  ∂p  T 

  ds =

 1  ∂u   p  ∂v   ∂u   ∂v    dp +   dT  +   dp +   dT  ...(8.94) T  ∂p T  ∂T  p  T  ∂p T  ∂T  p 

  ds =

 ∂v   1  ∂u  1  ∂u    ∂v    + p    dp +   + p  dT  ...(8.95) T  ∂p T T  ∂T  p  ∂T  p   ∂p T 

Differentiating the coefficient of dp with respect to T, and that of dT with respect to p gives  ∂  ∂u   ∂v   1  ∂u   ∂v   + p          + p   =    ...(8.96)  T ∂p  T  ∂T T ∂p  ∂T    v  ∂T ∂p This gives;

 ∂u   ∂v   ∂v       –T – v Tβ + pv K ...(8.97)  =   – p =   ∂T T  ∂T  p  ∂T T

8.8

EQUATION FOR ENTHALPY

Suppose

h = f (T, p) ...(8.98) ∂h dh =   dT + ∂  T p

 ∂h    dp ...(8.99)  ∂p T

8.12

Engineering Thermodynamics

But  ∂h    dT = Cp  ∂T  p



 ∂h  dh = Cp d T +   dp  ∂p T

Again

...(8.100) ...(8.101)

h = f (s, p) ...(8.102)

 ∂h   ∂h  dh =   ds +   dp ...(8.103)  ∂s  p  ∂p s



              

 ∂h   ∂s   ∂h   ∂h  =    –   ...(8.104) =   ∂p T  ∂s P  ∂p T  ∂p s  ∂h    = T  ∂s P

But

...(8.105)

 ∂s   ∂v            = –    ∂p T  ∂p s  ∂g   ∂h          = –=     v ...(8.106)  ∂p T  ∂p s From above three equations we get,  ∂h   ∂v           ∂p  = v – T  ∂T  ...(8.107) p  T Substitute Eqn. (8.105) in Eqn. (8.103), we get    ∂v  dh = CpdT +  v – T   dp  ...(8.108)  ∂T  p  

Therefore,

8.9

dh = Cp dT + [v – T β] dp.

...(8.109)

JOULE-THOMSON COEFFICIENT

The throttling process is described in Chapter 2. Generally the throttling process is an irreversible process, in which a gas passing through a restriction experiences a pressure drop. This is not similar to the expansion of gas through frictionless valve, a although in both the cases a pressure drop occurs. The expansion of fluid in the frictionless valve is considered as an internally reversible process. Let us consider a gas that passes through a porous medium from a high pressure region to a lower pressure one, in an insulated duct as shown in Fig. 8.1. The pressure and temperature on the high pressure side are arbitrarily

8.13

Thermodynamic Relations

chosen, and kept constant at p1 and T1. The pressure and the temperature of the gas measured are plotted in Fig. 8.2. Maintaining p1 and T1 constant, p2 is changed. The result is a set of discrete points, and they are plotted on the temperature – entropy diagram, as shown in Fig. 8.3. Thermometers T1

T2

Porous

Gas

Insulation

p1

p2

Pressure gauges

Fig. 8.1 Gas passes through a porous Inversion point 5

3

4

2

T 5

1

p

Fig. 8.2 T-p Diagram Inversion curve

Constant enthalpy lines

T

s

Fig. 8.3 T-s diagram

8.14

Engineering Thermodynamics

The specific enthalpy is the same at all points (1, 2, 3, 4 and 5) and a curve drawn through these points represents a constant enthalpy curve. However, it is to be noted that this curve does not represent the throttling process undergone by the gas, as the process is not reversible, and the intermediate states undergone by the gas cannot be represented by thermodynamic coordinates. It is not a curve of the throttling process, but the curve passes through points of equal enthalpy. Figure 8.2 shows that, if a throttling process takes place between p1 and T1 and p3, the temperature increases, whereas between (p1, T1 ) and at p3 there is a drop in the temperature. Therefore, there is a part of maximum enthalpy on the curve and this is called the inversion point. The initial pressure and temperature of the gas are set to new values, and by throttling to different states, a family of constant enthalpy curves are obtained for the gas, as shown in Fig. 8.3. The curve passing through the maxima of these constant enthalpy curves is called an inversion curve. The slope of the constant enthalpy at any point is the called Joule-Thomson coefficient and is given by

 ∂T   ...(8.110) mj =   ∂p h

The region inside the inversion curve where mj is positive is the cooling region the whereas the region outside the inversion curve where mj is negative is called heating region. In the cooling region, with a decrease in pressure, i.e., while throttling, the temperature decreases whereas in the heating region with throttling the temperature increases up to the inversion point. Consider Eqn. (8.32) dh = Tds + vdp If



s = f (T, p), then  ∂s   ∂s  dh = T   dT +   dp + vdp ...(8.111) ∂ T  p  ∂p T  ∂s    ∂s  = T   dT + T   + vdp   ∂T  p  ∂T T 

...(8.112)

 ∂s   ∂v  Using Maxwell relation,   =    ∂p T  ∂T  p and

∂s  Cp = T   ...(8.113)  ∂T  p

Then Eqn. (8.112) becomes,

  ∂s   dh = C p dT +  v – T     ∂T  p  

...(8.114)

8.15

Thermodynamic Relations



 ∂T  1  ∂vT   = mj =    – v ...(8.115) p C ∂ p  ∂T  p  h

For an ideal gas

pv = RT

R  v  ∂v  = =    ...(8.116)            ∂T  p  p   T 



 1 Tv v  –  = 0 ...(8.117) mj =   Cj v 1   

There is no change in temperature, when an ideal gas undergoes a JouleThomson coefficient (i.e., Throttling). For achieving the effect of cooling by the Joule-Thomson process, the initial temperature of the process must be below the point where the inversion curve intersects the temperature axis, i.e., below the maximum inversion temperature. For, most of the substances, the maximum inversion temperature is above the normal room temperature. Hence, the cooling can be achieved by the Joule-Thomson effect. However, in the case of hydrogen and helium, the gas is to be precooled below the maximum inversion temperature in the heat exchangers before they are throttled.

8.10 THERMODYNAMIC RELATIONS USING Cp AND Cv Consider the equation for small change in enthalpy dh = CpdT ...(8.118) Then,  ∂u   ...(8.119) Cp =   ∂T  p  ∂u  Cv =   .  ∂T  v

Case I We shall write a few equations in various forms, using the relation between the partial derivatives of temperature, specific enthalpy and specific entropy.  ∂h   ∂h   ∂s  or,         =    ...(8.120)  ∂T  p  ∂s  p  ∂T  p We know that  ∂h             = T ...(8.121)  ∂s  p

8.16

Engineering Thermodynamics

Substitute Eqn. (8.118) in Eqn. (8.120), we get  ∂h   ∂s            = T  ...(8.122) ∂ T  p  ∂T  p ∂s  Cp = T   ...(8.123)  ∂T  p Differentiating equation in both sides with respect to “p”, we get  ∂2s   ∂C p            ...(8.124)  = T  ∂p T  ∂p∂T  = T

∂  ∂s    ...(8.125) ∂T  ∂p 

We know that ∂s   ∂v  ...(8.126)           =   ∂T T  ∂T  p Substituting Eqn. (8.126) in Eqn. (8.125)  ∂C p  ∂  ∂v  Hence,       =T   ...(8.127) ∂T  ∂T  p  ∂p T

 ∂2 v   ∂C p  or,               = – T  2  . ...(8.128)  ∂p T  ∂T  p

Case II We shall write an equation relating between specific internal energy, temperature and specific entropy  ∂u   ∂T   ∂s           = – 1 ...(8.129)  ∂T  v  ∂s u  ∂u T  ∂u   ∂u   ∂s  or,          = –    ...(8.130)  ∂T  v  ∂s  v  ∂T  v We know that   Hence,

 ∂u    = T ...(8.131)  ∂s  p  ∂u   ∂s  Cv =   = −T  ...(8.132)  ∂T  v  ∂T  v

 2  Now,       ∂C v  = T  ∂ s  = − T ∂  ∂s  ∂T  ∂v T  ∂v T  ∂v ∂T 

...(8.133)

8.17

Thermodynamic Relations

 ∂p   ∂s  But,           =    ∂v T  ∂T  v

...(8.134)

∂C ∂  ∂p  So,        v  = − T   ...(8.135) ∂T  ∂T  v  ∂v T  ∂2 p  ∂C or,        v  = − T  . ...(8.136)  ∂T 2   ∂v T  

Case III Now by equations  ∂s   ∂s  C  – T  ...(8.137) p – Cv = T   ∂T p  ∂T  v Let

s = f (T, p) ...(8.138)  ∂s   ∂s  ds =   dT +   dp ...(8.139)  ∂T p  ∂p T

 ∂s   ∂s   ∂s  So,         = T    +   ∂T v  ∂T  p  ∂p T

 ∂p  ...(8.140)    ∂T  v

 ∂s   ∂p   ∂s   ∂s   –  = –    ...(8.141) or,      ∂T p  ∂T  v  ∂p T  ∂T  v  ∂p   ∂s  – T Hence, C  –   ...(8.142) p – Cv =  ∂T T  ∂T  v

 ∂s  ∂v  But,         = – p   ...(8.143) p ∂ ∂  T  T Now variable p, v, T we have ∂v   ∂T   ∂v                = − 1 ...(8.144)  ∂T v  ∂v  p  ∂p T  ∂p   ∂v   ∂p  ...(8.145) or,             = –  ∂T P  ∂v T  ∂T v Substituting both the equations 2  ∂v   ∂p  Cp – Cv = –T     . ...(8.146)  ∂T  p  ∂v T

8.18

Engineering Thermodynamics

Case IV Now since and

So,

1  ∂v   ...(8.147) b =  v  ∂T  p K =

1  ∂v    ...(8.148) v  ∂p T

 ∂v    = bv ...(8.149)  ∂T  p

 ∂p   1  and         = –   ...(8.150)  Kv   ∂v T From the equation we get  –1  Cp – Cv = (–T)(bv2)   ...(8.151)  Kv 

 T b2 v  Cp – Cv =   . ...(8.152)  K 

8.11 ISOTHERMAL COMPRESSIBILITY It is the relative change in volume with pressure at a constant temperature. It is denoted as K. The negative sign is used because specific volume decreases with the increase in the pressure; therefore, the partial derivative carries a negative value. The values of K can be obtained from the standard table of p–v–T data for any substance.

1  ∂v  K = –   ...(8.153) v  ∂p T

8.12 COEFFICIENT OF LINEAR EXPANSION AND VOLUME EXPANSION Consider a material which is heated upto a certain temperature. The material has tendency to change in its length, with respect to a change in temperature. This is known as the thermal expansion of the material. The particles or molecules of the material start moving, and there is a potential for separation. From the basics of physics, the coefficient of linear expansion (δ) is defined as the ratio between the degree of expansion and the change in temperature.

 1   ∂L   ...(8.154) δ =     L   ∂T  p

Thermal expansion is inversely proportional to the bond energy of the solid material.

8.19

Thermodynamic Relations

The coefficient of volume expansion is applicable to solids, liquids and gases. The coefficient of volume expansion is also known as volume expansivity, αp. It is defined as the ratio between the change in volume to the change in temperature, when the pressure remains constant.

 1   ∂v  αp =      v   ∂T  p

...(8.155)

The coefficient of volume expansion is three times greater than that of the coefficient of linear expansion.

αp = 3δT

...(8.156)

8.13 ISOTHERMAL COMPRESSIBILITY AND ADIABATIC COMPRESSIBILITY Another term, isothermal compressibility is the ratio between the change in volume with respect to pressure, when the temperature remains constant.

 1   ∂v  βT = –      v   ∂p T

...(8.157)

The reciprocal of isothermal compressibility is known as the isothermal bulk modulus. It is given as

 ∂p  βc = – v    ∂v T

...(8.158)

Adiabatic compressibility is the ratio between the change in volume with respect to pressure, when the entropy remains constant. It is given as  1   ∂v  βa =      v   ∂p s

...(8.159)

The reciprocal of adiabatic compressibility is known as the adiabatic bulk modulus.

 ∂p  βab = – v    ∂v s

...(8.160)

The velocity of sound (C) in a medium can be expressed, in terms of the adiabatic bulk modulus.

 ∂p  C2 =    ∂β s

...(8.161)



∂p C2 = –v2    ∂v s

...(8.162)



C2 = vβab

...(8.163)

It should be noted that the speed of sound of a gas is less than that for a liquid.

8.20

Engineering Thermodynamics

The values of the coefficient of volume expansion, isothermal compressibility and adiabatic compressibility are considered as the thermodynamic properties of a substance and these values can be obtained the from the standard data book for physical properties.

Solved Problems Problem 8.1 An ideal gas is filled in a container at 305 K and 0.8 m3/kg. The gas changes from its initial state to 310 K and volume 0.82 m3/kg. Calculate the change in the pressure of the gas. Solution We know that

p = f (T, v)

 ∂p   ∂p  dp =   +   ...(1) T ∂   v  ∂v T  RT  Taking the ideal gas equation for unit mass of gas ; p =    v   ∂p   RT     =   ∂T  v  v           

 ∂p  And         ∂v s ∂p   ∂p  Substituting the values of   and   in Eqn. (1)  ∂T   ∂v   RT   RT  dp =  –   v   v2  Change in temperature dT = 5 K Change in volume    dv = 0.2 m3/kg Substituting these values in dp;

0.287x   0.287 × 0.2  dp =  –   0.82   0.82 2 

dp = 0.0768 Pa. Problem 8.2 An ideal gas stored in a square container has increased in its pressure from 1 bar to 100 bar, when the temperature 20°C during a process. Determine (i) the workdone (ii) change in entropy (iii) heat transfer (iv) change in the internal energy of the gas per kg. Take the following data; (i) Volume expansivity     αp = 5 × 10–5 K–1

(ii) Isothermal compressibility βc = 8.6 × 10–12 m3/N (iii) Specific volume       v = 0.0012 m3/kg

8.21

Thermodynamic Relations

Solution (i) Work done during isothermal compression is w = ∫pdv



We know that the isothermal compressibility is  1   ∂v  βT =      v   ∂p T



v βT dpT = –(dv)T

Therefore, for the constant temperature process per kg of mass,

w =

(

vbT p 22 – p12

)

2 0.00012 × 8.6 × 10–12 × (100 2 – 12 ) w = 2 3 w = J/kg = 0.375 J/kg. 8 For determining the change in entropy, consider the Maxwells equation and the coefficient of volume expansivity.  ∂s   ∂v   1   ∂v  =  =     p T ∂ ∂    v   ∂T  p p           T  ∂s    = v αp  ∂p T If v and αp remain constant, then

(s2 – s1) = v αp (p2 – p1)

(s2 – s1) = 0.000112 × 5 × 10–5 × (100 – 1) (s2 – s1) = 0.0554 J/kg K.

The change in internal energy can be calculated using the first law; (u2 – u1) = q – w Calculating

q = T(s2 – s1)

q = 16.24 J/kg

(u2 – u1) = 16.24 – 0.375 (u2 – u1) = 15.865 J/kg.

Problem 8.3 In a scientific lab, steam at 600 kPa and 400 kPa was taken as a sample. Check the validity of the Maxwells relation  ∂s   ∂s    =  ∂ p  ∂T  p          T

p1 = 600 kPa and p2 = 400 kPa

8.22

Engineering Thermodynamics

Solution We know that  ∂s   ∆s   s1 – s2           =  =   .  ∂p T  ∆p  p  p1 – p2  Take the values of entropy of steam at 600 kPa and 400 kPa from the steam tables and substitute in the above equation. We will get 7.5662 – 7.3714  ∆s    = 600 – 400  ∆p 



 ∆s    = 9.69 × 10–4.  ∆p  Similarly, take the values of the volume and temperature of steam at 600 kPa and 400 kPa from the steam tables and substitute in the equation given below



 ∆v   ∂v     =   ∆T   ∂T 



 v1 − v2   ∂v    =   ∂ T    T1 − T2 

 v − v2  =  1   T1 − T2  = 9.69 × 10–4 Since both the values are equal, the Maxwell's relation holds good. Problem 8.4 Derive Clayperon-Clausius equation. Solution Let us consider the first law equation; dh = Tds + vdp During the phase change, pressure remains constant and hence dp = 0 Therefore,

dh = Tds

 dh  ds =    T  Integrating the above equation or,

 dh 



∫ds = ∫  T 



sf g =

h fg Tf

∂p   s fg Substituting sfg in  =  ∂T   v fg

   

8.23

Thermodynamic Relations

 ∂p   h fg We will get       =   ∂T   Tv fg

   

This equation is known as the clapeyron equation. For saturation, vf cv (b) The density is the minimum and cp = cv (c) The density is the maximum and cp = cv (d) The density is the maximum. 11. Which combination of the following statements is correct? (i) A gas cools upon expansion only when its Joule-Thomson coefficient is positive in the temperature range of expansion (ii) For a system undergoing a process, its entropy remains constant only when the process is reversible (iii) The work done by a closed system in an adiabatic is a point function (iv) A liquid expands upon freezing, when the slope of its fusion curve on the pressure temperature diagram is negative: (a) R and S (c) Q, R and S

(b) p and Q (d) p, Q and R.

Theory Questions 1. Define the Joule-Thomson coefficient. Why is it zero for an ideal gas? 2. Write the Maxwell’s relations and their significance. 3. Deduce the Maxwell’s relations and from the third relation, deduce the Clausius clapeyron equation. Also, apply this equation to the vaporization process for a pure substance. 4. Entropy is a function of any two properties, like p and V, p and T, etc. For a pure substance with the help of Maxwell’s equation, prove

8.27

Thermodynamic Relations

(i) Tds = Cv . dT + T [β/k] . dv (ii) Tds = Cp.dT – v . β . dp . T (iii) Tds = [kCv/β] . dp + [Cp/vβ] . dv 2  ∂v   ∂p  5. Prove that Cp – Cv = −T     . What are the facts that can be inferred  ∂T  p  ∂v T from the above equation?

6. Deduce the Maxwell's relations and the Clausius-Clayperon equation. 7. Derive the Tds equation when: (i) T and v are independent (ii) T and p are independent (iii) p and v are independent 8. Explain and derive the: (i) Joule-Thomson effect (ii) Clausius-Clapeyron equation.

Unsolved Problems  Cp  k 1. Determine the relation Tds =   C v dp +   dv. b  vb  2. Derive the following relations using the partial differentials:  ∂u  pv K – Tvb (i) =    ∂p T  ∂u   Tb  (ii)   =  –p  ∂v T  k  3. With the Jacobian method prove the following relations:

 vb   ∂h   = T +   ∂s   Cv k 

(i) 

4. A gas follows the Clausius equation of state. The initial pressure and temperature are p1 and T1 respectively. The final pressure and temperature are p2 and T2 respectively. If the enthalpy of the gas at the initial condition is known, find the change in the enthalpy of the gas. 5. Prove that the specific heat capacity (Cv) of real gas is a function of the temperature only. Take Van der Waal’s equation for real gas to solve the problem. 6. In an experimental observation, the densities of ice and water at 0°C are 915 kg/m3 and 1000 kg/m3 respectively. The latent heat of water at 0°C is –333.4 kJ/kg. Calculate the pressure to be applied, to melt the ice at –2°C.

8.28

Engineering Thermodynamics

7. The vapour pressure of a gas is 53.33 kPa at 60.6°C, and the normal boiling point is 80.1°C. Assume that the gas follows the ideal gas equation. What is the latent heat of vaporisation of the gas? 8. A rigid vessel is filled with water at 27°C and 0.1 MPa, and it is heated till the temperature rises to 50°C. Determine the pressure developed in the vessel, if the coefficient of volume expansion and isothermal compressibility of water are 2 × 10–4 K–1 and 5 × 10–4 MPa–1. 9. The latent heat of vaporisation of water at 100°C is 2257.0 kJ/kg. If the Cpf of water is 4.263 kJ/kg K and Cpg = 1.408 kJ/kg K, calculate the latent heat of water at 40°C. 10. Show that the slope of a constant volume process in a Mollier diagram is given by    

 vb   ∂h    = T +  ∂ s    Cv k 

11. Show that the Joule-Thomson coefficient for Van der Waal’s equation is  2 a( v − b )2 − RTbv2  m =    RTv3 − 2 a( v − b )2  12. Water has to be heated in a steam generator at a pressure of 1 MPa. Assume that the latent heat of vaporisation is 2257 kJ/kg and the normal boiling point of water is 100°C. Calculate the boiling point of water at 1 MPa. given by

13. Derive the expression for the enthalpy of a substance with respect to pressure, while the temperature is constant. Express in terms of measurable quantities. 14. Why does the fusion curve of water have a negative slope? Explain. 15. Water is boiled at 100°C on a hill top which is 2000 m above from sea level. Is it possible to boil?. If yes, justify.

 ∂b  ∂b  16. Prove the following relation;   =  . ∂ P   v  ∂p  17. Calculate the Joule-Thomson coefficient for water at 1 MPa and 250°C.  ∂p   7   ∂p  18. Prove that   =    . If the ratio of the specific heats is taken as 1.4.  ∂T s  2   ∂T  v 19. Derive the expression

 ∂p   ∂v  Cp = – T   .    ∂T s  ∂T  p



20. Derive the expression      

 ∂v   ∂p  Cv = T     .  ∂T s  ∂T  v

CHAPTER

9

PSYCHROMETRY

9.1 INTRODUCTION Psychrometry is the study of dry air and water vapour mixture. It should be noted that oxygen and nitrogen are the major constituents of air, and the minor are carbon dioxide, water, vapour, organ air is assumed to be a mixture of these two gases only. Psychrometry finds extensive application right from air conditioning to paper and textile industries involving hygroscopic materials like paper and textiles. The study of the properties of humid air is also useful in metrology and design of air conditioning equipment and cooling tower in power plants. In this chapter, the important terms used in the study of psychrometry, the instruments used to measure the air properties, different processes involved in the psychrometry and the use of psychrometry chart are discussed.

9.2

TERMS AND DEFINITION

A few important terms of psychrometry are given below: 1. Dry air 2. Dry bulb temperature 3. Wet bulb temperature 4. Dew point temperature 5. Absolute humidity 6. Relative humidity

9.2.1 Dry Air It is a mixture of oxygen, nitrogen, carbon dioxide, hydrogen and other gases like organ, helium, krypton, xenon. Dry air is assumed to have no water vapour The gravimetric and volumetric percentage of these gases are given in Table 9.1. But, it is impossible to get dry air without moisture. The sum of pressure of dry (pa) are and water vapour (pv) is equal to the barometric pressure (pb)

pb = pa + pv

9.2

Engineering Thermodynamics Table 9.1 Constituents of Dry Air (Gravimetric and Volumetric analyses) and Their Composition

Constituent

Volumetric

Gravimetric %

Molecular mass

Nitrogen

78.03

75.46

28

Oxygen

20.99

23.18

32

Argon

0.94

1.3

40

Carbon monoxide

0.03

0.06

44

Others

0.01

0.02

9.2.2 Dry Bulb Temperature The temperature measured by a thermometer with a dry bulb is called dry bulb temperature. This is purely free from the influence of moisture present in air. It is denoted as Tdb.

9.2.3 Wet Bulb Temperature When a thermometer is covered by a wet wick, or cloth and it measures the temperature of air then the temperature is called wet bulb temperature. It is denoted as Twb. The difference between the dry bulb temperature and wet bulb temperature is called wet bulb depression and it is a measure of the relative humidity.

9.2.4 Dew Point Temperature It is defined as the temperature of moist air, in which water vapour starts to condense continuously at constant pressure. The T-s diagram of the constant pressure cooling of moist air is shown in Fig. 9.1. The dew point is denoted as Tdp.

ps

1

Tdb pv

2

T

s

Fig. 9.1 Moist air in T-s diagram

Tdp

9.3

Psychrometry

The partial pressure of water vapour in the air is equal to the saturation pressure of water at the dew point temperature.

9.2.5 Specific Humidity or Humidity Ratio The amount of water vapour present in the moist air per unit mass of dry air is also known as absolute humidity or the humidity ratio. It is denoted as ω. ω =

mass of water vapour present in moist air in a given volume mass of dry air in same volume

mv v   = = mass of water vapour per kg of dry air ma v where,

mv = mass of water vapour in moist air



ma = mass of dry air



v = volume of given mixture



ω =

va vv

V  V = v= va   v, ma  mv 

Consider the perfect gas equation for water vapour

pvV = mv RvT ...(9.1)

Similarly, for dry air

paV = ma RaT ...(9.2)

Dividing Eqn. (9.2) by Eqn. (9.1)

pa V ma R a T pv V = m R T v

v

...(9.3)

ma R a pa = . ...(9.4) mv R v pv

The values of the gas constants for dry air and water vapour are taken as Ra =

8.314 = 0.287 kJ/kgK 28.9

Rv =

8.314 = 0.4615 kJ/kgK 18.05



pa ma × 0.622 ...(9.5) = pv mv

9.4

Engineering Thermodynamics

mv pv = 0.622 ma pa



...(9.6)



If the barometric pressure (p) is considered then p = pa + pv



pa = p – pv

or,

...(9.7)

pv mv = 0.622 ...(9.8) ma p – pv

So,

(

ω = 0.622



)

pv

( p – pv )

...(9.9)

9.2.6 Relative Humidity This is the ratio of actual partial pressure of water vapour present to the saturation pressure of water vapour at the same temperature and pressure. It is expressed in percentage. It is denoted as j. This is shown in Fig. 9.2. j=

Actual partial pressure of water vapour present in moist air taken Partial pressure of water vapour in saturated air taken at same pressure and temperature

ps T

Tdp

pv

s

Fig. 9.2 Partial pressure of water vapour in the T-s diagram

pv ...(9.10) ps The relative humidity varies place to place.

j =

9.2.7 Degree of Saturation The ratio of the mass of water present in the unit mass of dry air to the mass of water vapour present in the unit mass of dry air saturated at same temperature is called the degree of saturation or the saturation ratio. The degree of saturation is denoted as μ.

9.5

Psychrometry

Thus,



 pv  0.622  ( p – pv )  pv ( p – ps ) ( mv )T  m = ...(9.11) = = × ( ms )sat  ps  ps ( p – pv ) 0.622  ( p – ps )  

       μ = j ×

μ =

( p – ps ) ( p – pv ) ...(9.12)

pv . ...(9.12.a) ps

If the relative humidity j = 0, pv = 0 then μ = 0. If the air is saturated, then the relative humidity j = 1, pv = ps and m = 1 . It can be seen that μ also ranges from 0 to 1.

9.2.8 Carrier Equation for Partial Pressure of Water Vapour An empirical relationship was formulated by Carrier, to calculate the partial pressure of water vapour in the moist or humid air.         pv  = ( pvs )WBT –

9.3

( p – pv )WBT ( Tdb – Twb ) 1.8 . ...(9.13) 2800 – 1.3 ( 1.8 Tdb + 32 )

PSYCHROMETER

It is a device used to measure the dry and wet bulb temperatures of moist air. There are a few types of psychrometer available to record the above quantities. (i) Sling psychrometer (ii) Modified psychrometer (iii) Relative humidity sensor (iv) Dew point sensor (v) Hygrography recorder

9.3.1 Sling Psychrometer Figure 9.3(a) shows the components of a sling psychrometer. It consists of two thermometers. The arrangement is shown in Fig. 9.3(b). The thermometers contain mercury or alcohol. Both the thermometers are mounted on a wooden or steel plate. The bottom of the plate has a provision for a container filled up with water. One of the thermometers has a wick out

9.6

Engineering Thermodynamics

of its bottom the which will be kept in water. This enables the thermometer to measure the wet bulb temperature. At the top of the plate a handle is provided. The handle rotates the psychrometer. When the psychrometer stops its rotation the wet bulb temperature and dry bulb temperature are measured. This enables the thermometer to give an accurate reading of the wet bulb temperature, which would be affected by air velocity across the wet bulb. Swivel joint

Glass casing

Swivel handle

Sling psychrometer

Supporting frame

Dry bulb thermometer

Wet bulb thermometer

(a)

(b) Fig. 9.3 (a) and (b)

9.4

SPECIFIC ENTHALPY OF MOIST AIR

The total heat or specific enthalpy of moist air is generally determined as the sum of the specific enthalpy of dry air and water vapour, which is normally in the superheated condition in air water vapour mixture. Specific enthalpy hm = ha + ωhv

...(9.14)

where,

hv = specific enthalpy of water vapour



ha = specific enthalpy of dry air, kJ/kg



ω = specific humidity, kg/kg of dry air,

Cpa = specific heat of dry air, kJ/kgK Tdp = Dew point temperature, K

ha = Ca Tdb



hv = specific enthalpy of water vapour per kg of dry air

T. temperature, kJ/kg = (hw)dp + Cp (Tdb – Tdp )

...(9.15)

Therefore, hm = Cp Tdb + ωhω.

...(9.16)

a

a

9.7

Psychrometry

9.5

ADIABATIC SATURATION

An insulated device which is called the adiabatic saturator is shown in Fig. 9.4. Let us consider that ambient air steadily enters the adiabatic saturator. It passes through the water vapour sprayer. Now, assume that equilibrium is attained between the air and water droplets in the saturator. The saturated air and water vapour will leave the saturator, so that both the temperatures of the air and water vapour are the same. The temperature at which the mixture of air and water vapour leaving the saturator is equal to the water in the saturator, is called the adiabatic saturation temperature. 1

2 Insulated wall Saturate air leaving T1, 1 hair

Unsaturated air entering

2 1 Water filled in saturator (h1) Pump Entry of make up water

Fig. 9.4 Adiabatic saturation

Tdb

Saturated air 2

1

Tb Tdb

2

Tdb = Tb 2

1

2

2 Dew point of unsaturated air entering

Fig. 9.5 T-s diagram representing adiabatic saturation

Let us assume that the saturator is a control volume. Mass balance: The mass of air entering the saturator – ma1 The mass of air leaving the saturator – ma2 From conservation of mass; The mass of air entering the separator = mass of air leaving the separator

9.8

Engineering Thermodynamics

ma = ma

So,

1

...(9.17)

2

Mass of water entering the saturator = mw Mass of water leaving the saturator = mw

1

2

Mass of water consumed/absorbed from saturator = mw

Mass balance of water:

s

Mass of water entering the saturator    + Mass of water consumed from  = Mass of water leaving the saturator  the saturator  mω + mω = mω ...(9.18) Energy balance:

1

s

2

[(Enthalpy of air + Enthalpy of water)1 = [Enthalpy of air + Enthalpy of water]2 + Enthalpy of water in saturator]     (ma ha+ mwhw + mωh1)in = (maha + mωhω + mωh2)out

...(9.19)

Suppose the specific humidity is considered for water entering the leaving the Saturator, then (ha + ωhω + ωh)1 = (ha + ωhω + ωh)2

...(9.20(a))

ha1 + w1hw + w1h1 = ha2 + w2hw2 + w2h2 ...(9.20(b)) ha1+ w1(hw1 + h1) = ha2 + w2 (hw2 + h2) ...(9.20(c)) w1(hw1 + h1) = (ha2 + w2 (hw2 + h2) – ha1) ...(9.21) ω1 =

( ha2 – ha1 ) + w2 ( hw2 + h2 ) . ...(9.22) ( hw1 + h1 )

The adiabatic saturation temperature is denoted as Tas. The adiabatic saturator may be used to determine the specific humidity and relative humidity of moist air. The temperature of the air entering the saturator, and the specific humidity of the air entering are the influencing parameters of the adiabatic saturation temperature.

ma ωa + mb ωb = ma ωab – mb ωab ...(9.23)

ma ωa – ma ωab = mb ωab – mb ωb ma(ωa – ωab) = mb (ωab – ωb) wab – wb ma = wa – wab mb

...(9.24)

Applying the same to the energy balance equation and solving the equation we get hab – hb ma = . ...(9.25) ha – hab mb

9.6

PSYCHROMETRIC CHART

The thermodynamic properties of moist air can be shown in a relevant graphical representation, which is called psychrometric chart. In the psychrometric chart

9.9

Psychrometry

dry bulb temperature and specific humidity are taken as coordinates. The psychrometric chart is shown in Fig. 9.6.

Source : ASHRAE Fig. 9.6 Psychrometry chart

The chart contains. (i) The dry bulb temperature lines are shown as vertical line (X-axis) (ii) The specific humidity lines are shown as horizontal lines (Y-axis) (iii) The wet bulb temperature lines shown as inclined line (iv) The relative humidity curves shown with RH percentages (v) The dew point temperature lines shown unevenly spaced horizontal lines. (vi) The vapour pressure lines shown in the left side abscissa (vii) The specific volume lines are shown as inclined lines. (viii) The specific enthalpy lines uniformly spaced and parallel or coinciding with the wet bulb temperature. The specific enthalpy values may be obtained from the inclined scale above the saturation curve.

9.7

PSYCHROMETRIC PROCESSES

There are a few important psychrometric processes followed among many processes. They are

9.10

Engineering Thermodynamics

(i) Sensible cooling (ii) Sensible heating (iii) Humidification (iv) Dehumidification (v) Cooling and dehumidification (vi) Heating and humidification

9.7.1 Sensible Cooling The sensible cooling process is illustrated in Fig. 9.7(a). The process is shown in the psychrometry chart in Fig. 9.7(b).

ion rat u t Sa

Air out Air in

2

Cooling Cooling water in

h1 h2

Cooling water

(a)

1

1 = 2

Specific humidity



ve

1

cu r

2

Td2 DBT Td1

(b)

Fig. 9.7 (a) Sensible cooling (b) Sensible cooling in pschrometry chart

The sensible cooling of air is done by passing the air over a cooling coil through which cold fluid or water would be circulated. When the temperature on a cooling surface is above or equal to the dew point temperature of the surrounding air, the air will be cooled without any change in the specific humidity. This process is called sensible cooling. In this process the sensible heat in the air is removed. The air gets cooled along a constant specific humidity. Let us consider the following; Temperature of air at the inlet = T1

Temperature of air at the outlet = T2

Temperature of the cooling coil = T3

The temperature of air leaving at the outlet (T2) is greater than the temperature of cooling coil (T3). Dry bulb temperature

Amount of heat rejected in the cooling process

h = h1 – h2



h = Cpm(T1 – T2) ...(9.27)

...(9.26)

9.11

Psychrometry

where Cpm = humid specific heat = 1.002 kJ/kg K. Also, h = enthalpy change in dry air + enthalpy change in water vapour = Cpa(T1 – T2) + ω Cps (T1 – T2) ...(9.28) = (T1 – T2) [Cpa + ω Cps] [Cpa + ω Cps] is referred as humid specific heat and is denoted as Cp . m It is given in kJ/kg K. Therefore, h = (T1 – T2) Cpm = Cpm (T1 – T2) ...(9.29) = 1.002(T1 – T2) kJ/kg. ...(9.30)

9.7.2 Sensible Heating

h1 Air in

tu Sa

Air 1 Heating

tio

h2

2

1

2

Specific humidity

2

ra

1

nc urve

Sensible heating is the process in which the air is heated without any change in the specific humidity, but, the temperature changes. The heating of air is done by passing the moist air over a heating coil, through which hot fluid or water is circulated as shown in Fig. 9.8(a). The process is indicated in the psychrometric chart in Fig. 9.8(b).

T1 T2 Dry bulb temperature

(a)

(b)

Fig. 9.8 (a) Sensible heating (b) Sensible heating in psychrometric chart

The enthalpy of air during sensible heating is given by h = h2 – h1 ...(9.31) h = Cpm (T2 – T1) ...(9.32) h = 1.002 (T1 – T2) kJ/kg

9.7.2.1  Bypass Factor In the case of sensible cooling, the bypass factor is given as:

BPF =

(Dry bulb temperature of air at the exit – Coil temp.) (Dry buld temperature of air at the inlet – Coil temp.)

9.12

Engineering Thermodynamics

BPF =

( T2 – T3 ) ...(9.33) ( T1 – T3 )

Where T1 and T2 are dry bulb temperatures of air at the inlet and the exit In the case of sensible heating the bypass factor is given as:

BPF =

(Heating coil temp – Dry bulb temperature of air at the exit) Heating coil temp – Dry bulb temperature of air at the inlet)



BPF =

( T3 – T2 ) . ...(9.34) ( T3 – T1 )

9.7.3 Humidification

2 2

a tur Sa

2 1

1

1

Specific humidity, 

tio n

cu rv

e

Humidification is the process by which moisture is added to the air without affecting its dry bulb temperature. This process is represented by a straight line 1-2, as shown in Fig. 9.9. Let the initial and final values of the moisture content of air be ω1 and ω2 respectively. The latent heat of evaporation is hem. Then the change in enthalpy of the moist air in humidification is given Eqn.(9.35).

Td1 = Td2 Dry bulb

Fig. 9.9 Humidification process

The enthlpy change in the process is given as;

h2 – h1 = (ω2 – ω1)hem kJ/kg

...(9.35)

9.7.4 Dehumidification Dehumidification is the process, by which the moisture is removed from the air without affecting its dry bulb temperature. This process is represented by a straight line 1-2, as shown in Fig. 9.10.

h1 – h2 = (ω1 – ω2)hem kJ/kg

...(9.36)

The humidification and dehumidification processes are not commonly used in practical applications.

9.13

2

2

ra tu Sa

2 1

1

1

Specific humidity, 

tio

nc ur v e

Psychrometry

Td1 = Td2 Dry bulb

Fig. 9.10 Dehumidification process

9.7.5 Cooling and Dehumidification The process in which the air is cooled sensibly, and the moisture is removed from the air is known as cooling and dehumidification. It is one of the most commonly used air conditioning applications for the cooling purposes. It is used in many applications like air conditioning, cold storage, paper industry. Cooling and dehumidification done when the air at the given dry bulb is cooled below its dew point temperature. The steps involved in the cooling and dehumidification process are shown in Fig. 9.11 (a). a Entry air

b Leaving air

Cooling coil T < DPT of entry air

Fig. 9.11 (a) Schematic representation of the cooling and dehumidification process

The process of cooling and dehumidification is shown in Fig. 9.11(b). h1

2 1 2

Td2 = Td2

2 1

Specific humidity, 

h2 h2

Td1

Dry bulb

Fig. 9.11 (b) Cooling and dehumidification process in the psychrometry chart

9.14

Engineering Thermodynamics

Generally, cooling and dehumidification is done in three steps, which are (i) cooling the air below its dew point temperature (ii) allowing it to condense (iii) then heating the air to the required temperature. Water is sprayed from the top of the dehumidifier unit so that the air is cooled. Another way of cooling the air is by using the cooling coils, through which a refrigerant is circulated. When the air comes in contact with the cooling coil which is maintained at temperature below its dew point temperature, its dry bulb temperature starts decreasing. The process of cooling continues and at some point it reaches the value of the dew point temperature of the air. At this point, the water vapor within the air starts condensing on the surface of the cooling coil, and the moisture content of the air reduces. Further, the air is allowed to pass through a heating coil, so that the dry bulb temperature of the air is increased a litter. During the cooling and dehumidification process the dry bulb, wet bulb and the dew point temperatures of the air reduce. Similarly, the sensible heat and the latent heat of the air also reduce, leading to on overall reduction in the enthalpy of the air. The is enthalpy change of air during the combined cooling and dehumidification process

h = h1 – h2



h = (h1 – h2’) + (h2’ – h2) ...(9.38)

...(9.37)

It should be noted that, the heat removed in the process 1-2’ is sensible heat and the heat, in the process 2-2’ is the latent heat removed due to the condensation of vapour while reducing the moisture content.

9.7.6 Heating and Humidification The process by which the air is heated sensibly an the moisture is added to the air is known as heating and humidification. For example, during the winter season the heating of air is very much essential, but at the same time the moisture is added to the air. The heating and humidification process is carried out, by passing the air over a spray of hot water, which is maintained at a temperature higher than the dry bulb temperature of air, or by mixing the air and the steam. The arrangement is shown in Fig. 9.12 (a). a Entry

b Adding stem spray

Leaving

Fig. 9.12 (a) Arrangement of heating and humidification

In the heating and humidification process of air, the dry bulb temperature as well as the humidity of the air increase.

9.15

Psychrometry h2 h1 2 2

1

Td1

Specific humidity, 

h2

Td2 = Td2 Dry bulb

Fig. 12 (b) Heating and dehumidification in the psychrometry chart

The change in the enthalpy of air during heating and humidification is

h = h2 – h1



h = (h2 – h2’) + (h2’ – h1) ...(9.40)

...(9.39)

Here, the term (h2 – h2’) is the latent of heat added to the air, while increasing the moisture content, whereas (h2’ – h1) is the sensible heat.

9.8

SENSIBLE HEAT FACTOR

It is defined as the ratio of sensible heat to the total heat in the psychometric process, denoted as SHF. Let us take the sensible heat as SH and latent heat as LH.

The total heat = Sensible heat + latent heat SHF =

=

Sensible heat Sensible heat + latent heat SH . SH + LH

The sensible heat factor SHF for both (i) cooling and dehumidification and (ii) heating and humidification are given below SHF (cooling and dehumidification) = SHF (heating and humidification) =

( h1 – h2′ ) ...(9.41) ( h1 – h2 )

( h2′ – h1 ) ...(9.42) ( h2 – h1 )

Now the psychrometric processes differ from each other is illustrated in Fig. 9.13.

9.16

Engineering Thermodynamics

Sensible cooling

Reduce dry bulb temperature

Air at entry h1

Increase DBT

Dehumidification only

Reduce 

Tdb1

Increase 

Humidification only

Cooling and dehumidification

Reduce DBT and 



Increase DBT and 

Heating and humidification

Sensible heating

Fig. 9.13 Different psychrometric processes

9.9

MIXING OF AIR STREAMS

In psychrometry, there is an another important process considered which is known as adiabatic mixing. Stream 1

1

3 Mixed stream 1 and 2

Stream 2

2

Fig. 9.14 Mixing of two different air streams

1

h3

h2 3 2

1 3 2

Specific humidity

h1

Dry bulb temperature Tdb

Fig. 9.15 Mixing of air streams represented in psychrometry chart

9.17

Psychrometry

In adiabatic process more than one stream of air are mixed together. The properties of the streams of air are different. The specific application of such mixing of air stream is found in air conditioning, cold storage etc. Let us consider the air stream (1) enters through valve A, which has mass (m1) specific enthalpy (h1) and specific humidity (w1). Similarly another air stream (2) enters through valve B. Stream (2) has similar properties m2, h2 and w2. The mass of mixed air stream is m3. The specific enthalpy and humidity of air after mixing were h3 and w3 respectively. Let us consider that two different masses of air are to be mixed. Mass of stream 1 = m1 Mass of stream 2 = m2

Mass of mixed stream 1 and 2 = m3 Specific enthalpy of stream 1 = h1 Specific enthalpy of stream 2 = h2

Specific enthalpy of mixed stream 1 and 2 = h3 From conservation of mass;

Total mass entering the system = Total mass leaving the system

m1 + m2 = m3 ...(9.43)

Writing the conservation of water vapour;

Total mass of water vapour = Total mass of water vapour leaving the system entering the systems [m1ω1 + m1ω2] = m3 ω3 ...(9.44) Writing the energy balance equation;

The energy entering before the mixing of the stream = Energy leaving after the mixing of the stream.

m1h1 + m2h2 = m3h3

...(9.45)

Solving Eqns. (9.43), (9.44), and (9.45) we get         

m1 h3 – h2 = m2 h1 – h3 ...(9.46)

Similarly, we get   

m1 w3 – w2 ...(9.47) = m2 w1 – w3

Solved Problems Problem 9.1 The following observations were made during a testing of moist air Dry bulb temperature Tdb = 29°C Dew point temperature Tdp = 15°C

Total pressure = 1 bar

9.18

Engineering Thermodynamics

Determine the relative humidity and degree of saturation. Given Tdb = 29°C  or  302 K Tdp = 15°C  or  288 K Solution

Get the saturation pressure at Tdb and Tdp from stream tables. At 15°C At 29°C

pv = 0.01704 bar ps = 0.04004 bar

Relative humidity = ω =

pv 0.01704 = ps 0.04004

= 0.4255 pv p – ps × Degree of saturation = μ = p – pv ps =

1 – 0.04004 0.01704 × 1 – 0.01704 0.04004

= 0.4155. Problem 9.2 A thermometer measured the dry bulb temperature of atmospheric air as 25°C. A wet bulb thermometer measured the atmospheric air as 19°C. The total pressure is 1 bar. Calculate (i) The partial pressure of the water vapour (ii) Relative humidity (iii) Specific humidity (vi) Saturation ratio. Given Tdb = 25°C  or  298 k Tωb = 19°C  or  292 k

Total pressure p = 1 bar

Solution From the steam tables, the saturation pressure of vapour is obtained corresponding to the wet bulb temp Tωb = 19°C

pvs = 0.02196 bar.

The carrier equation is used to calculate the partial pressure of water vapour, corresponding to the dry and wet bulb temperatures

( p – pv )( Tdb – Twb ) 1.8 2800 – 1.3 ( Tdb – 32 )



pv = pvs –



pv = 0.02196 –

= 0.0175.

( 1 – 0.02196 )( 25 – 19 ) × 1.8 2800 – 1.3 ( 298 + 32 )

9.19

Psychrometry

pvs - Saturation pressure at wet bulb temperature. Similarly the saturation pressure of vapour corresponding to dry bulb temperature is obtained from steam tables.

ps = 0.03166 bar

(i) Dew point temperature: It is understood that the pressure of water vapour in the air is equal to the saturation pressure of vapour at dew point temperature. Therefore, corresponding to the pressure of water vapour (pv = 0.01979 bar) the temperature obtained from the steam table is Tdp = 16.8°C. (ii) Relative humidity: (iii) Specific humidity:

pv 0.01979 = = 0.6251 T = ps 0.03166 0.622 pv 0.622 × 0.01979 = = 0.01256 ω = 1 – 0.01979 p – pv

(iv) Saturation ratio:

μ =

=

pv ( p – ps ) × ps ( p – pv ) 0.01979 ( 1 – 0.03166 ) × 0.03166 ( 1 – 0.01979 )

= 0.5935. Problem 9.3 A sling psychrometer reads the values of the dry bulb and wet bulb temperatures are Tdb = 28°C, Twb = 18°C. The pressure measured is 740 mm of Hg. Calculate the following (i) Specific humidity (ii) Relative humidity at dry bulb temperature (iii) Dew point temperature 740 × 1.013 760 p = 0.9863 bar

Given Ambient pressure p = Total pressure

Tdb = 28°C = 301 K  ps = 0.03782 bar Tωb = 18°C = 291 K  ps = 0.02064 bar

pv = 0.02063 bar

(a) Partial pressure of water vapour pv = pvs – (b) Saturation pressure at dew point ps = ps –

( p – pv )(Tdb – Tωb ) 2800 – 1.3(Tdb + 32) ( p – pv )(Tdb – Tωb ) 2800 – 1.3(28 + 32)

9.20

Engineering Thermodynamics

(0.9863 – 0.02064)(28 – 18) 2800 – 1.3(28 + 32)



ps = psw –



ps = 0.01425 bar.

 ps   p – ps  (c) Degree of saturation =  ×   p – pv   pa  0.9863 × 0.03782 0.01425 × = 0.9863 – 0.01425 0.03782 = 0.3676 (d) Relative humidity:

Φ =

=

pv pa 0.01425 0.03782

= 0.3760. (e) Specific humidity of water vapour: ω =

0.01425 = 0.622 ×

0.622 pv ( p – pv )

0.01425 ( 0.9863 – 0.01425 )

= 0.009182. Dew point: Corresponding to pω = 0.009182 bar, the temperature obtained from steam table is 12°C Problem 9.4 In an experiment, the psychrometer obtained the following values. Dry bulb temperature: 32°C. Wet bulb temperature 22°C. The barometric pressure was 750 mm of Hg. Determine (i) The dew point temperature (ii) Relative humidity (iii) Degree of saturation (iv) Specific humidity (v) Specific volume and (vi) Specific enthalpy Given Dry bulb temperature = Tdb = 32°C Wet bulb temperature = Twb = 22°C

Barometric pressure = 750 mm of Hg Solution 750 × 1.013 760 = 0.999 bar Ambient pressure

p =

9.21

Psychrometry

Saturation pressure a t dew point ps = pvs –

( p – pvs ) × ( Tdb – Twb ) 1547 – 1.44 Twb

(0.99 – 0.02642) × (32 – 22) 1547 – 1.44 × 22 = 0.02006 bar

ps = 0.02642 –

From the steam tables corresponding to Twb, the saturation pressure obtained is pwb = 0.02642 bar. (a) Dew point temperature

The dew point temperature is found from the psychrometric table corresponding to the saturation pressure ps = 0.02642 bar. The dew point temperature is found to be: 22°C (b) The degree of saturation We know that the degree of saturation is

 pv   p – ps  j =     p – pv   ps 

   0.999 – 0.02642  0.02006 =    = 0.80949 0.999 – 0.02006 0.02462    = 0.7544 (c) The specific humidity Specific humidity

 pv  ω = 0.622    p – pv 

  0.02006 = 0.622 ×    0.999 – 0.02006  = 0.01274 kJ/kg of dry air.

(d) Specific volume

 RT   v =   p – pv 

=

0.287 × ( 273 + 32 )

( 0.999 – 0.02006 ) × 105

= 0.8941. (e) Specific enthalpy

h = 1.004 Tdb + ω [2528.9 + 1.884 (Tdb – Tdp)] = 1.004 × 32 + 0.01274 [2528,9 + 1.884 (32 – 19.5)]

= 64.646 kJ/kg of dry air. Problem 9.5 A certain space is occupied by humid air, which has a dry bulb temperature 30°C. The wet bulb temperature is 24°C. Obtain the following values

9.22

Engineering Thermodynamics

(i) Relative humidity (ii) Specific humidity (iii) Dew point at standard barometric pressure. Given Dry bulb temperature Tdb = 30°C = 303 K Wet bulb temperature Twb = 24°C = 297 K Solution hb ha



Fig. p. 9.1 Psychrometric process

(a) Relative humidity Corresponding to Tdb and (Tdb – Tωb) the point is located in the chart and we get the relative humidity value from the psychrometric table, as

j = 0.35.

(b) Specific humidity (ω) To find the specific humidity, first of all we should calculate the saturation pressure. The saturation pressure corresponding to Tdp is = 25°C (from psychrometric chart)

ps = 0.03166 bar

Therefore,

pω = j ps

= 0.35 × 0.03166 = 0.011081 bar So, the specific humidity ω = 0.622 =

pv p − pv

0.622 × 0.011081 1.013 – 0.01081

= 0.006878 kg/kg of dry air.

9.23

Psychrometry

(c) Dew point temperature The dew point temperature (T dp) is obtained from the steam table corresponding to the saturation pressure ps. i.e., Tdp = 2°C Tdb = 25. Problem 9.6 Moist air with a dry bulb temperature of 30°C and a wet bulb temperature of 15°C is cooled to 10°C, without affecting the moisture content. Determine the following in the sensible cooling process. (a) Wet bulb temperature after cooling (b) Relative humidity after cooling Given Dry bulb temperature       Tdb1 = 30°C = 303 K Wet bulb temperature      Tωb1 = 15°C = 288 K Dry bulb temperature after cooling Tdb2 = 12°C = 285 K Solution (a) Relative humidity after cooling The wet bulb temperature at point 2 is 5°C.

15° 5° 2

0°C

1

Specific humidity

Locate the dry bulb and wet bulb temperatures in the chart at Tdb1 and Twb1 respectively and mark the point 1. From point 1 draw a horizontal line to intersect the curve at point 2, where the dry bulb temperature lies at 10°C.

30°C

Fig. p. 9.2 Dry bulb temperature Tdb

(b) Relative humidity before cooling Find out the relative humidity curve which is nearer to point 1.

j1 = 16%

Relative humidity after cooling corresponding to point 2, obtained from the psychrometric chart is

j2 = 50%

Let us refer to the psychrometric chart, as shown is Fig. p. 9.2.

9.24

Engineering Thermodynamics

Problem 9.7 The moist air at 750 mm of Hg is heated in a heating coil. The following data were recorded in the heating process. Dry bulb temperature before heating Tdb1 = 20°C Dry bulb temperature after heating Tdb2 = 35°C Wet bulb temperature

Twb1 = 15°C

By pass factor of heating coil

(BPF) = 0.4.

hb ha a

Tda

b

Specific humidity

Calculate (i) The heating coil temperature (ii) The wet bulb temperature of the air leaving the coil (iii) The relative humidity of the air after heating and, (iv) The sensible heat added per kg of air.

a = b

Tdb

Fig. p. 9.3 Sensible heating

Given p = 750 mm of Hg,  Tdb1 = 20°C,  Tdb2 = 35°C,  BPF = 0.4 Solution (i) Heating coil temperature Tdb 3 – Tdb 2 Tdb 3 – Tdb 1



By pass factor =



0.4 =



0.4 (Tdb3 – 15) = (Tdb3 – 35)



0.4 (Tdb3 – 6) = Tdb3 – 35

Tdb 3 – 35 Tdb 3 – 15



0.6 Tdb3 = 29 29 Tdb3 = 0.6 = 48.3°C. (ii) Wet bulb temperature Locate the point 1 corresponding to the dry bulb temperature Tdb2, 35°C at point 2. The wet bulb temperature corresponding to point 2 is Twb2 = 19°C.

9.25

Psychrometry

(iii) Relative humidity of air after heating From the psychometric chart, the relative humidity of the air after heating is obtained at point.

j = 25%

From the psychrometric tables the specific enthalpy values h1 and h2 are obtained

h1 = 41 kJ/kg



h2 = 52 kJ/kg.

Added h2 – h1 = 57 – 41 = 16 kJ/kg Problem 9.8 In a combined heating and humidification process, moist air enters the heating coil with the dry bulb temperature of 20°C and 30% RH. After the process, the dry bulb temperature and the RH of air were found to be 40°C and 55% RH. The air passes through the heating coil at the rate of 350 m3/s. Calculate the following (a) Moisture added to the air (b) Heat added to the air, if it enters the processor at a rate of 350 m3/sec. (c) Heat added during the process (d) Heat added to the air per hours (e) sensible hat factor Given Tdb1 = 20°C

f1 = 30%

Tdb2 = 30°C

f2 = 55%

Solution (a) Moisture added to the air Locate point 1 corresponding to Tdb1 = 20°C, RH = 30% and find w1 = 0.0045 Similarly, locate point 2 corresponding to Tdb2 = 40°C, RH = 55% and find w2 = 0.026

w2 – w1 = 0.0215 kg/kg of air

(b) Heat added during the process To find the mass flow rate of air we should know the specific volume of air. The specific volume of air is obtained from the chart corresponding to Tdb1 = 20°C.

υ1 = 0.635 m3/kg

Therefore the mass flow rate of air =

Volume flow rate of air Specific volume of air

9.26

hb c

ha a

b

Tda

Specific humidity

Engineering Thermodynamics

Tda

Dry bulb temperature Tdb

Fig. p. 9.4 Heating and humidification process



m =

=

350 v1 350 kg/sec 0.835

(c) Heat added during the process = m(h2 – h1) = 419.16

The specific enthalpies at point 1 and 2 are obtained corresponding to Tdb1, and Tdb2 respectively, from the psychrometric chart.

h1 = 32 kJ/kg

h2 = 107 kJ/kg

(d) Heat added to air per hour =

m ( h2 – h1 )

3600 419.16(107 – 32) = 60 = 523.95 kJ/kg hi – h1 57 – 32 (e) Sensible heat factor: SHF = = = 0.33 . h2 – h1 107 – 32 Problem 9.9 The moist air is cooled in a cooling coil. The dehumidification air of is at a relative humidity of 60%. The dry bulb temperatures of air before and after cooling are 30°C and 15°C respectively. The wet bulb temperature at the exit of the cooling coil is 11. The volume flow rate of air is 540 m3/min. Calculate (i) The amount of moisture added or removed and (ii) The amount of cooling required. Given Dry bulb temperature before cooling = Tdb1 = 30°C = 303 K Dry bulb temperature after cooling = Tdb2 = 15°C = 288 K Wet bulb temperature after cooling = Tωb2 = 11°C = 284 K Volume flow rate of air ma = 240 m3/min Relative humidity of air j = 60%

9.27

Psychrometry

Solution (i) Amount of moist air added or removed: Let us first find the specific volume of air at initial dry bulb temperature Tdb1. v1 = 0.88 m3/kg

The specific volume We know the volume flow rate

= 240 m3/min

Therefore, the mass flow of air

=

Volume flow rate of air Specific volume of air

540 540 = = = 613.63 kg/min v1 0.88 = 613.63 kg/min The specific humidity at 30°C and 15°C are obtained from the chart.

ω1 = 16 × 10–3



ω2 = 6 × 10–3

40 h

30 2

 = 60 1

Specific humidity

kJ/kg

0°C 30°C Dry bulb temperature

Fig. p. 9.5 Cooling and dehumidification

Mass flow of moisture = mass flow of air × change in specific humidity = m(w1 – w2) = 613.63(16 – 6) × 10–3 kg/min

m = + 6.1363 kg/min.

(ii) Amount of cooling required: The cooling required is calculated by Q1–2 = m(h2 – h1) = 613.62 (30 – 70) = –24545.2 kJ/min. Problem 9.10 In an air conditioning plant, fresh air is mixed up with the returning air from the conditioned space. The fresh air drawn from the atmosphere has a dry bulb temperature of 32°C and a wet bulb temperature of 25°C. The fresh air is drawn at the rate of 100 m3/min. The returned air from the conditioned space has the dry bulb temperature 23°C, and relative humidity of 50%. Its volume flow rate is 200 m3/min. Determine the

9.28

Engineering Thermodynamics

(i) dry and web bulb temperatures (ii) specific humidity of the mixture. Given Tdb = 32°C, 1

V1 = 100

T

= 25°C,

wb1 m3/min Tdb = 23°C 2

RH = 50% V2 =  200 m3/min Solution The specific volumes of fresh are returned air are found from psychrometric tables, corresponding to the dry and wet bulb temperatures at the initial and final conditions of RH values of v1 = 0.8387 m3/kg and v2 = 0.8643 m3/kg V1 Mass flow rate of fresh air, m1 = v1 =

100 = 119.232 kg/min 0.8387

Mass flow rate of returned air = m2 =

V2 v2

200 = 231.4 kg/min 0.8643 m1 119.232 = 0.515. = m2 231.4

= The ratio

 m1 h3 – h2 w3 – w2  = =   m h1 – h3 w1 – w3               2 Locate the points 1 and 2 corresponding to the properties of fresh and returned air. Draw the diagram on a suitable scale. Locate the points 1, 2 in the diagram. Measure the length 1 and 2.          

m1 (1 – 3) . = m2 (3 – 2)

Measure the length 1 and 2 in the psychrometric chart. Now, mark point on the line 1-2 in the ratio of 0.515 times the length of 1-2. The point 3 lies on line 1-2. Corresponding to point 3 obtain the values of dry bulb temperature, and specific humidity. Dry bulb temperature = 27°C, Wet bulb temperature = 21°C specific humidity of the mixture = 13 g/kg of dry air.

9.29

Psychrometry

Objective Type Questions 1. For unsaturated air the dew point temperature is wet bulb temperature. (a) Equal to (b) Less than (c) More than (d) Cannot be ascertained 2. If the relative humidity is 100% then (a) Wet Bulb Temperature = Dry Bulb Temperature = Dew Point Temperature (b) Wet Bulb Temperature > Dry Bulb Temperature > Dew Point Temperature (c) Wet Bulb Temperature > Dry Bulb Temperature = Dew Point Temperature (d) Wet Bulb Temperature = Dry Bulb Temperature < Dew Point Temperature 3. During the sensible cooling of air the specific humidity (a) Increases

(b) Decreases

(c) Remains constant

4. During the sensible cooling of air the wet bulb temperature: (a) Increases

(b) Decreases

(c) Remains constant

5. The temperature of air recorded by a thermometer, when it is not effected by the moisture present in it, is called: (a) Wet bulb temperature (c) Dew point temperature

(b) Dry bulb temperature (d) None of these

6. On a psychrometric chart sensible cooling is represented by (a) Horizontal line (c) Vertical line

(b) Inclined line (d) None of these

7. The curved line on a psychrometric chart indicates: (a) Dry bulb temperature (c) Dew point temperature

(b) Enthalpy (d) Relative humidity

8. Absolute humidity is defined as: (a) Actual mass of air in a unit volume of air and water vapour mixture (b) Actual mass of water vapour in a unit volume of air (c) Actual mass of water vapour in a unit volume of air and water vapour mixture (d) Actual mass of air in a unit volume of water vapour. 9. If the air is saturated then: (a) Relative Humidity = 1 and Degree of saturation = 0 (b) Relative Humidity = 0 and Degree of saturation = 1 (c) Relative Humidity = 1 and Degree of saturation = 1 (d) Relative Humidity = 0 and Degree of saturation = 0

9.30

Engineering Thermodynamics

10. The sensible heat factor is defined as the ratio of: (a) Latent heat to sensible heat (c) Sensible heat to total heat

(b) Latent heat to total heat (d) Sensible heat to latent heat

11. If the relative humidity of air is 100%, the rate of evaporation of water will be: (a) Very high (b) Very low (c) Zero (d) It does not depend on relative humidity 12. The temperature of air recorded by the thermometer, when its bulb is surrounded by a wet cloth exposed to air is called (a) Wet bulb temperature (c) Dew point temperature

(b) Dry bulb temperature (d) None of these

13. During the adiabatic saturation process on unsaturated air, the (a) Dry bulb temperature remains constant (b) Wet bulb temperature remains constant (c) Dew point temperature remains constant (d) None of these 14. A low wet bulb temperature indicates (a) Very low humidity (b) Very cool air (c) Very low atmospheric temperature (d) Both (a) and (b) 15. In a psychrometric chart a vertical line indicates (a) Wet bulb temperature (c) Dew point temperature

(b) Dry bulb temperature (d) Relative humidity

16. The cooling of air without any change in its specific humidity is called: (a) Sensible heating (c) Humidification

(b) Sensible cooling (d) Dehumidification

17. The process of adding moisture to the air at constant dry bulb temperature is called: (a) Sensible heating (c) Humidification

(b) Sensible cooling (d) Dehumidification

18. During the sensible heating of air, the relative humidity: (a) Increases

(b) Decreases

(c) Remains the same

19. The wet bulb depression temperature is zero when the relative humidity of air is equal to (a) 0

(b) 0.5

(c) 0.75

(d) 1

9.31

Psychrometry

20. The psychrometer is a device used to measure (a) Dew point temperature (b) Dry bulb temperature of moist air only (c) Both dry bulb temperature and wet bulb temperature (d) None of these.

Theory Questions 1. Write Dalton’s law of partial pressure. 2. What is the composition of air? 3. Define dry air. 4. Define the dry and wet bulb temperatures. 5. Define the specific humidity of air. 6. Define the relative humidity of air. 7. Define the degree of saturation. 8. Write the Carrier equation for the partial pressure of water vapour. 9. Derive the expression for the specific enthalpy of moist air. 10. What is a psychrometer? What are the different psychrometers available? 11. With the help of a neat sketch, explain the working principle of a sling psychrometer. 12. What is adiabatic saturation? 13. What are the various psychrometric processes involved in the conditioning of air? 14. Define sensible cooling. 15. With the help of neat sketches explain sensible cooling. Write the expression for amount of heat removed during sensible cooling. 16. Define sensible heating. 17. With the help of neat sketches, explain sensible heating. Write the expression for the amount of heat added during sensible heating. 18. With the help of neat sketches, explain adiabatic humidification. 19. With the help of neat sketches, explain dehumidification by cooling. 20. With the help of neat sketches, explain how cooling and dehumidification is done. 21. With the help of neat sketches, explain how heating and humidification is done. 22. Derive the expression for the mixing of two air streams. 23. Sketch the sensible heating process on a skeleton psychrometric chart.

9.32

Engineering Thermodynamics

24. Explain the terms (a) Specific humidity (b) Dew point temperature. 25. What is adiabatic mixing and write the equation? 26. State Dalton’s Law of partial pressure. On what assumption is this law based? 27. What is specific humidity? When does it become the maximum? 28. Define dew point temperature. 29. Explain in detail the following: (i) Sensible heat and sensible cooling at w = constant, (ii) Cooling and dehumidification (iii) Heating and humidification 30. Explain the adiabatic saturation process using the T-s diagram, and derive an expression to determine the specific humidity of unsaturated air entering the adiabatic saturator. 31. Explain the following: Dew point, adiabatic saturation process, dry bulb temperature and relative humidity.

Unsolved Problems 1. Air at 20°C, 40% RH is mixed adiabatically with air at 40°C, 40% RH in the ratio of 1 kg of the former with 2 kg of the later (on dry basis). Find the final condition of air. 2. The atmospheric air at 30°C DBT and 75% RH enters a cooling coil at the rate of 200 m3/min. The coil dew point temperature is 14°C and the by-pass factor is 0.1. Determine (i) The temperature of air leaving the coil (ii) The capacity of the cooling coil in TR (iii) The amount of water removed (iv) Sensible heat factor for the process 3. The volume flow rate of air is 800 m3/min recirculated at 22°C DBT and 10°C dew point temperature is to be mixed with 300 m3/min of fresh air at 30°C DBT and 50% RH. Determine the enthalpy, specific volume, humidity ratio and dew point temperature of the mixture. 4. Atmospheric air at 1.0132 bar has a DBT of 32ºC and a WBT of 26ºC. Compute: (i) The partial pressure of water vapour (ii) The specific humidity (iii) The dew point temperature (iv) The relative humidity (v) The degree of saturation

Psychrometry

9.33

(vi) The density of air in the mixture (vii) The density of the vapour in the mixture (viii) The enthalpy of the mixture. 5. A room 7 m × 4 m × 4 m is occupied by an air-water vapour mixture at 8°C. The atmospheric pressure is 1 bar and the relative humidity is 70%. Determine the dew point temperature, humidity ratio, mass of dry air and mass of water-vapour. If the mixture of air-water vapour is further cooled at constant pressure until the temperature is 10ºC, find the amount of water vapour condensed. 6. A community hall of dimensions 15 m × 15 m × 6 m is provided air in it with a relative humidity of 60% at 1 atm and 30°C. Calculate (i) The partial pressure of water vapour in the air (ii) The specific humidity of air (iii) Mass of water vapour in the community hall and the (iv) Dew point. Take the values of the saturation pressure of water from the steam tables. 7. Air and water mixture enters an adiabatic saturator at 35°C and 1 atm pressure and leaves at 30°C and 1 atm pressure. Calculate the specific humidity, relative humidity and the dew point temperature of the entering air. 8. Air and water mixture occupies in a room. A thermometer measures the dry bulb temperature of the room as 40°C at 1 atm during the winter season. The wet bulb temperature of the room was also measured by the thermometer with a wet wick. The wet bulb temperature was found to be 25°C. If the temperatures were measured at 1 atm, calculate the specific humidity, relative humidity, and the enthalpy of the mixture. 9. An air conditioning unit provided in a room supplies 2 m3 of air at 10°C which is completely saturated with water vapour. If the air is mixed with 5 m3 at 30°C with a relative humidity of 40%. Find the specific humidity, relative humidity and the temperature of the room air after mixing. 10. An observation was made in a cooling device in which moist air enters at 35°C with 60% relative humidity. The air was required to leave the cooling device as saturated at 25°C. Assume that the flow rate of moist air entering the device was 250 m3/min at 1 atm. Calculate the amount of water condensed and rate of cooling required in the device. 11. For a particular application, the moist air has to undergo the process of dehumidification by cooling and supplied. The conditions of air entering the dehumidifier are 30°C, relative humidity (RH) 60% and 1 atm. After dehumidification, the air leaves as saturated, and the temperature is 20°C. Calculate the condensate removed and the tons of refrigeration. 12. An air conditioning plant supplies 40 m3/min of air at 10°C with 50% RH. The air mixes with the room air at 27°C DBT and 22°C WBT. If the mixing of air takes place at 1 atm, calculate the specific humidity, relative humidity, dry bulb temperature and the volume flow rate of air after mixing.

9.34

Engineering Thermodynamics

13. Air has to be warmed up a little by a heating coil in a chamber. The air inlet conditions are 20°C and 1 atm. The air exits at 30°C. The flow rate of air is 10 m3/min. Calculate the heat added to air in the heating chamber and the relative humidity of air in the exit. 14. A unit comprises cooling and reheating devices in it. The DBT of air entering the unit is 20°C with a relative humidity of 70%. The exit air leaves the unit with a DBT of 20°C and a RH of 70%. The volume flow rate of the air leaving the unit is 50 m3/min. Calculate (i) The rate of heat transfer in the cooling section (ii) Rate of heat transfer in the heating section (iii) The mass flow rate from the cooling section.

CHAPTER

10

FUELS AND COMBUSTION

10.1 INTRODUCTION Any material which possesses energy that can be used to generate heat energy or to perform a mechanical work is called a fuel. It may be released either by combustion of fuel with air, by chemical reactions or nuclear reactions. Fuel is an essential substance to produce (i) mechanical power in electrical power generation such as thermal power plants, (ii) in motive power such as internal and external combustion engines, and (iii) direct or indirect combustion applications. Fuels are obtained from conventional and alternative sources of energy. Conventional fuels are generally fossil fuels such as coal, petroleum crude oil which are produced from the decomposition of vegetable and organic matter under high temperature and pressure. Radioactive elements like uranium, thorium, plutonium, etc., are also included in conventional fuels. It is predicted that the fossil fuel reserves will be exhausted within this century.

10.2 CLASSIFICATION OF FUELS Generally fuels are organic in nature, and are composed of carbon and hydrogen. Fuels are classified into three types viz, (i) solids (ii) liquids and (iii) gases. They may be naturally available or artificially prepared.

10.2.1 Solid Fuels Solid fuel is a thick and dry substance which can burn and produce heat, carbondioxide and water vapor. Solid fuels generally contain carbon (C), Hydrogen (H), sulphur and a very small amount of incombustible elements like nitrogen (N), moisture and minerals. During combustion, we may find a very small quantity of moisture (H2O) and ash being formed. Examples of solid fuels include coal, coke, charcoal, biomass, municipal waste and industrial wastes. Coal is available in different grades, as peat, lignite, sub bituminous coal, bituminous coal and anthracite depending on their ash content. The heating value and the quality of the coal depend on the grade of the coal. The main source of electrical power generation in the world is only coal. It is also used for direct and indirect heating in industrial applications. Biomass is an alternative

10.2

Engineering Thermodynamics

source of energy. It is a renewable and largely available. It includes wood chips, agriculture residue such as rice husk, paddy straw, wheat straw, ground nut shell, plant wastes and animal wastes. Biomass is used directly or indirectly to generate heat and power. Municipal and industrial wastes are currently used for power generation.

10.2.2  Fuel Analysis The elemental composition of the solid fuels can be given with the help of two analyses (i) Proximate analysis and (ii) Ultimate analysis.

10.2.2.1 Proximate Analysis Coal analyses may be presented in the form of “proximate” and “ultimate” analyses, whose analytical conditions are prescribed by organizations such as the ASTM. A typical proximate analysis includes the study of the moisture, volatile matter, ash, and fixed carbon contents, of a fuel. (a) Moisture content: The moisture content is nothing but the water present in the fuel. When the fuel is burnt this moisture will be changed to vapour (steam), when the fuel reaches the boiling point of water. The moisture content is inevitable in any solid fuel. In order to determine the percentage of moisture in coal, a sample of coal with unit mass is heated to 150°C for 1 hour, in a crucible. After 1 hour the loss in weight in the fuel sample will be evaluated. The loss in weight will give the amount of moisture evaporated. This will help to calculate the percentage of moisture in the fuel. (b) Volatile matter: Any fuel that contains volatile matter that can evolve and ignite during the combustion process. The more the volatile matter, the lesser the ash content and vice versa. Suppose the known sample is heated to 950°C for about 7 minutes, the moisture and the volatile vapour will be eliminated from the crucible. By measuring the remaining weight of the mixture in the crucible, the weight loss of moisture and the volatile matter can be determined. With the available value of the moisture content, the percentage of volatile matter can be further calculated. (c) Ash content: It is the solid residue obtained at the end of the combustion process. Solid fuels generally have an ash content more than liquid fuels. If the unit mass of the fuel sample is heated to 720°C and allowed to burn completely. After burning, the mass of residue present in the crucible can be used to determine the percentage of ash content. (d) Fixed carbon: It is the material, other than ash, that does not vaporize when heated in the absence of air. It essentially contains carbon with fewer fractions of hydrogen, nitrogen, sulphur and oxygen. If the percentages of moisture, volatile matter and ash content are added together and deducted from 100%, then the remainder will indicate the percentage of fixed carbon in the fuel sample (coal).

10.3

Fuels and Combustion

10.2.2.2 Ultimate Analysis The hydrocarbon solid fuel is composed of carbon (C), hydrogen (H), oxygen (O), nitrogen (N), sulphur (S), moisture (M) and ash (A). The sum of these components is equal to 100%. A C–H–N–S analyzer is used to determine the carbon, hydrogen and nitrogen in the fuel.

The chemical compositions of different solid fuels are given in Table 10.1. Table 10.1 Ultimate Analysis and the approximate Calorific Value of Different Solid Fuels Fuel

C (%)

H (%)

O (%)

N (%)

S (%)

Ash content (%)

CV, kJ/kg

Wood

50

6

43.5

0.5

–

0.5

17300

Peat

60

6

30

0.9

0.1

3

19700

Lignite

67

5

20

1

1

6

21650

Bituminous

83.5

5

5

1.5

1

4.5

33500

Anthracite

90

2

3

0.5

0.5

4

34600

Coke

–

–

–

–

1.7 – 3.0

0.1 – 0.3

33600

Charcoal (Wood)

80

3

–

2.5

31350

3.5

10.2.3  Liquid Fuels Conventional liquid fuels generally include petroleum and other derivatives obtained from crude oil. Crude oil is a complex mixture of organic compounds, which include hydrocarbon, nitrogen, sulphur, paraffins and resins. The derivatives of crude oil such as gasoline, kerosene, diesel and residual oil are obtained during fractional distillation. Liquid fuels are mainly used in internal combustion engines, diesel power plants, aviation, and other industries. Due to the scarcity of crude oil and the huge demand for petroleum fuels, alternative liquid fuels such as alcohols and biodiesel have been introduced in the market after the Gulfway. Apart from these two alternative fuels some of the other possible alternative fuels such as reformulated gasoline, low sulphur gasoline, advanced petroleum based diesel fuels, Fishcer tropch fuels and pyrolytic oils are also in the research focus.

10.2.3.1 Crude Oil Crude oil varies from a light yellow to black color. From crude oil, various derivatives are obtained, starting from petrol to heavy oil. The properties of each derivative vary considerably. A large number of hydrocarbons are obtained from crude oil through a combination of chemical treatment and distillation. The main fuels obtained from crude oil are (i) Petrol (ii) Kerosene (iii) Diesel

10.4

Engineering Thermodynamics

and (iv) Heavy fuel oil. Table 10.2 gives the comparison of the properties of different liquid fuels that are commercially available in the market. Table 10.2 Properties of Different Liquid Fuels Commercially Available Fuel

Petrol

Diesel oil

Methanol

Ethanol

Biodiesel

Chemical formula

CnH1.87

CnH1.88

CH3OH

C2H5OH

C12 – C22 FAME

Molecular weight

110

170

32.04

46.07

~292

Distillation temperature, °C

60 – 230

340 – 480

64 – 66

78 – 79

315 – 350

Specific gravity@ 40°C

0.72 – 0.78

0.81 – 0.89

0.796

0.796

0.87 – 0.89

Kinematic viscosity @40°C

0.494

1.3 – 4.1

0.59

1.2

3.7 – 5.8

Higher calorific value, kJ/kg

47.3

44.8

22.7

29.7

16928 –17996 btu/lb

Lower calorific value, kJ/kg

44.4

43.4

19.93

28.865

15700 –16735

10.2.3.2 Biodiesel Biodiesel is a safe, biodegradable and renewable fuel. It is produced from natural renewable sources such as edible and non-edible vegetable oils, animal fats and algae, through the trans-esterification process. The oils and fats are filtered and preprocessed to remove water and contaminants. If free fatty acids are present, they can be removed or transformed into biodiesel using pretreatment technologies. The pretreated oils and fats are then mixed with an alcohol (usually methanol or ethanol) and a catalyst (usually sodium or potassium hydroxide). The oil molecules are broken and restructured into esters and glycerol. The methyl or ethyl esters and glycerol are separated from each other and purified. The methyl or ethyl ester of oil is called as biodiesel. Glycerol is obtained as a by-product in the process that is used in the production of pharmaceuticals and cosmetics. Upto 20% of biodiesel blended with diesel can be used as fuel in all diesel engines. It can be compatible with most storage and distribution systems. Even neat biodiesel can be used as fuel with little or no modification in diesel engines that have been manufactured since 1994. Biodiesel fueled convention diesel engines, offer reduces exhaust emissions like hydrocarbon, carbon monoxide, sulphates, polycyclic aromatic hydrocarbon, nitrated polycyclic aromatic hydrocarbon and particulate matter. Some of the problems associated with biodiesel as transport fuel include poor oxidation stability, poor cold flow properties, and biodiesel obtained from most of the feedstocks produce higher nitric oxide (NOx) emissions, when it is

Fuels and Combustion

10.5

used as fuel in combustion devices. In addition to these, the non edible feedstock available for production of biodiesel is considerably less, while the edible feedstocks and threat to food security.

10.2.3.3 Alcohols (i) Methanol: It is mainly produced from steam reforming of natural gas to form synthetic gas, which is then fed into a reactor vessel in the presence of a catalyst. Water vapour is also produced during the process. Synthetic gas commonly refers to a combination of carbon monoxide and hydrogen. Finally, the product is distilled to purify and separate methanol from water vapour. Methanol is also known as wood alcohol. It has been used as an alternative fuel in flexi fuel vehicles that run on M 85 (a blend of 85% methanol and 15% gasoline). It can also be used to produce methyl tertiary butyl ether (MTBE), an oxygenate which is blended with gasoline to improve octane, and increase the burning characteristics. Methanol is however, not commonly used in many countries. (ii) Ethanol: Ethanol is a clear colourless liquid, considered as an alternative fuel. It is produced by the fermentation and distillation of starch crops. Feedstocks for ethanol production include corn, barley and wheat. Ethanol can also be produced from cellulosic biomass materials, such as trees and grasses. It is used to increase the octane number and reduce the emission from gasoline engines. Ethanol at 10% is blended with 90% gasoline (E 10) in some countries, but it can be blended at higher concentrations to get E 85 and E 95. E-diesel a mixture of 15% ethanol and diesel-2 fuel, is found to be a viable fuel for diesel engines.

10.2.3.4  Other Possible Liquid Fuels (a) Reformulated gasoline: It generally uses an oxygenate such as, ethyl tertiary butyl ether(ETBE) or methyl tertiary butyl ether(MTBE) to maintain the performance and emissions of volatile organic compounds. (b) Low sulphur gasoline: It is presently produced by treating the petroleum feedstock at high pressure and temperature. Another possible method is a biosulphurisation technique using biological catalysts. (c) Advanced petroleum based diesel fuels: DME and DEE are referred as advanced petroleum based diesel fuels. Dimethyl ether (DME) can be produced from different feedstock and also from dimethoxymethane (DMM) and dioxymethylene dimethyl ether. Reformulated low sulphur diesel is also considered as an advanced petroleum based diesel fuel. Research is going on in this area, to find more possible methods of production. (d) Fishcer Tropch Diesel: Synthetic diesel or the fuel form gas to liquid is referred to as Fishcer Tropch Diesel. It is produced from natural gas, using a catalytic process.

10.6

Engineering Thermodynamics

(e) Pyrolysis fuels: Pyrolytic oils are obtained from different biomass materials and other organic substances from municipal waste and industrial wastes, through the pyrolysis process. In the pyrolysis process, the feedstock will be heated in an oxygen free reactor. The volatile matter of the feedstock will be condensed to get pyrolytic oil. The other products like pyrogas and char are also obtained in the process. The pyrolysis process is more suitable for feedstocks which cannot be converted into fuels by the transesterification process by which biodiesel is obtained, or fermentation by which ethanol is produced. The major drawback of pyrolytic oil is that, the complex carbon hydrogen chains present in it result in more problems, such as carbon deposit, injector choking, and erosion due to acidity, when they are used in combustion systems. (i) Character of a good liquid fuel: A good liquid fuel should have the following characteristics: (a) Higher calorific value (b) Low viscosity (c) Free from water or solid material in suspension (d) Lower density (e) Safe to handle (f) Easy to transport (ii) Advantages of liquid fuels: The advantages of liquid fuels over solid fuels are given below l

High calorific value

l

Clean and free from dust

l

Easy handling

l

Uniform combustion and easily fired

l

Continuous combustion

l

No ash formation

l

Elimination of wear and tear

(iii) Disadvantages: The disadvantages of liquid fuels over solid fuels are given below l

Production cost is high

l

They are explosive

l

Storage and transport costs are high.

10.2.4  Gaseous Fuels The mineral impurities present in crude petroleum are mainly concentrated in the liquid fuel. But a few are found in the form of gas, which are used as fuels. Gaseous fuels are mainly divided in to two types. (i) Conventional (ii) Renewable

Fuels and Combustion

10.7

(i) Natural Gas: Natural gas is a colourless, ordourless gaseous fuel. It mainly consists of methane. Natural gases are formed either with petroleum or produced by the reaction of water and hydro carbides in depth under high pressure and temperature. It is also called casing head gas. The gas stream produced from reservoirs contains natural gas and other materials. The gas is separated from liquids crude oil, condensate water and suspended solid. Contaminants are also removed. Hydrogen sulphide and sulphur components are also removed in the process. Natural gas is made in a compressed form (Compressed natural gas) or cryogenically cooled to the liquid state (Liquefied natural gas). (ii) Liquefied petroleum gas (LPG): LPG mainly comprises propane with other hydrocarbons such as propylene, butane and butylene. It is a good alternative fuel for both domestic and transport applications. It is produced as a by-product of natural gas processing and crude oil refining. It is generally produced by about 55% from natural gas purification, while the remaining 45% is produced by crude oil refining. (iii) Hydrogen: Hydrogen is a clean and environment friendly fuel. It can be easily produced. It is the most suitable alternative fuel for automotive vehicles, especially for gasoline vehicles. It can be produced by any of the following methods: l

Splitting of water into hydrogen and oxygen by electrolysis

l

Steam reforming that converts methane and other hydrocarbons into mixture of hydrogen and carbon monoxide, by reaction with steam over a nickel catalyst.

l

Steam electrolysis, that uses heat instead of electricity, to split water into carbon and hydrogen.

Hydrogen

methods:

can also be produced from renewable sources by the following

l

Chemicals and heat are used in multiple steps to split water into hydrogen by thermochemical splitting.

l

Sunlight and catalysts are used to split water by the photolytic process.

l

Sunlight and biological organisms are used to split water through biological or photological water splitting.

l

Biomass gasification uses selected microbes to breakdown biomass feedstock into hydrogen.

Higher production cost and flammability are the two major hurdles in using hydrogen as an alternative fuel widely in internal combustion engine applications. The volume of the storage tank required is large as the density of hydrogen is low. As a result the storage of hydrogen is difficult in a given space. Hydrogen has to be stored as a cryogenic liquid (at low temperature and pressure) or it has to be compressed. Hydrogen stored in the fuel cells is a possible solution to use hydrogen in vehicles safely. Fuel cell is a device in which electricity is produced from the electrochemical reaction between the hydrogen in a hydrogen containing fuel, and the oxygen present in the air.

10.8

Engineering Thermodynamics

(iv) Biogas: Organic materials can be converted into gaseous fuels by the anaerobic digestive process. Generally, biomass materials, such as cow dung, municipal wastes, and plant materials like water hyacinth, algae and certain types of grasses are used as feedstocks for biogas production. Biogas consists of a major proportion of methane, and the rest are carbon dioxide and hydrogen sulphide. Biogas can be produced in villages and rural areas as the feedstock are readily available and are cheap. The heating value is low and therefore the fuel consumption will be higher when it is used as a fuel in combustion systems. (v) Coal gas: It is a by-product obtained from the destructive distillation of coal. It is also called as town gas. The contents of coal gas are hydrogen, carbon monoxide and a few hydrocarbons. Since it is a product of the distillation of coal, its quality depends on coal and the process temperature. Its calorific value of it is in the average of 24000 kJ/m3. (vi) Producer gas: It is produced during an incomplete combustion of coke in current in the air. It is widely used in furnaces for various applications, like in glass industries, foundries etc. The calorific value is nearly 6500 kJ/kg. (vii) Blast furnace gas: It is obtained as a by-product from the blast furnace used in process of producing pig iron. The calorific value is about 3800 kJ/m3. Compositions of various gaseous fuels by ultimate analysis are shown in Table 10.3. The physical properties of some of the gaseous fuels are listed in Table. 10.4. Table 10.3 Chemical Composition of Different Gaseous Fuels Gas

CH4

CO

CO2

H2

N2

Cn Hm

Natural gas

85

0 – 0.45

0 – 0.8

0 – 1.8

0.5 – 8.4

–

Coal gas

35

8

2

45

6

4

Producer gas

4

23

5

6

62

–

Blast furnace gas

3

30

2

2

52

–

Table 10.4 Physical Properties of different Gaseous Fuels Property

CNG

LPG (Propane)

Chemical formula Molecular weight C:H:N, %

Hydrogen Producer gas

Coal gas

Biogas

CH4

C3H8

H2

CO + H2

CH4 + CO2

–

16.04

44.1

2.02

–

–

16.56

–

75:25:0

82:18:0

0:100:0

15 – 20:

32:17:51

20 – 40: 1 – 3:

42:2:56

10 – 15: Density, Flash point, °C

Blast furnace gas

1.07 –184.4

4.22 –73.33 to –101.11

– –

45 – 55 1.07 –

5 – 40 0.58 –

0.65–0.91 > 230

1.250 –

Fuels and Combustion

10.9

10.3 SOME IMPORTANT FUEL PROPERTIES 10.3.1 Density The relative density (generally referred as density) of a fuel is defined as the ratio of the weight of a given volume of the fuel to the weight of the same volume of water, both measured at 15°C and at a pressure of 1.013 bar. With an increase in temperature, the density of fuel decreases. The density of a fuel varies according to the physical conditions of its measurement.

10.3.2 Viscosity In a simple definition, it is said to be the resistance of the flow of fuel. The viscosity of a fuel is measured in units of kinematics viscosity on an absolute scale. The viscosity of fuels depends mainly on temperature. For an easy flow in pipes and normal operation of fuel in pumps, the temperature of the fuel should be maintained in the range of 60-70°C. One of the instruments used is the Redwood viscometer. The others are Saybolt, Fiscal and Angler viscometers.

10.3.3 Calorific Value or Heating Value of Fuels It is usually expressed as the Gross Calorific Value (GCV) or Higher Heating Value (HHV), and Net Calorific Value (NCV) or Lower Calorific Value (LCV). The unit of calorific value is given in kJ/kg. Fuels with higher specific gravity have higher carbon to hydrogen ratios. This causes the heavy oils to have lower gross calorific values on a weight basis.

10.3.3.1 Higher Calorific Value (HCV or GCV) The HCV or GCV is the quantity of heat liberated during the combustion of a unit mass of solid or liquid fuel under the condition that the water vapors are condensed at the room temperature of 15°C. For gaseous fuels or a cubic meter of gaseous fuel, it is under the condition that the water vapors are condensed at the room temperature of 15°C. The higher calorific value or gross calorific value of the fuel produces the maximum quantity of heat during the combustion of fuel. The higher calorific value (HCV) is also called the upper heating value. The HCV for solid and liquid fuels are determined at constant volume, whereas they are determined at constant pressure for gaseous fuels.

10.3.3.2 Lower Calorific Value The LCV or NCV differs from the higher calorific value by the heat of evaporation of the water vapour, that leaves with the combustible products. The heat of evaporation of water formed at atmospheric pressure is 2460 kJ/kg.

10.10

Engineering Thermodynamics

The lower calorific value is given by

LCV = HCV – the heat of evaporation

LCV = HCV – (Mass of steam formed per kg of fuel) × (heat of evaporation of water formed per kg of fuel)

The heat of evaporation (latent heat of evaporation) of water is generally taken at the standard temperature of 15°C.

LCV = HCV – 2460 × 9H2.

It should be noted that fuels should be compared, based on the net calorific value. The calorific value of coal varies considerably depending on the ash, moisture content and the type of coal, while the calorific value of fuel oils is much more consistent.

10.3.4 Flash and Fire Point The flash point of a liquid fuel is defined as the lowest temperature, at which an introduction of a spark or ignition source causes the fuel vapour to ignite under specified conditions. A high volatile fuel has a lower flash point. It is necessary that the flash point of a liquid fuel should be maintained greater than 339 K during its storage. Gasoline has a flash point below –40°C, which cannot be determined by any standard methods. The lowest temperature at which the fuel vapour continues to burn is called fire point.

10.3.5 Pour Point The temperature which is 2.8°C higher than that at which the fuel will cease to flow under specified conditions, is termed as pour point. It is the main indication of the flow of fuel. An increase in the viscosity and crystallization of wax causes the cessation of flow. Fuels with high densities and wax forming tendency have a higher pour point. This identifies the temperature below which the fuel can be pumped, only by providing some jackets in the fuel pipe lines. The pour point is one of the desirable properties when the fuel is used in cold climatic conditions.

10.3.6 Cloud Point This is the temperature that is higher than the pour point (usually by 5°C to 6°C), at which the oil becomes cloudy when it is cooled in a specific manner. A higher crystallization of wax or moisture content implies a higher cloud point. The cloud point is more important than the pour point for diesel fuel, because the wax crystals can choke the fuel filters and stop the flow even when the oil is above its pour point.

10.3.7 Ignition Temperature In a normal combustion process it should be noted that the fuel must be properly vaporised and mixed with oxygen in the surrounding air. The vapour of a liquid

10.11

Fuels and Combustion

fuel is formed by evaporation from the liquid surface, due to an envelope of a part of heat from the flame.

10.4 DETERMINATION OF THE HEATING VALUE OF A FUEL Let us consider 100 kg of fuel, whose contents are carbon (C), hydrogen (H), oxygen (O) and Sulphur (S).The assumption made is that, oxygen present in the fuel is already fully in combination with hydrogen. Water constitutes 8 parts by mass of oxygen and 1 part of hydrogen. Therefore, it is essential to deduct 1/8 of the oxygen present from the total hydrogen. That is [H – (O/8)] (a) The quantity of heat evolved due to combustion of carbon is C × 33800 kJ. (b) The quantity of heat evolved due to combustion of hydrogen is [H – (O/8)] × 144000 kJ. (c) The quantity of heat evolved due to the combustion of sulphur is S × 9270 kJ. Therefore, the total heat generated due to the combustion of 100 kg fuel is, = (a) + (b) + (c) = 33800 C + 144000 [H – (O/8)] + 9270 S kJ

...(10.1)

Per kg of fuel

HCV =

33800 C + 144000[H – (O /8)] + 9270 S kJ ...(10.2) 100

= 338 C + 1440[H – (O/8)] + 92.7 S kJ/kg

...(10.3)

The above equation is called Dulong’s formula. In this formula C is the mass percent of carbon, H is the mass percent of hydrogen, O is the mass percent of oxygen, and S is the mass percent of sulfur in the coal.

10.4.1 Lower Calorific Value The lower calorific value of a fuel can be calculated by the following expression

LCV = HCV – 2460 × steam formed kJ/kg

= HCV – ms × 2460 kJ/kg

...(10.4)

ms = Mass of steam formed in unit mass per kg of fuel.

If the LCV of hydrogen is taken as 121000 kJ/kg then LCV = 35000 C + 121000 (H – O/8) + 9160 S.

...(10.5)

10.4.2 Heating Value of Gaseous Fuels The calorific value of a gaseous fuel is determined by a separate chemical analysis. The composition of the fuel is first found out. The heat evolved during

10.12

Engineering Thermodynamics

the complete combustion of the given volume of gaseous mixture and in the subsequent cooling of the combustion products is the calorific value of gaseous fuel. The chemical analysis gives the volume of the individual gas present in the gaseous mixture. The calorific value of hydrocarbon such as CH4 and C2H6 may be calculated by adding the calorific values of carbon and hydrogen in a chemical reaction.

10.4.3 Experimental Method for Determination of the Heating Value The calorific value of fuels may be determined experimentally by two methods which are commonly available. The device used, is called the calorimeter. The calorimeters used are (i) Bomb calorimeter (ii) Junker’s calorimeter (iii) Boy’s calorimeter. The bomb calorimeter is used to measure the higher calorific value of solid as well as liquid fuels. The known weight of a substance or fuel is burnt in the calorimeter. The heat evolved during combustion is determined by measuring the rise in the temperature of the surrounding water in the calorimeter.

10.4.3.1 Bomb Calorimeter The set up of bomb calorimeter is shown in Fig. 10.1. It consists mainly of the following (i) Crucible (ii) Steel bomb (iii) Fuse wire (iv) Ignition rod (v) Thermometer and (vi) Vessel. In this arrangement, a closed shell called the Bomb is immersed in water. The water quantity is already measured. The top of the bomb has a valve which admits oxygen. The fuel substance is placed inside the bomb in a crucible. A few ignition rods support the crucible. The fuel substance is ignited with the help of the ignition rod fuse wires. The bottom end of the ignition rod is connected to the battery terminals. When the current passes through the ignition rod from the battery, the fuse wire is enabled to ignite the fuel. The heat liberated during the combustion is immediately absorbed by the water surrounded the bomb. Due to the rise in the temperature of water, a thermometer of the range of 0.01°C, shows the change in temperature of the water. Oxygen is admitted through the intake valve, and the exhaust gases of combustion will leave through the exhaust valve. The fuel is to be burnt at a constant volume and it is necessary to keep the bomb at high pressure. Initially, the sample of fuel or substance whose calorific value is to be determined, is weighed and kept in the crucible. The intake valve is opened to admit the oxygen inside the bomb till the pressure rises to 30 bar. Then the bomb is submerged in the water filled in a vessel. This vessel is highly insulated, so that radiation loss is avoided. The initial temperature of the water and the calorimeter is noted. When the content of the crucible reaches the steady state the fuel is ignited and allowed to burn. The heat liberated during combustion is gradually absorbed by the water surrounded it. The thermometer notes the rise in temperature.

10.13

Fuels and Combustion

Let us consider the following: (a) Mass of fuel substance taken in the bomb = mf . kg (b) Mass of water filled in calorimeter = mw kg (c) Mass of calorimeter = mc kg Exhaust gases out

Valve Shell or bomb Ignition rod Silica crucible Ignition wires Battery

Fig. 10.1 Bomb calorimeter

(d) Initial temperature of water in calorimeter = T1°C (e) Final temperature of water in calorimeter = T2°C (f) Higher calorific value of fuel = CV kJ/kg Heat released by the fuel = mass of fuel × calorific value of fuel = mf × CV kJ

...(10.6)

Heat absorbed by water = Total mass of the water and calorimeter × specific heat of water × rise in temperature = (mw + mc) Cw(T2 – T1) ...(10.7) Now we shall write the heat balance equation.

eat released by the fuel = Heat gained by water and fuel during H during combustion combustion calorimeter

mf × CV = (mw + mc) Cw(T2 – T1) CV =

( mw + mc ) C w ( T2 – T1 ) mf

Normally, the loss due to radiation is neglected.

...(10.8) kJ/kg

...(10.9)

10.14

Engineering Thermodynamics

10.5 BASIC COMBUSTION EQUATIONS A complete combustion is one in which the oxygen in the air mixes with the fuel and completely burns, so that there will be no oxygen remaining in the products of combustion. The main products of the complete combustion are CO2, SO2, NO2 and H2O. It is better to analyze the chemical reaction of combustion, in terms of the conservation of atoms rather than the conservation of the mass. (i) C + O2 → CO2

...(10.10)

1 kmol C + 1 kmol O2 ↔

1 kmol CO2

...(10.11)

12 kg C + 32 kg O2

44 kg CO2 ...(10.12)

→

Dividing the above equation by 12 1 kg C +

32 kg O2 12

→

1 kg C +

8 kg O2 3

→

44 kg CO2 12 11 kg CO2 3

(ii) 2C + O2 → 2 CO

...(10.13) ...(10.14) ...(10.15)

2 kmol C + 1 kmol O2 ↔ 2 kmol CO2 (10.16) 24 kg C + 32 kg O2

→ 56 kg CO ...(10.17)

Dividing the above equation by 24 1 kg C +

32 kg O2 24

→

56 kg CO 24

...(10.18)

1 kg C +

4 kg O2 3

→

7 kg CO 3

...(10.19)

(iii) 2H2 + O2 → 2H2O

...(10.20)

2 kmol H2 + 1 kmol O2 ↔

2 kmol H2O

...(10.21)

4 kg H2 + 32 kg O2

36 kg H2O ...(10.22)

→

Dividing the above equation by 4 1 kg H2 + 8 kg O2

→

4 kg H2O ...(10.23)

(iv) S + O2 → SO2

...(10.24)

2 kmol S + 1 kmol O2 ↔ 1 kmol SO2 ...(10.25) 32 kg S + 32 kg O2

→

64 kg SO2

...(10.26)

Dividing the above by 32 1 kg S + 1 kg O2 → 2 kg SO2 ...(10.27) From the above equations, the amount of oxygen required for each element in combustion is given in Table 10.5.

10.15

Fuels and Combustion

Table 10.5 Summary of the Basic Combustion Equations S.No.

Element

Molecular weight

Product obtained in the reaction

Product Quantity formed per kg of oxygen of reactant required (kg) (kg of O2/kg)

1.

C

12

CO2

3.67

2.67

2.

C

12

CO

2.33

1.33

3.

H2

2

H 2O

9

8

4.

S

32

SO2

2

1

While calculating the combustion parameters such as theoretical air, and excess air, the volume or mass of the constituents is considered.

10.6 AIR FUEL RATIOS FOR COMBUSTION There are two terms used for air fuel ratio calculations in combustion. They are (i) Theoretical air fuel ratio or Stoichiometric air fuel ratio and (ii) Excess air. By knowing the values of the theoretical air fuel ratio and excess air for the combustion of a particular fuel in a device, one can design the fuel and air supply systems accordingly.

10.6.1 Stoichiometric Air Fuel Ratio The theoretical or stoichiometric air per kg of fuel is the minimum quantity of air required, which provides oxygen for the complete combustion of the fuel. There is no oxygen in the end products when the complete combustion (Oxidation) is achieved with this theoretical quantity of air.

10.6.2 Excess Air It is the quantity of air supplied in addition (excess) the theoretical quantity of air required for complete combustion. This excess air will come out in the products of combustion as oxygen and nitrogen. In combustion the problem occurs only in supplying sufficient air. The knowledge of the air fuel ratio helps to know the excess or deficit of air supply.

10.6.3 Determination of Stoichiometric or Theoretical Air Fuel Ratio The combustion of simple hydrocarbon fuels forms carbon dioxide (CO2) from all the carbon and water (H2O) from the hydrogen. In general, the chemical formula of hydrocarbon fuel is given as CnHm.

10.16

Engineering Thermodynamics

       C n Hm +  n + m  O 2 → n CO 2 +  m  H 2 O 4  2

...(10.28)

In all the combustion processes, air is involved rather than pure oxygen. Therefore, the nitrogen in the air produces nitrogen oxides. Hydrocarbon fuels not only contain carbon, they also contain elements other than carbon and hydrogen. These elements may be transformed during combustion into other form. Therefore, combustion is not always complete, and the combustible gases contain unburned and partially burned products in addition to CO2 and H2O. Air is composed of oxygen, nitrogen, and small amounts of carbon dioxide, water vapor, argon and other elements. The major proportion in air is nitrogen, and then oxygen. The nitrogen and oxygen percentages on a volume and mass basis are given in Table 10.6. Table 10.6 Nitrogen and Oxygen in Air on Volume and Mass Basis Element

Volume basis

Mass basis

Nitrogen

79

77

Oxygen

21

23

From the above table it is clear that, for every mole of oxygen required for combustion, 3.76 mol of nitrogen is introduced in the combustion equation. Therefore, the equation can be written as follows m m m     C n Hm +  n +  ( O 2 + 3.76 N 2 ) → n CO 2 +   H 2 O + 3.76  n +  N 2 ...(10.29) 4 4  2 

From the above equation it is understood that for every mole of fuel burned, 3.76 (n + m/4) mol of air are required and 3.76 (n + m/4) + m/4 mol of combustion products are generated. The molar fuel air ratio for stoichiometric combustion is 1 I [3.76(n + m/4)]. In order to calculate the stoichiometric air fuel ratio, first write the combustion equation for complete combustion of the given hydrocarbon fuel. Then, balance the equation. Replace the O2 term in the reactants side by (O2 + 3.76 N2). Apply the value of the molecular weight for the term (O2 + 3.76 N2). This will give the amount of air required for complete combustion. An example is given below: Consider a hydrocarbon fuel C10H22. The theoretical air required for this fuel can be determined as follows C10H22 + O2 → CO2 + H2O m m m   C n Hm +  n +  ( O 2 + 3.76 N 2 ) → n CO 2 + H 2 O + 3.76  n +  N 2 4 2 4   Substituting n = 10 and m = 22 in the above equation we get 22  22 22    C 19 H 22 +  10 + H 2 O + 3.76  10 +  ( O 2 + 3.76 N 2 ) → 10CO 2 +  N2 4  2 4   

10.17

Fuels and Combustion

Taking the term 15.5 (O2 + 3.76 N2) and substituting the molecular weights of O2 and N2 respectively we get the mass of air required.

Mass of air = 15.5(32 + 3.76 × 28)



= 2127.84



Mass of fuel = C10H22 = (10 × 12 + 22 × 1) = 142

2127.84 142 Air fuelratio = 14.49 kg of air/kg of fuel.



Air fuel ratio =



Alternatively; the same value can be checked by balancing the equation; Writing the combustion equation for C10H22 for complete combustion; C10H22 + 15.5 O2 → 10CO2 + 11H2 O

C10H22 + 15.5(O2 + 3.76 N2) → 10CO2 + 11H2 O + 105.28 N2.

10.7 CONVERSION OF FUEL CONSTITUENTS GIVEN IN VOLUME TO MASS PERCENTAGE Suppose, a flue gas of a burnt fuel has CO2, CO, O2 and N2 in it and they are given in volume, then their volume can be easily converted into the mass percentage. The conversion is given in Table 10.7. The procedure is given below: 1. The volumes of each constituent A, B, C and D are given in m3 like x1, x2, x3, x4 respectively.

2. Multiply the volumes of these constituents by their corresponding molecular masses. y1, y2, y3, and y4.

3. The products of volume and molecular masses are added to find the total proportional mass. Table 10.7 Volumetric Analysis into Mass (Gravimetric) Analysis Component

Gas

Volume of constitute (m3) x

Molecular mass y

Product, xy

A

CO2

x1

y1

x1 y1

B

CO

x2

y1

x2 y2

C

O2

x3

y1

x3 y3

D

N2

x4

y1

x4 y4

% by mass xy ∑ xy

x1y 1 ∑ xy

x2 y 2 ∑ xy

x3 y 3 ∑ xy

x4 y 4 ∑ xy

10.18

Engineering Thermodynamics

Σxy = x1 y1 + x2 y2 + x3 y3 + x4 y4 ...(10.30)



4. Mass of gas constituent in % A =

x1 y 1 × 100 ...(10.31) Σxy

5. Mass of gas constituent in % B =

x2 y 2 × 100 ...(10.32) Σxy

6. Mass of gas constituent in % C =

x3 y 3 × 100 ...(10.33) Σxy

x4 y 4 × 100 ...(10.34) Σxy If we multiply the mass of the each constituent by 100, we will get the values in percentage of mass of the gas constituent. 7. Mass of gas constituent in % D =

10.8 EXCESS AIR SUPPLIED PER KG OF FUEL As mentioned earlier, excess air supports, in addition to the minimum air required per kg of fuel used in combustion, to approach conditions of stoichiometry. In most of the combustion systems, air supplied in excess quantity and fuel will be fully oxidized and the complete combustion occurs. The percentage of excess air is given as the following %Excess air =

Actual air supplied – Theoretical or stoichiometric air × 100 Theoretical or stoichiometric air ...(10.35)

10.9 QUANTITY OF CARBON IN FLUE GAS Flue gases from combustion systems normally contain carbon dioxide and carbon monoxide. The weight of carbon present in one kg of flue gas can be calculated from the amount of carbon dioxide and carbon monoxide present in it. Let us consider the general equation for the carbon reacting with oxygen during the combustion process

C + O2 → CO2 1 kg C +

...(10.36)

11 8 kg O2 → kg CO2 3 3

...(10.37)

7 4 kg O2 → kg CO 3 3

...(10.38)

Similarly,

1 kg C +

Therefore, 1 kg of CO2 contains 3 kg carbon. 7

3 kg carbon and 1 kg of CO contains 11

10.19

Fuels and Combustion

Hence, the weight of carbon in the flue gas mb = Weight of carbon in carbon dioxide × weight of carbon in carbon monoxide =

3 3 CO2 + CO. 7 11

...(10.39)

10.10 QUANTITY OF FLUE GAS PER KG OF FUEL BURNT The quantity of dry flue gas present per kg of fuel can be obtained by comparing the quantity of carbon present in the flue gases and the weight of carbon is the fuel. Let us assume

ma = Weight of carbon present per kg of fuel



mb = Weight of carbon present per kg of flue gases



m = Weight of flue gas per kg of fuel



m =

m =

Weight of carbon present per kg of fuel Weight of carbon present per kg of flue gases ma . ...(10.40) mb

10.11 FLUE GAS ANALYSIS Each constituent of dry flue gas in percentage volume can be obtained by the volumetric analysis. It can be done by an instrument called the Orsat gas analyser or Orsat apparatus. It is a small instrument used to analyse a gas sample (typically fossil fuel flue gas) for its oxygen, carbon monoxide and carbon dioxide content, within a fixed volume of a sample gas (100 cc). It works in a very simple method. The apparatus is shown in Fig. 10.2.

Fig. 10.2 Orsat apparatus

10.20

Engineering Thermodynamics

10.11.1 Construction The apparatus consists essentially of a calibrated water-jacketed gas burette connected by a glass capillary tubing to two or three absorption pipettes containing chemical solutions, that absorb the gases which are required to be measured. For safety and portability, the apparatus is usually encased in a wooden box.

10.11.2 Method of Analysis Initially, by arranging the rubber tube, the gas to be analyzed is drawn into the burette and flushed through several times. For easy calculation, 100 ml is withdrawn. Stopcocks are used to isolate the absorption burettes, the level of gas in the leveling bottle, and the burette is adjusted to the zero point of the burette. The gas is then passed into the caustic potash burette; it remains for about two minutes, and is then withdrawn, isolating the remaining gas via the stopcock arrangement. The process is repeated to ensure full absorption. After leveling the liquid in the bottle and the burette, the remaining volume of gas in the burette indicates the percentage of carbon dioxide absorbed. The same method is repeated for oxygen, using the pyrogallol, and for carbon monoxide using the ammonia cuprous chloride.

10.11.3 Advantages (i) Although advanced instruments are available now the Orsat apparatus is still considered as a reliable method of measurement (ii) It is relatively simple to use.

10.11.4 Limitations (i) It is not particularly accurate at detecting very low concentrations. (ii) Requires more time to measure each component of the sample.

10.12 FLAME TEMPERATURE When the fuel air mixture is burnt and the combustion products are obtained, the average temperature of the combustion products is known as the flame temperature. It can be given in four names; (i) theoretical flame temperature (ii) actual flame temperature (iii) adiabatic flame temperature and (iv) maximum adiabatic flame temperature. (i) Theoretical flame temperature: It is nothing but, the temperature at which fuel air undergoes complete combustion, and the entire heat of combustion goes to heat of the products of combustion.

10.21

Fuels and Combustion

(ii) Adiabatic flame temperature: In actual, the combustion is not fully complete at high temperatures because disassociation of reactants takes place. If the endothermic effect of disassociation reactions is considered, then the adiabatic flame temperature is more useful. The adiabatic flame temperature is defined as the maximum temperature that products can reach in a combustion process, assuming that the process is adiabatic, involving no work and negligible changes in the kinetic and potential energies. (iii) Actual flame temperature: In practical applications, there will always be some heat loss to the surroundings. Therefore, the adiabatic flame temperature is not very effective. In such cases, the resultant average temperature is taken, which is called as the actual flame temperature. (vi) Maximum adiabatic flame temperature: This is the maximum temperature that products can reach in a combustion process when the fuel is slightly in excess of the stoichiometric composition and the combustion process is adiabatic. For calculating the flame temperature, one has to write the heat balance between the reactants and the products in the combustion process. Let LCV : Lower calorific value of the fuel, kJ/kg ∆hf : Specific enthalpy of fuel above the reference state, kJ/kg ∆ha : Specific enthalpy of air above the reference state, kJ/kg ma : Mass of air supplied per kg of fuel, kg mg : Mass of flue gases per kg of fuel, kg ∆hg : Specific enthalpy of flue gas above the reference state, kJ/kg qd : Heat loss due to disassociation reaction of gases, kJ/kg qs : Heat loss due to surroundings, kJ/kg. Then the heat balance equation between the reactants and the products can be written as LCV + ∆hf + ma ∆ha = mg∆hg + qd + qs



...(10.41)

In this ∆hg can be calculated by the following = ∆h g T f C − Tr C ps T –0 ps ( T f –0 ) ( r )          

...(10.42)

Where, Tf = Flame temperature °C Tr = Temperature at reference state, 25°C Cps(T

f–

Cps(T

0) =

r – 0)

Mean specific heat of flue gases between Tf and 0°C

= Mean specific heat of flue gases between Tr and 0°C

Substituting Eqn. (10.42) in Eqn. (10.41) and rearranging the equation to get Tf : Tf =

LCV + ∆h f + ma ∆ha – qd – qs + mg Tr C ps (Tr –0) mg C ps (T f –0)



...(10. 43)

10.22

Engineering Thermodynamics

10.13 INTERNAL ENERGY OF COMBUSTION Consider the number of reactants that undergo the combustion process that ends up number of products as shown in Fig. 10.3. Let the internal energy of the reactants be Ur and the internal energy of the products is Up. Also, the enthalpy of combustion in the reaction is ∆H. Reactants at reference state (T0)

Products at reference state (T2)

Reactants at reference state (T1)

Fig. 10.3 Reactants and products in a combustion process

Then, the internal energy of combustion is the difference between the internal energy of the products and the internal energy of the reactants, including the enthalpy of combustion of the reaction. Upr = Up – Ur KJ

...(10.44)

The internal energy is the point function and not a path function. Therefore, the internal energy of a series of reaction depends on the end states and not the path followed. Hence, the equation for internal energy can be written as; U2 – U0 = (U2 – U1) – (U0 – U1)

...(10.45)

As we know that the internal energy is the function of temperature the equation can be re-written as     = U2 – U0

∑ mC v (T1 – T0 ) + ∑ mC v (T2 – T1 ) KJ p

...(10.46)

r

In terms of the enthalpy of formation and combustion, the internal energy for the above can be given as  U2 = – U 0 m  ∑ ∆h f + ∆h – pv      p

(



)p + ∑ ( h f + ∆h – pv )r  KJ ...(10.47) r



10.14 ENTHALPY OF COMBUSTION The enthalpy of the combustion of 1 mol of fuel is defined as the difference between the enthalpy of products at a reference state and the enthalpy of the reactants at the same reference state, when the reaction completes combustion with stoichiometric air.

Δhc = hp – hr

where,

Δhc = Specific enthalpy of combustion, KJ/kg



hp = Specific enthalpy of products, KJ/kg

hr = Specific enthalpy of reactants, KJ/kg

...(10.48)

10.23

Fuels and Combustion

Generally, the reference state is taken at 1 atm pressure and 25°C. In thermodynamic applications, specifically for combustion systems, the change in enthalpy is considered as an important factor, instead of absolute enthalpy. Therefore, the selection of the reference state for the calculation of the enthalpy of combustion is very important. For chemical reactions, the enthalpy of combustion depends on the phase in which the reactants and products exist. If the phase exists the in the solid phase, then it is indicated as “s”. Similarly, the liquid and gaseous phases are indicated as “l” and “g” respectively. The values of the enthalpy of combustion for various substances that exist in both liquid and gaseous states are given in Table 10.8. Table 10.8 Enthalpy of Combustion Values Chemical substance

Chemical formula

Molecular weight

Product(l) kJ/kmol

Product(g) kJ/kmol

Acetylene

C2H2(g)

26.038

–1300500

–1256400

Benzene

C6H6(g)

78.114

–3303800

–3171700

Carbon (graphite)

C(s)

12.011

–393790

–393790

Carbon monoxide

CO(g)

28.011

–283190

–283190

Ethane

C2H6(g)

30.070

–1560900

–1428800

Ethylene

C2H4(g)

28.054

–1411900

–1323800

Hydrogen

H2(g)

2.016

–286030

–242000

Methane

CH4(g)

16.043

–890900

–802800

Octane

C8H18(g)

114.23

–5515900

–5119500

Octane

C8H18(l)

114.23

–5474500

–5078200

Propane

C3H8(g)

44.097

–2221500

–2045400

10.15 ENTHALPY OF FORMATION The enthalpy of the formation of a chemical compound or a substance, is defined as the change in enthalpy between the formation of 1 mole of the compound or the substance from its elements, and all the elements in their standard states. Its symbol is ΔHf° or Δf H°. The superscript indicates that the process is carried out under standard conditions. The standard states for the different cases are as follows:

l In the case of a gas: The standard state is considered as a pressure of exactly



l Suppose a substance is present in a solution, then a concentration of exactly



l For a pure substance in a condensed state (a liquid or a solid): Pure liquid



l For

1 bar.

1 mole at a pressure of 1 bar is considered. or solid is taken as standard state

an element: the form in which the element is most stable under 1 atm of pressure and the specified temperature. (Usually 25°C or 298.15 K)

10.24

Engineering Thermodynamics

The enthalpy of formation is given by the following expression

Δhf = (hfo)compound – Σ(ni ho)elements KJ/kg

...(10.49)

If all the stable elements at the reference state have negligible values, then the above equation becomes

Δhf = (hfo)compound

...(10.50)

For calculation, the property values are obtained by first assigning all of the elements (such as N2, O2, C) in their chemically stable form, at the standard reference state, a value of zero (such as N2, O2, N2, C). During the conversion of a particular set of reactants to a particular set of products in a chemical process, the change in enthalpy is the same, even if the reaction takes place in one step or in a series of steps. This is because the enthalpy is a state function, which means that the change in enthalpy from a particular initial state to a particular final state is independent of the path.

Solved Problems Problem 10.1 Calculate the theoretical air fuel ratio of methane. Given Fuel : Methane chemical formula : CH4 Solution (i) Write the combustion equation for methane CH4 + O2 → CO2 + H2O m m m   C n Hm +  n +  ( O 2 + 3.76 N 2 ) → n CO 2 + H 2 O + 3.76  n +  N 2 4 2 4    n = 1; m = 4 Substituting n = 1; m = 4 4 4 4   CH 4  1 +  ( O 2 + 3.76 N 2 ) → n CO 2 + H 2 O + 3.76  1 +  N 2 4 2 4   Then the equation becomes CH4 + 2(O2 + 3.76 N2) → CO2 + 2H2O + 2N2. (ii) Take the second term in the reaction 2(O2 + 3.76 N2) which is used to determine the theoretical mass of air required. = 2 (O2 + 3.76 N2) Theoretical mass of air required for methane = 2 (O2 + 3.76 N2) Substitute the molecular weight of O2 and N2 we get = 2(32 + 3.76 × 28) = 274.56 kg

Mass of fuel = (12 + 4 × 1) = 16

10.25

Fuels and Combustion

274.5 16 Air fuel ratio for methane = 17.16. Air fuel ratio =

Problem 10.2 Formulate the combustion equation for n-Butane (C4H10) and calculate the following: (a) Theoretical air fuel ratio (b) 50% excess air to be supplied. Solution (a) Theoretical air fuel ratio combustion equation for n-Butane (C4H10) can be written as: m m m   C n Hm  n +  ( O 2 + 3.78 N 2 ) → n CO 2 + H 2 O + 3.76  n +  N 2 4 2 4    Substuting n = 4; m = 10 10   6.5 4 + = 4   C4H10 + 6.5(O2 + 3.76 N2) → 4 CO2 + 5H2O + 6.5 N2

Air fuel ratio =

Mass of air Mass of fuel

Mass of air = 6.5 (32 + 3.76 × 28)

= 892.3 kg

Mass of fuel = 4 × 12 + 10

= 58 kg

Air fuel ratio =

8923 58



Air fuel ratio =

8923 kg of air 58 kg of fuel



Air fuel ratio = 15.5

kg of air . kg of fuel

(b) If the excess air supplied is 50%, then the term (O2 + 3.76 N2) is multiplied by 1.5 with the existing combustion equation for C4H10: Therefore,

C4H10 + 6.5 × 1.5(O2 + 3.76 N2) → 4(CO2) + 5(H2O) + 24.4(N2) After balancing the equation, we get

Mass of air = 6.5 × 1.5(32 + 3.76 × 28)

= 1338.48 kg

Mass of fuel = 4 × 12 + 10 × 1 kg



= 58 kg

10.26

Engineering Thermodynamics

Therefore, 1338.48 58



Air fuel ratio =



Air fuel ratio = 23.07

kg of air . kg of fuel

Problem 10.3 In a particular combustion device Propane is burnt with 61% excess air. The air enters at 25°C. If the combustion is complete and the total pressure is 1.01 bar then find (i) The air fuel ratio (ii) Percentage of carbon dioxide by volume in the products. Given Excess air: 61% Pressure: 1.01 bar Temperature: 25°C or 278 K Solution (i) The air fuel ratio: If the excess air supplied is 61%, then the theoretical air is 161%. Writing the combustion equation for C3H8 m m m   C n Hm +  n +  ( O 2 + 3.76 N 2 ) → n CO 2 + H 2 O +  n +  N 2 4 2 4   Substituting n = 3 and m = 8

8 m   C 3 H8 +  3 +  ( O 2 + 3.76 N 2 ) → 3 CO 2 + 4 H 2 O +  n +  N 2 4 4  

Air fuel ratio =

kg of air kg of fuel

Taking the oxygen term = 5(O2 + 3.78 N2)

= 5 (32 + 3.78 × 28) = 686.4 kg of air

Mass of air required = 5 (32 + 3.78 × 28) Mass of fuel = (C3H8) = (3 × 12 + 8 × 1)

= 44 kg

686 kg of air 44 kg of fuel



Air fuel ratio =



Air fuel ratio = 15.6.

(ii) Consider the combustion equation after balancing No of mole of CO2 and the total mole of products in the equation are [NCO = 3 Nprod] = 3 + 4 + 3.05 + 30.27 = 40.32 2

10.27

Fuels and Combustion

The volume percentage of CO2 =

N CO2

N prod

3 × 100 25.8                =7.4%                =

Similarly,         NH

2O

=4

Then, the volume percentage of water vapour =

N H2 O N prod

4 × 100 40.32                    =9.9%. The volume percentage of water vapour =

Problem 10.4 The volumetric analysis of the dry products of combustion analyzed with the Orsat apparatus are obtained as follows; CO2 – 10%, CO – 1%, O2 – 3% and N2 – 86%. The products of combustion were obtained from the burning of ethane. Ethane has the chemical composition by mass as C = 84%, H2 = 14% and O2 = 2%. Calculate the (i) mass of dry flue gases formed per kg of ethane burned (ii) theoretical air fuel ratio and (iii) the percentage of excess air. Given Volumetric analysis of the combustible products: CO2 – 10%, CO – 1%, O2 – 3% N2 – 86%. Chemical composition: C = 84%, H2 = 14 y, and q2 = 2% Solution First convert the volume percentage of the components into mass percentage. Component

Volume percentage, x

Molecular weight, y

Product z = xy

Mass percentage = [z/Σz] × 100

CO2

10

44

440

14.80484522

CO

1

28

28

0.942126514

3

32

96

3.230148048

86

28

2408

81.02288022

O2 N2

The mass of carbon per kg of dry flue gas = mass of carbon from CO2                          + mass of carbon from CO. 3 3 × 0.148 + × 0.0094 11 7   = 0.04438 kg

        =       

The mass of dry flue gas formed per kg of ethane =

The mass of dry flue gas formed per kg of ethane =

Percentage of carbon Mass of carbon per kg of dry flue gas 0.84 0.04438

10.28

Engineering Thermodynamics

Mass of dry flue gas formed per kg of ethane = 18.92 kg. Percentage of N 2 × Percentage of carbon (ii) Mass of air supplied = 33 (% of CO + % of CO 2 in dry flue gas ) Per kg of ethane burned =

84 × 86 33 ( 10 + 1 )

Per kg of ethane burned = 19.0 kg of air/kg of fuel supplied. (iii) % of excess air =

Mass of actual air supplied per kg of ethane burned Theoretical mass of air supplied per kg of ethane burned

From the basic combustion equations we can get the theoretical air required C2H6 + O2 → CO2 + H2O 6  C 2 H6 +  2 +  ( O 2 + 3.76 N 2 ) → 2 CO 2 + 3 H 2 O = 3.5 N 2 4   

C 2 H6 + ( 3.5 O 2 + 3.5 × 3.76 N 2 ) → 2 CO 2 + 3 H 2 O = 3.5 N 2

Multiplying molecular weight of O2 and N2 and for C2 and H6 Quantity of air required = 217.28 kg Quantity of fuel = 30 kg  

Excess air supplied = Actual mass of air supplied –

Theoretical mass of air required

Excess air supplied = 19.0 – 11.53 kg



Excess air supplied = 7.47 kg



% of excess air =



% of excess air =



Mass of excess air supplied per kg of ethane Theoretical mass of air supplied per kg of ethane burned

7.47 × 100 11.53 % of excess air = 64.78%.

Problem 10.5 A fuel sample had 86% carbon and 13% hydrogen in it. It was found that 15% carbon in the fuel is converted in to CO even when 20% excess air was supplied. Calculate the following: (i) Minimum quantity if air required per kg of fuel (ii) Oxygen in the flue gas in kg per kg of fuel burned (iii) Weight analysis of the flue gas. Given Percentage of carbon in the fuel: 86% Percentage of hydrogen in the fuel: 13% Excess air : 20%

10.29

Fuels and Combustion

Solution (i) Minimum amount of air required: Oxygen required by the fuel = % C in the fuel × Oxygen required by carbon (a) = 0.86 ×              

=2.2933.

Molecular weight of oxygen Molecular weight of carbon

32 12

Molecular weight of oxygen Oxygen required by hydrogen = % H in the fuel × Molecular weight of hydrogen Oxygen required by the hydrogen (b) = 0.14 ×              

    =1.04

32 4

Oxygen required by the fuel = (a) + (b)               =3.33 Minimum air required by the fuel = Air required by the fuel =

Total oxygen reuired by the fuel % by mass of oxygen in the air

3.33 = 14.491 0.23

Air required by the fuel = 14.491 kg of air/kg of fuel burned (ii) Actual supply of air = Excess air % × minimum air required Actual supply of air = 1.2 × 14.491 Actual supply of air = 17.389 kg of air/kg of fuel burned. (iii) Oxygen in the flue gas: Chemical reaction

Mass of reactant

Oxygen consumed

Quantity of products of combustion (kg) CO

C→CO

(A) C% × 0.15 = 0.86 × 0.15

C→CO2

= 0.129

= 0.172

(B) C% × 0.85 = 0.7225

32 × (B) 12 = 1.9266

(C) 0.15 = 0.15

8 × (C) = 1.2

= 0.86 × 0.85 H→H2O

32 × (A) 24

CO2

H2O

O2

N2

56 ×A 24

= 0.301 44 ×B 12

= 2.694 0.15 × 9 = 1.35

[Table Cont...]

10.30

Engineering Thermodynamics

[Table Cont...] O2 supplied

(D)Actual supply of air × mass % of oxygen = 17.784 × 0.23 = 4.09

N2

(E)Actual supply of air × mass % of nitrogen = 17.784 × 0.77 = 13.69

Total O2 consumed A+B+C = 3.298

O2 supplied O2 used 4.09 – 3.298 = 0.792

Oxygen present in the product of combustion = Amount of oxygen supplied           – Oxygen used Mass analysis of flue gas Constituent

KJ/kg of fuel

% on total basis

% on dry flue gas basis

CO

0.301

1.598768

1.722264

CO2

2.694

14.30924

15.41454

O2

0.792

4.206724

4.53167

N2

13.69

72.71472

78.33152

H2O

1.35

7.170553

7.724438

Total

18.827

100

100

Dry flue gas = Total – water vapour

18.827 – 1.35 = 17.477

Problem 10.6 Methane is burned with 15% excess air during the combustion process. Determine the following if the combustion is complete and the total pressure is 100 kPa (a) air fuel ratio (b) volume percentage of the constituents of the flue gas (c) mass percentage of the constituents of the flue gas (d) dew point temperature Given Fuel: methane  Chemical formula : CH4,  Excess air : 15% Solution (a) Air fuel ratio : Write the combustion equation for methane CH4 + O2 →  CO2 + H2O CH4 + 2(O2 + 3.78 N2) →  CO2 + 2H2O + 2N2

(Referring solution for problem. 1)

10.31

Fuels and Combustion

Since excess air supplied is 15%, then the equation becomes CH4 + 1.15 × 2(O2 + 3.76 N2) →  CO2 + 2H2O + 3.76 N2 CH4 + 2.3(O2 + 3.76 N2) → CO2 + 2H2O + 0.6O2 + 8.648 N2

Mass of air supplied = 2.3 × [32 + 3.76 × 28]

= 315.744

Mass of fuel = CH4 = 1 × 12 + 4 × 1

= 16 315.7 16



Air fuel ratio =



Air fuel ratio = 19.73.

(b) Percentage volume of the constituents of flue gas: For calculating the percentage volume of the constituents of flue gas, take the number of moles of each constituent as in the table given below Component

No of moles in the constituent, N

CO2

1

0.082863772

8.286377

H2O

2

0.165727544

16.57275

O2

0.6

0.049718263

4.971826

N2

8.468

0.701690421

70.16904

Total

17.068

Y=

N ΣN

Y in %

100

(c) Volumetric analysis of the combustible products: CO2 – 10%, CO – 1%, O2 – 3% N2 – 86%. First convert the volume percentage of the components into mass percentage. Component

Volume percentage, x

Molecular weight, y

Product z = xy

Mass percentage = [z/Σz ] × 100

CO2

11.71783

44

364.6006

12.34901

H2O

35.1535

28

464.0371

15.71692

O2

3.51535

32

159.0984

5.388657

N2

49.61331

28

1964.733

66.54542

Total = Σz

2952.469

100

(d) Dew point temperature: To find the dew point temperature, first calculate the partial pressure of water vapour present in the fuel.

= PH2 O

          

N H2 O ΣN

× 100

10.32

Engineering Thermodynamics

2 = PH2 O × 100 12.06            

PH O = 16.58 kPa. 2

Obtain the saturation temperature of steam corresponding to the saturation pressure of 16.58 kPa, from the steam tables we get; Ts = 55.76°C. Problem 10.7 Calculate the adiabatic flame temperature of CH4 with 150% theoretical air in adiabatic control volume. The combustion equation of CH4 is given below CH4 + a(O2 + 3.76 N2) → x(CO2) + y(H2O) + 3.76 z(N2) For 150% theoretical air the equation becomes;, CH4 + 1.5 × 2(O2 + 3.76 N2) → CO2 + 2(H2O) + 3O2 + 5 × 3.76 (N2) Ccv = 0 = ΣNphp – ΣNrhr       p     r

Ccv = 0 = ΣNp[hf o + h(T)]p – ΣNr[hf o + h(T)]r.       p            r

where h(T) is the sensible enthalpy as a function of T. For each component; CO2 = 1 [–393520 + h(T)] H2O = 2 [–241820 + h(T)] O2 = 3 [h(T)] N2 = 18.8 [h(T)] CH4 = [– 74850] Substituting all these values in the combustion equation h(T) CO2 + 2[h(T)H O] + 3[h(T)O ] + 18.8[h(T)N2] = 802310 2

2

This equation cannot be solved in the present form. An approximate method can be used to solve this equation, in which all the components are assumed to be air. The number of moles in the equation is given below

CO2 = 1, H2O = 2, O2 = 3 and N2 = 18.8

Ntotal = 1 + 2 + 3 + 18.8 = 24.8 ΣNp[hf o + h(T)]p = NpmaCp(Tadiabatic – Tamb] Where,

ma = molecular mass of air kg/mol = 29 kg/mol

Cp

air



= Specific heat of air = 1.14 kJ/kg K

Tamb = 298 K. Substitute the values in the equation NpmaCp(Tadiabatic – Tamb) and equate to 802310 to get the value of Tadiabatic. Tadiabatic = 1275K.

10.33

Fuels and Combustion

Problem 10.8 In the combustion process of Octane, all the products of combustion are in the gaseous phase. Octane, as the reactant, is also in the gaseous form. The enthalpy of combustion at 25°C is –44812 kJ/kg. Calculate the internal energy of combustion. Given Fuel : Octane, chemical formula : C8H18 Enthalpy of combustion = –44812 kJ/kg; Reference temperature = 25°C Solution: Writing the equation for enthalpy of combustion;

∆ho = ∆uo + Ra T(Np – Na)

where,

∆ho = Enthalpy of combustion



∆uo = Internal energy of combustion

Ra = Gas constant

T = Reference temperature

Np = Number of moles of products in the reaction Na = Number of moles of reactants in the reaction ∆uo = ∆ho – RaT(Np – Na)



Consider the balanced combustion equation for Octane C8H18 + 12.5 O2 → 8CO2 + 9H2O In the above equation, number of moles of products and reactants are taken. Nr = 1 + 12.5 = 13.5 Np = 8 + 9 = 17 Np – Nr = 17 – 13.5 = 3.5 Substituting the values for ∆ho, Ra, T and (Np – Na) in the internal energy equation we get, 8.314 × 298 × 3 114 = – 44888 kJ/kg. = – 44812 +

Problem 10.9 Ethane undergoes complete combustion. All the products of combustion are in the gaseous phase. The enthalpy of combustion is –47590 kJ/kg at the reference state of 25°C. Find the enthalpy of combustion at 500°C. Take the mean specific heats of the reactants and the products at constant pressure in between 25°C and 5000°C as in the table given below C2H6

O2

CO2

H2O(g)

N2

2.8

0.889

1.049

1.987

1.066

Given Enthalpy of combustion of ethane = –47590 kJ/kg

Specific heats = Given in Table.

10.34

Engineering Thermodynamics

Solution We know that where,

ΔHc = Hp – Hr

ΔHc = enthalpy of combustion



Hp = enthalpy of products



Hr = enthalpy of reactants



Hp = H500°C – H25°C



Hr = H25°C – H540°C



ΔHc = H500°C – H25°C + ΔH25°C + H25°C – H540°C.

Writing the combustion equation for ethane with stoichiometric air; C2H6 + O2 → 2CO2 + 3H2O

Considering the molecular weight of each component to get the equation on mass basis in kg we get 30 kg C2H6 + 112 kg O2 → 88 kg CO2 + 54 kg H2O

Dividing the above equation by 30 to get 1kg of ethane. 112 30 54 88 kg C2H6 + kg O2 → kg CO2 + kg H2O 30 30 30 30  112  Otherwise for buring 1 kg of C2H6,   kg O2 is required and the reaction  30   88   54  kg H 2 O  . produces   kg CO2 and   30  30   For calculating Hr ;

H25°C – H500°C = ΣmrCp (25 – 500)

m = mass of the reactants in kg Cp = Specific heats of reactants at particular temperature, kJ/kg°C

 112  H25°C – H500°C = 1 × 2.8 × (25 – 500) +   × 0.889 × (25 – 500)  30  H25°C – H500°C = –1330 + (–1576.49) H25°C – H500°C = – 2906.49 kJ

Similarly, for Hp:

H500°C – H25°C = ΣmpCp (500 – 25)



H500°C – H25°C =



H500°C – H25°C = 3160.5 kJ/kg.

88  54  × 1.049 ( 500 – 25 ) +   × 1.987 ( 500 – 25 ) 30  30 

Substituting the values in the equation below to get ΔHc;

ΔHc = H500°C – H25°C + ΔH25°C + H500°C – H540°C

ΔHc = –2906.49 + 3160.5 – 47590 ΔHc = –44429.5 kJ/kg.

10.35

Fuels and Combustion

Problem 10.10 Octane is in the gaseous form. Calculate its enthalpy at 25°C and 1 atm for the following cases: (i) Water present in the products of combustion is in the gaseous phase (ii) Water present in the products of combustion is in the liquid phase Given Fuel: Octane Chemical formula: C8H18 Obtaining the enthalpy of formation for C8H18, CO2 and H2O from the tables; Value of enthalpy of formation for C8H18 = –249.95 Value of enthalpy of formation for CO2 = –393533 kJ/mol Value of enthalpy of formation for H2O (g) = –241.827 kJ/mol Value of enthalpy of formation for H2O (l) = –285.830 kJ/mol. We know that the combustion equation for C8H18 with stoichiometric air is; C8H18 + 12.5 O2 → 8CO2(g) + 9H2O (g) Solution (i) The water present in the products of combustion is in gaseous phase: Substitute the values of the enthalpy of formation of C8H18, CO2 and H2O(g) in the above equation ;

(∆hc) = 8(∆hf) CO2 + 9(∆hf) H2O – (∆hf) C8H18



(∆hc) = 8(–393.533) + 9(–241.827) – (–249.95)



(∆hc) = –5074.75kJ/mol

(ii) The water present in the products of combustion is in the liquid phase: Substituting the values for enthalpy of formation of C8H18, CO2 and H2O(l) in the above equation, we get; (∆hc) = 8(∆hf ) CO2 + 9(∆hf) H2O – (∆hf ) C8H18 (∆hc) = 8(–393533) + 9(–285830) – (–249.95) (∆hc) = –5470.88 KJ/mol.

Objective Type Questions 1. Which of the following contains the maximum percentage of carbon: (a) Peat (c) Anthracite

(b) Lignite (d) Bitumine

2. The calorific value of petrol varies from: (a) 30000 – 35000 (c) 40000 – 43000

(b) 35000 – 40000 (d) 40000 – 45000

3. The average calorific value of coal gas is (a) 24000 kJ/m3 (c) 30000 kJ/m3

(b) 34000 kJ/m3 (d) 20000 kJ/m3

10.36

Engineering Thermodynamics

4. Which of the following is used in the process of producing Pig Iron? (a) Producer gas (c) Natural gas

(b) Coal gas (d) Blast furnace gas

5. For the easy flow of fuel in pumps, the temperature of the fuel must be maintained in the range of (a) 100 – 120°C (c) 60 – 70°C

(b) 120 – 140°C (d) 90 – 100°C

6. When the volatility of the fuel is high, the flash point of the fuel will be: (a) High (b) Low (c) It does not depend on volatility (d) Zero 7. To find the calorific value of fuel the commonly used calorimeter is/are: (a) Bomb calorimeter (c) Boy’s calorimeter

(b) Junker’s calorimeter (d) All of the above

Theory Questions 1. What do you understand by a higher heating value and a lower heating value of a fuel? 2. Calculate the air fuel ratio by mass for the complete combustion of one kg of Octane (C8H18) with stoichiometric air. 3. Find the quantity of oxygen required to convert 1 kg of carbon monoxide into carbon dioxide. Define adiabatic flame temperature. 4. Why is excess air supplied in the combustion of solid fuels? 5. Indicate the methods to analyse the fuel gas composition. 6. Explain the terms: Sensible enthalpy, absolute enthalpy, enthalpy of formation and equivalence ratio. 7. Briefly explain the properties of solid, liquid and gaseous fuel. 8. Briefly explain how the adiabatic flame temperature for a given fuel air mixture gets affected by the equivalence ratio. 9. Describe the construction of the Orsat apparatus, using a neat sketch and explain how the mole fractions of the fuel gas constituents are determined. 10. Write short notes on the following: (i) Adiabatic peak flame temperature (ii) Excess air of combustion (iii) Stoichiometric A/F ratio (iv) Enthalpy of formation

10.37

Fuels and Combustion

Unsolved Problems 1. Petrol used in an engine contains 84% carbon and 16% hydrogen. The air supply is 80% of the theoretical requirement for complete combustion. Assuming that all the hydrogen is burned, and carbon burns partly to become CO and partly CO2, without any free carbon, find the volumetric analysis of the dry exhaust gases, and also find the percentage of the gross calorific value of the fuel lost due to incomplete combustion. Assume the gross calorific values as: C to CO2 = 35,000 kJ/kg; C to CO = 10,200 kJ/kg, H2 to H2O = 1,43,000 kJ/kg. 2. Write short notes on the following: (i) Adiabatic peak flame temperature (ii) Excess air of combustion (iii) Stoichiometric A/F ratio (iv) Enthalpy of formation. 3. Hexane(C6H14) is burnt with 80% theoretical air. In complete combustion produces CO2, CO, H2O and N2 in products. Calculate the air fuel ratio by mass and also the mass fraction of the constituents of the dry combustion products. 4. Methane(CH4) is burnt with atmospheric air. The analysis of the products on a dry basis is as follows: CO2 – 10.00% O2 – 2.37%, CO – 0.53%, and N2 – 87.1%. Calculate the air-fuel ratio, percentage of theoretical air and determine the combustion equation. 5. A hydrocarbon liquid fuel, having a hydrogen to carbon ratio of 0.174 by weight is burnt with 15% excess air. Determine the (i) the volume of air supplied at 30°C and 1 atm pressure (ii) the volume of flue gases at 300°C and 1 atm (iii) the mass of water formed per kg of fuel burnt. 6. A sample of natural gas is composed of CH4 – 92.5%, C2H6 – 4%, CO2 – 0.5% and N2 – 3%. The fuel is burnt with 20% excess air. Calculate (i) the volume of flue gases per cubic metre of the fuel burnt and (ii) the composition (iii) the molecular weight of the combustion products. 7. Coal from a particular coal mine is used as fuel in a boiler. The chemical composition of the coal is: C – 88%, H – 4.8%, N – 2%, S – 0.6 and O – 4.6%. The moisture and ash are 3% and 14.5% (as received) respectively. The coal consumption is 300 kg/h. Air is supplied at 27°C, and 1 atm is excess by about 25% than the theoretical. The flue gas leaves at 300°C and 1 atm. Find (i) the volume of flue gases that exit per minute (ii) flue gas composition (iii) power required by the blower to supply the air. 8. A gas burner is supplied with a gaseous fuel. The composition of the fuel is: H2 – 31%, CO – 24% and N – 45%. The composition of the flue gas is: CO2 – 12.1, O2 – 2.8% and CO-Nil. Calculate the excess air supplied per cubic metre for combustion.

10.38

Engineering Thermodynamics

9. Coal is supplied at the rate of 200 kg/h to the furnace of a boiler. The coal is composed of 60.5% C, 3.2% H, 1.4%N, 0.5% S, 9.4%O, 5% moisture, and 20% ash as dried. The orsat analysis of the flue gas is : CO2 – 13%, O2 – 5.5%, and CO – 1.2%. Find (i) the heat loss due to unburnt combustibles, if the dry refuse carries 20% combustibles (ii) the power required by the air blower (iii) the volume flow rate of the flue gas in cubic metre at 300°C and 1 atm. 10. A fuel gas has the chemical composition of: CO2 – 8.2%, O2 – 9.9%, CO – 0.1%, H2 – 0.4% and N2 – 81.4%. Calculate (i) the CO2 content of the dry theoretical flue gas (ii) the excess air supplied for combustion. 11. Calculate the following for Benzene; (i) theoretical air and (ii) adiabatic flame temperature. 12. The chemical composition of a gaseous fuel is as follows; H2 – 36.8, CH4 – 24.9%, C2H4 – 3.7%, C6H6 – 1.5%, CO – 17.4% CO2 – 3.4% and N2 – 12%. when it is burnt with air in a gas burner the CO2 analyser shows the CO2 in the flue gas as 10%. Compute (i) the composition of the flue gas (ii) excess air supplied in cubic metre of the fuel burnt. 13. A fuel analysis of coal shows 80% C and 5% H in it. When the coal is burnt, the composition of the flue gas is obtained as follows; CO2 – 12.5%, and O2 – 3.5%. Compute (i) the full composition of the flue gas, and (ii) air fuel ratio by weight. 14. Calculate the percentage of excess air supplied and the volume of flue gases if a fuel has the following flue gas analysis report; CO2 – 12%, CO – 1.4%, O2 – 3.4% and N2 – 83.2%. 15. The flue gas temperature and the pressure are 250°C and 1 atm respectively. Calculate (i) the percentage of excess air supplied (ii) the volume of the fuel gases in cubic metre per kg of fuel burnt. 16. Calculate the change in enthalpy when the theoretical products of the combustion of methane are cooled from 1400°C to 300°C. Give the answer in kJ/m3 of the combustible gases. Also, give in kJ/mol of methane. 17. Determine the change in enthalpy when the theoretical products of combustion of ethane are cooled from 1600°C to 300°C. Give the answer in kJ/m3 of the combustible gases. Also, give in kJ/mol of ethane. 18. LPG from a cylinder is burnt in a Bunsen burner. The air is mixed in the mixing tube. The fuel air ratio is 1:4. Calculate the theoretical air if the mixing tube carries 4% fuel in the fuel air mixture. 19. Calculate the theoretical air fuel ratio of the following: (i) Acetone (ii) Benzene (iii) Butane (iv)  Propane 20. Find the adiabatic flame temperature for the complete combustion of Propane(C3H8) with 150% theoretical air in an adiabatic control volume at 300 K.

CHAPTER

11

AIR STANDARD CYCLES

11.1 WHY AIR STANDARD CYCLES? Internal combustion (IC) engines play an important role in the civilized world. Generally they are operated by the petroleum fuels, used in the transportation, power, agricultural and industrial applications. These reciprocating engines generally follow the principle of air standard cycles. Such cycles are also called, gas power cycles. The analysis of the different thermodynamic processes involved in such engines is a complex problem. The knowledge on the analysis of the air standard cycles is useful to predict the maximum possible efficiencies of the internal combustion engines.

11.2 A THERMODYNAMIC CYCLE As described in Chapter 3, a thermodynamic cycle is one in which the working fluid of a system undergoes a number of processes that eventually returns the fluid to its initial state. Any device which is operated in a thermodynamic cycle, absorbs thermal energy from a source, rejects a part of it to a sink, and presents the difference between the energy absorbed and, energy rejected as work, is called a heat engine. In order to achieve this purpose, the heat engine uses a certain working medium, which undergoes the following processes: 1. A compression process where the working medium absorbs energy as work. 2. A heat addition process where the working medium, absorbs heat energy form a source. 3. An expansion process where the working medium transfers energy as work to the surroundings. 4. Heat rejection, when the working medium rejects energy as heat to a sink. If the working medium does not undergo any change of phase during its path through the cycle, the heat engine is said to operate in a non-phase change cycle. A phase change cycle is one, in which the working medium undergoes changes of phase. The air standard cycles, using air as the working medium are examples of non-phase change cycles. This definition applies to the combustion engines which use the chemical energy processed by a fuel.

11.2

Engineering Thermodynamics

11.3 AIR STANDARD CYCLE ASSUMPTIONS In principle, an air standard cycle is a cycle followed by a heat engine that uses air as the working medium. The analysis of energy interactions in the air standard cycle is the simplest and most idealistic. Hence, air standard cycles are also called ideal cycles, and the engines running in such cycles are called as ideal engines. For the calculation of heat and work transfer in an air standard cycle, it is essential to make some assumptions of important parameters. By assuming this, the outcome of the analysis will not be equal to that of the actual cycles. However, the analysis can give an idea of the performance. The analysis will be able to give an macro-level idea of the influencing parameters of the actual cycles. The following are the assumptions considered for the analysis: 1. The working medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes. 3. The processes are reversible. 4. Heat losses form the engine to the surroundings, are assumed to be zero in this analysis. 5. The working medium at the end of the process (cycle) is unchanged, and is in the same condition as at the beginning of the process (cycle).

11.4 THE CARNOT CYCLE This cycle was proposed by Sadi Carnot in 1824 and has a highest possible efficiency for any cycle. In the Carnot cycle, the piston is assumed to move in a cylinder without friction. It is also assumed that the cylinder wall is perfectly insulated. The piston and the cylinder assembly are shown in Fig. 11.1. Sink, T2 Source, T1

Insulated wall

Piston inside cylinder

Fig. 11.1 Carnot engine

11.3

Air Standard Cycles

Figures 11.2(a) and 11.2(b) show the p-V and T-s diagrams of the cycle respectively. 3

3

4

2

1

4 p

T 2 1 s

V

(a)

(b)

Fig. 11.2 (a) p-V diagram (b) T-s diagram

The Carnot cycle comprises two isothermal and two reversible adiabatic processes which are represented as follows l Isothermal

compression (Process 1-2)

l Reversible

adiabatic compression (Process 2-3)

l Isothermal

expansion (Process 3-4)

l Reversible

adiabatic expansion (Process 4-1)

For better understanding, let us consider the process 3-4. In this process, the piston assembly is brought in contact with the source, which is at high temperature. A certain amount of heat is added to the working substance, which is supplied by the source at constant temperature (Isothermal process). The pressure of the working substance decreases marginally from pressure p3 to p4 while the volume increases from V3 to V4. In an isothermal process the heat addition is given by Qs = Q3–4 = mRT3 ln

( V4 ) kJ ...(11.1) ( V3 )

Now, the cylinder assembly is made to be sealed, and hence, it acts as a perfect insulator. The hot air expands reversibly and adiabatically, and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p4 to p1 and the temperature reduces from T4 to T1 respectively. The volume increases from V4 to V1. The expansion work is given as W4–1 = Wc =

( p4 V4 – p1 V1 ) kJ ( g – 1)

...(11.2)

Now, the piston and cylinder assembly are brought in contact with a low temperature heat sink (T1). Heat is rejected from the cylinder assembly to the

11.4

Engineering Thermodynamics

sink at constant temperature (Isothermal compression). This is represented in process 1-2. The pressure of the working substance increases from pressure p1 to p2, while the volume decreases from V1 to V2. The heat rejection in the isothermal process is given by V  Qr = Q1–2 = mRT1ln  1  KJ ...(11.3)  V2  The working substance is now compressed reversibly and adiabatically in the cylinder by the movement of the piston. The pressure increases from p2 to p3, whereas the volume decreases from V2 to V3. The temperature of the working substance increases from T2 to T3. The compression work is given as

( p2 V2 – p1 V1 ) ...(11.4) KJ ( r – 1)

W1–2 = Wc =

W = heat supplied – heat rejected

          W = RT3 ln

V4 V − RT1 ln 1 V2 V3

...(11.5)

Let us consider the reversible adiabatic processes 2-3 1

         ( V3 ) = ( T2 ) v−1 ( V2 ) ( T3 )

...(11.6)

Similarly, in reversible adiabatic process 4-1 1

( V4 )  ( T1 )  r −1 =  ...(11.7)           ( V1 )  ( T4 )  Now, T1 = T2 and T3 = T4 Therefore,

( V3 ) ( V4 ) = ...(11.8) ( V2 ) ( V1 )

          or

( V4 ) ( V1 ) = ( V3 ) ( V2 )

         

Carnot efficiency = ηcarnot =

ηcarnot =

= r  (r = compression ratio)

...(11.9)

Work done Heat supplied mRT3 ln r − mRT1 ln r ...(11.10) mRT3 ln r

11.5

Air Standard Cycles

T3 – T1 T3



η =



η = 1 –

T1 . T3

...(11.11) ...(11.12)

From the above equation, it is understood that the thermal efficiency of the Carnot cycle is only a function of the maximum and minimum temperatures of the cycle. The efficiency will increase, if the minimum temperature (or the temperature at which the heat is rejected) is as low as possible. Also, it can be understood; that the Carnot efficiency will be equal to 1 if the minimum temperature is zero, which happens to be the absolute zero temperature in the thermodynamic scale. The heat engine must operate between the limits of the highest and lowest possible temperatures. That means, the engine should take in all the heat at as high a temperature as possible and should reject the heat at as low a temperature as possible. It is impossible to construct an engine which will work on the Carnot cycle. In such an engine, it would be necessary for the piston to move very slowly during the first part of the forward stroke, so that it can follow an isothermal process. During the remainder of the forward stroke, the piston would need to move very quickly as it has to follow an isentropic process. This variation in the speed of the piston cannot be achieved in practice. Also, a very lengthy stroke would produce only a small amount of work, most of which would be absorbed by the friction of the moving parts of the engine. Since the efficiency of the cycle, as given by Eqn. (11.12) is dependent only on the maximum and minimum temperatures, it does not depend on the working medium. It is thus independent of the properties of the working medium.

11.5 THE STIRLING CYCLE One of the drawbacks of the Carnot cycle is its low mean effective pressure. As a result of a low work output, the mean effective pressure is low. In order to achieve a higher mean effective pressure with full Carnot cycle efficiency theoretically, the Stirling cycle was introduced. The p-V and T-s diagram of the cycle are shown in Figs. 11.3 (a) and 11.3 (b) respectively. The adiabatic processes in the Carnot cycle are replaced with constant volume processes. The cycle consists of two isothermal and two constant volume processes. It has been proved that the heat addition and heat rejection during the constant volume processes are the same as found in the Carnot cycle. Although the Stirling cycle was used in hot air engines, it became obsolete because of the good application of the Otto and Diesel cycle. One of the major limitation found with the use of the Stirling cycle in an IC engine was the design and material compatibility of the heat exchanger to resist high temperature. Now-a-days, with the advancements in the materials and production technologies, high temperature resistant materials are introduced, which encourages using the Stirling cycle in practical applications.

11.6

Engineering Thermodynamics

3

3

p

4

T

4 2

2

1

1 s

V

(a)

(b)

Fig. 11.3 (a) p-V diagram (b) T-s diagram

Therefore, the efficiency of the cycle =

Work done Heat supplied

V  V  mRT3 ln  4  − mRT1 ln  1   V2   V3      = V  mRT3 ln  4   V3 

Here, V1 = V4 and V2 = V3

h =

T3 – T1 T =1– 1 T3 T3

...(11.13)

11.6 THE ERICSSON CYCLE Similar to the Stirling cycle, the Ericsson cycle has two isothermal processes but instead of two constant volume processes, two constant pressure processes are involved. The p-V and T-s diagrams of the cycle are shown in Figs. 11.4(a) and 11.4(b) respectively. It could produce a maximum mean effective pressure in a given specific volume, because of the higher specific heat capacity of the working substance. And also, it could produce a smaller pressure ratio for a given ratio of higher to lower specific volume. The cycle is more useful for gas turbine applications.

11.7 IMPORTANT AIR STANDARD CYCLES The cycles described in this section are air standard cycles applicable to reciprocating engines or piston engines. These engines are generally used in obtaining power from heat to operate stationary engines or automotive vehicles.

11.7

Air Standard Cycles

3

4

3

4

T

p

2

2

1

1 s

V

(a)

(b)

Fig. 11.4 (a) p-V diagram (b) T-s diagram

11.7.1 The Otto Cycle The Otto cycle was first proposed by a Frenchman, Beau de Rochas in 1862. However, it was first used in an engine built by a German, Nicholas A. Otto, in 1876. The cycle is also called a constant volume or explosion cycle. This is the equivalent air cycle for reciprocating piston engines using spark assisting devies. Figures 11.5(a) and 11.5(b) show the p-V and T-s diagrams respectively. 3

3

p 2

4

T

1

2

4

1

V

(a)

s

(b)

Fig. 11.5 (a) p-V diagram (b) T-s diagram

The following four thermodynamic processes occur in the Otto cycle: (i) Isentropic compression (Process 1-2) (ii) Constant volume heat addition (Process 2-3) (iii) Isentropic expansion (Process 3-4) (iv) Constant volume heat rejection (Process 4-1) Process 1-2: Isentropic compression uring this process, air is compressed isentropically. As a result the pressure D increases from p1 to p2. The volume decreases from V1 to V2. The temperature of air increases from T1 to T2.

11.8

Engineering Thermodynamics

We know that in this isentropic process

p1V1g = p2V2g  and also

T1V1g–1 = T2V2g-1 g–1

V  T           2 =  1  T1  V2  The compression work is calculated as W1–2 = Wc =

...(11.14)

( p1 V1 – p2 V2 ) kJ. ...(11.15) ( g – 1)

Process 2-3: Constant volume heat addition In this process heat is added to the compressed air at constant volume. As a result, the temperature of the compressed air increases from T2 to T3, while the pressure increases from p2 to p3. The volume remains unchanged. In this constant volume process, or,

p = Constant T

p3  p2             =   ...(11.16) T3  T2  The heat supplied, Qs in this process is given as

Q2–3 = Qs = mCv(T3 – T2) kJ

...(11.17)

Process 3-4: Isentropic expansion Now, the hot air expands isentropically and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p3 to p4, and the temperature reduces from T3 to T4 respectively. The volume increases from V3 to V4. p3V3g = p4V4g and also T3V3g-1 = T4V4g-1 V  T            4 =  3  T3  V4 

g–1



...(11.18)

The expansion work is given as W3–4 = Wc =

( p3 V3 – p4 V4 ) kJ. ( g – 1)

...(11.19)

Process 4-1: Constant volume heat rejection During this process, heat is rejected from the expanded air to the atmosphere at constant volume. The temperature drops from T4 to T1. The pressure decreases from p4 to p1, while the volume remains constant.

11.9

Air Standard Cycles

In this constant volume process p = Constant T

or,

p4  p1             =   . ...(11.20) T4  T1  The heat rejected, Qr per unit mass of charge is given by Q4–1 = Qr = mCv(T4 – T1)kJ ...(11.21) The air standard efficiency of the Otto cycle is given by η= 1 −

( T4 – T1 ) ( T3 – T2 )

  T4  − 1   T   T1             = 1 − 1   ...(11.22) T2   T3  1 −    T2   T Now,         1 = T2

g−1

g−1

 V2   V3  =  =   V1   V4 

          T4 =  V3  T3  V4 

g−1



V  Let us consider compression ratio  1  = r  V2  and

T4 ...(11.23) T3

...(11.24)

...(11.25)

V  Expansion ratio  4  = re. ...(11.26)  V3  Since the heat addition and heat rejection take place in constant volume, the compression ratio and expansion ratio are the same. Therefore, it can be written as  V2           V   1

And since

g–1

V  =  3  V4 

g–1

...(11.27)

T T T1 T4 = , we have 4 = 3 . T2 T3 T1 T2

T1  1             =   T2  r 

...(11.28)

g –1



...(11.29)

11.10

Engineering Thermodynamics

Substituting in the air standard efficiency equation h = 1 –



T1 T2

1 . ...(11.30) rg – 1 In the Otto cycle, the terms expansion ratio and compression ratio are considered equal. But in a real engine, these two ratios need not be equal, because of the valve timing, and therefore, the term expansion ratio is preferred sometimes. Equation (11.30) shows that the air standard efficiency of the Otto cycle is dependent on the compression ratio and specific heat ratio. But, it is independent of the heat added and the initial conditions of pressure, volume and temperature. Figure 11.6 shows the variation of the air standard efficiency with compression ratio for an Otto cycle for three different values the of the specific heat ratio (g). It can be observed from the figure, that the increase in efficiency is considerable at lower compression ratios. = 1 –



0.9  = 1.4

0.8 0.7

1.3 1.25

0.6 

0.5 0.4 0.3 0.2 0.1 0

0

4

8

16

12

20

24

28

rC

Fig. 11.6 Air standard efficiency of the Otto cycle for different compression ratios

11.7.1.1

Work Output and Mean Effective Pressure

The mean effective pressure (mep) is an important factor in an air standard cycle. It is defined as the ratio of the net work done and the swept or stroke volume. The unit of mean effective pressure is given in bar or kN/m2 or kPa. The mean effective pressure is that the constant pressure which is exerted on the piston for the entire expansion stroke, that will give the work equal to the work of the cycle. It is given by

mep =

W V1 – V2

The area under the p-V diagram that represents the net work done in the cycle.

11.11

Air Standard Cycles

Therefore, n et work done = Area A2341B – Area A21B Area A2341B = Work done in isentropic expansion during the process 3-4(We) Area A21B = Work done in isentropic compression during the process 1-2(Wc) 3 p

2 4 1 A

V

B

Fig. 11.7 Otto cycle

The expansion work (We) of the Otto cycle can be written as We =

p3 V3 – p4 V4 ...(11.31) g–1

Similarly the compression work can be written as

Wc =

p2 V2 – p1 V1 ...(11.32) g–1

The net work output of the Otto cycle is given as W = We – Wc

W =

p3 V3 – p4 V4 p2 V2 – p1 V1 – g–1 g–1

...(11.33)

From the p-V relationship p2 p3           = = rp p1 p4

...(11.34)

p3 p 4 = rp  (rp = pressure ratio) or,              = p2 p1

...(11.35)

Dividing the Eqn. (11.33), by p1V1 we get

W =

1 [ p3 V3 – p4 V4 ] [ p2 V2 – p1 V1 ] − ...(11.36) g –1 p1 V1 p1 V1



W =

1 [ p3 V3 ] [ p4 V4 ] [ p2 V2 ] − – + 1 ...(11.37) g – 1 p1 V1 p1 V1 p1 V1

Applying the compression ratio and pressure ratio expressions in the above equation, and simplifying it, we get

W =

p1V1 rp r g – rp – r g –1 + 1 ...(11.38) g–1

(

)

11.12

Engineering Thermodynamics



W =

p1V1  rp – 1 r g –1 – 1  ...(11.39)  g–1

(

)(

)

As mentioned earlier, the air standard efficiency of the Otto cycle depends only on the compression ratio. However, the pressures and temperatures at the various points in the cycle, and the net work done, all depend upon the initial pressure and temperature, and the heat input at constant volume heat addition process, besides the compression ratio. Now, considering the equation for mean effective pressure mep =



W ...(11.40) V1 – V2

Now V1 – V2 can be written as V1 – V2 = V2(r – 1) Therefore,

mep =

W V2 ( r – 1 )

mep =

p1 V1  r – 1 r g –1 – 1   ( g – 1)( r – 1)  p

or,

(

)(

)

...(11.41)

11.7.2 The Diesel Cycle This cycle, proposed by a German engineer, Dr. Rudolph Diesel is also called the constant pressure cycle. This is believed to be the equivalent air cycle for the reciprocating slow speed compression ignition(CI) engines. The p-V and T-s diagrams are shown in Figs. 11.8(a) and 11.8(b) respectively. 3 2

3

p 4

T

1

2

4

1

V

(a)

s

(b)

Fig. 11.8 (a) p-V diagram (b) T-s diagram

The cycle has the following four processes, which are the same as those of the Otto cycle except that the heat is added at constant pressure. 1. Isentropic compression (Process 1-2) 2. Constant pressure heat addition (Process 2-3)

11.13

Air Standard Cycles

3. Isentropic expansion (Process 3-4) 4. Constant volume heat rejection (Process 4-1) Process 1-2: Isentropic compression At the start of the cycle, the cylinder contains a mass “m” of air at the atmospheric pressure and volume indicated at point 1. The air is compressed isentropically and therefore the pressure increases from p1 to p2. As a result, the volume decreases from V1 to V2. The temperature of air increases from T1 to T2. We know that in this isentropic process; and also

p1V1g = p2V2g T1V1g–1 = T2V2g–1

T2  V1  =           T1  V2 

g− 1



...(11.42)

The compression work is calculated as W1–2 = Wc =

( p1 V1 – p2 V2 ) kJ . ...(11.43) ( g – 1)

Process 2-3 : Constant pressure heat addition In this process, heat is added to the compressed air at constant pressure. As a result, the temperature of the compressed air increases from T2 to T3, while the volume increases from V2 to V3. The pressure remains unchanged. In this constant pressure process, V = Constant T

or,          

V3  V2  =  T3  T2 

...(11.44)

The heat supplied, Qs in this process is given as

Q2–3 = Qs = mCp(T3 – T2)kJ. ...(11.45)

Process 3-4: Isentropic expansion Now, the hot air expands isentropically, and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p3 to p4, and the temperature reduces from T3 to T4 respectively. The volume increases from V3 to V4. p3V3g = p4V4g and also

T3V3g–1 = T4V4g–1          

T4  V3  =  T3  V4 

g –1



...(11.46)

11.14

Engineering Thermodynamics

The expansion work is given as

( p3V3 – p4V4 ) kJ . ...(11.47) ( γ – 1)

W3–4 = Wc =

Process 4-1: Constant volume heat rejection During this process, heat is rejected from the expanded air to the atmosphere at constant volume. The temperature drops from T4 to T1. The pressure decreases from p4 to p1, while the volume remains constant. In this constant volume process, or          

p = Constant T p4  p1  =  T4  T1 

...(11.48)

The heat rejected, Qr in this process is given by Q4–1 = Qr = mCv(T4 – T1)kJ ...(11.49) The thermal efficiency is given by h = 1 –

c v (T4 – T1 )

c p (T3 – T2 )

...(11.50)

  T4  – 1   T1  T 1   . = 1 –   1  γ   T3  T2  – 1    T2   

...(11.51)

Referring to the T-s diagram, Fig. 11.8(b), the difference in entropy between points 2 and 3 is the same as that between 4 and 1. Therefore, we get γ

T  T \         4 =  3  ...(11.52) T1  T2  Also from Eqn. (11.42), we get T T2

V   V1 

γ –1

1 2 =            = 

1 r

γ –1

. ...(11.53)

Substituting in the air standard efficiency equation, we get



  T γ   3  – 1 γ –1  T 1 1  h = 1 –     2   T3 γr –1    T2   

...(11.54)

11.15

Air Standard Cycles

Now let            

V3 T3 cut-off = = rc= cut – offratio ratio V2 T2 h = 1 –

1  rcg – 1   . r g –1  g ( rc – 1 ) 

...(11.55) ...(11.56)

When Eqn. (11.56) is compared with Eqn. (11.30), it is understood that the expressions are similar, except for the term in the parentheses for the Diesel cycle. It can be shown that this term is always greater than unity. V3 V3 V1 r = ×= V2 V4 V2 re

= rc Now,        

...(11.57)

where r is the compression ratio and re is the expansion ratio. Thus, the air standard efficiency of the Diesel cycle can be written as



  r g     –1  1   re   h = 1 – g –1   . ...(11.58) g r  g  r  – 1   re      

In practice, the diesel engine gives a better thermal efficiency than the Otto cycle engine, because the diesel cycle engine uses a higher compression ratio. The Otto cycle engines have compression ratios in the range of 7 to 12 while diesel cycle engines have compression ratios in the range of 16 to 22.

11.7.2.1  Net Work Output and Mean Effective Pressure The net work output of the diesel cycle is represented by the area which is shown in Fig. 11.8 (c). The net work output can be calculated as follows;

Net work output = Area under A2341B-Area under A21B

Area under A2341B = [Work during constant pressure process 2-3 + Isentropic expansion work during 3-4]

Area under A21B = Isentropic compression work during 1-2.

We know that the compression work done in a diesel cycle is given by

2 p

3 Net work output 4 1

A

V

B

Fig. 11.8 (c)

11.16

Engineering Thermodynamics

W1–2 = Wc =

( p1V1 – p2V2 ) kJ ( g – 1)



...(11.59)

W2–3 = p(V3 – V2) W3–4 = We =

( p3V3 – p4V4 ) ( g – 1)

Wnet = [W2-3 + W3-4 – W4–1] = p(V3 – V2) +

( p3 V3 – p4 V4 ) ( p1 V1 – p2 V2 ) – ( g – 1) ( g – 1)

W = p3 ( V3 – V2 ) +



= p2 ( V3 – V2 ) +

( p3 V3 – p4 V4 ) ( p1 V1 – p2 V2 ) – ...(11.60) ( g – 1) ( g – 1) ( p3 V3 – p4 V4 ) ( p1 V1 – p2 V2 ) – . ( g – 1) ( g – 1)

We know that [p2 = p3] (V3 – V2) = (rc – 1) Therefore,

W = p2V2 ( rc – 1 ) +

= V2



Substitute simplify it.



( p3 V3 – p4 V4 ) ( p1 V1 – p2 V2 ) – ( g – 1) ( g – 1)

p2 (rc − 1)( g − 1) + p3 rc − p4 r − ( p2 − p1 r ) ( g − 1)

  p4  p2 (rc − 1)( g − 1) + p3  rc − p3   W = V2  ( g − 1)  

  p r  −1− 1 p 2  

 r     

p p4 and 1 values in terms of re in the above equation, and p2 p3

  p4  p2 (rc − 1)( g − 1) + p3  rc − p3  W = V2   ( g − 1)  

  p r  −1− 1 p 2  

 r     

11.17

Air Standard Cycles

Swept volume Vs = V1 – V2

Then,

  p4   p1  p2 (rc − 1)( g − 1) + p3  rc − r  −  1 − p p 3   2  mep = V2   ( g − 1 ) ( V1 – V2 )  

 r   .   

11.7.3 The Dual Cycle In present day diesel engines, the heat addition does not follow the Diesel cycle as it is. Because of the nature of the combustion process, it is not possible to achieve heat addition at constant pressure. The process of heat addition takes place in two stages: (i) at constant volume and (ii) at constant pressure. This is known as the dual cycle or limited pressure cycle or mixed cycle. Practically, the performance of the actual cycles depends on two parameters; (i) the mean effective pressure and (ii) the maximum pressure. The mean effective pressure indicates the useful (average) pressure acting on the piston while the maximum pressure indicates the pressure which primarily affects the strength required of the engine structure. In the Otto cycle, the ratio of the mean effective pressure to the maximum pressure decreases when the compression ratio increases, or it can be said that for a given mean effective pressure the maximum pressure increases rapidly as the compression ratio increases. In a constant pressure heat addition cycle, the maximum pressure is low unless the compression ratio is quite high. Practically, the fuel would be injected at the compression stroke, if the combustion takes place at constant pressure. On the other hand, if the efficiency of the cycle has to be increased, the fuel supply must be cut off early in the expansion stroke. In both the cases, sufficient time is required for the fuel to burn. This problem can be overcome in such a way that the engine must be a slow speed type, so that the piston will move slowly for combustion to take place in spite of the late injection of the fuel. Particularly, in modern high speed CI engines, constant pressure combustion is not possible. Therefore, this requires the modern high speed diesel engines to run in such a way, that the combustion is nearly at constant volume (like in a spark ignition engines). However, the peak pressure is limited with respect to the strength of the engine, so the rest of the heat addition takes place at constant pressure in a cycle. This combined requirement results to form the dual combustion cycle. In this cycle, for high compression ratios, the peak pressure is not allowed to increase beyond a certain limit; considering the total heat addition, the rest of the heat is assumed to be added at constant pressure. Hence it is referred as limited pressure cycle. The p-V and T-s diagrams are shown in Figs. 11.9(a) and 11.9(b) respectively. The dual cycle constitutes the following five processes, in which the heat addition takes place in both constant volume and constant pressure processes, while heat rejection takes place in a constant volume process. The processes are given below: (i) Isentropic compression (Process 1-2)

11.18

Engineering Thermodynamics

3

4

4 3

p

2

T 5 1

2

5

1

V

s

(a)

(b)

Fig. 11.9 (a) p-V diagram (b) T-s diagram

(ii) Constant volume heat addition (Process 2-3) (iii) Constant pressure heat addition (Process 3-4) (iv) Isentropic expansion (Process 4-5) (v) Constant volume heat rejection (Process 5-1) Process 1-2: Isentropic compression At the start of the cycle, the cylinder contains a mass “m” of air at atmospheric the pressure and volume indicated at point 1. The air is compressed isentropically and therefore the pressure increases from p1 to p2. As a result, the volume decreases from V1 to V2. The temperature of the air increases from T1 to T2. We know that in this isentropic process; g

g

p1V1 = p2V2 and also

T1V1g–1 = T2V2g–1 V  T2 = 1 T  V2            1

g –1

The compression work is calculated as p V – p2 V2 kJ . W1–2 = Wc = 1 1 ( g – 1)

...(11.61)

Process 2-3: Constant volume heat addition In this process heat is added to the compressed air at constant volume. As a result, the temperature of the compressed air increases from T2 to T3, while the pressure increases from p2 to p3. The volume remains unchanged. In this constant volume process, p = Constant T or,         p3 =  p2    T3  T2 

11.19

Air Standard Cycles

The heat supplied, Qs1 in this process is given as

Q2–3 = Qs1 = mCv(T3 – T2)kJ

...(11.62)

Process 3-4: Constant pressure heat addition In this process heat is added to the compressed air at constant pressure. As a result, the temperature of the compressed air increases from T3 to T4, while the volume increases from V3 to V4. The pressure remains constant. In this constant pressure process, V = Constant T V V  or,         3 =  4  T3  T4 

The heat supplied, Qs2 in this process is given as

Q3–4 = Qs2 = mCp(T4 – T3)kJ

...(11.63)

The total heat supplied or heat addition = Qs = Qs1 + Qs2 Qs = mCv(T3 – T2) + mCp(T4 – T3). Process 4-5 : Isentropic expansion Now, the hot air expands isentropically and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p4 to p5, and the temperature reduces from T4 to T5 respectively. The volume increases from V4 to V5. and also

p4V4g = p5V5g T4V4g–1 = T5V5g–1

V  T5 = 4 T  V5            4 The expansion work is given as W4–5 = We =

g–1

( p4 V4 – p5 V5 ) kJ. ...(11.64) ( g – 1)

Process 5-1: Constant volume heat rejection During this process, heat is rejected from the expanded air to the atmosphere at constant volume. The temperature decreases from T5 to T1. The pressure decreases from p5 to p1, while the volume remains constant. In this constant volume process, P = Constant T p5  p1  =  or,         T5  T1 

11.20

Engineering Thermodynamics

The heat rejected, Qr in this process is given by Q5–1 = Qr = mCv(T5 – T1) kJ.

...(11.65)

The air standard efficiency is given by h = 1 −



c v ( T5 – T1 )

c v ( T3 – T2 ) + c p ( T4 – T3 )

  T  T1  4 – 1       T1  1 − =   ...(11.66)     T T4 3 T  – 1 T – 1 + g   3  2  T2 T   3  

p3 T3 Let         = = rp p2 T2 where rp is referred to as the explosion or pressure ratio. Also, let us take the cut-off ratio which is given by          

V4 V4 = = rc V3 V2 η = 1 −



1

(

rp rcg – 1 – 1

)

r g –1 rp – 1 + rp g ( rc – 1 )

11.7.4 The Lenoir Cycle The Lenoir cycle is of interest, because combustion (or heat addition) occurs without a compression of the charge. Figures. 11.10(a) and 11.10(b) show the p-V and T-s diagrams respectively. 2 2 p

T 1

3

3 1

V

s

(a)

(b)

Fig. 11.10 (a) p-V diagram (b) T-s diagram

The cycle comprises the following three processes (i) Constant volume process (Process 1-2) (ii) Isentropic expansion process(Process 2-3)

11.21

Air Standard Cycles

(iii) Constant pressure process (Process 3-1) (i) Constant volume process(Process 1-2) During this process it is assumed that the charge is ignited or heat is added. The addition of heat takes place at constant volume, so that the pressure rises from p1 to p2, and the volume remains constant. Heat supplied,

Qs = Q1–2 = mCr(T2 – T1).

...(11.67)

(ii) Isentropic expansion process (2-3) The hot working substance expands isentropically so that pressure decreases from p2 to p3 and the volume increases from V2 to V3. The temperature decreases from T2 to T3. Expansion work is given by We = W2–3 =

( p2V2 – p3V3 ) . ...(11.68) ( g – 1)

(iii) Constant pressure process (Process 3-1) During this process heat is rejected from the working substance to atmosphere at constant pressure. When the pressure remains unchanged, the volume decreases from V3 to V1 and the temperature falls from T3 to T1. Heat rejected,

Qr = Q3–1 = mCp(T3 – T1)

Now,

W = Qs – Qr



W = mCv(T2 – T1) – mCp(T2 – T1)

Thus,

h = 1 –

mC p ( T3 – T1 )

mC v ( T2 – T1 )

...(11.69)

...(11.70)

T  g  3 – 1  T1  = 1 – ...(11.71)  T2  – 1    T1  Considering process 1-2 T2 p2 = T p1            1 Process 2-3 g g          p2 V2 = p3V3 g

g

 V3   V3  p2 = or,          =    p3  V2   V1  p2  V3  =  p3  V1            

g

...(11.72)

11.22

Engineering Thermodynamics

Similarly, considering process 3-1 T V Since,         3 = 3 T1 V1 Comparing the equations we get; =

p2 T2 = p3 T1

p2 p2 T2           p = p = T ...(11.73) 3 1 1

\

Let, re = \

V  g  3 – 1 V h = 1 –  1 g  .  V    3  – 1   V1  

...(11.74)

V3 , be the volumetric expansion ratio. Then the equation becomes V1 h = 1 –

g ( re – 1 )

( reg – 1)

.

...(11.75)

Equation (11.75) indicates that the thermal efficiency of the Lenoir cycle depends primarily on the expansion ratio and the ratio of specific heats. The intermittent-flow engine, which powered the German V-1 buzz-bomb in 1942 during the World War II, operated on a modified Lenoir cycle. A few engines running on the Lenoir cycle were built in the late 19th century till the early 20th century.

11.7.5 The Atkinson Cycle This cycle is also referred to as the complete expansion cycle. It can be realized from the p-V diagrams of the Otto, Diesel and Dual cycles, that the end of expansion process does not reach the lowest possible pressure, i.e., atmospheric pressure. It is common for all practical engines that with the available exhaust opening during the exhaust stroke, the high pressure gases undergo a violent blow down process with the consequent dissipation of the available energy. This is essential to permit the gases to exit the engine due to a pressure difference, and hence, reduce the piston work to push out the gases. However, on a whole, as the area of the p-V diagram reduces, there will be a loss of the net work. If the end of expansion is allowed, till the atmospheric pressure and the heat rejection occur at constant pressure with modification in the Otto cycle, then the cycle is called the Atkinson cycle. The p-V an T-s diagram are shown in Figs. 11.11(a) and 11.11(b) respectively.

11.23

Air Standard Cycles 3 3 p

4 2

T

4 2

4 1

5

1

V

s

(a)

(b)

Fig. 11.11 (a) p-V diagram (b) T-s diagram

The following four thermodynamic processes occur in the Atkinson cycle. (i) Isentropic compression (Process 1-2) (ii) Constant volume heat addition (Process 2-3) (iii) Isentropic expansion (Process 3-4) (iv) Constant pressure heat rejection (Process 4-1) Process 1-2: Isentropic compression During this process air is compressed isentropically. As a result the pressure increases from p1 to p2. The volume decreases from V1 to V2. The temperature of air increases from T1 to T2. We know that in this isentropic process and also

p1V1g = p2V2g T1V1g–1 = T2V2g–1

T2  V1                =  T1  V2 

g− 1

.

The compression work is calculated as W1–2 = Wc =

( p1 V1 – p2 V2 ) kJ . ( g – 1)

...(11.76)

Process 2-3: Constant volume heat addition In this process heat is added to the compressed air at a constant volume. As a result, the temperature of the compressed air increases from T2 to T3, while the pressure increases from p2 to p3. The volume remains unchanged. In this constant volume process, p = Constant T p  p or,          3 =  2  T3  T2 

11.24

Engineering Thermodynamics

The heat supplied, Qs, in this process, is given as

Q2–3 = Qs = mCv(T3 – T2)kJ.

...(11.77)

Process 3-4: Isentropic expansion Now, the hot air expands isentropically, and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p3 to p4, and temperature reduces from T3 to T4 respectively. The volume increases from V3 to V4.

and also

p3V3g = p4V4g

T3V3g–1 = T4V4g–1

T4  V3  =  T V           3  4  The expansion work is given as

g –1

W3–4 = Wc =

p3 V3 – p4 V4 kJ . ( g – 1)

...(11.78)

Process 4-1: Constant pressure heat rejection During this process, heat is rejected from the expanded air to the atmosphere at a constant volume. The temperature drops from T4 to T1. The pressure decreases from p4 to p1, while volume remains constant. In this constant pressure process, V = Constant T V  V or,         4 =  1  ...(11.79) T4  T1 

The heat rejected, Qr per unit mass of charge, is given by = mCp(T4 – T1) and the thermal efficiency is given by

h = 1 −

mC p ( T4′ – T1 ) mC v ( T3 – T2 )



...(11.80)

T  gT1  4′ – 1   T1  = 1 − .  T3  T2  – 1  T2  T4′ V4′ = = rv ...(11.81) Now,        T1 V1 T3 As before,      = T2

p3 = rp p2

...(11.82)

11.25

Air Standard Cycles g

T2  V1  And        r g –1 = =  T1  V2  The efficiency, is therefore, given by h = 1 –



1 r

g –1

g ( rv – 1 )

( rp – 1)

Let us assume the expansion ratio as efficiency as h = 1 –



...(11.83)

V4′ ; we can rewrite the thermal V3′

1 g ( rv – 1 )

reg –1



( rvg – 1)

. ...(11.84)

The area under the p-V diagram is larger in the Atkinson cycle in comparison with the Otto cycle. Therefore, for the same compression ratio and heat input, the air standard efficiency will be higher for the Atkinson cycle. It is possible to construct an engine which will have complete expansion. However, this requires a larger stroke length. Therefore, it is not economical one to improve the engine efficiency.

11.7.6 The Brayton Cycle The Brayton cycle is also referred to as the Joule cycle or the gas turbine air cycle because all modern gas turbines work on this cycle. However, if the Brayton cycle is to be used for reciprocating piston engines, it requires two cylinders, one for compression and the other for expansion. Heat addition may be carried out separately in a heat exchanger, or within the expander itself. The p-V and the corresponding T-s diagrams are shown in Figs. 11.12(a) and 11.12(b) respectively. 3 2

3

p

T 2 1

4 V

(a)

4

1 s

(b)

Fig. 11.12 (a) p-V Diagram (b) T-s Diagram

The cycle has the following four processes which are the same as those of the diesel cycle except that the heat is rejected at constant pressure.

11.26

Engineering Thermodynamics

(i) Isentropic compression (Process 1-2) (ii) Constant pressure heat addition (Process 2-3) (iii) Isentropic expansion (Process 3-4) (vi) Constant pressure heat rejection (Process 4-1) Process 1-2: Isentropic compression At the start of the cycle, the cylinder contains a mass “m” of air at the atmospheric pressure and volume indicated at point 1. The air is compressed isentropically and therefore, the pressure increases from p1 to p2. As a result the volume decreases from V1 to V2. The temperature of air increases from T1 to T2. We know that in this isentropic process;

p1V1g = p2V2g

and also T1V1g–1 = T2V2g–1 T2  V1  =  T V             1  2 

g–1

The compression work is given as W1–2 = Wc =

( p1 V1 – p2 V2 ) kJ. ( g – 1)

...(11.85)

Process 2-3: Constant pressure heat addition In this process heat is added to the compressed air at constant pressure. As a result, the temperature of the compressed air increases from T2 to T3, while the volume increases from V2 to V3. The pressure remains unchanged. In this constant pressure process, or,

V = Constant T

V3  V2  =  T3  T2            The heat supplied, Qs in this process is given as

Q2–3 = Qs = mCp(T3 – T2)kJ.

...(11.86)

Process 3-4: Isentropic expansion Now, the hot air expands isentropically and therefore, work is obtained. Due to the expansion of hot air, the pressure decreases from p3 to p4 and temperature reduces from T3 to T4 respectively. The volume increases from V3 to V4. We know that in this isentropic process;

11.27

Air Standard Cycles



p3V3r = p4V4r and also

T3V3r–1 = T4V4r-1 V3  V2  =  T T            3  2 

g –1

The expansion work is given as W3–4 = Wc =

( p3 V3 – p4 V4 ) kJ ...(11.87) ( g – 1)

Process 4-1: Constant pressure heat rejection During this process, heat is rejected from the expanded air to the atmosphere at constant volume. The temperature drops from T4 to T1. The volume decreases from V4 to V1, while, pressure remains constant. In this constant pressure process, V = Constant T

or,

V4  V1  =  T T           4  1  The heat rejected, Qr is given by

Q4–1 = Qr = mCp(T4 – T1)kJ.

...(11.88)

The thermal efficiency is given by          

h= 1 −

( T4 – T1 ) ( T3 – T2 )



...(11.89)

  T4  – 1   T1  T   . = 1−  1   T  T3 – 1     2  T2  

         

  T4  – 1   T  T   . ...(11.90) = 1 − 1  1  T2   T3  – 1   T2    g−1

g−1

 p2  g  p3  g T2 Now,         = =  =  T1  p1   p4 

T3 T4

11.28

Engineering Thermodynamics

T2 And since      = T1

T3 T4 we have = T4 T1

Let us assume rp is the pressure ratio =

T3 T2

p2 p1

Substituting in Eqn. (11.90), the pressure ratio          

T1 T2

h = 1−



...(11.91)

1

= 1−

 p2     p1 

g−1 g

1

= 1− rp

g−1 g

This is numerically equal to the efficiency of the Otto cycle if we put, g−1

g−1 T1  V2  1 = =    T V r           2  1 

so that

h = 1 –

r r

g –1



...(11.92)

where r is the volumetric compression ratio. The Brayton cycle is applied in gas turbines and the air standard efficiency p depends on the pressure ratio 2 . p1 It is to be noted that if the Brayton cycle is used in reciprocating engines, V then the compression ratio 2 must be kept higher. Therefore, such engines V1 require lengthy connecting rod to provide a lengthy stroke that will make the working substance to expand to atmospheric pressure. In that case the shaft output (brake power) will be lower as a result of higher friction loss.

11.8 COMPARISON BETWEEN OTTO, DIESEL AND BRAYTON CYCLES The Otto, and Diesel cycles are applied in reciprocating engines, while the Brayton cycle is applied in gas turbines. The comparison between the Otto, Diesel and Dual cycles is given in Table 11.1.

11.29

Air Standard Cycles

Table 11.1 Comparison Between the Otto, Diesel and Brayton cycles S.No.

Description

Otto cycle

Diesel cycle

1.

Number of thermo- Four dynamic process

2.

Heat addition

Constant volume Constant pressure

Constant pressure

3.

Heat rejection

Constant volume Constant volume

Constant pressure

4.

Factor(s) affecting air standard efficiency

Compression ratio

Pressure ratio

5.

Nature of Applica- Spark ignition(SI) Compression tion engine ignition(CI) engine

Gas turbines

6.

Efficiency

Lower

7.

Compression and More efficient due More efficient due expansion to less friction to less friction and and turbulence turbulence

Less efficient due to more friction.

8

Mass flow rate to the system

Less

Less

High

9.

Need of application

Where less space is available and low power output is required

Where less space is available and low power output is required

Where large space is available and high power output is required

10.

Necessity of Not required intercooling during compression

Not required

May be required in multistage operation

11.

Presence of clear- May or may not ance volume be provided

May or may not be provided

Need not be provided

12.

Fuel air ratio

Higher

Four

Brayton

Compression ratio, expansion ratio and adiabatic index

Higher

May vary accord- May vary according ing to the need to the need

Four

Generally runs with maximum fuel air ratio

11.9 COMPARISON OF AIR STANDARD EFFICIENCIES The efficiencies of the Otto, Diesel, Dual, cycles are affected by parameters, such as, the heat supplied, maximum pressure or maximum temperature, net work output and heat rejected. Let us see how does the efficiency differs in between the three cycles in this section, by keeping are or more parameters constant. In such cases, let us assume that, the initial conditions of pressure and temperature are identical in all the comparisons.

11.9.1 Same Compression Ratio and Heat Supplied The same compression ratio means the initial and final volume (V1 and V2) are fixed in all the three cycles. Therefore, the altitude or heights of the p-V and T-s diagrams have an impact on the efficiency calculations. If we compare the p-V

11.30

Engineering Thermodynamics

and T-s diagrams of the Otto, Diesel and Dual cycles, it is understood that the addition of heat at constant volume results in the highest maximum pressure and temperature for the Otto cycle, while the addition of heat at constant pressure for the same compression ratio results in the lowest maximum temperature and pressure for the Diesel cycle. The maximum temperature and pressure for the dual cycle lie in between those for the Otto and Diesel cycles. The air standard efficiency of any of the cycles is given by heat rejected , heat added



h =



hotto = 1 –

1 r

g–1



  r g    – 1 1   re   hDisel = 1 – g –1    r  r g  – 1   re   



hDual = 1 –

g−

  rp rc –1 – 1    rp – 1 + rp g ( rc – 1 )   

(

)

From the efficiency equation for the three cycles, it is understood that although the value for the compression ratio (r) is the same, the other parameters such as the pressure ratio and expansion ratio in the Diesel cycle affect the denominator much, compared to those of the Otto cycle, and hence, the air standard efficiency for the Diesel cycle is the minimum while the efficiency is the maximum for the Otto cycle.

11.9.2 Same Maximum Pressure and Heat Supplied For the same maximum pressure and heat supplied, the heat rejected is the highest for the Otto cycle, followed by the Dual cycle and Diesel cycle. This is because to achieve the maximum pressure as that of the Diesel cycle, the CR should be increased for the same heat supplied in the case of the Otto cycle. This makes the cycle to expand low and hence the efficiency of the Otto cycle is the minimum. At the same time, the efficiency of the Dual cycle lies between that of Diesel and Otto cycles.

11.9.3 Same Maximum Temperature and Heat Rejected We know that the air standard efficiency is a function of the lowest and the highest temperatures as heat is directly proportional to temperature. It is apparent that, for the same maximum temperature and the heat rejected, either the compression ratio has to be increased or the heat supply must be higher for the Otto cycle. This makes the engine to practically experience more thermal

11.31

Air Standard Cycles

and mechanical stresses. The efficiency is the highest for the Diesel cycle, while it is the lowest for the Otto cycle. The efficiency of the Dual cycle lies between the two.

11.9.4 Same Compression Ratio and Heat Rejected In this case, the heat rejected for those cycles where the heat is rejected at a constant volume is the same. However, the heat added in the Diesel cycle is the highest, making the denominator of the efficiency equation higher. Hence, the efficiency is the lowest in the case of the Diesel cycle, followed by the Dual and Otto cycles.

11.9.5 Same Maximum Pressure and Output We know that the heat supplied is the sum of the work output and the heat rejected. Also, the efficiency depends on the ratio of the heat rejected to the heat supplied. For the same maximum pressure and the work output, heat rejection is higher in the case of the Otto cycle. For achieving the same maximum pressure and the work output, the compression ratio of the cycle has to be necessarily increased, which makes the cycle to have a higher heat rejection. The comparisons of the important factors of different air standard cycles is given in Table 11.2. Table 11.2 Summary of the Important Factors of Different Air Standard Cycles Factor

Carnot

Otto

Compression Adiabatic Isentropic process Heat addition Isothermal Constant volume General mRT3 ln equation for ( V4 ) heat addition ( V3 )

Expansion Heat rejection

mCvΔT

Adiabatic Isentropic process Isothermal Constant volume

General mRT1log equation for ( V1 ) heat rejection ( V2 )

mCvΔT

Air standard Lowest and Compresefficiency highest sion ratio depends on temperature

Diesel

Dual

Brayton

Lenoir

Isentropic Isentropic

Atkinson

Isentro- Nil pic Constant Constant volume Constant Constant pressure and then Conpressure volume stant pressure mCpΔT mCpΔT (temp. mCpΔT mCvΔT difference in constant pressure process)+ mCvΔT (temp. difference in constant pressure process) Isentropic Isentropic Isentro- Isentropic pic Constant Constant volume Constant Constant volume pressure pressure

Isentropic

mCvΔT

mCvΔT

mCpΔT

Compression ratio, expansion ratio and adiabatic index

Compression Pressure VoluCompresratio, expansion ratio metric sion ratio, ratio, cut-off ratio expansion Expansion and adiabatic ratio and ratio and index adiabatic adiabatic index index

mCpΔT

mCpΔT

Constant volume mCvΔT

Isentropic Constant pressure

11.32

Engineering Thermodynamics

Solved Problems Problem 11.1 In an Otto cycle the air that enters at 27°C and 1 bar is compressed adiabatically until its temperature rises to 477°C. If the peak temperature is 1227°C, calculate the air standard efficiency, compression ratio and the mean effective pressure. Assume g = 1.4 and Cv = 0.717 kJ/kgK for air. T Solution h = 1 – 1 T2 300 = 60% = 1 – = 0.6 750 1



V  T r – 1 r = 1 =  2  V2  T1  1

750  1.4 – 1 2.5 =  = 9.88  = 2.5  300  Heat supplied = Cv(T3 – T2) = 0.717(1500 – 750) = 537.75 kJ/kg Work done = h × Qs = 537.75 × 0.6 = 322.65 kJ/kg 3

p

2

4 1 V

Fig. p. 11.1



Work done Swept volume pV = mRT RT1 v1 = p1 pm =

287 × 300 0.861 m /kg = 1 × 10

r =



v2 =

V1 V2 0.861 = 0.087 m 3 /kg 9.88

11.33

Air Standard Cycles



Swept volume = v1 – v2 = 0.774 m3­ ­­ /kg

322.65 × 10 3 = 4.16 bar . 0.774 Problem 11.2 An engine working in an ideal Otto cycle has a compression ratio of 9 : 1. Air enters at 27°C and 1 bar before compression and 1200 kJ/kg of heat is added at the end of the compression stroke. Determine the peak temperature, pressure, work output per kg of air and air standard efficiency. Assume γ = 1.4 and Cv = 0.717 kJ/kg K for air.

pm =

 r1 T2 Solution       =  T1  r2

g –1

 0.4 = 9= 2.4 



T2 = 720 K



Qs = Cv(T3 – T2)



1200 = 0.717(T3 – 720)

T3 = 2393.6 K 3

p

2

4 1 V

Fig. p. 11.2 p-V diagram g

p2  V1  1.4 9= 21.67 =  =  p1  V2            p2 = 21.67 bar 1 1 1− = 58.47% hotto = 1 − g − 1 = 1.4 − 1 ) ( ) ( r r

Work output/kg of air = 1200 × 0.5847 = 701.64 kJ/kg.

Problem 11.3 In an Otto engine the air that enters at 27°C and 1 bar is compressed adiabatically to 12 bar. If the peak temperature is 1700 K. calculate (i) the heat supplied and the work done per kg of air (ii) the pressure at the end of adiabatic expansion (iii) the theoretical efficiency. Assume γ = 1.4 and Cv = 0.717 kJ/kg K for air.   Solution      T2 =  p2  T1  p1 

g –1 g

0.4

T2 = 300 × ( 12 ) 1.4 = 610 K

11.34

Engineering Thermodynamics 3

p

2

4 1



V

Fig. p. 11.3 p-V diagram

h = 1 –



T1 = 50.8% T2

Qs = Cv(T3 – T2) = 0.717 × (1700 – 610)

= 781.53 kJ/kg

h =

W Qs

W = 781.53 × 0.508 = 397 kJ/kgK          

p3 T3 = p2 T2



p3 = 1200 ×

1700 = 33.44 bar 610

g

p3  V1  p2 = =  V p p1              4  2 

 p1  p4 = p3 =   p2 

33.44 = 2.78 bar. 12

Problem 11.4 An engine operates on Otto cycle between pressures of 1 bar and 30 bar. The ratio of pressure at constant volume is 4. The temperature at the end of combustion and expansion is 200°C, and the law of compression and expansion is pV1.3 = Constant. If the engine now operates on the Carnot cycle for the same range of temperature, find the efficiency of the cycle. Solution

p1 = 1 bar



p3 = 30 bar

⇒

p3 = 4 p2

p

30

3 = 7.5 bar p2 = = 4 4

T2 = 473K

pV1.3 = Constant

11.35

Air Standard Cycles 3

2

p

4 1 V

Fig. p. 11.4 p-V diagram

Process 2 – 3 p2 p p T = 3 ⇒ T3 = 3 2 = 4 × 473 = 1892 K T2 T3 p2 Process 1 – 2 1.3 p1 × V1.3 1 = p2 × V2

V  ⇒ 1   V2 

1.3

=

p2 7.5 ⇒ r 1.3 = ⇒ r = 7.51/1.3 = 4.7. T1 1 0.3

 V2  T1V1.3-1 = T2V21.3 – 1 ⇒ T1 = T2 ×  =  1  V1 

T2 = r 0.3

For same range of temperature efficiency = 1 –

473 = 297.3 K 4.7 0.3

T1 287.3 = 84.3%. =1– T3 1892

Problem 11.5 An engine working on Otto cycle has a volume of 0.5 m3, pressure of 1 bar and temperature of 27°C at the commencement of the compression stroke. At the end of compression stroke the pressure is 10 bar. The heat added during the constant volume process is 200 kJ. Determine the (i) Percentage clearance (ii) Air standard efficiency (iii) Mean effective pressure (iv) Ideal power developed by the engine if runs at 400 rpm, so that there are 200 complete cycles per minute. p1 = 1 bar 

V1 = 0.5 m3

T1 = 300 K 

p2 = 10 bar

Solution

Qs = 200 kJ   N = 400 rpm



R = 1.4



p1V11.4 = p2V21.4

 V2    V          1 

1.4

=

p1 p2

11.36

Engineering Thermodynamics 3

p

2

4 1 V

Fig. p. 11.5 p-V diagram 1

1

 p  1.4  1  1.4 V1   × = 0.5 0.09653 m 3 V2 =  1  ×= p 10   2   Clearance volume, Vc = V2 = 0.09653 m3 V2 0.09653 Percentage clearance = = × 100 = 23.92% V1 – V2 0.5 − 0.09653 Compression ratio,

V1 0.5 = = 5.18 r = V2 0.09653

1 1 − g –1 1 – = 48.2% Efficiency = 1= r 5.181.4 –1 W­ net = h × Qs = 0.482 × 200 = 96.4 kJ

3 V­ s = V1 – V2 = 0.5 – 0.09653 = 0.40347 m Wnet 96400 = = 2.39 bar pm = Vs 0.40347 \  Mean effective pressure = 2.39 bar



N = 400 rpm

Engine makes 200 cycles in one minute. Time taken for one cycle, t =

60 s 200

Wnet 96.4 kJ p = = = 321.34 W 60 t s 200 Ideal power developed = 321.34 kW.

Problem 11.6 An engine works on the Diesel cycle with a compression ratio of 17 and cut-off ratio of 1.7. Calculate the air standard efficiency. Take γ = 1.4 Solution

1 1  rcg − 1  h = 1 − ( g – 1) . .   g  rc − 1  r h = 1 −

1.4   . 1 .  1.7 − 1  = 63.8% 1.4 – 1 ) 1.4  1.7 − 1  17 (  

1

11.37

Air Standard Cycles

Problem 11.7 Calculate the air standard efficiency of a diesel engine working with a compression ratio of 16 and expansion ratio of 10. Take γ = 1.4 Solution Given

r = 16   re = 10



rc =



h = 1 −

r = 1.6 re  g  . 1 .  rc − 1  g – 1) g   ( r  rc − 1  1

 1.4  1 . 1 .  1.6 − 1  = = 1 − 63.44%. 16( 1.4 – 1) 1.4  1.6 − 1  Problem 11.8 A diesel engine having a cylinder with a bore of 260 mm, stroke of 400 mm and a clearance volume of 1600 cc, runs with fuel cut-off at 5 % of the stroke. Calculate its efficiency if γ = 1.4 for air.

Solution Vs =

π 2 π D L = × 26 2 × 40 =21237 cc 4 4



V 21237 r = 1 = − s 1 – = 14.27 Vc 1600



rc =

V3 . V2

Cut-off volume = v3 – v2 = 0.05 Vs = 0.05 × 13.27 Vc = 0.6635 cc

v3 = 1.6635 Vc

rc = 1.6635

1 1  rcg − 1   h = 1 − ( g – 1) . .  g  rc − 1  r

1.4   1 . 1 .  1.6635 − 1  = = 1 − 61.37%. 1.4 – 1 ) 1.4  1.6635 − 1  ( 14.27  

Problem 11.9 Determine the air standard efficiency of a diesel engine with a compression ratio of 16 : 1 and air fuel ratio 50 : 1. The temperature of air before the compression is 50°C and calorific value of fuel used is 42 MJ/kg. Take γ = 1.4 and Cp = 1.005 kJ/kg K for air. Solution Consider the process 1-2 g –1

T2  V1  0.4 3.0314 = =  = 16 T1  V2            T2 = 979.15 K

11.38

Engineering Thermodynamics

Consider the process 2-3 2

3

4

p

1 V

Fig. p. 11.6 p-V diagram

Heat added/kg of air = Cp(T3 – T2) = T3 – T2 =

F × CV A

42000 = 835.82 K 50 × 1.005

T3 = 1815 K

rc =

V3 T3 = V2 T2



rc =

1815 = 1.853 979.15



1 1  rcg − 1   h = 1 − ( g – 1) . .  g  rc − 1  r

1 1 .  1.8537 1.4 − 1  62.11% . = 1 − ( 1.4 – 1) .  = 1.4  1.853 − 1  16

Problem 11.10 An oil engine works on the Diesel cycle, the compression ratio being 15. The temperature and pressure at the start of compression are 27°C and 1 bar respectively. 800 kJ of heat is supplied at constant pressure per kg of air, and it attains a temperature of 437°C at the end of adiabatic expansion. Find the air standard efficiency of the cycle. What would be the theoretical work done per kg of air. Also determine the mean effective pressure. Take Cv = 0.717 kJ/kgK and h = 1.4. Solution T2 = T1 r(g – 1) 300 × 15(1.4 – 1) = 886.25 K Heat supplied per kg (Qs) = Cp(T3 – T2) = 800 T3 – T2 =

800 = 796.81 1.004

T3 = 793.8 + 886.25 = 1683.06 K

hDiesel = 1 –

1  T4 – T1    g  T3 – T2 

11.39

Air Standard Cycles

= 1 –

1 1.4

  710 – 300   1683.06 – 886.25  

= 0.6325 = 63.25%

T3 1683.06 = 1.9 r= c = T 886.25 2 2

3

4

p

1 V

Fig. p. 11.7 p-V diagram

Heat rejected per kg (QR) = Cv(T4 – T1)

= 0.717(700 – 300) = 293.97 kJ/kg Work done per kg of air (W) = Qs – QR

= 800 – 293. 97 = 506.03 kJ/kg Mean effective pressure (pm) =

(

)

p1  gr g ( rc – 1 ) – r rcg – 1    – 1 r – 1 g ( )( )

(

)

1 1.4 × 151.4 ( 1.9 – 1 ) – 15 1.91.4 – 1   =  ( 1.4 – 1) ( 15 – 1)

= 6.06 bar. Problem 11.11 Air enters into an air standard Dual cycle at 1 bar and 27°C before compression, and has a compression ratio of 12. The maximum pressure and temperature of the cycle are 40 bar and 1600°C respectively. Determine (i) the temperature at the end of the constant volume heat addition (ii) cut-off ratio (iii) work done per kg of air and (iv) cycle efficiency. Assume Cp = 1.005 kJ/kgK Cv = 0.717 kJ/kgK and γ = 1.4 for air. Solution Consider the process 1-2 T2 g− 1 0.4 = r= 12= 2.7 T1

          T2 = 2.7 × 300 = 810 K

11.40

Engineering Thermodynamics

p2 γ = r= 12 1.4= 32.42 p1

          p2 = 32.42 bar 3 p

4

2

5 1 V

Fig. p. 11.8 p-V diagram

Consider the process 2-3 T3 p3 40 = = = 1.234 T2 p2 32.42

          T3 = 1.234 × 810 = 999.5 K rc =

V3 T3 1873 = = = 1.874 V2 T2 999.5

Heat supplied/kg of air = Cv(T3 – T2) Cp(T4 – T3)

= 0.717(999.5 – 810) + 1.005(1873 – 999.5) = 1013.74 kJ/kg Consider the process 4-5 γ–1

γ–1

r T4  V5  = =  =  T5  V4   rc            

0.4

 12  2.1  =   1.874 

T4 1873 T5 = = = 891.2 K 2.1 2.1 Heat rejected/kg of air = Cv(T5 – T1)

= 0.717 (891.2 – 300) = 423.89 kJ/kg Work output/kg of air = 1013.74 – 423.89 = 589.85 kJ/kg

hDual =

Work output 589.85 = = 58.12%. Heat input 1013.74

Problem 11.12 A compression ignition engine works on the dual combustion cycle. The pressure and temperature at the beginning of compression are 1 bar and 27°C respectively, and the pressure at the end of compression is 28 bar.

11.41

Air Standard Cycles

If 500 kJ of heat is supplied per kg of air during constant volume heating and the pressure at the end of adiabatic expansion is found to be 3.5 bar, find the ideal thermal efficiency and mean effective pressure. Assume Cp = 1.004 kJ/ kgK and Cv = 0.717 kJ/kgK. Solution Given In Process 1-2:

p1 = 1 bar, T1 = 300 K, p2 = 28 bar, p5 = 3.5 bar.

p1V1g = p2V2g

Now, we know that pV = mRT 105 × V1 = 1 × 287 × 300 V1 = 0.861 m3/kg. we know that

V2g =

p1 g V p2 1

V2 = 0.079 m3, V3 = 0.079 m3 Also

p5 = 3.5 bar,   V5 = 0.861 m3

Now, heat supplied per kg during constant volume (Qs) = Cv(T3 – T2)

500 = 0.717 × (T3 – T2)

where,

T2 = r(g – 1), T2 = 777.31 K

Thus solving equation 2 we get T3 = 1474.66 K In Process 2-3:

p3 T = 3 p2 T2 p3 =

T3 × p2 T2

= 53.12 bar Also

p4 = 53.12 bar. 4

3 p

2

5 1 V

Fig. p. 11.9 p-V diagram

In process 4-5:

p5V5g = p4V4g

V4 = 0.123 m3. Thus, we have

r =

V1 = 10.89 V2

11.42

Engineering Thermodynamics

rp =

p3 = 1.89 p2

rc =

V4 = 1.56 V3

  rp rvg – 1   hdual = 1 − ( g –1)  rp – 1 + rp g ( rc – 1 )  r   1



= 1 −

(

1 10.890.4

)

 1.89 – 1.561.4 – 1     0.89 + 1.89 × 1.4 × 0.56 

= 58.95%.

(

) (

 grp r g ( rc – 1 ) + r g rp – 1 – r rp rcg – 1 Mean effective pressure, pm = p1   ( g – 1)( r – 1)  pm = 1 × 10.025 = 10.025 bar.

)   

Problem 11.13 The pressure ratio is 1.5 in the process of isochoric heat supply for the cycle of an internal combustion engine, with a mixed supply of heat = 1200 kJ/kg and a compression ratio = 14. Find the thermal efficiency and temperature at the characteristic points of the cycle, if the initial parameters are 1 bar and 47°C and the working substance is air. Also, find the mean effective pressure. Given

p1 = 1 bar.

T1 = 273 + 47 = 320K. Solution

In Process 1-2: p1V1g = p2V2g



g

Thus, Also

V  V  p2 = p1  1  p1  1  V  2  V2  p2 = 40.23 bar

g

T1V1(g–1) = T2V2(g–1)

T2 = 320 × 140.4 T2 = 919.6 K. p2 T2 In Process 2-3:    p = T 3 3

T3 = 919.6 × 1.5 = 1379.4 K.

Also heat supplied = Cp(T4 – T3) + Cv(T3 – T2)

1200 = 1.004(T4 – 1379.4) + 0.717(1379.4 – 919.6)

T4 = 2246.25 K

11.43

Air Standard Cycles

Also we know that V3 = V4 and p3 = 40.23 bar p4 3 p

4

2

5 1 V

Fig. p. 11.10 p-V diagram

In Process 3-4:    

V4 T4 = V3 T3



V4 = 1.63 V3

1.63 V1 14 T5V5(g –1) = T4V4(g –1) =

T5 = 0.423 × T4 = 950.35 K.



Hence we have the following values r = 14; rp = 1.5; rc = 1.63. Hence,

 rp rvg – 1 1   h = 1 − ( g –1)   rp – 1 + rp g ( rc – 1 )  r  

= 1 −

(

1 1.40.4

)

 1.5 – 1.631.4 – 1     0.5 + 1.5 × 1.4 × 0.63 

= 62.34%.

(

) (

 grp r g ( rc – 1 ) + r g rp – 1 – r rp rcg – 1 Mean effective pressure, pm = p1   ( g – 1)( r – 1)  pm = 8.79 × 1 = 8.79 bar.

)   

Problem 11.14 A compression ignition engine works on the dual combustion cycle. The pressure and temperature at the beginning of compression are 1 bar and 27°C respectively, and the pressure at the end of compression is 28 bar. If 500 kJ of heat is supplied per kg of air during constant volume heating and the pressure at the end of the adiabatic expansion is found to be 3.5 bar, find the ideal thermal efficiency and mean effective pressure. Assume Cp = 1.004 kJ/kgK and Cv = 0.717 kJ/kgK. Given p1= 1 bar, T1 = 300 K, p2 = 28 bar, p5 = 3.5 bar.

11.44

Engineering Thermodynamics

Solution In Process 1-2 :

g

g

p1V1 = p2V2

Now, we know that pV = mRT 105 × V1 = 1 × 287 × 300 V1 = 0.861 m3/kg. p g we know that V2 = 1 V1g p2 V2 = 0.079 m3,  V3 = 0.079 m3



4

3 p

2

5 1 V

Fig. p. 11.11 p-V diagram

Also

p5 = 3.5 bar,  

V5 = 0.861 m3.

Now, heat supplied per kg during constant volume (Qs) = Cv(T3 – T2)



where,

500 = 0.717 × (T3 – T2)

T2 = T1r(g –1),  T2 = 777.31 K

Substituting T2 in equation for heat supplied and solving, we get

In Process 2-3:

T3 = 1474.66 K.

p3 T = 3 T2 p2



p3 =



T3 × p2 T2

= 53.12 bar Also In Process 4-5 :

p4 = 53.12 bar. g

g

p5V5 = p4V5

V4 = 0.123 m3. Thus, we have

r =

rp =

V1 = 10.89 V2 p3 = 1.89 p2

11.45

Air Standard Cycles

rc =

V4 = 1.56 V3

  rp rvg – 1   h = 1 – ( g –1)  rp – 1 + rp g ( rc – 1 )  r   1

(

)

 1.89 – 1.561.4 – 1  1 = 1 –   = 58.95%. 10.890.4  0.89 + 1.89 × 1.4 × 0.56 

(

) (

 grp r g ( rc – 1 ) + r g rp – 1 – r rp rcg – 1 p Mean effective pressure, pm = 1   ( g – 1)( r – 1) 

)   

pm = 1 × 10.025 = 10.025 bar.

Objective Type Questions 1. Compression ratio is the (a) Ratio between the initial volume and the final volume of the cylinder (b) Ratio between the final volume and the initial volume of the cylinder (c) Ratio between the final volume and the stroke length of piston (d) Ratio between the initial volume and the stroke length of piston 2. Clearance volume is provided (a) Above the piston (c) On the side of the piston

(b) Below the piston (d) None of these

3. Swept volume is the (a) Volume occupied by the piston when the piston sweeps from TDC to BDC (b) Volume occupied by clearance (c) Volume occupied by the cylinder (d) None of these 4. Total volume is the (a) Sum of the swept volume and clearance volume (b) Swept volume (c) Difference between the swept volume and the clearance volume (d) None of these 5. The Otto cycle is also called the (a) Constant volume cycle (c) Carnot cycle

(b) Constant pressure cycle (d) Dual cycle

11.46

Engineering Thermodynamics

6. The diesel cycle is also called the (a) Constant pressure cycle (c) Carnot cycle

(b) Constant volume cycle (d) Dual cycle

7. The dual cycle is also called the (a) Constant pressure cycle (c) Carnot cycle

(b) Constant volume cycle (d) Limited pressure cycle

8. In the brayton cycle the heat rejection takes place at (a) Constant pressure (b) Constant volume (c) Constant temperature (d) Constant entropy cycle 9. The mean effective pressure of the cycle is defined as the ratio between the (a) Work output and the swept volume (b) Swept volume and the work output (c) Heat supplied and the heat rejection (d) Heat rejection and the heat supplied 10. For an engine operating on an air standard Otto cycle, the clearance volume is 10% of the swept volume. The specific heat ratio of air is 1.4. The air standard cycle efficiency is (a) 38.3% (b) 39.8% (c) 60.2% (d) 61.7% 11. An ideal air standard Otto cycle has a compression ratio of 8.5. If the ratio of the specific heats of (y) is 1.4, what is the thermal efficiency (in percentage) of the Otto cycle? (a) 57.5

(b) 45.7

(c) 52.5

(d) 9

12. Which one of the following cycles has the highest thermal efficiency for the given maximum and minimum cycle temperatures? (a) Brayton cycle (b) Otto cycle

(c) Diesel cycle (d) Stirling cycle

13. Which one of the following cycles has both heat addition and heat rejection at constant pressure? (a) Brayton cycle (b) Otto cycle

(c) Diesel cycle (d) Stirling cycle

14. Which one of the following cycles is used in modern diesel engines? (a) Brayton cycle

(b) Otto cycle

(c) Diesel cycle (d) Dual cycle

15. Which one of the following cycles is used in gas turbines? (a) Brayton cycle

(b) Otto cycle

(c) Diesel cycle (d) Carnot cycle.

Theory Questions 1. Define the compression ratio. 2. Define the swept volume.

11.47

Air Standard Cycles

3. Why is the Carnot cycle not used in real time applications? 4. Write the air standard efficiency of the Otto, Diesel and Dual cycles. 5. Draw the p-V diagram of a Dual cycle. 6. Derive an expression for the efficiency of the Otto cycle in terms of the compression ratio. 7. Sketch the ideal Otto cycle on the p-V and T-s diagrams. Name all the processes. Write the expression for air standard efficiency. 8. Sketch Diesel cycle on the p-V and T-s diagrams, and write the expression for its air standard efficiency. Name all the processes. 9. Draw the p-V and T-s diagrams of the Dual cycle and derive the expression for its air standard efficiency. 10. Derive the expression for the work output and mean effective pressure of the Otto cycle. 11. Derive the expression for the work output and mean effective pressure of the Diesel cycle. 12. Derive the expression for the work output and mean effective pressure of the Dual cycle. 13. Draw the p-V and T-s diagrams of the Brayton cycle, and derive the expression for its air standard efficiency. 14. Derive an expression for the air standard efficiency of the Atkinson cycle. Draw the p-V and T-s diagrams also. 15. Derive an expression for the air standard efficiency of the Lenoir cycle. Draw the p-V and T-s diagrams also. 16. Derive an expression for air standard efficiency of the Stirling cycle. Draw the p-V and T-s diagrams also.

Unsolved Problems 1. A Carnot cycle works between the temperatures of 300 K and 700 K. Find the maximum work possible per kg of air. 2. An engine of 20 cm bore and 30 cm stroke works on the Otto cycle. The clearance volume is 1600 cu.cm. The initial pressure and temperatures are 1 bar and 30°C respectively. If the maximum pressure is limited to 24 bar, calculate (i) the air standard efficiency of the cycle, and (ii) the mean effective pressure of the cycle. 3. In an Otto cycle, the temperatures at the beginning and the end of isentropic compression are 43°C and 323°C respectively. Find the compression ratio and the air standard efficiency. 4. An engine working on the Otto cycle has a cylinder diameter 150 mm and length of the stroke 225 mm. The clearance volume is 0.00125 m3. Find its air standard efficiency assuming γ = 1.4. What is the temperature at the end of compression, if the initial temperature is 40°C ?

11.48

Engineering Thermodynamics

5. An air standard cycle has a compression ratio of 16, and begins at 1 bar and 20°C. The heat transferred at constant pressure is 312 kJ/kg, and it is equal to that at constant volume. Determine the temperature at all the cardinal points of the cycle and the thermal efficiency. Take Cp and Cv for air as 1.005 and 0.718 kJ/kgK respectively. 6. An air standard Diesel cycle has a compression ratio of 18 and the heat transferred to the working fluid per cycle is 1800 kJ/kg. At the beginning of the compression process the pressure was 0.1 MPa, and the temperature was 15°C. Determine: (a) The pressure and temperature at each point in the cycle. (b) The thermal efficiency. (c) The mean effective pressure. 7. One kg of air is taken through the Diesel cycle. Initially, the air is at 25°C and 1 bar. The compression ratio is 14 and the heat added is 1850 kJ/kg. Calculate the ideal cycle efficiency and the mean effective pressure. 8. The engines are to operate on the Otto and Diesel cycles with the following data: maximum temperature = 1400 K. exhaust temperature = 700 K, air is taken in at 1 bar pressure and 300 K temperature. Find the compression ratios, maximum pressures, air standard efficiencies, and rates of work output of the respective cycles for 1 kg/min of air supply. 9. An air standard Dual cycle has a compression ratio of 16, and, compression begins at 1 bar and 50°C. The maximum pressure is 70 bar. The heat transferred to the air at constant pressure is equal to that at constant volume. Estimate (i) the pressure and temperature at the cardinal points of the cycle (ii) the cycle efficiency (iii) the mep of the cycle (for air Cp = 1.005 kJ/kgK and Cv = 0.718 kJ/kgK). 10. A dual combustion air standard cycle has a compression ratio of 10. The constant pressure part of combustion takes at 40 bar. The highest and the lowest temperatures of the cycle are 1727°C and 27°C respectively. The pressure at the beginning of the compression is 1 bar. Determine (i) the pressure and temperature at key points of the cycle (ii) the heat supplied at constant volume (iii) the heat supplied at constant pressure (iv) the heat rejected (v) the work output (vii) the efficiency, and (viii) the mean effective pressure. 11. In a gas turbine plant working on the Brayton cycle with a regenerator of 75% effectiveness, the air at the inlet to the compressor is at 0.1 MPa, 30°C, the pressure ratio is 6, and the maximum cycle temperature is 900°C. If both the turbine and compressor each have an efficiency of 80%, find the % increase in the cycle efficiency due to regeneration. 12. In the Brayton cycle, the air enters the compressor at 1 bar and 25°C. The pressure of air leaving the compressor is 3 bar, and the temperature at the turbine inlet is 650°C. Determine per kg of air (i) the cycle efficiency (ii) the heat supplied to air (iii) the work output (iv) the heat rejected in the cooler, and, the temperature of air leaving the turbine.

Air Standard Cycles

11.49

13. Air enters the compressor of the gas turbine at 100 kPa and 25°C. For a pressure ratio of 5 and a maximum temperature of 850°C, determine the thermal efficiency of the Brayton cycle. 14. Calculate the efficiency and the specific work output of a simple gas turbine plant operating on the Brayton cycle. The maximum and minimum temperatures of the cycle are 1000 K and 288 K respectively. The pressure ratio is 6 and the isentropic efficiencies of the turbine and compressor are 85 and 90% respectively. 15. A gas turbine works on the Brayton cycle. The initial condition of the air is 25°C and 1 bar. The maximum pressure and temperature are limited to 3 bar and 650°C respectively. Determine the (i) cycle efficiency (ii) heat supplied and heat rejected per kg of air (iii) work output per kg of air and (iv) exhaust air temperature.

INDEX Absolute temperature 1.8, 1.9, 2.2, 2.5, 4.13, 5.1, 6.12 Absolute temperature scale 1.8 Absolute zero 1.9, 8.8, 11.5 Adiabatic flame temperature 10.20, 10.21 Adiabatic process 3.13, 3.27, 4.8, 4.15, 4.20, 6.17, 6.21, 9.17, 11.3, Adiabatic saturation process Adiabatic saturation temperature 9.7, 9.8 Adiabatic work Air standard cycles 11.1, 11.6, 11.31 Air standard efficiency 11.9, 11.12, 11.15, 11.25 Alcohol 9.5, 10.3, 10.5, ethanol 10.5 methanol 10.5 Ampere 1.2 Analysis 1.1, 2.8, 2.11, 10.2, 11.1, 11.2 proximate 10.2 ultimate 10.2, 10.3 Anthracite 10.1, 10.3 Ash content 10.1, 10.2 Available energy 5.1, 5.2, 5.4, 5.12, 5.15, 11.22 Availability 5.1, 5.2, 5.4, 5.12, 5.15, 11.22 non flow process 3.9, 5.5, 6.17 steady flow process 3.20, 5.4 Availability function 5.5, Avagdro’s law 2.4 Avagadro’s number 2.4 Base units 1.1, 1.2 Berthelot equation 7.10, 7.11, 7.19

Biodiesel 10.3, 10.4, 10.6 Biogas 10.8 Bituminous 10.1, 10.3 Blast furnace gas 10.8 Boyle’s law 2.1 Bourdon tube 1.7 Brake power 11.28 Brayton cycle 11.25, 11.28 By pass factor 9.24, Calorific value 10.3, 10.9, 10.11 Calorimeter 6.14, 10.12 bomb 10.12 separating 6.14, throttling 6.15, 6.27 Candela 1.2 Carbon dioxide 2.12, 6.1, 7.2, 9.1, 10.1, 10.8 Carbon monoxide 9.2, 10.4, 10.5, 10.8, 10.18, 10.20 Carnot cycle 4.15, 11.2, 11.3, 11.5, Carnot efficiency 4.9, 11.5 Carnot’s theorem 4.16 Carrier equation 9.5, 9.18 Celsius temperature scale Characteristic gas constant 2.3 Charls’ law 1.9, 2.2 6.23 Charcoal 10.1 Chemical equilibrium 3.5 formula 10.15 reaction 3.5, 10.1, 10.7, 10.14, 10.23 Clausius inequality 4.14, 4.16

I.2

Index

Clausius’ statement 4.4, 4.5 4.14 Clausius theorem 4.17 Clausius-Clapeyron equation 8.24 Clearance volume 11.29, 11.36 Closed system 3.2, 5.5, 6.17 Cloud point 10.10, Co efficient of performance 4.2, 4.3 Coal, 10.1, 10.8, 10.11 Coal gas 10.8 Coke 10.1, 10.3 10.8 Combustion 4.7, 5.2, 10.1 Compressibility chart 7.13 Compressibility factor 7.12, Compression ignition engine 11.12, 11.17 Compression ratio 11.4, 11.12, 11.15, 11.17 Compressor 1.7, 3.2, 3.18, 3.24 Condenser 3.20, 3.22, 6.16 Conduction 1.13 Conservation 1.13, 3.1, 3.7, 4.1, 9.7 Continuity equation 3.19 Control volume 3.2, 3.9, 3.19 Convection 1.13 Correction 7.1, 7.3, 7.5 Cooling and dehumidification 9.10, 9.13, 9.15 Cooling coil 9.10, 9.14 Critical pressure 6.4, 6.5, 6.9, 7.6 Critical temperature 6.4, 7.5, 7.6 Critical volume 6.4, 7.6 Crude oil 10.1, 10.3 Cut-off ratio 11.15, 11.20, 1131 Dalton’s law 2.8 Dead state 5.4, 5.6 Degradation of energy 5.4 Degree of saturation 9.4 Dehumidification 9.10, 9.12, 9.14 Degree of superheat 6.3, 6.5 Degree of saturation 9.4 Dehumidification 9.10, 9.12, 9.13, 9.16 Density 1.2, 1.7, 1.10, 3.3, 6.10, 7.1, 7.4, 10.8, 10.9 Dew point temperature 9.1, 9.2, 9.6, 9.14

Diathermic wall 3.5, 4.8, 4.9 Diesel cycle 11.5, 11.12, 11.15, 11.28 Displacement work 1.15 Dry air 9.1, 9.3, 9.4, 9.6 Dry bulb temperature 9.1, 9.2, 9.6, 9.9 Dryness fraction 6.2, 6.7, 6.9, 6.11, 6.14 Dual cycle 11.17, 11.17, 1130, 11.31 Efficiency 4.2, 4.4, 5.1 Electrical work 1.15, 1.16, 1.18 5.2 Energy 5.2, 5.4, 10.1, available 5.1, 5.2, 11.22 high grade 5.2, internal 1.12 low grade 5.2, potential 1.12, 10.21 unvailable 3.19, 3.21 Energy balance 3.19, 9.8 Energy equation 3.18, 3.20, 3.22, Enthalpy 2.4, 10.22 of combustion 10.22, 10.23 of formation 10.23, 10.24 Entropy 1.9, 2.4, 2.8, 3.11, 6.10, 8.11, 9.6, 9.8, 10.14 of ideal gas 2.4 principle 5.5, Equation of state 2.2, 6.17, 7.3, 8.11 Berthelot equation 7.10, 7.11 Dietrici equation ideal gas 2.1, 2.4, Redlich-Kwong equation van der Waals 7.1, 7.6 virial 7.9, Wohl 7.12 Equilibrum 3.4, 1.6, 6.3, 8.8 chemical 3.5 mechanical 3.5, 5.4 thermal 3.5 thermodynamic 3.5 Ericsson cycle 11.6 Evaporator 3.21 Excess air 10.15, 10.18 Extensive property 3.3 Expansion ratio 11.9, 11.10, 11.15

I.3

Index First law of thermodynamics 3.7, , 3.8, 3.11, 3.26, 4.1, 6.17, 8.4 for a closed 5.5 for a cycle for a steady flow process Fire point 10.10 Fishcer Tropch Diesel 10.5 Fixed carbon 10.2 Flame temperature 10.20, 10.21 actual 6.6, 10.20, 10.21, adiabatic 10.20, 10.21 average 10.20, 10.21 Flash point 10.9, 10.10 Flue gas 10.17, 10.18, 10.22 analysis 10.19 Flow work 1.17 Force of attraction 7.1, 7.2, 7.3, 7.4 Formation 6.2, 10.23 enthalpy of 2.4, 2.5, 2.13, 3.23, 6.8, 6.10, 6.11, 6.22, 9.8 Free expansion 1.15, 4.7 Fuel 3.18, 5.2, 10.1, 10.3 gas 10.6 liquid 10.3, 10.5, 10.9, 10.12 solid 10.1, 10.2 Fugacity

convection 1.13 radiation 1.13 Heating and humidification 9.14. 9.15 Heating value 10.11, 10.12 Helmholtz function 8.1, Humidification 9.12, 9.14 Humidity ratio 9.3 Hydrogen 2.12, 4.7, 6.5, 7.2 Hydrography recorder 9.6

Gas constant 2.3, 2.4, 9.3 characteristic 2.3 universal 2.3, 2.4 Gas equation, ideal 2.3, Gas mixtures 7.1 Gibbs function 8.3

Keenan function Kelvin Planck statement 4.4, 4.6 Kelvin temperature scale 1.8 Kilogram 1.2 Kinematic viscosity 10.4 Kinetic energy 1.12, 3.19, 5.2

Heat latent 1.14, 6.1, 9.15 sensible 1.14, 6.1, 9.10, 9.15 specific 2.5, 2.6, 6.11 Heat capacity 2.5, 6.13 at constant pressure 2.13 at constant volume 2.7 Heat engine 4.2, 4.6, 11.1, Heat pump 4.3 Heat transfer 3.7 conduction 1.13

Ideal gas equation 2.3 Ignition temperature 10.10 Internal energy 2.4, 10.22 Intensive property 1.8, 3.3 Inversion curve 8.14 Inversion temperature 8.15 Irreversible process 4.7 Irreversibility 5.1 Isentropic process 4.20, 6.21, 6.26 Isolated system 3.2 Isothermal compressibility 8.18, 8.19 Isothermal process 3.10, 6.21, 11.3 Joule-Thomson effect 8.15 Joule’s law 2.5

Law of thermodynamics first 3.7, 4.1, 3.26 second 1.13, 4.1, 4.11, 5.1 Latent heat 1.14, 6.5, 9.12 of fusion 6.1, 6.2 of sublimation 1.14 vaporization 1.14 Limited pressure cycle 11.17 Liquid fuel 10.3, 10.5 Liquefied petroleum gas 10.7 Low sulphur gasoline 10.3, 10.5

I.4 Maximum work 5.1 Maxwell’s equations 8.10 Mechanical equilibrium 3.5 Mechanical work 3.25, 5.2, 10.1 Mixture of gases 2.8, 2.13 Moist air 9.2, 9.6 Moisture content 10.2, 10.10 Mole 1.2, 2.8 Mole fraction 2.8, 2.9, 2.12 Molecular weight 2.3, 2.10 Mollier diagram 6.28 Natural gas 10.7 Nozzle 3.25 Open system 3.2, 3.18 Otto cycle 11.7, 11.10, 11.30 Paddle wheel work 1.16, 1.18 Partial pressure 2.8, 2.10, 9.5 Pascal Path 3.4, 4.17 Path function 3.6 Perpetual motion machine 3.8, 4.4 of the first kind 3.8, 4.6 of the second kind 4.4, Plane of angle 1.2, 1.3 Pour point 10.10 Point function 1.13, 10.22, Polytropic process 3.15, 6.22 Potential energy 1.12 Power 5.1, 11.28 Pressure 1.4, 4.18 absolute 1.5 measurement 1.6 partial 2.8 reduced 7.12 Process adiabatic 1.13, 3.27, 4.8, constant volume 1.9, 3.10, 6.18 constant enthalpy 6.28, 8.14 constant temperature 3.12, 3.27, 4.20, 6.24 constant pressure 1.10, 2.6, 2.7, 3.11, 3.27 irreversible 4.7, 8.12

Index isentropic 3.10, isothermal 4.20, 6.21, 6.26, 11.5 quasi static 3.5 Producer gas 10.8 Product 1.16, 2.4, 7.9 Property 1.8, 3.3, 3.4, 4.16, 6.7, 6.28, 8.1 intensive 3.3, extensive 1.7, 2.13, 3.3 thermodynamic 1.13, 3.3, 8.1 Proximate analysis 10.2, 9.5, Psychrometer 9.5, 9.6 sling 9.6 Psychrometric chart 9.8, 9.9 Psychrometric processes 9.9 Pure substances 6.9, 6.17 Quasi static process 3.5 Reciprocating compressor 3.24 Redlich-Kwong equation 7.8 Reduced properties 7.13 Reactant 10.16, 10.21, 10.22 Reference state 10.21 Reformulated gasoline 10.3, 10.5 Refrigeration 4.2 Relative humidity 9.4, 9.8 Reversibility 4.7, 5.1 Reversible process 3.5, 4.7 Saturated air 9.7 Saturation pressure 6.4, 6.6, 9.3 Saturation temperature 6.4, 6.10 Second law efficiency Second law of thermodynamics 1.13, 4.1, 4.4, 4.11, 5.1 Sensible cooling 9.10, 9.23 Sensible heating 9.11 Sensible heat factor 9.15, 9.26 Separating and throttling calorimeter Shaft work 1.15, 1.16 Sink 4.1, 4.4, 4.6, 4.9, 11.1 SI units 1.1, 1.2 Solid angle 1.2, 1.3

I.5

Index Source 4.1, 4.4, 4.6 Spark ignition engine 11.17 Specific enthalpy 6.2, 6.9, 6.11, 6.17, 6.28, 8.1, 9.6 entropy 6.8, 6.12, 6.13, 8.1 heat 2.5, 2.6, 6.3 volume 3.3, 6.2, 4.21, 6.7 Specific heat 2.5, 2.6, 2.7 at constant volume 2.6 at constant pressure 2.6 Specific humidity 9.3, 9.11 State 2.1, 2.2, 3.1, 3.3 Steady flow process 3.20, 3.26, 5.4, 6.28 Steady flow energy equation 3.18, 3.20 Steady state 10.12 Steam 1.1, 3.20, 3.20, 3.25 Steam table 6.7, 6.10, Steradian 1.3 Stirling cycle 11.5, 11.6 Stoichiometric air 10.15, 10.16 Super heat 6.2 System isolated 3.2 open 3.2, 3.18 Tds equation 8.5, 8.7 Temperature absolute 1.8, 1.9, 2.2, 2.5, 4.13, 5.1, 6.12 adiabatic saturation 9.7, 9.8 celsius 1.9 critical 6.4, 7.5, 7.6 dew point 9.1, 9.2, 9.6, 9.1 kelvin 1.8 measurement 1.9 Thermodynamic scale 11.5 wet bulb 9.2 Thermal equilibrium 3.5 Thermodynamic equilibrium 1.15, 3.5

Thermometer 1.8, 1.10 Theoretical air 10.15 Throttling process 6.27, 8.12 Ton of refrigeraion 4.2, 4.3 Torque 1.16 Triple point 1.8, 6.3 Turbine 3.23, 3.25 U-tube manometer 1.6 Ultimate analysis 10.3, 10.8 Universal gas constant 2.1 Unit 1.1 basic 1.1, 1.2 derived 1.2 supplementary 1.4, 5.2 Useful work 1.1 Vacuum 1.2, 1.5 Vapour pressure 8.9, 8.10 Virial expansion Viscosity 10.9, Volume 1.9, 1.10, 2.1, 2.3, 2.9 critical 6.4, 7.6 specific 3.3, 6.2, 4.21, 6.7 Volatile matter 10.2, 10.6 Volume expansivity 8.19, Water 1.2, 1.6, 1.13, 3.20, 6.1, 6.5 Watt Water vapour 6.2, 9.1, 9.3, 9.5 Wet bulb temperature 9.2, 9.5 Wet steam 6.1, 6.9, 6.14 Work transfer 3.6, 3.26, 4.1 Wohl’s equation 7.12, Wood 10.2, 10.5 Zeroth law 3.5