Thermal and Hydraulic Machines [2 ed.] 9788120344730

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Second Edition

Thermal and Hydraulic Machines

G.S. Sawhney

Thermal and Hydraulic Machines SECOND EDITION

G.S. Sawhney Professor and Head Mechanical Engineering Department Greater Noida Institute of Technology Greater Noida

New Delhi - 110001

2011

THERMAL AND HYDRAULIC MACHINES, Second Edition G.S. Sawhney © 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-4473-0 The export rights of this book are vested solely with the publisher. Second Printing (Second Edition)

...

...

...

November, 2011

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Mohan Makhijani at Rekha Printers Private Limited, New Delhi-110020.

Contents

Preface ......................................................................................................................................... xi Preface to the First Edition ...................................................................................................... xiii 1. BASIC CONCEPTS AND ZEROTH LAW OF THERMODYNAMICS ........................ 1–42 Introduction 1 Definitions 1 C1ncept of Perfect Gas 7 Specific Heat 8 Energy 9 Zeroth Law of Thermodynamics 10 Temperature 10 Pressure 11 Diagrams 13 Solved Problems 15 Objective Type Questions 28 State True or False 28 Multiple Choice Questions 30 Fill in the Blanks 38 2. FIRST LAW OF THERMODYNAMICS ................................................................... 43–86 Introduction 43 First law of Thermodynamics 43 iii

iv

Contents

Application of First Law of Thermodynamics Flow Process 47 Enthalpy 47 Stored Energy 48 Steady Flow Energy Equation 48 Limitations of First Law of Thermodynamics Perpetual Motion Machine 54 Solved Problems 54 Objective Type Questions 76 State True or False 76 Multiple Choice Questions 78 Fill in the Blanks 82

44

54

3. SECOND LAW OF THERMODYNAMICS ............................................................. 87–144 Introduction 87 Heat Reservoir 87 Heat Engine 87 Heat Pump 89 Refrigerator 90 Statements for the Second Law of Thermodynamics 91 Carnot Cycle 92 Carnot Theorem 94 Thermodynamic Temperature Scale 95 Clausius Inequality 96 Entropy and Available Energy 98 Solved Problems 103 Objective Type Questions 124 State True or False 124 Multiple Choice Questions 128 Fill in the Blanks 134 4. PROPERTIES OF STEAM AND THERMODYNAMICS ..................................... 145–173 Introduction 145 Definition 145 Properties of Steam 148 Steam Tables and Mollier Diagram 149 Dryness Factor Measurement 150 Solved Problems 152 Objective Type Questions 163 State True or False 163 Multiple Choice Questions 165 Fill in the Blanks 169

Contents

v

5. VAPOUR CYCLES ................................................................................................... 174–193 Introduction 174 Carnot Vapour Cycle 175 Rankine Cycle 176 Solved Problems 179 Objective Type Questions 187 State True or False 187 Multiple Choice Questions 189 Fill in the Blanks 190 6. THERMODYNAMIC CYCLES .............................................................................. 194–240 Introduction 194 Otto and Diesel Cycles 196 Engines 202 Indicated, Brake and Friction Power 207 Efficiencies 208 Solved Problems 208 Objective Type Questions 228 State True or False 228 Multiple Choice Questions 230 Fill in the Blanks 235 7. STEAM TURBINE .................................................................................................... 241–297 Introduction 241 Principle of Operation 241 Types of Steam Turbines 243 Advantages of Steam Turbine 245 Classification of Steam Turbines 245 Working of Steam Turbines 246 Differences between Impulse and Reaction Turbines 248 Compounding of Steam Turbines 248 Velocity Diagrams 250 Work, Power and Efficiencies 253 Definitions: Blade Velocity Coefficient, Blade Speed Ratio and Carry over Coefficient 254 Maximum Blade Efficiency 255 Maximum Work Done 257 Velocity Diagram for Parson’s Reaction Turbine 260 Reaction Turbine: Degree of Reaction 262 Work Done, Efficiency and Maximum Efficiency of Parson’s Reaction Turbine 265 Height of the Blade in Turbine 267 Stage Efficiency, Turbine Efficiency and Reheat Factor 268 Losses in Steam Turbine 269

vi

Contents

Governing of Steam Turbines 270 Important Formulae 273 Solved Problems 274 Objective Type Questions 292 State True or False 292 Multiple Choice Questions 294 Fill in the Blanks 295 8. GAS TURBINE ......................................................................................................... 298–346 Introduction 298 Gas Turbine Cycles 299 Comparison of Gas Turbine and IC Engine 300 Comparison of Gas Turbine and Steam Turbine 301 Brayton Cycle 301 Thermal or Air Standard Efficiency of Brayton Cycle 303 Work Ratio 306 Optimum Pressure Ratio 307 Intermediate Temperature 308 Maximum Work 309 Actual Gas Turbine cycle 310 Optimum Pressure Ratio for Maximum Cycle Thermal Efficiency 312 Methods to Improve Thermal Efficiency 313 Actual Brayton Cycle with Regeneration 316 Gas Turbine Cycle with Intercooling 318 Gas Turbine Cycle with Reheating 320 Important Formulae 322 Solved Problems 323 Objective Type Questions 343 State True or False 343 Multiple Choice Questions 344 Fill in the Blanks 345 9. COMPRESSOR ........................................................................................................ 347–401 Introduction 347 Uses of Compressed Air 348 Types of Compressors 348 Reciprocating Air Compressor 349 Clearance Volume 353 Volumetric Efficiency 354 Work Done: Air Compressor Cycle with Clearance 356 Effect of Increased Compression Ratio on Air Compressor Cycle 357 Actual Indicator Diagram for Reciprocating Compressor 357 Multistage Compressors 358 Two-stage Compression 360

vii

Contents

Three-stage Compression 360 Minimum Work of Compression 361 Pressure Ratio in Multistage Compression 362 Capacity Control of Compressor 362 Classification of Rotary Compressors 363 Comparison of Rotary Compressors with Reciprocating Compressors Root Blower 364 Vane-Type Blower 365 Screw Compressor 366 Centrifugal Compressor 366 Axial Flow Compressor 368 Comparison of Centrifugal and Axial Flow Compressors 368 Definitions 369 Input Work and Efficiencies in Rotary Compressor Cycle 370 Analysis of Centrifugal Compressor 373 Vane Shape and Characteristics 376 Slip Factor and Work Factor 378 Characteristic Curve, Surging and Choking 378 Surging and Choking 378 Stalling 380 Important Formulae 380 Solved Problems 382 Objective Type Questions 398 State True or False 398 Multiple Choice Questions 399 Fill in the Blanks 400

364

10. IMPACT OF JET ...................................................................................................... 402–436 Introduction 402 Force Exerted on a Stationary Flat Vane held Normal to Jet 403 Force Exerted on a Stationary Vane Held Inclined to the Jet 404 Force Exerted on Stationary Curved Plate 405 Force Exerted by a Jet on a Hinged Plate 407 Force Exerted by the Jet on Moving Flat Plate Held Normal to Jet 408 Force Exerted on a Moving Plate Inclined to the Jet 409 Force Exerted on a Moving Curved Plate Striking at Centre in Jet Direction 410 Jet Striking a Moving Curved Plate Tangentially at One End of the Plate 413 Important Results 417 Solved Problems 417 Objective Type Questions 432 State True or False 432 Multiple Choice Questions 433 Fill in Blanks 434

viii

Contents

11. HYDRAULIC TURBINES ........................................................................................ 437–499 Introduction 437 Classification of Hydraulic Turbines 437 Impulse Turbines 439 Work Done and Efficiency of a Pelton Wheel 442 Heads and Efficiencies of Pelton Wheel 445 Design Aspects of Pelton Wheel 448 Reaction Turbines 452 Francis Turbine 452 Inward and Outward Flow Reaction Turbines 455 Velocity Diagrams and Work Done by Water on the Runner of Reaction Turbines 455 Important Ratios of a Francis Turbine 458 Designing of Francis Turbine Runner 459 Comparison of Francis and Pelton Turbines 461 Degree of Reaction 463 Kaplan Turbine: Axial Flow Reaction Turbine 464 Flow and Work Done in Kaplan Turbine 466 Performance of Water Turbines 468 Characteristics Curves 471 Important Results 472 Solved Problems 473 Objective Type Questions 496 State True or False 496 Multiple Choice Questions 497 Fill in the Blanks 498 12. CENTRIFUGAL PUMPS ......................................................................................... 500–536 Introduction 500 Comparison with Reciprocating Pump 500 Main Parts of a Centrifugal Pump 501 Work Done 502 Heads and Efficiencies 504 Working Proportions 506 Multistage Centrifugal Pumps 507 Minimum Speed for Starting a Pump 508 Specific Speed of a Centrifugal Pump 509 Model Testing 510 Priming of a Centrifugal Pump 512 Main Characteristic Curves 512 Operating Characteristic Curves 513 Cavitation 513 Suction Height 514 Net Positive Suction Head 515

Contents

ix

Air Lift Pump 516 Jet Pump 516 Important Formulae 517 Solved Problems 519 Objective Type Questions 533 State True or False 533 Multiple Choise Questions 533 Fill in the Blanks 535 13. RECIPROCATING PUMPS ..................................................................................... 537–572 Introduction 537 Classification of Reciprocating Pumps 538 Single Acting Reciprocating Pump 538 Double Acting Reciprocating Pump 539 Discharge, Work and Power Input 540 Coefficient of Discharge and Slip 542 Effect of Acceleration of the Piston 542 Variation of Velocity and Friction 545 Indicator Diagram 546 Effect of Acceleration on Indicator Diagram 547 Effect of Friction on Indicator Diagram 547 Effect of Acceleration and Friction on Indicator Diagram 548 Separation of Flow of Water 549 Air Vessels 550 Friction Work Saved with Air Vessel 551 Important Formulae 555 Solved Problems 555 Objective Type Questions 569 State True or False 569 Multiple Choice Questions 570 Fill in the Blanks 572 Bibliography ............................................................................................................................. 573 Index ................................................................................................................................. 575–578

Preface

After receiving an overwhelming response from the readers, the book has been revised and enlarged in its second edition. The additional text has been introduced to the book, based on the syllabus requirement and examination pattern of the technical universities. I would like to thank my readers for their sincere and encouraging feedback and would request them for sending constructive suggestions and criticism, if any, for the further improvement of the text. Feedback can be emailed to [email protected].

GS. Sawhney

xi

Preface to the First Edition

Thermal and hydraulic machines are used vastly for the generation of electricity in thermal and hydraulic power plants. It is, therefore, essential that the engineering students of electrical, mechanical and other similar branches must have sound knowledge about these machines. This book is designed to meet the needs of such a course to provide a fundamental understanding of the principles of working and construction details of these machines. Though chiefly based on the latest syllabus (EME 309) ofUttar Pradesh Technical University, the effort has been made to cover the syllabi of several other universities as well. Based on my experience of teaching, I have endeavoured to present the systematic explanation of the basic concepts of the subject matter. A large number of solved problems and objective type questions with explanatory answers have been included to enhance the understanding of the underlying principles of theory. I wish to record my sincere thanks to my wife, Jasbeer Kaur for her patience shown throughout the preparation of the book. I am also thankful to my children Jasdev, Tejmohan, Pooja and Nandini for continuous encouragement extended to me to complete the book. I am also thankful to Dr. S. Prasad, Mr. K.L. Gupta, Mr. Bishnu Gupta, Prof. S.N. Sharan and Mr. Arvind Tiwari of GNIT, Greater Noida. I would appreciate students and teachers to send constructive suggestions and criticism with a view to enhance further the usefulness of the book. They can do so by emailing at [email protected] GS. Sawhney xiii

Basic Concepts and Zeroth Law of Thermodynamics

INTRODUCTION When a hot body is placed in contact with a cold body, the hot body cools down while the cold body warms up. The energy transferred from the hot body to the cold body as a result of temperature difference is called heat energy. The heat energy transferred is zero if both the bodies have the same temperature. This is the basis of zeroth law of thermodynamics. Heat is a transitory energy. It must not be confused with intrinsic (internal) energy possessed by a system. Similarly, work energy is also a transitory energy. Whenever heat or work transits a system, the state of the system changes. It is also observed that heat and work are two mutually convertible forms of energy. This is the basis of the first law of thermodynamics. It is also observed that heat never flows unaided from a hot body to a cold body. This is the basis of second law of thermodynamics. Applied thermodynamics is the science of the relationship between heat, work and the properties of a system. It is concerned with the means necessary to convert heat energy from available sources such as fossil fuel into work energy. The application of thermodynamics is extremely wide. Its principles are used in designing of energy converting devices. These devices will be discussed in later chapters but first some fundamental definitions must be made.

DEFINITIONS Thermodynamics: Thermodynamics is a branch of science dealing with energy and its transformation, specially transformation of heat into other forms of energy and vice versa. 1

-

Thermal and Hydraulic Machines

System: A system is any matter in space which is under analysis. The system may have a real or hypothetical boundary. Everything outside of a system is termed as surroundings. The boundary can be adiabatic which does not allow heat interaction between the system and surroundings or diathermic which allows the heat interaction between the system and surroundings. Universe = System + Surroundings Systems can be of three types as shown in Figure 1.1. An open system allows mass and energy interaction with the surrounding. A closed system allows only energy interaction. An isolated system allows neither mass nor energy transaction. Turbines, compressors and pumps are examples of the open system. Piston-cylinder assembly is an example of the closed system. Our universe is an example of an isolated system. ~ass ( ¥nergy Energy ~ Closed system Mass interaction >" 0 Energy interaction >" 0

Mass interaction = 0 Energy interaction >" O

FIGURE 1.1

Isolated system Mass interaction = 0 Energy interaction = 0

Types of systems.

The working fluid in a piston-cylinder assembly forms a closed system as shown in Figure 1.2. The mass of the closed system remains constant. The volume of the closed system need not to remain constant. The volume changes as the piston moves up and down in the cylinder. Only energy transfer (work) between the closed system and surroundings takes place due to the movement of the boundary of the system considered.

Cylinder i-======"'c:,:t--t--- Piston Surrounding System Energy

FIGURE 1.2

A closed system.

In the case of an open system, mass transfer also takes place along with energy transfer between the system and the surroundings. The boundary of the open system is known as control volume. The boundary of the system during the transfer of mass and energy may or may not change. The open system can be one-flow or two-flow boundary system. Air leaving a compressed air cylinder [Figure 1.3(a)] can be considered a one-flow boundary system. Here, the boundary is not changed during the mass transfer (compressed air). However, if we consider air escaping from a balloon [Figure 1.3(b)] to the surroundings, the boundary shape also changes. This is an example of moving boundary one-flow open

Basic Concepts and Zeroth Law of Thermodynamics

-

system. In the case of an air compressor as shown in Figure 1.3(c ), air is taken from the surroundings and it is sent out of the system after compression. Hence mass and energy transfer is taken place across the boundary. It is an example of two-flow open system.

Boundary (moving)

i--- Boundary

Compressed : air cylinder

(constant) Energy

++-+-- - Energy

_____________________ ~---~,

(a) Constant boundary-One-flow open system

(b) Moving boundary-One-flow open system

High pressure air

Air compressor

(c) Two-flow open system FIGURE 1.3

Open system.

A system can be considered as a closed system while its sub-systems may form individually open systems. The reason is that the subsystems are inter-connected. A thermal power plant consisting of a boiler, a turbine, a condensor and a pump as shown in Figure 1.4, can be considered to be a closed system. Heat transfer (through the boiler and the condensor) and work transfer (through the turbine and the pump) take place between the system (power plant) and the surroundings through the boundary. No mass transfer takes place through this boundary. However, every subsystem (boiler, turbine, pump and condensor) is an open system as there is a mass and energy transfer across their boundaries.

A state: A state of a system indicates the specific condition of the system. Properties like pressure, temperature, volume, etc. can define a state of a system. Process: A system undergoes a change due to energy and mass interaction. The mode of the change of a system is called process. It may be constant pressure (isobaric) or constant volume (isochoric) or constant temperature (isothermal), adiabatic or isoentropic. Path: A path is the locii of various intermediate states passed through by the system during a process ..

-

Thermal and Hydraulic Machines Heat input

,---------

Pump work

r-- ---~ :

~

-~,-'CJ

r

I

I I I

I I I

-+-: I Turbine !... _ _ _ _ _ _ _ _ _ _ _ _1 I work I

I

: Pump :_ - - - - - ~

-~,

:

r

I I I I I

I

: ~---~: I

I

: .........,,_ _ _~:

__________ ,__ s;_oa!"''"---' ___________ : Heat rejection

FIGURE 1.4

Thermal power plant (closed system).

State and path functions: The state function is independent of the path while the path function is dependent on the path. To understand this, consider a person travelling from a point x to a point y. There can be various routes to reach a point y from a point x, but the travelling distances would be different for different routes. If a car is used, work done and fuel consumed are dependent on the route and the mileage of the car. However, the locations of points x and y are fixed and they are independent of the routes. Hence, the positions x and y are point or state functions while fuel consumed and work done are path functions. State functions are represented by a point on a graph while path functions by an area. Also

f (state

f

function) = 0 while (path function) -: ;:. 0. Mathematically, a state function is an exact differential while a path function is an inexact differential. If z = f(x, y), then z is an exact differential for dz

=(

!~ }x !; ) +(

dy

and

if z

--

d)'dX

if z

--

dXd)'

Now let us find out whether v = RTIP is a state function where, v = specific volume, T = temperature, P = pressure and R = gas constant. Differentiating w.r.t. T, we get

dV

-

c)T

R -

P

Basic Concepts and Zeroth Law of Thermodynamics

and

-

R

p2

Similarly, differentiating w.r.t. P, we get

av aP

RT

-=--

P2

and

Hence --=--

therefore v is a state function.

Cycle: A cycle is a sequence of processes undergone by the system so that the initial and final states are the same. Thermodynamics properties remain unchanged on the completion of a cycle. Thermodynamic equilibrium: It is a state of a system when its state does not change and its properties remain constant. A system is said to be in thermodynamic equilibrium if it is in state of mechanical, thermal or chemical equilibrium. Equilibrium is a state wherein there is no tendency for a change. In other words, the rate of the process is zero, where the rate is the ratio of driving force to resistance. Hence, the rate can be zero when either driving force is zero or resistance is infinitely large. Equilibrium is true when driving force is zero. It is false when resistance is infinitely large. We are interested only in true equilibrium when driving force is zero. What is a mechanical equilibrium? In a piston and cylinder arrangement, the piston stops moving when the pressure inside the cylinder and surroundings is the same. This is called mechanical equilibrium. What is thermal equilibrium? When two bodies in contact attain the same temperature and heat stops transferring, it is said to be in thermal equilibrium. What is chemical equilibrium? Once chemical potential of two phases are equal, the net rate of the mass transfer between phases is zero, and thus it is called chemical equilibrium. Reversible process: In a reversible process, the system can come back to the original state on removal of the factors responsible for the occurrence of the process. On reversal of the process, no trace of occurrence is left and the system follows the same path back to the initial state. It is a frictionless process. Irreversible process: In an irreversible process, the system cannot come back to the original state on removal of the factors responsible for the occurrence of the process. Friction and dissipative effects are the causes for irreversibility.

-

Thermal and Hydraulic Machines

Quasi static process: It is not possible for a system to attain equilibrium in finite time. For the sake of study and analysis, certain assumptions can make a system akin to a system in equilibrium. When a system changes its state very slowly under the influence of very small differences of temperature and pressure, the process is called quasi static process. Quasi means 'almost' and static means 'non-dynamic'. Hence, almost non-dynamic process is a quasi static process. A reversible process is always a quasi static process.

System analysis: Macroscopic (visible to the naked eye) and microscopic (minute) approaches are the two approaches for analyzing systems. In the macroscopic approach, the structure of matter is not considered. A complete system is considered and the state is found out with measurable properties. Classical thermodynamics adopts the macroscopic approach. In the microscopic approach, the constituents and microsystem of the system are analyzed. Statistical thermodynamic adopts the microscopic approach for analysis. The result of macroscopic approach analysis is equal to the summation of microscopic approach analysis. Concept of continuum: A substance is composed of a vast number of molecules. Most engineering systems are concerned with the macroscopic or bulk behaviour of a substance rather than the microscopic or molecular behaviour. In most cases, it is convenient to think of a substance as a continuous distribution of medium or a continuum. However, there are certain instances, for example, (rocket explosion at very low pressure) in which the concept of a continuum is not valid. In the concept of continuum, the substance is considered free from any kind of discontinuity. As the scale of analysis is large, the discontinuity of the order of intermolecular spacing or the free mean path is negligible. State: Matter is found in three states-solid, liquid and gas. Gases have some special properties like low density, no definite volume and easy compressibility. Properties: Properties are observable characteristics of a system like pressure, temperature and volume. Intensive properties such as pressure and temperature are independent of mass. Extensive properties such as volume, entropy and enthalpy depend upon mass. Boyle's law: Boyle's law gives a relation between pressure and volume of a gas. For a given mass of gas at constant temperature (T), volume (V) is inversibly proportional to pressure (P): 1 V oc p Charles' law: Charles' law gives a relation between volume and temperature. For a given mass of a gas at constant pressure, volume is directly proportional to absolute temperature: V

oc

T

where T is in kelvin

Law of pressure: The law of pressure gives a relation between pressure and temperature. For a given mass of a gas at constant volume, pressure is directly proportional to absolute temperature: P oc T where T is in kelvin

Basic Concepts and Zeroth Law of Thermodynamics

CONCEPT OF PERFECT GAS

-

The perfect gas equation states that volume, pressure and temperature of a given mass of a gas are interconnected as

Pv=RT where R

= universal gas constant =

8314 J/lanol K, here kmol

v = volume

=

1000 mol

per mol

The gas equation can also be written as

PV = mRT where m = mass (=nM) V = volume (=nv) R = gas constant (= R /M) n = number of moles (1 mole = 6.023 x 1023 molecules) M = molecular weight. Another form of the gas equation is Pv

= RT

where v is specific volume (Vim). As per the kinetic theory, pressure and temperature of gas are proportional to "square of root mean square (rms) velocity (c)", i.e. p

2 = 31 pc-2 = 3

. energy) KE (k"met1c

where p is density. 1 Mc 2 T=---

3

R

If isothermal lines are drawn for Pv = RT, we get curves which have maxima and minima (Figure 1.5). However, at critical point (CP) and above it, the maxima and minima on isothermal curves do not exist. A gas can be liquefied below the critical temperature (Tc). Above the critical point a gas cannot be liquefied by pressure. If the critical temperature is above room temperature, a gas is easily liquefied and stored. A domestic gas cylinder contains 14 kg LPG which is mainly butane gas. Butane can be liquefied at high pressures as its critical temperature is above room temperature. In a liquefied state, a small cylinder can hold sufficient gas. However, hydrogen can be kept in the liquefied state below critical temperature (-240°C). It is the best fuel and it is used in the liquid state in rockets. It requires either very large storage capacity at room temperature or economical cooling facility to keep it below the critical temperature before it can be used for automobiles.

-

Thermal and Hydraulic Machines

p

Liquid vapour region 1~ - -+-- - - Liquid region V

FIGURE 1.5

PV diagram.

A real gas does not obey the ideal gas equation Pu = RT for all pressures and temperatures. The ideal gas equation is based on two assumptions that molecules do not exert intermolecular attraction and the volume occupied by the molecules is negligible, which is not fully correct at all pressures and temperatures. At a very low pressure (pressure tending to zero) and a high temperature (temperature tending to infinity), the real gas obeys very nearly the ideal gas equation. Another deviation from ideal gases is that most of the real gases get liquefied at low temperatures.

Compressibility factor: The compressibility factor (z) is a measure of deviation of real gas from ideal gas behaviour:

Pu

Compressibility factor (z) = RT The value of z is one for ideal gas. For real gases, z can be read from P and z charts for different isothermal lines. Pu v P Actual volume Actual pressure - - - - - - - - - - - - - - -----"---RT RT IP RT IP Ideal volume Ideal pressure

z- -

If attractive forces predominate, z is less than unity (P actual < Pictea1). When repulsive forces predominate (high pressure), z is greater than unity. As a real gas deviates from the perfect gas equation, the van der Waals equation overcomes this problem. The equation is as follows:

where

a b

= =

constant to take care of attraction amongst the molecules constant to take care of volume of molecules

SPECIFIC HEAT Specific heat is the heat required to raise the temperature of a unit mass by unity. Mathematically, Q(heat)

=

c(specific heat) x m(mass) x flT

-

Basic Concepts and Zeroth Law of Thermodynamics

or c = -Q- mx!iT If m = 1 and !iT = 1, then Q = c If 1 kg of gas is compressed and its temperature is raised by 1°C then 0

c=---=0 mx!iT Similarly, if a compressed gas is allowed to expand and its temperature falls, then heat (Q) is given to maintain its original temperature. Now we have c

= _Q_ = infinity mxO

Hence, the specific heat of a gas can have value from zero to infinity. Hence, we define the specific heat of gas at constant volume and at constant pressure (cv and cp). cP (specific heat at constant pressure) is greater than cv (specific heat at constant volume). Heat at constant volume is fully used for heating gas by 1°C while heat at constant pressure is used both for heating gas and doing work (P x dV) as gas expands against atmospheric pressure while heating. We can write cpdT

= cvdT + PdV

As PV = RT, PdV = RdT at constant pressure. Therefore, the above equation becomes cPdT

or

cp -

Cv

= cvdT + RdT =R

ENERGY Work (W) and heat (Q) are energies and equivalent to each other by the relation W=JQ

where J is joule's constant. J = 1 if W and Q have the same unit otherwise J = 4.2 J/cal when W is in joules and Q is in calories. Work done by a system depends not only upon initial and final states but also upon the path adopted by the process. The area on a PV diagram under the process

J

2 1

P dV is

work. Similarly, f PdV gives work on a PV diagram for a cyclic process. During the free expansion, work done by a gas is zero as expansion does not take place against atmospheric pressure.

-

Thermal and Hydraulic Machines

The energy can be defined as the capacity to do work. Energy possessed by a system which can cross its boundary is called transit energy. Energy possessed by a system within its boundary is called stored energy. Potential energy (PE), kinetic energy (KE) and internal energy ( U) are stored energies. Total stored energy is the sum of PE, KE and U. Heat energy is transit energy as it can go out and come in the system depending on temperature. Work is also transit energy which can go out or come in the system due to difference in any intensive property other than temperature. The heat and work are not system properties but appear as a boundary phenomenon, i.e. they are observed at the boundary of the systems. Both are path functions and hence they are inexact differentials. The magnitude of heat transfer and work transfer depends upon the path followed by the systems during the process. Work is the area under the process or area enclosed in the cyclic process on a PV diagram. Similarly, heat is the area under the process or area enclosed in the cyclic process on a TS (temperature entropy) diagram.

ZEROTH LAW OF THERMODYNAMICS If two bodies A and B are in thermal equilibrium with a third body C separately, then two bodies A and B shall also be in thermal equilibrium with each other (Figure 1.6).

's' 0 ____,. \V ~ ~ © By Zeroth law

-c----

Thermal equilibrium

FIGURE 1.6

Thermal equilibrium

Zeroth law.

Temperature Temperature is an intensive property of a system and it requires reference states for calibration. The boiling point and the freezing point of water are acceptable reference states. The thermometer is a temperature measurement system which can show some change in its characteristic (termed thermometric property) due to heat interaction taking place with the body whose temperature is being measured. Centigrade (Tc), Fahrenheit (T1), and kelvin (Tk) temperatures are interrelated as follows:

Tc _ T1 - 32 = Tk - 273.15 100 180 100 Temperature in Rankine is equal to temperature in Fahrenheit plus 459.67. In the mercury scale thermometer, the length of the mercury column (l) is proportional to temperature (T) and temperature is given by T = lO0(lT - lo) 1100 - lo

Basic Concepts and Zeroth Law of Thermodynamics

where

IT= length at T

-

10 = length at freezing point l 100 = length at boiling point Similarly, in the constant volume thermometer, pressure varies with temperature and temperature is given by T = lO0(PT - P0 )

Piao - Po where PT = pressure at T

P O = pressure at freezing point P 100 = pressure at boiling point

Similarly, in the resistance thermometer, resistance (R) varies with temperature and temperature is given by T = lO0(RT -

Rwo -

Ra) Ra

where RT = resistance at T R0 = resistance at freezing point R 100 = resistance at boiling point

In thermocouple, emf ( electromotive force) induced is proportional to temperature difference between hot and cold junctions. Bismuth-antimony and copper-iron are a common pair of metals to form junctions. Therefore emf= aT + bT2 where a and b are constants. Very high temperatures are measured by pyrometers. The pyrometer can be total radiation pyrometer or disappearing filament pyrometer. The radiated energy is proportional to the fourth power of temperature of a hot body.

Pressure The standard atmospheric pressure is defined as the pressure produced by a column of mercury 760 mm high. Patm

= pgh =

(13.6

X

10 3)

X

9.8

X

760 1000

= 1.01 bar (bar = 10 5 pascal)

111111

Thermal and Hydraulic Machines

Pressure of the system can be higher or lower than atmospheric pressure . Pressure is measured by the manometer. A manometer is a U-tube containing mercury with one end opens to atmosphere and the other one is connected to a system/vessel (Figure 1.7). If pressure in a vessel (Pabs) is higher than Paim, mercury is forced up the limb that opens to atmosphere. If pressure in a vessel (Pabs) is lower than atmospheric pressure, mercury is forced into the limb connected to the vessel. Higher than atmospheric pressure is known as gauge pressure while lower pressure than atmospheric pressure is called vacuum pressure.

T

hv

............ ..t

Pabs = Patm + gauge pressure

= Patm + Pm X hg x g h 9 = gauge pressure height

FIGURE 1.7

Pabs = Patm -

vacuum pressure = Patm - Pm X hu X g h0 = vacuum pressure height

A U-tube manometer.

Note: NTP is normal temperature and pressure. Normal temperature is 0°C and normal pressure is 760 mm of Hg. STP is standard temperature and pressure. Standard temperature is taken as 15°C or 25°C depending upon the geographical location and standard pressure is 760 mm of Hg. Water is about 1300 times heavier than air. Pressure rises swiftly above the atmospheric pressure as we descend inside water. The pressure increases by one atmosphere for every 10 m of depth inside water. On land, if we climb to a height of 150 m the change in pressure will be slight and indiscernible. At the same depth under water, our blood vessels will collapse and our lungs will be compressed dangerously. It is impossible to go beyond a depth of about 72 m without assistance of diving suits connected to an air pump. The average depth of oceans is about 4 km. The pressure at this depth is equivalent to about 400 atm which is sufficient to crush anything. A diver (Figure 1.8) inside water experiences the same pressure as the surrounding water. We are made largely of water which is almost incompressible. However, it is gases inside our body, particularly inside our lungs that create problem while going down in water. The gases inside the body of the diver compress when he descends in water and the compression becomes fatal at some point. To avoid this, diving suits are used which are connected to an air pump at the surface by a long hose to develop suitable air pressure inside the body of the diver. The 'squeeze' occurs when the air pump at the surface fails which results into the loss of pressure in the suit. The air leaves the suit with such a force that the hopless diver dies instantly.

111111

Basic Concepts and Zeroth Law of Thermodynamics

FIGURE 1.8

Diver.

We breathe air which contains 80% nitrogen. Under pressure (in deep water), nitrogen gas gets dissolved in our blood. If pressure is changed too rapidly when we ascend from deep water the nitrogen dissolved in the blood begins to liberate in the same manner of a freshly opened bottle of Coca Cola. The bubbles of nitrogen clog the blood vessels and the flow of the blood stops. The stoppage results into the deprivation of oxygen to the tissues of our body. The deprivation causes pain so excruciating that we are prone to bend double in agony. In diving, this condition is called bends. The bends are the occupational hazards for pearl divers and caisson workers (men working in enclosed dry chambers built on river beds).

DIAGRAMS Processes give different curves on PV diagrams, PT diagrams, TS diagrams and hS diagrams ( enthalpy-entropy). On a PV diagram, the isobaric process (P = constant) is shown as a straight horizontal line while the isochoric process ( V = constant) is shown as a vertical line. However, the isothermal process (T = constant) is depicted as a hyperbolic curve. Refer to Figure 1.9. V= constant

p

p

p P= constant

V

V

(a) Isobaric process

(b) lsochoric process

FIGURE 1.9

V

(c) Isothermal process

PV diagrams.

On a VT diagram, the isochoric process appears as a horizontal line and the isothermal process appears as a vertical line. However, the isobaric process is shown as an inclined line. Refer to Figure 1.10.

-

Thermal and Hydraulic Machines

T= constant

V

V

V= constant

T

T

T

(a) lsochoric process

(b) Isothermal process FIGURE 1.10

(c) Isobaric process

VT diagrams.

On a PT diagram, the isobaric process is shown as a horizontal line, the isothermal process appears as a vertical line, and the isochoric process as an inclined line (Figure 1.11 ).

T= constant

p

p

p

P= constant

T

T

T

(a) Isobaric process

(b) Isothermal process FIGURE 1.11

(c) lsochoric process

PT diagrams.

On a TS diagram (Figure 1.12), the isothermal process is shown as a horizontal line, and the isentropic process as a vertical line. However, the isobaric process is shown as a curve with slope = TlcP and the isochoric process is shown as a curve with slope = Tieu. Since cP > cv, the slope of an isochoric process is greater than the slope of isobaric process on the T-S diagram.

T

T = constant

T

s

s (a) Isothermal process

(b) lsentropic process FIGURE 1.12

(c) Isobaric process

(d) lsochoric process

TS diagrams.

On a PV diagram, the isothermal process and the adiabatic process are curves as shown in Figure 1. 13. The slope of an isothermal process is smaller than that of an adiabatic process.

-

Basic Concepts and Zeroth Law of Thermodynamics T1 T2

p

p

~ ' T•

~ • ooosta"I

'°"stam

V

V

(a) Isothermal process FIGURE 1.13

(b) Adiabatic process PV diagrams.

On a enthalpy-entropy (hS) diagram, the isentropic process appears as a vertical line while the constant enthalpy process like throttling as a horizontal line. Isobaric, isochoric and isothermal processes are also shown in Figure 1.14. The slope of an isochoric curve is greater than that of an isobaric curve.

S= constant h

h

h

= constant

s

h

s

s

(a) Constant enthalpy process

(b) lsentropic process

J)v,

qT T3> T2> T1

V2> V1

h

(c) Isobaric process

h

s (d) lsochoric process FIGURE 1.14

s (e) Isothermal process

hS diagrams.

SOLVED PROBLEMS 1. Find the pressure difference shown by a manometer, showing the difference of a mercury column of 100 mm of Hg in the limbs. Take PHg = 13.5 x 10 3 kg/m 3 and g = 10 m/s 2 •

Pressure difference

= PHg

x g x h

..

Thermal and Hydraulic Machines

13.5

X

103

X

10

lOO

X

1000

13.5 kPa

2. A tank is filled with water. Find the gauge pressure at a depth of 2 m from the top. Take g = 10 m/s 2 and Pwater = 1000 kg/m 3 •

Gauge pressure =

Pwater

x g x depth

= 1000 X 10 = 20 kPa

X 2

3. Find the absolute pressure of gas if a manometer reads a gauge pressure of 50 kPa and atmospheric pressure is 100 kPa.

Absolute pressure = atmospheric pressure + gauge pressure 100 + 50 150 kPa

4. Find the absolute pressure of gas if vacuum pressure is 60 kPa and atmospheric pressure is 100 kPa.

Absolute pressure = atmospheric pressure - vacuum pressure

= 100 - 60 = 40 kPa 5. A container has an absolute pressure of 40 kPa. The area of the lid is 1000 mm 2 • What force is required for opening the lid? The container has lesser pressure than atmospheric pressure. The force has to be applied to open the lid which is the pressure difference on the lid multiplied by the area of the lid.

Force = pressure difference x area

=

(Patm -

pcontainer) X

1000

3

= (100 - 40)

= 60

X

10

A

X

lO00

X

lO00

kN

6. A manometer shows a gauge pressure of 50 kPa of a gas when atmospheric pressure is 100 kPa. What will be the gauge pressure indicated by the manometer in space where residual atmospheric pressure is 50 kPa? Pabs

= Patm +

Pgauge

100 + 50 150 kPa

Basic Concepts and Zeroth Law of Thermodynamics

In space

= Pres atm + P gauge 150 = 50 + Pgauge P gauge = 100 kPa P abs

or

-

7. A mercury manometer shows a gauge pressure of 100 mm of Hg. In case it is replaced by a water manometer, what gauge pressure in mm of water will be shown by it? Take g = 10 m/s 2 • pgauge = PHg

g

X

13.7

X

10

13.6 kPa p gauge = Pwater X g 1

13.6 hwater

X

10 1000 13.6

10

X

100 kPa 1000

X

X

hwater

hwater 1000

hwater

X

X

hg

X

1000

10 1360 mm 1.360 m

8. Gas A and B have absolute pressures of 220 kPa and 100 kPa. What will be gauge pressure shown by the manometer that is put between gas A and B? Gas A

Gauge pressure = absolute pressure of gas A - absolute pressure of gas B

= 220 - 100 = 120 kPa 9. Gas A has a vacuum pressure of 20 kPa and gas B has a gauge pressure of 20 kPa. What will be gauge or vacuum pressure shown by the manometer.

Take Patm = 100 kPa. P, = Pabs of gas A = Patm - Pvacuum

=

100 - 20

= 80

kPa

P 2 = Pabs of gas B = Patm + Pgauge

=

100 + 20

=

120 kPa

-

Thermal and Hydraulic Machines

A

Gauge pressure of the manometer = P1

P2

-

= 80 - 120 = -40 kPa Therefore, the manometer will show a vacuum pressure of 40 kPa.

10. A balloon is immersed in the sea. How deep is it to be immersed so that the volume is reduced to half of its size? Take r water = 1000 N/m 2, g = 10 m/s 2 and P attn = 100 kPa. The volume of the balloon is full at atmospheric pressure, i.e. 100 kPa. The volume will reduce to half in case pressure becomes double, i.e. 200 kPa. The increase of pressure by 100 kPa (200 - 100 = 100 kPa) will be given by the depth of water.

p 100

X

g

= Pwater X

103 = 1000

10

X

depth

X

depth

X

Therefore, Depth

=

100 X 103 1000 x 10

=

10 m

11. A balloon bursts if its volume becomes double. Find the height it climbs before it bursts if atmospheric pressure reduces by 0.1 kPa/km climb? Take P atm at the earth surface = 100 kPa and temperature remains constant. The volume of the balloon will increase with climb as atmospheric pressure will reduce with climb. The volume will be double when atmospheric pressure will be half, i.e. when atmospheric pressure is 50 kPa.

Atmospheric pressure at any height = atmospheric pressure at the earth surface-fall of atmospheric pressure for height (h) Mathematically

Pres attn=

100 - 0.1

X

height (h)

=

100 - 0.1

X

h

50 or

h = 50

X

10 = 500 km

12. Water boils at a temperature of 100°C and a pressure of 100 kPa. What will be pressure inside a cooker in case water boils at 123 °C?

Basic Concepts and Zeroth Law of Thermodynamics

Substituting the values in the above equation, 100 P2 ----=-~~ or 100 + 273 123 + 273

=

100 373

or

..

P2

400

P-400x100 2 373

or

= 107.23 kPa 13. Find the work done for process a-b, c-d and e-f as indicated on the following PV diagram (Figure 1.15).

30

!f

20 + - - - + - - - - - , , < - - - + - - - - - I

10 e'

-

C'

~--+----+--+-----
m3 h3

I I I I I

I

Y2 _______________________ : Adiabatic mixing

FIGURE 2.12

Adiabatic mixing.

If fluid is the same, then T3 -_ m 1I; + m 2 T2

m3

Throttling:

Throttling (Figure 2.13) is a process in which a fluid passes through a restricted opening under isoenthalpic condition (Lih = 0). Pressure drop is achieved without work or heat interaction. Also KE and PE remain constant. Temperature may drop or increase during the throttling process. Throttling is used in the throttling calorimeter for measuring the dryness factor of wet stream. It is also used in the refrigeration cycle (window air conditioner and refrigerator) to throttle the high pressure liquid refrigerant to low pressure evaporator to extract heat.

r-----------------------1 I I

I ________________ _ L

Restricted opening

FIGURE 2.13

Throttling.

IIEII

Thermal and Hydraulic Machines

Applying the SFEE on the throttle, we get

h, = h2 where p 1 >> p 2 and Q

=

0, W

=

0.

LIMITATIONS OF FIRST LAW OF THERMODYNAMICS The first law of thermodynamics has the following limitations: 1. The first law of thermodynamics does not differentiate between work and heat. It assumes full convertibility of one with the other. Though full convertibility of work into heat is possible as work is high-grade energy, but full convertibility of heat (lowgrade energy) into work (high-grade energy) is not possible. 2. It does not explain the direction of a process. It permits (even theoretically) heat transfer from a low-temperature body to a high-temperature body which (we all know) cannot take place in general life.

PERPETUAL MOTION MACHINE The perpetual (continuous) motion machine of the first kind (PMM-1) is a hypothetical device conceived based on violation of the first law of thermodynamics. Figure 2.14( a) shows a device which is continuously producing work (W =t:- 0) without any supply of energy (Q = 0). This device is a PMM-1 as it violates the first law of thermodynamic as well as the law of conservation of energy. Similarly, Figure 2.14(b) shows a device which is continuously generating heat (Q =t:- 0) without any other form of energy (W = 0) supplied to it. Hence it is a PMM-1.

(b)

(a)

FIGURE 2.14

Perpetual motion machine.

SOLVED PROBLEMS 1. Calculate the work done by a gas as it is taken from state a to b, b to c and c to a. Process ab is a constant pressure process (Figure 2.15). Work done by the system is the area under line ab: Wah =

100

X (4

- 2) = 200 kJ

Process be is a constant volume process and

Wbc

= 0.

..

First Law of Thermodynamics

C

200

a, Q. _,,,_

Q

100

b

a

0

2

4

V(cm 3)

FIGURE 2.15

PV diagram (Problem 1).

Process ca has volume reducing, and work done on the system is the area under line ac:

Wtr = -[½(200 -100)

Work done in the cycle

=

-[100 + 200]

=

200 - 300

=

X

=

(4 - 2) + 100 X (4- 2)] -300 kJ

-100 kJ

The negative sign shows that work is done on the system which is also evident as the cycle is traced anticlockwise.

2. A system is taken through a cyclic process. It absorbs 100 kJ in process be the system is compressed adiabatically, and in rejects 120 kJ. If 70 kJ work is done during process be, find the system at b and c, the internal energy at a is 150 kJ. Also system during process ca (Figure 2.16).

of heat in process ab, process ca the system the internal energy of find work done by the

p

V

FIGURE 2.16

Cyclic process (Problem 2).

Process ab is isochoric and work done is zero. Heat supplied is used for increasing internal energy: Li Uab = Qadd = 100 kJ Ub - Ua = 100 kJ ub - 150 = 100 kJ ub = 250 kJ Process cb is adiabatic, i.e. Q = 0. Q = DU+ W (volume decreasing) 0 = Li ucb - 70 Therefore, uc - ub = 70 Uc - 250 = 70 Uc= 250 + 70 = 320 kJ

Therefore,

-

Thermal and Hydraulic Machines

Process ca is isobaric and volume is increasing. Therefore, work will be done by the system. Qea = LiUca + Wea - Qea = ( Ua - UJ + Wea

- 120 = (150 - 320) + Wea Wea= (170 - 120) kJ

= 50 kJ 3. Find out work done in processes ab, be, cd and da. Calculate also work done in complete cycle abcda. Process ab is an isobaric process and work done by the system is the area under line ab (Figure 2. 17): 14

a

12

ca a..

10

6

8

0...

6

b

'

4 2

d

C

2

4

3

5

6

V(m 3)

FIGURE 2.17

=

Processes ab, be, cd and da (Problem 3).

12

X

(6 - 1)

=

60 kJ

Process be is isochoric and W6e = 0. Process cd is isobaric and volume is reducing. Therefore, work done on the system is the area under line cd: Wed= -Pd(Ve - Vd) = -2(6 - 1) = -10 kJ

Process da is isochoric and Wda

=

Cyclic work

0.

= Wab + Wed = 60 - 10 =

50 kJ

Work is positive as evident from the cycle which is traced clockwise.

First Law of Thermodynamics

..

4. A system undergoes state A to states B and C and returns to A. If UA = 0, Us = 30 J and the heat given to the system in process BC is 50 J, then determine (i) internal energy at state C, (ii) heat given into the system in process AB, (iii) heat given into the system in process CA , and (iv) net work done in the complete cycle (Figure 2.18). 90 ----------------- C

ca

i

50 J

60

Q..

30

,B Ua= 30

,

2

0

3

V(m 3)

FIGURE 2. 18

PV diagram (Problem 4).

Process BC is isochoric and hence Wsc = 0.

Qsc = !J..Usc + Wsc 50 or

=

Uc - Us

=

Uc - 30

Uc= 80 J

Now for process CA, the volume is decreasing:

WcA = - [area under line CA]

=- [

=- [ =

½x AB x BC + area under AB]

½ 2 X

X

-(60 + 60)

60 + 30 =

X

2]

-120 J

For process AB: WAB = area under AB = 60 J

QAS = Ll UAS + WAS

= (Us - UA) + WAs = (30 - 0)

+ 60

= 90

J

For process CA:

QcA

= Li UcA + WcA = (UA - Uc) + WCA = (0 - 80) - 120 = -200

J

-

Thermal and Hydraulic Machines

QcA is negative and hence heat 200 J is rejected by the system. Work done in the whole cycle is the area enclosed = 60 J. Work is positive as the cycle is traced clockwise.

5. As shown in the Figure 2. 19, 100 J of heat is given in taking a system from state A to state C along path ADC and 50 J of work is done by the system. (i) If the work done by the system is 15 J along path ABC, then how much heat is given in taking the system from A to C? (ii) How much heat will be absorbed or given out if the work done on the system along curved path from C to A is 15 J, and (iii) If U8 - UA = 40 J, then how much heat will be absorbed in each of the processes AB and BC? p

V

FIGURE 2.19

Q

Process ADC:

=

PV diagram (Problem 5).

100 J, W

50 J

=

Q = flUA c + W flUA c

=

Q - W

100 - 50

=

= 50 J W

Process ABC:

=

15 J, fluA c

=

50 J

Q =flu+ W =

W

Process CA (curved):

50 + 15

= -

65 J

=

15, fl UcA

Q = flUcA + W = - 50 - 15 Process AB:

fl

uab

WABC QAB

= -

-50 65 J

= 40 J WAB ( ·: Wsc = WAB = 15 =

= flUAB +

=

0)

WAB

= ( U B- UA)

Process BC:

=

+ WAB

= 40

+ 15

= 55

J

QABC = 65 and QAB = 55 Qsc

= QABC - QAB = 65 - 55 = 10 J

6. The values of internal energy of a system at states A, B and C are 10 J, 30 J and 200 J, respectively (Refer to Figure 2.20). Heat 5 J is released in process DA. Determine

m

First Law of Thermodynamics

(i) internal energy of state D, (ii) heat released in process CD, (iii) heat released in process AB, and (iv) heat absorbed in process BC.

Process DA is an isochoric process. Therefore, W = 0. Q = - 5 J (Given) DUDA = Q -

w

C

60

20 10 + - - - + - - - - - - ~ A

0

FIGURE 2.20

2

6

Internal energy of a system at states A, B and C (Problem 6).

UA - UD = -5 UD = UA + 5 = 10 + 5 = 15 J

or Process CD: WCD

= Area under line CD 1

BD x BC + area under line BD

2

1

-

X

2

= Q =

(6 - 2)

80 + 80 fl

=

X

(60 - 20) + 20

X

(6 - 2)

X

(6 - 2)

160 J

UCD + WCD

= (UD - Uc) + WCD = (15 - 200) + 160 = - 185 + 160 = - 25 J Process AB:

llUAB

= Us - UA

J WAB = Area under line AB = 30 - 10 = 20

= -1

X

= 20

+ 40

2

(20 - 10)

X

(6 - 2) + 10

= 60

Since the volume is decreasing and hence WAB

=-

60.

-

Thermal and Hydraulic Machines

Q = LiUAs + WAS 20 - 60

=

= -

40 J

Process BC: It is an isochoric process and hence W = 0. Li Use= Uc - Us = 200 - 30

= 170 J Q = Li U + W

Therefore,

= 170 J 7. Calculate the increase in internal energy of 1 kg of water at 100°C when it is converted into steam at 100°C and atmospheric pressure (I 00 kPa). Pwater = 1000 kg/m 3, Psteam = 0.6 kg/m3 and hconversion = 2.25 X 106 J/kg Volume will increase when water is converted into steam. vsteam

vwater

vsteam -

vwater

1

1

Psteam

0.6

= -=

m 3 = 1.67 m 3

1

1

Pwater

1000

= 1.67 - 0.001 "' 1.67 m 3

Work done =

Patm (Vsteam -

100

Vwater)

1.67 kJ

X

167 kJ

Q

=

heat given for conversion

=

2.25 x 10 6 J

= 2250 kJ Q=LiU+W or

Li U = Q - W = 2250 - 167 = 2083 kJ

8. In a nozzle (Figure 2.21 ), air at 827°C and atmospheric pressure enters with negligible velocity (C 1) and leaves at a temperature of 27°C. Determine outlet velocity (C 2) if LiQ = 0 and Liz = 0. Take cP = 1 kJ/(kg K).

= 827°C

C2 =? T2 = 27°C

+273

+273

C1 =0

T1

FIGURE 2.21

A nozzle (Problem 8).

..

First Law of Thermodynamics

C2 = ~2cp(T2 -T.) ~2 X 10 3 (1100- 300) 1265 mis

~16 X 105

9. An air compressor (Figure 2.22) takes shaft work of 200 kJlkg and compression increases enthalpy by 100 kJlkg of air. Cooling water picks up 90 kJlkg of air of heat while cooling. Determine the heat transferred from the compressor to the atmosphere.

-::;:::==:=::i --water out Q water = Heat extracted

by water

Wc=200I I

I I

\

I

:

'!

I

: Q~

= Heat transferred

h,

FIGURE 2.22

Q

=

An air compressor (Problem 9).

heat given out by the system

=

Q0 + Qw

Applying SFEE on the compressor,

Q + h1 +

Tc2 + gz =

Wc

+ h2 +

Tc2 + gz

2

z 1 = z 2, C 1 = C2, Wc = -200, h 2

Q= (h 2

-

h 1)

-

-

h1

100

200

= 100 - 200 = - 100 kJ = Qa + Qwater -100 = Q 90 Q = - 10 kJ Q

0

or

-

0

10. The inlet and outlet temperatures for a flow through a nozzle are 400 K and 300 K (Figure 2.23). Mass flow of air is 1.0 kgls. If initial velocity is 300 mis, determine exit velocity and the ratio of inlet to exit area of the nozzle. cP = 1 kJl(kg K).

T, = 400K C1 = 300 m/s

FIGURE 2.23

Flow of air through a nozzle (Problem 10).

..

Thermal and Hydraulic Machines

Applying the SFEE on the nozzle, we get

Q + m ( h1+

½c/ + g z, )

= W + m ( h2 +

cf + g z2)

Q = 0, W = 0, z 1 = z 2 , m 1 = m 2 = m = 1.0

= cp(T1 1

T2 )

-

X 10 3

(400 - 300)

= 105

cf =

2 X 10 5

+

cf =

2 X 10 5

+9

X

10 4

= 29 X 104 C2 = 451 mis

or

Volume of air at the inlet= Volume of air at the outlet Area at the inlet

x velocity of the inlet = Area at the outlet x velocity at the outlet

Mathematically

or

~= C2 = 451 A2 C1 300

Therefore,

A 1 : A2

::

~!-2 1

1.5 : 1

11. An air compressor compresses air at 0.1 MPa and 27°C by ten times the inlet pressure. During compression the heat loss to the surroundings is estimated to be 5% compression work. Air enters the compressor with a velocity of 40 mis and leaves with 100 mis. Inlet and outlet cross-sectional areas are 100 cm 2 and 20 cm 2 , respectively. Estimate the temperature of air at the exit and power input to the compressor (Refer to Figure 2.24). (UPTU: July 2002) r----------1

: :

,

:

-+--

__ ; CD ', -------- '.:,._~ P1 =0.1 MPa T1

P2 = 1 MPa A2 = 20 cm 2

C:1 = 100 m/s

0=0.05 We

= 27°C

C1 = 40 m/s A 1 = 100 cm 2

FIGURE 2.24

Compression of air (Problem 11 ).

-

First Law of Thermodynamics

First we find out v 1 (specific volume) at the inlet so as to find mass flow (m 1).

287

X

300

0.1

X

105

= 0.861 m3/kg (100

volume/s

X

10- 4 )

X

40

0.861

= 0.4646 kg/s m 1 = m 2 = m = 0.4646 kg/s (20

X

10- 4 )

X

100

- - - - - - - = 0.4305 m3 /kg 0.4646

or

1

X

106 X 0.4305 287

1500 K or (1227°C) Applying the SFEE on the compressor, we get

= 0.4305 [ 1

X

2 10 3 (1500 - 300) + 100 ; 402 )

0.95 WC= 621 kW or

We= 590 kW

12. A system moves from state 1 to state 2 on the TS diagram (Figure 2.25). T 1 = 330 K, T2 = 440 K, internal energy at state 1 = 170 kJ, internal energy at state 2 = 190 kJ, entropy S 1 = 0.23 kJ/K, entropy S2 = 0.3 kJ/K. Find work done by the system.

-

Thermal and Hydraulic Machines

S1

S2

Entropy (S) FIGURE 2.25

TS diagram (Problem 12).

Heat given to the system, Q = area under line 1-2

Q = T1(S2

-

S1) +

1

2 (S2 -

S1) (T2

-

T 1)

= 330(0.3 - 0.23) + _!_(0.3 - 0.23) (440 - 330) 2

= 26.95 kJ Q=!lU+W 26.95 = (190 - 170) + W

W = 26.95 - 20

or

=

6.95 kJ

13. A centrifugal air compressor delivers 15 kg of air per minute. The inlet and outlet conditions are as follows: At the inlet: velocity 5 mis, enthalpy 5 kJlkg. At the outlet: velocity = 7 .5 mis, enthalpy = 17.3 kJlkg. Heat loss to cooling water is 756 kJ. Find (i) the power of the motor required to drive the compressor, and (ii) the ratio of the inlet pipe diameter to the outlet pipe diameter if the specific volumes at the inlet and the outlet are 0.5 and 0.15 m 3lkg (Refer to Figure 2.26). Cg) C2 = 7.5 m/s h2 = 17.3 kJ/kg

G)

C1 = 5 m/s

h1

FIGURE 2.26

= 5 kJ/kg

Centrifugal air compressor (Problem 13).

Applying the SFEE, we get

Q

=

Wc +

m[(h

2 -

h1)+

½(Ci

- cf)+g(z2

-756 =WC+ ~[(17.3- 5) + .!..(7.5 2 -5 2 ) 60 2 Wc

= -54.6 kJls

X

10-3 ]

-

z,)]

First Law of Thermodynamics

-

The power of the motor to drive the compressor is 54.6 kW. Flow rate is constant, i.e. A1C1

m=-V1

~

or

A =

Similarly,

A=-2 C2

I

60

X

0. 5 = 0.025 m 2 5

mv2

= 0.005

15 0.15 x60 7.5 m2

or

= 2.236 14. Gas leaving a turbine enters a jet pipe with an enthalpy of 915 kJlkg and leaves with an enthalpy of 800 kJlkg (Figure 2.27). The inlet velocity is 300 mis. Find the exit velocity. G)

FIGURE 2.27

Change in enthalpy of gas (Problem 14).

Applying the SFEE, we get

h + I

c,2 2

cf

=h +

2

2

cJ- cf= =

2(h 1

-

h2 )

2(915 - 800)

= 230

X

10 3

= 23

Ci = 23

X

10 4 + 9

=

32

X

10 4

=

565.7 mis

C2

X

X

10 3 X

10 4

10 4

15. An inverter claims to invert a device which works in a cycle taking heat 50 J, 30 J and - 40 J, and giving net work of 60 J. Check the claim.

-

Thermal and Hydraulic Machines

I.Q

50 + 30 - 40

=

40 J

=

I. W = 60 J

Since I, W > I, Q, the device violates the law of conservation of energy. The claim of the inverter is incorrect.

16. A system undergoes a cyclic process through four states 1-2, 2-3, 3-4 and 4-1. Find the values of xi, x 2 , Yi, y 2 and y 3 in Table 2. 1. TABLE 2. 1

Four states of a cycle process

Process

Heat transfer (kJ/min)

Work tranfer (kW)

Change of internal energy

1-2

800

5.0

Y1

2-3

400

XI

600

3-4

-400

X2

h

4-0

0

3

Y3

(UPTU: Dec. 2005) Process 1-2: Q=W+!).U 800 = 5

Yi

=

X

60 + Yi

800 - 300

=

500 kJ/min

Process 2-3: Q=!lU+W 400 = 600 +

Xi

xi = -200 kJ/min For a cyclic process I.Q=I.W 800 + 400 - 400 + 0 = 300 - 200 + x 2 + 3 or

Xz

= 800 - 280 = - 520 kJ/min

Process 3-4: Q=!lU+W

= y 2 + 520 Y 2 = -920 kJ/min

- 400 Process 4-0:

Q=!lU+W

= 180 + fl U y 3 = -180 kJ/min

0

X

60

First Law of Thermodynamics

-

17. 0.8 kg/s of air flows through a compressor in steady state conditions. The properties of air at the entry are: pressure 1 bar, velocity 10 m/s, specific volume 0.95 m3/kg and internal energy 30 kJ/kg. The corresponding values at the exit are 8 bar, 6 m/s, 0.2 m3 /kg and 124 kJ/kg. Neglecting the change in PE, determine the power input and the pipe diameters of the entry and exit.

(UPTU: Carry over Aug. 2005-2006) Q

+

M[ h, + V1 (b) V2 < V1 8. A cyclic process performs positive work if the curve on a PV diagram traces

(a) clockwise

(b) anticlockwise

9. Throttling is a process which has constant _ _ __ (a) enthalpy (b) velocity 10. A perpetual motion machine violates _ _ __ (a) energy equation (b) first law of thermodynamics 11. A nozzle is a converging type if fluid has _ _ _ _ flow.

(a) subsonic (b) supersonic 12. A nozzle is a diverging type if fluid has _ _ _ _ flow. (a) subsonic

(b) supersonic

13. For adiabatic mixing of two fluids, the heat interaction is _ _ __

(a) unity

(b) zero

14. Evaporation of water and freezing of water are examples of an _ _ _ _ process.

( a) isochoric

(b) isobaric

15. Internal energy remains constant during an _ _ _ _ process.

(a) adiabatic

(b) isothermal

16. Q-W remains constant for _ _ _ _ processes.

(a) reversible

(b) all

..

First Law of Thermodynamics

17. Heat generation in a boiler takes place at constant _ _ __

(a) temperature

(b) pressure

18. A pump is used to increase the pressure of _ _ __

(a) liquid

(b) vapour

19. Enthalpy is the summation of internal energy and _ _ _ _ energy.

(a) free

(b) flow

20. At constant pressure, heat supplied is used for increasing _ _ __

(a) internal energy

(b) enthalpy

Answers State True and False 1. True

2. False (Heat added is positive and heat rejected/released is negative.) 3. False (Workdone by the system is positive and work done on the system is negative.) 4. False (In free expansion, work is zero as the expansion of the system has not been

carried out against atmospheric pressure.) 5. True (Clockwise is positive and anticlockwise is negative.) 6. True

7. True (Q - W is equal to ~u (u 2 change from state 1 to state 2.)

-

u 1) which is constant for a process undergoing

8. True 9. True (Q =

~u +

W)

10. False (As the system in a cyclic process comes back to initial position, hence change

of internal energy is zero for all processes.) 11. False (Internal energy depends upon temperature only. Hence internal energy remains

constant for isothermal process.) 12. True (Q

= ~U +

W. As for adiabatic process Q

= 0, therefore

W

=-

~U)

13. True (Q = ~U + W. As W(=Pdv) is zero for an isochoric process, all the heat is used for increasing internal energy, i.e. Q = ~ U)

= ~U + W. As for an isolated system Q = 0 and W = 0, internal energy remains constant, i.e. ~U = 0)

14. True (Q

15. True (Q

=

~U

+ W and

~U

= 0 for a cyclic process. Hence

Q

=

W for a cyclic

process.) 16. False (If the volume of a system increases, work is done by the system and if volume

decreases work is done on the system.)

-

Thermal and Hydraulic Machines

17. True 18. True 19. True

20. True (Pv° 21. True

=

(PV1 =

22. True (Pv Ps. Hence

is

JPdV)

~w, > ~W2 as work

-

Thermal and Hydraulic Machines

.

22. (a) (Work or heat 1s the area enclosed = nr 1 r2 = n x X

10 3 = 3.14

X

100

X

(300-100) x 10---6 x (300-100) 2 2

100 X 10-2 = 31.4 J)

23. (d) [area= (2P - P) (2V - V) = PV]

24. (b) 25. (d) (Q - W = ll U) 26. (d) 27. (a)

28. (b) 29. (c) (For a cyclic process Q - W = 0.) 30. (d) (Q = W for an isothermal process.) 31. (a) (W = - ll U for an adiabatic process.) 32. (c) (For an isochoric process W = 0, therefore Q = llU.)

Fill in the Blanks 1. (b)

2. (b)

3. (b)

4. (b) (Fluid for expansion requires increasing area.) 5. (a) (Fluid for compression requires decreasing area.) 6. (a)

7. (b)

(b) 14. (b)

11. (a)

15.

18. (a)

10.

8. (a)

9. (a) 13. (b)

(b)

(b) 16. (b)

19. (b)

20. (b)

12.

17. (b)

Second Law of Thermodynamics

INTRODUCTION The first law of thermodynamics cannot explain non-occurrence of certain processes as well as the direction of a process. Feasibility of a process, the direction of a process, and grades of energy (low and high) are clarified by the second law of thermodynamics. Other things like maximum possible efficiency of a heat engine; coefficient of performance of a heat pump and a refrigerator, and the concept of a temperature scale, which is independent of physical properties, are also explained by the second law of thermodynamics.

HEAT RESERVOIR Heat reservoir is a system/body having extremely large heat capacity. It is capable of absorbing or rejecting finite amount of heat without any change in temperature. In this respect the atmosphere, rivers and seas are reservoirs from which we can extract or dump any amount of heat without changing temperature. Source is a heat reservoir at higher temperature from which heat is extracted without change of its temperature. The sun is a "source" heat reservoir. A sink is a heat reservoir capable to absorb any amount of heat without change of its temperature. The atmosphere or surroundings is a "sink" heat reservoir.

HEAT ENGINE A heat engine is a device used for converting heat into work (Figure 3 .1 ). It is possible to convert work into heat directly but the heat engine is required to convert heat into work. 87

-

Thermal and Hydraulic Machines T, Source

w

Sink

T2

FIGURE 3.1

Heat engine.

A heat engine can be defined as a device operating in a cycle between a high-temperature source and a low-temperature sink and producing work. The heat engine receives heat (Q 1) from the source and transforms some portion of heat into work (W) and rejects balance heat (Q 2) to the sink. Heat and work have been categorized as two forms of energy-low grade and high grade. Conversion of high-grade energy into low-grade energy can be carried out fully and spontaneously without aid of any device. However, complete conversion of low-grade energy into high-grade energy is impossible and non-spontaneous. We require a device like a heat engine or gas turbine plant to convert low-grade energy (heat) into high-grade energy (work). However, conversion from low-grade to high-grade energy cannot be achieved fully. A gas turbine plant (Figure 3.2) consists of a boiler, a turbine, a heat exchanger and a compressor. Heat (Qadd) is added in the boiler, expansion takes place in the turbine producing work ( We), the heat exchanger cools the fluid by extracting heat (Qrej) and the compressor compresses the fluid when compression work (We) is given to it by external source.

Oadd

Boiler

Compressor

Turbine

Heat exchanger

Cold water

FIGURE 3.2

Gas turbine plant.

The efficiency of a heat engine is to convert as much heat into work as possible: W . Net work output Effi1c1ency 1J = - - - - - - = Heat supplied Qactct

..

Second Law of Thermodynamics

A gas turbine plant is also designed to extract as much work from the supplied heat.

w = we -

WC

where We= expansion work from turbine We

= compression work given to compressor.

Also Therefore,

-- we - WC --

Qadd - Qrej

Qadd

Qadd

Qrej TJ= 1 - -

(3. I)

Qadd

It can be seen from Eq. (3.1) that efficiency can be increased by reducing heat rejected.

HEAT PUMP A heat pump is a device used for extracting heat from a low temperature body and sending it to a high temperature body while operating in a cycle [Figure 3.3(a)]. Transfer of heat from a low temperature body to a high temperature body is a non-spontaneous process. However, it is possible with the help of a heat pump which uses external work supplied to it. A heat pump is used in cold regions where temperature of the surroundings is low and room temperature is to be kept at higher temperature [Figure 3.3(b)]. A heat pump picks heat (Q 2) from the surroundings (low temperature T2) and delivers heat (Q 1) to the room which is at high temperature (T1) using external work ( W). High temperature Body

T1 _,,-_/',,,,,,

/

Surrounding

',,,,',,, (T2)

i'j

---w

02

01

Body

Work by electricity

T2

Low temperature (b)

(a)

FIGURE 3.3

Heat pump.

Coefficient of performance: The coefficient of performance (COP) is defined as the ratio of desired effect to external work supplied for getting the desired effect:

-

Thermal and Hydraulic Machines

COP = Desired effect Work supplied

Desired effect for a heat pump is to supply heat Q 1 to the hot body or room.

Q,

(3.2)

COPheat pump = W

However

Therefore, Eq. (3.2) becomes (3 .3)

REFRIGERATOR A refrigerator is a device similar to a heat pump, but the desired effect is to extract heat as much as possible from the cold body/space and rejects to a high temperature body/ surroundings. The desired effect of a refrigerator is heat (Q2) removed from cold space. In a domestic refrigerator and an air conditioner, heat (Q2 ) is removed from the refrigerator or room as shown in Figure 3.4 by supplying work (W) and heat (Q 1) is rejected to the surroundings.

Q2 COPrefrigerator = W

or

(3.4)

COP refrigerator= Q, _ Q2 Cold space 02 ( T2) ,,...----,,,"'-> Surroundings (T,)

Body T1

__,,·······...

Surrounding

··•·· ... (T2)

:·./

Compressor

(a) Refrigerator

FIGURE 3.4

Electricity

(b) Refrigerator

02

Work by electricity

(c) Air conditioner

Working of a refrigerator and an air conditioner.

Since Q 1 > Q 2 (as Q 1 = Q2 + W), COP of a heat pump is always greater than COP of a refrigerator.

..

Second Law of Thermodynamics

COPheat pump=

COP refrigerator

Q, -Q2

= Q, _ Q2

[from Eq. (3.3)]

[from Eq. (3.4)]

Therefore, COP heat pump - COP refrigerator =

or

Q, -Q2 =1 Q, -Q2

COP heat pump = 1

+ COP refrigerator

(3.5)

STATEMENTS FOR THE SECOND LAW OF THERMODYNAMICS Clausius statement: The Clausius statement for the second law of thermodynamics states that it is impossible to have a device that while operating in a cycle produces no effect other than transfer of heat from a body at low temperature to a body at higher temperature. A non-spontaneous process such as transfer of heat from a low temperature body to a high temperature body can be realised when some other effects such as external work requirement are bound to be there. Heat pumps and refrigerators use work (electrical energy) to transfer heat from low to high temperature surrounding. Kelvin-Planck statement: The Kelvin-Planck statement for the second law of thermodynamics is that it is impossible for any device operating in a cycle to produce net work while exchanging heat with bodies at a single fixed temperature. No cyclic engine can convert whole heat into work. It is impossible to build a heat engine which has 100% efficiency. There is degradation of energy in a cyclic heat engine as some heat has to be degraded or rejected to a low temperature body. There has to be atleast two heat reservoirs (source and sink) for a heat engine to perform. There is an equivalence between the Kelvin-Planck statement and the Clausius statement. Any system based on violation of the Kelvin-Planck statement leads to violation of the Clausius statement and vice versa. Violation of the Clausius statement leads to violation of the Kelvin-Planck statement. As shown in Figure 3.5(a), device A is violating the Clausius statement and it is tranferring heat Q2 from the sink to the source without any work. If a heat engine is made to work in parallel to it, we get a composite system (Figure 3.5(b)) which produces work interacting with one reservoir. Violation of the Kelvin-Planck statement leads to violation of the Clausius statement (Figure 3.6). Device B is violating the Kelvin-Planck statement and it is producing work taking heat from one reservoir. If a heat pump (HP) is made to work using work output of device B, we get a composite system which extracts heat from the sink and delivers to the source without aid of external work. The composite system violates the Clausius statement.

..

Thermal and Hydraulic Machines T,

T,

0@

W=0,-0:!

/

Composite system

T2 (b)

(a)

Figure 3.5

Violation of the Clausius statement.

T,

T,

02 W=O

0

@

Composite system

02 T2

T2

(a)

(b)

FIGURE 3.6

Violation of the Kelvin-Planck statement.

Perpetual motion machine:

The perpetual motion machine (PMM) of second kind is a machine which violates the Clausius or Kelvin-Planck statement of the second law of thermodynamics. Figure 3.7(a) shows PMM II, which extracts heat Q2 from the cold body (T2) and delivers to the hot body (T1) without aid of work (W = 0). Figure 3.7(6) shows PMM II, which takes heat (Q) from the hot body (T 1) and converts fully into work (W) without any rejection of heat to the cold body. Hot

T, T,

Hot

W=O

PMM-11 (a)

FIGURE 3.7

PMM-11 (b)

Perpetual motion machine.

CARNOT CYCLE The reversible process, as the name suggests, can come back to the original state through the same path on removal of factors affecting the change. All processes are attempted to

-

Second Law of Thermodynamics

reach close to a reversible process in order to give the best performance. However, all practical processes are irreversible as these cannot attain their original state or follow back the path due to friction or dissipation of energy. The Carnot cycle is a reversible thermodynamics cycle comprising of four reversible processes (Figure 3.8): p

T Oactct

2

We

We

We

3

We

3

4 Orej

s

V

(a) PV diagram

(b) TS diagram

FIGURE 3.8

1. Reversible 2. Reversible 3. Reversible 4. Reversible

Carnot cycle.

isothermal heat addition (Qactct)-process 1-2 adiabatic expansion process giving work output (We)-process 2-3. isothermal heat rejection (Qrej)-process 3-4. adiabatic compression using external work (Wc)-process 4-1.

~Q= Qadd - Qrej =

~w

we -

(3.6) WC

we - WC

Net work

11 = Heat added

=

Qadd

Substituting Eq. (3.6) in the above equation, _ Qadd - Qrej _ l

11 -

------=- -

Qadd

- -

Qrej

Qadd

The analysis of each process of the Carnot cycle can be done for heat and work as given below:

Process 1-2:

It is an isothermal process which means flU

converted into work.

where

r

=

· rat1·0 -- V2 compress10n

= 0.

Complete heat is

-

Thermal and Hydraulic Machines

Process 2-3: It is an adiabatic process which means ilQ = 0. Therefore, expansion work is achieved by reduction of internal energy, i.e. We= -ilU

we -

W

or (c) Process 3-4:

=

e

P2V2

-

~V3

mR(T2 -T3 )

--=---=--------''----=- -

r-1

---=--------''-

-

r-1

mR(I; - T3) ·: (T

y-1

= I

T ) 2

It is an isothermal process, i.e. ilU

-

Qrej -

=

0. Therefore,

Qrej =

W34 •

V3 = mRT3 ln V2 V4 ½

P 3 V3 ln -

= mRT3 ln r (d) Process 4-1:

It is an adiabatic, i.e. ilQ

=

mR(T1

-

r-1

=

T4 )

0. Therefore, We =

mR(T1

-

= -

ilU.

T3 )

r-1

The Carnot cycle is not practical for the following reasons: 1. Frequent change of the cylinder head to make insulating for an adiabatic process and diathermic for an isothermal process. 2. It is practically impossible to achieve isothermal heat addition. 3. Reversible adiabatic expansion and compression are impossible. 4. Reversible isothermal processes are very slow processes while reversible adiabatic processes are fast processes. Speed fluctuation in the revolution of an engine is not possible. The Carnot cycle can also be operated in reverse direction, i.e. anticlockwise. In that case, work is negative, i.e. work has to be done on the system to get the desired output (cooling of space or heating of room). The reversed Carnot cycle is also called Carnot refrigeration cycle.

CARNOT THEOREM The Carnot theorem states that a Carnot heat engine has efficiency greater than that of any other heat engine operating between the same temperature limits. To prove the Carnot theorem, consider heat engines A and B operating in parallel between two reservoirs at temperatures T 1 and T2 (Figure 3.9(a)). Heat engine A is the Carnot heat engine while heat engine B operates irreversibly. Assume 1Js > 1JA which will give Ws > WA. Now we reverse heat engine A to operate as a heat pump (Figure 3.9(b)) and it can take work WA from heat engine B. Now we get a composite system, which takes heat from reservoir T2 and it converts it fully to work output (Ws - WA). The composite system

-

Second Law of Thermodynamics

violates the Kelvin-Planck statement. Hence our assumption is wrong and the Carnot heat engine has efficiency greater than any other heat engines.

~

Composite

:_e_j_o_,~--A --- }.O, ~ system , I

HP

HE

-~:f------r~---~a

Wa-WA

I

o, ~

I

I

T2

(b)

(a)

Carnot theorem.

FIGURE 3.9

The corollaries (deductions) of the Carnot theorem are as follows: I. Efficiency of all reversible engines operating between the same temperature limits is the same. 2. Efficiency of a reversible engine does not depend on the working fluid in the cycle.

THERMODYNAMIC TEMPERATURE SCALE A temperature scale which is independent of the property of thermometric substance is defined as a thermodynamic temperature scale. The thermodynamic temperature scale is developed based on the fact that the efficiency of a reversible heat engine does not depend on the working medium, but it depends on the temperature of two reservoirs in which it is operating. Q2

I - QI

11carnot =

-2!_

or

=

f(T

Q2

(Ti, T2)

=

T ) = t,

1/f (7;) l/f(T2)

2

By choosing suitable equivalent value of function, we may write as follows:

-2!_ - 2i_ Q2

T2

where T1 and T2 are in kelvin. Hence heat flow is proportional to temperature of the reservoir. Now consider a series of reversible engines operating and producing equal work W while operating between a series of reservoirs (Figure 3.10). As work output is equal for each engine, we can write: W = Qi - Qz = Qz - Q3 = Q3 - Q4 = Q4 - Qs = T1 -

T2 = T2

-

T3 = T3

-

T4 = T4

-

T5

FIGURE 3.10

Thermal and Hydraulic Machines

T,

A series of reversible engines operating between a series of reservoirs.

The difference between the temperatures of successive reservoirs is equal, which can be made small or large depending upon the requirement of the temperature scale. Heat interactions in a reversible engine are proportional to the absolute temperature of the source and the sink. Hence efficiency of a heat engine (HE) and COP of a heat pump (HP) can be written in terms of reservoir temperatures T1 and T 2 , i.e.

Sh_ =Ii

C

0 PHP =

__2i_ QI = QI -Qz Ti -Tz

0 P,ef =

_!i_ Qz = QI -Q2 Ti -T2

C

CLAUSIUS INEQUALITY For cyclic processes, let us see the ratio of change of heat to temperature: In a reversible cycle (Figure 3 .11)

AB= CD or

Second Law of Thermodynamics T

s FIGURE 3.11

Reversible cycle.

or

or

In an irreversible cycle (Figure 3.12)

s FIGURE 3.12

Irreversible cycle.

AB< CD QI< _ Q2

or

I;

or

or The Clausius inequality states that ,( dQ T

I. j'

=

0 for a reversible process

,( dQ < 0 for an irreversible process T

2. j'

,( dQ > 0 for an impossible process 3. j' T

T2

-

-

Thermal and Hydraulic Machines

ENTROPY AND AVAILABLE ENERGY

Entropy is a measure of disorder of the system. The greater is disorder, the higher is entorpy. Entropy is highest in the gaseous state and lowest in the crystalline solid state. Liquid has entropy more than solid and less than gas. Disorder also increases with irreversibility of the process. A few examples of the increase of disorder or entropy are as follows: 1. A cup on a table (ordered state) falls onto the floor and breaks in small pieces

( disordered state). 2. The pieces of the jigsaw game start off in a box in an ordered arrangement in which they form a picture. On shaking, the pieces will take up another arrangement in which the pieces do not form a proper picture as they are now in a disordered state. 3. The computer memory is in a disordered state in the beginning. When we feed information, computer memory changes from the disordered state to an ordered state. It is necessary to use a certain amount of electric energy to do so. The electric energy is also used for running a cooling fan and dissipation of heat from the computer which increases the degree of disorder in surroundings. The degree of disorder in surroundings is greater than the degree of order in the computer memory. Hence the total disorder of the universe increases. 4. The universe is expanding after the big bang. The universe is moving from an ordered state (before the big bang) to a disordered state. 5. We consume food (ordered state) and a small part of energy from it is used for useful purposes (thinking process-ordered state), thereby increasing the disorder. All heats are not equally valuable for converting into work. Heat at higher temperatures has greater possibility of conversion into work than heat at lower temperatures. The quantity

di

J

is the same for all reversible processes between state 1 and state 2.

It is independent of the path and hence it is a state function. Hence dQ is an exact T differential of entropy (s), i.e. ds. We can say dQ = TdS which means entropy increases

with heat addition and decreases with heat removal. Entropy for a reversible process is defined by the following relation: dS

=

(dQ) T

reversible

To find entropy for a irreversible process, the actual process is substituted by an imaginary reversible process. The change of entropy from state 1 to state 2 for an imaginary reversible process and an actual irreversible process would be the same. Entropy is a function of heat and temperature, which shows the possibility of conversion of that heat into work. The increase in entropy is small when heat is added at high temperature and greater when heat is added at lower temperatures. Therefore, heat having

Second Law of Thermodynamics

-

higher entropy has lower possibility for conversion into work. Similarly, heat having lower entropy has higher possibility for conversion into work. Actually entropy is zero at absolute zero temperature, and it is impossible to achieve absolute zero temperature. Hence convenient temperature is selected at which entropy is given arbitrary value of zero. The selected temperature for steam is 0°C while for NH 3 , Fe-12 and CO 2, it is - 40°C. We determine the change of entropy from these points. Therefore, we cannot measure absolute value of entropy. Entropy change for various systems is as follows: 1. Entropy change of an ideal gas for a closed system is: From the first law of thermodynamics dQ = dU + dW TdS =

dT + PdV

CV

(3.7)

We know that P=

RT V

Therefore, Eq. (3.7) becomes dV Tds = cv dT + RT V

J2 dS = i2 cv dTT +i2 RdVV I

S2

-

I

I

T2 V2 S 1 = cv ln - + R ln -

Vi

1i

(3.8)

For constant volume process V2 = V1• Therefore, Eq. (3.8) becomes S2

-

S1 =

Cv

ln

T2

(3.9)

1i

2. Entropy change for an open system is: h=u+Pv dh = du + Pdv + vdP

= dQ +

vdP

cPdT = TdS - RT

r S2

-

dS =r I

S1

C

P

= Cp ln

pdP (·:

Pv = RT)

dT _ r RdP T I P Tz

1i

- R ln

Pz

Pi

(3.10)

-

Thermal and Hydraulic Machines

For constant pressure processes P 2 = Pi. Therefore, Eq. (3.10) becomes

3. Entropy change for a polytropic process is

S2 From PV

=

S1 =

-

Cv

1n T2 +RlnV2

(3 .11)

Vi

1i

RT and pyn = constant

Therefore, Eq. (3 .11) becomes I

S2 -

s,

=

Cv

T2 + ln T,

Cv

(r- 1) ln (

T,T.21 ) ~

(":R

=

Cv(Y-

1))

I

(n - r)

T2 - - - =cln v 7i n - 1

4. Entropy change for an isoentropic process is zero while for an isothermal process, it is Q T

Entropy generation:

For a reversible process, entropy change (dS)R is equal to dQ and T for an irreversible process, entropy change (ds) is more than reversible process. Hence for an irreversible process dQ dS > T dQ (3.12) dS = + Sa T

where

Sa

It is clear from Eq. (3 .12) that Sa process.

=0

= entropy

generation

for a reversible process and Sa > 0 for an irreversible

Entropy increase: The principle of entropy increase states that the entropy of an isolated system will always increase, i.e. !).S ~ 0 for an isolated system. Our universe is also an isolated system in which all processes take place. If we take an individual system in the universe which receives heat dQ at temperature T from the surroundings (temperature T8 ),

Second Law of Thermodynamics

then change of entropy of the universe is: LiSuniverse

=

LiSsystem

+

LiQ

LiQ

T

½

-

LiSsurroundings

(Change of entropy for a system is positive as heat LiQ enters while change of entropy for the surrounding is negative as heat leaves).

LiSuniverse

= LiQ [ _!_ -

L'.iSuniverse

>

T

_!_] T,

As

As change of entropy dS

=

0

(di) R, heat interaction dQ = Tds for a reversible process.

Hence JTdS is the area under curve on a TS diagram which is equal to the heat interaction for the process. We know that in a cyclic process fdQ

= fdw

f TdS = f pdv

or

We know that in a reversible adiabatic process, heat interaction is zero, i.e. dQ = 0 or TdS = 0 or dS = 0 or S = constant. Hence a reversible adiabatic process is also called is entropic ( constant entropy) process. However, an is entropic process need not be an adiabatic process as the entropy during a particular process may be kept constant by heat transfer to or from the system. We have already seen that LiQ = TdS. Also dQ = cv dT for constant volume and dQ = cP dT for constant pressure.

or Similarly or

dQ

TdS

= Cv dT

T

dT dS

(for constant volume)

Cv dQ T cP

= TdS = Cp dT -

dT dS

(for constant pressure)

m

Thermal and Hydraulic Machines

Since cP > cv, the constant volume line has a greater slope than the constant pressure line on a TS diagram. Third law: The entropy of pure substance approaches zero at absolute zero temperature. This is called the third law of thermodynamics. Available energy: The second law of thermodynamics prohibits the complete conversion of low-grade energy into high-grade energy. A portion of energy that can be converted into work is called available exergy or exergy and the rest is unattainable energy or anergy. Therefore,

Energy = exergy + anergy At each temperature (1), heat has two portions viz., available heat energy capable of doing work and unavailable heat energy which is rejected to the surroundings (T0). dW 1J = - = 1 dQ

or

dW=

T0

1i

dQ(l - ~) To

= dQ - dQ -

7i

T,

dQ = dW + dQ....Q.

or

(3.13)

7i

T,

From Eq. (3.13), it is clear that heat rejection = ;

x dQ. However, I

dQ T,

= dS

Therefore, heat rejection = T0 dS Hence, unavailable energy

=

anergy

=

heat rejection

=

T 0 dS

As the surroundings temperature (T0 ) is known, change of entropy (dS) is a measure of unavailable heat energy. Entropy generation in a closed system is: (LiS)total = sgeneration = (LiS)system

S

generation

= m(S2

_

s,) +

+

(LiS)surroundings

Qsurroundings Tsurroundings

Entropy generation in an open system is: (LiS)total = sgeneration = (LiS)system

+

(LiS)surroundings

Second Law of Thermodynamics

Sgeneration

=

(S2 -

S,)

+

Qsurroundings

(So -

S,)

+ T

-

surroundings

For steady flow conditions, inside change of entropy (S2 entropy at the inlet and the outlet (S0 - S 1) will remain. Sgeneration

=

(So - S,)

+

S 1) is zero and only change of

-

Qsurroundings Tsurroundings

Irreversibility: Entropy change (dS) is a point function and hence it is a thermodynamic property. However, entropy generation is not a thermodynamic property as it is dependent on the path which the system has followed. Irreversibility is i

= wrev_

W

=

To

Sgeneration

where wrev is maximum work which can be extracted by a reversible engine. Exergy is maximum possible theoretical work which can be extracted from a system and environment before they come to dead state. At the dead state, they have energy but no exergy. Maximum work can be obtained when a process takes place in reversible manner. Almost all processes are irreversible. Exergy (availability) cannot be conserved like energy. Exergy gets destroyed due to irreversibility in process. Irreversibility is equal to the product of entropy generated and temperature. As irreversible processes are continuously increasing, it means unavailable energy is also gradually increasing. This is called law of degradation of energy. Exergy or availability of energy is a measure of the state of a system from the state of environment/surrounding. Therefore, exergy is an attribute of a system and environment together. If the environment is specified, a numerical value can be assigned to availability in terms of property value for the system only. Therefore, exergy can be regarded as the property of the system.

SOLVED PROBLEMS 1. A heat engine, a heat pump and a refrigerator are working between two reservoirs at

temperatures of 600 K and 300 K. Find the efficiency of heat energy, COP of the heat pump and the refrigerator. 1J of the heat engine

=

1-

Tz Ti

1-

300 600

1 2

m

Thermal and Hydraulic Machines

1i COP of the heat pump = - Ti - T2 600 600 - 300 = 600 = 2

300 COP of the refrigerator =

Ti - T2

300 600 - 300

300 = 1 300

2. A heat engine, a heat pump and a refrigerator reject 50 kJ, 125 kJ and 150 kJ heat, respectively. If each one get heat 100 kJ, find 17 of the heat engine and COP of the heat pump and the refrigerator. Heat engine:

Q1 = 100 kJ and Q2 = 50 kJ W = QI - Q2 = 100 - 50 = 50

50

W

11 Heat pump:

QI = 100

=

=

0.5

Q2 = 100 and Q1 = 125 kJ W = Q, - Q2 = 125 - 100 = 25 kJ

COP = Q, W

Refrigerator:

= 125 = 5 25

Q2 = 100 and Q1 = 150 kJ W = Q, - Q2 = 150 - 100 = 50 kJ

Q2 100 COP=-= = 2 W 50

3. A Carnot engine has an efficiency of 0.5. Find COP of a refrigerator working within the same temperature limit.

11 of the heat engine= 0.5 = 1 - T2

1i

or or

T

---1.

1i

= 0.5

T2 = 0.5 T 1

Second Law of Thermodynamics

COP of the refrigerator =

1i -

=

T2

0.5

7i -

-

=l

7i 0.5 7i

4. Determine the heat to be supplied to a Carnot engine operating between 800 and 400 K and producing 100 kJ of work.

17=l-T2

1i 1 -

w

11

400 = 0.5 800

= 0.5

Q.dd

100

= 0.5

Qadd

or

Qadd

=

100 = 200 kJ 0.5

5. A refrigerator operates on a reversed Carnot cycle between 900 and 300 K. If heat at the rate of 3 kJ/s is extracted from the low temperature space, find the power required to drive the refrigerator.

or

QI

900

3

300

Q1 = 9 kJ/s W= Q, - Q2

= 9 - 3 = 6 kJ/s = 6 kW Hence the power to derive the refrigerator is 6 kW. 6. A cold storage plant of 20 tonnes of refrigerator capacity operates between 200 and 300 K. Determine the power required to run the plant if plant has half COP of a Carnot cycle. (Take 1 tonne refrigeration = 3.5 kW.) COP of the Carnot cycle

=

T2 Ti -T2

200 = - - - =2 300- 200

-

Thermal and Hydraulic Machines

COP of the plant =

1

2 (COP)camot 1

-

2

X

2 = 1

Q2 = 20 tonnes of refrigeration =

COP

=

20 Q2

w

3.5

X

=

=

70

w

70 kW =

1

W= 70 kW

or

Hence, 70 kW power is required to run the plant. 7. 300 kJ/s of heat is supplied at a constant temperature of 500 K to a heat engine. The heat rejection takes place at 300 K. The following results are obtained: (i) 210 kJ (ii) 180 kJ (iii) 150 kJ Classify which of the result is reversible, irreversible or impossible result. (i)

I. dQ

300

210

500

300

---

T =

0.6 - 0.7

=

-0.1 < 0

Hence the cycle is irreversible. (ii)

I. dQ

300

180

500

300

---

T =

0.6 - 0.6

=

0

Hence the cycle is reversible. (iii)

I. dQ T

300 500

150 300

= 0.6 - 0.5 = 0.1 > 0

Hence the cycle is impossible. 8. A steam turbine plant is as shown in Figure 3.13, between working in temperature range 500 K and 300 K. Enthalpy at various states are as follows: (a) State 1 (b) State 2

= 700 kJ/kg = 2200 kJ/kg

Second Law of Thermodynamics

( c) State 3 = 1500 kJ/kg (d) State 4 = 500 kJ/kg

-

Condenser --- O:!

I---------~ @

Pump

FIGURE 3.13

Working of a steam turbine plant (Problem 8).

Verify the Clausius inequality.

Q,

= h2 - h, =

Q2

=

h4 - h3

2200 - 700

=

1500 kJ/kg

1500 - 500

=

1000 kJ/kg

=

1500

1000

500

300

-----

= 3 - 3.33 = -0.33 kJ/kg K < 0 Hence the Clausius inequality is proved.

9. Determine the change in entropy of the universe if a copper block of 1 kg at 150°C is placed in sea water at 25°C. The heat capacity of the copper block is 0.393 kJ/(kg K). LiSuniverse

=

LiSblock

+

LiSwater

T 1 = 150°C + 273 = 423 K T 2 = 25°C + 273 = 298 K LiSblock

= mC =

Heat given to water

Tz

In -

Ti

298 1 x 0.393 x l n 423

= -0.138 kJ/K = Heat lost by the block = +1 X 0.393 X 125

-

Thermal and Hydraulic Machines

= 49.13 kJ Q

L'.iSwater

=

49.13 = WS

T

0.165 kJ/K

=

2

L'.iSuniverse = -0.138 + 0.165 = 0.027 kJ/K = 27 J/K 10. A cold body at temperature T 1 is brought in contact with a high temperature reservoir

at temperature T2 • The body comes in equilibrium with reservoir at constant pressure. Considering the heat capacity of the body as C, show that entropy change of the universe can be given as

L'.iSuniverse = L'.iSreservoir + L'.iSbody The body will heat up to temperature of T2 from T 1• T2 LiSbody = mC ln -

1i

Heat lost by the reservoir = heat gained by body

= mC(T2 L'.iSreservoir =

T 1)

-

- mC(T2 - T1) T 2

LiS .

universe

=

m

C 1 T2 n T, I

mC(T2 - T1) T 2

11. A heat engine works between starting temperature limits of T 1 and T2 of two bodies. Working fluid flows at the rate of "m" kg/s and the specific heat at constant pressure

is cP" Determine the maximum obtainable work till the bodies attain the same temperature. Let final temperature = T 3 As the engine works, heat from the body is taken till its temperature falls to T 3 • ASbody!

L1

=

f

T3

1i

dT mcpT

= mep ln

T3

1i

Second Law of Thermodynamics

Similarly, the cold body will attain temperature T 3 from T2 •

J mcP -dTT T3

LiSbody 2 =

=

-

T3 mcP ln -

T2

T2

For the maximum work, the process must be reversible and entropy change is to be zero.

+

LiSbodyl

T3 mcP lnI;

LiSbody2

=0

T3

+ me ln- = 0 P

T2

T T ln --1. + ln 2 = 0 I; T2

or

T T ln --1. x ---1.. = 0 (= ln 1 ( ·: ln 1 = 0) I; T2

or

Therefore, or Maximum work = heat given by the first body - heat taken by the second body Maximum work = mcp(T1

-

T 3) - mcp(T3 - T 1)

= mcp(T1

-

2T3

= mcp(T1

-

2~I;T2

+ T2) + T2 )

12. Determine anergy or unavailable energy if a heat engine is working between the temperature limit of 1000 and 300 K. Heat delivers to engine is 1000 J and work output is 400 J. LiS

= QI + I;

Here

or

Q2 T2

= heat supplied to the engine Q2 = heat rejected by the engine QI - Q2 = W Q2 = QI - W Q1

=

1000 - 400

= 600

J

a

Thermal and Hydraulic Machines

-1000 1000

600 300

--+-

!).S

1 J/K Unavailable energy = T2 x !).S

= 300

X

= 300 J 13. A closed system executed a reversible cycle 1-2-3-4-5-6-1 consisting of six processes (Figure 3.14). During processes 1-2 and 3-4, the system receives 1000 kJ and 800 kJ of heat respectively at constant temperature of 500 K and 400 K, respectively. Processes 2-3 and 4-5 are adiabatic expansions in which the temperature is reduced from 500 K to 400 K and from 400 K to 300 K, respectively. During process 5-6, the system rejects heat at a temperature of 300 K. Process 6-1 is an adiabatic compression process. Determine the work done by the system during the cycle and thermal efficiency of the cycle. 012 =1000 kJ

T

2

500 K

034 =800 kJ

400 K

4

300 K

6

5

056

s FIGURE 3.14

A closed system with six processes (Problem 13).

Process 1-2:

Q12

= T1(S2 - S1)

1000 = 500(S2 or

S2

-

Process 3-4:

or

Process 5-6:

S1

=

S 1)

-

2 kJ/K

= T3 (S4 - S3) 800 = 400(S4 - S3) Q34

S4

-

S3

Qs6

=

2 kJ/K

= T6(Ss = 300[(S2

= 300

X

4

S6) -

S 1) + (S4

=

-

S 3)]

1200 kJ

Heat added = Q 12 + Q 34 = 1000 + 800 = 1800 kJ Heat rejected = Q 56 = 1200 kJ

Second Law of Thermodynamics

W =

1111

Qadd - Qrej

1800 - 1200 = 600 kJ Therefore,

W 600 1 --=--=-

11

Qadd

1800

3

= 33.34% 14. A reversible heat engine operated between two reservoirs at temperatures of 600°C and 40°C (Figure 3 .15). The engine drives a reversible refrigerator which operates between reservoirs at temperature of 40°C and -20°C. The heat transfer to the heat engine is 2000 kJ and net work output of the combined engine-refrigerator plant is 300 kJ. Evaluate the heat transfer to the refrigerator and the net heat transfer to the reservoir at 40°C. (UPTU: Dec. 2005) 600°C

w,

Heat engine

,-......,-~• W3 = W1 = W2= = 360 kJ

-20°C

FIGURE 3.15

T1

A reversible heat engine system (Problem 14).

= 600 + 273 = 873 K

T2 = 40 + 273 = 313 K = T 3

T4 = -20

1JHE = =

Wi Q,

w,

+ 273

=

253 K

I; -Tz

i; 873-313 560 = 873 873

= 1JHE = 0.641 = 0.641 X 2000 = 1283 kJ

llfl

Thermal and Hydraulic Machines

W2 = W1

Q2

=

W3 = 1283 - 360 = 923 kJ

-

w,

Q, -

2000 - 1283

=

717 kJ

=

5535 kJ

T4 253 -~-=---T3 - T4 313 - 253

CO Pref

253

= -

COP

=

Q4

=

Q3

=

=

4.22

=

4.22

X

4.22

60

Q4 W2 923

=

3895 kJ

W2 + Q4 = 923 + 3895 4818 kJ

Heat transfer to the refrigerator (Q4 ) = 3895 kJ Heat tranfer to the reservoir at 40°C

=

Q3 + Q2

15. Obtain the COP of the composite refrigerator system in which two reversible refrigerators A and B are arranged in series in terms of the COP of refrigerator A and COP of refrigerator B only (Figure 3.16). (UPTU: 2003-2004 and May 2008)

Refrigeration

Refrigeration

T3

FIGURE 3.16

Composite refrigerator system (Problem 15).

Let COP of the composite system = C, COP of refrigerator = A and COP of refrigerator = B.

C =

A

B

1111

Second Law of Thermodynamics

Q,B - Q2B

or

Q1A

= (~ +1) Q2B

QIB

=

1

B Q2B

= (: +1) Q2B

Now Q2A = Q,B Therefore,

or

or or

1

(1 +A) (1 + B)

C

AB

- +1

C

or or

l+AB+A+B -AB AB

1

or

C=

AB A+B+l (COP) A (COP) A

(COP)composite

= (COP) A + (COP) B+ 1

16. Which is more effective way to increase the efficiency of a reversible heat engine (i) to increase the source temperature T 1 while sink temperature T2 kept constant or (ii) to decrease the sink temperature by the same amount while source temperature is constant. (UPTU: 2006-2007) T2

1]=1-Tj

Case 1:

Reduce the temperature T2 by dT: '11

dT T2_- _ -1 __ _ 1j

•mow -

=

(l _

= 11 + -

T2 ) + ..!!!._ T 1j

dT

1j

llfl Case 2:

Thermal and Hydraulic Machines

Increase the temperature T 1 by dT n

-1-

'!new -

Ti

T2

+ dT

=1 -

= 1- T2 (1 -

Ti

dT]

Ti

As

Hence efficiency increases more when temperature of source is increased as compared to lowering temperature.

17. A metal block of 5 kg and 200°C is cooled in a surrounding of air which is at 30°C. If the specific heat of metal is 0.4 kJ/kg K, calculate the following to (a) entropy change of block and (b) entropy change of the surroundings and universe. (UPTU: 2006-2007)

(.::iS)block

=m

=5

X

X

CP

X

0.4

X

303 log 473

= -0.89 kJ/K Heat absorbed by the atmosphere is

Q=

m

x

cP

=5

X

0.4

x (T 1

= 340 kW

X

-

T2)

(473 - 303)

1111

Second Law of Thermodynamics

Now for surroundings Q

340 T2 - 303

---

( LiS)surroundings

1.12 kJ/K (LiS)universe

= (L'.iS)block + (LiS)surroundings = -0.89 + 1.12 = 0.23 kJ/K

18. A cyclic heat engine operator between a source temperature of 800°C and a sink temperature of 30°C. What is the least rate of heat rejection per kW net output of the engine? (UPTU: 2007-2008)

1017

= 0.718

w

0.718

given W

=

1 kW

=

10 3 W

Q, 103

or

1.4

0.718

X

103

1.4 kW

Q2 = Q, - W =

1.4 - 1

=

0.4 kW

19. A fluid undergoes a reversible adiabatic compression from 0.5 MPa, 0.02 m 3 to 0.05 m 3 according to the law pv 13 = constant. Determine the change in enthalpy, internal energy, entropy and heat of work transfer during the process. (UPTU: 2006-2007)

or

Thermal and Hydraulic Machines

P2

= 0.5

X

0.02 )1.3 (-0.05

=0.152MPa (a) Work for adiabatic process

W=

=

J\V1 -

P2V2

1

V -

1.52 X 105 X 0.05 - 5 X 105 X 0.02 1.3-1

= - 8 kJ Negative sign means that work is done on the system. (b) Change in internal energy Q = llU + W Q = 0. Therefore, For adiabatic process llU = -W

= 8 kJ (c) Change in enthalpy !lh = llU +

(P2V2 - P1v1)

= 8 kJ + 2.4 kJ = 10.4 kJ (d) Change in entropy

Q = 0 and /lS =

Q

T

!lS = 0.

20. At a place where surroundings are at 1 bar 27°C, a closed rigid thermally insulated tank contains 2 kg air at 2 bar, 27°C. This air is then churned for a while, by a paddle wheel connected to an external motor. If it is given that the irreversibility of the process is 100 kJ, find the final temperature, and the increase in availability of air. Assume for air Cv = 6.718 kJ/kg K. (GATE: 1997)

Surroundings

Insulated Q = 0

Second Law of Thermodynamics

1111

No heat flow from system and entropy change of surroundings is (L'.iS)surroundings

Q

= - =0 To

Irreversibility I= T0(LiSsys + LiSsur)

= To or

X

LiSsys

I

100

T0

300

- - -

(LiS)sys

= 0.33 kJ/K

But we know

0.33 = 2

X

T2

0.718 log 300

T2 = 3783 K

Increase of availability is LiEavail = m(LiE - ToliS)

3783

= 2[0.718(3783 - 300) - T0 log - ] 300

= 12.54 kJ 21. An iron cube at a temperature of 400°C is dropped into an insulated bath containing 10 kg water at 25°C. The water finally reaches a temperature of 50°C at steady state. Given that the specific heat of water is equal to 4186 J/kg K. Find the entropy change for the iron cube and the water. Is the process is irreversible? If so, why? (GATE: 1996) Entropy change of cube is

mccpc

T3

= mcCpc log Ti Heat lost = Heat gained X (T, - T3) = mw X Cpw(T3 (LiS)cube

- T2)

where T 1 = initial temperature of cube, T2 = initial temperature of water and T 3 = final temperature (me x cpc) (400 - 50) = 10 x 4186 x (50 - 25)

a

Thermal and Hydraulic Machines

10

X

4186

X

25

350 (LiS)cube

= 2990 J/K

50 + 273

= 2990 log 400 + 273 = -2990 X 0.734 = -2195 kJ/K

(LiS)water

=

T3

mw Cpw

log -T 2

10

X

4186 log

323 298

= 3372 kJ/K (LiS)iotal

= (LiS)cube + (LiS)water = -2195 + 3372 = 1177 kJ/K

Since (LiS) 101a1 > 0, the process is irreversible.

22. A certain mass of a pure substance undergoes an irreversible process from state 1 to state 2, the path of the process being a straight line on TS diagram (Figure 3.17). Calculate the work interaction. Some properties at the initial and final states are: T 1 = 330 K, T 2 = 440 K, U 1 = 170 kJ, U2 = 190 kJ, H 1 = 220 kJ, H 2 = 24 kJ and S 1 = 0.23 kJ/K and S2 = 0.3 kJ/K where T, U, Hand S represent temperature, internal energy, enthalpy and entropy, respectively. (GATE: 2000)

~

::::,

'§ T1

t--------41F--------;

CD

a.

E

i!

S1 Entropy_

FIGURE 3.17

Q 1_2

= Area

S2

TS diagram (Problem 22).

under line 1-2

Second Law of Thermodynamics

= 300(0.3 - 0.23) +

1

2 (0.3

1111

- 0.23) (440 - 330)

= 26.95 kJ ~U,_2 = U2 - U, =

190 - 170

=

20 kJ

Q,_2 = ~U,_2 + W,_2

or

W,_2 = Q,_2 - ~U,_2

= 26.95 - 20 = 6.95 kJ 23. One kilomole of an ideal gas is throttled from an initial pressure of 0.5 mPa to 0.1 mPa. The initial temperature is 300 K. The entropy change of universe is (a) 13.38 kJ/K (b) 401.3 kJ/K (c) 0.0446 kJ/K (d) - 0.0446 kJ/K

(GATE: 1995) S2

-

T2 P2 S 1 = c log - - R log -

Pi

i;

p

-

P2

= -R log -

Pi

(~S)universe

=

0.5 +8.314 log O.l

=

(~S)sys

+

=

(~S)sys

+0

=

+13.38 kJ/K

(~S)sur

= 13.38 Option (a) is correct. 24. For two cycles coupled in series, the top cycle has efficiency of 30% and the bottom cycle has efficiency of 20% (Figure 3.18). The overall combined cycle efficiency is (a) 50% (b) 44% (c) 38% (d) 55%

(GATE: 1996)

or

a

Thermal and Hydraulic Machines

o,

Heat exchanger

//////////////////////

FIGURE 3.18

Low temperature

Two cycles coupled in series (Problem 24).

Q3 - = 0.8 or Q3 = 0.8, Q2 = 0.8 x 0.7, Q 1 = 0.56 Q 1 Q2

or

Combined

=

Q,

= 0.44 or 44% Option (b) is correct.

25. A cyclic heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%, the heat rejected per cycle is (a) 16f kJ

(b) 33½ kJ

(c) 37½ kJ

1J =0.75

W

50

Q,

Q,

= - =-

50 Qi

=

0.75 =

50

X

3

4

(d) 66f kJ

Second Law of Thermodynamics

1111

= 200 =662/3 kJ 3 Q2 = Q, - w = 200 -50 3 200-150 3 = 50 =163. 3 3 Option (a) is correct. 26. The operating temperature of a cold storage is -2°C. Heat leakage from the surrounding is 30 kW for the ambient temperature of 40°C. The actual COP of the refrigeration plant used is one fourth that of an ideal plant working between the same temperature. The power required to drive the plant is (a) 1.86 kW (b) 7.72 kW (c) 7.44 kW (d) 18.6 kW

271 271 -- -- . (COP) theore11ca1 - 313 _ 271 42 1 271 Q2 30 (COP)actual = - X =- =4 42 W W or

30x4x42 W=---=18.6kW 271

Option (d) is correct. 27. If a heat engine gives an output of 3 kW when the input is 10,000 J/s, then the thermal efficiency of engine will be (a) 20% (b) 30% (c) 70% (d) 76.7% (Civil Services: 1995) 3xl03 1J = Q = 10,000 W

= 0.3 or 30% Option (b) is correct. 28. A heat engine is supplied with 250 kJ/c of heat at a constant fixed temperature of 227°C. The heat is rejected at 27°C. The cycle is reversible if the amount of heat rejected is (a) 273 kJ/s (b) 200 kJ/s (c) 180 kJ/s (d) 150 kJ/s.

(Civil Services: 1995)

Thermal and Hydraulic Machines

300 =1--=0.4 500

0.4 = W QI

= QI -

Q2 =0.6 =

0.6

= 1-

Q2

QI

but Q 1 = 250 kJ/s

QI Q2

Q2

QI

X

250

=

150 kJ

Option (d) is correct. 29. The efficiency of a reversible cycle process undergone by a substance as shown in TS diagram (Figure 3.19) is (a) 0.4 (b) 0.55 (c) 0.66 (d) 0.80.

(Civil Services: 1994)

1500 ~

)

ci. 1000

E

~

500

2

3

4

5

Entropy, kJ/K

FIGURE 3.19

TS diagram (Problem 29).

The substance is operating between 1500 K and 500 K. And the efficiency of a reversible cycle depends on only temperatures. Tz 1J = 1 - -=1 -

Ti

2 = 1500 3 500

= 0.6 Option (b) is correct. 30. Given that path 1-2-3, a system absorbs 100 kJ heat and does 60 kJ work while along the path 1--4-3 it does 20 kJ work (Figure 3.20). The heat absorbed during cycle 1-2-3: (b) -80 kJ (c) -40 kJ (d) 60 kJ. (a) -140 kJ

(Civil Services: 1994)

m

Second Law of Thermodynamics p 3

V

PV diagram (Problem 30).

FIGURE 3.20 Q1-2-3

= 1).U13 + w,-2-3

100 = LiU13 + 60

LiU13

=

Qt-4-3

= LiU13 + W,_4_3

Q,_4_3 =

I 00 - 60

40 + 20

=

=

40

60 kJ

Option ( d) is correct. 31. The block diagrams of two systems are given in Figure 3 .21. Giving proper reasons indicate: (i) name of the system (i.e. HE, RE or HP) and (ii) type of cycle is possible or impossible and reversible or irreversible. (UPTU: May 2008)

~

~

~ 1000 kJ

~ 1000 kJ

700 kJ

600 kJ

30°c

27°C

FIGURE 3.21

Block diagrams of two systems (Problem 31 ).

As machines are taking heat from one reservoir and rejecting part of heat to the second reservoir with output of work, machines are engines in both cases.

Machine 1

Machine 2

Q,

= 1000 kJ T 1 = 500 K

Q,

Q2 = 1000 - 700 = 300 kJ

Q2 = 1000 - 600 = 400 kJ

T2 = 27 + 273 = 300

T2 = 30 + 273 = 303

= 1000 kJ T 1 = 900 K

-

Thermal and Hydraulic Machines

Applying Clausius inequality:

300 300

Applying Clausius inequality: Q2 1000 ---- -

T2

-

= 2-1

900

1.11

= 1> 0

400 303 1.32

= -0.21 < 0

Hence it is impossible process.

Hence process is possible but irreversible.

OBJECTIVE TYPE QUESTIONS State True or False 1.

The first law of thermodynamics cannot explain non-occurrence of certain processes (True/False) as well as the direction of processes.

2.

Heat and work are energy, and they have no difference in grade.

3.

High and low-grade energy is classified according to potential energy. (True/False)

4.

All spontaneous processes proceed in one direction.

5.

A spontaneous process cannot be made to proceed in the reverse direction without aid. (True/False)

6.

The second law of thermodynamics rules out the possibility of a spontaneous process proceeding in the reverse direction. (True/False)

(True/False)

(True I False)

7. A thermal reservoir at a low temperature to which heat is rejected is called source. (True/False) 8.

A sink is a thermal reservoir to which a heat engine rejects heat.

(True/False)

9.

A thermal power plant is a heat engine.

(True/False)

10. A heat engine is a device which receives heat from a source and rejects heat to a

sink while undergoing a cyclic process and performs work.

(True/False)

11. A heat reservoir has large heat capacity and can receive or reject heat without

change of temperature. 12. The surrounding is generally used as a sink for a heat engine.

(True/False) (True /False)

13. The surrounding is used as a low temperature body in a heat pump.

(True /False) 14. Heat supplied to room is desired effect for a heat pump.

(True/False)

15. Heat rejected to the surrounding is the desired effect for a refrigerator.

(True /False) 16. COP of a refrigerator is greater than COP of a heat pump working within the same

temperature limits.

(True/False)

m

Second Law of Thermodynamics

17. Performance of a heat engine can be given by COP.

(True/False)

18. A heat pump is required to extract heat from room to bring down the temperature.

(True/False) 19. A heat pump is also used during the warm season for heating a room.

(True/False) 20. A heat pump is used in winter for heating room.

(True/False)

21. The main purpose of a refrigerator is to extract heat from cold storage space.

(True/False) 22. The surroundings are a high temperature reservoir for a domestic refrigerator.

(True/False) 23. The thermal efficiency of a heat engine is the ratio of work output to the heat added.

(True/False) 24. The second law gives the possibility of a spontaneous process to be reversed itself

unaided.

(True /False)

25. The second law of thermodynamics prohibits the possibility of designing of heat

engine with 100% efficiency.

(True/False)

26. In an isothermal process all heat is converted into work.

(True/False)

27. The difference of COP of a heat pump and a refrigerator is unity.

(True/False)

28. The Kelvin-Planck statement permits the possibility of a device producing work by

drawing heat from a single source.

(True/False)

29. A device can work as per the Clausius statement but violates the Kelvin-Planck

statement. 30. The Clausius and the Kelvin-Planck statements are equivalent.

(True /False) (True /False)

31. The Clausius statement permits heat to flow from a low temperature body to a high

temperature body without interaction of any other energy. 32. All spontaneous processes are irreversible.

(True/False) (True/False)

33. In a reversible cycle, all processes constituting the cycle are reversible.

(True/False) 34. A process which can proceed forward or reverse direction without violating the

second law is a reversible process.

(True/False)

35. Violation of the Kelvin-Planck statement does not lead to violation of the Clausius

statement of the second law of thermodynamics.

(True/False)

36. A Carnot cycle must have more than one reservoir.

(True IF alse)

37. A Carnot cycle has two adiabatic and two isobaric processes.

(True IF alse)

38. A Carnot cycle is a hypothetical device and consists of only a reversible process.

(True /False) 39. A Carnot heat pump and a refrigerator work on a reversed Carnot cycle or refrigeration

cycle.

(True/False)

-

Thermal and Hydraulic Machines

40. The COP of a Carnot heat pump and refrigerator is the highest.

(True/False)

41. The COP of a normal refrigerator is higher than that of a Carnot refrigerator. (True/False) 42. A Carnot refrigerator will require minimum energy for desired effect. (True/False) 43. The performance of other engines is compared with a Carnot heat engine as it is used as standard of perfection. (True/False) 44. The performance of an actual refrigerator and a heat pump cannot be compared with a Carnot heat pump and a refrigerator as standard of performance since these are hypothetical. (True/False) 45. As a heat pump and a refrigerator work on a reversed Carnot cycle, they cannot be used as replacement for each other with some modification. (True/False) 46. Desired effect for a heat pump will be more than that of a refrigerator as work input is also added up with heat removed from a cold body while rejecting to hot body. (True/False) 47. The Carnot engine has greater efficiency than any reversible engine between given temperature limits. (True/False) 48. The Carnot engine can have different efficiencies depending upon temperature limits. (True/False) 49. The Carnot engine can have different efficiencies depending upon working medium. (True/False)

50. All reversible engines have the same efficiency working between the same temperature limits. (True /False) 51. The Carnot engine will have greater efficiency with increase of source temperature. (True/False) 52. The Carnot engine will have increased efficiency with lowering of sink temperature. (True/False) 53. The efficiency of the Carnot engine depends upon source temperature and sink can have any lower temperatures. (True/False) 54. Lowering sink temperature will be a more effective way to increase the efficiency of the Carnot engine as compared to increasing source temperature. (True/False) 55. The COP of a refrigerator is the ratio of heat supplied to a hot body to work input. (True /False) 56. The COP of a heat pump is the ratio of heat extracted from a cold body to work (True/False) input. 57. A Carnot pump with COP = 4 is reversed to work as a heat engine. The efficiency of the heat engine will be 25%. (True/False) 58. If the COP of a refrigerator is 4, then the COP of a heat pump will be 5. (True /False)

m

Second Law of Thermodynamics

59. If a refrigerator with COP= 5 is given 5 kW energy, then heat extracted from cold

space will be 25 kW. (True/False) 60. The Clausius inequality is not based on the second law of thermodynamics. (True/False) 61. Considering temperatures T 1 and T 2 vary but the ratio of T 1 to T 2 remains the same. The efficiency of a Carnot engine will increase with lowering of T 2 (sink temperature).

(True/False) 62. The COP of a refrigerator will slightly fall as surroundings temperature increases in day time. (True/False) 63. The difference of the COP of a heat pump and a refrigerator will remain unchanged

even when temperatures of source and sink are changed.

(True/False)

64. The efficiency of a Carnot engine is 33.34%. If its cycle is reversed to work as

a heat pump, then the COP of heat pump will be 3. 65.

j Tds = j Pdv

(True/False)

for a cyclic process.

(True/False) (True/False)

66. Entropy is not conserved. 67. A reversible adiabatic process differs from an isentropic process.

(True/False) 68. Spontaneous processes occur such as to decrease the entropy of the universe. (True/False) 69. The system having lower entropy is more prone to a spontaneous process.

(True/False) 70. For a reversible engine Q1 + Q2 = 0, where T 1 and T2 are source and sink temperatures I; T2 and Q 1 and Q 2 are heat added and rejected. (True/False) 71. For an irreversible engine taking Q 1 heat from T 1 source and rejecting Q 2 heat from T sink then Q1 2 ' T.l

+ Q2 < 0.

£dTQ 72. Whenever a system undergoes a cyclic process y

73.

j d;

has the same value for all reversible processes. is an exact differential.

7 4. ( dQ ) T

(True/False)

T2

~

0.

(True/False) (True/False) (True/False)

reversible

75. dQ = dS is true for all reversible processes. T

(True/False)

76. Change of entropy is a state function and does not depend upon the path.

(True/False) 77.

dQ

T

< dS for an irreversible process.

(True/False)

-

Thermal and Hydraulic Machines

78. Entropy change during melting is equal to the latent heat of freezing divided by 273 K. (True/False)

79. Entropy change during evaporation is equal to the latent heat of vaporization divided by 373 K. (True/False) 80. Entropy changes when a hot body is dropped in cold liquid.

(True/False)

81. If two gases at the same temperature and pressure are mixed, entropy does not change. (True /False) 82. Entropy change between two states remains constant irrespective of the path (both reversible and irreversible). (True/False) 83. If a refrigerator is working in an isolated room where temperature is increasing, (True/False) then the COP of the refrigerator will decrease. 84. If two refrigerators with each having COP = 4, works in parallel, then the COP of such arrangement will remain the same. (True/False) 85. If a heat pump with COP = 5 and a refrigerator with COP = 4 are working in parallel, then the COP of both working as refrigerator will have COP = 4.

(True/False) 86. If an engine is working between 800 and 400 K and supplied heat 100 kJ, then unavailable energy will be 40 kJ. (True/False) 87. If a system is working within the temperature limit of 600 and 300 K and change of entropy is 1.5, then unavailable energy will be 400 kJ. (True/False) 88. If a system has energy 400kJ.

=

600 kJ and anergy

=

200 kJ, then it has an exergy of (True /False)

89. If the difference of unavailable energy between 300 K and 200 K is 100 J, then (True/False) change of entropy of the system will be unity. 90. Degree of irreversibility of a system will be 450 J, in case surroundings temperature (True/False) is 300 K and entropy generation is 1.5 J/K. 91. If work output of an engine = 0.3Q + a where Q constant, then the efficiency of engine will be 30%.

=

heat supplied and a is a (True/False)

Multiple Choice Questions

1. The second law of thermodynamics defines (a) heat (b) work ( c) enthalpy

(d) entropy

2. For a reversible adiabatic process, change of entropy is (a) minimum (b) zero (c) positive

( d) negative

3. The net entropy change in any irreversible process is (a) positive (b) zero ( c) infinite

(d) unity

4. For any reversible process, the change in entropy of the system and surroundings is (a) unity (b) negative (c) positive (d) zero

m

Second Law of Thermodynamics

5. Isentropic flow is

(a) irreversible adiabatic

(b) reversible adiabatic

( c) isothermal

( d) isoenthalpic

6. A Carnot cycle has two adiabatic and two other processes which are ( a) isochoric (b) isothermal ( c) isobaric ( d) polytropic

7. The efficiency of a Carnot cycle depends upon (a) working substance ( c) temperature of the source only

(b) temperature of the sink only ( d) temperatures of the source and sink

8. The efficiency of a Carnot cycle depends upon

(a) work output only (c) amount of heat rejected

(b) amount of temperature rejected (d) ratio of work output to heat added

9. The efficiency of a Carnot engine is (d) 7; -Tz

i; 10. The efficiency of a Carnot cycle can be 100% only if sink temperature can be

(a) 0°C

(b) 0°F

(c) -200°C

(d) 0 K

11. In a reversible cycle, the entropy of a system will

(a) increase (c) remain constant

(b) decrease (d) depend on working substance

12. The Kelvin-Planck statement of the second law of thermodynamics is about (b) conservation of heat (a) conservation of mass ( d) conversion of work into heat ( c) conversion of heat into work 13. In the device shown in the figure

(a) PPM I violating the second law of thermodynamics (b) PPM II violating the Kelvin-Planck statement ( c) PPM II violating the Clausius statement 14. In the device shown in the figure (a) PPM I violating the second law of thermodynamics

(b) PPM II isolating the Kelvin-Planck statement

-

Thermal and Hydraulic Machines

( c) PPM II isolating the Clausius statement

15. The Clausius statement is that (a) PMM II can be devised (b) PMM I can be devised ( c) heat cannot flow from cold body to hot body unaided 16. The Kelvin-Planck statement prohibits work output from a heat engine if the (a) Difference between source and sink temperatures < 200°C (b) Difference between source and sink temperatures < 300°C ( c) Heat interaction is there with the source only 17. A heat engine is a device that (a) Generates heat from work (b) Converts full heat into work ( c) Converts a portion of heat into work and rejects the remaining 18. A heat pump is a device that (a) Pumps heat out of a room (b) Supplies heat to a room extracting from surroundings with work supplied ( c) Delivers heat to a room from surroundings without assistance T,

HE

T1

T1

HP

Refrigerator

T2

T2

T2

Figure (a)

Figure (b)

Figure (c)

19. The efficiency of heat engine shown in Figure (a) is (a)

Q, Q, -Q2

(b) Q,

+ Q2 Q,

20. The efficiency of the heat engine shown in Figure (a) can also be given as (a) _T._,:z; - Tz

(b)

:z; -Tz :z;

(c)

:z; + Tz :z;

21. The COP of the heat pump shown in Figure (b) is

(a)

Q, -Q2 Q,

(b)

Q2 Q, -Q2

(c)

Q, Q, -Q2

m

Second Law of Thermodynamics

22. The COP of the heat pump shown in Figure (b) is (a)

___2i_

(b)

~-½

__!i_

~-½

23. The COP of the refrigerator shown in Figure (c) is

24. The COP of the refrigerator in Figure (c) can be given as (b) ~ -T2 ~

25. A heat engine will have greater efficiency in case source temperature (T1) or sink temperature (T2) is varied (a) T 1 + ~T

(b) T 1

-

~T

(c) T2 + ~T

(d) T2

-

~T

26. Heat engines A and B are working on the Carnot cycle and have air and steam as working fluid, respectively. Choose the correct answer. (c) 11A = 11s

(a) 11A > 11s

27. The efficiency of a heat engine having half the efficiency of a Carnot engine in temperature limits of 600 and 300 K is (a) 0.75

(b) 0.5

(c) 0.25

28. The efficiency of a Carnot engine as compared to other engines will be (a) minimum

(b) maximum

(c) equal

29. A refrigerator has half the COP of a Carnot refrigerator in temperature limit of 350 and 300 K. The COP of the refrigerator is (a) 6

(b) 3

(c) 4

30. A heat pump has half the COP of a Carnot heat pump operating between 360 and 300 K. The COP of the heat pump is (a) 3

(b) 4

(c) 6

31. The COP of a Carnot heat pump is 5. The COP of a Carnot refrigerator within the same temperature limit is (a) 6

(b) 4

(c) 5

32. The efficiency of a Carnot engine is 25%. What is the COP of a heat pump working between the same temperature limit? (a) 3

(b) 4

(c) 5

33. A reversible process can be replaced with a series of (a) reversible adiabatic and isothermal processes (b) reversible isobaric and isothermal processes ( c) reversible isochoric and isothermal processes

m

Thermal and Hydraulic Machines

34. Two bodies of equal mass and material at T 1 and T2 (T 1 > T2) are used as a source and a sink for a Carnot heat engine. If T3 is final temperature then

35. A Carnot engine has Q 1 = 1000 kJ, T 1 = 1000 Kand T2 = 300 K. Q2 and W will be (a) 300 kJ and 700 kJ (b) 400 kJ and 600 kJ (c) 200 kJ and 800 kJ

36. A Carnot engine takes 800 kJ and 700 kJ heat from two sources at 800 and 700 K, respectively. If the sink is at 300 K, then heat rejected and efficiency are (a) 400 kJ and 50%

(b) 600 kJ and 60%

(c) 500 kJ and 55%

37. A Carnot engine (1] = 0.5) runs a Carnot refrigerator having COP = 5. The ratio of the heat added to a heat engine to the heat extracted from cold space by the refrigerator is (a) 0.1

(b) 0.5

(c) 0.4

38. If a Carnot heat pump and a refrigerator are working within the same temperature limits, then (a) COPHP = COPRef (b) COPRef > COPHP (c) COPHP > COPRef

39. A Carnot heat pump and a refrigerator are working within the same temperature limit. If the COP of the heat pump is 6, then the COP of the refrigerator is (a) 5

(b) 7

(c) 4

40. Energy, exergy and anergy are related as under (a) Energy = Exergy + Anergy (b) Energy = Exergy - Anergy ( c) Exergy = Energy + Anergy

41. For an adiabatic process, the change of entropy is (a)!!,.S~0

(b)!!,.S::;0

(c)!!,.S=0

42. When a system undergoes a process, the entropy generation is (a) Sc

=0

(b) Sc

~

0

(c) Sc < 0

43. During throttling, the change of properties is (a) dh = 0, dS > 0

(b) dh = 1, dS > 0

( c) dh = 0, dS = 0

44. When the entropies at the inlet and the outlet of an adiabatic turbine are S I and S 2 , then (a) S 2 = S 1 (b) S2 > S 1 (c) S 1 > S 2

m

Second Law of Thermodynamics

45. Combining du (a) du

=

=

dQ - dW and ds

Tds + pdv

=

(b) du

(di) R , the relation is =

pdv - Tds

(c) du

=

Tds - pdv

46. What will be the change of entropy of a reservoir if 1000 kJ of heat is supplied to

it at 200 K? (a) 3 kJ/K

(b) 4 kJ/K

(c) 5 kJ/K

47. What will be the degree of irreversibility of a system if Sg = 5 kJ/K and surroundings

at 300 K? (a) 2000 kJ

(b) 1500 kJ

(c) 1200 kJ

48. If two Carnot heat engines (work output each = co) work in parallel within the same

temperature limit, then total work will be (a) co (b) 2m

(c) m/2

49. If two Carnot heat engines are used in series with the temperatures of the source and the sink as T 1 and T2 (individually work output of each is co for T 1 and T2), then

output in series for each engine is: (b) 2m

(a) co

(c) m/2

50. If 5 Carnot heat engines are operating in series within the source ( 1000 K) and the

sink (500 K). The temperature difference across each heat engine having equal work output is (a) 500 K (b) 100 K (c) 200 K 51. Four Carnot heat engines are operating in series in between 800 and 400 K. Temperature gradient is the same across each engine. If the first engine takes 80 J heat, then work

output for each engine and total output are (b) 20 J and 30 J (a) 10 J and 40 J

(c) 9 J and 36 J

52. An inventer claims to develop a device which takes a stream of fluid (entropy =

4 kJ/K) and gives out two streams of fluid (entropy K). Comment on his claim. (a) Claim is feasible (b) Claim is unfeasible ( c) Insufficient data to comment

=

2 kJ/K and entropy

=

3 kJ/

53. An inventor claims to develop a magic tube which takes two streams of fluid (S 1 = 3 and S 2 = 5 kJ/K) and merges the streams to one stream (S3 = 6 kg/K). Comment

on his claim. (a) Claim is feasible (b) Claim is unfeasible ( c) Insufficient data to comment 54. The system at state A has entropy = 4 kJ/K and at state B has entropy = 3 kJ/K. Comment on the feasibility of the process from state A to state B.

(a) Spontaneous

(b) Impossible

( c) Slow

1111

Thermal and Hydraulic Machines

55. Two Carnot heat engines are working in series (work output is equal) within the temperature limit of 600 and 200 K. If heat supplied by the source (600 K) is 30 J,

then the temperature of the reservoir between engines and heat rejected at the sink (200 K) are (a) 300 K and 25 J (b) 400 Kand 10 J (c) 425 K and 30 J 56. A Carnot heat engine is driving a refrigerator with a source = 800 K and a sink = 400 K. If the heat engine takes 40 J from the source and the refrigerator extracts

5 J from the sink, the net work output will be (a) 20 J (b) 15 J

( C) 10 J

57. If a system is working between 800 and 400 K and supplied heat = 200 kJ, the

unavailable energy is (a) 150 kJ

(b) 125 kJ

(c) 100 kJ

58. If a system works between 600 and 300 K and entropy change is 1.2 kJ/K, then

unavailable energy is (a) 360 kJ

(b) 400 kJ

(c) 300 kJ

59. If degree of irreversibility is 600 J at temperature of surroundings (300 K). The

entropy generation is (a) 2 kJ/K

(b) 2.5 kJ/K

(c) 3 kJ/K

Fill in the Blanks 1. Work is

(a) high

2. Heat is (a) high

grade energy. (b) low grade energy. (b) low

3. All spontaneous processes move in

(a) one 4. A source is a reservoir at

directions.

(b) both

- - - temperature.

(a) high

(b) low

5. A sink is a reservoir at - - - temperature.

(a) high

(b) low

6. Surroundings are an ideal _ _

(a) source 7.

(b) sink

fdS for a system is _ __ (a) zero

(b) > 0

Second Law of Thermodynamics

8.

j di

m

for a system is _ _ __ (b) < 0

(a) zero 9. If Q1

= heat supplied at T 1 and Q2 is equal to _ _ __

=

heat rejected at T2 for a Carnot cycle, then Q1

10. The efficiency of a Carnot cycle with source (a) 1 - -~

(b) 1 -

=

T 1 and sink = T2 is

T2 ~

T2

11. A process which can proceed in a forward or reverse direction without violating the

second law is _ _ _ _ process. (a) homogeneous

(b) reversible

12. As per the Kelvin-Planck statement, a heat engine has to interact with _ _ __

reservoir(s). (a) one

(b) atleast two

13. If a hypothetical heat engine can produce work interacting with one reservoir, then

its efficiency is _ _ __ (a) infinity

(b) 100%

14. If the temperature of a source is T 1 and that of a sink is T2 , then the COP of a heat

engine is _ _ __ (b)

15. If the temperature of a source is T 1 and that of a sink is T2 , then the COP of a

refrigerator will be _ _ __ (b)

16. If heat interaction

= LiQ at temperature

T, then entropy change for a reversible

process is _ _ __ (a) T x LiQ

(b) L1Q T

= entropy generation for a system interacting with surroundings at T0 , then degree of irreversibility is _ _ __

17. If Sg

(a)

sg To

-

Thermal and Hydraulic Machines

18. If ilQ = heat interaction, ilS = entropy charge and T is - - - (a) ilQ = DS T

19. For an impossible process, (a) zero

=

temperature, then their relation

(b) ilQ < DS T

0

20. The entropy of our universe will always _ _ __ (a) increase (b) decrease 21. The change of entropy of a heating body (C = heat capacity) from T 1 to T2 is T (a) Clog ---1.

1i

22. The unavailable energy will ____ with lowering of sink temperature. (a) increase

(b) decrease

23. The COP of a refrigerator is _ _ _ _ if the COP of a heat pump is 5, working within the same temperature limits. (a) 4

(b) 6

24. The efficiency of a heat engine will ____ in case source temperature is increased. (a) increase

(b) decrease

25. The efficiency of a heat engine will ____ in case sink temperature is decreased. (a) increase

(b) decrease

26. If Q = latent heat at temperature T, then change of entropy is _ _ __ (a) Q

X

T

(b) Q T

27. If 600 kJ is rejected to surroundings at temperature 300 K, then change of entropy is (a) 2 28. PPM II is a device which (a) violates

(b) 0.5 the second law of thermodynamics. (b) supports

29. The refrigeration cycle is (a) similar

to a Carnot cycle. (b) reversed

30. A Carnot engine can have (a) the same efficiency

depending upon the temperature limits. (b) different efficiencies

Second Law of Thermodynamics

m

31. Change of entropy is a ____ function. (a) state (b) path 32. A portion of energy which can be converted into work is called _ _ __ (a) anergy (b) exergy 33. Exergy ____ be conserved like other energies. (a) can (b) cannot 34. Exergy (a) can

be regarded as a property of the system. (b) cannot

35. At the dead state, a system can have _ _ __ (a) exergy (b) energy 36. Entropy of steam is measured from _ _ __ (a) 0 K (b) 0°C

Answers State True or False 1. True 2. False (Energy is graded as high and low. Work is high-grade energy and heat is lowgraded energy.) 3. False (High-grade energy can be converted fully into low-grade energy while lowgrade energy cannot be converted fully into high-grade energy.) 4. True (All processes proceed to increase entropy.) 5. False (Interaction of other energy is required to reverse the process.) 6. False (Lays down the condition of interaction of other energy to reverse.)

7. False (Heat is rejected to the sink at low temperature.) 8. True 9. True (Both are devices to convert heat into work.) 10. True 11. True

12. True

13. True 14. True 15. False (Heat extraction from cold space is desired effect for the refrigerator. Heat rejected to surroundings has no significance to the refrigerator.) 16. False (COPHE > COPref.) 17. False (A heat engine is required to convert heat into work. It has nothing to do with supply of heat or extraction of heat.)

-

Thermal and Hydraulic Machines

18. False (A heat pump is required to supply heat to a hot body (room) by extracting heat

from surroundings.) 19. False (A heat pump is meant for heating. In warm weather, heating is not required.)

20. True 21. True 22. True (For a refrigerator cold space is a low temperature body and the surrounding

is a high temperature body.) 23. True

(11 = ~ ) Qadd

24. False (A spontaneous process cannot be reversed unaided.) 25. True 26. True (For an isothermal process ~U

=

0. Therefore Q

=

W.)

27. True (COPHP = COPref + 1) 28. False (A heat engine can work between the source and the sink. It cannot work with

one reservoir. Such device is known as PMM-11.) 29. False (The Clausius and Kelvin-Planck statements are equivalent. Violation of one

statement leads to violation of other.) 30. True 31. False (Any device working as violation of the Clausius statement is known as PMM11.) 32. True 33. True (If even one process of a cycle is irreversible, then cycle becomes irreversible.) 34. True 35. False (Both statements are equivalent.) 36. True 37. False (A Carnot cycle has two adiabatic and two isothermal processes.) 38. True 39. True 40. True 41. False 42. True 43. True 44. False 45. False

m

Second Law of Thermodynamics

46. True (Q 1 = Q2 + W where Q2 is heat from cold body and Q 1 is heat rejected to hot

body.) 47. False (All reversible engines have the same efficiency operating between the same

temperature limit.) T 48. True (1] = 1 - ---1. and as T 1 and T 2 will vary, efficiency will also vary.)

Ii

49. False (Efficiency of a Carnot cycle does not depend on working medium.) 50. True 51. True (1]

=

52. True (1]

=

T2

1 - - . When T 1 increases, 1] will increase.)

Ii

T2

1 - - . When T 2 increases, 1] will increase.)

Ii

53. False 54. True [1] 1

T2 - ~T

=1-

Ii

T ) ~T T2 = ( 1----1. +-=1]+a;1]2 =1Ii Ii Ii + ~T

1-

=

(l _Ii

T2 )

+ ~T X

T;

Ii

= 1] + aT2 Ii

Hence 1] 1 > 1]2.] 55 _ False (copref = Heat removed from cold body) Work input 56 _ False (coPHP = Heat supplied_ to hot body) Workmput

Ii Ii -

= 4.

57. True (COPHE

=

58. True (COPHP

= COPref + 1)

T2

59. True (COPref = 5 = Q2

w

3 1 - 4

=

-

Thermal and Hydraulic Machines

60. False

T2 T2 61. False (1] = 1 - -;;; , hence 1] depends upon which is constant.) 11

7i

T2 62. True (COP ref= - - - and hence COP will fall when T 1 increases.) 7i - T2 63. True (COPHP = COPref + 1)

7i - T2 7i 64. True (1] = - - - and COPHP = - - - which is reverse of 1] and equal to 3.) 7i 7i - T2 65. True 66. True 67. True (A reversible adiabatic process is isentropic but isentropic may not be adiabatic.) 68. False (Spontaneous processes are irreversible and generate entropy. Entropy of the

universe is continuously increasing.) 69. True (A system with lesser entropy has heat at higher temperature which is more

valuable and prone to a spontaneous process.) 70. True ( I: dQ = 0 for reversible processes.) T

71. True ( I: dQ < 0 for irreversible processes.) T 72. False (

73. True (

74. True

f d; =O)

f di

(J J( d.;tJ

75. False ( ( 76. True 77. True

is a state function and does not depend upon the path.)

dS =

di tv ds) =

1111

Second Law of Thermodynamics

78. True

[.1s

79. True

[.1s = 373 hfg ]

=

hsf ]

273

80. True [(LiS)universe = (L'.iS)body + (L'.iS)liq] 81. False (Entropy increases for all processes except for a reversible adiabatic process.) 82. True (Change of entropy is a state function.)

T2 83. True (COP ref = ---=---- with T 1 increasing COP will fall.) 1i - T2 84. True 85. True 86. False (Heat rejected

=

unavailable energy

87. False (Unavailable energy 88. True (Exergy

=

=

=

300 x LiS, A 2

90. True (Degree of irreversibility 91. True (W

=

dW 0.3Q + a, dQ

Q 1 x T2

1i

=

100 x 400 800

=

50 kJ)

T0 LiS = 300 x 1.5 = 450 kJ)

energy - anergy

89. True (Anergy A 1 X 1 = 100)

=

=

=

600 - 200

=

200 x Li S; (A 1

=

400) -

A 2 ) = (300 - 200) LiS = 100

T0 x LiS = 1.5 x 300 = 450 J)

=

0.3

= 1])

Multiple Choice Questions 1. (d)

2. (b)

3. (a)

5. (b)

6. (b)

7. ( d) ( 1] = 1 - T2

8. (d)

(11-w)

(c)

9. (d) ( 1] = 1i 12. ( C) (QI

)

Ii

Qadd

11.

4. (c)

=

~ T2)

W + Q2)

13. (c)

14. (b)

15. (c)

16. (c)

17. (c)

18. (b)

19. (c)

20. (b)

21. (c)

22. (a)

23. (b)

24. (c)

25. (d)

-

Thermal and Hydraulic Machines

27. (b)

26. (c) (11 does not depend upon medium)

(11 = 1 -

T2

1i

=1 -

300 600

=_!_) 2

28. (b) T2 300 29. (b) (COPcarnot ref= - ~ = = 6 1i - T2 50

30. (a) (COPHP =

1i 1i _ T2

:. COPactual =

360 1 = 360 _ 300 = 6, COP actual = 2

1

2

6 = 3)

X

6 = 3)

X

31. (b) (COPRef = COPHP - 1 = 5 - 1 = 4) 32. (b)

(11 = 1i

1i

and COPHP = 1i

T2

1i -

T2

=- 1- =4) 0.25

33. (a) 34. (c) (See solved example of Question 11.) 35. (a) (11 = 1- T2 = 1- 300 = 0.7 = W or W = 0.7 x QI = 0.7 x 1000 = 700; 7i 1000 Q, Q2 = QI - W = 1000 - 700 = 300 kJ) 36. (b) (Q; = Q;

X T~

7i

= 800

X

300 = 300 kJ, Q'; = Q'; 800

= 700

X T~,

X 300

= 300

700

T1

kJ)

Q2 = Q; + Q'; = 300 + 300 = 600 kJ 11 =

Q, - Q2 1500 - 600 3 Q, = 1500 = 5 = 60 %) W

37. (c) (11 = 0.5 = -Q or W = 0.5 Qadd, COP= 5 =

Q2 W

add

Q2 = 05 Q •

QQ:d = 0.4)

add

38. (c)

39. (a)

40. (a)

41. (a)

42. (b)

43. (a)

44. (b)

45. (c)

dQ

46. (c) (dS = -

T

1000 = = 5 kJ/K) 200

47. (b) (dQ = TdS = 300 x 5 = 1500 kJ) 48. (b)

m

Second Law of Thermodynamics

49. (c) (W = Q 1 - Q2 • If one engine is operating and W = (Q 1 - Q 3) + (Q 3 - Q 2) and if two engines are operating W = W 1 + W2 but W 1 = W2 • Work output of each engine

= ~) - T 2 = T2 - T3 = T3 - T4 = T4 - T 5 = T 5 - T6. The temperature difference 1000 - 500 = lOO K)

50. (b) (T 1 =

5

51. (a) (T, - T2 = T2 - T3 = T2 - T4 = T5 - T6 = 800 Q1 - Q2 1i - T2 = -100 -W = ---'--'-------'-== ---'-----=-

Q,

1i

W = 4 W 1, = 4

X

Q,

52. (a) (fl.S = S2

-

53. (b) (fl.S = S2

-

w,

800

=

1

-x 80 8

=

~ 400 = 100; 10 J;

10 = 40 J)

= (2 + 3) - 4 = 1 > 0. Hence feasible) S 1 = 6 - (3 + 5) = -2 < 0. Hence impossible)

S1

54. (b) [Since SB < SA, hence process A to process B is impossible, i.e. SB - SA = 3 4 = - 1 kJ/K is impossible] 55. (b) T 1

-

600-200 = 200 K 2

T2 = T2 - T 3 =

T2 = 600 - 200 = 400 K. Now

30 or 600

= -Q3

200

or Q3

Ii

56. (b) (17HE = 1 - -

1i

= 10

Q,

1i

=

Q2

T2

=

Q3

T3

J)

400 1 WE WE =- =- -800 2 Q 40

- 1- -

400 Q2 Q2 5 or WE= 20 J; COP = 800 _ 400 = 1 = W = W = W or ref ref ref output = WE -

Wref

Wref

= 5 J. Net work

= 20 - 5 = 15 J)

57. (c) (unavailable energy = Q1 x T2 = 200 x 400 = 100 kJ) 1i 800 58. (a) (unavailable energy = fl.S x T2 = 1.2 x 300 = 360 kJ) 59. (a) (Degree of irreversibility = fl.S x T2 = 600

or

fl.S

= 600/300 = 2 kJ/K)

-

Thermal and Hydraulic Machines

Fill in the Blanks

2. (b) 6. (b)

3. (a) 7. (a)

8. (b)

10. (b)

11. (b)

12. (b)

13. (b)

14. (a)

15. (b)

16. (b)

17. (b)

18. (b)

19. (b)

20. (a)

1. (a)

5. (b) 9. (a) ( Q, = Q2 )

Ii

4. (a)

T2

21. (a) 22. (b) [ unavailable energy = Q1 x

i

which will decrease on lowering of T2

23. (a) (COP ref= COPHp- 1) 24. (a) [ 1J =1 -

25. (a) [ 1J =1 -

i; i;

1J will increase with increase of 7i ]

1J will increase with decrease of T2 ]

26. (b) 27. (a)

[ 0rej

=!ls X T

0 ,

28. (a) 30. (b) [ 1J =1 -

31. (a) 34. (b)

hence !ls= 600 = 2kJ/K] 300 29. (b)

i;

1J will be different for different Ii and T2 ] 32. (b)

33. (b)

35. (b)

36. (b)

]

Properties of Steam and Thermodynamics

INTRODUCTION Steam is the most common working substance employed as working fluid in stream engines, steam turbines and atomic power plants for power generation. It acts in these applications as a transport agent for energy and mass interactions. Steam is a pure substance . It can be easily converted into any of three states, i.e. solid, liquid and gas. It has capability to retain its chemical composition and homogeneity in liquid and gaseous phases. A pure substance has chemical homogeneity and constant chemical composition. Water is a pure substance as it meets both the above requirements. A substance cannot be a pure substance if it undergoes a chemical change . In this chapter, we will learn about the properties of steam as a pure substance and how to change the properties of steam during a process to attain the desired effect, i.e. maximum work for a given heat intake.

DEFINITION Pure substance: A pure substance is a substance which has constant chemical composition and chemical homogeneity. Water is a pure substance. Any substance which undergoes a chemical reaction, cannot be termed pure substance. Sensible heating: Sensible heating of a substance is heating in a single phase. The heat is used for raising the temperature of the substance. 145

-

Thermal and Hydraulic Machines

Latent heating: Latent heat of a substance is the heat that causes its phase change without raising its temperature.

Boiling point: Boiling point is the temperature at which vapour pressure is equal to atmospheric pressure and phase change takes place from liquid to gas. Melting point: Melting point is the temperature at which phase changes from solid to liquid when heat supplied is equal to latent heat. Saturation state: Saturation state is the state at which its phase transformation takes place without any change in pressure and temperature. Therefore, there can be saturated solid state, saturated liquid state and saturated vapour state. Saturation pressure: Saturation pressure is the pressure for a given temperature at which a substance changes its phase. The saturation pressure for water at 100°C is one atmospheric pressure. Saturation temperature: Saturation temperature for a given pressure is the temperature at which a substance changes its phase. Triple point: The triple point of a substance is the state at which the substance can coexist in solid, liquid and gaseous phases in equilibrium. The triple point for water is 0.01 °C. Critical point: The critical point of a substance is the state at which the substance can coexist in two phases for last time in equilibrium. It is the point of highest pressure and temperature at which water and vapour coexist for last time. The substance has vapour phase only above this point. The critical point of water has a pressure of 22.12 MPa and a temperature of 374.15°C. Dryness fraction: Dryness fraction is the ratio of the mass of vapour to the mass of both liquid and vapour in any liquid-vapour mixture region. The liquid having temperature less than saturation temperature corresponding to a given pressure is called subcooled liquid or compressed liquid. Steam having temperature more than saturation temperature corresponding to any given pressure is called superheated steam. Refer to Figures 4.1 and 4.2. When ice is heated, ice will have sensible heating up to 0°C along AB. Here ice will start melting along BC at constant temperature. CD shows sensible heating of water. DE shows evaporation of water at constant temperature. EF shows sensible heating of steam. The above temperature vs heat as well as temperature vs entropy diagrams have been drawn for one atmospheric pressure. On the similar line, T-S variation of water can be obtained for other pressures. If all points of D (saturation liquid side) and E (saturation vapour side) are joined, a bell type curve is obtained with critical point (CP) at top (Figure 4.3). The saturated liquid line and the saturated vapour line come closer as pressure

111111

Properties of Steam and Thermodynamics

and temperature increases. These lines meet at the critical point. The region enclosed between the saturated liquid line and the vapour line is wet steam (liquid-vapour) region. The region on the right of the saturation vapour line is the superheated region. The region left to the saturation liquid line is liquid (water) region. F

100

0

D Vaporisation

----------------------~--'-----' E

Pr= 1 aTM

0°

B Fusion - - - - - - -,.-----------1 C I I

I I

Solid---.:.--- Liquid -------.,.- Liquid~,.--Vapour---. , vapour , I I A "'-----+----+----+-.--La-t-en-t-~+-~Superheating--. heating fusion

heat vaporisation

Heat-

FIGURE 4.1

Temperature vs heat. F

T

D E

Q)

5 "§

Pr

1 aim

Q)

a.

E

~

0

B

-------

A

C

Entropy (S)

FIGURE 4.2

Temperature vs entropy.

The wet steam is the region between the saturated liquid line and the saturated vapour line. Dryness factor (x) is zero on the saturated liquid line and one on the saturated vapour line. Dryness at other point in the wet steam region is given by equal dryness lines. DiEi shows heat of vaporisation (h1g) which decreases with increase of pressure. CiDi shows sensible heating and (Tm - Tei) gives degree of subcooling at temperature Ci. Similarly, EiFi shows sensible heating and (TFi - TEI) gives degree of superheating at temperature Fi.

-

Thermal and Hydraulic Machines

TF2

-----

Saturated CP Saturated liquid lin~ ------v/ur line-@ p 2 > P,

Ltq_uid _____ region

~

@

p

®

2

~

\

P,

\ \

·~=-------'---,--> P and T1 =

r;.

It is evident that dryness x 2 > x;.

23. True

24. True 25. False (Sensible heating and latent heating take place.) 26. False (Rankine cycle has a pump instead of a compressor.) Multiple Choice Questions

1. (a)

2. (b)

3. (a)

4. (b)

5. (a) 6. (b) (If condensor pressure is decreased, then only efficiency increases.) 7. (a)

8. (b)

9. (a) (State must have entropy matching the entropy of saturated liquid line at high (boiler) pressure.)

10. (c) (Wet steam has water and dry steam mixture and the specific volume is large resulting large compression work and low net work output.) 11. (a) 12. (a) (Higher inlet temperature to the boiler will give higher efficiency.) 13. (b)

14. (c)

15. (b)

16. (a)

m

Vapour Cycles

Fill in the Blanks 1. (b)

2. (b)

3. (a)

4. (b)

5. (b)

6. (b)

7. (a)

8. (a)

9. (a)

10. (b)

11. (b)

12. (b)

13. (b)

14. (b)

15. (b)

16. (a)

m Thermodynamic Cycles

INTRODUCTION Thermodynamic cycles can be classified based on their utility as power cycles and refrigeration cycles. Power cycles can be vapour power cycles and gas power cycle/air cycles. Vapour power cycles have been explained in Chapter 5. In this chapter, we will study air-standard cycles. An air-standard cycle is a thermodynamic cycle working with air as the working substance with certain assumptions. The assumptions are as follows: I. 2. 3. 4. 5. 6. 7. 8.

Air as working substance behaves as a perfect gas. The mass and composition of working substance do not change. All processes are reversible. Heat transfer from combustion is considered from an external source. The engine operates as a closed system. Specific heats remain constant. KE and PE remain constant. Heat loss to surroundings is nil.

Otto and Diesel cycles are gas power cycles with air as working substance. Otto and Diesel cycles are modified forms of the Carnot cycle in order to make the cycles realistic. Engines are generally designed based on Otto and Diesel cycles. 194

m

Thermodynamic Cycles

Certain terms used for the analysis of Otto and Diesel cycles are as follows: I. Top dead centre (TDC): It is the highest point to which the piston can reach during a stroke (Figure 6. I). 2. Bottom dead centre (BDC): It is the lowest position to which the piston can reach during a stroke (Figure 6. I). 3. Swept length (1): It is the distance between TDC and BOC . 4. Clearance length (1): It is the distance between TDC and the cylinder top. 5. Clearance volume (V): It is the product of the area of the piston and the clearance length. 6. Swept volume (V): It is the product of the area of the piston and the swept length. 7. Compression ratio (r): It is the ratio of the sum of the swept volume ( V) and clearance volume (V) to the clearance volume (VJ r= V,+Vc =1+ V,

=1+1_

Ve Ve ( The compression ratio (r) for SI (spark ignition) is in the range of 5: 1 to 11: 1 while for compression ignition (Cl), it is in the range of 12: 1 to 25: I.

TDC--

m ,

'

---:: }

I

'

II I I \ \

I

I I I I I

'

I

' \

I

FIGURE 6.1

·-*-~~-----------~~ le

I I

'I \

I

I

I

I I

\

'' '

I I I

Piston positions in TDC and BDC.

The cylinder head block and the engine block are joined together as illustrated in Figure 6.2 . These blocks get worn out (shaded portion) over a period of time and need to be machined resulting in reduction of clearance length (le). The compression ratio (r) increases due to machining of the cylinder head.

Clearance length Uc) Piston

Pi ston Engine block

Cylinders

FIGURE 6.2

Cylinder head block joined with engine block.

-

Thermal and Hydraulic Machines

OTTO AND DIESEL CYCLES

The Otto cycle consists of two isochoric and two adiabatic processes. The Otto cycle is also known as a constant volume cycle. Reciprocating spark ignition engines are based on this cycle. Four processes of the Otto cycle are shown in Figure 6.3. p

T

3

3

4

M

2

Oadd

2

,4 rej

I I

_v~

: O,ei I

s

V

(a) PV diagram

(b) TS diagram

Otto cycle.

FIGURE 6.3

I. Process 1-2:

Reversible adiabatic compression

2. Process 2-3:

Constant volume heat addition

3. Process 3-4:

Reversible adiabatic expansion

4. Process 4-1:

Constant volume heat rejection

Heat and work interactions of the Otto cycle are as follows : 1. Process 1-2: ,Q2 = ,W2 + (U, - U2), ,Q2 = 0

,w2

=

U2 -

u,

cv(T2 - T,)

=

2. Process 2-3: 2Q3

= (U3 -

= 0) = cv(T3 -

U2) (As 2W3

3. Process 3-4: 3W4 = (U3 - U4 ) (As

T2)

3Q 4

= 0) = cv(T3 - T4 )

4 W3

= 0) = cv(T 1 - T4 )

4. Process 4-1: 4Q1

= (U1

-

U4 ) (As

The efficiency of the Otto cycle can be worked out as follows:

1J

=

net work heat added c,.(T3

-

Q add

=

T2 ) c"(T3

-

Qrej

Qadd

c"(T4

-

T2 )

-

'fi)

m

Thermodynamic Cycles

For an adiabatic process, we have PvY

Combining this with

= c.

V1

T2 = Tiry -

where r = -

t,

Pv

T

= R,

we get

. . = -V4 =compression rat10

V2

V3

T3= T4rr - t

11 - 1 (T4 - I;) = I - rY- '(T4 - I;)

1

Therefore, the efficiency of the Otto cycle depends upon the compression ratio (r) and the adiabatic index (y). Efficiency increases if the compression ratio increases (see Figure 6.4).

r

FIGURE 6.4

Variation of 1] with r.

The maximum work from the Otto cycle depends upon the compression ratio at the highest temperature (T3) and lowest temperature (T 1): W=

dW dr

Qadd -

Qrej

0 for maximum work w.r.t. compression ratio. On solving, we get I

r

= ( ~ )2(y-I)

Also The Diesel cycle consists of two reversible adiabatic process, one constant pressure process and one constant volume process. The Diesel cycle has four processes as shown in Figure 6.5. Let us discuss these processes one by one.

-

Thermal and Hydraulic Machines

I.

Process 1-2:

2. Process 2-3:

Reversible adiabatic compression Constant pressure heat addition

Heat addition is stopped at point 3 which is known as cutoff point. Cutoff ratio

(/J) = ViVz. 3. Process 3-4: Process 4-1:

4.

p

2

Reversible adiabatic expansion Constant-volume heat rejection T

3

3

.----11....+-~

2 4

O,ei

O,ej

s

V

(a) PV diagram

(b) TS diagram

Diesel cycle.

FIGURE 6.5

Heat and work interactions in various processes of the Diesel cycle are as follows: I. Process 1-2: As

Q2 = 0 W2

= cu(T2

-

T1) where r

Also

. ratio. = -v, =compression V2

JdU + JPdv

2. Process 2-3:

Jd(U + Pv) =

Also

3. Process 3-4: As

Jdh

=

cp(T3 - T2)

T3 = T1rY- lf3 3W4 = U3 - U4 3Q4 = 0, = Cv(T3 - T4)

4. Process 4-1:

4Q1 = U 1

-

U4

= Cv(T, - T4)

where f3

= -V3 = -T3 =cutoff ratio Vz

Tz

m

Thermodynamic Cycles

Also

and

v4

v1

V2

V2

V3

=-X-=rX

V3

/3

The efficiency of the Diesel cycle can be found out as follows:

11 =

net work Q..dd

= 1-

cu( T4 - Ti ) cp(T3 - T2 )

=1-

1 (T4 -

Ti)

Y (T3 - T2) =1-

1

r

(7t/3Y - '.fi) ('.fi/3ry-l - :z;rr-1)

1 [ 13r - 1]

=1- ry-1 r(/3-1) 1 = 1 -y-- 1x k r

13r -

where

1

k=---.

r 1Joual > 1Joiesel·

-

Thermodynamic Cycles p

3

V

FIGURE 6.8

Equal compression and heat input.

2. Cycles for equal maximum pressure and heat input are shown in Figure 6.9. The Otto cycle is 1-2-3-4, the dual cycle is 1-2"-2'"-3"-4" and the Diesel cycle is 12'-3'-4'. As per the area enclosed, we can say 17Diesel > 17ouaI > 170110· p

V

FIGURE 6.9

Equal maximum pressure and heat input.

3. Cycles for maximum pressure and temperature are shown in Figure 6.10. Otto cycle is 1-2-3-4, the dual cycle is 2-2"-2'"-3-4 and the Diesel cycle is l-2'-3-4. As per area enclosed in the cycles, we can say 17Diesel > 17ouaI > 17011o· p P3 = Maximum pressure T3 = Maximum temperature

V

FIGURE 6.10

Equal maximum pressure and temperature.

In the Sterling cycle (regenerative cycle), a regenerator is used, which stores the rejected heat energy during the heat rejection process and supplies the same during the heat addition process. It consists of two isothermal and two constant volume processes as shown in Figure 6.11.

Iii

Thermal and Hydraulic Machines p

V

FIGURE 6.11

Sterling cycle.

ENGINES Internal combustion (IC) engines: IC engines are those engines in which fuel is burnt inside a cylinder. Petrol and diesel engines are internal combustion engines as fuel is burnt inside the cylinder of these engines. External combustion (EC) engines: EC engines are those engines in which fuel is burnt outside the cylinders. The steam engine is an external combustion engine. Internal combustion engines can be 4-stroke engines or 2-stroke engines. In 4-stroke engines, the cycle is completed in four strokes or two revolutions of the crankshaft. In 2-stroke engines, the cycle is completed in two strokes or one revolution of the crankshaft. 4-stroke engines can be petrol engines in which the fuel used is petrol or diesel engines in which the fuel used is diesel. Petrol engines work on the Otto cycle while diesel engines work on the Diesel cycle. 4-stroke engines have inlet and outlet valves which are opened and closed by the camshaft, which runs at half speed of the crankshaft. Spark ignition (SI) engine has a carburettor for making air - fuel mixture and a spark plug to ignite the mixture. Four strokes of SI engines (Figures 6.12 and 6.13) are as follows: Suction stroke (Process 0-1): The piston moves from TDS to BOC creating vacuum inside the cylinder. The inlet valve opens and air - fuel mixture enters into cylinder.

I.

Inlet valve

@/

Ai_r-petrol/ @ mixture 1

~

TDC

~

~

--------- -------------------

~-BDC

Suction

Spark plug

Outlet valve

-----TDC-

J

-,---.L:::....---,-f-------- ------------------

CfiJ-u-il-------Compression

FIGURE 6.12

®

--------BDC

Combustion

Four strokes of SI engine.

CfiJU Exhaust

-

Thermodynamic Cycles p

4

I

I

o ,ej

0 '-,--------4-----=--I

V

FIGURE 6.13

2.

PV diagram for 4-stroke SI engine.

Compression stroke (Process 1-2): The piston moves from BOC to TDC. Both valves are closed. Air - fuel mixture is compressed.

3. Combustion and power stroke (Processes 2-3 and 3-4): The air - fuel mixture is ignited before the piston reaching TDS. Both valves remain closed. There is a rise of temperature and pressure due to combustion at constant volume. The temperature is around l 800-2000°C and pressure is around 30-40 bar. Due to high pressure, a force acts on the piston and the piston moves from TDS to BOS. Power is obtained from this stroke. At the end of this stroke, heat is rejected to surroundings at constant volume (Process 4-1 ). 4.

Exhaust stroke (Process 1-0): The inlet valve remains closed and the exhaust valve opens. Burnt gases are pushed out through the exhaust valve.

Compression ignition (CI): Engine has a fuel injection pump to inject fuel at high pressure through the injector in the cylinder. It has no spark plug or carburettor as combustion takes place by compressing air (no air - fuel mixture) at high pressure and then injecting diesel through the injector at very high pressure so that spontaneous combustion takes place. Cl engine works on the diesel cycle. The four strokes of a Cl engine (Figures 6.14 and 6.15) are as follows: I.

Suction stroke (Process 0-1): The piston is at TDC, the inlet valve is open and outlet valve is closed. The piston moves from TDC to BOC creating vacuum in the cylinder. The air is sucked in.

2.

Compression stroke (Process 1-2): Both inlet and outlet valves are closed. The piston moves from TDC to BOC compressing the air. The temperature and pressure of air increase. The compression ratio reaches 12-25. The pressure of air is around 60 bar and the temperature is about 600°C. Temperature is sufficient for auto ignition of fuel. Diesel is injected by the fuel injection pump at high pressure when the piston is about to reach TDC.

3. Combustion and power stroke (Processes 2-3 and 3-4): The combustion of fuel takes place at constant pressure. The fuel enters at point 2 and fuel is cut off at point 3. Hot gases now expand pushing the piston downwards towards BOC. Power is obtained in this stroke. Both valves are closed during the stroke. At BOC heat is rejected at constant volume (Process 4-1 ).

Iii

Thermal and Hydraulic Machines

Inlet valve

j

Injector

@

Outlet valve

I

~

TD:i,

~ .........

TDC~

oo

n. . . . . . . . . CiJ n. . . . . . . . BDC

Suction

Compression

FIGURE 6.14

~~

D BDC

Combustion

OD

~

gases

u

Exhaust

Four strokes of Cl engine.

p 2

3

4

V

FIGURE 6.15

4.

PV diagram for four strokes of Cl engine.

Exhaust stroke (Process 1-0): Inlet valve remain closed while exhaust valve opens. The piston moves towards TDC pushing the burnt gases out of cylinder through exhaust valve.

In a 2-stroke petrol engine, the cycle is completed in two strokes of the piston or one revolution of the crankshaft, the shape of the piston is having deflector-like shape at top to deflect the incoming air - fuel mixture to top thereby, preventing its escape through the exhaust port (Figure 6.16). The cylinder body has three ports (Note: No valves as in a 4-stroke engine) which are opened in turn by the piston. The inlet port is opened when the piston is at TDC . When the piston moves down, it first opens the exhaust port and closes the inlet port. Later on the piston opens the transport port. The cycle is completed (Figures 6.17 and 6.18) as follows: I. Ignition and induction: The piston is almost at TDC towards the end of the compression stroke. The exhaust port and transfer port are covered by the piston while the inlet port is uncovered. The compressed charge is ignited by spark. Combustion of air - fuel mixture takes place. Temperature and pressure increase in the combustion chamber of the cylinder. Vacuum is simultaneously created below the piston in the crankcase and fresh charge is inducted in the crankcase through the inlet port.

-

Thermodynamic Cycles

Deflector

p

Exhaust Inlet :opens

closes

Inlet closes V

FIGURE 6.16

FIGURE 6.17

2-stroke SI engine.

SP

2-stroke SI cycle.

Scavenging Transference >c:---ttl~S_P...._

t

Burnt gases IP

Ignition and induction

Expansion and compression

FIGURE 6.18

Exhaust and transference

Compression and intake

2-stroke cycle.

Expansion and compression: It is expansion of high pressure and temperature gases after combustion and compression of fresh charge in the crankcase. The piston moves down due to expansion of gases after combustion and power is developed. The downward movement of the piston closes the inlet port and compresses fresh charge in the crankcase to about 1.4 bar. The moving piston uncovers the exhaust port when about 80% expansion has taken place. The burnt out gases being higher in pressure than atmospheric pressure escape out from the cylinder through the exhaust port. 3. Exhaust and transference: It is exhaust of burnt out gases and transference of compressed charge from the crankcase to the cylinder when the piston is almost at BDC. At this point, the piston uncovers the transfer port and transfers the slightly compressed charge from the crankcase to the cylinder through the transfer port. The deflector like shape on the piston helps in pushing out the burnt out gases through the exhaust port. This process is known as scavenging. 4. Compression and intake: When piston moves upwards from BDC, it first covers the transfer port to stop transfer of fresh charge. A little later it covers the exhaust port to stop exhaust and to start compression. The compression is complete near TDC, and the inlet port is uncovered for fresh intake in the crankcase. 2.

-

Thermal and Hydraulic Machines

A 2-stroke diesel engine works similar to a 2-stroke spark ignition engine and differences are as follows : I . A diesel engine has an injector to inject fuel instead of a spark plug. 2. Air is compressed and combustion takes place due to autoignition.

The comparison of a 4-stroke engine and a 2-stroke engine is given in Table 6.1. TABLE 6.1

Compari son of a 4-stroke engine and a 2-stroke engine

4-stroke engine

2-stroke engine

1. Cycle completed in four strokes or two revolutions of the crackshaft 2. One power stroke in two revolutions of the crankshaft 3. Engine is heavy for the same power 4. Heavier flywheel required as one powerstroke in two revolutions 5. One combustion in two revolutions Therefore, lesser cooling and lubrication is required 6. Camshaft and valves at inlet and outlet 7. High initial cost 8. Used in heavy vehicles 9. High thermal efficiency 10. High volumetric efficiency

1. Cycle is completed in two strokes or one revolution of crankshaft 2. One power stroke in one revolution of the crankshaft 3. Engine is light 4. Light flywheel is required

5. More cooling and lubrication is required

6. Ports instead of valves and camshaft 7. Low initial cost 8. Used in light vehicles 9. Low thermal efficiency 10. Low volumetric efficiency

The comparison of a spark ignition engine and a compression ignition engine is given in Table 6.2. TABLE 6.2

Comparison of a spark ignition engine and a compression ignition engin e SI engine

1. 2. 3. 4.

Based on the Otto cycle Low compression ratio (range 5 to 11) Petrol used as fuel Air-fue l mixture is compressed in the cylinder 5. Carburettor makes air - fuel mixture 6. Ignition by a spark plug

7. Combustion is isochor ic 8. Low compression and eng ine is light 9. High engine speed (8000-6000 rpm) IO. Low thermal efficiency I I. Low maintenance cost and high running cost 12. Used in light veh icles

CI engine 1. 2. 3. 4.

Based on the Diesel cycle High compression ratio (range 12 to 25) Diesel used as fuel Only air is compressed

5. Fuel injection p ump is required 6. High pressure fuel from the injection pump is injected through the injector in the cylinder and autoignition takes p lace 7. Combustion is isobaric 8. H igh compression and engine is heavy 9. Low speed (400-3500 rpm) IO. High thermal efficiency II. High maintenance cost and low running cost 12. Used in heavy vehicles

-

Thermodynamic Cycles

INDICATED, BRAKE AND FRICTION POWER

Combustion takes place in the cylinder and power is taken from crankshaft. The shaft work available is less than total energy released inside the cylinder due to frictional and others losses. Indicated power is the power available inside the cylinder and provided to the piston. Indicated power is measured from an indicator diagram (similar to PV diagram) obtained using an indicator mechanism. Indicated power = Energy in fuel - Energy loss in exhaust, coolant and radiation Brake power (BP) is the power available at crankshaft. Brake power is measured by dynamometers.

Brake power = Indicated power - Energy loss in friction and pumping 2nNT

60 ;

.

.

where T is torque and N is rpm

Friction power (FP) is the power lost due to friction and other reasons.

Friction power = Indicated power - Brake power Mean effective pressure (mep or Pm) is the average pressure per stroke and it can be obtained from an indicator diagram. The indicator diagram indicates displacement of the piston on x-axis and cylinder pressure on y-axis. P

=

mep

m

=

Enclosed area of indicator diagram x Indicator spring constant Length of diagram

-------------

Indicated power (IP) is the power generated per unit time: IP= where n = number of cylinders Pm = mean effective pressure

= area of the piston L = length of the stroke

A

N= rpm K

= I for a 2-stroke engine _!_ for a 4-stroke engine 2

nxPm xAxLxNxK

60

-

Thermal and Hydraulic Machines

EFFICI ENCi ES

Mechanical efficiency ( 17mech) is the ratio of brake power (BP) to indicated power (lP): BP

IP - FP

IP

17mech =

IP

=

Volumetric efficiency (hv) is the ratio of the actual volume of the charge (Vactual) admitted during suction stroke reduced to NTP to the swept volume of the piston (Vswept):

11 = v

vactual V.wept

Brake thermal efficiency is the ratio of brake power (BP) generated to fuel energy used: hbrake

- m

thermal -

BP c

X

f

where m1 =

mass of fuel

c = calorific value of fuel

Indicated thermal efficiency can be given as follows: IP

11 indicated

thermal

=

m f

X C

The knocking in an St engine is ignition of air - fuel mixture before spark reaches it. lsooctane content in fuel for St engines retards autoignition while normal heptane accelerates autoignition. The knocking in an St engine increases with increase in the compression ratio and decrease in speed. The ignition quality of fuels in St engines is determined by octane number rating. Higher octane of fuel will decrease tendency of knocking. The knocking tendency in a Cl engine is increased with decrease of the compression ratio. ln an St engine, an ignition coil is used to generate high voltage for the spark plug. lt can be appreciated that in a multicylinder St engine, spark at particular order is given to each cylinder so that power obtained is smooth in revolution of the crankshaft. Hence a distributor is used to obtain required firing order of spark plugs so that spark is obtained at right moment in each cylinder.

SOLVED PROBLEMS 1. An engine cylinder has a piston of 0.12 m2 and contains gas at a pressure of 1.5 MPa.

The gas expands according to a process which is represented by a straight line on a pressure - volume diagram. (Figure 6.19) The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.3 m. (UPTU: 2005)

m

Thermodynamic Cycles p P, I I

I I I

I I I

I I I

,a

P2 ----,--------------- 2

---vs - - • a'

FIGURE 6.19

V

2'

PV diagram (Problem 1).

Vs = Swept volume = A x /stroke = 0.12 x 0.3 = 0.036

Work done = area of triangle 12a + area of rectangle 2aa'2'

=

1

l

X

(1500 - 150)

X

10 3

X

(0.036) + 150

X

10 3

X

0.036

= (24300 + 5400) kJ = 29700 kJ 2. Suction pressure and temperature for a theoretical diesel cycle are 1 bar and 27°C. The pressure at the end of the compression stroke found to be 24 bar. The maximum temperature of the cycle is limited to l 200°C (Figure 6.20). Determine (a) cutoff ratio (b) net output of the cycle (c) thermal efficiency (UPTU: 2005) p 2

24 bar

3

4

1 bar

--------------------

1

V

FIGURE 6.20

PV diagram (Problem 2).

For air: R Cp

= =

0.287 kJ/(kg K) 1.005 kJ/(kg K)

Process 1-2:

Thermal and Hydraulic Machines

= 0.718 kJ/(kg K) = t.4

Cv

r Adiabatic process P1

= 1 bar

P2

T 1 = 27°C = 300 K

= 24 bar

T2 =?

For an adiabatic process:

0.4 (24)1. 4

or Process 2-3:

T2

300 x 2.479

=

=

743.8 K

Isobaric process 1200 + 273 743.8

½ ½ -J3T2

V2

/3 = Qadd =

1.98 Cp(T3 - T2)

= 1.005

1.005 x (1473 - 743.8)

=

729.2

X

= 732.85 1Joiesel

1

(/3r -

1)

rr-1

r(/3 -

1)

=1---x 1

-1- - 9.67°.4

(1.981.4 - 1) 1.4(1.98 - 1)

X ----

= 1_ __ 1 x_l._6__ 2.47 = 1 -

Now Therefore,

X

1.4 X 0.98

0.47 = 0.53

11 = Wnet

= 0.53

X

732.85

= 388.4 kJ

a

Thermodynamic Cycles

Note:

Similar problems can be solved for the Otto cycle with only difference is heat

1 I - r=T. r Also in case mep is to be calculated, then

add

= cu(T3 -

T2) and 11

=

wnet

mep =

v,

where V, = swept volume.

3. An engine of 250 mm bore and 375 mm stroke works on the Otto cycle. The clearance volume is 0.00263 m3 • The initial temperature and pressure are 50°C and I bar. If the maximum pressure is limited to 25 bar. Find (a) the air standard efficiency of the cycle and (b) the mean effective pressure for the cycle (Figure 6.21 ). (UPTU: 2000) p 2

3

4

V

FIGURE 6.21

PV diagram (Problem 3).

Given: P 1 = I bar, T 1 = 50°C and P 3 = 25 bar, L = 375 mm, D Ve = 26.3 X 10- 4 m3 1rD 2

v, or

Compression ratio

V,

(r)

X

L

4

= 184

X

10- 4 m 3 (184 + 26.3) X 10- 4 26.3 X 10-4

= =8

P2 = P,

11 = 1 -

(~r

=

X

=

1-

= I - 0.43 = 0.57

rY= l x 8 14 = 18.38 bar 1 (8)04

=

1 -

1

2.29

250 mm,

Thermal and Hydraulic Machines y-1

(~)r

Also

= (18.38)~::

T2 = 323 x (18.38).2 85 as ( ·: T,

= 323 X 2.29 = 740.5 K T3

Qadd

Wnet

=

= = = = = =

mep

T2( ~)

=

740.5

X

c:~

273 + 50°C

=

=

323 K)

8)

1007 K cv(T3 - T2) 0.718(1007 - 740.5)

191.3 kJ/kg

11 X Qadd 0.57 X 191.3

=

109 kJ/kg

wnet 109 = -- = ---- =

Vs

184

X

10- 4

4

0.59 x 10 kPa

4. A Carnot engine working between 400°C and 40°C produces 130 kJ of work. Determine the following:

(a) Thermal efficiency (b) Heat added (c) Entropy change during heat rejection T1

=

400 + 273

T2

=

40 + 273

11 = 1 -

7i

673 K

=

313 K

=

313 673

1 -

T2

= 0.535 130

130

Qadd -- 0.535

= 243 kJ

Heat rejected = Qrej = = 243 - 130 Qrej

= 0.535

= 113 kJ = T2 x !).S

Qadd -

Wnet

Thermodynamic Cycles

113

= 3 13 x t:.S

..

113 t:.S = = 0.361 kJ/(kg K) 313

5. If an engine works on the Otto cycle between temperature limits 1450 K and 310 K, find the maximum power developed by the engine assuming the circulation of air per minute as 0.38 kg (Figure 6.22). p 3

4

2

V

FIGURE 6.22

PV diagram (Problem 5).

For maximum power T2 = T4 = ~I; T3 where and

= maximum temperature T 1 = minimum temperature

T3

T2 = T4 = .JI450 x 310 = 670.4 K

W = cu[(T3 - T2) - (T4 - T1)] Cv

= 0.718[(1450 - 670.4) - (670.4 - 310)] = 0.718(779.6 - 360.4) =

W

0.718(419.2) = 301 kJ/kg

= m X

=

W

=

0. 3 S X 301 kW 60

1.9 kW

6. A 4-stroke diesel engine has LIA ratio of 1.25. The mean effective pressure is found with the help of an indicator equal to 0.85 MPa. If the engine produces indicated process of 35 HP while it is running at 2500 rpm, find the dimension of the engine. (UPTU: 2004)

Pm= 0.85

X

10 3 kPa

1111

Thermal and Hydraulic Machines

IP= 35 x 0.746 kW= 26.11 kW n

pm

X

A

X

L

X

N

X

X

K

=--~-------60

n = 1 for 1 cylinder and K = 1/2 for 4-stroke. Therefore,

850 26.11=

X

A

L

X

X

4

26.11 850

7r

-x D 2 4

X

X

10- 3

1.25 D = 1.474

X

10- 3

X

X

1

60

L = 1.474

D2

2500

2 ---------=-

AL= volume= 7r -

X

1.474

X 4 1.25

D3 = - - - 7r X

X

X X

60 X 2 2500

10- 3 = 1.502

1.474

X

X

10- 3

10-3

D= 1.145 x 10- 1 m = 11.45 cm L D

1.25

L = 1.25 x 11.45 = 14.31 cm

7. A 4-cylinder diesel engine of 4-stroke type has stroke to bore ratio as 1.2 and the cylinder diameter is 12 cm. Estimate indicated power of the engine using the indicator diagram arrangement. The indicator card shows the diagram having an area of 30 cm2 and length as half of the stroke. The indicator spring constant is 20 x 10 3 Nm/m 2 and the engine is running at 2000 rpm. Also find out the mechanical efficiency of the engine if 10% of power is lost in friction and other losses. Given:

LID = 1.2, D = 12 cm

Therefore,

L = 1.2 x 12 = 14.4 cm

1 Length of the indicator diagram = - x stroke 2

=

1

2

x 14.4 = 7.2 cm

Area of the indicator diagram Pm= mep = - - - - - - - - - - " - - - x Spring constant Length of the indicator diagram

m

Thermodynamic Cycles

-

(~ X ~)

7.2

10

= 8.333 IP

n

4

X

X

X

(20

X

8.333

=

103

X

10 3 )

10 5 N/m 2

Pm xAxLxN 60

= = 90,400 Friction power loss

X

X

105

X

7r

4

X

k

X

(0.12)2

X

0.144

X

1 2

X

2000

60 W 0.1 x 90,400

=

9040

IP - FP IP

1J mechanical

90,400 - 9040 90,400

= 0.899 or 89.9% 8. An engine with bore 7 .5 cm and stroke 10 cm has a compression ratio of 6 to 1. To increase the compression ratio, 5 mm is machined off from the cylinder head face. Calculate the new compression ratio. Compression ratio (r) =

Ve+ Vs

6.1 or

l l

....!....

= 5. 1

C

or

l

=

C

l~

=

O.l m

5.1

=

19.6 mm

Clear length after machining

=

19.6 - 5

=

14.6 mm

~

New compression ratio = 1 + l' C

=

1 +

100 14.6

= 1 + 6.85 = 7.85 9. The power output of an IC engine is measured by a rope dynamometer. The diameter of the brake pulley is 700 mm and the rope diameter is 25 mm. The load on the tight

Thermal and Hydraulic Machines

side of the rope is 50 kg mass and the spring balance reads 50 N. The engine running at 900 rev/min consumes the fuel of calorific value 40,000 kJ/kg at a rate of 4 kg/hr. Assume g = 9.81 m/s 2 • Calculate (a) brake specific fuel consumption, and (b) brake thermal efficiency. (GATE: 1997) (W - S),r(D + d)N Brake power= - - - - - - -

6000

where W = weight on tight side, S = spring balance reading, D d = diameter of the rope, N = rpm.

=

diameter of the pulley,

(50 X 9.81 - 50),r(0.7 + 0.025)900 Brak e power = - - - - - - - - - - - - 60,000

= 15.03 kW . . kg of fuel/hr Brake specific fuel consumption = - - - - BHP(kW)

=-

4

15.03

=0.26 kg/kW hr

BHP Brake thermal 1J = - - - - - - - m I x calorific value

15.03 =

4 3600

X

0.31

or

15.03

44,000 = 4

X

X

44

3600 X

103

31%.

10. The minimum pressure and temperature in an Otto cycle are 100 kPa and 27°C. The amount of heat added to air per cycle is 1500 kJ/kg. Determine the pressure and temperature at all points of the air standard Otto cycle. Also calculate the specific work and the thermal efficiency of the cycle for a compression ratio of 8 : 1 (Take cv(air) = 0.72 kJ/kg Kand cPlcv = 1.4) (Figure 6.23). (GATE: 1998) p

V

FIGURE 6.23

PV diagram (Problem 10).

m

Thermodynamic Cycles

Process 1-2: -P2 =

Pi

P2

(Vj- JV =(r)

v

V2 =

P,(8)'- 4

=

100

X

(8) 14

= 1838 kPa Also

0.4

1838 )1.4 =300 X ( - 100

= 689 K Process 2-3:

or

T3

=

Q, 689 + ~

=

689 +

CV

1500

0.72

= 2773 K T3 P 3 = P 2 x T2 = 1838

= 7394 kPa Process 3-4:

~ = (:: J= P4

=7394

X

(ff

1 Jl.4 (S

= 402 kPa

2773 X

689

1111

Thermal and Hydraulic Machines

0.4

)1.4 402 =2773 X ( 7394 = 1207 K Heat rejected

Qrej

= cv(T4

-

T,)

= 0.72(1207 - 300) = 653 kJ/kg Work =

Qadd - Qrej

= 1500 - 653 = 847 kJ/kg

1J = -

W

Qadd

847

= = 0.565 1500

11. A large diesel engine runs on a stroke cycle at 2000 rpm. The engine has a displacement of 25 litres and a brake mean affective pressure of 0.6 mn/m2• It consumes 0.018 kg/s (calorific value = 42,000 kJ/kg). Determine the brake power and the brake thermal efficiency. (GATE: 1999) (pm)(l X A) X N

Break power= - - - - - - x n 60 l x A = displacement volume = 25

X

10-3 = 0.025 m 3

n = 1/2 for 4 stroke 0.6 X 103 X 0.025 X 2000 Break power= - - - - - - - - - 2 X 60 = 250 kW Break power Break efficiency 1J = - - - - Heat supplied 250

=------

0.018

=0.331

X

42,000 or

33.1%

-

Thermodynamic Cycles

12. A diesel engine develops a brake power of 45 kW. Its indicated thermal efficiency is 30% and the mechanical efficiency is 85%. Take the calorific value of the fuel as 40,000 kJ/kg and calculate (a) the fuel consumption in kg/hr and (b) the indicated specific fuel consumption. (GATE: 2000)

Given:

BP

17thermal =

0.3 and

17mech =

0.85, BP

4.5,

=

11mech

=

IP

BP 4.5 IP= - - = =5.29kW 11mech 0.85

or

IP

IP

11thermal

= _H_t____ = - - ea given

m1 x c1

5.29 0.3=-----mf X 40 X 106 mf

= 0.44

X

10-3 kg/s

0.44

X

10-3

=

X

3600

=

1.58 kg/hr

Indicated specific fuel consumption

= l.SS =0.298 kg/kW hr 5.29

13. In a spark ignition engine working on the ideal Otto cycle, the compression rated is 5.5. The work output per cycle (i.e. area of PV diagram) is equal to 23.625 x 10 5 x Ve joules, where Ve is clearance volume in m3 • The indicated mean effective pressure is (a) 4.295 bar (b) 5.25 bar (c) 86.87 bar (d) 106.3 bar (GATE: 2001)

Compression ratio

r

or

V

Work

V

Ve

= 5.5

= 5.5 Ve

= Pmean (V = 23.625

pmean

X

- Ve)

10 15

X

Ve

23.625 X 105 X Ve = ------V - Ve

1111

Thermal and Hydraulic Machines

=

=

23.625

X

105

23.625

X

105

X

Ve

4.5 = 5.25 bar

Option (b) is correct. 14. An ideal air standard Otto cycle has a compression ratio of 8.5. If the ratio of the specific heats of air ( v) is 1.4, what is the thermal efficiency (in percentage) of the Otto cycle? (a) 57.5 (b) 45.7 (c) 52.5 (d) 95

(GATE: 2002)

1

11 = 1 =1 -

r

v-1

1 (8.5)'-4-1

1

=1 - - - -

(8.5)°"4

= 1 - 0.425 = 0.575 or 57.5% Option (a) is correct. 15. An engine working on air standard Otto cycle has a cylinder diameter of 10 cm and stroke length of 15 cm. The ratio of specific heats for air is 1.4. If the clearance volume is 196.3 cc and the heat supplied per kg of air per cycle is 1800 kJ/kg, the work output per cycle per kg of air is (d) 973.5 kJ (a) 879.1 kJ (b) 890.2 kJ (c) 895.3 kJ

(GATE: 2004) Volume swept V8

=

,r

4

x d

2

x L

1C

= - X (10)2 X 15 4

= 1178 r

cc

= Compression ratio

=

Ve+ V,

Ve

Thermodynamic Cycles

196.3 + 1178 =----=7 196.3

-

1 11otto =1-v=ir

1 =1- - 71.4-1

= 0.541

or

54.1 %

w

1]= -

Q

or

W=71xQ

= 0.541

X

1800

= 973.5 kJ Option (d) is correct.

16. A 4-cylinder petrol engine has a swept volume of 2000 cm 3 and the clearance volume in each cylinder is 60 cm 3 • If the pressure and temperature at the beginning of compression are 1 bar and 24°C and the maximum cycle temperature is 1500°C, the air standard efficiency will be (b) 59% (c) 61% (d) 63% (a) 58% (GATE: 2005)

. . V,+Vc Compression ratio r = V C

2000+60 X 4 =----4 X 60

= 9.333 11otto = 1 -

1 ,v-1

1 =1 - - - - (9.333)'-4-l

= 0.59 or 59% Option (b) is correct.

17. For the previous problem of the petrol engine the mean effective pressure (Figure 6.24) will be (a) 4.83 bar (b) 5.83 bar (c) 6.83 bar (d) 8.83 bar

Iii

Thermal and Hydraulic Machines p 3

4

2

V

FIGURE 6.24

T1

=

24°C

PV diagram (Problem 17).

307 K

=

v-1

T2

= I;

(

V2 )

=307

'1;

= 307

x r

v-t

(9.333) 04

X

= 726 K 1500°C (Given)

T3

1500 + 273 1773 K a

pm

T

1773

T2

726

= - 3 = - - = 2.44 =Pi = 1

X

X

r

a -

1[rv-l - 1]

X --

9.33

V

X

r - 1

-1

1

2.44 [(9.33)'- 4- 1 1.4 - 1 9.33 - 1

1]

= 5.83 bar Option (b) is correct. 18. A diesel cycle takes air at 1.0 bar and 300 K and compresses it to 16 bar. Heat is added till its temperature becomes 1700 K. Calculate (a) work from cycle, and (b) air standard efficiency (Figure 6.25). (UPTU: 2007-2008) Process 1-2:

Compression ratio = r =

(p

V, )1/v _!_= _2.

V2

Pi

=(16) 111 .4 =7.24

..

Thermodynamic Cycles p

2

3

4

V

FIGURE 6.25

PV diagram (Problem 18).

v-1

Also

T2

I;

=(P2 )-;Pi

T2 = 300

X

(16)0.4/1. 4

= 616 K Process 2-3: Given:

T3

= 1700 K

V3 T3 1700 - = - = - = 2.78=/3 V2 T2 616

Process 3-4:

= (16. 1700 2

=

_l_)v-1 2.78 =

=2

850 K

1.005(1700 - 616)

11111

Thermal and Hydraulic Machines

= 1089.4 kJ/kg q2 = cu(T4 - T1)

= 0.718(850 - 300) = 0.718

X

550

= 395 kJ/kg

=1-

395 1089.4

= 1 - 0.363 = 0.637 W = q, - q2

= 1089.4 - 395 = 694.4 kJ/kg 19. An engine working on a direct cycle has air intake conditions of 1 bar and 310 K and compression ratio of 17. Heat added at high pressure is 1250 kJ/kg. Make calculation for the maximum temperature of the cycle, net power output and thermal efficiency (Figure 6.26). (UPTU: 2006-2007) p

2

3

V

FIGURE 6.26

q,

PV diagram (Problem 19).

=

1250 kJ/kg

Process 1-2: T For adiabatic process: --2.

I;

)v- = 17°.4 =3.1 = (v, ___.!... 1

V2

T2 = 3.1 x 310 = 963 K

m

Thermodynamic Cycles

Process 2-3: q1

=

1250

=

cp(T3

= 1.005(T3

T2)

-

963)

-

T3 = 2007 K

/3

=

cutoff ratio

=

V3 V2

-

T3 T2

- -

2007 963 - 2 .3

- -- -

Process 3-4:

For adiabatic process:

T3 = T4

=

(V )v-t 4

V3

(Vi_ . v2 Jv-l V2

V3

[as

Vi =

Vi]

=(;f 17 = (2.3 or

)o.

4

=2.17

2207 T4 = - - = 1017 2.17 q2 = cv(T4

-

T 1) = 0.718(1017 - 310)

= 507.5 kJ/kg q2 507.5 11 = l - -;;; = l - 1250

= 0.7084 W = 11 X q, =

0.7084

=

885.6 kJ/kg

X

1250

20. In a Diesel cycle, the compression ratio is 10 and cutoff ratio is 3. If the initial temperature is 300 K, find other temperaturs. Assumes v = 1.4.

-

Thermal and Hydraulic Machines

r

T2 is retated to T 1 by compression ratio, i.e. T2 = T1 rv-- 1• T2 is related to T3 by cut-

off catio, i.e. T3 = /JT2 • Also T4 is celated to T3 by celation T3 =

T2

=

300

X

10 14- 1 = 300

X

r.(;

10° 4

= 753.6 K T3

=

/3

T2

X

=

3

X

753.6

= 2260 K 2260

T3

(;f- en· 2260

= - - = 1396K 1.618

21. For an Otto cycle shown in Figure 6.27 has T 1 = 300 K, T2 and T4 = 900 K. If cv = 0.718, find work and efficiency.

=

800 K, T 3

=

2100 K

p

V

FIGURE 6.27

PV diagram (Problem 21).

Work = q 1

q2

-

= cu(T3 - T2) - Cv(T4 - T1) = cu[(T3 - T2) - (T4 - T 1)] = 0.718[(2100 - 800) - (900 - 300)] = 0.718(1300 - 600) qi

1J

= = =

502.6 J/kg 0.718

X

1300

933.4 kJ/kg w

502.6

qi

933.4

= - = - - = 0.54

22. If in an Otto cycle, the product of pressure and volume at each point starting from initial is: (a) p 1v 1 = 700 kJ/kg, (b) p 2v 2 = 1200 kJ/kg, (c) p 3v 3 = 4200 kJ and (d) p 4 v 4 = 1400 kJ. Find work. Assume R = 0.286 kJ/kg K and cv = 0.718 kJ/kg K

Thermodynamic Cycles

Work = cu[(T3 - T4) - (T2 - T 1)]

= RT

PV

But

-

0.718 = - - [(4200 - 1400) - (1200 - 700)] 0.286 0.718 = - - [1800 - 500] 0.286

= 3.26 kJ/kg 23. For a Diesel cycle, the following data were observed. Air inlet pressure and temperature = 1 bar and 300 K. Compression ratio = 20, cutoff ratio = 2. Calculate the temperatures at all points and cycle, net power output and thermal efficiency of the cycle. (UPTU: May 2008) From initial temperature T 1, we can find out other temperatures as under: (a) T2 = T 1(r)v-- 1 (b) T3 = T2

X

/3

Now T 1 = 300 K, hence we have T2 = T1

X

= 300

(r) v--l X

(20)!4-l

= 994.3 K also

T3

=

/3 T2

=

2

X

994.3

= 1988.6 K also

Now

T3

1988.6

(;f-(~f = 791.64 K qadd = q, = Cµ(T3 = 999.6 kJ/kg qrej = q2 = cv(T4 -

T2)

= 1.005(1988.6 - 994)

T,)

= 0.718(791.64 - 300)

-

Thermal and Hydraulic Machines

= 353 kJ/kg w = q, - q2 = 999.6 - 353 = 966.6

1J

=1 -

q2

qi

=1 -

353 999.6

=0.647

OBJECTIVE TYPE QUESTIONS State True or False 1. An air-standard cycle with assumptions is a gas power cycle.

(True/False)

2. An air-standard cycle is assumed to be a closed air cycle.

(True/False)

3. Heat transferred is assumed to be through combustion in the cylinder. (True/False) 4. If the cylinder head is machined off, the compression ratio decreases. (True/False) 5. The Otto cycle consists of two isobaric and two adiabatic processes. (True/False) 6. The Diesel cycle consists of one isochoric, one isobaric and two adiabatic processes. (True/False)

7. Spark ignition engines are based on Diesel cycle.

(True I False)

8. Compression ignition engines use diesel as fuel.

(True/False)

9. Diesel has high autoignition temperature.

(True I False)

10. Petrol engines are based on the Otto cycle.

(True I False)

11. Petrol has low self-ignition temperature.

(True /False)

12. The efficiency of the Otto cycle decreases with increase of compression ratio. (True/False) 13. For maximum work from the Otto cycle with the highest temperature = T 3 and the

lowest temperature= T 1, T2 and T4 are equal and the value is

.JTi T

3 •

(True I False) I

T3)2 Diesel cycle (b) diesel cycle > Otto cycle ( c) otto cycle = Diesel cycle 46. An SI engine working on the Otto cycle has the compression ratio 5.5, the work output/cycle on a PV diagram is 23.625 x 10 3 x fie joules (fie = clearance volume). The indicated mcp is (a) 4295 bar (b) 5250 bar (c) 106.3 bar

Thermodynamic Cycles

Fill in the Blanks 1. In the Otto cycle, heat addition and rejection is an ____ process. (a) isochoric (b) isobaric 2. In the Diesel cycle, heat is added by an ____ process. (a) isobaric (b) isochoric

3. A 2-stroke IC engine has (a) valves 4. A 4-stroke IC engine has (a) valves

(b) ports (b) ports

cycle. 5. An SI engine is based on the (a) otto (b) diesel

6. A CI engine is based on the (a) otto

cycle. (b) diesel

7. An SI engine has a ____ compression ratio in comparison to an CI engine. (a) higher (b) lower 8. An SI engine has (a) higher

speed in comparison to CI engine. (b) lower

9. The intake of an SI engine is ____ during the suction stroke. (b) air - fuel mixture (a) air 10. The intake of a CI engine is ____ during the suction stroke. (b) air - fuel mixture (a) air 11. A 2-stroke IC engine has an inlet port, a ____ port and an exhaust port.

(a) transfer

(b) midport

12. The exhaust port is located ____ to TDC in a 2-stroke IC engine. ( a) nearest (b) farthest 13. The inlet port is located ____ to BDC in a 2-stroke IC engine. ( a) nearest (b) farthest 14. The transfer port is located below the ____ port in a 2-stroke IC engine. (a) exhaust (b) inlet 15. A crank also performs the operation of ____ of charge in the crankcase before its transfer to the cylinder in 2-stroke CI engine. (a) transfer (b) precompression

16. High octane SI fuel has ____ tendency of knocking. (a) high (b) low

-

Thermal and Hydraulic Machines

17. For the same compression, the Otto cycle has a ____ efficiency than that of the Diesel cycle. (a) higher (b) lower 18. A Carnot engine has a ____ efficiency than that of the Otto or Diesel cycle. (a) lower (b) higher

19. The efficiency of the Otto cycle ____ with increase of the compression ratio. ( a) increases (b) decreases 20. The efficiency of the Diesel cycle ____ with increase of the cutoff ratio. ( a) increases (b) decreases 21. Knocking in an SI engine is ____ by reduced turbulence of air - fuel mixture in the cylinder. (a) encouraged (b) discouraged 22. For the same heat input and maximum pressure, the thermal efficiency of the Otto cycle is _ _ _ _ than that of the Diesel cycle. (a) higher (b) lower

Answers State True or False 1. True 2. True 3. False (Heat transfer is considered from an external source instead of combustion in the cylinder.) 4. False (In an actual engine, the cylinder head block and engine block are joined together to form the cylinder of the engine. Machining of any of two reduces the clearance length.) 5. True 6. True

7. False (An SI engine uses petrol only as petrol gives readily air - fuel mixture which has high self-ignition temperature and can be ignited by spark.) 8. True (Diesel has lower self-ignition temperature and undergoes spontaneous combustion when diesel is sprayed in highly compressed and hot air.) 9. True 10. True 11. False (Petrol has high self-ignition temperature to avoid any self-ignition of mixture before spark reaches it, otherwise knocking will take place if any self-ignition starts.) 12. False (The efficiency of the Otto cycle increases with increase of compression ratio

as 1J -- 1 - -1- )

,r-1

-

Thermodynamic Cycles

13. True 14. True 1 15. False (Diesel cycle efficiency = 1- ,r _1 k and k factor > 1)

16. False (Both are used in IC engines.) 17. True 1 18. False (1] = 1- ,r-i k and k =

/JY -1 r(/3 - l) . As k increases with cutoff (/3),

1]

decreases.)

1 19. True (1] = 1- y-=T x k and as r increases, the negative quantity decreases.) r 20. True (High pressure for fuel is required to be injected in compressed air.)

21. True (Fuel - air mixture is taken in the suction stroke.) 22. False (Compression ratio is more in a CI engine. As for self-ignition of diesel, air is to be at high pressure and temperature.) 23. True (There is one power stroke in 2 revolutions of the crankshaft. Energy is stored in power stroke which is to be used in remaining 3 strokes. Hence a heavy flywheel is to store more energy.) 24. True (A 2-stroke engine has one power stroke per revolution while a 4-stroke engine has one power stroke per two revolutions.) 25. False (A 2-stroke engine has ports which are covered and uncovered by the piston side.) 26. True (Precompression of charge takes place in the crankcase when the piston moves down from TDC during the power stroke.) 27. True (The transfer port connects the crankcase to the cylinder at a place between the inlet port and the outlet port. The precompressed charge from the crankcase is transferred to the cylinder when the piston is moving down in power stroke and uncovers the transfer port.) 28. True (Inlet and outlet valves are to be opened once in two revolutions of the crankshaft. The camshaft is designed to run at half speed of the crankshaft to operate inlet and outlet valve mechanism.) 29. False (The camshaft runs at half speed of the crankshaft.) 30. True 31. False (IHP > BHP and !HP-friction horse power= BHP. Also

= BHP/IHP) 32. False (IHP is power actually generated in the cylinder while power available at the crankshaft is BHP = (IHP-FHP)) 11mechanical

-

Thermal and Hydraulic Machines

33. False (Volumetric efficiency in a 2-stroke engine is lower as charge is lost in scavenging,

i.e. pushing out the burnt gases out of the cylinder through the exhaust port.)

34. True 35. True 36. True (The piston has deflector-like shape and pushes incoming precompressed charge

through the transfer port towards top of the cylinder, thereby preventing charge rushing out of the exhaust port.) 37. True (Combustion cannot be isochoric as the piston moves down as diesel is being

sprayed and spontaneously undergoing combustion.) 38. True (Spark is given to the compressed fixed volume of air and fuel mixture which

burns spontaneously at constant volume). 39. True 40. False (As compression is higher in a CI engine as compared to an SI engine, the

stroke length and pressure against which the piston is moving. TDC is higher resulting into lower speed.) 41. False (Since stroke length is more, the size of the cylinder of a CI engine is more.

Higher compression means more pressure requiring thicker wall cylinder, piston and other parts. Therefore, a CI engine is heavy and can be used for heavy vehicles.) 42. True (The average temperature at which heat can be added to the system is higher

in the Carnot cycle as compared to the Otto and Diesel cycles. Therefore, the efficiency of the Carnot cycle is maximum.) 43. False (IP = BP + FP) 44. True (Extreme positions are called TDC and BDC.) 45. True 46. True 47. True 48. True 49. False (Ports are covered and uncovered by the piston while moving from TDC to

BDC and vice versa.) 50. Tree (Due to the volumetric efficiency of a 4-stroke engine is higher than that of a

2-stroke engine.) 51. True [In case mixture gets ignited before spark reaches the mixture. (Piston is yet

to reach TDC) power will be used firstly to stop the upward movement of the piston abruptly and remaining power for acceleration of the piston towards BDC. The effect will be knocking Gerking) and wastage of power.] 52. True 53. True 54. True

Thermodynamic Cycles

Multiple Choice Questions 1. (b)

2. (b)

3. (c)

4. (b)

5. (a)

6. (b)

7. (a)

8. (b)

-

9. (c) 10. (b) (Inlet and outlet are to be opened once in 2 revolutions of the crankshaft.) 11. (c)

13. (c)

12. (a)

14. (a) (Lesser compression means shorter piston movement resulting higher speed.) 15. (c)

16. (c)

17. (c)

18. (c)

19. (a) 20. (c) (Work output depends upon the heat added. Since fuel is lost to some extent in

scavenging, heat added is smaller in a 2-stroke engine in comparison to a 4-stroke engine for work output resulting into lower efficiency of a 2-stroke engine.) 21. (c) (Some charge in a 2-stroke engine is wasted in scavenging.) 1 1 22. (a) (11otto = 1- y=-i"" and 1Joiesel = 1- y=-i"" x k where k > 1. Therefore, for the same r r r (compression ratio) 1] Otto > 1Joiese1)

24. (c) 28. (c) 29. (b) (Friction power remains constant.) 23. (b)

25. (a)

26. (b)

27. (a)

31. (b)

30. (c)

[/3 = V2 V1

32. (b) (11

=

1 -,r-1 -1-

1-) = = (1-7 o.4

33. (c)

= 1+ Vs = 1+ 400 = 40

Ve

1]

=

1- -

,r-

Qactct =

37. (b) (

1 1

=

1-

1 + 10

1

11°.4

=

100 0 _616 = 162 kW)

/3 = ~: = :~ =

1.5)

10

1 - 0.33

34. (a)

36. (a) (r

= 20 = 2,

T= V4 V2

= 80 = 8]

= 0.67) 35. (b)

=

11

1 - 0.383

= 0.617 =

W Qadd

10

-

Thermal and Hydraulic Machines

38. (a)

= cp(T3 - T2) = I x (800 - 500) = 300 kJ/kg) 40. (b) (Qrej = cu(T4 - T1) = 0.72(400 - 300) = 72 kJ/kg) 41. (a) (W = Q add - Q rej = 300 - 72 = 228 kJ/ K)

39. (a)

(Qadd

42. (a) 43. (c) (The octane number is parts of isooctane in 100 parts of fuel.) 44. (a) (A 4-stroke engine will require a heavier flywheel as there is one power stroke in 2 revolutions of the crankshaft.) 45. (b) (Work of the Diesel cycle is higher as shown in the diagram by the shaded portion .) p

4

V

46. (a) (mep =

mep =

area swept vol

23.625

X

105

X

Ve

vs

1 V Ve -and 1 + _£ = 5.5 or Ve vs 4.5

23.625 X 105 = 5250 bar) 4.5

Fill in the Blanks 1. (a)

2. (a)

3. (b)

4. (a)

5. (a)

6. (b)

7. (b)

8. (a)

9. (b)

10. (a)

11. (a)

12. (a)

13. (a)

14. (a)

15. (b)

16. (b)

1 17. (a) (1Jotto = 1 - - - and r Y- l '

1

1Joiesel

-= 1 - r1 k where k > r

1Jotto >

1JoieseI)

18. (b) 19. (a)

20. (b)

21. (a)

22. (a)

Steam Turbine

INTRODUCTION A steam turbine is a rotary machine which is required to convert steam energy having high pressure and temperature into mechanical power. In steam turbine, its operation is fully dependent on the dynamic action of the steam. The steam is firstly expanded in a set of nozzles from high inlet pressure to exit pressure, thereby the pressure energy of the steam is converted into kinetic energy in the nozzles. These nozzles are fixed to the casing of the turbine. The high velocity steam is passed over the blades (curved vanes) of the turbine. The steam follows the curved surface while passing over the blades and the direction of the steam flow changes while flowing over the blades. This results into the change of momentum of the steam which develops a resultant force acting on the blades. The blades are attached to the rotor of the turbine which is free to rotate. The force on the blades causes the rotor to rotate and performs useful work. The ring of nozzles fixed to the casing and the ring of moving blades fixed to the rotor form a pair and it is called a stage or a turbine pair. The steam turbine is a prime mover in which the potential energy of the steam ( enthalpy) is transformed into kinetic energy, which is further transformed into mechanical energy by causing the rotation of the turbine shaft. A steam turbine is used in many fields in industry which mainly include power generation and transportation (shipping).

PRINCIPLE OF OPERATION In a steam turbine, the energy in steam as pressure energy is used for producing rotary motion, which is usually used to run electric generators in power house. The steam is discharged in the 241

-

Thermal and Hydraulic Machines

form of a high velocity jet from a nozzle which impinges upon blades mounted on a wheel, whereby the wheel is caused to rotate. In order to convert the pressure energy of steam as efficiently as possible into kinetic energy, nozzles (called Laval nozzles) are used comprising of an inlet portion, a constricted throat and a gradually widening outlet as shown in Figure 7. I A( a). As a result of the constriction of the passage followed by the widened outlet portion (also called diffuser) the pressure energy of the steam flowing through this nozzle is converted into kinetic energy. The higher is the velocity of steam, there is larger force in the jet while exiting from the nozzle and impinging on the blades. The force on the blades also depends upon how much the steam is deflected from the original path. The wheel is provided with a set of blades which can deflect the steam into very nearly the circumferential direction as shown in Figures 7. I A(b) and (c) and hence steam exerts on the wheel a force in circumferential direction, causing the wheel to rotate. In order to achieve the fullest possible utilisation of steam energy, a number of successive stages are usually arranged one behind the other. All these wheels in different stages are mounted on the same shaft and they all therefore revolve with the same speed.

High pressure and low velocity steam

Low pressure and low velocity steam

(a) Nozzle Nozzle

Moving F. d Moving blades ixe blades blades

Nozzle

Motion of rotor (b) Wheel with blades

FIGURE 7. l(A)

'°'

(c) Cut section showing rows of nozzles, moving blades and fixed blades

Principle of steam turbine.

In a reciprocating engine, the pressure energy of steam is used to overcome external resistance so that steam expands when piston moves, thereby doing external work. The dynamic action of the steam therefore does not exist in the steam engine. However, steam turbine does not operate in such a manner. The turbine depends upon the dynamic action of the steam. The steam is caused to fall in pressure in a nozzle. The pressure (enthalpy) energy is converted into kinetic energy. The high velocity steam coming out from the nozzle impinges on the blade of the turbine. The blade changes the direction of motion of the steam, thereby causing the change of momentum of the steam. A force therefore acts on the blade which is equal to the change of the momentum of the steam. As shown in Figure 7.1 B, the nozzle forms a high velocity jet and impinges on the

Steam Turbine

blade (bucket). The moving bucket converts the kinetic energy of the steam into mechanical work. When bucket velocity ( Vb) is zero, then the inlet velocity ( V1) and outlet velocity (V2) of steam jet is equal which gives maximum force on the bucket. However, no mechanical work is done in this situation as the bucket velocity is zero (work= F x Vb)- As the bucket is allowed to speed up, the outlet velocity of steam jet decreases, thereby resulting into fall of force and in an increase of mechanical work. The steam jet performs maximum work when bucket speed is half of the inlet speed of the steam. In this condition, the outlet speed of the jet is almost zero. Impulse turbine works on this principle and this is characterised by the constant pressure of the steam from inlet to outlet of the blade. If the pressure of the steam at outlet is lesser than that at inlet to the blade, the drop in pressure suffered by the steam during its flow through the moving blade causes a further generation of kinetic energy within the blade which adds to the propelling force. The turbines using this principle are called impulse-reaction turbines. r----,...-Bucket or blade V ~

F

F

V2

V1

= V2

(a) Maximum force (start)

(b) Force and work (normal)

(a) Maximum work (maximum load)

Q)-"

0

....

.... 0

~$

Vb = Blade velocity v, = Inlet velocity V2 = Outlet velocity 0

0.5

1

vb

V, (d) Variation of force and work with blade velocity

FIGURE 7.1(8)

Principle of steam turbine.

TYPES OF STEAM TURBINES The steam turbines can be: (a) Impulse turbines (b) Reaction turbines Impulse turbines have following features: (a) The turbine works on the principle of impulse in which the kinetic energy is obtained by the steam after passing through a fixed nozzle and this energy is used to exert the force on moving blades fixed onto a rotor.

m

Thermal and Hydraulic Machines

(b) The pressure of steam remains constant when it moves on the moving blades. ( c) The mechanical work is obtained by changing the momentum of the steam when it moves on the blades. (d) Turbines such as DeLaval, Curtis and Rateau are impulse turbines. The reaction turbines have following features: (a) There is a continuous drop of steam pressure while the steam passes over the moving blades. (b) The moving blades are suitably designed such that the steam is made to expand, thereby moving blades also act like nozzles. ( c) The motive force acting on the moving blades is the sum total of (i) the change of momentum of steam entering the moving blades with some definite velocity and (ii) the drop of pressure of the steam due to nozzle type shape of the blades which results into further change of momentum, thereby generating reactive force. (d) Reaction turbine is generally having axial flow of the steam, i.e. the steam flows over the moving blades in direction parallel to the axis of the rotor of the turbine. (e) Lungstorm reaction turbine is the only reaction turbine which has radial flow instead of axial flow. The steam enters in this turbine along the axis of the turbine rotor and then it flows towards the rotor's circumference (Figures 7.2 and 7.3). Steam from boiler

Steam from boiler

(a) Impulse turbine

(b) Reaction turbine

FIGURE 7.2

Impulse and reaction turbine.

(a) Impulse blades (symmetrical)

FIGURE 7.3

(b) Reaction blades (asymmetrical)

Blades of impulse and reaction turbine.

Steam Turbine

ADVANTAGES OF STEAM TURBINE The advantages of steam turbine are: (a) It rotates at high speed and it has a great range of speeds. (b) It is compact. Hence, it has a low weight to power ratio. It therefore needs less floor area. ( c) It has perfect balance. Hence, its running is vibration free. ( d) It has much higher thermal efficiency as compared to steam engine. (e) It has least balancing problem as it has no reciprocating part. (t) It needs no internal lubrication as there is hardly any rubbing part in the turbine.

(g) The power generation in steam turbine takes place at uniform rate. Hence, no fly wheel is required. (h) Overloading is possible at the expense of slight reduction in overall efficiency. (i) It has low initial cost as well as maintenance cost. G) It is highly suitable for running electrical generators which can be directly coupled to it.

CLASSIFICATION OF STEAM TURBINES Steam turbines can be classified in the following ways: (a) According to the principle of working: (i) Impulse turbine (ii) Reaction turbine (b) According to the direction of steam flow: (i) Axial

(ii) Radial (iii) Tangential ( c) According to the number of pressure stages: (i) Single stage (ii) Multistage ( d) According to the method of governing: (i) Throttle governing (ii) Nozzle governing (iii) By-pass governing (iv) Combination of throttle-by pass and nozzle-by pass governing (e) According to the heat drop process: (i) Non-condensing

1111

Thermal and Hydraulic Machines

(ii) Condensing (iii) Regeneration (t) According to the steam conditions at inlet:

(i) Low pressure up to 2 bar (ii) Medium pressure up to 50 bar (iii) High pressure above 50 bar (iv) Supercritical pressure above 224 bar (g) According to their usage: (i) Stationary with constant speed (ii) Stationary with variable speed (iii) Non-stationary. These are used in marine and locomotive engines

WORKING OF STEAM TURBINES In impulse turbine, the steam coming out at a very high velocity from a fixed nozzle, impinges on the blade which changes its direction. The resultant reaction force rotates the rotor. In pure reaction turbine, the high pressure steam accelerates on the nozzle-shaped blades, thereby the velocity of the steam increases. The reaction force on the blades provides the rotary motion. The impulse blades are symmetrical in shape whereas the reaction blades are asymmetrical. The shape of reaction blades are thicker at one end which helps in providing a suitable passage for steam to expand. The nozzle and blades arrangement for an impulse turbine is shown in Figure 7.4. The variation of pressure and velocity of steam while passing through the blades is also shown in this figure. The drop of pressure in the nozzle gradually generates increased velocity in the nozzle. In moving blades, the pressure of the steam remains constant and velocity gradually decreases as it is used up for work output.

Velocity blades /

/

Pressure of steams

Steam flow

Velocity of steam at exit 1-+----+-+--C-0-n-denserl pressure (a) Blades arrangement

(b) Pressure and velocity variation

FIGURE 7.4

Impulse turbine.

Steam Turbine

The reaction turbine has both the stationary and moving blades which are arranged as shown in Figure 7.5. The variation of pressure and velocity is also shown in this figure while steam passes through the stationary and moveable blades. The pressure gradually falls and velocity gradually increases when steam passes through the stationary blades (guide vanes). The steam with increased velocity enters the moveable blades. Here, both pressure and velocity fall when steam passes through the moveable blades, thereby steam performs work on the rotor. Fixed blades

Moving blades

Fixed blades

Moving blades

Steam flow

Direction of rotation (a) Blades arrangement

Velocity of steam / at entrance

(b) Pressure and velocity variation

FIGURE 7.5

Reaction turbine (two stage).

Most turbines have axial flow except Lungstorm turbine which has radial flow. Maximum efficiency of impulse turbine results when the speed of entering steam is twice that of the speed of the moving blades. In order to achieve maximum efficiency, impulse turbine must operate either at extremely high speed or the pressure drop must be carried out in a number of stages. Most of the turbines are operated at 3000 - 3600 rpm and they have a number of stages. This speed range helps in direct coupling of turbines to the electric generators. Most of the power plant turbines are of impulse-reaction types with first one or two stages consisting of impulse type turbines as they are suitable for low volume steam at the high inlet pressure. Nearly 20 - 30 stages are commonly used with reaction turbines.

Ill

Thermal and Hydraulic Machines

Differences between Impulse and Reaction Turbines The comparison of impulse turbine and reaction turbine is given in table 7. I. TABLE 7. 1

Impulse turbine and react ion turb ine Reaction turbine

Impulse turbine 1. Steam completely expands in the nozzles.

1. No nozzle but the shape of moving blades is such that they act as nozzle.

2. Pressure remains constant during flow over moving blades .

2. Pressure changes while the steam is moving on the moving blades.

3. Relative velocity of steam while moving on the moving blades remains constant.

3. Relative velocity of steam changes as the steam expands while moving over the moving blades.

4. Pressure on both ends of the moving blade is same.

4. Pressures at the inlet and exit of the moving blades are not same.

5. Number of stages required is less for a given pressure drop.

5. Number of stages required is more for a given pressure drop.

6. The blade efficiency curve is not flat.

6. The blade efficiency curve is flat. Hence it has part load economy.

7. The steam velocity is high.

7. The steam velocity is not high which gives turbine a relative low speed.

COMPOUNDING OF STEAM TURBINES To increase the thermal efficiency and to reduce the size of the plant, the present trend is to increase the steam pressure and temperature. The pressure in the range of 100 - 140 bar is commonly used in power plants. The velocity of the steam will be very high, incase entire pressure drop from the boiler pressure (I 00 - 140 bar) to condenser pressure (0 .1 bar) is carried out in a single stage nozzle. Due to high steam velocity, the turbine speed has to be high which is not useful for practical purposes as (i) reduction gearing has to be used between the turbine and electric generator and (ii) there is also risk of structural failure of blades at high speed. Hence steam velocity is generally controlled by using stages. The method of reducing the rotor speed is called the compounding of steam turbine. Each method of compounding consists of multiple rotor system keyed to the common shaft with a number of stages arranged in series. The compounding can be done by (i) velocity compounding, (ii) pressure compounding and (iii) the combination of pressure and velocity compounding.

Velocity compounding: In this type of compounding, the steam expands in a set of nozzles from high pressure (boiler pressure) to low pressure (condenser pressure), thereby converting complete pressure energy into kinetic energy. There are two or more rows of moving blades. There is a row of fixed guide blades in between the moving blades. The function of fixed blades is only to guide the steam coming from first moving blades row to next moving blades row as shown in Figure 7.6. The enthalpy of steam drops at nozzle of first stage, thereby heat energy is converted into kinetic energy. The kinetic energy of the steam at nozzles is successively absorbed

Steam Turbine

in stages by the rows of moving blades and finally the steam is exhausted from the last row of blades. Curtis turbine uses such velocity compounding. The variation of pressure and velocity of the steam along the axis is also shown in Figure 7. 7. The specific volume of the steam remains constant during flow. Hence blade height is same in all rows. Nozzle

Moving blades

Stationary blades

.

Moving blades

Stationary blades

.

I

Pressure of steam at inlet

Initial velocity /

lT FIGURE 7.6

Velocity compounding in impulse turbine: principal. ~~~~~-~-~+-- Casing Fixed blades

Stearne:::=>

Moving blades

--------i--Rotor

Shaft

FIGURE 7.7

Velocity compounding: arrangement.

Pressure compounding: In this compounding, the total pressure of the steam is not dropped at first stage but it takes place in successive stages. The steam is partially expanded in first ring of nozzles and velocity gained is absorbed in the first ring of moving blades. The steam exiting in second stage is again expanded in second ring of nozzles and velocity gained is absorbed in the second ring of moving blades. This arrangement continues till the pressure has been reduced to the condenser pressure. Since the steam is partially expanded in each ring of nozzles, the steam velocity is not high which ensures low turbine velocity. In this arrangement, the turbine is provided

ml

Thermal and Hydraulic Machines

with one row of nozzles (fixed blade type) at the entry of each row of moving blades as shown in Figure 7 .8. The total pressure drop of the steam does not take place in a single row of nozzles but it is divided among all the rows of fixed blades which work as nozzles as shown in Figure 7.8. As the pressure of steam gradually decreases, the specific volume of the steam gradually increases. Hence the blade height has to be increased towards the low pressure side. Nozzle

Moving blades

Nozzle

Moving blades

Bioler pressure

Initial velocity Condenser pressure

FIGURE 7.8

Pressure compounded impulse turbine.

Pressure and velocity compounding: This is a combination of pressure and velocity compounding. The total pressure drop of the steam is divided into a number of stages as done in pressure compounding, and the velocity obtained in each stage is also absorbed in several stages using velocity compounding. This arrangement requires fewer stages, thereby a compact turbine can be designed for a given pressure drop. The velocity and pressure drop along the axis of turbine are shown in Figure 7 .9. The blade height in the second stage must be greater than that of the first stage as the pressure is reduced (specific volume increases) after each stage. This method has the advantage of pressure compounding which gives higher pressure drop in each stage, thereby fewer stages, and the advantage of velocity compounding which gives reduced velocity after each stage.

VELOCITY DIAGRAMS In order to determine the force acting on blades, the rate of change of momentum of steam across the moving blades has to be found out. In order to find power developed by the steam, it

Steam Turbine Nozzle

Moving blades

\

I I I Q)

/

ai :i

/

= u, al~

l e.

t t

I I

Fixed blades

Moving blades

Nozzle

Moving blades

\

\

\

\ \

\

\

\

\

\\

\

\

\

\

Fixed blades

Moving blades

Nozzle

-

\

. ,r

\

Outlet

/

Initial velocity ,.-----First stage-------.i.------Second stage-----.,

FIGURE 7.9

Condenser pressure

Pressure and velocity compounding.

E

V,,

=

V, 2 (BC= BO)

if no friction C

(a) Inlet and exit velocity diagram

FIGURE 7.10

(b) Combine velocity diagram

Velocity diagram of impulse turbine.

-

Thermal and Hydraulic Machines

is required to determine the flow rate of the steam and the speed of the movable blades. To ensure that steam should enter and leave the blades without any shock, suitable inlet and outlet angles of the moving blades have to be provided. The force, power and angles can be found out with the help of velocity diagrams at the inlet and outlet of the moving blade as shown in Figure 7 .10. This velocity diagram, as shown in the figure, is valid for impulse turbine only as in the reaction turbine, the relative velocity of steam increases when steam flow over the moving blades. Hence the velocity diagram of the reaction turbine is different from impulse turbine in which the relative velocity of steam remains constant. The notations used in constructing the velocity diagrams are:

u V1 V2 V w, V w2 Vfl Vfl Vr, Vr2 m

d = h = 0 = t/J

=

a =

/3 =

the velocity of the blade (tangential) absolute velocity of steam at inlet to moving blade absolute velocity of steam at outlet to moving blade tangential component of Vi tangential component of V2 axial component of V1 (flow velocity at the entry) axial component of V2 (flow velocity at outlet) relative velocity of steam w.r.t. moving blade at entrance relative velocity of steam w.r.t. moving blade at exit mass of steam flowing over blades mean diameter of drum height of blade inlet angle of moving blade exit angle of moving blade exit angle of nozzle (absolute angle of steam at the inlet) absolute angle of steam at outlet

The steam jet from nozzle impinges the blade at an angle of ato the plane of rotation. The blade velocity is u. The tangential component of steam velocity (Vw 1) contributes towards the work output as it is in the direction of blade velocity (u). The relative velocity of steam at inlet (Vr,) makes an angle 0with the plane ofrotation of the wheel. The inlet tip angle of the blade has to be equal to 0 in order to avoid any shock on the blade when steam enters the blade. The axial components ~ 1 and ~ are responsible for the flow in and out of the steam in axial direction. In case, there is no friction between the moving blades and the steam, the relative velocity of steam at inlet and outlet will remain constant in magnitude, i.e. Vr 1 = Vr2 • The absolute velocity of steam at exit (V2) can be found out by taking the vector summation of blade velocity (u) and the relative velocity ( Vri) at exit. Similarly, the absolute velocity at entrance can be found out by taking vector summation of blade velocity (u) and the relative velocity (Vr 1) at entrance. The absolute velocity and relative velocity at exit are inclined at angle /3 and angle t/J to the plane of rotation. The steam enters in the second ring of fixed blades with this angle /3. The combined inlet and outlet velocity diagram is drawn on the same blade velocity (u) as shown in Figure 7 .11. The work done on the blade is resulted due to forces acting on the blade along the direction of blade movement. The tangential velocity of steam at inlet causes a force to act on the blades in the direction of motion.

Steam Turbine

The axial velocity of steam (also called velocity of flow) at inlet determines the mass of steam flowing through the blades. The axial thrust on the rotor is also caused by this flow velocity. The absolute velocity at inlet and outlet of the blade decides the inlet and outlet angles when the turbine is running at a constant speed.

E

V, 1 < V, 2 (BC< BO) FIGURE 7.11

C

Velocity diagram of reaction turbine.

If there is no friction on the blades, then we have Vr 1= Vrz · However, there is always a certain loss of velocity of steam during the flow of steam over the blade which is given by the velocity coefficient (k). The velocity coefficient is: k

= V,2 V,1

The velocity coefficient (k) is always less than one as Vr 1 > Vrz· It has been seen that the relative velocity of steam in impulse turbine remains either constant (Vr 1= Vri) if blade is smooth or the relative velocity at the outlet is smaller than at the inlet (Vr 1 > Vr2 ) due to friction acting on steam while flowing over the blade surface. However in reaction turbine, the steam expands as steam flows over the moving blades, whereby the velocity of steam increases as it flows over the moving blades. Hence the relative velocity ( Vr2 ) at outlet is greater than the relative velocity ( Vr 1) at the inlet, i.e. Vr2 > Vr 1. The velocity diagram for reaction turbine is shown in Figure 7.11.

WORK, POWER AND EFFICIENCIES Work and power: As per Newton's second law, the force (F) acting on the blades is equal to the rate of change of momentum of the steam as it flows over the blade surface. Hence we have: F = rate of change of momentum = (mass flow rate of steam) x ( change in tangential velocity of the steam at inlet to outlet) = m [Vw1 - (-Vw2)]

= m [Vw1 + Vw2] =m Vw where

Vw= Vw 1 + Vw2 Work done (W)

Power developed P

= Force x Distance moved per second = (m Vw) x u =

m·Vw·U

1000

kW

in direction of blade motion

Thermal and Hydraulic Machines

Blade or diagram efficiency: This is specified for a single blade stage of an impulse turbine. It is the ratio of energy developed per stage to kinetic energy supplied per kg of steam and it is given by: Work done on blades per kg of steam T]b = Kinetic energy supplied per kg of steam

2·Vw·U

=---"---

v,2 l

Stage or gross efficiency: The stage efficiency is the ratio of work done per kg of steam to the theoretical enthalpy drop in the nozzle per kg of steam.

V,,)

(b) Impulse turbine stage

(V,2 :,; V,,)

P1 = Inlet to fixed vane P2 = Outlet to fixed vane P3 = Outlet to moving vane

Entropy (c) Enthalpy drop

FIGURE 7.18

Reaction turbine stage.

The degree of reaction turbine stage is defined as the ratio of enthalpy drop in moving blades to the enthalpy drop in that stage of the turbine.

m

Steam Turbine

. Degree o f react10n or

Enthalpy drop in moving blades Enthalpy drop in the stage

= Rd = ----"-'------"------"---

=

R d

~

1111, + 111½

The enthalpy drop in the moving blades in reaction turbine is utilised in increasing the relative velocity of the steam from inlet to outlet. 2

111½

= v,'.2 - Yi

2

2

The total enthalpy drop in the stage is 11h 1 + 11h 2 which is equal to work done by the steam in this stage. 11h I + l1h2 = W = (VWI + Vw2) x U

R= d

~

!111, + M2 2

½ - Yi

2

=-~2~_

(Vw, + Vw 2 )xu

(7.6) The velocity diagram of a reaction turbine is shown in Figure 7 .19.

C

FIGURE 7.19

V,.2 = V w1 + Vw2

~

Velocity diagram of reaction turbine.

cosec