Thermodynamics & Heat Engines [1, 7 ed.] 8185444021

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'1 11 I� ll)If)I)YN11)1 I(�S 1\NI) 1 Ill�1l'l l�Nf,INl�S \T()IJ. I R. YADAV

CENTRAL PUBLISHING HOUSE ALLAHABAD

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Thermodynamics & Heat Engines Volume I

(THERMODYNAMICS) (SI Units)

R. YADAV B.Sc. Engg., M.E., Ph.D., F.l.E. Professor & Former Head Mechanical Engineering Department, M.N.R. Engineering College Allahabad-2n 004 AND

SANJAY, B.E, M.E. Sr. Engineer, TATA Project·L1rD., Hyderabad AND RAJAY, l:S.Tech. Engineer, B.H.E.L., Hardwar

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Central Publishing House 18-C, Sarojini Naidu Marg, . Allahabad•211 G01

© COPYRIGHT 1991 Publisher and Author First Edition 197 5 1981 Second Edition· Revised Edition 1985 .· 1986 Fourth Edition (Sli � Fifth Revised Edition 1988 Reprinted : 1991 1994 Reprinted 1997 Sixth. ReviseJ Editicn . Seventh Revised Editio� lOOO · • 2001

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2002 -. 2003

Reprint

ISBN 81-854,44-02-1

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Printed in India ai the Halcyon .Press, 18-C, Sarojin i Naidu Marg. Allahabad . .and Published by Central 'Publishing House, 18-C. Sarojini Naidu Marg. Allah,1h.i •

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$.J,1cvrc;.,1,u._ CAIL-- g P:REFACE 'I'O SIXTH REVISED (SI) EDITION Updating and change are the essence ofa popular text book. Keeping this in 'view the present edition is an attempt in this direction. In this edition a thorough revision has been made. Some new articles have been added. All figures have been replaced by new ones with more informations and larger in size .. Some more typical high standard numericals have been solved and also added in excercise section. The book has been presented in royal size. I hope _ that this new edition will prove more meaningful and useful to students. Allahabad Engineers Day, 1997

R. Yadav Sanjay Rajay

PRE-FACE TO THE SI EDITION Most of the engineering institutions in India and abroad have switched over from MKS or FPS to SI (Systeme Internationale) System of Units' and rest are in transition iaage. In view of this, this book has been thoroughly revised in SI units to cater to the needs · of student · community and to keep pace with-the new developments. Chapters 7 and 9 have been renamed according to their CQntents. Chapters Io; 11, 12 and 13 have been thoroughly revised. In the preparation of this edition, the author has consulted a large pumber of books which are listed in the References. The author sincerely acknowledges them. Any mistakes and error found i,IYlhe tes� will be appreciated if it is brought to the notice of the author. Finally. the author wishes to thank his wife Meera and sons Sanjay and Rajay who endured cenam hardness due to his engagement in completing this work. 19th June. 1986 Alhthabad

R. Yadav

• PREFACE TO THE FIRST EDITION This book has been written to suit the courses followed in engineering colleges, A.M.I.E., U.P.S.E., U.P.P.S.C. and other similar institutions. As there is no book available in the market that covers the full syllabus the present attempt is expected to fill this gap. Because of the compass to this book involves the subject has been divided into two parts, of which volume I covers the syllabus of 'Thermodynamics' while volume II covers the syllabus of 'Heat Engines'. I have attempted in this book to present the matter in a simple lucid and precise manner keeping in view the student's difficulty in solving problems and hints have been given where they are likely to go wrong. A novel feature of this book is that a large num�er·of problems have heen taken from most of the leading universities of India and abroad, U.P.S.C. and U.P.P.S.C. and solved. I have also taken the liberty of adding, at the end of each chapter a number of viva voce, theoretical and numercial questions wih answers. It is hoped that they will add to the usefullnessiOf the book for the students for whom it is primarly meant. I take this opportunity to express my gratitude to Or. B.K. Gupta, Professor and Head, Mec.:h. Engg. Deptt. M.N.R. Engg. College, Allaryabad; Sri. P.N. lv1askara, Reader in mech. Engg. Deptr., M.N.R. Engg. College, Allahabad to my colleagues and well wishers for their encouragement. .

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I am indebted to the various authors whose books I have consulted in the preparation ot this book. · And last but not the least, credit goes to my wife Meera without whose co-operation I could not even think of going for this venture. And lastly, I would appreciate suggestions and criticism of the views expressed in this book with a view to its improvement. IIT Delhi

R. Yadav

.lune 9th, 1975

. �'

CONTENTS l. BASIC CONCEPTS AND DEFINITIONS : Thermodynamics; Dimensions and units; Mass and weight; Thermodynamics system; surrounding and universe; Types and system; Macroscopic and microscopic point of view; Concept of continuum; Density and specific weight; Specific volume; Specific Gravity; Pressure; Thermodynamic equilibrium; Property; State; Path; Process; Cyclic Process; Quasi- static 'process; Reversible Process; Irreversible process; Energy; Work; Power; Forms of .works during Quasi-static process; Nonquasi static Foi:ms of works; Work as a path Function; Heat; Singn corivenfion of various energies in their algebric sense; Mechanical Equivalent of Heat. 1-45

. AND ZEROTH LAW OF THERMODYNAMICS : Concept of (y TEMPERA temperatur�; Equality of temperature; Zeroth Law of Thermodynamics; Measurement of temperature; Temperature scale; Liquid thermometers (Mercury); Limitations of liquid Thermometers; Gas.._ thermome,ters; Principles of gas thermometers; Constant volume gas thrmometer; Electrical resistance Thermometers; Thermoelectric thermo�ek:rs; Radiation Pyrometer; Segar Cone; Optical pyrometer; The International temperature scale; Practical · ·· aspect of temperature measurement. 46-65

TURE

®

iDEAL (PERFECT) AND REAL GASES : Concept of an ideal Gas (Macro-analysis); Boyles law;• Charles' Law; Characteristic E uation of gas; Avagadro's hypothesis (Universal gas constant); Enthalpy of an ideal gas; Specific heats o an ideal Gas; Kinetic theory of gases (Microscopic analysis of perfect gas); Mean free path; Deviation of rea! gases from ideal gases; Critg state; Equation of state; Vander Waal' s equation of state;· Limitations of Vander Waal's equation; Reduced coordinates; Compressibility of factor and law of corresponding states; Other equations of for real gases P.V.T. surface; P.V.T. surface of an.ideal gas; Representation of partial de:-ivatives on a P.V.T. surface. 79-155

state

·@IRST LAW OF THERMODYNAMICS : First law of thermodynamics; Internal energy; Internal energy of a perfect gas; Application of First law to a closed system (Non-flow • processes); Flow processes and control volume; Flow energy or flow work: Energy accompanying mass in flow process; Steady and unsteady flow processes; First law of thermodynamics applied to 9pen system; Mechanical work in a steady flow system; Steady flow process for an id� Continuity equation; Euler and.bernoullie equations; Throttling. process; Application of steady flow energy equations; Unsteady flow process; Filling I 26-2 12 :,rocess; Emptying process. /sECOND LAW OF THERMODYNAMICS : Limitations of First Law and essence of Second Law; Thermal reservoir; Heat engines; Thermal efficiency of heat engines; Heat pump and its efficiency; Available and unavailable Energy; The dead state; Statement of Second Law; Kelvin and Clausius statement; Reasons of irreversibility Carnot cycle; Corollary I : Maximum reversible engine efficiency (Carnot's theorem); Corollary 2: All reversible engines have same efficiency (Carnot's theorem); Corollary 3: Thermodynaic temperature, Absolute temperature scale or Kelvin temperature scale; Corollary 3: JJnattainubility of negative absolute temper,llurc; Corollary 3: Unattainability of asolute zero

l

temperature; Corollary 4: Clausius inequality; Corollary 5: Entropy; Corollary 6: Temperature entropy diagram; Entropy changes for an ideal gas during proc.,esses: Carnot cycle on T.S. diagram; Physical interpretation of entropy. Causes of irreversi�ility. 2 I 3-298 6. GENERAL THERMODYNAMIC RELATIONS AND THIRD LAW : Introduction; Some mathematical preliminaries; definitions; Differential relations for U.H.G. and F; Some important properties expressed in differential form; Generalised relations for Cp Cv Kand �; Relations for internal energy and enthalpy; The various T.ds - equations; clapeyron equation; Gas tables; Third law of thermodynamics. 299-323 /.AVAILABILITY. IRREVER_SIBILITY AND "EQUILIBRIUM, Introduction; Availability of heat; Availability of a clpsed system; Availability function of the closed system; Av.ailability of steady flow system; availability function of open system; irreversibility for closed open system; Effectiveness; Second La� analysis of the power plant; CJ1emical potential; Criteria of equilibrium; Types of equilibrium. 324-358 �

OPERTIES OF PURE SUBS�ANCE : Pure substance (steam); Phase�; Phase Transformation (Formation of steam) at constant pressure; Saturation pressure and temperature of steam: Enthalpy changes during formation of steam; West steam (two phase Mistures); Properties of steam; Steam property diagrams; Processes of vapor (steam); Constant pressure processes; Adiabatic process (reversible and irreversible); Isothermal process; Polytropic process; Throttling process; method for determining the dryness fractions; Super critical pressure of steam; Super saturated flow of steam. 359-432 9. NON-REACTIVE GAS MIXTURES AND PSYCHROMETRY : Introduction, Some basic definitions for gas gixtures; P.V.T. relationship for mixtures of ideal gases; Properties of mixtures of ideal gases; Entropy changes due to mixing; Mixtures of perfect Gases at different initial pressures and temperatures; Hetrogeneous system of several phase�; Mixture of gas and vapours; Psychometric terms; West bulb temperature; Adiabatic saturation temperature; Enthalpy of moist air; Psychrometric chart; Psychrometric process; Adiabatic mixing of Air streams; Sensible heating and cooling process; By pass factor : Humidification and dehumidification processes; Cooling and humidification; Heating and f humidi ication : Heating and dehumidification : Cooling and dehumidification process; 433-500 Summary of psychometric processes; application of psychometrics. , �AS POWER CYCLES : Introduction; Air standard cycles; Some definitions for pist�nder arrangement; Carnot cycles; Otto cycle; Diesel cycle; Dual cycle; Comparison between Otto, Diesel and dual cycle; Variatipns between the Air standard Otto cycles and actual cycles; Effect of variable specific. heat and dissociation on air standard Otto Cycle; Variation between the air cycles and the actual cycle of diesel engines. 501-574 11. VAPOR POWER CYCLES : ln_troduction; The Carnot vapour cycle; Rankine cycle; Effects of operating condition on efficiency; Principle of increasing the thermal efficiency; Method of increasing the thermal efficiency; Deviation of actual cycle fr.om theoretical cycle; Thermal efficiency ratio and S.S.C.; R�uirement of an ideal working

/

Fluid; Regenerative feed heating cycles; The reheat cyc�Binary-vapor cycle; Nuclear power plant cycles; Solar enery-power plant. Geothermal and OTEC power plants. 575-639 12. DIRECT ENERGY CONVERSION : Introduction; Thermoelectric converters: Thermionic converters; Magnetohydrodynamic generator(M.D.H. ); Solar cell power plants; Fuel cells; Fuel cell system under development; Hydrogen-oxygen fuel cells; Directo oxidation hydrocarbon fuel cells; Indirect oxidation fuel cells. 640-667 13. FUELS AND COMBUSTION : Introduction; Classification of Fuels; Mole. Kilogram and Mole Fraction; Complett: combustion equation (or stiochiometric equation); Air-fuel Ratio; The gas analysis by volume, Orsate apparatus; Calculation of the mass c,f dry flue gases per kg of fuel of known carbon content fro� Orsate est; Convcn,ion of gravimetric analysis to volumetric analysis and vice-versa; Minimum air required in kg per kg of solid or liquid fuel. Minimum quantity of air required _for the complete combustion of one cubic meter of gaseous fuel; Determination of percentage of carbon in fuel burning to C(h or CO from Orsate analysis or volumetric composition; Determination of the percentage of carbon in fuel burning to CO2 or CO from gravimetric Analysis of fuel gases; The enthalpy and internal energy ofreaction; Calorific (or heating) value offuel; Higher and lower calorific Value of fuel; Experimental determination of the calorific value of fuel by Bomb Calorimeter; Experimental detenni'nation oflhe calorific value of gaseous Fuel, Junker's gas calorimeter; adiabatic-com-bustion temperature; Maximum explosion pressure; Entropy changes for reacting mixtures. 668-721 References

722

Appendices

724

A

NOMENCLATURE area

a

sound velocity

B

constant

C

velocity

C

constant

D

diameter

E

total_ energy

F ff h J

force

total enthalpy, kJ enthalpy, kJ/kg

· · mechanical equivalent of heat

K

constant

m

mass, mass flow rate

N

speed

n

index

p

power

Q.

total heat, kJ

q R

gas constant

r

s

heat, kJ/kg radius

total entropy, kJ/K

s

entropy, kJ/kg K

T

absolute temperature, K

t

temperature,

u

·c

total internal energy, kJ

u

internal energy, kJ/kg

V

volume, m 3

V

specific volume, m 3/kg

w w

total work, kJ

specific work, kJ/kg

1 Basic Concepts and Definitions Thermodynamics is the.science that deals with energy interactions between material systems. In other words,-it is the science which deals with the transformation of energy of all kinds from one Jorm to another. In fact, one very excellent definition of thermodynamics is that it is the science of three "E" namely energy, entropy and efl.Milibrium)Generally speaking, thermodynamics is primarily concerned with two forms of energy heat and work. Cam_ot, Joule, Kelvin and Clausius were the main scientists who developed this branch of science in the past century. In 1876, J.W. Gibbs advanced classical thermodynamics to such an extent that in its scope it could be applied to almost any physical or chemical phenomena. The science of thermodynamics is basec;t on the four laws of thermodynamics known as zeroth, first, second and third laws. Although, the formulation of these laws is simple but their applications are remarkably extensive. And surprisingly, there is no mathematical proof for. these laws. These laws were· deduced from experimental observations and are based on logical reasonings.

· The application of thermodynamics is extr�mely wide. Its 12rinciples are used_jn the designing of energy converting devices, such as steam engines, internal combustion engines, steam and gas turbines.fuel cells, thermoelectric and thermionic generators . It is also used in refrige�ors, air-conditio'lf!}, heat transfer, phase equilibrium or �n equilibrium, etc. The credit for proving that mechanical work can be convened to heat energy goes to James Joule and for converting heat into mechanical work to James Wau, the inventor of steam engine.

When we consid�r matter from the macroscopic point of view, the subject is called classical thermodynamics and when from microscopic point of view, it is cal led statistical thermodynamics. In this volume, we will deal mainly with the subject of classical thermodynamics. In this chapter some basic concepts and definitions of various terminologie'S related with the study of ther;nodynamics are presented. 1.1

.l..

Dimensions and U nits

Dimensions and units are the main tools of engineering. The dimensions of a physical uantity m_fil' be defined as the properties in terms of qua1ity not of mngnitude, J2y which a physical ua!1tity 111ay be described. Length, area and volume arc all different dimens10ns w 1ch describe certain measurnble characteristics of an ob_1cct. For example, an area can be regarded as a length squared and a volume as a length cubed.

(I.1) ( 1 :2) The right hand side of these equations shows dimensions of Lhe left hand side. These equations state that the dimensions of an area and volume are equivalent to the (A ) = (L2) ( V) = (/))

Thermodynamics and Heat Engines

2

dimensions of length squared and length cubed respectively. The fundamental dimensions are the time, length mass and force. A unit is a definite scandard by which a dimension is co be measured,for instance, in this book we will Joi/ow the S.l. units. The following are dimensions and the units commonly used in engineering-

(i) Time ( 8). The fundamental unit of time is the second (s). It is defined as 1/886,400 part of a mean solar day.

(ii) length (L ). The fundamental unit of length is metre (m). It is the distance between two marks on a plantinum-iridium rod at 0°C, kept in a vault at the International Bureau of Weights and Measures at Sevres, France.

(iii) Mass (M). The fundamental unit of mass is the kilogram mass (kg), defined as the mass of a lump of platinum iridium als� kept at Sevres, France. Kilogram mass will

be denoted as kg in this book.

(iv) Force (F). The fundamental unit of force is newton (N). The best way to define force is the Newton's Second Law of Motion. According to this law, force is proponional to the produce of mass and acceleration. Thus

F a (m) (a)

F = . _!__ (m) (a) gc

( 1 .3)

Where F = Force m = mass

a = acceleration

1 . · constant. - = a proporuona1Jty gc Let us find 1he dimension �nd unit of gc

(a) In the SJ. System, the unit of force is new11m (N) defined as that force which will cause a mass of 1 kg to accelerate l m/s. 2 So, by substituting these values in equation (1 .3), we have

or

gc = 1

(.

Writing dimensions, we have

kg . m ) 2 newton .s

So in this book, gc will not be wrilten since gc = I in S.I. unit Therefore, the form of force is expressed as F = ma

(l .4)

Basic Co11cepts and Defi11itions

3

j For S. I. units, see the appendix. · (b) In M.K.S. System, the unit of force is kg1. This is defined as that force which , will cause a mass of 1 kg to accelerate 9.80665 m/s2 so that •

I

m 1 kg1 = _.!_ (1 kg) (9.80665 2) . Kc s

g c = ( 9.80665 kg.m 2 kg1 .s

or

,=

9.81

kg.m

} kg1 .cm J

(1.5)

(c) In the C.G.S. system, the unit of force is dyne, It is defined as that force which will cause a mass of 1 gram to accelerate I cm/s2 so that 1 Dyne = or

Writing dimensions, we have

:c

{l gm) ( l e;)

g c = 1 (gm.cm2) dyne.s [g c] = [;

(1.6)

;2]

(ti) In the F.P.S. system, the unit of force is the pound force (Ib1 ) . It is defined as thatforce which will cause a mass of 1 pound to accelerate 32.1739 ft/s2 so that l lb1 = _!_ j l/b ,,. ) &2. 1739ft /s 2 } gc

or

..

Also, Writing dimensions, we have

g c= ( 32. 179

lb,,J \) lb1 .s

gc = 1 .0 slug ft/lb1.s2

. (1 .7)

(gc] = [; }]

Introduction of gc reveals that force is a fundamental dimension. gc is called the dimensional constant which includes four basic dimensions mass, length, force and time.

(e) A complete dimensional system can be built up with only three basic dimensions. For example, force can be expressed in tenns for only 3 fundamental dimensions [8], [l] and [M]. According lO this, Newton's Second Law becomes

F = (m) (a)

[F ]

=

[Mo�]

4

Thermodynamics and Hear Engines

Writing units, we have

or

• 1 newton = (1 kg) (11 7

2)

1 newton = 1 kg.m -2s

• ( 1 .8)

(/) gc as a Unit Conversion Factor :

As already discussed 1 kilogram .force will ac_celerate a kQogram mass at the rate of 9.80665 m!s2 and I newton will acclerate -1 kilogram mass at the rate of 1 m/s2 • So, by implication. 1 kg, = 9.80665 newton = 9.81 newton. Since kg1 and newton are both units of force, we have· [Fl = [Fj

9 _80665 newton = 1 kg!

( l .9)

From equations_( l .8) an9 (1 .9), we have

9.80665 kg .m = I kg.m2 = 2 N.s kg! .s

( 1 . 1 0)

In equation (1 . 10), the left hand side is equal to gc . And since the absolute value of gc is unity as is obvious from the right hand side of the equation ( I . I O) g., can be considered as a unit conversion factor becaus� by unity we can multiply or divide any expression without changing the value. The main point to be noted is that the multiplication or division must be done by the composite factor 9.80665 kg-m/kg1 .s2 and not by numerical value of gc To check the unit and dimension, whether, balanced or not, gc plays an - important role and Newton's Second Law of Motion is not dimensionally balanced unless gc is put in the denominator.

1.2 Mass and Weight. Mass is thcfpieasure of quantity of matter as well as the ..m-operty call¢ inertia. It is true that mass is related to weight but the two terms are · different. Weight signifies the sense (,f force and it refers to the force exerted by gravity on Lite mass. Weight is proportional to mass at a particular location but the proportionality factor changes from place to place. The weight of an object varies from place to place but its mass remains (if the velocity is constant) constant. As we know, � value of acceleration due to gravity decreases as we go abov� or below the surface of the earth, so it follows that the weight of an object will decrease above and below the surface of the �- For example, the weight of an object on the moon will be 1/6 that of the earth because the value of g on the moon is 1/6 that of the earth.

On the earth (i:e. sea level) that value of mass and weight is �e same but with a different_sense. According to Newton's Law. Force a (m) (a)

•

Busic Concepts and Deji11itio11s

Force d�e to gravity

5

i.e. weight = mass X occeleralion due to gravity gc

or

W

= mXg gc

( L l l:

Notc : Conversion Table from M.K.S. to SJ. units is1 given at the end of book. kg x (�;s ) · IN ) = �--F '\ kg.m/N s 2 2

Now,

( 1 . 12

From equation ( 1 . 1 1), we have w/g = mfgc In S.I., w = mg ·

Note that F = (m) (a) is only valid in SJ. unit (ie. unit of force is Newton) and C.G.S. unit (i.e. unit of force is Dyne) because gc is equal to unity in these two cases. It is obvious from tl'e equation ( 1 . 12) that I kg will produce 9.8 1 N in gravitational field. S.I. units of various parameters used in thermodynamics are tabulated as follows in table 1 . 1 . TABLE 1.1 Parameiers Length Mass Force Time ·volume ✓Pressure

Velocny · Spt>cific volume Den�ily Heat energy. \\·ork energy Power

Relations

Unlt

metre (m) kilogram (kg) newton (N) second (s) m3 newton/metr�2 (N/m2) or pascal cPa)

m/s m 3/kg kg/m 3 joule (J) joule (J) or newton-metre (N m) J/s or watt (W)

I m = 100 cm 1 kg = 1 000 g 1 kN = 1 03 N -.../standard atmospheric pressure = 75 cm of Hg = 1 .01 325 barometric (bar) 1 bar = 105 N/m2 = 105 Pa 1 N/m2 = I Pascal (Pa) I kN/m2 = 103 N/m 2 1 MN/m2 = 106 N/m 2 1 kJ = l ()()()J

. i k.W = l u3.W '

I

I MW = 106W Table Co11t.i .

6

Thermodynamics and Heat E1ji11es

Temperature

kelvin (K)

I J/s = 1 W T = t°C +273 K

Problem 1.1 Calculate the force exerted in M. K.S. and SJ. unit by 10 kg mass because of gravity at a place where the acceleration due to gravity is 9.80665 m/s.2 Solution. We know that

or

O x 9.80665 = lO kg F = (m) (a) = (m) (a) = I t 9.80665 gc gc

Ans.

10 x 9.80665 = 98 _ 5 N 066 Ans. I Problem 1 .2. A certain mass is weighed at the poles where acceleration due to gravity g = 9.85m/s2 by a spring balance which was calibrated at sea level where g = 9.80665 m/s.2 The spring balance reads 500 Newton. Find the mass of the body. F

=

Solution. The spring scale registers Newton hence the body is under the influence of gravitational force of 500 Newton is the weight of the mass at sea level. From Newton's Law

or

m =

5 oo x l 9.85

=

50.761 kg

1 .3 Thermodynamic System, Surrounding and Universe; (i) System. The term system is defined as a prescribed region or space offinite quantity of matter surrounded by an envelope called the boundary. The boundary may be a real physical surface, such as the walls of a vessel, I. C. engine cylinder, etc. or it may be an imaginary 'surface enclosing some matter such as steam, gas, vapor, etc. The boundary may be fixed or it may be moving, as when a system containing a gas is compressed or expanded. Thus we can say thaJ the system may be a quantity of steam, mixture -of vapor and gas or an l.C. engi-1ie cylinder or its conlenls.

In a broad sense, the sytem is defined as a specified region wherein changes due to transfer of mass or energy iJr bol/J are 10 be studied. In a system, it is not necessary that th�_ .yolume or shape should remain fixed.

(ii) Surrounding. The space and mailer external to the thermodynamic system and outside boundary is called the sUTrounding. (iii) Universe. When system and surrounding . are put toKelher it is called universe.

In order• to understand these, consider a piston cylinder arrangement shown in Fig. 1 . 1 in whr,:h 'the system, boundary and surrounding are described. In this case, there are three fu:ed and' one movable boundary formed by the bottom surface of the piston. The gas contained in the cylinder and enclosed by the boundaries is the system. All the matter and space external to these four boundaries are known as the surrounding. It is to be noted that all transfers of mass and energy between the system and sunounding are evaluated at Lile boundary.

r

Basic Concepts and Definitions

7

-surrounding Fixed

and real boundaries

System (Gas) I

: I I I

Movable real boundary

L. _ - - �- - - _ J Cyfinder

1.1. Repescntation of Systan, Boundary and Surrounding.

1.3.1. Types or .System. There are three types of systems � (i) closed (ii) open and (iii) isolated systems used in thennodynamics. (i) .Closed System. A �d_ a _closed system if the mass within the ..boundary ot the system remains.constant and only 1he energy (heat and work)-ma✓ tr®Sfer ac,:pss it..s boundary. Such a closed system is shown in Fig. 1 .2. in which the gas

•

is confined between the piston and the cylinder. The mass of the gas is constant within the system even though the gas may be expanded or compressed because it is closed all sides. A variatiqn in the volume of the gas may occur. But the transfer of heat and work _energy· are tajcing place in the closed system. Other examples of closed system include a free·booy aod a PQint mass as used in mechanics.

from

,

· (ii) Open System. A system is called an open syste"!. if the mass as well as the _energy transfer across its bowii!iines. Such open system is shown in Fig. 1.3 in which

an

the gas is entering and coming out of the system while transfer of heat and work energy is also taking place. The net amount of mass within the system may also vary with time.

In the closed system, the analysis is focused on a fixed mass of matter, whereas in the open system, the analysis centres about a region in space through which matter flows. It is frequently useful to think of a fixed region in space, called a control volume, through which mass, momentum and energy may flow. The surface of the control volume is called the control surface. The control volume may be stationary or may be moving at a.constant velocity. If no mass transfer across the control surface takes place the control volume is identical with a closed system. It should be noted that the terms closed system and open system are sometime used as the equivalent of the terms system -U-txed mass) and control volume (involving a flow of mass). Control volume is a very useful concept for the study of fluid flow. Examples of the open systems are compressors, turbines, nozzles, diffusers, steam engines, I.C. engines., etc. (iii) Isolated System. A_sjstem is called an isolated system if neither mass nor energy transfers across its boundaries. Example of isolated system is fluid enclosed in a perfectly insulated c_!osed vessel (thermos flask)

1.4 Phase. // a q_uantit)' of matter is homogeneous OF uniform throughout in physical structllre and chemical composition it is termed as a phase. Homogeneity of physical structure means that the material is either all solid or all liquid or all gas. Homogeneity of chemical composition includes one or more homogeneous chemical

8

Thermodynamics and Heat Engines

High pressure gas Boundary ►

w Control volume (C.V.)

Low pressure gas Fig. 1 .2 Closed System Fig. 1.3 Ope n System species. Thus a piece of iron, the gas contained in a vessel and a liquia contained in a pot will be called, one phase, A mixture of gases is one phase ; a system consisting of a liquid and gas is two phases (i.e. a liquid phase and a gas phase) and a mixture of solid, liquid and gas is three phases. A system consisting of two miscible liquid is said to be in a single liquid phase and system consisting of two immiscible liquids forms two liquid phases. On •.he basis of phases the system may be classified as (i) homogeneous system and

(ii) heterogeneous system.

(i) Homogeneous System. If a system consists of single phase it is called homogeneous system. Examples of homogeneous system are mixture of air and water vapor, solution of NH3 in water, water and nitric acid, octane and heptane, etc.

(ii) Heterogeneous System. If the system consists of more than one phase, it is called heterogeneous system. Examples of heterogeneous systems are water and steam, ice and water, water and oil, etc.

1.5

Macrosct-pic and M icroscopic Point of View

If the analysis of a thermodynamic system is explained by the measurable properti(!s such as pressure, volume, etc . , of the system, it is termed as macroscopic point of view. The microscopic point of view focuses on th_e statistical behaviour of mass consisting of numerous individual molecules and correlates macroscopic properties of the matter with molecular caofignration and with intermolecular forces,,. Thus lh£__microscoiru; point of iew deals with the time avera e behaviour of the thermodynamic system. The difference between two approaches may be explained a"!' that pressure does fluctuate with time owing to the random motion of t 1e molecules. Th . concept emplily�bil�redicts that tbe-averag_e tehaviour of the moleculeremains uniform .although the behaviour of the molecule does r-ot. Thus, � microscopic point of view d_eals with the structures of the system and time average behaviour.

-

1 .6

---

-

-

Concept of Continuum .

"The concept of continuum is very useful in dealing with classical thermodynamics. From the continuum point of view, the matter is seen as being distributed through space and not, as in the particle view, localised. Matter exists in "big chunks" having mass, energy and momentum, as do particles, but it also possesses some additional continuum characteristics·, such as volume, density and temperature. In other words, continuum idealisation treats the substance - as b in conti uous disregarding the action of individual · · - - rrwleculeS:From the continuum stand point_ we can speak ofproperties at a point. For example, let us consider the d� of a fluid at a poin!_L which is surrounded by a s,mall size volume t.V and let the mass contained by this volume be t.m. It is obvious from the 5 raph as shown in Fig. 1 .4 that if the volume t. v' becom�aller and smaller, the ➔ Region Mass �m Volume o.V Fluid

f I

of discrete particles

Region of continuum

Limit

P = /lV ➔ AV ➔

Fig.--l .4. Density at a Point

t:.v

t:.m

"Kv

IO

Thermody11amics 011d Heal E11gi11es

density i.e. ratio Am/AV deviates very much from its asymptotic value at Av' and the value of Am/AV is either very large or very small. Thus, the volume AV' of the point P is the limit below which there is a large variation in density, but at Av' and slightly above this, the density may be referred to as density as a point. Mathermatically, density at a point may be expressed as L1m Lim P = ..1 V ➔ ..1 V ' L1 V

(1.13)

Where ..1V' is the smallest volume about point P in which the fluid can be called in continuum.

J]Je continuum idealisation is therefore not useful for volumes which are too small. U:...the_matial variations m ..1V are too great, a limit may not exist aiid the continuum _!!ledlocwould be invalid.

Other continuum conceP-ts inch1de,. ..mas�,;'flow rate. electric curren mass flux and the press__pre exerted by a fluid_on the walls of a container. In the_f .Qntinuum -analysis, calculus is_emplQy� For example, the mass of a body can be expressed as p V. m =J Vol. d

The above discussion leads into say that the continullllL� is ext:remel'.l small c;ompared.Jo the dimension of 1he installation. But this is not true in the case of ballistJc missil�supersonic rockets and earth satellites. At about 200,000 ,m altitooe (ffigh vaccum) the mean molecular path of a molecule may be as large as 3.5 m, i.e. approximately equal to the vehicle itself. In such circumstances (high vaccum) the fluid can not be treated as continuum. The microscopic point of view can be applied for the study. Except for a few exceptio�s. we can conclude that thermodynamics is a continuum science.

· 1.7

Density and Specific Weight-

(i) Density. The average density of a system is the ratio of its total mass to its total volume. The density of a continuum of a particle is . Lim Am ; kg/m 3 p = AV ➔ AV' ..1 V

Here, the volume .1V must contain a large number of molecule and yet it must be small compared to the dimension of the system, so that, the volume .:1 V is the smallest volume capable of maintaining the continuum model. Here,.:1V' is called dontinuum volume as discussed earlier. It is also known as mass density. (11) Specific weight. Specific weight is the weight per-unit volume. It may be expressed in N/m3 • It is also known as weight density. Let w be the weight corresponding to a certain mass m, then

/

Basic Concepts and Definitions Dividing both sides by volume Pw

= g.!!.. = p.g. c

Where Pw is the weight density and p is the mass density.

11

(1. 14)

Numerically, specific weight is the same as mass density if g/g c is unity. On the � glgc is unity in M.K.S. unit.

1 .8

Specific volume

The specifc volume v is defined as the volwne per unit mass and may be expressed in

rrf/kg. Thus, it is reciprocai of the density. Or

(1.15)

v = .!_ ; m �kg

Specific Gravity. The specific gravity of a substance is defiMd as the ratio of the densiry of the substance to the density of a given substance (water). 1 .9.

Pressure

The pressure exerted by a system is defined as theforce exerted Mrmal to unil area of the boundary. When a fluid is contained within a vessel, the pressure exerted on the vessel is equal to the mean change of momentum of the molecules exerted perpendicular to the confining bounday per unit area, and per unit time. Hence, pressure of fluids is analogous to normal stress in solids. From the continuum point of view the pressure at a point is the force per unit area in the limit; where the area tends to be very small i.e. approaching zero. Thus. p

=

Lim

L\ A ➔ L1A '

AF

L\A

N/m 2

( 1 . 1 6)

Where AF in N ·is the force normal to the area L1A and L1A' (m 2) is the smallest possible area capable of maintaining the continuum model.

A fluid is defined as a substance in which the shear stresses arc zero whenever it is at rest relative to its container. When a fluid is at rest, only normal stresses exist but when a fluid is in motion shear and tangential stresses exist in addition to compressive stresses. • When a fluid is in equilibrium the pressure becomes a scalar point function i.e. it becomes independent of direction, this is called static pressure When a fluid is not in equilibrium, the pressure may vary according to direction. Ex� ples of this are ideal or inviscidfluid.

In the case of an ideal fluid, shear stresses arc absent even if there is relative motion within the fluitJ. Under such a condition, pressure is independent or direction whether L'le ideal 11uid is in motion or at rest.

The following tenns are generally associated with the measurement or pressure. (a) Atmospheric pressure (.p01,,.) . This is the pressure exerted by the atmospheric air. The standard atmospheric pressure is defined as the pressur�· produced by a column of mercury 760 mm high, the mercury density being 1 3.595 1 kg/m3 and the acceleration due to gravity being 9.80665 m/s2, the standard atmospheric pressure i s 1 0 1 .325 kN/m2 or 1.01 325 bar or 76 cm of Hg. or l .0 1 325X 1 05 Pa . I

12

Thermodynamics a11d Heat E11gi11es

(b) Gauge Pressure (pgaug�). The pressure measured from the gauge and inst.--ument such as Bourdon and manometers is called the gauge pressure. The gauge actually measures the difference between the fluid pressure and the pressure of atmosphere surrounding the gauge.

P > Patm

Patm

•

l Gas f

p

(b)

Vacuum P < Parm

p

(c)

P > Patm

(d)

Fig. 1 .5 Pressure �auges (a) Bourdon guage, (b) Open U-tube indicating guage pressure

__ J

(c) Open U-tube indicating vaccum, (d) Closed U-tube indicating absolute pressure

1-1�

13

Basic Concepts and Definitions

(c) Absolute Pressure (pabs). The pressure measured from the level of absolute zero pressure is called absolute pressure. Absolute zero pressure will only occur when the moleculer momentum of fluid is zero. Such a situation can only arise if there is a perfect vacuum. A state of perfect vacuum is treated as reference point

hence

N /m2 (1.17) Pabs = Pa1m + Pgauge (p (d) Vacuum Pressure vac) . If the pressure of fluid to be meas�..,d is less than atmospheric pressure, the gauge will read on the negative side of the atmospheric pressure is known as vacuum, rarefaction or negative pressure. hence N Jm2 P • = P °'"' + P 1auge (1.18) The various pressure gauges are shown in Fig. 1.5. JU,

The relationship of pressure is shown in Fig. 1 .6

Absolute pressure (g reater that' atmospheric) Pao. "" Pgaug11 + Patm

P > Patm Gauge pressure ( ?gauge ) (Difference between absolute and atmospheric

Atmospheric i, ,1:1ssure

P = Patm Vacuum gauge pressur� (Pvacuum) ( Dirterence beM-een atmosoheric and absolute ) p < Parm

(Patm)

Absolute pressure /Less than atmospheric)

Pabs. vay a measuring device (such as a manometer, pitot tube, etc:) moving with the same velocity as the fluid stream.

·(ii) Impact or Stagnation or Total Pressure. Stagnation pressure is defined as the force per unit area perpendicular to the direction of the force when the fluid is brought to rest reversibly. For a constant density fluid the impact pressure is given by-

= P-1 = Ps111t +

£!f! ; Nllm 2

(1.19)

2 Where C is ·the fluid velocity (mis) and p is the density of fluid (kg/,rz 3). The difference between the stagnation and static pressure is the velocity pressure. The stagnation pressure is measured by an instrument called pitot tube Po

(iii) Velocity (Dyna.-nic) Pressure. The velocity or dynamic head is termed as velocity pressure. Thus, velocity pressure 2

p = fC /2 ;

Nlm 2

(1 .20)

Fig. I .7 shows the measurement of different types of pressure. p > Patm

Gas _.

c ....

.

.

pipe

3

-

· -- · -- ·--· - Pitot

tube

i

Velocity head

Patm

Static

pressure

♦

Parm

Total .

pressure·

Fig. 1 .7. Types of Pressure

In fluid flow through a duct, the pressure measured at the wall is static pressure because the velocity at the wall is zero. If the fluid particles move parallel to the centre line of the duct, the static pressure is u�iform across any section of the duct. The pressure measured at the centre of the duct is the stagnation pressure. 1 .9.1. Pressure as the Result of a Depth of Fluid. Consider the case of a

Basic Co11cepts and Defi11itions

15

homogeneous fluid of density p in static equilibrium. There exists a pressure difference between two points that are separated by distance h in the vertical direction. Hence, the mass of cylinder of fluid can be equated tc, the difference between forces due to pressure at · the two ends of the cylinder, so that the fundamental hydrostatic relationships governed by p,,,,,,. A = pghA

or p1,.,,,. = pgh .

N/m2

(1.21)

where p is in kg/m3 and h is in metre.

The height to which the column of liquid will rise will depend upon the liquid used.

It is to be noted that the hydrostatic pressure is the same to all points in a horizontal plane and varies only with depth.

In order to measure pressure slightly different from atmospheric pressure, a manometer shown in Fig. 1.8 is used. The pressure. is determined according to the hydrostatic formula given by the equation (1.22). The liquid used in manometer may be mercury, water, or alcohol, etc. Since the manometer fluid is in equilibrium, the pressure along a horimntal line xx is the same for either branch cf manometer. Thus,

(1.22)

. P + Pi gh 1 = p,. + P2 gh2 ' Patm

)(

hi;. . . ..1easuring Pressure by means of Manometer.

Where p is the absolute pressure in the bulb, p,. is the atmospheric pressure exerted on liquid free surface and p 1 and p2 _ are the densities· of the fluid in the bulb and the · manometer respectively. If

p1 c

76 - 71

5 cm of Hg '

5 x 101.3

2 = 6.66. kN/m Ans. 76 Problem 1.6. A forced fan supplies air to the furnace of a boiler at a head of 40 mm. of water column. Determine the absolute pressure of air supply if the barometric reading is 760 mm of Hg.

I•

=

=

=

Solution. Since it is forced fan hence the absolute pressure will be above atmospheric. We know that p = p1 gh1 = �h2 13.6 mm of water _column = 1 mm of Hg 1 x 4o = 2 94 mm of Hg :. 40 mm of water column = . 1 3 .6

2 Hence Pabs = 160 + 2.94 = 762.94 mm of Hg = 762 ·94 x 101.3 = 101 .69 kN/m A ns. 760 . .A>roblem 1.7. Air having density 1 kg/m3 is flowing at speed of 60 mis thrnugh a ·ctuct The static pressure measured is 400 kN/m2. Find the stagnation or impact pressure. Solution.

p., = p..,, +

rr 2 2

t::::::._

l x 60 400 + --2 x 1000 2

=

=

401.8 kN/m 2

Ans.

Problem 1.8. Convert the following :

(a) (b)

(c) (d)

Hg.

400 kN/m2 abs to kN/m2 gauge.

50 cm Hg vacuum · to cm Hg abs and kN/m2 abs.

60 kN/m2 to cm Hg vacuum.

30 cm Hg gauge to cm Hg abs and to bar. Assume barometer r�ds 76 cm of

Basic Co11cepts a11d Defi11itions

17

Solution. 16 cm of Hg = 101.3 kN/m2 (a) Pga,,ge = Pabs - Patm = 400 - 101.3 = 398.7 kN/m 2 gauge

(b) Pabs = Pa1m. - Pvac = 16 - 50 = 26 cm of Hg abs

=

26 x 1 0 I .3 = 34.655 kN/m2 abs 76

Ans.

(c) Pvac = Patm - pabs = 101 .3 - 60 = 4 1 .3 kN/rn2 vacuum

Ans

(d) Pabs = Patm + P8 auge = 16 + 30 = 106 cm of Hg = I06 x IOI.3 = 141 .286 kN/m2 abs • 76

Ans Problem 1.9. A vessel is shown in Fig. 1 .9 in which there are two compartments at different pressure. 1be pressure gauge A reads 4 bar gaug� and B reads 2 bar gauge. The barometer reads 760 mm ofHg, calcula� the reading of gauge C. _ X

y

B

C

A

Fig. 1 .9.

. Solution. We know that the pressure gauge reads the gauge pressure, i.e. it reads the difference 10 pressures between the inside and outside region in which the dial is located. Let X and Y be the absolute pressure in bar in both compartments 760 mm of Hg = 1 .013 bar

:. Y - l .013 = reading on gauge A = 4 bar gauge

(1 .23)

(1.24)

Y - X = reading on gauge B = 2 bar gauge

(1.25)

X - 1 .013 = reading on gauge C

From (1 .23), Y = 5.1013 bar abs

Putting the value of Y in (1 .24), we·have

or

5.01 3 - X = 1 .2

X = 3.8 1 3 bar

Putting the value of X in ( 1 .25); the reading of the prt'ssure. gat.1g� = 3.8 1 3 - 1 .0 1 3 = 2.8 bar gauge.

The absolute pressure for reading C is given by = 2.8 + 1 .0 1 3 = 3.8 1 3 har

.,..--

e,

Ans

18

Thernwdynamics and Heat Engines

Problem 1.10. Generally. the atmospheric pressure air is measured by mercury barome:er. It is calibrated for 0°C. The density of mercury and length of the scale change with temperature. For actual pressure of air, the values read on barometric scale should be corrected for 0°C. The correction fonnula is given by Bo = B ( 1 - 0.000172t)

Bo = corrected barometer pressure at 0°C

B = actual height of mercury column of barometer at t°C.

For this case, calculate the absolute prt�ssure if the vacuum gauge on the condenser fitted in steam turbine condenser reads 65 cm of Hg. The barometer reads 75 cm of Hg at a temperature of 20°C in the boiler house. Suddenly, if the barometric pressure drops to 66 cm of Hg and pressure in condenser is constant, find the vacuum gauge pressure. Solution. We have

Bu == i1 (I ·- 0.0(XH72t) = 75 (1 - 0.000172 x 20) = 74.742 cm of Hg

:. Pabs

= Pa1m - Pva,: =

74.742 - 65 = 9.742 cm = 9 ·742 x 1 .0 1 3 = 0.00989 bar 76

. Let IJ'0 ::: new barometric reading reduced to 0°C

Ans

:. B'0 = 66 (1 - 0.000172 x 20) = 65.7 cm of Hg

Hence, the new vacuum gauge reading

P,ac == 65.7 - 9. 742 = 55.958 cm of Hg.

1 .10

Thermodynamic Equilibrium

Ans

A system is said to be in a stale oj thermodynamic equilibrium if the value of the property is the sa'!'!le at all points in the system. Consider an isolated system, i.e., any gas enclosed in a container in which neither mass nor energy is transferring. The measured pressure and temperature at various points of the system may initially vary with �ime but a final siage win came when there is no variation of pressure and temperature with time and between two points cf the system. This final steady state, i.e. invariable property v. ith time and uniform property throughout the isolated system is called a state of thermodynamic equilibrium. In other words, an isolated system always reaches in course of time a state of tr.ermodynamic equilibrium and can never depart from it spontaneou,;ly.

lt is to be noted that the thermodynamic equilibrium is a complete equilibrium and it includes the following equilibr;um. (a) M-x:hanica! equilibriunr.:·

(b) Chemical equilibrium (c) Thermal cquiitbrium

(d) Electrical equilibrium (a ) Mechanical equilibrium : A system is said to be in a state of mechanical eauilihrium !f there exists no unbalanced for�e either in the interior of the system or

Basic Conrepts and Definitions

19

b-etween the system and surrounding . This i s only possible i f pressure i s the samt. throughout the system which, in tum, is also equal to th.at of surroundings. For example, if the pressure is not uniform throughout the system, internal changes in the state of the system will take place till the mechanical equilibrium is restored. Similarly, if the weight on the piston fro!TI piston-cylinder arrangement as shown is Fig. 1 . 1 is removed, the system boundary will expand until the mechanical equilibrium is again restored.

.The work interaction in each case is represented by the area underneath

37

Basic Con�epts and Definitions

the corresponding path. From the figure it is clear that the work is different in each case because the path depends on the nature of the process. This suggests that work is not a property or a state function rather it is a path function, and an infinitesimal increment of work is an inexact differential. Hence, a system does not possess work but work is a mode of transfer of energy and this transfer of energy occurs only at the boundaries of the system during a change of state of the system. Hence work is expressed by

p

t

A

2

Fig. 1.23 Work Depends on Path

W 1 -2 �

f: c5W am not by W

2

- W1

1 .29. Net Workdone by A System. In many engineering situations, different forms of work transfer occur simultaneously during a process execured by a system. Under such circumstances, the total or net work done by the system would be equal in the algebric sum of these given as follows. Wtotal = wdisplacement + wshear +Wslirring +Welec1rical + magnetic + ... .... . . . Problem 1.18. Calculate the work done when the volume changes from 4m3£:, Q� through a non-flow quasi-static process in which the pressure p is given by p = (4V - 5) bar

w

Solution.

W 1 - 2 = f •2=&pdV

But p = (4V - 5) X 1 05 N/m 2 • 1 =4

8

5 f4 2 W 1 - 2 � s •l=S (4V -_5) x 10 dV = v - s v ] x 10 • 1 =4 L 2 4 5

5 4 42 4 82 = [( � _ 5 x 8 ) - ( � - 5 x 4)] x 10 = 76 x 1 0 5 J

Ans.

Problem 1.19. Calculate the workdone in a piston cylinder arrangement dming an

,_

-'O expansion process where the process is given by the equation p = (V2 + 6V) bar Assume that during the expansion process the volume changes from lm3 to 4m3• Solution. p = (V2 + 6V) x 1 05 N/m 2 ' 2 . s 6V V 2=4 = - + -- x 10 w 1 _ 2 = f • pdV = J • 2! =4=I ( v 2 + 6V ) x 10 s dV 2 3 • ! =I [

=

t

3

+ 6 �•' ) - h' + 6 � 1 ')] x 1 0' = £,i x 10 '1

]

Ans.

Problem 1.20. In a piston-cylinder arrangement, the non-flow reversible process is given by V= 200/p m 3 where p is in 1 bar. Find the workdone when the pressure increases from 1 bar to 10 bar. Indicate whether the process is compression or expansion. 200 3 Solution . Given m . v = p 200 = 20 m 3 3 200 V 1 = - = 200 m aro V 2 1 10 .

Also

W 1 _2

=

I

p =

•2=:1.0

• 1 =200

2oo x 10 5 N/m 2 V

(200/V ) x 1 0 dV 5

=

200 x 10 [1og e V ] 5

'.!l

an

-

= - 4.6052

7 x 10 J

Ans.

The negative sign shows that the process occuring is the compression process. This is also obvious from the decrease in volume or increase in pressure during lhe process. Problem 12.1. A capacitor has a capacitance of 1 micro farad. How much work is required to change it to the level of 1000 volts ? -6 Solution. W 1 . 2 = - .!.._ C. V 2 = - .!.._ X 10 (1000) 2 = - 0.5 Joules A ns. 2 2 1.30 Heat. Heat is an energy interaction between the system and surroundings. Assun 1.! that two closed systems at different temperature are brought into cont.act with each other so that they have a common boundary. Due to temperature difference, energy in the form of heat is transferred from the higli temperature system to the low Lt:mperature system. This process is conti@ed until thermal equilibrium is attained. Here, heat is not observed but instyad heat transfer is inferred through termperature changes or through . ome similar effects. This heat transfer may be regarded as energy interaction between two c!t)sc-tl systems in which there is no work interaction. In the words of Obert Heat is energy transferred, without transfer of mass, across the boundary of system, because of temperature difference between system and surroundings. The unit of heal is Joule (J) in S .I. system A · adiabatic process is defined as one in which no heat i_nteraction occurs. This

39

Basic Concepts and Definitions

'l)k}.-l i:J adiabatic condition exists if lh e system i s surrounded b y an enclosure impermeable to heat A walrwhich is impermeable to the flow of heat is an adiabatic wall whereas a wall which permits the flow of heat is a diathermal wall.

1.31 Difference between Heat and Work

These are some similarities and dissimilarities between heat and work. They are as follows :-

S i milarities

(a) Heat and work transfer are the energy interactions. (b) Heat and work are both transient phenomena. They are not possessed by the system but either or both ci:oss the system boundary when a system undergoes a change. (c) Bot.Ii heat and work are boundary phenomena and are observed only at th� boundary of the system. Both represent energy crossmg the boundary of the system. Once the heat or work transfer is over no heat or work is present, only, the · the energy, i.e. ..___ re.mlt of heat or work transfer. (d) l_:!eat and work are not the property of the system. (e) Both heat and work arc path functions and inexact differentials. So instead of dQ or dW these are expressed as SQ or .5W.

Dissim ilarities

(a) Heat transfer-� the energy ir.teractions due to temperature difference only. All other enregy interactions are termed as work transfer. (b) Heat is a low grade energy while work is a high grade energy. The following example will clearly demonstrate the difference between heat and work. _ In order to understand the difference between heat and work consider a gas contained in a rigid vessel (Fig. 1.24) on which the resistance coils are wound. When current flows through the resistance coils, the temperature of the gas increases. Now question arises whether heat or work crosses the boundary of the system. Resistance coil

/

System

___..___-

Battery (a)

1

I I I I

I

b�u�d�� _ _

-=-=:\ _ _

7 I

Gas

I 1 ,..._________, 1 L _ _ _ _ _ _ _ _ _ _ .J

lb) fig. 1.24. Difference between Heat and Work.

In the first case(Fig. 1 .24a), we consider the gas as the system. Here, r.he er.ci'gy ::rnsses the boundary ot the-system because the temperature of the walls is high0r than the

40

Thernwdynamics and Heat Engines

temperature of the gas. Therefore, this energy transfer is regarded as the heat crossing the �undary of the system. In the second case (Fig. 1 .24b), we consider the vessel and the resistance heater as the system. Electrical energy crosses the boundary of the system. It can be shown similar to the case discussed about Fig. 1.16 that the electrical energy crossing the boundary of the· system is the work. 1.32. Sign Convention or Various Energies in Their Algebric Sense . Q is positive Heat received by system Q is negative Heat rejected from system Heat neither received Q=0 nor rejected : Internal Energy increases : tlU is positive Internal Energy decreases : t!U is negative Internal Energy neither Increases nor decreases : tlU = 0 Workdone by system : W is positive Workdone on system : W is.negative NO wordone on or by system : W = 0 Problem 1.22. A pump delivers 1200 litres of water per hour against a constant head of 100 metres. Calculate the workdone per hr in heat units. Solution. The mass of 1 litre of water is 1 kg. Therefore, the quantity of water raiser per hour is 1200 x 1 = 1200 kg: Workdone by pump/hr = Force per hr x distance. = m x g x h = 1200 x 9.81 x 100 = 1 1 77200 1 Problem 1.23. Compressed helium gas kept in a bottle is used to inflat an inelastic flexible balloon, originaly folded completely flat to a volume of 0.6m3 • If the barometer reads 76 cm Hg, calculate the amount of work done upon the atmosphere by the balloon. Sketch the system showing the boundary before and after the process. Comment on the result. Solution. Refer to Fig. 1 .25. The finn and dotted lines show the boundary of the system before and after the process. The displacement work is given by wd..p =

f:

Ballom

pdV +

f:

pdv · = p .1v + o

Bottle

= 1.01325 X 105 X 0.6 = 0.60795 X 105 J = 60.795 kJ

A ns.

The workdone is positive because it is done by the system. There is no displacement work don� by the bottle as the wall of the bottle is rigid. It has been assumed that the pressu�e m the balloon is atmosp�eric at all times since the balloon fabric is light, �e�uc and unstressed. Under the crrcumstance of elastic and stressed balloon during the filling process the workdone by the gas would be greater than the present value by tht:.

•

41

Hasic Concepts and Vejinitions

Final volu me of balloon

-j/

V2 = 0.6 m3

Balloon initially flat

Valve

Bottle

'

--+-

Helium Gas

Fig. 1 .25

amount equal to the workdone i n SI.retching the balloon. But i t i s to be noted that if the system includes. both bottle and gas as the system the displacement work would be as calculated above. 1.33 Vari,::.n·, . hermodynamic Process

Various thermodynamic processes used are - given below

(i) Constant pressure process (p = c) (ii) Constant volume process (v = c) (iii) Isothermal process (pv = c, T = c) · (iv) Adiabatic process (pvY = c) (v) Polytropic process (pvn = c) ( vi) Free expansion process ( u= c)

Application of these process is are given in chaptei: 3.

Viva Voce and Theoretical :

EXERCISES

Define the term thermodynamics. Define dimensions and units. Define the units for time, length and mass. Define force in general and in tenns of different system of units. Give the dimensions and units i n S.I. system o f the following. (a) Velocity, (b) Acceleration, (c) Work, (d) Energy, (e) Power (f> Momentum, (g) Pressure, (h) Discharge, (i) Temperature, (j) Density. 1 . 6. DifferentiaLC between mass and weight.

1.1 1 .2. 1 .3 . 1 .4 . 1 .5 .

42

Thernwdynamics and Heat Engines

1 .7 . Explain thermo.jynamic system, surrounding and universe. Distinguish between closed, open, isolated, homogeneous and heterogeneous systems. Illustrate with examples. (B.U . 1970) 1 .8. What do you mean by phase of a system ? 1 .9. Discuss the macroscopic and microscopic point of view 0f thcnnodynamics . . I 1 0. Discuss the concept of continuum in thermodynam ics. 1 . I I . Define the following terms. (a) density, (b) specific weight, (c) specific volume, (d) pressure, (e) atmospheric pressure, (f) absolute pressure (g) gauge pressure, (h) vacuum pressure, (z) static pres. Prove that the weight of a body at an elevation z above sea level is given hy. w = mg [r/ (r + z) ] 2 when r is the radius of earth at equator 1 .27. Choose whether the following is open or closed system ? (a) Water pump, (b) Pressure cooker (c) Water wheel. (cf) Thermometer surrounded by a medium at high temperature, (e) Automobile engine, (f) Boiler including all piping and radiators, (g) Air compressor, (h) A dash pot consisting of a cylinder, piston and containing liquid, 1 . 28 . Indicate the number of phases in the following-(a) Pure water, (b) Water alcohol mixture, (c) coffee (d) water oil mixture. (e ) mild steel at room temperature. (f) liquid oxygen at room temperature. : . : ';). I f z is a function of three properties x,y and z such that dU = Mdx + Ndy + Pdz whcrr M. N and P are functions of properties x,y and z . Prove that the following conditions are essential for U. to be an exact differential.

oP oy

oN oM oM oz

oP oN

oM

OX OX oy 1 .30. Draw the diagrammatic sketches of the following systems, in each case, exphasizing the system, boundary, surrounding and the direction of heat and work. Write whether the sys�em is open or closed and indicate the energy conversions. (a) Mercury in glass thermometer surrounded by a medium at high temperature. (b) A perfectly insulated constant pressure chamber containing a mixture of ice an, water constantly agitated by a motor driven stirrer. (c) A non-insulated nozzle receiving gas at a high pressure with negligible velocity and expanding it down to a lower pressure. Assume no loss due to friction or turbulence during the process. (cf) Inflation of a flat tyre by forcing air into the tyre with tyre walls as the system (R.U.1 970 A) boundary, the tyre walls being non-conducting. · (R.U. 1970 A) l . 3 1 . Justify the statement that work and heat are not properties. 1 .32. Distinguish clearly between the following, giving examples wherever necessary : (a) Closed system and open system. (b) Heat and work. (c) Point functions and path functions. (cf) Enthalpy and internal energy. [R.U. 1 970 (s)1 Numerical 1 .33. On the surface of the moon, the gravitational effect is 1/9.2 of that on the earth's surface. Calculate the weight of the mass of a solid. Ans. m/6.2 where m is the mass. 1 .34. A man weighs 760 N at a location where acceleration due to gravity g = 9.8 m/s2 What will be the.weight of thy man on a planet where local g = 2.45 m/s2 ? Ans. 1 90 N 1 .35. A ball with an initial kinetic energy of 700 J is shot vertically from a cannon and the ball rises to an altitude of 250 m. Calculate the mass of the ball. Ans. 2.854 kg

44

Thernwdynamics a_nd Heat Engines

1 )6. A force of 390 N is applied on a mass of 2 kg along an inclined surface at a location where g = 9.81 m/s2 • If the force and the inclin'!d make a 30 angle with the vertical, determine the acceleration of the mass if the force is acting (a) upwards, (b) downwards. 1 .37. The pressure of steam generated in a boiler as recorded by the Bourdon pressure gauge was 9.2 bar. Calculate the absolute pressure of steam if the barometer reads 774 mm of Hg. 1 .38. The pressure of the oil delivered by an oil pump as recorded by a pressure guage was 9 bar gauge. What must be the height of an oil column which corresponds to this pressure ? The density of oil is 0.9 gm/cm. 3 1 .39. The gas used in a gas engine trfr l was tested. The pressure of gas supply was 7 cm of water column. Find the absolute pressure of the gas if the barometricpressure is 760 mm of Hg. 1 .40. · A barometer reads 76 cm of Hg. What would be the absolute pressure if (a) a pressure gauge connected to a turbine inlet reads 28 bar and (b) a vacuum gauge connected in the exhaust line of the same turbine reads 70 cm of Hg ? 1 .4 1 . Whi!e measuring the air pressure in an air chamber, a liquid of specific gravity 0.8 shows height of 4.25 cm. To what pressure is this equivalent in cm of water ? What in the absolute pressure in Pa ? 1 .42. A vertical composite liquid column with its upper end exposed to aur.ospllere comprises of 54 cm of Hg (sp. gravity of Hg = 1 3.6), 65 cm of water (sp. gravity of water is 1) and 80 cm of oil (sp, gravity of oil is 0.8). Calculate the pressun:: (a) at L'1e bottom of the column,(b) at the enter surface of oil and water and (c) at the inter surface of water and Hg. 1 .43. A non flow reversible process occurs for which p = 3 V2 + 1/V where p is in bar and V in m3 • Wpat will be the work done when V changes from 0.5 m3 to 1 m3 ? 1 .44. The level of the water in an enclosed water tank in 50 m above the ground. The pressure in the air space above the water in the tank in 140 kPa. The average density of water is 1000 kg/m3 . Calculate the pressure of water at the ground level. 1 .45. · A diver descends 1 00 m to a sunken ship. A container is found with a pressure guage reading 100 kPa (gauge). Atmospheric pressure is 1 00 kPa. What js L'le absolute pressure of the gauge in the container expressed in kPa ? The density of water is 100 kg/m3 • Ans : 1 181 kPa 1 .46. A certain engine is supplied with 1 .59 kg of fuel per minute. The potential energy stored per kg of fuel is equivalent to 24,000 kJ/kg. If 20 per cent of the energy supplied to engine is converted into mechanical energy, find out the rate at which the engine is developing power. 1 .47. At a speed of 50 km/h, the resistance to motion of a car is 900 N. Neglecting losses, calculate the power of the engine of the car at this speed. Determine also the heat equivalent of workdone per minute in the engine. 1 .48. Calculate the height of a column of water equivalent to atmospheric pressure of 1 bar if the water is at 1 5°C. What is the height if the water is replaced by mercury ? 1 .49. If all the atmosphere is liquified what will be the equivalent height of this liquid air in metre of water at 15 °C if the earth surface is assumed flate ?

Basu· Concepts and Definitions 1 .50. Find the mass of air in kg of the atmosphere surrounding the earth if the pressure is I bar everywhere on the surface. Assume that the earth is a sphere of 12850 km. l . 5 l . A piston-cylinder arrangement is containing a fluid at I O bar, the initial volume being 0.05m3 • Find the work-done by the fluid when it expands reversibly for the following cases. (a) at constant pressure to a final volume of 0.2 m3 . '-{b) according to linear law to a final volume of 0.2 m3 and a final pressure of 2 bar. (c) according to a law pV = constant to a . final volume of O. l m3 (d) According to a law p\13 = constant to final volume of 0.06m3 .

(e) according to law p = (A/V3 ) - (B/V) to a final volume 0. lm 3 and a final pressure of 1 bar wher A and B are constanL Sketch all processes on the p - V diagram. Ans. 1 50 kJ, 90 kJ, 34.7 kJ, 7.64 kJ,17.67 kJ. Hint : Where necessary, calculate the constant with the given data. 1 .52. A cylinder piston arrangement is containing a fluid at a pressure of 3 bar and with ..,,, specific volume of 0. 1 8 m3/kg allows the fluid to expand reversibly to a pressure 'Of 0 .6 bar according to law, p = c/v2 where c is a constant. Detennine the workdone by the fluid on the piston . Ans. 29.84 kl. 1 .53 . A special manometric fluid of specific gravity of 2.98 is used to measure a pressure of 2.2 atm at a locatin where the barometric pressure is 750 mm Hg. What heigh. will the manometer fluid register ? Sp. gravity = 1 .0

1 00 cm

T

70 cm

84 cm

Sp. g ravity = 1 3.54

Fig. 1 .26.

1 .54. With reference to Fig. 1.26, find (Pa - Pb) in N/m2.

1 75 cm

2 Temperatur_e and Zeroth Law of Thermodynamics This chapter deals with the concept of temperature, zeroth law of thermodynamics and basic principle of various thermometers used in the measurement of temperature. 1.1.

.Concept of Temperature.

The concept of temperature arises from the sensory perception of hotness or coldness. The housewife thinks of temperature as how hot she must set the oven to cook the roast. The computer engineer thinks of temperature as how cool he mus� keep the transistors and integrated circuit to- make them work reliably. The plasma physicist views temperature as a measure of kinetic energy of the molecules or electrons. And the astronomer views temperature as a measure of the radiant energy emission from the stars. Truely speakings such physiological sensation is insufficient for precise knowledge of temperature. But all these diverging concepts have one thing in common; they all relate to energy or energy tran')fer, hence clearly mark temperature as a therrnodynami.c property. Actually, the word temperature is derived from the Latin word "Temperature " meaning proper mixing or tempering implying attainment of thermal equilibrium. According to :Maxwell "temperature of body is its thermal state considered with reference to its ability to communicate heat to other bodies. In a precise way, temperature is 1ha1 property of a system which determines whether or not it i.1 i11 1/1er111al 0int (Fig.1.2), Using eq. (2.1), we get. t1 = ax1 + b and r2 = ax2 + b

a�

1 00°C --

Fig. 2.2. Mercury Thermometer.

50

Thernwdy11amics a11d Heat E11gi11es

or

The fixed points assigned to the ice and steam point in Centigrade (or Celcius) are 0°C and 100°C and in Fahrenheit scales, these values are 32°F and 212°F respectively. Thus in Centigrade scale, t1 - t2 = 100°C and t1 = 0°C hence the above equation takes the form of t"C =

100(�\ � x-,- x 1}

100 (

·L - L ;

L $ - L '·

(2.3)

)

Note lha1 Eq (2.3) defines the centigrade temperature on the scale of this particular thermometer. A second thermometer is constructed from different kind of glass and using a different liquid and is calibrated in the above manner. Let L', and L'i be the reading of their fixed points. The new thermometer is placed in contacl with the same body for which the first thermometer gave the temperature I. Let t' be the temperature of same bcdy by second thermometer, then ( will re given by �

t ' = loo /L' - Ll )

f

(2.4) Generally, it is found that t' i.s not equal to t even though these are temperatures of the same body. This is because of the fact that the two liquids donot expand in the same manner. In addition different variety of glasses have different expansion coefficients and the method of construction also affects the readings. This drawback is applicable to all thermometers which.are based upon single thennometric property. The use of two fixed points was found unsatisfactory and later abandoned because Qf (a) the difficulty of achieving equilibrium between pure ice and ice and air saturated water and (b) extreme sensitiveness of the ste.:m point to the change in pressure. \L;

- LI

(ii) Method in Use af'ter 1954 :

Since 1954 only one fixed point has been in use i.e. the triple point of water. It is the state at which ice, liquid water and vapor coexist in equilibrium. The temperature at which this state exists is arbitrarily assigned a value of 273.16 degrees kelvin or 273.16 K. Let x, be the thermometric property and the body is pl�ced in contact with the triple point of water whose temperature be 11 then as per eq. (2.2) t = ax and t, = ·a x, •

or

a = !.!_ = 273.16

x,

t

x,

= 273.16 .:!..

�omparison o f Thermometers

(2.5)

x,

Applying the method in use after 1954 to the following thermometers the mperature recorded by them will be given by (a)

Liquid thennometer ;

t (L) = 273.1 6 !::.._ L,

(2.6)

Temperature and Zeroth Law of Thermodynamics (b) (c) (d) (e)

Thermocouples :

�-

Electric resistance thermometers : _.;-

Constant pressure gas thermometers

Constant volume gas thermometers :

t (e )

=

t (R) .

=

t (V)

=

t (p)

=

273. 16 -=e,

273.16 .!£. R,

273.16 �

v,

273.16 l!_

P,

51 (2.7)

(2.8)

(2.9)

(2.10)

If above thermometers are employed to measure the temperature of a system simultaneously, it will be observed that Ltiere is a considerable difference among the readings due to the reasons stated above. The minimum variation is found in the case of gas thermometers and that is why gas is c�sen as standard thermometric substance. 2.6. Liquid Thermometers (Mercury) Out of the many liquid thermometers, the mercury thermometers is the most commonally used in scientific work for ordinary temperature. This is because of the fact that mercury has many advantages as the thermometric fluid over other fluids. It consists of a cylindrical glass bulb (Fig. 2.2) containing mercury to which a graduated capillary be of uniform bore is attached. The upper point (steam point) and the lower point (ice point) are set by using hypsometer and ice respectively. Since ice and steam points � fixed, hence Re F - 32 C - = --- = 80 100 180 C = F - 32 = Re ..... (2. 1 1) )r 4 5 9 For ordinary thermometers, "Vender" glass is in use while for high temperature measurement "Juna 16" and Boro Silicate are used. As pointed out, thermometers are generally used in the rough work but when they are employed for precision work many errors have to be taken into consideration and these have to be corrected. Lim itation of Liquid Thermometers. The main drawback of liquid thermometers is that, they use different liquids such as mercury and alcohol and are calibrated as stated above, they do not agree at any temperature except 0°C and 100° . This is due to the fact that expansion of the different liquids is neither regular nor similar. Thus, such thermometers give an arbitrary scale of temperature. In addition, the mercury thermometer needs many corrections. It is hardly suited for precision work because its nearest accuracy is of 0.05°C. 2 .7 Gas Thermometers (Standard or Absolute Scale of Temperature). The Advantages of Gas as Thermometric Substar;ce Over Liquids : We have already seen that the thermometers which are based on a single property are not entirely satisfactory for temperature scale. The gas thermometer fills this gap. The advantages possessed ty gas as a thermometric substance qver liquids are (i) For the same increase of temperature gases expand much more than liquids. For 1 legrce rise in temperature, a pennanent gas expands twenty times as much as mercury.

52

Thermodynamics and Heat Engi11es

Gas thermometers, will Jead to a negligible correction. (ii) The expansion of the gases is more regular over a much wider range of temperature compared to liquids. For example. a permanent gas when not subjected to high pressure, maimains its state-and behaves in the same manner at very high or low temperature whik men:wy is limited by its freezing and boiling point (iii) Gases have much lQ.wer thermal capacity than liquids, it is thus more sensitive to gain or loss of heal {iv) The gases possess vecy low specific gravity and as obtainable in the same purit) all over the world. A perfect dry state of gas is essential for this purpose. (v) The scale furnished by different gases are almost identical. This is because of tht: fact that the volume and pressure coefficients of all permanent gases nearly equal. Thu me.ans that the tempentture recorded is independent of thennometric substance. That � why thermometels are taken as primary standards and all the other thennometers are compared with them for �l>ration and hence called secondary thennometers. Pr.nclp� of' Gu Thermometers. The use of gas as a thermometric substance is based on ihe expansion m contraction of the gas which is a function of temperature. In other words, it is based on perfect gas law i.e. the combination of Boyle's and Charles' laws which is given as

gas

or

Pl v , P2V2 --· "" -- = constant T1 T2

(2.12:

It is obvious from the equation (2. 12) that pv product of all gases is a function of temperature alone and is independent of the thermometric property. Thus. gas may be used as a thennome1ric property to measure temperature and thermometers using gas as thennomt;tric property will give a standard scale of temperature or an absolute scale of temperature. Equation (2. 12) indicates two ways of measuring tempenure with gas thennometer :(i) Keeping the pressure consta11t and observing the change m volume-constant pressure thermometer. (iz) Keeping the volume constant and noting the change in p�sure-