a NC a eC g Pre,Applied Thermodynamics For Engineering Technologists Fith Edition The late T.D.EASTOP A. McCCONKEY BSc,PhD.CEng, BSc, PhD, CEng, PIMehieRCLESE, — FIMecE: Formeriy Head ofthe School Formerly Head ofthe ‘of Ensinccring at Deparment of Mechanical and Woverhampion we Industrial Engineering at Dundee Collegeof Tecnoloyy COMPLIMENTARY COPY RAYMOND TUAN PEARSON MALAYSIA SDN BHD «soow) Higher Education io Sy nan Banagin §, "rama Dest Deol 75230 Cheng Bara, Mela ‘reuran 065121 #16 UP 012-6082060, ‘Eni raymond @pearsonedomany ro gh «andan «New Yor Bin Sn Func Ta Seago Srp ong Kong Ses pe Yom sabes cy un nh“anbugh Ge Hisex C2021 ‘and Associated Companies thnoaghou the word st arom the World Wide Web a: huputwoepeasonede com (© Longman Group UK Lined 1963, 196, 1978, 1986, 1983, ‘lights reserved no pst of thi pbliction may be epoca, Stored in eval system, or transmit nay form ot by an ‘means, lectreni, mechanical, photocopying, cording, or terwise ‘without cite the prior writen permission ofthe Publishers ra ence peiting vesicted copying nthe United Kingdom ised by ‘he Copyright Licensing Ageny Lu, 90 Toenam Cou Road, ondan, WIT LP ist publishod 1963 Second Eton 1969 ‘Third Ration 1978 Four Eon 1986 Pil Edion 1993, ISBN 042-09193-4 Brith Library Cataloguing in Publiston Data ACP record fr this book is aval rom the British Library Library of Congress Catalogng-in-Puiiation Data ast, ED, (Thomas D) ‘Apolo tsodyamics for engagering technologist. Bey, A McConkey Shed Pc Ineloes bibliographical references an inde. L'Themodyaamie. I. MeConke, A (Allan, 1927 nn ‘Ti2gsii2 1995 eaiaor tae ‘Set by 6 n 1012p Monotype Lasereomp Times 569 Produced by Pea Education Asa Pe Lid Pine in Singapore (COS) nu 07 05 05 08 0%Contents Preface Acknowledgements Nomenclature 1) Introduction and the First Law of Thermodynamics 4 12 13 14 Ls 16 17 18 Heat, work, and the system Units ‘The state of the working fuid Reversibility Reversible work Conservation of energy and the First Law of Thermodynamics ‘The non-fiow equation ‘The steady-flow equation Problems 2. The Working Fluld Liquid, vapour, and gas ‘The use of vapour tables The perfect gas Problems Reversible non-fow processes Reversible adiabatic non-flow processes Polytropie processes Reversible low processes Irreversible processes Nonsteady-flow processes Problems 4 The Second Law 44 ‘The heat enginecontents 42. Entropy 43° The T-s diagram 44 Reversible processes on the T-s diagram 45. Entropy and irreversibility 46 Exergy Problems 5 The Heat Engine Cycle S41 The Carnot eyele 52 Absolute temperature scale 53 The Carnot eyele for a perfect gas 54 The constant pressure cycle 53. The air standard eyele 56 The Otto cycle 57 The diesel cycle 58 The dual-combustion cycle 59 Mean effective pressure 5.10 The Stirling and Ericsson eyeles Problems 6 Mixture 61 Dalton’ law and the Gibbs~Dalton law 62 Volumetric analysis of a gas mixture 63 The molar mass and specific gas constant 64 Specific heat capacities of a gas mixture 65 Adiabatic mixing of perfect gases 66 Gas and vapour mixtures 67 The steam condenser Problems 7 Combustion TL Basie chemistry 72° Fuels 73 Combustion equations 74 Stoichiometric air-fuel ratio 71S Exhaust and fue gas analysis 16 Practical analysis of combustion products 17 Dissociation 78. Internal energy and enthalpy of reaction 79° Enthalpy of formation 710 Calorific value of fuels TAL Power plant thermal efficiency 112 Practical determination of calorific values 713 Air and fuel-vapour mixtures Problems 93 9 109 us 21 125 12s 127 128 130 133 135 136 138 a1 3 145 vr 147 150 131 137 162 166 170 130 " 2 Steam Cycles &1 The Rankine cycle 82 Rankine eycle with superheat 83 The enthalpy-entropy chart 84 The reheat eycle 85 The regenerative eycle 86 Further considerations of plant efficiency 87 Steam for heating and process use Gas Turbine Cyck Problems ‘The practical gas turbine eycle “Modifications to the basic eycle Combustion ‘ditional factors Problems Nozzles and Jet Propulsion 104 02 103 wos 10s 106 107 108 109 1010 Nozzle shape Critical pressure ratio Maximum mass fow [Nozzles off the design pressure ratio ‘Nozzle efficiency ‘The steam nozzle Stagnation conditions Jet propulsion ‘The turbojet “The turboprop Problems Rotodynamic Machinery td 2 3 114 us 116 17 8 us 11.10 Rotodynamic machines for stoam and gas turbine plant ‘The impulse steam turbine Pressure and velocity compounded impulse steam turbines ‘Axial-flow reaction turbines Losses in turbines Axialflow compressors Overall efficiency, stage efficiency, and reheat factor Polytropic efficiency Centrifugal compressors Radial-ow turbines Problems Positive Dieplacement Machines 124 122 Reciprocating compressors Reciprocating compressors including clearance vilcontents 1% “ 6 123 124 125 126 129 Multistage compression Steady-flow analysis Rotary machines Vacuum pumps Air motors Problems Reciprocating Internal-combustion Engines 14 132 23 Ba 135 136 87 a3 ne Bao 131 2 Bo 134 1s Four-sroke eycle ‘Twostroke cycle Other types of engine Criteria of performance Engine output and efficiency Performance characteristics Factors influencing performance Real eycles and the air standard eycle Properties of fuels for IC engines Fuel systems ‘Measurement of air and fuel fow rates Supercharging Engine emissions and legal requirements ‘Alternative forms of IC engines Developments in IC engines Problems Refrigeration and Heat Pumps 1d 142 43 144 las 146 147 148 149 4.10 14 1412 Reversed heat engine cycles ‘Vapour-compression cycles Refrigerating load ‘The pressure-enthalpy diagram Compressor type ‘The use of the flash chamber Vapour-absorption eyeles Gas cycles Liquefaction of gases Steamet reftigeration Reffigerants Control of refrigerating capacity Problems Peychrometry and Air-conditioning 1st 152 153 154 Paychrometric mixtures Specific humidity, relative humidity, and percentage saturation Specific enthalpy, specific heat capacity, and specitic volume of moist air Airconditioning systems 396 406 sul 412 4166 w Contents 155 Cooling towers 553 Problems 556 Heat Transfer 561 16.1 Fourier’s law of conduction 562 162 Newton's law of cooling 565 163 The composite wall and the electrical analogy 568 164 — Heat flow through a cylinder and a sphere 32 165 General conduction equation 37 166 Numerical methods for conduction 584 167 Two-dimensional steady conduction 387 168 One-dimensional transient conduction by finite difference 593 169 Forced convection 399 16.10 Natural convection 610 1641 Heat exchangers 613 16.12 Heat exchanger effectiveness os 16.13 Extended surfaces er 16.14 Black-body radiation 633 1615 The grey body 634 16.16 The Stefan—Boltzmann law637 oT 1617 Lambert's law and the geometric factor 9 1618 Radiant interchange between grey bodies 63 16.19 Heat transfer coeficent for ra 49 1620 Gas radiation 650 1621 Further study 651 Problems 652 ‘The Sources, Use, and Management of Energy 663 174 Sources of energy supply, and energy demands 664 172 Combined cycles 610 173 Combined heat and power (co-generation) on 174 Energy management and energy audi 680 175. The technology of energy saving 688 176 —Altemative energy sources 696 177 Nuclear power pla i) Problems 701 Index 08Preface to the Fifth Edition ‘This book aims to give students of engineering a thorough grounding in the subject of thermodynamics and the design of thermal plant. The book is comprehensive in its coverage without stcrificing the necessary theoretical ra; the emphasis throughout is on the applications of the theory to real fprocesses and plant, The objectives have remained unaltered through four previous editions and continuing interest in the book not only in the UK but ‘Also in most other counties in the English-speaking world has confirmed these “objectives as suitable for students on a wide range of courses, ‘The book is designed as a complete course text for degree courses in mechanical, aeronautical, chemical, environmental, and energy engineering, tngincering science, and combined studies courses in which thermodynamies fand related topies are an important part of the cucriculum, Students on Technician diploma and certificate courses in engineering will also find the book suitable although the coverage is more extensive than they might requir 'A numberof lecturers in universities and polytechnics in the UK were asked for comments on the book before the fifth edition was prepared; the consensus twas that the balance ofthe book was broadly correct with only minor changes needed, but a more modern format was thought to be desirable, "The fith edition has therefore been completely recast in a new style which ‘will make it more attractive, and easier to use. The opportunity has also been taken to rearrange the chapters in what seems to be a mote logical order. “Throughout the book the emphasis snow on the effective use of energy resources and the need to protect the environment. The chapter on energy sources, tse and management (Ch. 17), has been improved and extended; it now includes fa more extensive coverage of combined heat and power and a new section on fenergy recovery, including a brief mention of pinch technology. The material fon gas turbines, steam turbines, nozzles, and propulsion (Chs 8-10) has been rewritten in a more logical format giving a more general treatment of blade design while sil stressing the differences in design procedures for steam and igas turbines. In the chapter on refrigeration (Ch. 14) more emphasis is given {o the heat pump and to vapour-absorption plant. A new section on refrigerants discusses the vitally important question of the thinning ofthe ozone layer due to CFCs; examples and problems in this chapter now use refrigerant 134aface to the Fith Edition instead of refrigerant 12, and tables and a reduced seale chart for RI34a are included by permission of ICI. Analysis of exhaust gases, emission control for IC engines, and the greenhouse effect are also included [A new sign convention for energy transler across a system boundary has ‘come into general use in recent years and has therefore been introduced inthis ‘book. The convention is to treat both work and heat crossing @ boundary as [postive when itis transferred from the surroundings to the system, Als, there has been an international agreement to standardize symbols used for heat and ‘mass transfer and the symbols in this text have been chosen accordingly. For example, the symbol for heat transfer coefcient is a, that for thermal ‘conductivity 4, that for dynamic viscosity m, and that for thermal diffusivity x. “Molar quantities are now distinguished by the overscrpt,~ Thanks are due to Dr Y.R. Mayhew for many helpfl discussions on the use of physical quantities, units, and nomenclature. Inthe chapter on combustion (Cl, 7) the section on dissociation has been rewritten to conform with the use of @ standard thermal ‘equilibrium constant as tabulated in the latest edition of Rogers and Mayhew's Thermodynamic and Transport Properties of Fluid, ‘While preparing this new edition T have been ever conscious ofthe loss of my co-author and colleague for so many years, Allan McConkey, who died {vst after the publication ofthe previous edition in 1986. I would ike to dedicate this edition to Allan with deep affection and gratitude for a long and fruitful collaboration. TDR 1992Acknowledgements We are grateful to Blackwell Publishers for permission to include extracts from the Rogers and Mayhew Thermodynamic and Transport Properties of Fluids (SI Units) (4th ed), 1988, Figure 154 is reproduced by permission of the Chartered Institution of Building Services Engineers, copies of the chart (size A3) for record purposes may be obtained from CIBSE, 222 Balham High Road, London ‘SWI2 9BS. Figure 14.14 is reproduced by permission of ICI; Table 141 is an textract with some interpolated values of thermodynamic properties of HFA 1134a by permission of ICI, Runcorn, Cheshire. ‘The following sources have been drawn on for information: Figures 13.21 ‘and 1322 are adapted from The Internal Combustion Engine in Theory and Practice by CF. Taylor, MIT Press. Section 13.13 includes material adapted from Exhaust Emissions Handbook published by Cussons Ltd. The data for Fig. 12.30 was provided by JS. Milne ofthe Department of Mechanical and Industral Engineering, Dundee College of Technology, from an original test carried out by him.Nomenclature air-fuel ratio; area velocity of sound; acceleration; non-low specific exergy bottom dead centre British Standard Biot number steady-flow specific exergy brake mean effective pressure brake power velocity; constant; thermal capacity combined heat and power ‘compression ignition coefficient of performance calorie value discharge coeficient specific heat capacity molar heat capacity specific heat expacity at constant pressure specific heat capacity of air per unit mass of dry ait specific heat capacity at constant volume ‘molar heat capacity at constant pressure molar heat capacity at constant yolume bore; diameter ‘missive power; energy eccentricity force; geometric factor fuel injection Fourier number ftition factor; frequency {ition power ieeadiation {ross calorific value Grashof number jgravitational acceleration‘enthalpy; fundamental dimension of heat hydrocarbons enthalpy of reaction specific enthalpy specific enthalpy of reaction ‘molar enthalpy ‘molar enthalpy of reaction specific enthalpy of a saturated liquid specific enthalpy of vaporization specific enthalpy of a saturated vapour electri current internal combustion intensity of radiation indicated mean effective pressure indicated power current density radiosity Colbura factor for heat transfer ‘equilibrium constant standardized equilibrium constant isentropic index for steam; blade velocity coeficient stroke; fundamental dimension of length length; characteristic linear dimension fundamental dimension of mass Mach number molar mass mass flow rate rotational speed net calorific value non-dispersive infra-red Nusselt number number of transfer units polytropic index; amount of substance; number of eyinders; nove are length ‘octane number perimeter performance number Prandtl number absolute pressure; blade pitch ‘mean effective pressure brake mean effective pressure indicated mean ellective pressure pressure loss heat ate of heat transfer rate of heat transfer per unit area rate of heat transfer per unit volumeNKR SE SSEEEE specific gas constant; thermal resist thermal capacities ‘molar gas constant reheat factor Reynolds number radius; expansion ratio pressure ratio compression ratio entropy; steam consumption spark ignition Stanton number specific entropy specific fuel consumption absolute temperature; torque; fundamental dimension of time top dead centre temperature; fundamental dimens thickness temperature difference {rue mean temperature difference arithmetic mean temperature difference logarithmic mean temperature diflerence internal energy; overall heat transfer coefficient intemal energy of reaction specific internal energy specific internal energy of reaction ‘molar internal energy molar internal energy of reaction volume :ate of volume flow specific volume ‘work; brake load rate of work transfer, power temperature on any arbitrary scale dryness fraction; nozzle pressure ratio; length height above a datum level ‘number of stages 08; radius; ratio of| of temperature; blade Greek symbols er angle of absolute velocity; heat transfer coefficient; absorptivity for radiation blade angle; coeticient of cut ratio of specific heats, ¢y/¢, film thickness degree of reaction emissivity; effectiveness ofa heat exchanger expansionNomenclature egenauenns Subscripts as. g ge Sie a et ele eiate 75 geecc efficiency; dynamic viscosity thermal diffusivity thermal conductivity; wavelength kinematic viscosity ensity; reflectivity Stefan-Boltzmann constant time; shear stress in a fluid; transmissivity relative humidity; angle specific humidity; solid angle Peroentage saturation air standard dry air; atmospheric; arora; absolute velocity absolute velocity at inlet absolute velocity at exit black body brake thermal blade velocity cold fuid; compressor condensate; convective; critical value; clearance dew point; diagram ary bulb exit; exhaust fin; fuid saturated liquid; fuel; films low velocity change of phase at constant pressure saturated vapour; gases {ross hot fluid; high- heat pump. intercooler indicated thermal inlet; a constituent in a mixture; inside surface; intermediate; indicated; mesh point; injector mesh point; jet low-pressure stage mechanical normal net stagnation condition; overall; ouside; ero or reference condition essure stageBagi at agenegr length x length has units ‘of m?; specific volume = volume/mass has units of m’/kg Force, energy, and power Newton's second law may be written as force oc mass x acceleration for & body of constant mass, ie kona ay ‘where m is the mass of a body accelerated with an acceleration a, by a force Fi; is a constant In a coherent system of units such as SI, k= 1, henoe ‘The SI unit of force is therefore kg m/s*. This composite unit is called the newton, N, ig. 1'N isthe force required to give a mast of | kg an acceleration of m/s? It follows that the SI unit of work (= foree x distance) is the newton metre, Nim, As stated earlier heat and work are both forms of energy, and hence both can have the units of kg m?/s? or Nm. A general unt for energy is introduced by giving the newton metre the name joule, J12 Unite ie, 1 joule, 3= 1 newton > I metre or = Nm The use of additional names for composite units is extended further by introducing the wat, W, as the unit of power, ie L watt, W= 13/8 1N m/s Pressure ‘The unit of pressure (Force per unit atea) is N/m? and this unit is sometimes called the pascal, Pa, For most cases occurring in thermodynamics the pressure expressed in pascals would be a very small number; a new unit is defined as follows: bar = 10° N/m? = 10° Pa ‘The advantage of using a unt such as the bar is that itis approximately equal to atmospheric pressure, In fact the standard atmospheric pressure is exactly 1101325 bar. {As indicated in section 1.1, i€ i often convenient to express a pressure as head of a liquid. We have: Standard atmospheric pressure ‘Temperature ‘The variation ofan easily measurable property of a substance with temperature ‘can be used to provide a temperature-measuring instrument. For example, the length of a column of mercury will vary with temperature due to the expansion ‘and contraction ofthe mercury. The instrument can be calibrated by marking the length ofthe columa when it is brought into thermal equilibrium with the ‘vapour of boiling water at atmospheric pressure and again when it sin thermal equilibrium with ice at atmospheric pressure. On the Celsius (or Centigrade) scale 100 divisions are made between the two fixed points and the zero is taken atthe ice point. ‘The change in volume at constant pressure, or the change in pressure at constant volume, of fixed mass of gas which snot easily liquefied (e.g. oxygen, nitrogen, helium, etc.) can be used as a measure of temperature. Such an instrument is celled a gas thermometer. It is found for all gases used in such thermometers that if the graph of temperature against volume in the ‘constant pressure gas thermometer is extrapolated beyond the ie point co the point at which the volume ofthe gas would become zero, then the temperature ff this point is —273°C approximately (Fig. 1.10). Similarly ifthe graph of temperature against pressure in the constant volume as thermometer is ‘extrapolated to zero pressure, then the same zero of temperature is found. An absolute zero of temperature has therefore been fixed, and an absolute scale of temperature can be defined, Temperature on the absolute Celsius scale can be 7Introduction and the Firt Law of Thermodynamics Fig. 110. Graph of temperature against volume fora as, Volume ‘ol seam | “Temperatur/"C) ‘obtained by adding 273 to all temperatures on the Celsius scale; this scale is called the Keloin sale. The unit of temperature is the degre kelvin and is given the symbol K, but since the Celsius scale which is used in practice has a diferent ‘ero the temperature in degrees Celsius is given the symbol C(eg. 20°C = 293 K ‘approximately; also, 30°C ~ 20°C = 10K), In this text eapital T is used for absolute temperature and small for other temperatures. In Chapter 5 an absolute scale of temperature will be introduced as a direct consequence of the Second Law of Thermodynamics. It is found that the gas thermometer absolute scales approach the ideal scale as a limit, Also, with regard to the practical absolute temperature scale, there is an internationally agreed working scale which gives temperatures in terms of more practicable and more accurate instruments than the gas thermometer (see rel 13). Multiples and sub-multipl ‘Maultipies and sub-multiples ofthe basic units are formed by means of prefixes, and the ones most commonly used are shown in the following table: ‘Multiplying factor Prefix. Symbol One millon million tea ‘One thousand million giga. = G ‘One million, 10° megs M ‘One thousand, 10* kilo k ‘One thousandth, 10" milli om One millionth, 10° micro ye ‘One thousand millionth nano One million milionth pico p For most purposes the multiplying factors shown in the above table are suficient For example, power ean be expressed in either megawatts, MW, or kilowatts, EW, or watts, W.In the measurement of length the millimetre, me, the metre, m, and the kilometre, km, are usually adequate. For areas, the difference in size13 Fig. 11 State ofa ‘working fd on a property diagram 0 ofthe working Fluid ‘between the square millimetre, mm?, and the square metre, mis large (a factor fof 10%), and an intermediate size is useful; the square centimetre, em*, is recommended for limited use only. For volumes, the difference between the cubic millimetre, mm?, and the cubic metre, m?, is much too great (a factor of 10°), and the most commonly used intermediate unit is the cubic decimetre, «dm, which is equal to one-thousandth of a cubic metre (ie. | dm? = 10"? m) ‘The cubic decimetre can also be called the litre, I dma? = 10° m? ie Litre, (Note, for very precise measurements, 1 litre = 1.000028 dim*,) ‘Certain exceptions to the general rule of multiplying fuctors are inevitable ‘The most obvious example is in the case of the unit of time. Instead of the centisecond, kilosecond, or megasecond, for instance, the minute, hour, day, te. are used, Similarly, a mass flow rate may be expressed in kilograms per hous, kg/by if this gives a more convenient number than when expressed in kilograms per second, kg/s. Also the speed of road vehicles is expressed in kilometres per hour, km/h, since this is more convenient than the normal unit of velocity which is metres per second, m/s. The state of the working fluid In all problems in applied thermodynamics we are concemed with energy transfers to oF from a system. In practice the matter contained within the boundaries of the system can be liquid, vapour, or gas, and is known as the working fluid, At any instant the state of the working fuid may be defined by ‘certain characteristics called its properties, Many properties have no significance in thermodynamis (electrical resistance), and will not be considered. The thermodynamic properties introduced in this book are pressure, temperature, specific volume, specific internal energy, specific enthalpy, and specific entropy. It has been found that, for any pure’ working fluid, only two independent properties are necessary to define completely the state of the fluid. Since any {wo independent properties suffice to define the state ofa system, itis possible to represent the state ofa system by a point situated on a diagram of properties. For example, a cylinder containing a certain fiuid at pressure p, and specific volume », is at state 1, defined by point 1 on a diagram of p against 0 (Fig. 1.11(a)) Since the state is defined, then the temperature ofthe fui, 7, is re ri r e o °Introduction and the Firat Law of Thermodynamics fixed and the state point can be located on a diagram of p against T and T against v (Figs 1.11(b) and 111(c)) At any other instant the piston may be moved in the elinder such that the pressure and specific volume are changed to pzand ¢. State2can then be marked on the diagrams. Diagrams of properties are used continually in applied thermodynamics to plot state changes. The most important are the pressure-volume and temperature-enteopy diagrams, but enthalpy-entropy and pressure-enthalpy diagrams are also used frequently. Fig. 112. Reversible and irreveribleprocesses 10 1m section 1.3 it was shown thatthe state of Mud can be represented by a point located on a diagram using two properties a coordinates. When a system, Changes state in such a way that at any instant during the process the state Point can be located on the diagram, then the proces is suid to be reversible ‘The fid undergoing the process passes through a continuous series of equilibrium states. A reversible process between two slates can therefore be drawn as a line on any diagram of properties (Fig. 1.12(a)}- In practice, the fuid undergoing a process cannot be kept in equilibrium in its intermeiate states and a continuous pth cannot be traced ona dagram of properties. Sich real processes are called reversible processes. An irreversible proces is usually represented by a dotted line joining the end states to indicat that the intermediate sates are indeterminate (Fig .12(b)) ‘A more rigorous definition of reversibility is as follows: When a lui undergoes a reversible process both the fluid and its surroundings ‘cam always be restored to their original state ‘The eriteria of reversibility ate as follows: (a) Thoprocess must be frictionloss. The uid itself must have no internal ition and there must be no mechanical friction (eg. between cylinder and piston) (b) The difference in pressure between the uid and its surroundings during the process must be infinitely smal. This means tha the process must take place infinitely slowiy, since the force to accelerate the boundaries of the system is infinitely smallFig. 113 Fh ina clindse undergoing 8 compresion (€) The difference in temperature between the uid and its surroundings during the process must be infinitely small, This means that the heat supplied or rejected to of from the Aid must be transferred infinitely slowly It is obvious from the above criteria that no process in practice is truly reversible. However, in many practical processes avery close approximation to an internal reversbity may be obtained. In an internally reversible process, although the surroundings ean never be restored to their original state the uid itself st all mes in an equilibrium state and the path of the process can be ‘exactly retraced to the initial state. In genera, processes in cylinders with a reciprocating piston are assumed to be internally reversible as a reasonable ‘approximation, but processes in rotary machinery (eg. turbines) are known to be irreversible due to the high degree of turbulence and scrubbing of the fi Reve work Consider an ideal frictionless fivid contained in a cylinder behind « piston. ‘Assume that the pressure and temperature of the fuid are uniform and that there is no friction between the piston and the cylinder walls. Let the cross-sectional area ofthe piston be A, lt the pressure ofthe fluid be p, let the pressure of the surroundings be (p + dp) (Fig. 1.13). The force exerted by the piston on the fluid is pA. Let the piston move under the action of the Torce exerted a distance dl to the left. Then work done on the fluid by the piston is given by force times the distance moved, (ra) x al ie. Work done, dW —pav hese dV is «small inerease in volume. The negative sign is necessary because the volume is deoreasing (Or for a mass, m, AW = mp do where v is the specifi volume. This is only true when criteria (a) and (b) hold as stated in section 1.4 ‘When a fluid undergoes a reversible process a series of state points can be joined up to form a line on a diagram of properties. The work done on the fluid during any reversible proosss, W, is therefore given by the area under the line ofthe process plotted on a p-» diagram (Fig. 1.14), ie pa mmf rae =m shaded are on Fa 138) (12) nInroduetion and the First Law of Thermodynamics ig. 14 Work done ina compression proses ‘hon pcan be expresied in terms of v then the integral, [2 pd, ean be evaluated Example 1.1 Unit mass of a fluid at a pressure of 3 bar, and with a specific volume of 0.18 m!/kg, contained in a cylinder behind a piston expands reversibly to a pressure of16 bar according toalaw p = c/o, wherecisa constant. Calculate the work done during the process Solution Refering to Fig. 1.15 ig. 115 Pressurespeific volume diagram for Example 11 premure(N/a?) w= am [pdr -m shade ae) -o(-il 3 x 0.187 = 0.0972 bar (m?/kg)? 2 os catia aoe therefore 402 m3. w= —o0972 x 105(1 tN my O18 0409, . = 29840 N m/ke ie. Work done by the fuid = +29840 N m/kg, 12Fig. 116 Reversible expansion process on spe diagram | Fig. 117 Reversible cycle on ap diagram Example 1.2 1.5. Reversible work ‘When an expansion proces ake place reversibly (oe Fig. 1.16) th integral 3p, is positive, ie inf nae [A process from right to left on the p-p diagram is one in which there is a work input to the fuid (i. Wis positive), Conversely, a process from left to right is fone in which there is a work output from the fluid (ie. Wis negative) ‘When a fluid undergoes a series of process and finally returns to it initial state, then itis suid to have undergone a thermodynamic cycle. A eycle which consists only of reversible processes is a reversible cycle. A cycle plotted on a diagram of propertis forms a closed figure, and reversible cycle plotted on a ‘pro diagram forms a closed figure the area of which represents the net work ofthe cycle, Forexample,a reversible eyele consisting of four reversible processes 102,210 3,3 to 4, and 4 to 1 is shown in Fig. 1.17. The net work input is qual to the shaded area, Ifthe cycle were described in the reverse direction (is. 1 t0 4,4 to 3, 3 to 2, and 2 to 1), then the shaded area would represent ‘et work output from the system. The ruleis thatthe enclosed area ofa reversible eyele represents net work input (ie. net work done on the system) when the cycle is described in an anticlockwise manner, and the enclosed area represents work output (ie. work done by the system) when the eyele is described in a clockwise manner. w 1 (shaded area on Fig. 1.16) Unit mass of a certain fluid is contained in a eylinder at an initial pressure (020 bar. The fluid is alowed to expand reversibly behind a piston according toa law pV = constant until the volume is doubled. The fluid is then cooled 13Introduction and the Fist Lew of Thermodynamics reversibly at constant pressure until the piston regains its original position; heat is then supplied reversibly withthe piston firmly locked in position until the pressure rises tothe original value of 20 bar. Calculate the net work done by the fui, for an intial volume of 0.05 m’ Solution Referring to Fig. 1.18 Fig. 118 Figure for Example 12 a ta q y : Ly 2 an : 105 a Votonea?) therefore v,)* _20 amp) =a shee noni -F ™, = J» tom equation (1.2 area 128A ie Wa J Sa¥ whee = 9:77 20 005 bar mt therefore hs = 10" «20 001 | {] = =10° 20 x ooms(t aa) = =s0000 Nn 005 a1 Way = area 32BA3 = py(Vy — Va) = 108 % 5 x (0.1 ~ 0005) 25000Nm ‘Work done from 3 to | is zero since the piston is locked in position, Therefore Wha + Was = (enclosed area 1231) $0000 + 25000 = ~25000 Nm Hence the net work done by the Buid is +25000 N m. Net work done thas beon stated above that work is given by ~{ p do fora reversible process ‘only. It can easily be shown that —f pdv is not equal to the work done if @Fg. 19 Compartments with sining partitions Fig. 120. Ireversible process on Poe diagram 16 1.6. Conservation of ergy and the firet law of tharmodinam I 2p OR 4 proces is ireversible For example, consider a cylinder, divided into a number of compartments by sliding partitions (Fig 119). Initially, compartment A filled with a mass of fui at pressure p,. When the sliding partition I is removed 4uickly, then the Nuid expands to fil compartments A and B. When the system settles down to a new equilibrium state the pressure and volume are fixed and the state can be marked on the p-V diagram (Fig. 120). Sliding partition 2is now removed and the fluid expands to occupy compartments A, B, and C. ‘Again the equilibrium state can be marked on the diagram. The same procedure can be adopled wit partitions 3 and 4 until finaly the fuid isa p» and occupies volume V when filling compartments A, B, C, D, and E. The area under the curve 1-2 on Fig. 1.20 is given by f3 pV; but no work has been done (apart from the negligible work required to move the partitions), No piston has been ‘moved, no turbine wheel has been revolved; in other words, no external force thas been moved through a distance. This isthe extreme case ofan irreversible process in which f pdV has a value and yet the work done is zero. When fluid expands without a restraining force being exerted by the surroundings, as in the example above, the process is known as fre expansion. Free expansion is highly ireversible by criterion (b), section 14, In many practical expansion ‘provesses some work is done by the fuid whichis less than § p do and in many ‘practical compression processes work is done whichis greater than jp do. Its ‘important to represent all irreversible processes by dotted lines on a pv diagram ‘asa reminder that the area under the dotted line does not represent work. Conservation of energy and the First Law of Thermodynamics ‘The concept of energy and the hypothesis that it can neither be ereated nor destroyed were developed by scientists inthe early part ofthe nineteenth century, 16Introduction and the Fist Law of Thermodynamics Example 1.3 Solution Fig. 121 Steam plant for Example 13 16 tnd became known as the Principle of the Conservation of Energy. The First Law of Thermodynamics is merely one statement of ths general principe with Particular reference to thermal energy, (ie. heat), and mechanical energy, (ie. work) ‘When a system undergoes a complete thermodynamic cycle the intrinsic ‘energy ofthe system isthe same at the beginning and end of the eyele. During the various processes that make up the eycle work is done on or by the Aid ‘and heat is supplied or rejected; the network input can be defined as FW, and the net heat supplied as .Q, where the symbol represents the sum for a complete cycle. Since the intrinsic energy of the system is unchanged the First Law of ‘Thermodynamics states that: When a system undergoes a thermodynamic eyele then the net heat supplied to the system from its surroundings plus the net work input to the system rom its surroundings must equal zero ‘That is Eo+Ew a3) {In a certain steam plant the turbine develops 1000 kW. The heat supplied to the steam in the boiler is 2800 kJ/kg, the heat rejected by the steam to the cooling water in the condenser is 2100 ki/kg and the feed-pump work required to pump the condensate back into the boiler is SW. Calculate the steam flow rate ‘The eyce is shown diagrammatically in Fig. 1.21. A boundary is shown which ‘encompasses the entire plant. Strictly, this boundary should be thought of ‘as encompassing the working fuid only. For unit mass low rate ¥ dg = 2800 — 2100 = 700 ks kg Let the stem flow be vi kg/s. Therefore 40 = 700m kW and F-dW=5~ 1000. ~995 kW Tortie Ps os. oy [endear1.7 The non-fow equation “Then in equation (1.3) Edo+ yaw =o ie 7007-995 =0 therefore 421 kg/s ie, Steam mass flow rate required = 1421 kg/s The non-flow equation In section 1.6 itis stated that when a system possessing a certain intrinsic energy is made to undergo a eycle by heat and work transfer, then the net heat supplied plus the net work input is 2er0. This is true for a complete cycle when the final intrinsic energy of the system is equal to its initial value. Consider now a process in which the intrinsic energy of the system is finaly preater than the initial intrinsic energy. The sum of the ret heat supplied and the net work input has increased the intrinsic energy of the system, ie. Gain in intrinsic energy = Net heat supplied + net work input When the net effect is to transfer energy from the system, then there will be a loss in the intrinsic energy ofthe system. ‘When a fluid is notin motion then its intrinsic energy per unit mass is known 1s the specific internal eneray ofthe fuid and is given the symbol w, The specitic internal energy ofa fluid depends on its pressure and temperature, and is tse! 1 property. The simple proof that specific internal enerey is a property is given ref, Lo, The internal energy of mass, m, of a Quid is wetten as U, ie. mu = U. ‘The units of internal energy, U, ate usually written as kl ‘Since internal energy is @ property, then gain in internal energy in changing fom state 1 fo state 2 can be written Us — Uy, Also, gain in internal energy = net heat supplied + net work input, in Uso Um hed haw ‘This equation is true for a prooess or series of processes between state 1 and state 2 provided there i no flow of Hud into or out of the system. In any one nnon-fiow process there will be either heat supplied or heat rejected, but not ‘both; similarly there wil be either work input or work output, but not both Hence, U-u, Q+W fora nonlow process or, for unit mass o+w um 4) "7Introdvetion and the First Law of Thermodynamics 18 Example 1.4 Solution Example 1.6 Solution ‘This equation is known as the non-low energy equation. Bquation (1.4) is ‘very often written in diferential form. For a small amount of heat supplied dQ, 4 small amount of work done on the uid 41, and a small gain in specific internal energy du, then dQ +4W = du as) In the compression stroke of an internal-combustion engine the heat rejected to the cooling water is 45 kJ/kg and the work input is 90 kJ/kg. Calculate the change in specific internal energy of the working fuid stating whether ities gain or a loss, 4s bake (ve sign since heat is rejected). W = 901d /kg Using equation (1.4) O+Wau—m 454 90=u; uy therefore uy =m, = 45g ile, Gain in internal energy = 45 KI/ke In the cylinder of an air motor the compressed air has a specific internal ‘energy of 420 kd kg atthe beginning ofthe expansion and a specifi internal ‘energy of 200 Kd /kg after expansion. Calculate the heat flow to or from the cylinder when the work done by the air during the expansion is 100 k5/kg, From equation (14) + Wau uy ie, Q-100= 200-420 therefore = 1201 /kg ie, Heat rete by the wir ~ +120 kg Tis important to note that equations (1.3), (14), and (15) are true whether cor nt the proces is reversible These are energy equations For reversible non-flow processes we hav, fom equation (1.2) wen nav or in diferent form dW = —mpdv18 Fig. 122. Stendy-fow open system Boundaer., Int rit] Fig. 123. Section at int to the system | 41.8. The stendy-flow equation Hence for any reversible non-flow process for unit mass, substituting in ‘equation (1.5) dQ = du + pdo (16) ‘or substituting in equation (1.4) @ wm) pd (17) Equations (1.6) and (1.7)can only be used for ideal reversiblenon-flow processes. The steady-flow equation In section 1.7 the specific internal energy of a fluid was said to be the intrinsic energy of the uid due to its thermodynamic properties, When unit mass of a ‘uid with speciic internal energy, u is moving with velocity C and is a height Z above a datum level, then it possesses a total energy of u + (C2/2) + Zg, ‘where C#/2is the kinetic energy of unit mass ofthe uid and Zg is the potential nergy of unit mass of the uid Inmost practical problems the rate at which the fluid flows through a machine Cr piece of apparatus is constant, This type of flow is called steady flow: Consider a uid flowing in steady low with a mass flow rate, i through @ piece of apparatus (Fig. 1.22), This constitutes an open system as defined in section 1.2 The boundary is shown cutting the inlet pipe at section 1 and the ‘outlet pipe at section 2. This boundary is sometimes called a control surface, and the system encompassed, @ control volume, Boundary 2 Boundary" Let it be assumed that a steady rate of low of heat Q units is supplied, and that H isthe rate of work input on the fuid as it passes through the apparatus. Now in order to introduce the fluid across the boundary an expenditure of ‘energy i required; similarly in order to push the fuid across the boundary at ‘exit, an expenditure of energy is required. The inlet section is shown enlarged in Fig. 123. Consider an element of fuid, length f, and let the cross-sectional area of the inlet pipe be A,. Then we have Energy required to push element across boundary (Pdi) x 1 = py x (volume of uid element) 9Introduction and the First Law of Thermodynamics therefore Energy required for unit mass flow rate of uid ‘where 0 isthe specific volume of the Muid at section 1 Simitarly it ean be shown that Energy required at exit to push unit mass flow rate of fluid ueross the boundary Pat ‘Consider now the energy entering and leaving the system. The energy entering the system consists of the energy of the lowing Mud at inlet C (uy +S 2, ( is ‘) the energy term rip, 2, the heat supplied Q, and the rate of work input, W. The energy leaving the system consists of the energy of the lowing uid at the outlet section ss Ez) and the energy term rip,0,. Since there is steady low of fui into and out of the system, and there ae steady flows of heat and work, then the energy entering sis cacly onl ery ng ct cg +e zarnn) rds on(ut Fs 20400) a (1s) In nearly all problems in applied thermodynamics, changes in height are negligible and the potential energy terms can be omitted from the equation. ‘The terms in w and po occur on both sides ofthe equation and always will do so in a flow process, since a uid always possesses a certain internal energy, and the term po always occurs at inlet and outlet as seen in the above proot. ‘The sum of specific internal energy and the po term is given the symbol h, and is called speefc enthalpy, +p «sy ‘The specific enthalpy of a Nuid isa property of the Bui, since it consists of the sum ofa property and the product of two properties, Since specific enthalpy is a property like specific internal energy, pressure, specific volume, and temperature, it can be introduced into any problem whether the process is a flow process or & non-flow process. The enthalpy of mass, m, ofa fluid ean be ‘written as H (ie. mi = H), The units of hare the same as those of nternal energy. ‘Substituting equation (1.9) in equation (1.8) (i+ ze)sorwen(n+Eez0) cao) ie. Specific enthalpy, h Equation (1.10) is known as the steady-flow energy equation. In steady flow the rate of mass low of uid at any section is the same as at any other section.“The steedy-flow equation Consider any section of cross-sectional area A, where the fuid velocity is C, then the rate of volume flow past the section is CA. Also, since mass flow is volume flow divided by specific volume cA Mass flow rate, i= A= pCa ay where v isthe specific volume at the section and p the density at the section ‘This equation is known as the eontinulty of mass equation ‘With reference to Fig, 1.22 Example 1.6 In the turbine of a gas turbine unit the gases flow through the turbine at 1 7kg/s and the power developed by the turbine is 14000 kW. The specitic enthalpies of the gases at inlet and outlet are 1200K)/kg and 260 kI/kg respectively, and the velocities ofthe gases at inet and outlet are 60 m/s and 150 m/s respectively. Calculate the rate at which heat is rejected from the turbine. Find also the area of the inlet pipe given that the specific volume of the gases at inlet is 0. m?/kg. Solution A diagrammatic representation of the turbine is shown in Fig, 1.24. From ‘equation (1.10), neglecting changes in height ni(in +P) 404 W oni(n+ 2) Fig. 124 Gas turbine for Example 16 For unit mass flow rate Ch _ 60 Oke Kinetic energy at inlet = Ct = 9 majg2 = 0" a 2 ag 1800 N m/kg = 18 kS/kg a 2 1.25 ky/kg (since C2 = 2.5C,) Kinetic energy at outlet =! = 2.5? x (kinetic energy at inlet) aIntroduction and the First Law of Thermodynamics Example 1.7 Solution Fig. 1.25. Air compressor for Example 17 Also W = ~14000 kW. Substituting in equation (1.10) 17(1200 + 1.8) +0 ~ £4000 = 17(360 + 11.25) therefore Q=-1193 KW ie, Heat rejected = +1193 kW ‘To find the inlet area, use equation (1.11) i. at oe ant ther Inet a, om Air flows steadily atthe rate of 04 kg/s through an air compressor, entering at 6 m/s with a pressure of 1 bar and a specific volume of O85 m/ke, and leaving at 45m/s with a pressure of 69 bar and a specific volume of (0.16 m?/ke. The specifi internal energy ofthe ar leaving is 88 ki/kg greater than that of the air entering. Cooling water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 59 KW. Caleulate the power required to drive the compressor and the inlet and outlet pipe cross-sectional In thisprolemitismre convenient to wit the Now equtionasinequation 1, citing the Z terms, ' 2 a ie a(ut Fenn) +04 Wan(n +S +n) A diagrammatic representation of the compressor is shown in Fig. 1.25. Note that the heat rejected across the boundary is equivalent to the heat removed by the cooling water from the compressor. For unit mass flow rate: Sig 8 J/kg = 0018 KI/ke Ain 2 Ai ota Problems Ch 45x45 2 J/kg = 10.1 J/kg = 0.0101 Ki /kg ivy = 1x 108 x 08S Paty = 69 x 108 % 0.16 = 110400 5/5 yy = 88K kg Also = ~S9kW 110453 /kg a_o Now 0+ ries — 1) + (rar nse +( 2) le 94 Wm o4(ss + 104 ~85 + 010 ~ a8) thereore 1044 kW (Note that the change in kinetic energy is negligibly small in comparison with the other terms.) ie, Power input required = 104.4 kW From equation (1.11) Ay 0.057 m* ie, Inlet pipe cross-sectional area = 0.057 m* Similarly 1p = AXONS _ 014m? as ie. Outlet pipe cross-sectional area = 0.014 m* Tin Example 1.7 the steady-flow energy equation has been used, des fact that the compression consists of suction of air, compression in a closed cylinder, and discharge of air. The steady-flow equation can be used because the eyele of processes takes place many times in a minute, and therefore the ‘average effet is steady flow of air through the machine. Problems ‘A osrtain uid at 10 ba i contained ine eyinder behind a piston, the initial volume being 005 m°, Caleulate the work done by the Buid when it expands reversibly: {i at constant pressure toa inl volume of 02 ms (i) cording toalinea law toa nal volume of0.2 manda final pressure of? bar; 23Introduction and the Fist Law of Thermodynamics 24 13 14 15 16 uw (ii) according toa law pY (iv) according to a law po? = constant to final volsme of 0.06 ms () according toa lam, p = (4/¥)—(B/V), toa final volume of 0.1m? and a Binal pressure of I bar, where 4 and 8 ae constants, ‘Sketch all processes on a p-v diagram. (150000 N m; 80000 N m; 34700 Nm; 7640'N my; 19200 N mm) ‘kg of a Bud is compresed reversibly according to a law pw =0.25, where p isin ‘bar and vis in m/kg. The final volume is £ of the iil volume. Calculate the work done on the Bui and sketch the process on a p-v diagram. (24660 8 m) (005 m? of a gas at 69 bar expands reversibly ina cylinder behind a piston according to the law po! = constant, until the volume is 008 m°, Caleslate the work done by the gas and sketch the process ona p-V diagram, (15480. mp 1 ig of a uid expands reversibly according to 4 tineat lw from 42 bar to 1.4 bar; ‘he intial and final volumes are 0008 m? and 02m The Bud is then cooled reversibly at constant pressure, and finaly compressed reversibly according to a law ‘pe constant back to the inital conditions of 42 bar and OOM”. Calculate the ‘work done in each proces and the net work of th eyle Sketch the eye on a pv siagram, (-480 Nm; 41120; +1885 Nm; —1515 Nm) ‘A Dud at 0.7 bar occupying 009 m? is compressed reversibly to a presure of 3 bar According toa law pu" = constant. The Mud it then heated reversibly at constant volume until the pressures 4 ar; the specie volume i then 0. m3 /k.A yeversibe expansion according to a law po? = constant restores the Auid to is nial state, Sketeh the ‘yee on a po diagram and calculate (Gj) the mass of fud present, (i) the value of in the Dist proces; (i) the net work ofthe eye. (00753 kg; 1887; ~640 Nm) [A Suid is heated reversibly at a constant pressure of 10S bar wnt it has a specific ‘volume of. m/kg- tis then compressed reversibly according 1 law po = constant to a pressure of 42bur, then allowed to expand reversibly according to a law "7 = consant, and is finally heated at coastant volume back to the intl conditions, ‘The work done in the constant pressure proses is —S1SINm, and the mass of fuid preset is O2kg. Calculate the net workf the eyele and sketch the eyele on & po siagram. (2781Nm) In an ir compressor the compression takes place at @ constant internal energy and SD1d of heat are rejected to the cooling water for every kilogram of air. Caleuate ‘the work inpt forthe compression stroke per kilogram of a, (soxs/ig) Inthe compression stroke of a gas engine the work done on the gas by the piston is T01U/kg and the heat rejected to the cooling water is 42k)/kg. Calculate the ‘change of specific internal energy stating whether it again or alo. (28 kag gain) ‘A mass of gas at an initial pressure of 28 bar, and with an internal energy of 1500, is contained in @ welkinsulated cylinder of volume 006m The gus i allowed 10133 ‘expand behind a piston unt its internal energy i 1400; the law of expansion is po! = constant. Calulate: ) the work done (i) the inal volume; (il) te ial pressure (1001; 0148 m?; 459 bar) “The anses in the cylinder of an jnternal combustion engine have a speci internal energy of 800 kJ/kg and specie volume of 006 mk at the beginning of expansion, ‘The expansion ofthe gas may be assumed to take place according to a reversible law, po"* = constant, from $3 bar to 14 bar. The speciic internal nergy after expansion ir 250 K/kg. Calulate the heat rejected to the lind cooling water pet Selogram of gases during the expansion stroke (10443) ‘A scam turbine receives a stam flow of 1.35 kg/s and the power output is $00 kW. ‘The het los from the casing is negligible. Caleulate (3) the chango of specie enthalpy across the turbine when the veloc ‘and exit and the difrence in clevation are negligible; (i) the change of specific enthalpy across the turbine when the velocity at entrance is (ens, the velosity at exit 860 m/5, and the inlet pipe is 3m above the exhaust Pipe (3101 /kg; 433 3 /ke) [A stady flow of steam enters & condenser witha specie enthalpy of 2300 kI/kg anda velocity of 3501m/s The condensate leaves the condenser with aspecitic enthalpy ‘of 160 kd/kg and a velocity of 101m/s. Calculate the heat tanser to the cooing uid po klogram of steam condense. (2199 13/88) |A turbine operating under steady-flow conditions recives steam at the following State: pressure, 138 bar; specie volume 143m/kg, specie internal energy 2590 KJ/ks, velocity 30m/s. The state of the steam leaving the turbine is as follows pressure 0.35 bat, specie volume 437!m*/kg, speci intemal energy. 23601 /kg. ‘oloeity 90 m/s, Hest is rejected to the surroundings atthe rate of 025 kW and the rate of steam flow through the turbine is 038 kg/s. Calculate the power developed by the erbine (102.72W) [A nozdeis a device fr increasing the vlocty ofa steadily owing uid. At the inet toa certain nozzle the specie enthalpy of the Mud is 302530/kg and the velocity {= Gos. At the exit fom the novle the specific enthalpy is 790 ki /kg. The nozle 's horizontal and there sa negligible eat los from it, Cale (@) the velocity ofthe uid at ext; (ii) the rate of flow of fuid when the inlet are is 01m? and the specie volume at ink is 0.19 >; (i) tho exit atea ofthe nozzle when the specie volume atthe nozzle exit is 0.5 m3/kg {688 mjs; 31.6 kg/s; 00229 m?) References LL easror rp and cxorr 0x 190 Energy Eieieney Longman, 22 DOUGLAS Jf, GASIOREK J Mand SWAFFIELD 3A 1986 Fladd Mechanies 2nd edn ‘Longman 25Introduction and the First Law of Thermodynamics 1a 14 BS 1041 Temperarure Measurement HMSO; Section 21 1985 Guide to Seletion and Use of Liguid:i-gass Thermometers; Pat 31989 Guide to Selection and Use Of Indusrial Resistance Thermometers; Part 4 1966 Thermocouples; Part $1972 Radiation Pyrometers; Part 7 1988 Gulde to Selection and Use of Temperature-tne Records oGERs © ¢ and uaviteW vx 192 Englnering Thermodynamics, Work and Heat Transfer 4th edn Longman24 2 The Working Fluid In section 1.3 the matter contained within the boundaries ofa system is defined as the working fluid, and it is stated that when two independent properties of the fluid are Known then the thermodynamic state of the fid is defined. In thermodynamic systems the working fluid can be in the liquid, vapour, or ‘gaseous phase, All substances can exist in any one ofthese phases, but we tend to identify all substances with the phase in which they are in equilibrium at atmosphere pressure and temperature. For instance, substances such as oxygen ‘and nitrogen are thought of as gases; H,O is thought of as liquid or vapour (je. water or steam); mercury is thought of as a liquid. AI these substances can exist in diferent phases: oxygen and nitrogen can be liquefied; H, can become a gas at very high temperatures; mercury can be vaporized and will fact as a gas if the temperature i raised high enough. Liquid, vapour, and gas Consider a p-v diagram for any substance. The solid phase is not important in engineering thermodynamics, bing more the province of the metallurgst ot physicist. When a liquid is heated at any one constant pressure there is one fixed temperature at which bubbles of vapour form in the liguid; this phenomenon is knowin as boiling. The higher the pressure of the liguid then the higher the temperature at which boiling occurs. It is also found that the volume occupied by 1kg of a boiling liquid at a higher pressure is slightly larger than the volume occupied by 1 kg of the same quid when itis boiling at a low pressure. A series of boiling-points plotted on a p-v diagram will ‘appear as a sloping line as shown in Fig 2.1. The points P, Q, and R represent the boiling-points of aliquid at pressure pp, po, and py respectively ‘When & liquid at boiling-point is heated further at constant pressure the ‘additional heat supplied changes the phase of the substance from liquid to ‘vapour; during this change of phase the pressure and temperature remain ‘constant. The heat supplied is called the specific enthalpy of eaportzation, It is found that the higher the pressure then the smaller is the amount of heat required, There isa definite value of specific volume of the vapour at any oneThe Working Fluid Fig. 21 Boling points plotted on a poo diagram Fig. 22. Points of complete vaporization plotted on a ov diagram 28 a) pressure, a the point at which vaporization is complete; hence a series of points such as P,Q, RY can be plotted and joined to form a line as shown in Fig. 22. ‘When the two curves already drawn are extended to higher pressures they form a continuous curve, thus forming a loop (see Fig. 23). The pressure at which the turning point occues is called the critical pressure and the turning point itself called the critical point (point C on Fig. 2.3). It can be seen that atthe critical point the specific enthalpy of vaporization is zro, The substance existing at a state point insde the loop consists of a mixture of liquid and dry ‘vapour and is known as a wet tapowr. A saturation state is defined as a state atwhich a change of phase may oeur without change of pressure or temperature. Hence the boiling-points P,Q, and R are saturation states, and a series of such boiling-points joined up is called the saturated liquid line, Similarly the points P,Q, and R’, at which the liquid is completely changed into vapour, are saturation states, and a series of such points joined up is called the saturated ‘pour line. The word ‘saturation’ as used here refers to energy saturation. For ‘example, a slight addition of heat to a boiling liquid changes some of it into a vapour, and it is no longer a liquid but is now a wet vapour. Similarly when 2 substance just on the saturated vapour line is cooled slightly, droplets of liquid will begin to form, and the saturated vapour becomes a wet vapour. A saturated vapour is usualy called dry saturated to emphasize the fact that no liquid is present in the vapour in tis state. Lines of constant temperature, called isothermals, can be plotted on a pv diagram as shown in Fig. 24, The temperature ines become horizontal betweenFg.23. Wet loop plotted on @ p-edagram Fig. 24 Tsothermals fora vapour plated on apes diagram 2.1 Liquid, vapour, and ‘the saturated liquid line and the saturated vapour lin (eg. between P and P’, ‘Qand QR and R’), Thus there isa corresponding saturation temperature for each saturation pressure. At pressure pp the saturation temperature is T;, at presure po the saturation temperature ie T,, and at pressure py the saturation temperature is T,. The critical temperature ine Tp ust touches the top of the loop at the critical point C. ‘When a dry saturated vapour is heated at constant pressure its temperature rises and it becomes superheated The diference between the actual temperature ‘of the superheated vapour and the saturation temperature at the pressure of the vapour is called the degree of syperheat, For example the vapour at point (Fig. 24) is superheated at po and T,, and the degree of superheat is Ty — Ty In section 1.5 it is stated that two independent properties are sufcient to define the state of a substance. Now between P and P’, Q and Q', R and R’ ‘the temperature and pressure are not independent since they remain constant for a range of values of r. For example, a substance at po and T; (Fig. 24) could be a saturated liquid, a wet vapour, ora dry saturated vapour. The state cannot be defined until one other property (eg. specific volume) is given. The condition or quality of a wet vapour is most frequently defined by its dryness 29‘The Working Flu 22 Jraction, and when this is known as well as the pressure or temperature then the state of the wet vapour is flly defined Dryness fraction, x the mass of dry vapour in 1k of the mixture (Sometimes a wetness fraction is defined as the mass of liguid in 1 kg of the ‘mixture, ie. wetness fraction = 1 — x.) [Note that for a dry saturated vapour x = 1, and that fora saturated liquid ‘The distinction between a gas and a superheated vapour isnot rigid, However, at very high degres of superheat an isothermal line on the p-» diagram tends to become a hyperbola (ie. po = constant). For example the isothermal T, on Fig 24 is almost a hyperbola, An idealized substance called a perfect gas is assumed to have an equation of state po/T = constant. It ean be seen that when a line of constant temperature follows a hyperbolic law then the equation po/T = constant is satisfied. All substances tend to obey the equation o/T = constant at very high degrees of superheat. Substances which are thought ofus gases (eg. onygen, nitrogen, hydrogen etc, are highly superheated at normal atmospheric conditions. For example, the critical temperatures of oxygen, nitrogen, and hydrogen are approximately ~119, ~147, and ~240°C respectively. Substances normally existing as vapours must be raised to high temperatures before they begin to act as a perfect gas. For example, the critical temperatures of ammonia, sulphur dioxide, and water vapour are 130, 157, and 374.15°C respectively. ‘The working fvid in practical engineering problems is either 2 substance which is approximately a perfect gas, or a substance which exists mainly as liquid and vapour, suchas seam and the refrigerant vapours, For the substances which approximate to perfect gases certain laws relating the properties can be assumed. For the substances in the liquid and vapour phases the properties are not related by definite laws, and values of the properties are determined empirically and tabulated in a convenient form, The use of vapour tables ‘Tables are available for a wide variety of substances which normally exist in the vapour phase. The tables which will be used in this book are those arranged by Rogers and Mayhew (ref. 21), which are suitable for student use. For more comprehensive tables for steam, ref, 2.2 should be consulted. The tables of Rogers and Mayhew are mainly concerned with steam, but some properties of refrigeranis are also given. Saturation state properties ‘The saturation pressures and corresponding saturation temperatures of steam are tabulated in parallel columns in the rst table, for pressures ranging from (0.006 112 bar tothe critical pressure of 221.2 bar. The specific volume, internal energy, enthalpy, and entropy are also tabulated for the dry saturated vapour‘Table 21 Extract from tables af properties of Fg. 28 Points ‘dented on a 2.2 Tho use of vapour tables ? 4“ & WM (bar) CC) _(mfAw) Chafee) (ike K) 034 72-4647 302-2472 3022828 2600 9H 67457725, % at each pressure and corresponding saturation temperature. The sux gis used to denote the dry saturated stage. A specimen row from the tables is shown in Table 2.1. For example at 034 bar the saturation temperature is 72°C, the specific volume of dry saturated vapour, vat this pressure is 4649 m? kg, the internal energy of dry saturated vapour, tis 2472 kJ/kg, andthe enthalpy of ny saturated vapout hy is 2630 kJ/kg. The steam isin the state represented by point A on Fig. 2, At point B dry saturated steam at a pressure of 100 bar and saturation temperature 311°C has a specific volume, yf 001802 m/s, internal energy, uy of 2545 kl/kg and enthalpy, hy of 2725 KI/kg. aos 7 ‘Specie yolume/(m? ha) ‘The specie intemal energy, specific enthalpy, and specie entropy of saturated liquid are also tabulated, the sui {being used for this state. For fxample at bar and the corresponding saturation temperature. 143.6°C, Saturated water has a specie internal eneray, 1, of 605 KJ/ke, and a speci enthalpy hy of 605 KI/ke. This state corresponds to point C on Fig. 25. The Specie Volume of saturated water, is tabulated in a separate abl, but itis usually negligibly small in comparion withthe speci volume of the dry saturated vapour, and is variation with temperatures vey small the saturated liquid line on a p-e diagram is very nearly coincident with the pressure axis Jn comparison with the width ofthe wet loop (s® Fig. 25). AS seen from the table, values of, vary from about 001 m3/kgat 001 °C to about O00 m? kg at 160°C; as the pressure approaches the crtial value the increase of i tore marked, and at the eritcal temperature of 374.15°C the value of o i (0.00317 mk ‘The change in specific enthalpy from hy toh, is given the symbol hy. When saturated water is changed to dry saturated vapour, from equation (1), Q+ Wau mau‘The Working Fuld Also —W is represented by the area under the horizontal line on the p- diagram, ie W=—(0,-09p therefore = (ym) + PY — 0) (4+ Pm) ~ (ue + Bed) From equation (19) w+ po “The heat required to change a saturated liquid to a dry saturated vapour is called the specific enthalpy of vaporization, fg In the case of steam tables, the specific intemal energy of saturated liquid is taken to be zero at the triple point (i. at QOL°C and 0.006 112 bar). Then since, from equation (19), = w+ po, we have (0.006.112 x 10* x 0,001 0002 10° ‘hat 001°C and 0,006 112 bar = 0+ ‘where vat 001°C is 0001 0002.m*/kg, 112 x 10-* kg ‘Thisis negligibly small and hence the zero for enthalpy may be taken at 01°C. [Note that at the other end of the pressure range tabulated in the frst table the pressure of 221.2 bar is the etitical pressure, 374.15°C is the critical temperature, and the specific enthalpy of vaporization, hy, is zero. ie A Properties of wet vapour For a wet vapour the total volume of the mixture is given by the volume of liquid present plus the volume of dry vapour present. Therefore the specitic volume is given by volume ofliquid + volume ofdry vapour ‘otal mass of wet vapour Now for I kg of wet vapour there are «kg of dry vapour and (1 — x) kg of liquid, where x is the dryness fraction as defined earlier. Hence, peal x) box ‘The volume of the liquid is usually negligibly small compared to the volume of dry saturated vapour, hence for most practical problems aExample 2.1 Solution Example 2.2 Solution 22. Tho use of vapour tables ‘The enthalpy of a wet vapour is given by the sum of the enthalpy of the liquid plus the enthalpy of the dry vapour, ie haa) bay therefore baht xh) ie hehe thy 22) Similarly, the internal energy of a wet vapour is given by the internal energy ‘ofthe liquid plus the internal energy of the dry vapour, 1 xe + xy (3) A+ xu, — 4) es) Equation (2-4) can be expressed in a form similar to equation (22), but equations (2.3) and (24) are more convenient since 1, and u are tabulated and the difference, u, ~ uy, is not tabulated in ref. 2.1, Calculate the specific volume, specific enthalpy, and specific internal energy of wet steam at 18 bar, dryness fraction 09. From equation (21) therefore 09 0.1104 = 0.0994 m?/kg From equation (22) h therefore him 8854 (09 x 1912} From equation (2.3) 1 — x) + x, itty 2605.8 kl kg (1 = 09)883 + (09 x 2598) = 24265 kJ/kg Calculate the dryness fraction, specifi volume and specific internal energy of steam at 7 bar and specific enthalpy 2600 iI /ke. AUT bar, hy = 2764 kI kg, hence since the actual enthalpy is given as 2600 KI/kg, the steatn must be in the wet vapour state. Fromequation (2.2) h = y+ xg, ie, 2600 = 697 + x2067‘The Working Fluid Table 22. Extract fom {ables of properties of fuperbeated steam at oar (saturation ‘emporature 2124°C) therefore ‘Then from equation (21) p= xn = 0921 x 02728, 2515 m?/kg From equation (23) (1m +a therefore = (1 = 0:921)696 + (0921 « 2573) ie w= 2020KI/kg Properties of superheated vapour For steam in the superheat region, temperature and pressure are independent ‘properties. When the temperature and pressure are given for superheated steam then the state is defined and all the other properties ean be found, For example, steam at 2bar and 200°C is superheated since the saturation temperature at 2ar is 1202°C, which is less than the actual temperature. The steam in this ‘state has.a degre of superheat of 200 — 120.2 = 798 K. The tables of properties ‘of superheated steam (ref. 21) range in pressure from 0.006 112 bar tothe ertial pressure of 221. bar, and there is an additional tabie of supercritical pressures ‘up to 1000 bar. At each pressure there is a range of temperatures up to high 109 ie 3051 — 98 33251 — 33178) (Note the negative sign inthis ase sine at 1S bar, 422°C is lange than hat 20 bar, 452°C.) Then frat 185 bar, 432°C 3525.1 ~ (07 x 73) = SORT ke Sketch a pressure-volume diagram for steam and mark on it the following points, labelling clearly the pressure, specific volume and temperature ofeach point. (a 20ber, 37‘The Working Fuld (b) ¢=2124°C, » = 0.09957 m°/ke (c) p= 10 bar, hm 2650 KI kg (A) p= 6 bar, b= 316641 /ke Solution Point (a): At 20 bar the saturation temperature is 212.4°C, hence the steam is superheated at 250°C, Then from tables, v= 0.1115 m*/kg. Poin (b): At 2124°C the saturation pressure i 20 bar and p, is 09957 m/e. ‘Therefore the steam is just dry saturated since » = Point (c): At 10-bat, hy is 2778kI/kg, therefore’ the steam is wet since 2650 kJ/kg. Since the steam is wet, the temperature is the saturation temperature at 10 bar, ie. = 179.9°C. The dryness fraction can be found from equation (2.2), heahyt thy therefore 2650 ~ 763 _1887 _ 995, 2018 ~ 201 ‘Then from equation (2.1) 19 = 0937 x 0.1944 = 0.182 m?/kg Point (d): At 6 bar, fi, is 2757 kJ/kg, therefore the steam is superheated, since it is given that h = 3166 kJ/kg. Hence from tables at 6 bar and h = 3166 kI/ke, ‘the temperature is 350°C, and the specific volume is 0.4743 m*/kg. ‘The points (a), (b), (c), and (@) can now be marked on a pv diagram as shown in Fig, 28 Fig 28. Solution for Example 25, a i ho 2 [mre ree ay < ‘ ‘Spee volume?) 38Example 26 Solution Example 2.7 Solution 23 2.3. The porfoct gas Calculate the internal energy for each ofthe four states given in Example? (a) The steam is superheated at 20 bar, 250°C, ie, w= 2681 KI /kg (b) The steam is dry saturated at 20 bar, ie ou 600 KI kg (c) The steam is wet at 10 bar with x= 0937, Therefore = x)u+-xu,_ from equation (2.3) (1. 0937)762 + (0.937 x 2584) = 2470 15/kg (4) The steam is superheated at 6 bar, 350° ie w= 2881 KI /kg Using the properties of ammonia given in ref. 2.1, caleulate! (i) the enthalpy at 1.902 bar, dryn (ii) the enthalpy at 857 bar, 60°C. (@) From equation (22) 3s fraction 0.95; ha het he, ‘Therefore, at 1.902 bar, A i) AC 8.570 bar the saturation temperature is 20°C so the ammonia at 60°C is superheated by (60 — 20) = 40 K. It is therefore necessary to interpolate to find the enthalpy, 40 © (iso72 — 14626) 462.6 + 5 x ( ) = 15703 kak, ‘The perfect gas The characteristic equation of state [At temperatures that are considerably in exoess ofthe evtical temperature of 8 fuid, and also at very low pressures, the vapour of the fuid tends to obey the equation Po constant = R T‘The Working Fluid No gases in practice obey this law rigidly, but many gases tend towards it An imaginary ideal gas which obeys the law is ealled a perfect gas, and the equation, po/T = R, is called the characteristic equation of state of a perfect gas. The constant, R, is called the specific gas constant, The units of R are N m/kg K. oF Kd/kg K. Bach perfect gas has a different specific gas constant ‘The characteristic equation is usually written po=RT es) or for 4 mass, m, occupying a volume, ¥ pV =mRT 6) ‘Another form of the characteristic equation can be derived using the amount of substance (sometimes called the mole). The amount of substance is defined by the 1971 General Conference of Weights and Measures (CGPM) as follows: ‘The amount of substance of a system is that quantity which contains as many elementary entities as there are atoms in 0012 kg of eatbon-I2; the clementary entities must be specified and may be atoms, molecules, ions, cleetrons, or other particles, or specific groups of such particles. ‘The normal unit symbol used for the amount of substance is ‘mol’. In ST it is convenient to use "kmol’ ‘The mass of any substance per amount of substance is known as the molar en where m isthe mass ands the amount of substance. The normal units used Tor m and n ate kg and kmol, therefore the normal unit for fis kg/km. Relative masses of the various elements are commonly used, and physicists and chemists agreed in 1960 to give the value of 12 to the isotope 12 of carbon (this led to the definition of the amount of substance as above), A scale is thus ‘obtained of relative atomic mass oF relatioe molecular mass (e.g. the relative ‘atomic mass ofthe element oxygen is approximately 16; the relative molecular ‘mass of oxygen gus, O,, is approximately 32). ‘The relative molecular mass is numerically equal to the molar mass, i, but is dimensionless. ‘Substituting for m from equation (2.7) in equation (2.6) gives Z an! PY = niRT or ik = Pe [Now Avogadro's hypothesis stats thatthe volume of I mol of any gas is the same as the Volume of | mol of any other gas, when the gases are at the same temperature and pressure. Therefore V/n i the same for all gases at the same value of p and 7: That is, the quantity pV'/n7 is a constant forall gases. This constant is called the molar gas constant and is given the symbol, R, ~ v je R= R= or pv = nk 2.8) Be or (28)Example 2.8 Solution Example 2.9 7 es) “The value of R has been shown to be 8314.5 N m/kmol K, ‘From equation (29) the specific gas constant for any gas ean be found when the molar mass is known, eg, for oxygen of molar mass 32 kg/kmol, the specific 488 constant s BOS as eNmitek [A vessel of volume 0.2.m? contains nitrogen at ,013 bar and 15°C. 10.2 kg of nitrogen is now pumped into the vessel, calculate the new pressure when the vessel as returned to is initial temperature. The molar mass of nitrogen is 28 kg/kmol, and it may be assumed to be a perfect gas. From equation (29) Specific gas constant, R= From equation (26), for the initial conditions Kam RT therefore _ pi, _ 1.013 x 108 x 02 RT, 29695 x 288 5 + 273 = 288K "The mass of nitrogen added is 0.2 kg, hence m, = 0.2 +0237 = 0437 kg. ‘Then from equation (2.6), for the final conditions pale but ¥y = ¥; and T; = 7, therefore taRT, _ 0437 x 20695 x 288 10 x02, 237 kg kT ‘A certain perfect gas of mass 0.01 kg occupies « volume of 0.003 m? at a pressure of 7 bar and a temperature of 131°C. The ga i allowed to expand ‘anil the pressure is I bar and the final volume is 0.02 m?, Calculate: (a) the molar mass of the gas; (i) the final temperature a‘The Working Fluid 42. Solution (i) From equation (26) PAY, = mRT, therefore ih _ 7x 108 x 0.003, mT, 001 x 404 where T, = 131 +273 = 404 K. ‘Then from equation (29) R shrdoe fous SHS sexgynmal 5297 16ke/kmol __Nolarmass = 16k From equation (28) P2¥a=mRT, therfore rafal 10 O02 ay mR O01 x 520 ie, Final temperature = 384.5 ~ 273 = 115° Specific heat capacity ‘The specific heat capacity of a solid or liquid is usually defined as the heat required to raise unit mass through one degree temperature rise, We have 440 = med, where m is the mass, dT isthe increase in temperature, and ¢ is the specific heat capacity. For a gas there are an infinite number of ways in Which heat may be added between two temperatures, and hence a gas could have an infinite number of specific heat capacities. However, only two specific heat capacities for gases are defined; the specific heat capacity at constant volume, ¢,, and the specific heat capacity at constant pressure, cy ‘The definition must be restricted to reversible non-low processes, since irreversbilities can cause temperature changes which are indistinguishable from ‘those due to reversible heat and work quantities. Specific heat eapacitis can ‘be introduced more rigorously as properties of «fluid, We have in the limit o-), st o-G), ‘A more rigorous treatment is given in ref. 23,23 The perfect gas We can write c,4T_ for a reversible non-flow process at constant pressure (2:10) and 4Q.=e, dT for a reversible non-flow process at constant volume Qu) For a perfect gas the values of ¢, and e, are constant for any one gas at all [pressures and temperatures. Hence integrating equations (210) and (2.11) we Ihave for a reversible constant pressure process Q= melt —T) (212) {or a reversible constant volume prooess mme,(Ts~ 3) 213) For real gases, ¢, and, vary with temperature, but for most practical purposes ‘a suitable average value may be used, Joule’s law oule’s law states that the internal energy ofa perfect gus i a function ofthe absolute temperature only, jeu = ((T)-To evaluate this function let unit mass of perlet gus be heated at constant volume. From the non-ow energy ‘uation, (13) dQ +dW = du Since the volume remains constant then no work is done, ie. rr {At constant volume for a perfect gas from equation (2.11), for unit mass, dQ =c,a7 ‘Therefore, dQ nT+K where K is a constant oule’s law states that w= (7), hence it follows that the internal energy ‘varies linearly with absolute temperature. Internal energy ean be made zero at any arbitrary reference temperature, For a perfect gas it can be assumed that =O when T = 0, hence the constant Kis zer0, 0, therefore 6. 4T, and integrating ie. Specific internal energy, u = ¢,7" for a perfect gas (214) ‘or for mass, m, of a perfect ga, Internal energy, U = me, (215) 43,‘The Working Fluid In any provess fora perfect gas, between states I and 2, we have from equation 213), Gain in internal energy, Us ~ Uy = me\(Ts—T) (216) The gain of internal energy for a perfect gas between two states is always given by equation (2.16), for any process, reversible or irreversible. Relationship between the specific heat capacities Let a perfect gas be heated at constant pressure from T, to Ts. From the non-flow equation (1.4), Q-+ W= (U3 ~ Uy). Also, for a perfect gas, from ‘equation (2.16), Uz ~ Uy = me,(Ts ~ 7). Hence, Q+ W=me(T,—T) In a constant pressure process the work done is given by the pressure times the change in volume, i. W= —p(V,—V,) Then using equation (2.6), pV, =mRT; and p¥, = mRT;, we have We -mR(7, ~ 7) ‘Therefore substituting Q~mR(T, ~ T,) = me(Ts ~T) therefore Q=mic, + RET) [But for @ constant pressure process from equation (2.12) Q=me,(T—T) Hence by equating the two expressions for the heat flow, Q, we have ames + RYT = T) = me,(T ~ Ti) therefore gtRae, or (217) Specific enthalpy of a perfect gas From equation (19), specific enthalpy, h = u + po For a perfect gas, from equation (2.5), po = RT. Also fora perfect gas, from Joules law, equation (214), u = ¢,T Hence, substituting ee THRT=(G4 RIT23 Tho porfect g But from equation (2.17) Gp G=R oF G+ R= ey ‘Therefore, specific enthalpy, h, for a perfect gas is given by hee? (218) For mass, m, ofa perfect gas H=me,T 219) (Note that, sinoeit has been assumed that u = O at T'= 0, then k= Oat T= 0.) Ratio of specific heat capacities ‘The ratio ofthe specific heat capacity at constant pressure to the specific heat ‘capacity at constant volume is given the symbol 7 (gamma), (229) Note that since ¢, — ¢= R, from equation (2.17), it is clear that ¢, must be ‘greater than c, for any perfect gas. 1¢ follows therefore thatthe ratio, ¢,/¢, = 7, is always greater than unity. In general, 7 is about 14 for diatomic gases such as carbon monoxide (CO), hydrogen (H), nitrogen (N,), and oxygen (O.) For monoatomic gases such as argon (A), and helium (He), is bout 1.6, and {or triatomic gases such as carbon dioxide (CO,), and sulphur dioxide (SO,), ‘ris about 13, For some hydrocarbons the value ois quite low (eg. for ethane (CoH), 7 = 1.22, and for isobutane (CyHyq) y = 1.1} Some useful relationships between ¢,, ¢,, R, and can be derived. From equation (2.17) R Dividing through by ¢, ae 221 5D 21) Also from equation (2.20), c, = 7¢y, hence, substituting in equation (2.21) oR e 222) 15D (22) 45‘The Wor ing Fluid Example 2.10 Solution Example 2.11 Solution ‘A certain perfect gas has specific heat eapacities as follows: p= O86 KI/kgK and c= 0657 KI/kg K CCalevlate the gas constant and the molar mass ofthe gas. From equation (2.17) ie, R= 0846-0657 From equation (29) 189 ky kg K = 189 N m/kg K sa14s in om 189) = 44g /kmol ‘A perfect gas has a molar mass of 26kg/kmol and a value of y= 1.26. CCaleulate the heat rejected: () when unit mass of the gas is contained in a rigid vessel at 3 bar and 315°C, and is then cooled until the pressue falls to 1.5 bar; (ii) when nit mass flow rate of the gas enters a pipeline at 280°C, and flows steadily to the end of the pipe where the temperature is 20°C. Neglect changes in velocity of the gas in the pipeline. From equation (29) R_ sis ; BSS _ sos wm/kgk From equation (221) R 3198 G1) 1051261) Jao ike K ‘Also from equation (2.20) therefore ey = ep = 1.26 % 1.229 = 1.548 KIRK (i) The volume remains constant for the mass of gas present, and hence the specific volume remains constant, From equation (25), T, and p20; = RTsaa 22 Problems Therefore since v, = v; we have 15 Pa 588 3 > 3 where T; = 315 + 273 = $88 K ‘Then from equation (2.13) Heat supplied perkgofgas = c,(T, — T,) = 1.229(294 — 588) = 1.229 x 294 = —361 KI/ke +361 kg (ii) From the steady-flow energy equation, (1.10), alee ct cit eer a(n D)rorw (hse) In this case we are told that changes in velocity are negligible; also there is no ‘work done. "Therefore we have 94K ie. Heat rejected per kilogram of ga vhs +O =ihy For a perfect gas, from equation (218) hao? therefore te, T, ~ T}) = 1 x 1.548(20~ 280) = —403 kW ic. Heat rejected per kilogram per second = +403 kW ‘Note that itis not necessary to convert ¢, = 280°C and ¢, = 20°C into degrees Kelvin, since the temperature difference (t, ~ f,)is numerically the same as the temperature diflerence (7, ~ 73). Problems Note: the answers to these problems have bosn evaluated using the tables of Rogers find Mayhew (re 2.1). The values of Ry cy, cy And for aie may be assumed (0 beasgiveminthe tables (ie = 0.287 kI/kg Kc, = 1005 Kk Kc, = 07181 keg K; and y= 14). For any other perfect gas the values of R ey, cy, and i required, must be calelated from the information given inthe problems te valve of Rie given in the tables (ref 21] ‘Complete Table 24 (p48) wsing steam tables. Insert a dash for ielevant tems, and interpolate where necessary. (see Table 26 p. $0) [A vessel of volume 003 m* contains dry saturated stcam at 17 ba. Calculate the ‘mass of steam i the vessel and the enthalpy of this mass (0287 kg; 718) “7‘Tho Working Fluid ‘Table 24 Data for Problem 2.1 2a 24 26 te se : Degree of (a CO) (mike) x _superbeat(hiyka)_tka7hw) oo 361 20 299 5 ses 188 2400 au 09 a3 as 3 aw 5 ous 10 sais 250 L601 saa os. 2 095 23 300 abo ‘The completed table is given on p. 50 as Table 26. ‘Steam at Thar and 250°C enters a pipeline and flows along it at constant pressure Tr the steam rejects heat study to the surroundings, at what temperature will droplets ‘of water begin to form in the vapour? Using the sletdy-flow energy equation, and neglecting changes in velocity ofthe steam, calculate the heat rejected per Kilogram of seam flowing. (195°C: 191 3/ke) (005g of stem at 15 bar is contained ina rigid vessel of volume 00076 n°. What isthe temperature ofthe steam? Ifthe vessel is cooled, at what temperature wil the seam be jost dry saturated? Cooling is continued until the pressure inthe Yess ie 11 bar; eaeulate the fina drynese fraction ofthe steam, and the heat ejected between ‘the initial and the final state, (230°C; 1914°C; 0857; 18.545) Using the tables for ammonia given in ef, 21, calculate: (Gj) the specific enthalpy and specie volume of ammonia. 09; (i) the spciic enthalpy and specie volume of ammonia at 13°C saturated; (Gi) the spciic enthalpy of ammonia at 7529 bar, 30°C. (1251 Kk 1.397 m? kg; 1457 Kl/kg, 0.1866 m/e: 14965 keg) (0717 bar, dryness faction Using the property value for refrigerant HFA. [3a given in Table 25, calculate (i) thespeciic enthalpy and specific volume of HFA 13a at ~8°C, dryness fraction O85; (i) the spcitic enthalpy of HPA 134a at $7024 ba, 35°C. (259.6 Lk, 00775 m2; 32825 kd ke)Problems ‘Table 25 Data for Problem 26 Superteat Vales deree sapere saturation ¥ sox p ay aauneC Un Eee OCCEEE CO) toa) wee) OB) tare) io 200si ome S698 HABE 50868 “3 2usnt Oue —9346 29177 31205 20 S704 0036 12692 306222895 27 Tho relative molecular mass of carbon dioxide, COs, is 44. In an experiment the value of» for CO; was found to bo 1.3. Assuming thal CO, i a perfect fas, calculate the spesific gus constant, R, and the speci: heat eapacis at eamstant presse and constant volume, ¢ and (0.189 kJ/kg K; 0819 KJ/kg K; 063 k3/ke K) 28 Calelate the internal energy and enthalpy of 1 kg of ai occupying 00S m* at 20 bat. he internal energy is increased by 120) as the ai is compressed to 0 bar, calculate the new volume occupied by 1 kg of thea. (2501 kk 350. ke 0.0296 m") 29 Oxygen, Os, at 200 bar isto be stored in ste vesel at 20°C. The capacity of the vessel i O04 m?, Assuming that O, is a perfect as, calculate the mass of oxygen that can be stored in the vessel. The vessel i protected against exesive pressure by «fusible plug which will melt ifthe temperature rises too high. At what temperature ‘tthe pg melt to limit the presse inthe vesel to 240 bas? The molar mass of ‘oxygen is 32 kg/kmol, (105g; 786°C) 240 When a certain perfect gas is heated at constant presture from 15°C to 95°C, the heat required is 11364i/kg, When the same gas is heated at constant volume between ‘the same temperatures the heat requied is 808 kJ/kg Calculate cy, e, 9, R and the ‘molar mass ofthe ea (142 Ky kg K; 101 Kd /keg K; 1405; 41 Kika K; 2028 kg/kmel) 2.11 In an air compressor the pressures a let and outlet are I bar and 5 bar respectively. ‘Tho temperature of the air at inlet is 15°C and the volume at the beginning of ‘ompresion is three times that at the end of compression. Calculate the temperature fof the si at outlet and the increase of internal energy pe kg ofa. (207°C; 138 ka/kg) 242A quantity ofa cata perfect as is compressed from an inital state of 68S m?, ‘hae to a final state of 004m’, 39 bar. The specific heat at constant volume is (0724 kJ/kg K, and the specie het a constant pressures 1.020 kl eg K. The oberved temperature rise is 146K. Calculate the specific gas constant, R, the mass of gas present, andthe increase of intl energy ofthe pi. (0.256 kd kg K; 0.11 kg 11.6343)‘The Working File Table 26. Solution to oblem 2. on ? £ * : « Problem 2.1 on p.47 pour (a) mg) x superheat (kik) Ga/k) om «9% 261, 1 ° 260 aoe » ang oms9s7 ° 21992600 5 11803865, gst dis 2471 2 18 Otel ays 2162400 4 gogo 2a aa os a3 278 oss 2900 216s 3 mm O76 6s 23652681 5 2 0182 SLT 226 130500087 19.2 3353007 1828) L601 186 bn 3 382-2476 = OKITS Os 2s 2296 as 207 ones 2032505 500 Lise is Bor 2308 520096 1643 me Bs References 2A ROGERS 6 Fe and waviEw YR 1987 Thermodynamic and Transport Properties of Fluids 4th edn Blackell NATIONAL ENGINEERING LABORATORY Steam Tables 1964 HMSO. oGHks @ rc und waview vx 1992 Enginering Thermodynamics, Work and Heat ‘anger &th edn Longman 22 23 5034 Shot Reversible and Irreversible Processes In the previows chapters the energy equations for non-flow and flow processes are derived, the concepts of reversibility and ireversibility introduced, and the properties of vapours and perfect gases discussed. It is the purpose of this chapter to consider processes in practice, and to combine this with the work of the previous chapters Reversible non-flow processes Constant volume process In a constant volume process the working substance is contained in a rigid vessel, hence the boundaries of the system are immovable and no work ean be done on or by the system, other than paddle-wheel work input. It will be ‘assumed that ‘constant volume" implies zero work unless stated otherwise "From the non-flow energy equation, (14), for unit mass, O+ Wau — a, Since no work is done, we therefore have Q=m—u Gy or for mass, m, of the working substance Q=Us-U; G2) {Allthe heat supplied in «constant volume process goes to increasing the internal nergy. ‘A constant volume process for a vapour is shown on a p-v diagram in Fig. 3.1(a) The inital and final states have been chosen to bein the wet region ‘and supetheat region respectively. In Fig. 3.1(b) a constant volume provess is shown on a p-r diagram for a perfect gas. For a perfect gas we have from equation (2.13) Q=me(Ts—T)and Irreversible Process Fig. 31 Constant ’ ” volume proces for a ‘vapour and a perfect os ‘ ie o : Constant pressure process ‘I-can be seen from Figs 3.1 (a) and 3.1(b) that when the boundary ofthe system is inflexible as in a constant volume proces, then the pressure rises when heat is supplied. Hence for a constant pressure process the boundary must move ‘against an external resistance as heat is supplied; for instance afuid in a cylinder ‘behind a piston can be made to undergo & constant pressure process. Since the piston is pushed through a certain distance by the force exerted by the fuid, then work is done by the fluid on its surroundings From equation (1.2) for unit mass w = few for any reversible process ‘Therefore, since pis constant, We of “d= —Plo2—,) From the non-flow eneray equation, (1.4), oww Hence for a reversible constant pressure process Q = (us ~ m4) + (Os ~ 04) = (us + 72) — (uy + D4) Now from equation (1.9), enthalpy, = u + po, hence, Q=ha— hy (63) (oF for mass, m, of a fluid, Q=1,-H, Ga ‘A constant pressure process for a vapour is shown on a p-o diagram in Fig. 32(a}. The inital and final states have been chosen to be in the wet region and the supetheat region respectively. In Fig. 32(b) a constant pressure process for a perfect gas is shown on a p-v diagram, For a perfect gas we have from 52.Fig. 32 Constant pressure process fora ‘apour and aperect gs Example 3.1 Solution 21. Reversible non-flow processes \ Ly ~ A \ = equation (2.12), Q=me,(T—T) [Note that in Figs 3.2(a) and 3.2(b) the shaded areas represent the work done by the fuid, (oy — 01) ‘A mass of 0.05 kg of a fluid is heated at a constant pressure of 2 bar until the yolume occupied is 0.0658 m*, Calculate the heat supplied and the work done: (]) when the fuid is steam, intially dry saturated; (i) when the Bud is air, initially at 130°C, (i) Iotialy the steam is dry saturated at 2 bar, hence, hy = gat bar = 2707 kd ke Finally the steam is at 2 bar and the specific volume is given by 0.0658 5 = 0658 = 1 316 mk 005 a. Hence the steam is superheated finally. From superheat tables at 2bar and 1.316m?/kg the temperature of the steam is 300°C, and the enthalpy is hy = 30721 /kg. ‘Then from equation (3.4) Q= Hy — Hy =m(h; hy) = 0.05(2072 — 2707) jie, Heat supplied = 005 x 365 = 1825 3 ‘The process is shown on a po diagram in Fig. 33, =< W = plo, ~v,) = shaded area at 2 bar = 0.8856 m?/kg, and v, = 1.316m"/kg. Therefore 2 x 10*(1.316 ~ 08856) = — 86080 Nm/kg Now #4 w