Thermodynamics

ENGINEERING THERMODYNAMICS TTA 005 Steven J. Thorpe Department of Aeronautical & Automotive Engineering Loughborough University Leicestershire LE113TU Copyright This report may not be reproduced in whole or in part without the written permission of the author ENGINEERING THERMODYNAMICS NOTES FOR A FIRST YEAR COURSE CONTENTS 1.Thermodynamics Fundamentals 2.Air Standard Cyclesfor IC Engines 3.Heat Transfer 4.Tutorial Sheets 5.Past Exam Papers ENGINEERING THERMODYNAMICS Section 1 : Thermodynamic Fundamentals Introduction and definitions General properties of gases Daltons law of partial pressures Specific heats First law of thermodynamics, non flow processes Work and reversibility Heat and reversibility Steady flow energy equation for open systems Stagnation properties and flow measurements Second law of thermodynamics Entropy List of symbols Main thermodynamic equations Engine performance definitions and equations 1 THERMODYNAMICS Introduction The science of Thermodynamics is a formalisation of principles governing the relationships between the properties of a system and its heat and work exchanges with its surroundings. A knowledge of thermodynamics is essential for the analysis of 'heat engines' - machines which convert chemical or nuclear energy into useful work. Examples of other applications of this science are: Steam turbines, Boilers, Cooling Towers and Heat Exchangers. Much of the subject is concerned with the behaviour of matter. On a microscopic scale matter is composed of a large number of small particles and studies conducted at a microscopic level are called 'Statistical Thermodynamics'. Engineers are usually concerned with 'classical' or 'engineering' thermodynamics which is the study of the general behaviour of relatively large quantities of matter (i.e. making observations on a macroscopic scale relative to the human senses, when matter is perceived to be homogeneous and continuous). However, it is often useful to keep in mind the microscopic structure of matter when trying to comprehend new phenomena. 2 Some Important Definitions SystemRegion in space containing a quantity of matter whose behaviour is to be investigated. A system is fully defined by a.Fluid Type b.Its boundary c.Mass of Fluid FluidA substance which always continues to deform when subjected to, shear stresses, e.g. a gas (air), a vapour (steam), a liquid (water) or any, combination of these. BoundarySeparates the system from the surroundings and can be either tangible or imaginary. It is across the boundary that Work and Heat are . transferred. SurroundingsOutside the system (i.e. outside the boundary) but sufficiently close to be affected by changes within the system. Closed SystemNo movement of matter into or out of the system (non-flow process) Open SystemMatter crosses the boundary (flow process). PropertyA quantity describing the system, e.g. a linear dimension, colour or pressure. From the standpoint of thermodynamics many properties are irrelevant and it can be a practical difficulty deciding which properties are of interest. Further, the thermodynamic properties of many fluids are not independent of each other. 3 StateCondition of a system as defined by the relevant properties, e.g. the air in a room could be described by its volume, pressure and temperature. More complex formulations might involve concentrations of constituent gases. Two independent thermodynamic properties are required to define the state of the system. Heat (Q)Is the transitory energy exchanged due to a difference in temperature between a system and its surroundings. If the direction of heat transfer is into the system, the quantity is said to be positive; whilst if heat is transferred from the system to the surroundings it is said to be negative. Note that heat is never contained in a system and is not a property. Work (W)Is the transitory energy transfer due to a movement of a part of the boundary under the action of a force. Work transfer from the system to the surroundings is. defined as positive. Note that work is never contained in a system and is not a property. Internal EnergyThe energy contained within a system. Internal energy is a property. (U)The internal energy will be affected by changes in temperature and any chemical or nuclear reactions. EquilibriumA system is in thermodynamic equilibrium when no changes of state are taking place, i.e. no changes of matter, energy or chemical composition. (A system is most easily described when it is in a state of equilibrium, since all the fluid properties are steady and uniform) SpecificThese are per unit mass. For example specific work output is the work Quantitiesoutput per kg of fluid in the system. EnergyMolecules may absorb energy in more than the kinetic vibration form, i.e. atoms may revolve about the centre of a molecule or they may vibrate with respect to one another. The latter two forms of energy may not have any effect on the measured temperature - although they may vary with it. We describe the energy associated with the vibration, spin and translational movement of the constituent particles as Random Energy. Other forms of energy are, for example, Potential, Kinetic, Chemical and Nuclear. Process. Transition of the system from one state to another. Whilst changing states a real system cannot be at equilibrium. However, if the process is sufficiently slow, the system may be assumed to pass through a series of equilibrium states. 4 Cycle. A closed, system completes a cyclic process or cycle when it passes through a series of states in such a way that its final state is equal in every respect to its initial state. 5 General Properties of Gases Introduction A gas is composed of a collection of particles which move randomly at high speed (typically 1000mIs at room temperature). The term particle is being used to describe the smallest part of any gas which still retains the properties of the substance. Two points should be noted: i)If the substance is a chemical compound, for example Propane which has chemical formula C3H8, then the particle comprises several atoms chemically bonded together and is called a molecule. Hence, each propane molecule would comprise 3 atoms of Carbon and 8 of Hydrogen. ii) When the gas is a chemical element - (Oxygen, Hydrogen, Nitrogen etc) it is most likely to be encountered in diatomic form, i.e. each particle is a molecule which contains two atoms. However, at high temperatures, such as those found in combustion products, these gases can exist in monatomic form - each particle being a single atom. In addition atoms of the Inert Gases ( Argon and Neon for example) are reluctant to form chemical bonds with other atoms and so these are always found in monatomic form. Kinetic Theory The Kinetic Theory of Gases is an attempt to theoretically model the behaviour of any gas by analysing the motion of its particles. The analysis is limited to Perfect Gases which are hypothetical substances in which: a) Particles are perfectly rigid, hence all collisions are perfectly elastic (no loss of momentum) b) The total volume occupied by the particles is negligible when compared with the total volume occupied by the gas - (resulting in a large mean free path) c) There are no attractive or repulsive forces between the particles By considering the change in momentum of particles as they bounce off the walls of a container, simple mathematical manipulation yields VNkT32p =Where P=absolute pressure T=absolute temperature V=container volume N=number of particles in volume V K=constant 6 Boyle and Charles carried out constant temperature and constant pressure experiments respectively, and deduced two laws which are consistent with the kinetic theory, i.e. pV= constant (Boyles Law) and V/T=constant (Charles Law) Characteristic equation of state (Perfect Gases) The mass, m, of a given amount of gas must be proportional to the number of molecules/atoms present. Hence the Kinetic Theory yields pV=mRT where mR= 32kN and R, the characteristic gas constant, has a different value for each gas(e.g. for air R = 287 J/kg K) Real gases The Kinetic Theory, Boyle's Law and Charles' Law all describe a Perfect Gas. However this is a hypothetical entity since our assumptions about the behaviour of molecules are not strictly true. At first sight i) The assumption that collisions between particles are perfectly elastic seems reasonable since no loss in pressure is observed when a gas is kept in a closed vessel for long periods of time (i.e. No loss of particle momentum). ii) The assumption that particles are of negligible size seems reasonable since a gas is readily compressed. iii)The assumption that attractive forces between particles are negligible seems reasonable since a gas readily expands to fill all the available space. One set of substances which do closely obey the perfect gas law are the permanent gases (i.e. O2, H2, N2 etc) but only at high temperatures and low pressures. At low temperatures and high pressures the particles are much closer together, thus assumptions ii) and iii) are less valid, and deviations from pv = RT can be observed. 7 Relative Atomic/Molecular Mass ( ARand MR) It is usual to express the mass of an atom of a particular substance relative to the Unified Atomic Mass, u rather than in absolute terms. The atomic mass scale in common use attributes a particular Isotope of Carbon (C-12) with a value AR = 12. Hence the Unified Atomic Mass kg 10 x 66 . 112atom 12 - C of Mass u27 = = Most elements have values of ARwhich can be considered with reasonable accuracy to be integer values (e.g. H = 1, N = 14, 0 = 16). Relative molecular mass is also expressed relative to u and so, for example, Oxygen (O2) has a value MR=32 whilst Propane (C3H8) has a value of 44. Avogadros Law Avogadro stated that Equal volumes of all gases contain the same number of particles when at the same temperature and pressure. Proof From Kinetic Theory we have( ) T / pV k23N =Kilogram-mol The term Mol (or Mole) is used as a measure of the amount of substance and represents a fixed number, NA, of molecules or atoms. However, in order to be meaningful it must be pre-fixed by a unit of mass which using SI units results in the kilogram-mol (often written as kgmol or kmol). In this case NA, which is known as Avogadro's Constant, has a value of 6.022 x 1026 and is specifically chosen so that 1 kgmol of carbon (relative atomic mass = 12) has a mass of 12kg. Clearly, the molar mass M, of any substance must be numerically equal to its atomic mass or its molecular mass - whichever is appropriate. It follows that 1 kgmol of a particular gas always has the same mass, whilst its volume, vo varies with p and T. i.e. n . M m = where m=mass(kg) M=molar ass (kg/kgmol) n=number of mols (kgmol) e.g.1kgmol of C=12kg 8 Universal Gas Constant Considering 1kgmol of substance which has molar mass M and noting that generallypVo = mRT we can write pvo= MRT or pvo/T = MR where nVvolume molar vo= =Hence, for any particular gas pvo/T always has the same value since both molar mass, M and R are constants. But Avogadro states that for given values of pressure and temperature vo always has the same value, irrespective of which gas is being considered. Consequently pvo/ T and hence MR is always constant and we write MR = Ro whereRo is the universal gas constant and has value 8314 J/kgmol K. Note:The value of R for any gas can be calculated from a knowledge of its relativeatomic or molecular mass and the value of Ro. Various forms of equations of state It should now be clear that the equation of state can be listed in many forms, some of the most common ones are: pv=RT(v = specific volume, i.e. per unit mass) pV=mRT pvo=MRT (vo = molar volume, M = molar mass) pvo=RoT pV=nRoT(n = Total number of kgmols) 9 Daltons Law of Partial Pressures Dalton's Law states that the pressure of a mixture of gases is equal to the sum of the partial pressures, pi of the i constituents. (Note The partial pressure of any individual gas is the pressure which it would exert when occupying the same volume as the mixture and at the same temperature). This law, which is based on experimental results, can be predicted using the kinetic theory as follows VT R npo ii =where pi= partial pressure of gas i and ni=number of kgmols of gas i Gas AGas BGas CMixture V,T ++=p n V,T pA nA V,T pB nB V,T pC nC Accounting for each constituent we have =io iiiVT R np Bur during mixing the total number of molecules (or atoms when dealing with a monatomic gas) and hence the number of kgmols remains unchanged.Hence = n niwhen n = number of kgmols of mixture, giving pVT R npio iii= = Molar Fraction From above we have VT Rn pandVT Rn po oi i= = By comparison, the partial pressure of any constituent nnwhere pnnpi ii = is the molar fraction of constituent i 10 Volumetric Analysis Avogaro's Law could be written as At a given temperature and pressure the volume occupied by a gas is directly proportional to the number of molecules present - with the constant of proportionality being the same for all gases. It follows that each constituent in a mixture of gases occupies a fraction of the total volume which is equivalent to its molar fraction. Specific Heats The specific heat, c, of a substance is defined as the quantity of heat required to cause unit temperature rise in unit mass. Hence, if q is the heat transfer per kg of fluid K kg / JdTdqc = However, the amount of heat absorbed depends upon the nature of the process, and not just the change in temperature. In practice we use values of c which relate to two particular processes namely, those which take place at either constant volume or constant pressure.These give: vvdTdqc |.|

\|= (= Specific heat at constant volume) and ppdTdqc |.|

\|= (= Specific heat at constant pressure) Relationships between Cp and Cv It can easily be shown that cp and cv are related by the equations ( ) K kg / J R c cv p= where R=Characteristic Gas Constant or multiplying by the molar mass, M we get ) K kgmol / J ( R c co v po o= also, the ratio v pc / c occurs so often that it has its own symbol ( ) ( ) 0 . 1 c / C c / covop v p> = = vc / R 1 = For air,is about 1.4 at room temperature 11 Temperature Effects Unfortunately the values of cp and cv for a particular gas vary slightly with temperature. Usually their values can be accurately represented by a polynomial expansion in T which allows straightforward mathematical manipulation of any equations in which they appear. When considering processes in which temperature changes are modest it is usually possible to assume that cp and cv are constant and have values corresponding to the average temperature. First Law of Thermodynamics When analysing a closed system we draw an imaginary boundary around a fixed mass of fluid. This boundary can be distorted but no fluid is allowed to cross it, either into or out of the system. By considering the conservation of energy during any process from state 1 or state 2 we arrive at the First Law u w q A = or, considering infinitesimal values ofdu dw dq , u and w , q = A where q=specific heat transfer. (By convention this has a positive value when heat is supplied to the system) w=specific work, and is positive when the system does work on the surroundings u=specific internal energy.This can be considered to be thermal energy and is associated with the \kinetic and PotentialEnergies of the nuclear, atomic and sub-atomic particles. For a cycle or cyclic process,0 u = and hence 0 w q = 12 Work and Reversibility Reversible Process We have met the non-flow energy equation, q-w = u = u2-u1 and noted that we can use specific properties because the end states are in thermodynamic equilibrium. For a particular class of processes which are reversible we imagine the system to pass through a continuous series of equilibrium states. In this case we consider any infinitesimal part of the process and can write dq - dw = du (Once again we can use specific values if desired since the system properties are always uniform) Work During a Reversible Process The analysis of a reversible process can be carried out by considering it as a series of infinitesimal processes - the overall changes which occur between the end states being calculated by integration. In particular, the work done by a closed system can be expressed in terms of the fluid pressure and volume as follows Consider a small expansion from one equilibrium state to another which occurs because of an infinitesimally small change in the externally applied force. i.e F is slightly smaller than pA. The piston moves a short distance, dx 13 The work done by the system is dW = pAdx = pdV where dV is the resulting small change in volume. If the process continues in this way through a series of equilibrium states, it can be represented as a continuous line on a p-V diagram. The total work done = =2121pdV dW W (This is equal to the total area under the p-V curve between states 1 and 2). A reversible process is so called because the system may be returned to its original state by simply reversing the work transfer, i.e. in this case = 12pdV W However, in order to achieve this the force differences (pA compared with F) must be infinitesimally small in order that the movement of the piston is infinitesimally slow. This. ensures that no pressure gradients or eddies are set up within the system and so its properties can be considered uniform. i.e. A reversible process is a hypothetical ideal. In any real process = pdV W the work done by a system, for example, is always less than the area under the p-V curve. As suggested above, a reversible process is an ideal one which delivers the maximum possible work = pdV Wrev It follows that the work done during a real compression is always greater than the minimum value, Wrev , required if the process is reversible. 14 Heat and Reversibility Heat Reservoir That part of the surroundings which exchanges heat energy with the system is called a heat reservoir and is either a source or sink of heat. A reservoir is usually so large that any heat transfer does not change its temperature. If heat transfer takes place due to a large temperature difference between the system and surroundings there will be a temperature gradient within the system fluid. This non-equilibrium condition occurs because the rate of heat flow is so high as to cause the fluid near the boundary to be hotter than that in more remote regions. However, if an infinitesimal temperature difference can be maintained between the system and surroundings reversible heat transfer takes place and the direction of heat flow can be reversed by an infinitesimally small change in source temperature. Heat Transfer for a Closed System Referring to the definition of the specific heats cp and cv, (above) it can be seen that the heat transfer during constant pressure and constant volume processes are, respectively = = dT c q and dT c qv p For all other types of process involving a closed system q is calculated by finding the values of w and u (see below) and then applying the first law. Calculation of u (no chemical or nuclear reactions) Consider the heating of a gas at constant volume 1st Law gives du dw dq = but( ) du dT c dq 0 dwv= = = hence =21v 12dT c u A It can be shown that u is a function of temperature only, and so this relationship holds good for all processes - not just those for which the volume remains constant 15 Enthalpy Consider a closed system which exchanges heat with the surroundings and also does work such that the pressure remains constant 1st Law gives pdv du dq = hence( ) pv d du dq + = sinceconstant p = hence| | ( ) pv u d dT c dqp+ = = The group of properties (u + pv) occurs so often in thermodynamics. that it is given its own name - Enthalpy, with symbol h . Generally, if any fluid undergoes a process from state 1 to state 2, the change in enthalpy is always given by = 21p 1 2dT c h h Since h is a combination of properties, it is itself a property. Although introduced here for convenience, h is encountered most often when considering an open system - see below. Note that the internal energy and enthalpy defined above ignore chemical and nuclear reactions. When combustion occurs for example, additional terms to account for the change in chemical energy have to be included Reversible Adiabatic Process A process which occurs often in Engineering in general and Internal Combustion (Piston and Gas Turbine) Engines in-particular is. the adiabatic process, i.e. one in which no heat transfer occurs between the working fluid and the surroundings. Since q = 0, any work done by thesystem is at the expense of internal energy (closed systems) or the sum of enthalpy kinetic energy (open systems). An adiabatic process which can also be considered to be reversible is often termed Isentropic (constant entropy) and is described by the equation constant pV = For a perfect gas, we can obtain (using the equation of state) 121121122112VVTTTTVVpp||.|

\|=||.|

\|||.|

\|=||.|

\|=||.|

\| 16 also 1V p V pw or1V p V pw1V V p V V p1VconstantVdVconstant pdV w2 2 1 1 2 2 1 111 1 112 2 22- 12121== =

== = Alternatively for a perfect gas ( )( ) ( )1T T mRw or1T T RwT T Cu u w2 1 2 12 1 v2 1 == = = Isentropic Efficiency For compression process w > wrev 1 21 2actualrevisenT TT Tww'= = where T2is the value based on an Isentropic process, given by( )( ) / 11 2 1 2p / p x T ' T= For expansion process w < wrev 1 21 2revactualisenT TT Tww' = = Note: The temperature at the end of an irreversible process will always be higher than that for a reversible process over the same pressure ratio 17 Polytropic Process The compression/expansion process may not be adiabatic. However, providing it can be assumed to be reversible, it can be approximated by a polytropic process defined by constant pVn= where n is the polytropic index Hence an adiabatic process is a special case with = n We can therefore write ( )1 nT T R1 nV p V pwTTpp2 1 2 2 1 11 nn1212==||.|

\|=||.|

\| or ( )1 nT T mR1 nV p V pw2 1 2 2 1 1== Special Cases n=0constant pressure processn= adiabatic process n=1constant temperature processn= constant volume process Note:that for n=1 (isothermal process) cannot find work from equation above but use ( ) gas perfectVVIn RT orVVIn V pVVIn V p w12122 2121 1||.|

\|||.|

\|=||.|

\|= 18 Flow Process (Open Systems) So far we have only discussed processes in which no fluid crosses the system boundary. When we developed the First Law with reference to a closed system we noted that although the total energy of the system E = K.E + P.E + U, the kinetic and potential energies of the bulk fluid were of little interest since they remained constant. We now consider open systems, in which the kinetic and potential energies can vary between the system inlet and exit, and will again use the First Law in order to derive an appropriate energy equation Steady Flow Energy Equation (SFEE) Flow processes may be steady or non-steady. We confine our attention to steady processes in which the mass flow rate and all fluid properties (e.g. p, T, u, v, c etc) are constant, with time, at any point in the system. In order to analyse an open system we construct an imaginary closed system which is assumed to undergo a process during which it deforms from state a) to state b) in time dt. The amount of heat input is dQ and shaft work done by the system = dW. Clearly, a steady flow process is made up of a whole series of such non-flow processes and it should be noted that the energy stored within the shaded area, E is always constant. 19 First Law For any system E W Q A = (1) a)System Energy, E At time, t, taking energy in mass dm=E1 Total system energy = E + E1 But 1211 1z . g . dm2C. dm u . dm E + + = At time (t + dt) Total system energy = 2E ' E +where 2222 2z . g . dm2C. dm u . dm E + + = b)Work Shaft work done by system = +dw Meanwhile the system does Flow Work in order to displace its boundary at position 2, i.e. dm v pV pdx A p dW2 22 22 2 2+ =+ =+ = Similarly the surroundings do work on the system equal to p1v1dm or, adhering to our sign convention, the system does work given by dm v p dW1 1 1 = 20 c)Heat Transfer Heat transferred from surroundings to system =dQ Energy Equation:Substituting for Q, W and E in equation (1) ( ) ( ) ( )1 2 2 1E ' E E ' E dW dW dW dQ + + = + + i.e. ( )2 2 1 1v p v p dm dW dQ + ||.|

\|+ + ||.|

\|+ + =1211 2222gz2Cu dm gz2Cu dm Letting qand w be specific heat flow and specific work, i.e. dmdWw anddmdQq = = and rearranging gives ( ) ( ) ( )1 221221 1 1 2 2 2z z g2C Cv p u v p u w q +||.|

\| + + + = Noting that (u + pv)is specific enthalpy, h, we can write the steady flow energyequation (SFEE) as ( ) ( )1 221221 2z z g2C Ch h w q +||.|

\| + = or ( ) ( )1 221221 2 pz z g2C CT T c w q =||.|

\| + = in terms of rate quantities, we may write ( ) ( )

=||.|

\|+ = 1 221221 2 pz z g2C CT T c m W Q Equation for Continuity of Mass 21 For any steady flow process, such as a fluid passing through the pipe above, the properties at any point do not vary with time and the mass flowing into the system must be. equal to the mass flowing out, i.e. 2 2 1 1 2 1V V or m m = =giving 2 2 2 1 1 1C A C A p p = wheremand Vare the mass and volumetric flow rates. Stagnation Properties Adiabatic flow in a streamtube Consider the adiabatic flow along a streamtube between stations 1 and 2. In order to obtain the relevant energy equation for this system we note a)Flow is adiabatic - hence Q = 0 b )There is no movement of a system boundary i.e. W = 0 c)Potential Energy of the bulk fluid is constant (z1 = z2) The steady flow energy equation simplifies to ( ) 02C CT T c21221 2 p=||.|

\| + Hence p211p222C 2CTC 2CT + = +or, more generally, constantC 2CTp2= + (2) 22 Second Law of Thermodynamics One form of the Second Law states No engine operating in a cycle can convert all the heat energy it absorbs into useful work Consequently, every engine must reject some heat energy at some point in its thermodynamic cycle.It follows that, even if there are no mechanical or pumping losses, no engine could be 100% efficient.This can be demonstrated by considering the Carnot cycle which gives the best possible efficiency for an engine whose working cycle operates between fixed maximum and minimum temperatures. Carnot Cycle It is possible to link a series of processes together in order to form a thermodynamic cycle.If such a cycle is to be reversible, or ideal, each individual process must be reversible.The original conception of a reversible cycle was devised by Sadi Carnot in 1824 with reference to a closed system operating with a series of non-flowing processes (e.g. a piston in a cylinder). Heat is transferred between the engines and two reservoirs maintained at temperatures TH and TC.The cycle comprises four processes 1-2Isothermal compression at a temperature of TC 2-3Adiabatic compression 3-4Isothermal expansion at a temperature of TH 4-1Adiabatic expansion Considering unit mass and using the convention that both q1 and q2 have positive values, the Carnot efficiency is given by 12 11cqq qqw = = (1) 23 to find q1 and q2 ||.|

\|= =34H 34 1vvn 1 RT w q (2) Similarly ||.|

\|= =21C 12 2vvn 1 RT w q (3) Combining (1),(2) and (3) gives ||.|

\|||.|

\|||.|

\|=34H21C34Hcvvn 1 RTvvn 1 RTvvn 1 RT But generally, for an adiabatic process 1baabvvTT||.|

\|= Hence, for processes 2-3 and 4-1 114123HCvvvvTT ||.|

\|=||.|

\|= giving 34211423vvvvorvvvv= = hence HCHC HcTT1TT T == Thus, the efficiency of the Carnot cycle depends only upon the temperatures of the hot and cold reservoirs, TH and TC.100% efficiency can only be achieved if TH is infinitely greater than TC (impossible) or if TC = 0 (also impossible) 24 Clausius Inequality Whenever a system undergoes a cycle,( )T / dQis zero if the cycle is reversible or negative if irreversible. i.e.( ) 0 T / dQ =if reversible ( ) 0 T / dQ < if irreversible Entropy We now introduce the property called Entropy and note that we are usually interested in the amount by which it changes during a particular process, rather than its absolute values.By way of introduction it is useful to draw comparisons with the work equation for a closed system undergoing a reversible process: pdv dw = Work is produced as a result of a pressure gradient across the system boundary causing a change in volume. In a similar way heat transfer occurs as a result of a temperature gradient across the system boundary and causes a change in entropy.For a reversible process: ds T dq = hence = =2121Tdqs or ds T q Significance of Entropy An entropy term is often to be found in thermodynamic relationships but there are two instances where it has particular significance. a)Entropy Principle The First Law merely states that energy is conserved if it is transformed;it does not state whether the transformation is possible or not.Generally, there are preferred directions of energy flow and the entropy principle can be used to predict whether a particular process (which is known to satisfy the First Law) is actually possible.It states that For any process the entropy of the Universe remains constant or increases Here the term universe is used to describe the system plusthat part of thesurroundings affected by changes within the system.(Note: if the process isreversible, the entropy of the universe is constant otherwise it increases. 25 b)Calculation of Heat Transfer We have seen that for a closed system undergoing a reversible process the areaunder the p-V line is equal to the work done.In a similar way, the area under the corresponding T-s line would yield the heat transfer.By way of example, we re- calculate the efficiency for the Carnot cycle which can be represented on aTemperature Entropy diagram as follows 12 1cqq q = n but = = s T ds T q A(for constant temperature process) hence,HCHC HcTT1s Ts T s T ==AA An Calculation of Entropy Changes Generally,dw du dq + = but for reversible processpdv dw and Tds dq = = henceor , pdv du Tds + = dvTpTduds = = Integrating gives dvTpTdus s21211 2 + = In order to proceed further we need to know the variation of u with T and Tpwith v.For a perfect gas , for example, this is straightforward. dT c du andvRTpv= = giving = = 2121v1 2vdvRTdT . cs s 26 In order to perform this integral we need to know the variation of vc with T, with the most simple case being vc = constant.Also, although the theory was developed by considering a reversible process, the above equation is valid for all processes involving a perfect gas (Entropy is a property and so s depends only upon the end states and is dependent of the path between them). Taking a perfect gas and assuming vc is constant then ||.|

\|+||.|

\|= 1212v 1 2VVn . RTTn . c s s [Entropy in terms of T and V] Alternative forms can be derived. The change in entropy in terms of T and p can be found by substituting ||.|

\|||.|

\|=||.|

\|122112TT.ppVV ||.|

\|+||.|

\|= 122112v 1 2TT.ppn RTTn . c s s

||.|

\|||.|

\|+||.|

\|=121212vppn RTTn RTTn . c SinceR c cv p+ =then ||.|

\|||.|

\|= 1222p 1 2ppn RTTn c s s In terms of p and V, substitute for ||.|

\|12TTto obtain ||.|

\|+||.|

\|= 12p12v 1 2VVn cppn c s s From the above, we can write the change in entropy for various processes. 27 Isentropic 1 2s s but0VVnppncs s1212v1 2=||.|

\|+||.|

\|= 1VV.pp1212=||.|

\|||.|

\|

constant V p V p1 1 2 2= = Reversible constant pressure ||.|

\|= =12p 1 22 1TTn c s sp p Reversible constant volume ||.|

\|= =12v 1 22 1TTn c s sV V Reversible isothermal ||.|

\| =||.|

\|= =12121 22 1ppn RVVn R s sT T Datum Value for Entropy Data Entropy is usually taken to be zero at some datum condition, say 0.However this does not usually matter since we are only interested in changes between state points 1 and 2 since ( ) ( )1 20 1 0 20 1 0 2 1 2s ss s s ss s s s s s =+ = = 28 List of Symbols Cvelocity m/s op pc , cSpecific heat at constant pressure kJ(kg, K). kJ(kgmol K) ov vc , cSpecific heat at constant volume kJ(kg, K). kJ(kgmol K) EEnergy kJ GGravitational acceleration m/s2 (normally 9.81 m/s2) h,HEnthalpy, kJ/kg,kJ mMass kg MMolar mass kg/kgmol nNo. of moles kgmol or polytropic index pPressure kN/m2 or bar or Pa q , QHeat transfer kJ/kg, kJ RCharacteristic or Specific Gas Constant kJ(kgK) oRUniversal Gas Constant kJ(kgmol k)=8.3143 s , Sentropy kJ(kgK), kJ/K TTemperature K u , UInternal Energy kJ/kg, kJ w, WWork transfer kJ/kg, kJ zHeight above datum m Ratio of specific heats pDensity kg/m3 Note:1 bar=105Pa = 100kN/m2 29 MAIN THERMODYNAMIC EQUATIONS (NOTE that there are alternative forms for many of these equations) Definition of the mol M . n m = M = 2 for H2,12 for C etc Perfect gas equation mRT pV = Idealisation for gases Gas constantsM / Ro R = Ro = 8314.3J/kgmol K Daltons Lawp .ntnpii|.|

\|=( )ntni = molar fraction Definition of cv ( )v vdT / dq c=Specific heat at constant volume Definition of cp ( )p pdT / dq c =Specific heat at constant pressure Relationship between kR c , cv p ( )( ) units molar R c cunits massR c co voopv p= = Derived from 1st Law Definition of o ov p v pc / c c / c = = = Specific Heats Ratio 1st Law for non-flow (closed) 1 2 2 1 2 1U U W Q = True for any non-flow process Definition of non-flow reversible work =212 1pdV WWork due to deformation of boundary Q for constant P Q for constant V ( )( )1 2 v 2 11 2 p 2 1T T mc QT T mc Q = = Only true for these 2 processes Change in U for a perfect gas ( )1 2 v21v 1 2T T mcdT c m U U == General expression. Only if cvis constant Definition of enthalpy pv u h + = Always true Change in H for a perfect gas ( )1 2 p21p 1 2T T mcdT . C m H H == General expression. Only if cp is constant Isentropic process constant s0 Qconstant pV== = Isentropic = reversible adiabatic For perfect gases only ( ) ( )( ) ( )( )( ) ( )( )( ) ( ) = ===1 / v p v p Wp / p T / Tv / v T / Tv / v p / p1 1 2 2 2 111 2 1 212 1 1 22 1 1 2 These follow from C pV = and perfect gas equation 30 Polytropic process i.e. general non-flow reversible processes Aof instead n with aboveasEquation0 s0 Qgeneral InV constant for np constant for 0 nisothermal for 1 nadiabatic for nt tan cons pVn== ===== Difference between n and due to heat transfer For isothermal 2 1 2 - 1W Q=for isothermal process ( )1 2 2 1v / v n . constant W = constant = mRT or v p or v p2 2 1 1 1st Law flow (open system) ( ) ( ) ) ( ) ( | |1 22122 1 2 2 1 2 1z z g 2 / C C h h m W Q + + = Can be written in terms of mass or mass flow rate Definition of Stagnation temperature Ts Definition of Stagnation pressure Ps ( )p2sc 2 / C T T + = ( )( ) 1 /s sT / T . p p= When C=o, Ts = T Stagnation is also called total. T = static temperature Isentropic Efficiency for compressor ( ) ( )1 2 1'2 c , isenT T / T T = nCompressor ( )( ) 11 2 1'2p / p x T T=Isentropic Efficiencyfor turbine ( ) ( )1'2 1 2 t , isenT T / T T = nwhere '2Tis the outlet temperature for an isentropic process Turbine ( )( ) 11 2 1'2p / p x T T=Thermal Efficiency supplied Q / Wnet th = n= W Wnet Carnot Engine( Reversible engine operating in a cycle) ( )H cH c H c , thT / T 1 orT / T T = = n TH = Hot Res Temp TC = Cold Res Temp Reversible heat pump ( )( )c H H . P . HT T / T . P . C =Reversed Carnot Engine Entropy, s = Tds q Reversible process only Change in entropy ( ) ( )( ) ( )( ) ( )12p 1 2 v1 2 1 2 p1 2 1 2 v1 21 2V / Tds q n . c p / p . n . c orp / p n . R T / T . n . c orV / V n . R T / T . n . cTdv / p T / du s sT / q s s = + = =+ =+ = = Only if T is constant General equation Perfect gas with cpconst Change in entropy for any process. As written, gives specific entropy with units same as cp, cv, R. Clausius Inequality ( )le irreversib ifreversible if 0 T / dQ> The air standard cycles for reciprocating engines and gas turbines are nowconsidered in sections 2 and 3. 34 2.2Reciprocating Engines Two main classifications as follows: a)Spark Ignition (SI) [petrol] b)Compression Ignition (CI) [diesel] Comparison of the main features SI EngineCI Engine Fuel TypePetrol, Gasoline, Natural GasDiesel Oil IgnitionElectrical Discharge Compression Temperature Compression RatioTypically 9.0:1Typically 18.0:1 Fuel System Carburettor or Low Pressure F. Injection High Pressure Fuel Injection Mixture in CylinderHomogenousStratified Load ControlQuantity Governed (Throttle) Quality Governed (AFR Control) Advantages High Specific Power Relatively Low Cost High Thermal Efficiency Note that both SI and CI engines can be 1.Turbocharged 2.4 stroke or 2 stroke 3.Water or air cooled 35 The reciprocating engine employs a cycle consisting of a succession of non-flow processes.Consequently, analysis is done by simple application of the non-flow energy equation to each process in turn. The reciprocating engine has two main advantages. i)Each process is fairly slow and more reversible than those in gas turbines. ii)It can be run with stoichiometric mixtures (giving very high temperatures typically 2,300C) since the cylinders are not constantly exposed to hot gases. One major disadvantage is that piston engines tend to be bulky for a given air mass flow rate and hence heavy for a given power output.This makes them unsuitable for larger aircraft where typical propeller shaft power can be of the order of 9,000 kW. Three important Air Standard Cycles form the basis for all reciprocating engines- namely, the Otto, Diesel and Dual Cycles.These are considered below. The 4- stroke operating cycle The 2- stroke operating cycle 36 2.2.1Otto Cycle This is the theoretical heat cycle upon which calculations for a spark ignition (SI) engine are based and has a similar form for both 2 and 4 stroke cycles. ProcessHeat FlowWork Done *a-1Induction ( ) ( ) + a 1 1v v P1-2 Adiabatic Compression( ) ( ) 1 2 vT T C2-3 Constant Vol Heat Addition ( ) ( ) + 2 3 vT T C3-4 Adiabatic Expansion ( ) ( ) + 3 4 vT T C4-1 Const Vol Heat Rejection( ) ( ) 4 1 vT T C*1-aExhaust ( ) ( ) 1 a 1v v P*Note:A 2-stroke cycle is defined by points 1-4 whilst a 4-stroke cycle also includes an additional loop a-1 and 1-a.However these induction and exhaust strokes are assumed to involve no net work or heat transfers. p32a14V1234TSvolce ' Clvolume swept37 Thermal Efficiency Supplied Heat TotalOutput Work NetTH = n ( ) ( )( )2 3 v3 4 v 1 2 vT T CT T C T T C = ( ) ( )( )2 31 4 2 3T TT T T T =Hence

=1 T T1 T TTT12 31 421THn (1) defining compression ratio v3421rvvvv=

=we have 2314 1v4312TTTTandrTTTT= = = Substituting into (1) 1v 21THr11TT1

= =n (2) This is the air standard efficiency of the Otto Cycle and it will be noted that it has the same form as the Carnot efficiency.However, there are two points to note a)The THn is less than that for a Carnot Cycle operating between the sametemperature limits since 2T is not the maximum temperature. b)The efficiency can be expressed in terms of compression ratio vr only and this relationship is described by the graph below.(However, specific work output increases with the upper temperature 3T 38 It should be noted that although the efficiency of the Otto Cycle appears tocontinuously increase with compression ratio the efficiency of a real SI engine islimited by knock.Even when using high octane fuels, compression ratios used in practice do not normally exceed 10:1.Even when knick is not a limiting factor, the combined detrimental effects of heat transfer, dissociation and variable specific heats result in reducing thermal efficiency at compression ratio greater than about 14:1 2.2.2Diesel Cycle This theoretical cycle forms the basis for calculations on Compression Ignition (CI) engines.It is similar to the Otto Cycle except that heat is supplied at constant pressure and not at constant volume.However, it should be noted that the modern CI engines are more accurately represented by the Dual Cycle.In CIengines, fuel is injected into the cylinder towards the top of the compression stroke, hence pre-ignition cannot occur and so increased compression ratios can be used.In any case, compression ignition of common fuel oils is not usually possible for values of vr less than about 12:1 ProcessHeat FlowWork Done 1-2 Adiabatic Compression( ) ( ) 1 2 vT T C2-3 Constant Pres Heat Addition ( ) ( ) + 2 3 pT T C ( ) ( ) + 2 3T T R3-4 Adiabatic Expansion ( ) ( ) + 3 4 vT T C4-1 Const Vol Heat Rejection( ) ( ) 4 1 vT T C 39 Thermal Efficiency input Heat TotalOutput Work NetTH = nHence ( ) ( )( )2 3 p1 4 v 2 3 pTHT T CT T C T T C = n (3) We can express all other temperatures in terms of say 2T as follows For 1TNoting that compression ratiothen vvr21v = 1v11221r1vvTT

=

= For 3TPoint 3 is the heat supply cut-off point and we define the cut-off ratio 2 3 cv v r =Hence c2323rvvTT= = For 4T 1vc1422314334rrvvxvvvvTT

=

=

= but 1vcc1vc24233424rrxrrrTTTTxTTTT=

= = Substituting 1T , 3Tand 4Tinto equation (3) and simplifying gives ( )

=1 r1 rr11cc1vTHn(4) It can be seen that the efficiency of the diesel cycle depends upon the cut-off ratio, cr(i.e. the quantity of heat added) as well as upon vrand Since the term in [ ] is always greater than the unity the Diesel Cycle always has a lower efficiency than an Otto Cycle for a given compression ratio.In theory, Their efficiencies are equal when the cut-off ratio cr equals unity, but this example is trivial since it represents zero heat input! However, as mentioned above, practical CI engines run at much higher compression ratios (12:1 < vr < 20:1) than SI engines and so, in general have higher efficiencies. 40 2.2.3Dual or Mixed Cycle The behaviour of many modern reciprocating engines is best represented by a combination of the Otto and Diesel Cycles, called the dual or mixed cycle.In this case part of the het is added at constant volume and the remainder at constant pressure. ProcessHeat FlowWork Done 1-2 Adiabatic Compression( ) ( ) 1 2 vT T C2-3 Constant Vol Heat Addition ( ) ( ) + 2 3 vT T C3-4 Constant Pres Heat Addition ( ) ( ) + 3 4 pT T C ( ) ( ) + 3 4T T R4-5 Adiabatic Expansion ( ) ( ) + 4 5 vT T C5-1 Constant Vol Heat Rejection ( ) ( ) 5 1 vT T C 41 Net Work Output( ) ( ) ( )1 5 V 3 4 p 2 3 vT T C T T C T T C + =Heat Added( ) ( )3 4 p 2 3 vT T C T T C + =Thermal Efficiency ( )( ) ( )3 4 P 2 3 v1 5 vTHT T C T T CT T C1 + = n ( )( ) ( )3 4 2 31 5T T T TT T1 + =(5) Again( ) ( ) ratio off - Cutvvr , ratio n Compressiovvr34c21v= =and defining 2 3 pp p r= .Also, expressing all temperatures in terms of say 2TFor 1T 1v 21r1TT

= For 3T p2323rppTT= = For 4T p c233424r x rTTxTTTT= = For 5TFirstly, 1vc1vc112233411415445rrr1x 1 x rvvxvvxvvvvvvTT

=

=

=

=

= Hence 1vp c244525rr rTTxTTTT= = Substituting into equation (5) gives ( ) ( )

+ =1 r r 1 r1 r rr11c p pc p1vTHn 42 It will be noted that when1 rp = this reduces to the same expression as for the Diesel Cycle.The graph below compares the T-S diagram for an Otto, Diesel and DualCycle which all have the same heat input and compression ratio( )2 1v v Since all processes are reversible, the heat input is equal to the area under the T-S curve (i.e. under process 2-3 for Otto and Diesel Cycles and 2-3-4 for Dual Cycle).Similarly the heat rejected is given by the area under curves 4-1 (Otto and Diesel) or 5-1 (Dual).It can be seen that for the same compression ratio, the cycle efficiencies decrease in the order Otto, Dual, Diesel because the Otto Cycle rejects the least amount of heat and the Diesel Cycle the most. 2.2.4Comparison of real cycles with the air standard cycle A P-V diagram comparing a real and air standard cycle is shown below 43 It is clearthat there is a substantial difference between the cycles, the main reasons being as follows a)Working fluid not air b)Specific heats not constant c)Compression and expansion processes not isentropic d)Dissociation e)Combustion not at constant volume/pressure or at TDC f)Valves do not open and close at TDC/BDC The area of the P-V diagram is proportional to the work done and a real cycle will only typically have just over half the area compared to the ideal cycle.Also, the peak cycle pressure and temperature predicted for the ideal cycle will be much higher than for the real cycle (mainly due to variable vC , and dissociation).The compression and expansion processes are not adiabatic but can be approximated well with a polytropic process, constant PVn= , where n is typically 1.32 for the compression and 1.28for the expansion. Improvements to the air standard cycle predictions can be achieved by using dissociation charts which cater for variations in specific heat ( with both temperature and composition) and dissociation.However, whilst dissociation charts produce improved predictions compared to the air standard cycles, they still do not cater for finite combustion rates, valve timing etc and air standard cycles and dissociation charts have been replaced by complete engine simulation models. 2.2.5Criteria of Engine Performance Thermal efficiency is a useful measure of engine running costs, but size, complexity and capital costs are also important.For example, size (i.e. weight) is particularly important for transport and so an important criterion becomes the power output from a given size of engine.This is quantified using Mean Effective Pressure (MEP) 2.2.6Mean Effective Pressure (MEP) Work ratio is very useful in indicating cycle susceptibility to Irreversibilities but is usually used in cycles involving steady flow processes (e.g. gas turbines).In reciprocating engines it is not always easy to isolate the (+)ve work and so the similar parameter MEP is used. Brake Mean Effective Pressure, MEP or mp , is the pressure which, when acting on the piston over one stroke, can produce the net measured work of the cycle. Area of rectangle equals areas within p-v loop i.e.( ) W pdv v v p1 2 m= = 44 2.3.Gas Turbine Engines We now discuss two Air Standard Cycles used for predicting the performance of gas turbines.The Joule or Brayton Cycle describes the processes involved in the operation of most engines whilst the Ericsson Cycle is included as a method of achieving the Carnot Efficiency. Both cycles are analysed by applying the Steady Flow Energy Equation to each process and, as usual, velocity terms are assumed to be negligible. 2.3.1Joule (or Brayton) Cycle (Flow) ProcessHeat FlowWork Done 1-2Isentropic Compression ( ) ( ) 1 2 pT T C2-3 Const Press Heat Addition ( ) ( ) + 2 3 pT T C 3-4 Isentropic Expansion ( ) ( ) + 3 4 pT T C4-1 Const Pres HeatRejection ( ) ( ) 4 1 pT T C 45 Thermal Efficiency, Input Heat TotalOutput Work NetTH = n ( ) ( )( )| || | 1 T T1 T TTT1T TT T T T2 31 4212 31 4 2 3 = = Expressing pressure ratio 23141p4312p4312TTTTandrTTTTgives rpppp= = == = Hence n1p 21THr11TT1||.|

\| = = Note Since pr equals vr , the efficiencies of the Otto and Brayton cycles are the same.However, it is much more convenient to define pressure ratio( )prfor a gas turbine than the compression ratio( )vr . 2.3.2Ericsson Cycle Theoretically the efficiency of a gas turbine can be improved to equal the Carnot Efficiency by employing constant temperature work phases (as in the Stirling Cycle) and raising the gas temperature between the compressor and turbine at constant pressure. An isothermal compression is achieved by using an infinite number of stages and intercooling -i.e. this represents heat rejected.Similarly, an isothermal expansion would have an infinite number of stages with reheat (heat input).It is shown below that the heatrequired between the compressor and turbine (2-3) can be completelyfurnished by the heat rejected (4-1) using a perfect heat exchanger. 46 Process (all reversible)QW 1-2Isothermal Compression ( ) ||.|

\|'21n 1vvL RT ( ) ||.|

\|'21n 1vvL RT2-3 Const Press Heat Exchange ( ) ( ) + '2 3 pT T C3-4 Isothermal Expansion ( ) +||.|

\|3'4n 3vvL RT ( ) +||.|

\|3'4n 3vvL RT4-1 Const Pres HeatExchange ( ) ( ) 1'4 pT T C Note1'4 3'2'2 1'4 3Q Q , T T and T T Since= = =Hence

=+=3' 4n 3' 21n 13' 4n 3' 34' 34 ' 12THvvL RTvvL RTvvL RTQw wn Since ' 2 ' 2 1 1 1 4' 3 ' 2v p v pandp pandp p = = = 47 Then 3' 4' 21' 431' 2vvvvandpppp= = Giving ( ) efficiency Carnot equals TT131TH = n 48 THERMODYNAMIC MODULE TTA005 LABORATORY COURSEWORK FULL THROTTLE PISTON ENGINE PERFORMANCE TEST OBJECTIVES The objectives of this laboratory are that students will: I.Gain familiarity with some of the measurement and experimental techniques used in piston engine testing. II.Learn how various fundamental laws and principles introduced during Thermodynamics lectures can be used in practice. III.Gain a better understanding about the practical engineering issues, which influence engine performance. BACKGROUND The phases between the initial design concept and the manufactured piston engine involves a combination of a wide range of human skills and computing power to ensure that the product will prove of practical interest. Further work is carried out to investigate the behaviour of the engine at various applied loads; this is known as Engine Performance Testing. EQUIPMENT The equipment comprises of a standard 1.4 litre automotive engine, which is mounted on a test bed and connected to a water-cooled dynamometer. Precise details of the engine are given on the attached sheet. INSTRUMENTATION Instrumentation is available to enable the accurate measurement of Engine Speed, Fuel Flow Rate, Air/Fuel Ratio and Engine Torque. Further instructions on how to use the various pieces of equipment will be given at the start of the laboratory session. PROCEDURE I.Normally you will be required to take a reading of the ambient pressure and temperature. However for the purpose of this experiment, the valves for these two parameters are set. See attached sheet. II.Start the engine and allow it to warm up. Adjust the throttle to a setting, which is on or close to fully open. Increase the dynamometer load to the maximum value, which can be achieved without causing the engine to run in an unstable way. III.Record the values of Engine Speed (engine speed values already entered in results table but if any different then correct accordingly), Time taken to consume the prescribed amount of fuel, Air/Fuel Ratio, and Dynamometer Torque on the attached experimental results table. IV. Reduce the dynamometer load so that the engine speed increases by 200rpm (this will be the rate of increase until the engine speed reaches 3000rpm, after this point the engine speed will be increased by 100rpm). Take a new set of readings. V. Repeat (IV) until the dynamometer is having no braking effect or until the engine has reached the maximum speed of about 4700rpm as set by the Laboratory Supervisor. 49 ANALYSIS I.Consider each set of readings and calculate Brake Power, Volumetric Efficiency, Thermal Efficiency, Brake Specific Fuel Consumption and Brake Mean Effective Pressure (see attached Formulae Sheet). II.Produce graphs which show how each of these parameters varies with engine speed. REPORT Write a report, which clearly describes your work. You should include this handout in an Appendix and refer to it as necessary. Give a concise description of the equipment and instrumentation and describe any matters which were relevant to the general conduct of the experiment and which have. already not been covered. Also a Schematic drawing of the experimental layout will be required. Finally, produce a discussion, which clearly describes WHY each of the experimental data follows the trends shown on your graphs. PITFALLS The main area you are likely to make a mistake, which will carry through your caldulations, will be the use of correct units. Pay particular attention to this. REPORT LAYOUT Refer to the departmental notes on report writing, this you should have if not, you can obtained a copy from the department office, or on the learn server. To access it on the learn server, log on, and go to the department, select the course you are studying, select general model this is under the code TTZOO, and finally select notes on report writing. On a whole a good write up should take the form of: -Title page -Contents page -List of symbols/abbreviations -Introduction -Aims -Objectives -Description and experimental procedure -Equipment used -Theory - Evaluation/discussion of results -Conclusion -Appendices The above layout is only for thought and you should refer to the departmental publication on report writing. 50 ENGINE DETAILS ENGINE - ROVER TYPE - K SERIES PORT INJECTION TYPICAL VEHICLE APPLICATION - ROVER 1.4 NUMBER OF CYLINDERS - 4 CAPACITY - 1.397 LITRES BORE - 75.0mm STROKE - 79.0mm COMPRESSION RATIO - 9.5 MAX POWER OUTPUT - 70kW at 6250 rev/min MAX TORQUE - 123Nm at 4000 rev/min FUEL DATA Density of fuel (petrol) 3fuelm / kg 741 = p Lower Calorific Value of Fuel kg / kJ 42000 LCVfuel = AMBIENT CONDITIONS K 288 Tbar 1 Paa== 51 EXPERIMENTAL RESULTS No. Engine Speed (RPM) Fuel Flow Rate Time for 50ml (secs)Air/Fuel (By Mass) Dynamometer Torque (Nm) I1800 22000 32200 42400 52600 62800 73000 83100 93200 103300 113400 123500 133600 143700 153800 163900 174000 184100 194200 204300 214400 224500 234600 244700 52 FORMULAE SHEET Brake Power, bp (kW) ( ) ( )310 / xT 60 / x 2 Nx bp t = Volumetric Efficiency( ) %voln ( ) | |( ) | |air cfuelvolx 120 / N x AxLxnx t / V Fx / Appn = Where ( )( )( )( ) J/kgK Constant GassticCharacteri RK eTemperatur Ambient TN/m pressureAmbient PR TPkg/m air of Densitya2aaa 3air==== = p Thermal Efficiency( ) %THn ( )fuel fuelTHxLCV x t / Vbppn = Brake Specific Fuel Consumption,( ) kg/kWhr b.s.f.c. ( )3600 xbpx V/t b.s.f.c.fuelp= Brake Mean Effective Pressure,( ) bar b.m.e.p. ( ) ( )5c310 x AxLxn x 120 / Nbpx10b.m.e.p. = 53 NOMENCLATURE, SYMBOLS AND UNITS A=Piston Cross-section Area (m2) F / A =Air / Fuel Ratio by Mass bp =Brake Power (kW) . c . f . s . b =Brake Specific Fuel Consumption (kg/kWhr) b.m.e.p. =Brake Mean Effective Pressure (bar) L=Stroke (m) fuelLCV =Lower Calorific Value of Fuel (kJ/kg) cn =Number of Cylinders N =Engine Speed (rev/min) t =Time to Consume V (s) T =Dynamometer Load (Nm) V =Volume of Consumed Fuel (m3) voln =Volumetric Efficiency (%) THn =Thermal Efficiency (%) fuelp =Density of Fuel (kg/m3) 54 ENGINE PERFORMANCE 1.ENGINE PARAMETERS For any one cylinder, the following dimensions are of particular interest: DBore (diameter LStroke u Crank angle Connecting rod length cV Clearance volume dV Displacement volume Top dead centre (TDC) of an engine refers to the crankshaft being in a position such that o0 = u . The volume in this position is minimum and is often called the clearance volume ( )TDC cV V = . Bottom dead centre (BDC) refers to the crankshaft being at o180 = u .The volume is maximumat BDC.The difference between the maximum and minimum volume is the displacement or swept volume( )s dV or V : L D4V V V V2TDC BDC d st= = = The compression ratio( )cr is defined as the ratio of the maximum to minimum volume: cscs ccBDCcVV1VV VVVr + =+= = Modern spark ignition (SI) engines have compression ratios of 8 to 11, while compression ignition (CI) engines have compression ratios in the range of 12 to 24. 2.WORK Work( ) W output by a reciprocating Internal Combustion (IC) engine is generated by the gases in the combustion chamber of the cylinder.It is the result of the gas pressure force on the moving piston inside the cylinder: = = PdV dx PA Wp 55 where volume displaced: Vdistance displaced: xarea facepiston: Appressure gas : P For unit mass of gas m within the cylinder.The specific work is: = kg / kJ Pdv W where volume specific : v 2.1Indicated work( )iW Specific work W is equal to the area under the process lines on thev P diagram, and is called indicated work( )iW 2.2Brake work( )bW Work delivered by the crankshaft is less than indicated work due to mechanicalfriction.Actual work available at the crankshaft is called brake work( )bW kg / kJ W W Wf i b = where fW :specific work lost due to friction The ratio of brake work at the crankshaft to indicate work in the combustion chamber defines the mechanical efficiency of an engine: % 100 x W / Wi b m = n Mechanical efficiency will be in the order of 75% to 95%at high speed enginesoperating at wide-open throttle. 56 3.MEAN EFFECTIVE PRESSURE (mep) An average mean effective pressure (mep) is defined by dv / W mep = where W: specific work of one cycle dv :specific displacement volume Mean effective pressure is a good parameter to compare engines for design or output because it is independent of engine size and/or speed. If brake work is used, brake mean effective pressure is obtained. d bv / W bmep = Indicated work gives indicated mean effective pressure: d iv / W imep = 4.TORQUE AND POWER Torque (T) is a good indicator of an engines ability to do work.It is defined as force acting at a moment distance and has units N.m.It is related to bmep by: t 4 / V . bmep Td= 4-stroke engines bmep is used because torque is measured off the output shaft Power is defined as the rate of work of the engine.If N is the engine speed in rpm W60N 2. T bp |.|

\|=t The ratio of the brakes to indicated [power define the mechanical efficiency mn 57 Brake power (bp) and Torque (T) are normally measured with adynamometer or brake. 5.AIR-FUEL and FUEL-AIR RATIO Energy input to an engine Q comes from the combustion of a hydrocarbon fuel. Air is used to supply the oxygen needed for this chemical reaction. Air-fuel ratio (A/F) and Fuel-air ratio (F/A) are parameters used to describe the mixture ratio: a f a ff a f am / m m / m A / Fm / m m / m F / A = == = where f am / m mass of air and fuel, respectively f am / m mass flow rate of air and fuel, respectively The ideal or stoichiometric A/F for many petrol engine fuels is about 15:1 Evidence ratio is defined as the actual ratio of fuel-air to ideal or stoichiometric fuel-air. ( ) ( )( ) ( )act stoichstoich actF / A / F / AA / F / F/A o== For stoichiometric mixture = 1 6.SPECIFIC FUEL CONSUMPTION Specific fuel consumption is defined by the rate of fuel consumed per unit output power. Brake power gives brakes specific fuel consumption: hr . kw / g b / m bsfcp f= Indicated power gives indicated specific fuel consumption 58 hr . kw / g i / m isfcp f= It also follows that efficiency mechanical bsfc / isfcm= = n 7.ENGINE EFFICIENCIES 7.1Thermal Efficiency thn A thermal efficiency is defined as the power output divided by the thermal energy input and can have either indicated or brake thermal efficiency defined by: ( )( ) % 100 x LCV . mf / bp )% 100 x LCV . mf / ip )i thi th==nn where fmmass flow rate of fuel (kg/s) LCVlower calorific value of the fuel (kJ/kg) It follows that i th b th m) / ) n n n = Typicalb th) n is engine CI 45%engine SI % 35)b th==n 7.2Volumetric Efficiency This is defined as the actual volume flow rate into the engine divided by the rate at which volume is displaced by the piston. % 100 Nx . V / m . ns a a volp n= 59 where

( )( )( )( ) rev/s speed engine: Nm volumeswept total : Vengines strokefour for 2 n cycle. per srevolution of number : nkg/m conditions catmospheri at evaluated density air :(Kg/s) enginetheinto air of rateflow mass: m3s3aa=p Standard values of surrounding air pressure( )oPand temperature( )oTcan be used to find density. 3o o aoa om / Kg RT / PK 298 Tp 01 . 1 P===p at standard conditions,3am / Kg 181 . 1 ~ p Typical values of volumetric efficiency for an engine at wide-open throttle (WOT) are: % 90 % 75vol ~ n 60 Section 3:Heat Transfer 1.Introduction 2.Fouriers law of conduction 3.Newtons law of cooling 4.Conduction through a composite wall 5.Heat flow through a cylinder: radial heat transfer 61 SYMBOLS AArea pCSpecific heat at constant pressure vC Specific heat at constant volume D Diameter hSurface heat transfer coefficient hi Inside surface heat transfer coefficient ho Outside surface heat transfer coefficient k Thermal conductivity kMean thermal conductivity K Temperature Length Characteristic length scale Thickness of a wall Q Heat transfer rate r Radius R Thermal resistance TR Total thermal resistance T Temperature mT A Logarithmic mean temperature difference (LMTD) UFluid velocity Overall heat transfer coefficient uDynamic viscosity Nu Nusselt Number khLNu =Pr Prandtl Number kC .Prpu=ReReynolds Number up L URe =62 HEAT TRANSFER 1.Introduction 1.1What is it? Heat is energy in transition under the motive force of a temperature difference.Heat transfer (H.T.) deals with the study of the rate at which such energy is transferred. 1.2What is it for? For two types of engineering problems Design of boilers and heat exchangers: Promotion or maximising of the rate of heat transfer with the minimum possiblesurface areas and temperature difference. Design of thermal insulation structure Prevention or minimising of the heat transfer rate e.g. lagging of jet pipes, wall of domestic building, cooling of combustor wall etc. 1.3What is the mechanism of H.T.? There are three modes of heat transfer Conduction Convection Radiation Conduction i)Heat is transferred on a molecular scale with no movement of macroscopic portions of matter relative to one another. ii)Vibrational energy from more energetic molecules to those with lower energy. iii)It is the predominant method of HT in solids. iv)Good electrical conductors are usually good thermal conductors (High K) v)Higher density, higher elasticity higher K Convection i)H.T. is caused by the motion of a fluid. ii)This motion of a fluid is resulted from density difference (natural buoyancy forcesnatural convection or free convection). e.g. heat transferred from a hot-plate to the atmosphere. iii)Also the fluid movement can be caused by external forces (e.g. by a pump) forced convection.e.g. domestic fan-heater for room heating. 63 Radiation i)All matter continuously emits electro-magnetic (EM) radiation (i.e. transfer of thermo energy) unless its temp is absolute zero. Radiation is the H.T. by EM waves. ii)The higher the temp, the greater is the energy radiated. iii)This mode of H.T. does not require medium for its propagation i.e. no physical material is required (vacuum medium is all right) 2.Fouriers Law of Conduction We are only interested in one dimensional steady-state heat flow.Temperature varies in one direction only and does not change with time. Fouriers Law is an empirical law based on observation dxdT. A . k Q =(1.1) whereQ= rate of heat flow in x direction (W) A= cross-sectional area normal (right angle) to the direction of heat flow ( )2m dxdT= temperature gradient (K/m) k = coefficient of thermal conductivity (W/m.K.) It is the heat flow per unit area per unit time when the temp. decreased by one degree in unit distance. 64 The following table shows the thermal conductivities of some materials at temperature 300K SubstanceThermal conductivity k(W/mK) Pure copper386 Aluminium229 Steel55 Concrete0.9 -1.4 Building brick0.35 -0.7 Wood0.15 0.2 Asbestos0.163 Rubber0.15 Cork board0.043 Air0.0262 For Const k From Fig 1.2 and equation (1.1) ( )( )1 21 2TTxxx xT TA . k Q AdT . k dx QdT . kA dx QdxdT. A . k Q2121 = = = = 65 For variable k Let us assume consts) areb and (a b aT k = =k varies linearly with T ( )dxdT. A . b aT Q + = Integrating between 1 and 2 ( ) ( ) ( )

+ = 1 22122 1 2T T b T T .21A x x Q Hence ( ) ( )( )1 21 2 1 2x xT T. b2T T a. A Q

++ = Mean conductivity ( )

++= b2T T ak1 2 ( )( )1 21 2x xT TA . k Q = Notes: Good thermal conductors: high thermal conductivity k such as pure metals (homogenous solids, high density, high elasticity) Good thermal insulators: low thermal conductivity k such as asbestos, cork board etc ( porous, cellular, fibrous, non-homogenous materials) A good thermal conductor is a bad thermal insulator and vice versa. 66 Worked Example 1 The inner surface of a plain brick wall is at 40oC and the outer surface is at 20oC.Calculate the rate of HT per m2 of the surface area.The brick is 250mm thick.The K for the brick is 0.52 w/m.k. Solution: Refer to Figure 1.3 and from Fouriers Law ( )( )( )( )) m / W ( 6 . 4110 x 25040 20x 52 . 0AQk const. forx xT TA . k Q231 21 2= = = Figure 1.3 67 3.Newtons Law of Cooling Heat Transfer through Boundary Layers Heat transfer:Fluids solid surface Solid surface fluids Described by Newtons Law of Cooling The law states: ( )A 1 AT T . A . h Q =(3.1) Where T1the wall surface temperature TAthe temperature of the bulk fluid A hAsurface heat transfer coefficient or film coefficient (W/m2K) hA:depends on the wall surface characteristics and the fluid properties (such as: 1c , k , , p u etc) and the velocity of the moving fluid. It is normally determined by experiments. Similarly ( )2 B BT T . A . h Q = (3.2) Fouriers Law for the wall heat transfer LT TA . k Q1 2 =(3.3) Heat flow rate is constant through each boundary layer and the wall due to steady-state heat flow assumption. 68 From the equations (3.1), (3.2), (3.3) Rearranging them, A . hQT TA1 A= (3.4) L .A . kQT T2 1= (3.5) A . hQT TBB 2= (3.6) To sum up the above equations ( ) ( ) ( )( )||.|

\|+ += + + = + + B AB AB AB 2 2 1 1 Ah1kLh1A . T TQA . hQkAL QA . hQT T T T T T (3.7) 4.Conduction through a composite wall. 4.1Composite wall not including surface heat transfer 69 Heat flow rate is constant through each layer wall Q Q Q Q3 2 1 = = = For layer 1: ( )A k. Q T TT T. A . k Q11B A1A B1= = Also A . kQ T TA . kQ T T33D C22C B= = Summing up the above equations, giving ( )D A332211332211D AT T .k k kAQk k k.AQT T||.|

\|+ +=

+ + = Definition: T . A . U Q A = U :overall heat transfer coefficient( ) K . m / W2 For a composite wall == + +=n1 iii332211k U1k k k1U (n number of composite walls) 70 Electrical analogy Thermal resistance From Ohms Law: RVResistance ElectricalVoltage Potentiall current Electrical = = Similarly Resistance ThermalT Difference eTemperaturQ flow HeatA= For the same composite wall problem, an alternative method can be used. Layer 1: ( )||.|

\|= ==A . kT TA . kT T T TA k Q11B A11A B1A B1 1 Similarly ||.|

\|=||.|

\|=A . kT TQ ;A . kT TQ33D C322C B2 In these equations, each denominator can be regarded as a thermal resistance for each layer. i.e. ||.|

\|=||.|

\|||.|

\|=A . kR andA . kR andA . kR333222111 For overall thermal resistance (in series) D A332211T3 2 1 TT T T ) (potential difference etemperatur totalA . k A . k A . kRR R R R resistance Total = + + = + + =A By using Electrical Analogy ( )332211D ATD Ak k kT T . ART TQ + +== For overall heat transfer coefficient U 71 332211k k k U1T . A . U Q + + = = A 4.2Composite wall including surface heat transfer coefficients From Newtons Law: ( )1 filmfluid theof resistance thermal ... .......... ..........A . h1RA . h1T TT T . A . h Q1f 11A 1A 1 1||.|

\|=||.|

\|= = Similarly 2 filmfluid theof resistance thermal ... .......... ..........A . h1RA . h1T TQ2f 222 D||.|

\|=||.|

\|= 72 ( )131 i111tt2 33221112 12 3322111h1k h1U1ARAR1U t coefficien transfer heat Overallh1k k k h1T T . AR . totalTQA h1A k A k A k A h1Resistance Total+ + = == ||.|

\|+ + + += = + + + + = = A Worked example: A furnace wall consists of 125mm thick refractory brick and 125mm thick insulating firebrick separated by an air gap.The outside wall is covered with a 12mm thickplaster.The inner surface of the wall is at 1100oC and room temperature is 25oC. Calculate Q(heat loss) per m2 of the wall surface. Solution: Resistance of the refractory brick ( ) W / K 0781 . 01 x 10 x 6 . 1125A . kl3= = = 2m 1 A Area : Note = =Resistance of the insulating firebrick ( ) W / K 417 . 03 . 010 x 125A . kl3= = = 73 Resistance of the plaster ( ) W / K 0857 . 014 . 010 x 12A . kl3= = = For fluid film between outside wall and air: Resistance of air film on the outside wall ( ) W / K / 0588 . 0171A . hl= = = Hence, total resistance ( ) K/W 0.8 gap) (air 16 . 0 0588 . 0 0857 . 0 417 . 0 0781 . 0 RT= + + + + = = Heat flow rate ( )( )2tB Atm per lossheat of rate. .......... .......... kW 344 . 1w 13448 . 025 1100RT TRTQ==== = A For interface temperature 3 2 1T , T , Tand outside wall temperature 4T Layer 1:Refractory brick C 995 T 13440781 . 0T 1100QO11= == Similarly C 220 T 1344417 . 0T TQC 780 T 134416 . 0T TQO33 2O22 1= === == For air film C 1048 T134417 / 125 TQO44=== 74 5.Heat Flow Through a Cylinder Radial Heat Transfer 5.1Heat transfer through a single cylinder Assumptions: one- dimensional flow temperature varies radially only The length of the cylinder is Fouriers Law drdTA . k Q =(r any radius) The area normal to the heat flow r 2 A t = ( r . 2t = circumference) drdT. k . . r . 2 Q t = ) stateflow steady ( . const : QT T : T Forr r : r Foro io 1 Integrating both sides 75 ( )||.|

\| = = ioei oTTrrrrlogT T . k . . 2QdT . 2 . k drr1. Qoioitt or ( )||.|

\| = ioni orrT T . k . . 2Qt It can be seen that the heat transfer rate depends on ratio of the radii i or / rinstead of thedifference) r / r (i o. Also for thermal resistance of the cylinder ( )( )

=k . . 2r / r InT TQi oi ot Thermal resistance( )k . . 2r / r InRi o t= 5.2Composite tube (including surface heat transfer) 76 ( )w fT T . A . h Q = h:surface heat transfer coefficient fT :fluid temperature wT :wall temperature It should be noted that h is a function of many parameters(wall roughness,k . . T . p uand fluid. For inner cylinder film ( )( ) resistance filmh r . 21Rh . r . 2QT Ti 1ii 1fi witt= = For outer cylinder film ( )( ) resistance filmh r . 21Rh . r . 2QT To 4oo 4wo fott= = For a composite wall ( )( )( )( )( )( )( )( )( )( )( )( )33 4 n3 wo3 4 n3 wo 322 3 n2 32 3 n2 3 211 2 n2 wi1 2 nwi 2 1k . 2r / r lT Tr / r lT T k . . 2Qk . 2r / r lT Tr / r lT T k . . 2Qk . 2r / r lT Tr / r lT T k . . 2Qtttttt= == == = to sum up the above equations ( ) ( ) ( )||.|

\|+ + + + = =o 4 33 4 n22 3 n11 2 ni 1fo fih r . 21k . 2r / r lk . 2r / r lk . 2r / r lh . r 21Q T T T t t t t tA From Fourniers Law and electrical analogyresistance totalRTQt = A 77 ( ) ( ) ( )||.|

\|+ + + + =o 4 33 4 n22 3 n11 2 ni 1th r . 21k . 2r / rk . 2r / rk . 2r / rh . r 21R Total In general ( )

+ + == o on1 n nni no ni ith . r1kr / rh . r121R :length of cylinder of tube ih :inside film coefficient oh :outside film coefficient For a different form ( )( )( )

+ + =

+ + = = ====o o n nni no ni io o n nni no ni ixXX Xttx x Xxx xr h1kr / rr h1r h1kr / rr h12r . 2U1A . UR1RTQradius chosen any isr r 2 At coefficien transfer heat overall UT A . U Q 78 Worked example: Material A is a five times better insulating material than material B i.e.A Bk 5 k =i.e. A is a better insulator, but a poorer conductor) For Case 1:Material A is inside of material B ( ) ( )( ) ( )( )( )( )1 Casethefor resistance totalk . 2774 . 0505 . 0 / 075 . 0025 . 0 / 05 . 0k . 21R R Rk . 5 . 205 . 0 / 075 . 0k . 2r / rRk . 2025 . 0 / 05 . 0k . 2r / rRAnnAB A tAnB1 o nBAnA1 o nA= =

+ = + = = == = For Case 2:Material B is inside of material A For thermal resistance ( ) ( )( )( ) ( )( )2 casefor resistance totalk . 2544 . 0R R Rk . 205 . 0 / 075 . 0k . 205 . 0 / 075 . 0Rk 5 . 2025 . 0 / 05 . 0k . 2025 . 0 / 05 . 0RAA B'tAnAnAAnBnB= = + = = == = 2 and 1 casein T sametheforRTQtAA = % 70774 . 0544 . 01 casefor resistance Thermal2 casefor resistance ThermalRR2 casefor lossHeat1 casefor lossHeatt't= = = =2. casethan insulation better 30% is1 Case79 4. Radiation Radiation is a form of electromagnetic energy transmission which requires no transfermedium. The amount of energy transferred depends on the absolute temperature of the bodyand the radiant properties of the surface. Dark bodies found to emit or absorb a considerable high amount of radiant heatenergy as compared with light bodies. 4.1Black Body Is a body which emit or absorb the maximum level of the radiant energy. The energy transfer due to radiation from the surface of a black body can be calculated from 4AT Q o = where o :Is the Stefan-Boltzmann constant( ) )4 2 8K m / W 10 x 67 . 5 T :The body absolute temperature (K) A:The body surface area (2m ) Q:The rate of heat transfer (W) Net heat transfer between two black bodies: The rate of heat transfer from surface 1: 41AT Q o = The rate of heat transfer from surface 2: 42AT Q o = The net heat transfer: ( )4241 2 1T T A Q Q Q = = o In practice, a combination of heat transfer by radiation and convection is typical. 80 Worked example 4.1 In a storage radiator the surface of the core has an area of 0.5 m2 and temperature of 200C. If the outer cover has the same surface area and a temperature of 60C, calculate the rate of heat transfer taking place. Assume the surfaces to act as black bodies and the convective heat transfer coefficient between the surface of the core and the air to be 10 W/m2K Take the ambient temperature to be 20C. Solution: The rate of heat transfer is the sum of the heat transfer by radiation and convection from the core to the outer cover. i.e. C RQ Q Q + = heat transfer by radiation ( )( ) ( ) ( )W 4 . 1070273 60 273 200 5 . 0 X 10 X 67 . 5T T A Q4 4 84241 R=+ + = =o heat transfer by convection ( )( ) W 900 20 200 5 . 0 X 10Ta T A h Q1 c C= = = therefore the rate of heat transfer W 4 . 1970 900 4 . 1070 Q = + = 81 4.2Grey Body Radiators These are real bodieswhich absorb or emit radiant energy at a lower level than a black body.The radiant energy emitted by a real surface is less than that of the black body and is given by: A . T Q4co = where c : is a radiative property of the surface called emissivity The net rate of radiation heat exchange between two grey bodies at temperatures 1T and 2Trespectively is given by: ( )||.|

\| + = 11 1/ T T A Q2 14241 ( ) 1 ; T T A Q; 1 For42412 1< == = 82 Worked Example 4.2 An un-insulated steam pipe passes through a room in which the air and walls are 25oC.The outside diameter of the pipe if 70mmand its surface temperature and emissivity are 200oC and 0.8 respectively.If the coefficient associated with free convection heat transfer from the surface to the air is 15 W/m2K, what is the rate of heat loss from the surface per unit length of pipe? Solution: Heat loss from the pipe is by convection to the room air and by radiation exchange with the walls. ( )( )DL A withRadiation T T A QConvection T T hA QQ Q Q4341 R2 1 CR C= = =+ = ( )( ) ( )( )4341 2 1T T DL T T DL h Q + = The heat loss per unit length of pipe is then ( )( ) ( )( )m / W 998 421 577298 473 07 . 0 x 10 x 67 . 5 x 8 . 0 25 200 07 . 0 x 15LQ4 4 8= + = + = Comments: Note that temperature may be expressed in two ways (i.e. in oC or K units) whenevaluating the temperature difference for a convection (or conduction) heat transfer. However, temperature must be expressed in Kelvins (K) when evaluating a radiation heat transfer. 8 . 01 Assume12= ==c cc 83 Section 5:Tutorial Sheets 84 1.Perfect Gases 1/1 Calculate the mass of: i)A molecule of Nitrogen, N2 ii)An atom of Oxygen iii)A molecule of Octane( )18 8H C (Ans: 46.48x10-27 , 26.56x10-27 , 189.24x10-27 kg) 1/2 Calculate the Molar Mass and the Characteristic Gas Constant for: i)Molecular Hydrogen ii)Molecular Oxygen iii)Argon (Inert Gas of relative atomic mass = 40) iv) Air (Assume to be 21% O2 and 79% N2 by volume) (Ans: 2.0, 32.0, 40.0 and 28.84 kg/kgmol; 4157.0, 259.8, 207.9 and 288.3 J/kgK) 1/3 Using the data given in question 2, calculate the composition of air by mass. (Ans: 23.3% O2, 76.7% N2) 1/4 A tank containing a quantity of air is fitted with a "U" tube type manometer and athermometer, which show readings of 0.5 MPa and 27C, respectively. If a barometer situated close by reads 742 mmHg, calculate the absolute pressure and density of the air in the tank. (Data: Density of Mercury=13,600kg/m3 g=9.81 m/s2 R=287 J/kgK) (Ans:598kPa,6.945 kg/m3) 1/5a)Consider 1kgmol of O2 at pressure and temperature 1.0 Bar and 0oC, and calculate i)the overall volume ii)the specific (by mass) volume b)Would these results have been any different if we had considered a different gas say H2? (Ans:a)22.7m3,0.709m3/kg b)22.7m3,11.350m3/kg ) 85 2.Daltons Law 2/1 Two metal tanks, each of volume 2:5 m3; contain 4,kg of Oxygen and 5 kg ofNitrogen, respectively. All the contents of one tank are pumped into the otherand then sufficient time is allowed for the temperature to reduce back to theambient value of 300 K. Consider the gas mixture and calculate: i)the absolute pressure ii)the molar fractions of the two constituents (Ans: 3.029 Bar, 0.412, 0.588) 2/2 Combustion of a hydrocarbon fuel in Oxygen gives the following reaction: Fuel + Oxygen 7CO2 + 8H2O + 17.6 O2 i)Calculate the volumetric analysis of the products (Ans: 21.5%, 24.5%, 54.0%) ii)Calculate the ultimate analysis (by mass) of the products (Ans: 30.3%, 14.2%, 55.5%) iii)What is the density of the products if the combustion chamberpressure and temperature are 30 bar and 1600 K, respectively? (Ans: 7.02 kg/m3) 2/3 Consider air saturated with water vapour (which at low pressures can beconsidered to be a perfect gas). Given that the pressure and temperature of the air/water mixture is 1.0 Bar and 15C, calculate the percentage by mass of water given that: Molar mass of dry air = 28.84 kg/kgmol K Partial Pressure of water vapour at 15C = 1780 Pa (Ans: 1.10%) 86 3.Specific Heats 3/1At 300 K the value of for air is 1.40 whilst at 1070 K this reduces to 1.33. Calculate the corresponding values of ovC and opC in terms of J/kgmol K. (Ans: 300 K:20785 and 29099 J/kgmol K 1070 K:25194 and 33508 J/kgmol K) 3/2Given that the molar mass for air is 28.84 kg/kgmol, re-calculate the values of Cv and Cp at 300 K, in terms of J/kg K. (Ans: 720.8 and 1009 J/kg K) 3/3a)The temperature of 5kg of air is increased from 273 K to 773 k. Assuming constant values of Cp.and Cv of 1005 and 718 J/kg K, calculate the heattransfer if the process is carried out i)at constant volume (Ans: 1.795 MJ) ii)at constant pressure (Ans: 2.513 MJ) b)Why is the answer to part ii) above, greater than that for part i)? 87 4.First Law for Closed Systems 4/1 A system receives 100 kJ of heat whilst it does work of an amount 125 kJ on. the surroundings. Is this possible? Why? 4/2 The internal energy of a system increases by 50 kJ whilst the system is receiving 40 kJ of work. How much heat is transferred and in what direction? 4/3 A system works in a cycle and performs 100 kJ of work. What is the change in internal energy and how much heat has to be transferred into the system? 88 5.Work and Heat Transfer Equations Closed Systems 5/1 A closed thermodynamic system contains a perfect gas. Derive expressions for the work and heat transfer for each of the following REVERSIBLE processes (Cp and Cv are to be assumed constant). a)Constant Pressure process(Show that your answers are consistent with the first law) b)Constant Volume process(Again, demonstrate first law consistency) c)Isothermal process(i.e. Temperature is constant) d)Adiabatic process - i.e. q = 0(Formula pv= constant) e)Polytropic process(Formula npv= constant where n= ) 89 6.Numerical Problems Closed Systems 6/1 The pressure and-temperature of the-air in a cylinder are-1x105N/m2 and 50C. The air is compressed according to the law pv1.3 = constant until the pressure is8.25x105N/m2. The volume of air initially is 0.042m3.Find(a) the mass of air in kg,(b) the temperature at the end of the compression,(c) the work done in compression,(d) the heat transfer between the gas and its surroundings. For air = 1.4, Cv= 0.718kJ/kgK,R=0.287 kJ/kgK (Ans: 0.045 kg, 526 K, -8.79 kJ, -2.09 kJ) 6/2 0.34m3 of gas at 10.0 Bar and 130C expands reversibly and adiabatically until its pressure is 1 Bar after which it is compressed isothermally to its original volume.

Find the final temperature and pressure of the gas and the change in internal energy. The specific heats of the gas at constant pressure and volume are 1.005 kJ/kg K and 0.718 kJ/kg K. (Ans: 208.6 K, 5.176x105N/m , -410 Id) 6/3 A cylinder fitted with a piston contains air at 1.0 Bar and 17C. The gas is compressed according to the law PVn = const., until the pressure is 4 bar when the specific volume is found to be 28% of the initial value. Heat is then added to the air at constant pressure until the volume is doubled. The same amount of heat is nowremoved from the air at constant volume. Determine the value of the index n in the compression process.Find also(a) the overall change in internal energy/kg of air and(b) the final pressure of the air. R = 0.287 kJ/kgK, Cp = 1.005kilkgK (Ans:1.09,-68kJ/kg,1.2 x 105N/m2 90 7.Numerical Problems Open Systems 7/1 A vehicle is fitted with a forward facing engine .intake and is run on a day when ambient pressure and temperature are 1.0 Bar and 20C. Calculate thestagnation temperature and stagnation pressure in the intake system for forward speeds of 25, 50, 75 and 100 m/s (57, 114, 170 and 227 mph, respectively).kgK / J 1005 Cp = (Ans: 293.31, 294.24, 295.80 and 97.98(K) 1.004, 1.015, 1.034 and 1.061 (Bar)) 7/2 Air enters a centrifugal compressor at a pressure and temperature of 1.0 Bar and 15C and leaves at 2.0 Bar and 95C. If the mass flow is 55 kg/min., find: a)The actual power required to drive the compressor (73.7kW) b)The power required to produce the same pressure ratio if thecompression had been frictionless (58.1kW) c)The rate of increase of enthalpy due to friction (15.6kW) (Neglect kinetic energy changes and assume that y = 1.4 andCp = 1.005kJ/kgK) 7/3 9kg of air per minute enters a nozzle with negligible velocity and expands from a pressure of 4.13x105Pa to 2.27x105Pa. The temperature falls from 900C to750C in the process. Assuming= 1.4 and Cp = 1.005kJ/kgK find a)The velocity of the air leaving the nozzle (549m/s) b)The velocity which would have been reached if the process had been frictionless (609m/s) c)The nozzle efficiency [=Actual Enthalpy Change/Ideal EnthalpyChange] (0.813) d)The nozzle exit area (0.353x10-3m2) 7/4 The flow rate through a turbine is 11300 kg/h and the inlet and exit velocities are 1830m/min and 7610m/min respectively. If the initial and final enthalpy values are 2790 kJ/kg and 2090 kJ/kg and the heat loss through the casing amounts to 36.7 kJ/sec; calculate the shaft power generated. (2137kW) 91 8.Second Law of Thermodynamics 8/1An inventor claims to have devised a steady flow compressor which requires no shaft power input. He claims that carbon dioxide at 13.6 bar and 49C can becompressed to 20.4 bar, where it will emerge at -6C, simply by transfer of energy as heat from his device. His patent application states that the device will handle 2kg of carbon dioxide per second and is driven by a `cold source' at -95C towhich the heat transfer rate is 92.4 kW. Carry out a critical analysis of hisinvention: a)by investigating whether the claims are consistent with the first law, and b)by calculating the total entropy change during the process in order to see if the second law has been violated. (For CO2 take R = 0.190 kJ/kgK and Cp = 0.84kJlkgK) 8/2A heat engine operates in a Carnot Cycle, exchanging heat energy with hot and cold reservoirs of temperature 1000K and 400K. If the heat flow rate from the hot reservoir is 50 kW, calculate the rate of heat rejection (do not resort to acalculation of cycle efficiency in order to do this). Is it possible to have an engine which operates between the same two reservoirs, also receiving heat at a rate of 50 kW, but rejecting 10% less heat than thatcalculated above? Briefly explain your answer by making reference to: i)The Entropy Principle, and ii)The concept of Carnot Efficiency 92 9.Air Standard Cycles 9/10.5 Kg of air at 1 Bar and 15C is compressed isothermally to 10 Bar. It is then heated at constant pressure and finally expanded isentropically to its original condition. Calculate the mean effective pressure and the efficiency of the complete cycle. ) 295 . 0 , Bar 065 . 1 MEP (4 . 1 , K kg / kJ 287 . 0 R= == = 9/2In a gas turbine plant the pressure ratio is 6.0. Calculate for an inlet temperature of 15C and a maximum temperature of 1200K. a)The heat supplied per Kg of air. b)The net work output per Kg of air. =1.4,Cp = 1.0, KJ/Kg K.(a = 719 KJ/Kg, b = 289 KJ/Kg). 9/3Show that the ideal MEP of the Otto cycle is given by ( )( )( )( ) 1 1 rr 1 prv1v v1 o wherep = Pressure at beginning of compression. rv = Compression ratio. = Ratio of maximum pressure to pressure at the end of compression. 9/4A quantity of gas at a pressure of 1 Bar in a cylinder has a volume of 0.14 m3 at a temperature of 67C. It is taken through the following cycle of operations: a)Isothermal compression until the pressure is 15 Bar b)Heat addition at constant volume, the temperature being raised by 1100K. c)Heat addition at constant pressure, the volume being increased by 50%. d)Isentropic expansion to its original volume e)Cooling at constant volume so that the gas is brought to its initial state. Find the pressure and temperature at the end of each stage of thecycle4 . 1 = ( ) Bar 51 . 2 p , K 858 T , Bar 5 . 63 p , K 2160 T , Bar 5 . 63 p , K 1440 T , Bar 15 p , K 340 T5 5 4 4 3 3 2 2= = = = = = = = 93 9/5An engine working on the constant volume cycle has a compression ratio of 6 to 1 and both the compression and expansion are according to the law 3 . 1pv = const. The conditions at the commencement of compression are, pressure 0.98 Bar, temperature 323 K.The maximum pressure at the end of constant volumecombustion is 29.0 Bar.Assuming that the working fluid is air, estimate: a)The mean effective pressure.' b)The heat supplied per Kg during combustion. c)The heat passed to or from the cylinder jackets per Kg of air during both compression and expansion, stating the direction of flow in each case. ( ) expansion during received 158KJ/Kg n, compressio during rejected 55KJ/Kg KJ/Kg, 746 , Bar 24 . 5 4 . 1 , KgK / KJ 287 . 0 R , KgK / KJ 718 . 0 Cv= = = 9/6A Diesel engine has a compression ratio of 14 to 1 and the fuel is cut off at 0.08 of the stroke.If the relative efficiency is0.52, estimate the mass of fuel of calorificvalue.43,200 KJ/Kg which would be required per hour for 1 kw output.(.272 Kg/Hr). 9/7An air-standard Diesel cycle has a compression ratio of 15 and the heat transferred to the working fluid per cycle is 1850 KJ/Kg.At the beginning of the compressionprocess, the pressure is 1.013 Bar and the temperature is 15oC. Determine: a) The pressure and temperature at each point in the cycle. b) The thermal efficiency. c)The mean effective pressure K Kg / KJ 718 . 0 c , 4 . 1v = = ( ) 13.4Bar 55.1%, 1444K, Bar, 5.082692K, 851K, Bar, 44.9 9/8A compression ignition engine working on the dual combustion cycle has acompression ratio of 10.5 to 1, and two-thirds of the heat of combustion is liberated at constant volume, the remainder at constant pressure. The maximum pressure is 44.1 Bar and the pressure and temperature at the start of compression are 0.93 Bar and 49oC. The index of compression and also of expansion is 1.33 and the working fluid may be assumed to be air throughout the cycle. Find: a) The thermal efficiency of the cycle. b)The mean effective pressure. (52.3, 5 Bar). 94 10.Heat Transfer 10/1A pipe containing a gas at a temperature of 195C has an external radius of 75mm. It is lagged to a radius of 150mm with asbestos of conductivity 0.