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THERMAL, ENGINEERING - I, (N–SCHEME), , N. IYANARAPPAN,, , M.E., M.I.S.T.E., , !, !, !, !, !
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Unit – I, Chapter 1., , BASICS OF THERMODYNAMICS, , 1.1 Introduction, Thermodynamics is a field of science, which deals with the, energies possessed by gases and vapours. It also includes the, conversion of these energies into heat and mechanical work, and, their relationship with properties of the system. A machine which, converts heat energy into mechanical work is known as heat, engine., 1.2 Definitions and Units, a) Mass (m) : Mass is the amount of matter contained in a, given body. Mass does not change from place to place. It is, expressed by the symbol 'm'. In SI system, the unit of mass is 'kg'., b) Weight (W) : Weight is the amount of force acting on, the mass of a body due to gravitational acceleration. Since the, gravitational pull varies with the distance of the body from the, centre of the earth, therefore weight of the body will also vary with, its position on the earth's surface. It is expressed by the symbol 'W'., In SI system, the unit weight is N (newton) or kN., c) Force (F): According to Newton's second law of motion,, the applied force is proportional to mass and acceleration. Force is, expressed by the symbol 'F'. In SI system, the unit of force is, N (newton) or kN., Mathematically, Force, F ∝ ma or F = kma, Where, k is a constant of proportionality, For the sake of convenience, the unit of force adopted is such, that it produces a unit acceleration to a body of unit mass i.e. k=1, ∴ F = ma = mass × accelerati on, Unit – I , ,, , 1.1
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d) Density (ρ) : The density of a substance is defined as, the mass per unit volume of the substance. It is expressed by the, symbol 'ρ'. In SI system, the unit of density is ' kg/m3 ', mass, m, =, Density, ρ =, ( kg / m3 ), volume V, Density is also known as mass density or specific mass., Density of water is 1000 kg/m3 ., e) Specific weight (w) : Specific weight of a substance is, defined as the weight per unit volume of the substance. It is, expressed by the symbol w. In SI system, the unit of specific weight, is N/ m 3, , weight W mg, =, =, (N/ m 3 ), volume V, V, Specific weight of water is 9810 N/ m 3 ., Specific weight, w =, , f) Specific volume (v) : Specific volume of a substance is, defined as the volume occupied by unit mass of the substance. It is, the reciprocal of density. It is expressed by the symbol v. In SI, system, the unit of specific volume is m 3 /kg., Specific volume, v =, , volume V, =, ( m 3 /kg), mass, m, , g) Specific gravity (s) : Specific gravity of a substance is, defined as the ratio of the density (or specific weight) of that, substance to the density (or specific weight) of a standard, substance. For liquids, water is taken as standard substance ., Specific gravity is expressed by the symbol 's'. Since it is the ratio, of two same quantity, it has no units., Specific gravity,, density (or specific weight) of the given substance, s=, density (or specific weight) of the standard substance, Unit – I , ,, , 1.2
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1.3 Pressure, The pressure is defined as the normal force per unit area., Pressure is expressed by the symbol 'p'., Force F, =, Pressure, p =, (N/ m 2 ), Area A, Unit of pressure : The unit of pressure depends upon the, units of force and area. In SI system, the practical unit of pressure, is N/ m 2 , N/m m 2 , kN/ m 2 , bar, Pascal (pa), etc., 1 Pa = 1 N/ m 2 ; 1 Mpa = 1 × 106 N/ m 2 ., 1 bar = 1 × 105 N/ m 2, Gauge pressure, Absolute, pressure, , Atmospheric, pressure, , Vacuum pressure, Atmospheric, pressure, , Absolute, pressure, , Fig.1.1 Pressure relationship, , Atmospheric pressure (patm) : It is the pressure exerted, by the air on the earth's surface. The value of atmospheric pressure, at Mean Sea Level is 1.03125 bar or 760 mm height of mercury in, a barometer., Atmospheric pressure, patm = 1.01325 bar, = 1.01325 × 105 N/ m 2, = 101.325 kN/ m 2, = 760 mm of Hg, =10.34 m of water, Absolute pressure (pabs) : The actual pressure in any, system is known as absolute pressure., Gauge pressure (pg) : It is the pressure recorded by the, pressure gauge. All the pressure gauges read the difference, between the actual pressure and the atmospheric pressure., Absolute pressure = Atmospheric pressure + Gauge pressure, Pabs = patm + pg, Unit – I , ,, , 1.3
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Vacuum pressure (pg) : The pressure which is below the, atmospheric pressure is known as vacuum pressure or negative, pressure. Vacuum gauges are used to record this pressure. In this, case,, Absolute pressure = Atmospheric pressure – Vacuum pressure, 1.4 Temperature, Temperature is a thermodynamic property, which, determines the degree of hotness or the level of heat intensity of a, body. It may also be defined as a measure of velocity of fluid, particles. The Temperature of a body is measured by an instrument, known as thermometer. The commonly used scales for measuring, the temperature of a body are :, 1. Celsius or centigrade scale, 2. Fahrenheit scale, Celsius or centigrade scale : This scale was first used by, Celsius in 1742. In this scale, the freezing point of water is taken, as zero and the boiling point of water is taken as 100 o C . The space, between these two points has divided into 100 equal divisions and, each division represents one degree Celsius., Fahrenheit scale : In this scale, the freezing point of water, is taken as 32oF and the boiling point of water is taken as 212oF. The, space between these two points has divided into 180 equal divisions, and each division represents one degree Fahrenheit., The relationship between Celsius scale and Fahrenheit, 5, scale is given by ; C = ( F − 32), 9, 1.5 Absolute temperature, The zero readings of Celsius and Fahrenheit scales are, chosen arbitrarily for the purpose of simplicity. But, whenever the, value of temperature is used in equations relating to fundamental, laws, true zero or absolute zero temperature should be used., Absolute zero temperature is the temperature below which the, temperature of any substance cannot fall., Unit – I , ,, , 1.4
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The absolute zero temperature is taken as –273 o C in the, case of Celsius scale. The temperature measured from this zero, temperature is called absolute temperature. In Celsius scale, the, absolute temperature is called degree Kelvin (K). In SI units,, degrees Kelvin is not written as o K but only K., , ∴ K = oC + 273, , Note : For thermodynamics calculations, all the temperature, values must be converted into Kelvin (K) before substitution in the, formula. For example, if the temperature is given as 100 o C , then, the absolute temperature = 100 + 273 = 373 K, 1.6 Standard Temperature and Pressure (S.T.P), The conditions of temperature and pressure of any gas, under standard atmospheric conditions are termed as Standard, Temperature and Pressure (S.T.P). S.T.P Values are,, Standard temperature = 15 o C = 288K, Standard pressure = 760 mm of Hg = 101.325 kN/ m 2 ., 1.7 Normal Temperature and Pressure (N.T.P), The conditions of temperature and pressure of any gas, under normal atmospheric conditions are termed as Normal, Temperature and Pressure (N.T.P). N.T.P Values are,, Normal temperature = 0 o C = 273K, Normal pressure = 760 mm of Hg = 101.325 kN/ m 2 ., 1.8 Heat, The heat is defined as the energy transferred across the, boundary of a system due to the temperature difference between, the system and the surroundings. It is represented by the symbol, Q. In SI system, the unit of heat is joule (J) or kilo-joule (kJ)., If m kg of substance is heated from an initial temperature, T1 to a final temperature T2 , then the heat transfer is given by,, Q = m.C .(T2 − T1 ), , where, C = specific heat of the substance (J/kg.K), •, •, , If Q is positive, heat is supplied to the system (heating process), If Q is negative, heat is rejected from the system ( Cooling process), Unit – I , , ,, , 1.5
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1.9 Specific heat capacity (C), The specific heat capacity of a substance is defined as the, quantity of heat transfer required to raise or lower the temperature, of the unit mass of the substance through one degree. It is, represented by the symbol C. In SI system, its unit is given as, J/kg.K or kJ/kg.K., Specific heat capacity at constant volume (Cv): It is, defined as the quantity of heat transfer required to raise or lower, the temperature of the unit mass of the substance through one, degree when the volume remains constant. It is represented by the, symbol Cv. When a gas is heated or cooled at constant volume, the, heat transfer is given by,, Q = m.Cv.(T2 − T1 ) kJ, where, Q = Heat transferred (kJ), m = Mass of the gas (kg), Cv = Specific heat capacity at constant volume, T1 = Initial temperature of the gas (K), , T2 = Final temperature of the gas (K), Specific heat capacity at constant pressure (Cp): It is, defined as the quantity of heat transfer required to raise or lower, the temperature of the unit mass of the substance through one, degree when the pressure remains constant. It is represented by, the symbol Cp. When a gas is heated or cooled at constant volume,, the heat transfer is given by,, , Q = m.C p .(T2 − T1 ) kJ, where, Cp = Specific heat capacity at constant pressure, Note : C p, , is always greater than Cv . When a gas is heated at, , constant pressure, the volume of the gas increases. Thus work is, done by the gas by expanding. Hence heating in constant pressure, gives a higher value for the specific heat than heating in a constant, volume., Unit – I , ,, , 1.6
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Adiabatic index (γ): The ratio of the two specific heat, capacities remains constant and is called as adiabatic index. It is, represented by the symbol γ., Cp, Adiabatic index, γ =, Cv, For air: C p =1.005 kJ/kg.K;, , Cv = 0.718 kJ/kg.K; γ=1.4., , 1.10 Work, Work is defined as the product of force (F) and the distance, moved (x) in the direction of force. It is represented by the symbol, W. In SI system, the unit of work is N-m or joule (J)., Work done by the gas, , Pressure, , p1, , 1, , 2, , p2, , V1, Volume, , dV, , V2, , Fig.1.2 Work done by the gas, , Consider a piston and cylinder arrangement filled with a gas, as shown in the fig. Let the gas expands from state 1 to state 2., Let, p – Intensity of pressure of the gas on the piston (N/ m 2 ), A – Area of cross section of the piston ( m 2 ), Force on the piston, F = p. A, Consider a small distance dx moved by the piston. Then, the work done by the gas is given by,, dW = Force × distance = p.A.dx, , dW = p.dV ( A.dx = dV ; Change in volume), Unit – I , ,, , 1.7
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Total work done during the process 1 to 2 is given by,, 2, , , , 2, , dW =, , 1, , p.dV, 1, 2, , W1 − 2 =, , p.dV, , (N - m), , 1, , Hence, for any process, the mechanical work done is given by,, 2, , W=, , p.dV, , (N - m or J), , 1, , •, •, , If W is positive, work is done by the system (gas) and the, process is called expansion process, If W is negative, work is done on the system (gas) and the, process is called compression process., , 1.11 Power, Power may be defined as the rate of doing work or work, done per unit time. It is represented by the symbol P. In SI system,, the unit of power is Watts (W)., 1 W = 1 J/s = 1 N-m/s, 1 kN-m/s = 1kJ/s = 1 kW, 1 kW-s = 1 kJ; 1 kW-hr = 3600 kJ., 1.12 Thermodynamic system, boundary and surroundings, System : The thermodynamic system may be defined as a, definite area or a space where some thermodynamic process is, taking place., Surroundings : Anything outside the boundaries which, affects the behaviour of the system is known as surroundings., Boundaries : The system and the surroundings are, separated by the system boundary. The system boundary may be, real or imaginary., 1.13 Types of thermodynamic systems, Thermodynamic systems may be classified as follows:, 1. Closed system 2. Open system and 3. Isolated system, Unit – I , ,, , 1.8
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1. Closed system, A closed system permits the transfer of heat and work, across its boundaries; but it does not permit the transfer of mass., The mass of the working substance in a closed system remains, constant. The system boundary is determined by the space, occupied by the working substance., Example :, Weight, , Piston, , System (Gas), , Cylinder, Surroundings, , System boundary, , Fig.1.3 Closed thermodynamic system, , The piston and cylinder arrangement shown in the figure, is an example of closed system. The gas in the cylinder is, considered as system. If the heat is supplied to the cylinder, the, temperature of the gas will increase and the piston will move., As the piston moves, the boundary of the system moves. In, other words, the heat and work energy crosses the boundary of the, system during this process, but there is no addition or loss of the, original mass of the working substance., 2. Open system, Heat, H.P Air out, , H.P Air in, , System (Air Compressor), , Work, , Motor, , System boundary, , Fig.1.4 Open thermodynamic system, , Unit – I , ,, , 1.9
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In this system, the mass of the working substance crosses, the boundary of the system. Heat and work may also cross the, boundary. The mass within the system may not be constant during, the process. An open system may be called as control volume., Example :, The compressor unit shown in the figure is an example of, open system. In this system, the low pressure air enters the, compressor and leaves the high pressure air. Thus the mass of, working substance crosses the boundary of the system. The work, crosses the boundary of the system through the driving shaft and, the heat is transferred across the boundary from the cylinder walls., 3. Isolated system, A system which is not influenced by the surroundings is, called an isolated system. In an isolated system, there is no mass,, heat or work transfer takes place. This is an imaginary system., Example : An open system with an universe as its, surrounding is an example of an isolated system., 1.14 State of a system, The state of a system is the condition of the system at any, particular moment which can be identified by the statement of its, properties. The number of properties required to describe a system, depends upon the nature of the system., 1, , Pressure, , p1, , 2, , p2, , V1 Volume, System, (Gas), p1, V1, T1, , V2, , Cylinder, Piston, 1, , 2, , Fig.1.5 State of a system, , Unit – I 1.10, ,
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Consider a system (gas) enclosed in a cylinder and piston, arrangement as shown in the figure. Let the equilibrium state of, the piston at position 1 is represented by its properties p1 , V1 , and, , T1 . When the system expands, the piston moves towards right and, occupies the final position 2. The final equilibrium state is, represented by the properties p2 , V2 , and T2 . The pressurevolume (p-V) diagram indicating the initial and final states is also, shown in the figure., 1.15 Properties of a system, The quantities, which identify the state of a system, are, called properties. The thermodynamic properties of a system may, be classified as, a) Extensive properties, b) Intensive properties., a) Extensive properties : The properties of system, whose, value for the entire system is equal to the sum of their values for, the individual parts of the system, are called extensive properties., These are dependent on the mass of the system., Example : Total volume, total energy, total mass, etc., b) Intensive properties : The properties which are, independent on mass of the system. These properties remain same, in all individual parts of the system., Example : Temperature, pressure, specific volume, etc., 1.16 Thermodynamic process, Thermodynamic process is a path of change in state of a system, from one equilibrium state to another equilibrium state. The, different thermodynamic processes are:, • Constant volume process, • Constant pressure process, • Constant temperature process, • Isentropic or reversible adiabatic process, • Polytropic process, • Hyperbolic process, • Free expansion process, • Throttling process, Unit – I 1.11, ,
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1.17 Thermodynamic cycle or cyclic process, 2, , 3, , 2, , 3, , A, 2, , 4, , 1, , Volume, (a) Cyclic process, , Pressure, , B, Pressure, , Pressure, , 1, , Volume, (b) Closed cycle, , 1, Volume, (c) Open cycle, , Fig.1.6 Cyclic process, , When a process or processes are performed on a system in, such a way that the final state is identical with the initial state,, then it is said to be thermodynamic cycle of cyclic process., In fig.(a), 1-A-1 and 2-B-1 are processes whereas 1-A-2-B-1, is a thermodynamic cycle or cyclic process, Thermodynamic cycle may be classified as, (a) Closed cycle and, (b) Open cycle, (a) Closed cycle : In a closed cycle system, the working, substance is recirculated again and again in the system. Fig.(b), represents the closed cycle system gas turbine plant., (b) Open cycle : In an open cycle system, the working, substance is exhausted to atmosphere after expansion. Fig.(c), represents an open cycle gas turbine plant., 1.18 Point function and path functions, p1, , 1, , Pressure, , c, a, , b, 2, , p2, , V1 Volume, , V2, , Fig.1.7 Point and path functions, , Unit – I 1.12, ,, , 4
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Point function : A function, whose value is independent, of the path followed by the system, is known as point function. The, values depend only on the initial and final state of the system., Thermodynamic properties are point functions, since for a given, state, there is a definite value for each property., Example : Pressure, temperature, volume, etc., Path function : A function, whose value depends on the, path followed by the system, is known as path function., Example : Work done, heat transfer, etc., Consider a system undergoing a change of state from state, 1 to state 2., Let, p1 , V1 = Pressure and volume of system at state 1, , p2 , V2 = Pressure and volume of system at state 2, Change in pressure, dp = p2 − p1, Change in volume, dV = V2 − V1, The change of state from 1 to 2 may be carried out through, a number of paths such as 1-a-2, 1-b-2, 1-c-2, etc. It can be noticed, that the change in pressure or change in volume is the same, irrespective of the path followed by the system. Therefore, these, properties are point functions., Work done during the process 1-a-2 = Area 1-a-2- V2 - V1 -1, Work done during the process 1-b-2 = Area 1-b-2- V2 - V1 -1, Work done during the process 1-c-2 = Area 1-c-2- V2 - V1 -1, Thus the work done during a process depends on the paths, followed by the system. So, it is a path function., 1.19 Energy, The energy is defined as the capacity to do work. A system, possesses the following two types of energies:, a) Stored energy, b) Transit energy (or energy in transition), a) Stored energy : it is the energy possessed by a system within, its boundaries. The potential energy, kinetic energy and internal, energy are the example of stored energy., Unit – I 1.13, ,
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Potential energy : it is the energy possessed by a system, by virtue of its position above the ground level, Potential energy, PE = mgz, where, m = mass of the system, g = Acceleration due to gravity, z = Height of the system above the ground level, Kinetic energy : It is the energy possessed by a system, by, virtue of its mass and velocity of motion., 1, 2, Kinetic energy, KE = mv, 2, where, v = Velocity of the system., Internal energy : It is the energy possessed by a body or, a system due to its molecular arrangement and motion of the, molecules. It is generally represented by U., b) Transit energy or energy in transition : It is the energy, possessed by a system which is capable of crossing boundaries. The, heat, work and electrical energy are the examples of transit energy., 1.20 Law of conservation of energy, It states that, energy can neither be created nor destroyed,, but it can be transferred from one form to another form i.e. the total, energy in any system remains constant., In thermodynamics, it states that, the total heat, transferred in a system must be equal to the sum of the external, work done and the change in internal energy., Heat transfer = Work done + Change in internal energy, Q = W + ∆U, 1.21 Thermodynamic equilibrium, A system is said to be in equilibrium, if it does not tend to, undergo any change of state on its own accord. Any further, change must be produced only by external means., Unit – I 1.14, ,
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A system is said to be in thermodynamic equilibrium, if it, satisfies the following three requirements of equilibrium., Mechanical equilibrium : A system is said to be in, mechanical equilibrium, when there is no unbalanced forces acting, on any part of the system., Thermal equilibrium : A system is said to be in, mechanical equilibrium, when there is no temperature difference, between the parts of the system or between the system and, surroundings., Chemical equilibrium : A system is said to be in, mechanical equilibrium, when there is no chemical reaction within, the system., 1.22 Zeroth law of thermodynamics, It states that, when two systems are each in thermal, equilibrium with a third system, then the two systems are also in, thermal equilibrium with one another. This law provides the basis, of temperature measurement., A, , B, , C, , Fig.1.8 Zeroth law of thermodynamics, , As shown in the figure, systems B and C are in thermal, equilibrium with a third system A. Now, the systems B and C are, also in thermal equilibrium with each other., 1.23 First law of thermodynamics, It states that, when a system undergoes a thermodynamic, cycle, then the net heat supplied to the system is equal to the net, work done by the system., Mathematically,, , dQ = dW (or) dQ − dW = 0, , [ − cyclic integrals, represents the sum for a complet cycle], Unit – I 1.15, ,
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Consider a system undergoes a change of state from state, 1 to state 2. This law when applied to this process,, U1 + Q = U2 + W, , Q = W + U2 − U1 = W + ∆U, where, Q = Heat transferred, W = Work done, ∆U = Change in internal energy, 1.24 Correlation of First law of thermodynamics, 1. Perpetual Motion Machine (PMM) of the first kind is impossible, A perpetual motion machine (PMM) of the first kind is an, imaginary machine which delivers work continuously without any, heat input. In this case,, Q = 0; ∴ W must be equal to zero., But PMM of the first kind delivers work continuously, which is impossible. It creates energy without any input and thus, violates the first law of thermodynamics. Hence a perpetual motion, machine of the first kind is impossible., 2. If a closed system is isolated from its surroundings, there, is no change in internal energy of the system., For the isolated system, Q = 0 and W = 0, By first law of thermodynamics, Q = W + ∆U, 0 = 0 + ∆U, ∆U = 0 i.e. U 2 = U 1, , 1.28 Second law of thermodynamics, The second law of thermodynamics may be defined in many, ways. The two common statements are Kelvin-Planck statement, and Clausius statement., Kelvin-Planck statement, (i) It is impossible to construct a heat engine working on cyclic, process, whose only purpose is to convert all the heat energy, supplied to it into an equal amount of work, (or), Unit – I 1.16, ,
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(ii) No heat engine which is working on cyclic process can convert, more than a small fraction of the heat energy supplied to it into, useful work. A large part of it is necessarily rejected as heat., Explanation :, Source, T1, Source, Qs, Qs, , W = Qs - QR Heat, engine, T1 > T2, , W = Qs Heat, engine, , QR, , (a) Impossible heat engine, by second law, , Sink, T2, (b) Possible heat engine, by second law, , Fig.1.9 Kelvin-Planck statement, , Figure shows a heat engine working on a cyclic process. It, receives heat (Qs ) from a heat source and delivers work (W) equal to, heat received. By second law of thermodynamics, all the heat received, cannot be converted into useful work and a part of heat received is to, be rejected to a low temperature reservoir (sink). Hence a heat engine, should have two heat reservoirs at different temperatures for, converting continuously heat energy into useful work., Clausius statement, (i) Heat can flow from a hot body to a cold body without any, assistance. But heat cannot flow from a cold body to a hot, body without any external work., (or), (ii) It is impossible to construct a machine working on a cyclic, process whose only purpose is to transfer heat from a cold, body to a hot body., Explanation :, Consider a heat source at higher temperature T1 and a heat, sink at a lower temperature T2 . By Clausius statement heat cannot, flow from the sink to heat source without any external work., Unit – I 1.17, ,
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Source, T1, , Source, T1, , Q, , T1 > T2, , Q1=W+Q2, W=Q1-Q2, , Heat, pump, , Heat, pump, , T1 > T2, , Q, , Q2, , Sink, T2, , Sink, T2, , (a) Impossible system, , (b) Possible system, , Fig.1.10 Clausius statement, , But heat can flow from sink to the source with the help of, an external work through a device such as heat pump., 1.29 Combination of Kelvin-Planck and Clausius statement, Consider a system consisting of a heat engine and heat pump, working in the same temperature range as shown in the figure., Heat source, T1, T1 > T2, Q1, , Q2, Heat, pump, , Heat source, T1, , Heat W=Q1-Q2, engine, , Q2, , Q2, , Q1, , Heat, pump, , Heat, engine, , Q2, , Q2, , T1 > T2, , Q2, , Heat sink, T2, , Fig.1.11 Equivalence of Kelvin-Planck and Clausius statement, , The heat engine receives heat Q1 from the source and rejects, heat Q2 to the heat sink and develops a net work W. The net work, done is given by, W = Q1 − Q2 . This follows Kelvin-Planck statement., The heat pump transfers heat, , Q2, , from the low, , temperature sink to high temperature source without any work, input. This violates the Clausius statement., Unit – I 1.18, ,
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The heat engine and the heat pump can be combined by, eliminating the sink. The heat rejected by the engine (Q2 ) is taken, by the pump. The equivalent system is shown in fig.(b). This, system acts an engine transferring heat from only one reservoir, and converting it into equivalent amount of work. This violates, Kelvin-Planck statement., Hence violation of Clausius statement also violates KelvinPlanck statement., Consider another system as shown in fig.(a). The heat, engine receives a heat of Q1 from the source and converts in into, equivalent amount of work W. This violates Kelvin-Planck, statement., Heat pump transfers a heat of Q2 from the sink and, supplies a heat of Q2 + W to the source. It receives a work of W, from the heat engine. It is not contrary to Clausius statement., An equivalent system for this is shown fig. (b). This system, transfers heat Q2 from the sink at the lower temperature to the, source at higher temperature without the aid of external work., This violates the Clausius statement., From the above examples, we can understand that both, Kelvin-Planck and Clausius statement of second law or, thermodynamics are same. Violation of one statement will violates, the other statement., 1.30 Perpetual Motion Machine (PMM) of the second kind, A perpetual motion machine (PMM) of the second kind, receives heat continuously from a single reservoir and converts it, into equivalent amount of work. Thus it gives 100% efficiency. This, violates the second law of thermodynamics. Hence a PMM of the, second kind is impossible to construct., , Unit – I 1.19, ,
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1.31 Perfect gases, A perfect gas may be defined as a state of substance, whose, evaporation from its liquid state is complete, and strictly obey gas, laws under all conditions of temperature and pressures. In actual, practice, there is no gas which strictly obeys the gas laws over the, entire range of temperature and pressure. But the real gases which, are ordinarily difficult to liquefy, such as oxygen, nitrogen,, hydrogen and air may be considered as perfect gases within certain, temperature and pressure limits., 1.32 Laws of perfect gases, The behaviour of perfect gas is governed by the following, gas laws:, • Boyle's law, • Charles’s law, • Joule's law, • Regnault's law, • Avagadro's law, a) Boyle's law, Boyle's law states that, the absolute pressure of a given, mass of a perfect gas is inversely proportional to its volume, when, the temperature remains constant., 1, or pV = Constant, Mathematically, p ∝, V, i.e. p1V1 = p2V2 = C, where, p1 = Original pressure of the gas (N/ m 2 ), , V1 = Original volume of the gas ( m 3 ), p2 = Final pressure of the gas after change of state (N/ m 2 ), V2 = Final volume of the gas after change of state ( m 3 ), T1 = T2 = T = Constant temperature of the gas, b) Charles’s law, Charles’s law states that, the volume of a given mass of, perfect gas is directly proportional to its absolute temperature,, when the pressure remains constant., Unit – I 1.20, ,
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Mathematically, V ∝ T, i.e., , (or), , V, = Constant, T, , V1 V2, =, =C, T1, T2, , c) Gay-Lussac law, Gay-Lussac law states that, the absolute pressure of a, given mass of perfect gas is directly proportional to its absolute, temperature, when the volume of the gas remains constant., p, Mathematically, p ∝ T (or), = Constant, T, p, p, i.e. 1 = 2 = C, T1, T2, d) Joule's law, Joule's law states that, the change of internal energy of a, perfect gas is directly proportional to the change in temperature., Mathematically, dE ∝ dT (or) dE = m.C.dT = m.C(T2 − T1 ), where, m = Mass of the gas, C = A constant of proportionality, known as specific heat., e) Regnault's law, Regnault's law states that, the two specific heats C p and, , Cv of a perfect gas do not change with the change in temperature,, i.e. C p and Cv of a gas always remain constant., f) Avagadro's law, It states that, equal volumes of all gases, at the same, temperature and pressure, contain equal number of molecules., Mathematically, M .v = Constant, where, M = Molecular weight of the gas, v = Specific volume of the gas., M.v is known as molar volume and it is represented as Vmole ., , The molar volume of any gas at N.T.P [ o C and 1.01325 bar] is, given by Vmole = 22.4 m3 / kg.mole, , Unit – I 1.21, ,
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1.32 General gas equation, The general gas equation can be obtained by combining, Boyle's law and Charles’s law., 1, 1, (or) V ∝, According to Boyle's law, p ∝, ……. (1), V, p, According to Charles’s law, V ∝ T, 1, T, Combining (1) and (2) V ∝ and T (or) V ∝, p, p, i.e. pV ∝ T (or) pV = CT, , ……. (2), , The more useful form of general gas equation is, p1V1, pV, = 2 2 = Constant, T1, T2, 1.33 Characteristic equation of gas, It is the modified form of general gas equation. If the volume, (V) in the general gas equation is replaced by specific volume (v),, then the constant C is represented by another constant R., Thus the general gas equation becomes, pv = RT, where, R = Characteristic gas constant or gas constant., For any mass m kg of gas, the characteristic equation becomes,, Q v = V , p. V = RT, , m, m , pV = mRT, , p=, , m, RT, V, , Q m = ρ, V, , The unit of gas constant R may be obtained as follows:, p = ρRT, , pV N m 2 × m, N-m, =, =, = J/kg.K, mT, kg × K, kg × K, 3, , R=, , In SI units, the value of R for air = 287 J/kg.K = 0.287 kJ/kg.K, 1.34 Universal gas constant (Rmole), The product of molecular weight (M) and the characteristic, gas constant (R) is same for all gases. This constant is known as, universal gas constant. It is expressed as Rmole or Ru., Unit – I 1.22, ,
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or C p − Cv = R, Dividing throughout by Cv ,, , Cp, R, −1 =, Cv, Cv, R, γ −1 =, Cv, R, or Cv =, γ −1, , REVIEW QUESTIONS, 1., , Explain the relationship between absolute, atmospheric pressure and gauge pressure., , pressure,, , 2., , What is absolute zero of temperature., , 3., , Define specific heat capacity of gas., , 4., , Why specific heat of air at constant pressure is greater than, specific heat at constant volume?, , 5., , Define the terms: system, boundary and surroundings., , 6., , State and explain the three types of thermodynamic systems., , 7., , What are intensive and extensive properties? Give examples., , 8., , Distinguish between point function and path function., , 9., , Explain the various forms of energies of a system., , 10. State and explain the law of conservation energy., 11. State and explain the zeroth law of thermodynamics., 12. State the first law of thermodynamics and explain the same., 13. Explain the correlations of first law of thermodynamics., 14. State and explain the second law of thermodynamics., 15. Explain the Kelvin-Planck statement of the second law of, thermodynamics., Unit – I 1.24, ,
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16. Explain the Clausius statement of the second law of, thermodynamics., 17. Explain perpetual motion of the second kind is not possible., 18. State and explain the laws of perfect gas., 19. Derive the characteristic equation of perfect gas., 20. Derive the relationship C p − Cv = R ., 21. What is universal gas constant? Explain its relationship with, characteristic gas constant., , POINTS TO REMEMBER, 1. Absolute pressure = Atm. Pressure ± Gauge pressure, 2. Absolute temperature = Temperature in o C + 273, 3. Heat transfer, Q = mC(T2 − T1 ), 4. Ratio of specific heats, γ =, , Cp, Cv, , 2, , 5. Work done, W =, , p.dV, 1, , 6. By first law of thermodynamics, Q = W + ∆U, 7. Boyle's law p.V = C, V, 8. Charles’s law , =C, T, p, 9. Gay-Lussac's law = C, T, pV, pV, 10. General gas equation 1 1 = 2 2 = C, T1, T2, 11. Characteristic gas equation p.V = m. R.T ; p = ρRT, 12. Molar volume, Vmole = M.v, 13. Universal gas equation p.Vmole = Rmole.T, R, 14. C p − Cv = R; Cv =, γ −1, , Unit – I 1.25, ,
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SOLVED PROBLEMS, Example 1.1, One litre of petrol weighs 7.5N. Calculate its specific, weight, density and specific gravity., Given data, Weight, W = 7.5 N, Volume, V = 1 litre = 0.001 m 3, To find, 1. Specific weight, w, , 2. Density, ρ, , 3. Specific gravity, s, , Solution, Specific weight, w =, , Weight, = W = 7.5, Volume V, 0.001, , = 7500 N/ m 3 = 7.5 kN/ m 3, Density, ρ =, , w 7500, =, = 764.526 kg/ m 3, g 9.81, , Specific gravity, s =, , Density of petrol 764.526, =, Density of water, 1000, , = 0.764526, Result, 1. Specific weight, w = 7.5 kN/ m 3, 2. Density, ρ = 764.526 kg/ m 3, 3. Specific gravity, s = 0.764526, Example 1.2, A gauge fitted to a compressor records a pressure of 26.26, kN/ m . Compute the corresponding absolute pressure in kN/ m 2 ., 2, , Given data, Gauge pressure, pg = 26.26 kN/ m 2, To find, Absolute pressure, pabs, Solution, Assume, atmospheric prssure, patm = 101.325 kN/ m 2 ., Unit – I , , P1.1
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pabs = patm + pg =101.325 + 26.26 = 127.585 kN/ m 2, Result, Absolute pressure, pabs = 127.585 kN/ m 2, Example 1.3, A vacuum gauge fitted to a condenser reads 745mm of, mercury. Find the absolute pressure in the condenser in Pa. The, barometer reading is 762mm of mercury., Given data, Gauge pressure, pg = 745 mm of Hg, Barometric pressure, patm = 762 mm of Hg, To find, Absolute pressure, pabs, Solution, Absolute pressure, pabs = patm − pg, = 762 – 74 = 17 mm of Hg, We know that, 760 mm of Hg = 101325 pa, 101325, ∴ pabs = 17 mm of Hg =, × 17 = 2266.48 Pa, 760, Result, Absolute pressure, pabs = 2266.48 Pa, Example 1.4, If a pressure gauge reads 70cm of mercury, what is the, gauge pressure in (a)kg/c m 2 , (b) N/ m 2 ., Given data, Gauge pressure, pg = 70 cm of Hg, To find, Gauge pressure in (a) kg/ m 2 ; (b) N/ m 2 ., Solution, (a) 76 cm of Hg = 1.0332 kg/ m 2, 1.0332, × 70 = 0.9516 kg/c m 2, ∴ 70 cm of Hg =, 76, Unit – I , , P1.2
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(b) 76 cm of Hg = 101325 N/ m 2, 101325, × 70 = 93,325.658 N/ m 2, ∴ 70 cm of Hg =, 760, Result, Guge pressure, pg = 0.9516 kg/c m 2 = 93,325.658 N/ m 2 ., Example 1.5, 5 kg of steel, specific heat capacity 480 J/kg.K is heated, from 15 o C to 100 o C . How much heat is required?, Given data, Mass, m = 5 kg, Specific heat, C = 480 J/kg.K, Initial temperature, t1 = 15 o C ; ∴ T1 =15 + 273 = 288 K, Final temperature, t2 = 100 o C ; ∴ T2 = 100 + 273 = 373 K, To find, Heat required, Q, Solution, Heat required, Q = m.C.(T2 − T1 ), = 5 × 480 × (373 − 288), = 204,000 M = 204 kJ, Result, Heat required, Q = 204,000 J = 204 kJ, Example 1.6, o, , 6 kg of air is heated at constant volume from 25 o C to 100, C . Find the heat transferred. Take Cv =0.718 kJ/kg.K, , Given data, Mass, m = 6 kg, Initial temperature, t1 = 25 o C ; ∴ T1 = 25 + 273 = 298 K, Final temperature, t2 = 100 o C ; ∴ T2 = 100 + 273 = 373 K, Specific heat capacity, Cv = 0.718 kJ/kg.K, To find, Heat transferred, Q, Unit – I , , P1.3
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Solution, Heat required, Q = m.Cv .(T2 − T1 ), , = 6 × 0.718 × (373 − 298) = 323.1 kJ, Result, Heat transferred, Q = 323.1 kJ, Example 1.7, An iron casting of mass 10 kg has an original temperature, of 200 o C . If the casting loses heat to the value of 715.5 kJ, what, is the final temperature? Specific heat of cast iron = 477 J/kg.K, Given data, Mass of the casting, m = 10 kg, Original temperature, t1 = 200 o C ; ∴ T1 = 200 + 273 = 473 K, Heat transferred, Q = –715.5 kJ ( Q Heat lost), Specific heat of cast iron, = 477 J/kg.K = 0.477 kJ/kg.K, C, To find, Final temperature, T2, Solution, Heat transferred, Q = m.C.(T2 − T1 ), −715.5 = 10 × 0.477 × ( T2 − 473), , T2 − 473 =, , −715.5, 10 × 0.477, , 715.5, = 323 K, 10 × 0.477, T2 = 323 K (or) t2 = 323 − 273 = 50 o C ., T2 = 473 −, , Result, Final temperature, t2 = 50 o C (323 K), Example 1.8, At a speed of 50 km/hr, the resistance to motion of a car is, 900 N. Neglecting the losses, what is the power output of the engine, of the car at this speed?, Unit – I , , P1.4
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Given data, Speed, v = 50 km/hr =, Resistance = Force, F = 900 N, To find, Power output, P, Solution, Power output, P =, , Workdone Force × distance travelled, =, Time, time, , 900 × 5000, =12500 N−m/s = 12.5 kN−m/s = 12.5 kJ/s, 3600, ∴ Power output, P = 12.5 kW, , =, , Result, Power output, P = 12.5 kW, Example 1.9, A closed system receives 105 kJ of heat while it performs, 130 kJ of work. Determine the change in internal energy., Given data, Heat supplied, Q = 105 kJ, Work done, W = 130 kJ, To find, Change in internal energy, ∆U, Solution, By first law of thermodynamics, Q = W + ∆U, ∴ ∆U = Q − W = 105 − 130 = −25 kJ, The negative sign indicates that there is a loss of internal energy., Result, Change in internal energy, ∆U = −25 kJ., Example 1.10, During the compression stroke of an engine, the work done, on the working fluid in the engine cylinder is 70 kJ/kg and 40, kJ/kg heatis rejected to the surroundings. Determine the change, in internal energy., Unit – I , , P1.5
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Given data, Work done, W = −70 kJ/kg (` Q Work is done on the system), Heat transferred, Q = −40 kJ/kg ( Q Heat is lost), To find, Change in internal energy, ∆U, Solution, By first law of thermodynamics, Q = W + ∆U, ∴ ∆U = Q − W = −40 − (−70) = 30 kJ, Result, Change in internal energy, ∆U = 30 kJ, Example 1.11, A gas at a pressure of 3 bar and temperature of 27 o C, occupies a volume of 1.5 m 3 . If the gas constant is 287 J/kg.K,, determine the mass of the gas., Given data, Pressure, p =, Volume, V =, Temperature, t =, Gas constant, R =, , 3 bar = 3 × 105 N/ m 2, 1.5 m 3, 27 o C ; ∴T = 27 + 273 = 300 K, 287 J/kg.K, , To find, Mass of the gas, m, Solution, By characteristic gas equation, p.V = m. R.T, ∴m =, , p.V 3 × 105 × 1.5, =, = 5.2264 kg, R.T, 287 × 300, , Result, Mass of the gas, m = 5.2264 kg, Example 1.12, A gas whose original pressure, volume and temperature, was 120 kN/ m 2 , 0.125 m 3 and 30 o C respectively is compressed, such that its pressure and temperature are raised to 600 kN/ m 2, and 70 o C respectively. Determine the new volume of the gas., Unit – I , , P1.6
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Given data, Initial pressure, p1 = 120 kN/ m 2, Initial volume, V1 = 0.125 m 3, Initial temperature, t1 = 30 o C ; ∴ T1 = 30 + 273 = 303 K, Final pressure, p2 = 600 kN/ m 2, Final temperature, t2 = 70 o C ; ∴ T2 = 70 + 273 = 343 K, To find, Final volume, V2, Solution, By characteristic gas equation,, , V2 =, , p1V1 p2V2, =, T1, T2, , 120 × 0.125 × 343, = 0.0283 m 3, 303 × 600, , Result, Final volume, V2 = 0.0283 m 3, Example 1.13, A mass of air has an initial pressure 1.3 MN/ m2 , volume, 0.014 m3 and temperature 135 o C . It is expanded until its final, pressure is 275 kN/ m 2 and its volume becomes 0.056 m3 ., Determine (a) mass of air and (b) the final temperature of the air., Take R = 0.237 kJ/kg.K., Given data, Initial pressure, p1 = 1.3 MN/ m 2 = 1300 kN/ m 2, Initial volume, V1 = 0.014 m 3, Initial temperature, t1 = 135 o C ; ∴ T1 = 135 + 273 = 408 K, Final pressure, p2 = 275 kN/ m 2, Final volume, V2 = 0.056 m 3, Gas constant, R = 0.237 kJ/kg.K, To find, (a) Mass of air, m, , (b) Final temperature of air, t2, , Solution, By characteristic gas equation, p1 .V1 = m. R.T1, Unit – I , , P1.7
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p1 .V1 1300 × 0.014, =, =0.18822 kg., R.T1, 0.237 × 408, We know that, p2. .V2 = m. R.T2, p .V, 275 × 0.056, ∴Final temperature, T2 = 2 2 =, = 345.23 K, m. R, 0.18822 × 0.237, T2 = 345.23 K (or) t2 =345.23 − 273 = 72.23 o C, ∴ Mass of air, m =, , Result, (a) Mass of air, m = 0.18822 kg, (b) Final temperature , T2 = 345.23 K ( t2 = 72.23 o C ), Example 1.14, A vessel of capacity 5 m3 contains air at a pressure of 12, bar and temperature of 30 o C . Additional air is pumped into the, vessel untile the pressure rises to 30 bar absolute and temperature, rises to 65 o C . Determine the mass of air pumpted in and also, express this quantity as a volume at 1.02 bar and 25 o C ., Given data, Initial pressure, p1 = 2 bar = 2 × 105 N/ m 2, Initial volume, V1 = 5 m 3 = V2, Initial temperature, t1 = 30 o C ; ∴ T1 = 30 + 273 = 303 K, Final pressure, p2 = 30 bar = 30 × 105 N/ m 2, Final temperature, t2 = 65 o C ; ∴ T2 = 65 + 273 = 338 K, To find, (a) Mass of the air pumped, m, (b) volume of air at 1.02 bar and 25 o C ., Solution, (a) Mass of the air pumped,, m = Final mass of air (m2 ) − Initial mass of air (m1 ), By characteristic gas equation, p1 .V1 = m1 . R.T1, , p1 .V1 2 × 105 × 5, =, = 11.499 kg, R.T1, 287 × 303, We know that, p2 .V2 = m2 . R.T2, , ∴Initial mass of air, m1 =, , Unit – I , , P1.8
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p2 .V2 30 × 105 × 5, =, = 154.629 kg, R.T2, 287 × 338, Mass of air pumped, m = m2 − m1, ∴ m2 =, , = 154.629 − 11.499 = 143.13 kg., (b) Volume of air at p =1.02 bar and t = 25 o C ., Pressure , p =1.02 bar = 1.02 × 105 N/ m 2, Temperature, t = 25; ∴ T = 273 + 25 = 298 K, We know that, p.V = m. R.T, m. R.T 143.13 × 287 × 298, ∴V =, =, =119.904 m 3, 5, p, 1.02 × 10, Result, (a) Mass of air pumped, m = 143.13 kg, (b) Volume of air, V = 119.904 m 3, Example 1.15, A gas having molecular weight 28 occupies a volume of, 0.15 m3 at a pressure of 2 bar and a temperature of 20 o C . Find, the mass of the gas and the volume of the gas at 0 o C and 1 bar, pressure. Find also the density of the gas at 0 o C and 1 bar, pressure. The value of universal gas constant Ru= 8314, J/kg.mole.K, Given data, Molecular weight, M = 28, Volume, V1 = 0.15 m 3, Pressure, p1 = 2 bar = 2 × 105 N/ m 2, Temperature, t1 = 20 o C ; ∴ T1 = 20 + 273 = 293 K, Pressure, p2 = 1bar = 1 × 105 N/ m 2, Temperature, t2 = 0 o C ; ∴ T2 = 0 + 273 = 273 K, Universal gas constant, Ru = 8314 J/kg.mol.K, To find, (a) Mass of the gas, m (b) Volume of gas, V2 (c) Density of gas, ρ 2, , Unit – I , , P1.9
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Solution, By characteristic gas equation,, , p1V1 p2V2, =, T1, T2, , p1V1 .T2 2 × 105 × 0.15 × 273, =, = 0.2795, p2 .T1, 293 × 1 × 105, R, 8314, Gas constant, R = u =, = 297 J/kg.K, M, 28, We know that, p2 .V2 = m. R.T2, V2 =, , p2 .V2 1 × 105 × 0.2795, =, = 0.3447 kg, R.T2, 297 × 273, m 0.3447, Mass, Density of gas, ρ 2 =, =, =, =1.2333 kg/ m 3, Volume V2 0.2795, Result, (a) Mass, m = 0.3447 kg, (b) Density, ρ 2 =1.2333 kg/ m 3, ∴Mass, m =, , PROBLEMS FOR PRACTICE, 1. The pressure of steam inside a boiler, as measured by pressure, gauge, is 1 N/m m 2 . The barometric pressure of atmosphere is, 765 mm of mercury. Find the absolute pressure of steam in N/, m 2 , kPa, bar and N/m m 2 . [Ans: pabs = 1.102 N/ m 2 , 1102, kPa, 11.02 bar, 1.102 N/mm2], 2. In a condenser of a steam power plant, the vacuum is recorded, as 700mm of mercury. If the barometer reading is 760 mm of, mercury, find the absolute pressure in the condenser., [Ans: pabs = 7998 N/ m 2 ], 3. Calculate the quantity of heat required to raise the, temperature of a steel forging of mass 180 kg from 300 K to, 1265 K. The specific heat of steel = 0.49 kJ/kg.K, [Ans: Q = 85113 kJ], 4. At a speed of 40 km/hr the resistance to motion of a car is 1, kN. Neglecting losses, determine power output of the engine at, this speed., [Ans: P=11.11 kN], Unit – I P1.10
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5., , A car a mass of 1500 kg and its engine develops 20 kW when, travelling at a speed of 65 km/hr. Neglecting losses,, determine the resistance of motion., [Ans: F = 1.1077kN], , 6., , A closed system receives 105 kJ of heat while it performs 130, kJ of work. Determine the change in internal energy., [Ans: ∆U = −25 kJ], , 7., , A closed system receives an input of 1 kW−hr and the increase, in internal energy of the system is 1500 kJ. Determine the, heat loss from the system. [Ans: Q = 2100 kJ], , 8., , A gas occupies a volume of 0.1 m 3 at a temperature of 20 o C, and a pressure of 1.5 bar. Find the temperature of the gas, if, it is compressed to a pressure of 7.5 bar and occupies a volume, [Ans: t2 = 313 o C ( T2 = 586 K)], of 0.04 m 3 ., , 9., , A gas whose original pressure, volume and temperature were, 140 kN/ m 2 , 0.1 m 3 and 25 o C respectively is compressed, such that its new pressure is 700 kN/ m 2 and its new, temperature is 60 o C . Determine the new volume of the gas., [Ans: V2 = 0.02233 m 3 ], , 10. A mass of gas has an initial pressure of 1.1 MN/ m 2 , volume, 0.015 m 3 and temperature 150 o C . It is expanded till the, pressure and volume of the gas becomes 250 kN/ m 2 and 0.06, m 3 respectively. Determine (a) mass of the gas and (b) the, final temperature of the gas. Take R =0.287 kJ/kg.K, [Ans: m=0.1359 kg; t2 =111.55 o C ( T2 =384.55 o C )], 11. A vessel of capacity 3 m 3 contains air at a pressure of 1.5 bar, and a temperature of 25 o C . Additional air is now pumped, into the system until the pressure rises to 30 bar and, temperature rises to 60 o C . Determine the mass of air, pumped in and express the quantity as a volume at a pressure, of 1.02 bar and temperature of 20 o C ., [Ans: m = 88.91 kg; v = 73.3 m 3 ], 12. A vessel of capacity 5 m 3 contains 20 kg of an ideal gas, having a molecular mass of 25. If the temperature of the gas, is 15 o C , find its pressure., [Ans: 3.83 bar], Unit – I P1.11
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Unit – I, Chapter 2., , THERMODYNAMIC PROCESSES, OF PERFECT GASES, , 2.1 Introduction, When a system changes its state from one equilibrium, state to another equilibrium state, then the path of successive, states through which the system is passed, is known as, thermodynamic process. If the process is assumed to take place, sufficiently slow so that the properties in the intermediate states, are in equilibrium state, then the process is called quasi−static or, reversible process. If the process takes place in a such a manner, that the properties at the intermediate state are not in equilibrium, state, then the process is said to be non−equilibrium or irreversibel, process., The different thermodynamic processes include the, following:, 1. Constant volume (Iso−choric) process, 2. Constant pressure (Iso−baric) process, 3. Constant temperature (Iso−thermal) process, 4. Hyperbolic process, 5. Reversible adiabatic (Isentropic) process, 6. Polytropic process, 7. Free expansion process, 8. Throttling process, The following notations are used in this chapter., p1 = Initial pressure of the gas (N/ m 2 ), , V1 = Initial volume of the gas ( m 3 ), T1 = Initial temperature of the gas (K), H1 = Initial enthalpy of the gas (kJ), U1 = Initial internal energy of th egas (kJ), S1 = Initial entropy of the gas (kJ/kg), p2 = Final pressure of the gas (N/ m 2 ), V2 = Final volume of the gas ( m 3 ), Unit – I 2.1, ,
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T2 = Final temperature of the gas (K), H2 = Final enthalpy of the gas (kJ), U 2 = Final internal energy of th egas (kJ), S2 = Final entropy of the gas (kJ/kg), , C p = Specific heat at constant pressure (kJ/kg.K), Cv = Specific heat at constant volume (kJ/kg.K), γ = Ratio of specific heats, 2.2 Change in internal energy during a process ( ∆U ), , Pressure, , p2, , 2, V=C, , p1, , Volume, , 1, , V1=V2, , Fig.2.1 Constant volume heating, , Consider a system (gas) as shown in figure, which is to be, heated at constant volume from state 1 to state 2., p1 , V1 , T1 = Initial pressure, volume and temperature of the, gas respectively before heating., p2 , V2 , T2 = Final pressure, volume and temperature of the, gas respectively after heating., By first law of thermodynamics, Q = W + ∆U, 2, , W=, , p.dV = 0 [`Q There is no change in volume , i.e. dV= 0], 1, , ∴ Q = 0 + ∆U = m.Cv .(T2 − T1 ), (or) ∆U = m.Cv .(T2 − T1 ) …………. (1), 2, , Mathematically, change in internal energy, ∆U = m.Cv . dT, 1, , For unit mass, ∆u = Cv .(T2 − T1 ) = Cv . dT, The equation (1) can be used for all the processes of a, perfect gas between the temperature range T1 to T2 ., Unit – I , ,, , 2.2
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2.3 Enthalpy, Enthalpy is the sum of the internal energy and the product, of pressure and volume (p.V). It is represented by the symbol H., Mathematically, enthalpy, H = U + p.V, Since ( U + p.V ) is made up of properties, enthalpy (H) is, also a property. For unit mass, specific enthalpy, h = u + p.v, where, u = specific internal energy, v = specific volume, 2.4 Change in enthalpy during a process (∆H ), , Pressure, , p1=p2 1, , p=C, , 2, , W, V1, , Volume, , V2, , Fig.2.2 Constant pressure heating, , Consider a system (gas) as shown in figure, which is to be, heated at constant pressure from state 1 to state 2., p1 , V1 , T1 =Initial pressure, volume and temperature of the gas, respectively before heating., p2 , V2 , =Final pressure, volume and temperature of the gas respectively, after heating., T, 2, , By first law of thermodynamics, Q = W + ∆U, , W=, , 2, , 2, , 1, , 1, , p.dV = p dV [`Q p =, , …..(1), , constant], , W = p.[V ]12 = p.(V2 − V1 ) = p2 .V2 − p1 .V1, , ∆U = U2 − U1, Substituting (2) and (3) in (1), Q = W + ∆U = p2 .V2 − p1 .V1 + U2 − U1, , ….. (2), ….. (3), , Q = (U2 + p2 .V2 ) − (U1 + p1 .V1 ), Q = H2 − H1 = ∆H = Change in enthalpy, ∴ ∆H = Q = m.C p (T2 − T1 ), ….. (4), The equation (4) can be used for all the processes of a, perfect gas between the temperature range T1 to T2 ., Unit – I , ,, , 2.3
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2.5 Entropy (S), Entropy is defined as a function of quantity of heat with, respect to temperature. Entropy of a substance increases when heat, is supplied to it and decreases when heat is rejected from it. Entropy, is a form of unavailable energy. It is represented by the symbol S., 2.6 Change in entropy ( dS ), In a reversible process, the increase or decrease of entropy,, when multiplied by the absolute temperature gives the heat, absorbed by the working substance., Mathematically, dQ = T . dS, 2, , Temperaturre (T), , T2, , 1, , T1, , S1, dS, Entropy (S), , S2, , Fig.2.3 T−S diagram, , Consider a gas which is heated from state 1 to state 2., There is a change in entropy of the gas. If we consider a small, amount of heat addition dQ to the gas at an absolute temperature, T, then change in entropy is given by,, dQ, dS =, T, This heating process can be plotted on a graph, by taking, entropy in x−axis and absolute temperature in y−axis. This graph is, known as temperature − entropy (T−S) diagram. The area under the, curve in T−S diagram gives the heat transfer during the process., As shown in the figure, consider an elemental strip of width ‘dS’., Area of elemental strip = Heat transfer through the strip, dQ, T .dS = dQ (or) dS =, T, Change in entropy during the process 1−2 is given by,, 2, , 2, , dS =, , , , 1, , 1, , dQ, T, , Unit – I , ,, , 2.4
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The constant volume process is represented by a vertical, straight line in the p−V diagram. It is represented as a curve in the, T−S diagram., (b) Work done (W), Workdone during the process is given by,, 2, , W=, , p.dV, 1, , But, dV = 0 [ Q There is no change in voluem], ∴W = 0, (c) Change in internal energy ( ∆U ), The change in internal energy during the process is given by,, 2, , , , ∆U = m.Cv . dT = m.Cv (T2 − T1 ), 1, , ∴ ∆U = m.Cv .(T2 − T1 ), (d) Heat transferred (Q), By first law of thermodynamics,, Q = W + ∆U = 0 + ∆U [ Q W = 0 ], ∴ Q = m.Cv .(T2 − T1 ), (e) Change in enthalpy ( ∆H ), The change in enthalpy during the process is given by,, 2, , , , ∆H = m.C p . dT = m.C p (T2 − T1 ), 1, , ∴ ∆H = m.C p .(T2 − T1 ), (f) Relation between p,V and T, By characteristics gas equation,, , p1 .V1 p2 .V2, =, T1, T2, , p1, p, = 2, T1, T2, (g) Change in entropy (dS), The change in entropy during the process is given by,, V , T , dS = m. R. ln 2 + m.Cv . ln 2 , V, 1, T1 , But, V1 = V2 ; ∴, , Unit – I , ,, , 2.7
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(g) Change in entropy (dS), The change in entropy during the process is given by,, dQ, dS =, = 0 ( Q dQ = 0 ), T, ∴ dS = 0 (or) S1 = S2, 2.12 Comparison of isothermal and adiabatic processes, Isothermal process, 1. It follow the law p.V = C, , Adiabatic process, It follows the law p.V γ = C, , 2. In this process, temperature In this process,, remains constant, remains constant., , entropy, , 3. It is a very slow process so that It is a very rapid process so, to, maintain, a, constant that no heat transfer takes, temperature., place., 4. There is no change in internal There is a change in internal, during the process., energy during the process., 5. There is a heat transfer during There is no heat transfer, the process., during the process., 6. There is no change in enthalpy There is a change in, during th eprocess., enthalpy during the process., 2.13 Polytropic process [ p.V n = C ], (Constatn volume), n=, , Pressure, , n=0 (Constant pressure), n=1(Constant tem.), n=1.2 (Polytropic), , n=γ, (Adiabatic), , Volume, Fig.2.8 Polytropic process, , Unit – I 2.15, ,
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The polytropic process follows the law p.V n = C ∞ , where, ‘n’ is a constant. All the processes are special cases of polytropic, process having different values of ‘n’. ∞, • When n=0, p.V 0 = C ; or p = C . It is called constant, pressure process., • When n = ∝ , p.V ∝ = C ; or V = C . It is called constant, volume process., • When n =1, p.V = C . It is called constant temperature process., •, , When n = γ , p.V γ = C . It is called adiabatic process, , •, , When n=n, p.V n = C . It is called polytropic process., , (a) p−V and T−S diagram, p2, , 1, , 2, n, , W, , p1, , V1, , Temperaturre (T), , Pressure, , pV =C, , 2, , V2, , Volume, , 1, , Q, , S1, , Entropy (S), , (b) T-S diagram, , (a) p-V diagram, , Fig.2.9 p−V and T−S Polytropic process, , (b) Work done (W), Workdone during the process is given by,, 2, , W=, , p.dV, 1, , For a adiabatic process, p.V n = C (or) p =, , C, Vn, , Substitute the value of p in the integral,, 2, , W=, , 2, , C, , V, , .dV = C, n, , 1, , V, 1, , 2, , dV, n, , 2, , , , = C V − n .dV, 1, , V2, V, , − V1− n +1 , W = C, , = C, − n +1, − n + 1 1, , , − n +1, , − n +1, , Unit – I 2.16, ,, , S2
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(e) Change in enthalpy ( ∆H ), The change in enthalpy during the process is given by,, 2, , , , ∆H = m.C p . dT = m.C p (T2 − T1 ), 1, , (f) Relation between p,V and T, i) Relation between pressure and volume, In polytropic process, p1V1 n = p2V2 n, 1, , n, , p1 V2 , V2 p1 n, =, =, , (or), p2 V1 , V1 p2 , ii) Relation between temperature and volume, In polytropic process, pV n = C, ….. (1), p.V, ….. (2), From gas equation,, =C, T, Dividing equation (1) by (2), , p.V n, V n .T, = C (or), =C, V, p.V , , , T , , V n −1 .T = C, , ….. (3), , Applying the initial and final conditions,, V1 n −1 .T1 = V2 n −1 = C, , or,, , T2 V1 , =, , T1 V2 , , n −1, , (or), , T1 V2 , =, , T2 V1 , , n −1, , iii) Relation between temperature and pressure, In polytropic process, pV n = C, ….. (1), p.V, From gas equation,, =C, T, pn .V n, = C ….. (2), Raising both sides to the power n,, Tn, Dividing equation (2) by (1),, pn .V n, n, , T . p.V, , n, , =C;, , pn, n, , T .p, , = C (or), , Unit – I 2.18, ,, , pn −1, Tn, , =C
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Applying the initial and final conditions,, , p1n −1, T1n, , =, , p2 n −1, T2 n, , p , (or) 1 , p2 , , n −1, , T , = 1 , T2 , , n, , n, , n −1, n, , n, , n −1, n, , p T n −1, T p , ∴ 1= 1, (or) 1 = 1 , p, T, T2 p2 , 2 2, p T n −1, T p , ∴ 2= 2, (or) 2 = 2 , p, T, 1, 1, T1 p1 , , , , , , (g) Change in entropy (dS), The change in entropy during this process is given by,, V , T , ∆S = dS = m. R. ln 2 + m.Cv . ln 2 (or), V, 1, T1 , , p , T , dS = m. R. ln 1 + m.C p . ln 2 (or), p2 , T1 , p , V , dS = m.C p . ln 2 + m.Cv . ln 2 , V1 , p1 , 2.14 Free expansion process, Free expansion of gas occurs, when it expands suddenly, into a vacuum space through a large orifice. During this process,, there is no change in temperature., Consider two chambers A and B separated by a partition, C. Both the chambers are well insulated so that no heat transfer, takes place. The chamber A contains a gas and the chamber is, completely vacuum. If the partition C is removed, the gas expands, freely and occupies the full volume of the chambers A and B., Insulated wall, , A, , Partition (C), , B, , Fig.2.10 Free expansion process, , Unit – I 2.19, ,
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During this process, there is no expansion of boundary., Hence no work is done i.e. W=0. Since the chmabers are well, insulated, there is no heat transfer takes place i.e. Q = 0., By first law of thermodynamics, Q = W + ∆U, 0 = 0 + ∆U, ∴ ∆U = 0 i.e. U 2 − U1 = 0 ; (or) U 2 = U 1, There is no change in internal energy., 2.15 Throttling process, Insulated wall, , p1 V1 U1, , High, pressure, , Orifice, , Low, pressure, , p2 V2 U2, , Fig.2.11 Throttling process, , When a gas or vapour expands and flows throug a small, passage (like small orifice, partially opened valve, etc.), the process, is called throttling. During throttling process, no heat transfer, takes place and no work is done. Hence there is no change in, internal energy. W = 0 ; Q = 0 ; ∆U = 0 ., 2.16 To prove adiabatic process follows the law p.V γ = C, By first law of thermodynamics, Q = W + ∆U, In an adiabatic process, Q = 0., ∴ W + ∆U = 0, or, ∆U = −W, m.Cv .dT = − p.dV, ….. (1), By gas equation, p.V = m. R.T, Differentiating, p.dV + V .dp = m. R.dT = m.(C p − Cv ).dT ….. (2), Dividing (2) by (1), m.(C p − Cv ).dT, , m.Cv .dT, , =, , p.dV + V .dp, − p.dV, , Unit – I 2.20, ,
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7., , What is meant by adiabatic process? Sketch the adiabatic, process on p−V and T−S diagram., 8. Derive an expression for the work done during adiabatic, process., 9. Explain the differences between isothermal and adiabatic, process., 10. Derive an expression for the ratio of temperature in terms of, ratio of compression when a gas follows adiabatic process., 11. Prove that an adiabatic process follows the law p.V γ = C ., 12. Derive an expression for work done during a polytropic process., 13. Derive the expression for the heat supplied and change in, internal energy while a gas is expanding according to the law, p.V n = C, 14. Derive an expression for polytropic specific heat., , POINTS TO REMEMBER, 1., , ln( A) = log e ( A) = 2.3 log10 ( A), , 2. For air: C p = 1.005 kJ/kg.K, Cv = 0.718 kJ/kg.K, γ = 1.4, , p, log10 2 , p1 , 3. For adiabatic process, γ =, V, log10 1 , V2 , , p , log10 2 , p1 , 4. For polytropic process, n =, V , log10 1 , V2 , γ −n, 5. Polytropic specific heat, Cn = Cv , , n −1, , Unit – I 2.23, ,
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SOLVED PROBLEMS, Example 2.1, A mass of 2.25 kg of nitrogen occupying 1.5 m3 is heated, from 25 o C to 200 o C at constant volume. Calculate the initial and, final pressures of the gas. Molecular weight of nitrogen is 28., Universal gas constant is 8314 J/kg.mole.K, Given data, Mass of nitrogen, m = 2.25 kg, Initial volume, V1 = 1.5 m 3, Initial temperature, t1 = 25 o C ; T1 = 25 + 273 = 298 K, Final temperature, t2 = 200 o C ; T2 = 200 + 273 = 473 K, Constant volume heating: V2 = V1 =1.5 m 3, Molecular weight, M = 28, Universal gas constant, Ru = 8314 J/kg.mole.K, = 8.314 kJ/kg.mole.K, To find, 1. Initial pressure, p1 2. Final pressure, p2, Solution, , Ru 8.314, =, = 0.29693 kJ/kg.K, M, 28, By gas equation, p1 .V1 = m. R.T1 ; p2 .V2 = m. R.T2, m. R.T1 2.25 × 0.29693 × 298, =, ∴ Initial pressure, p1 =, V1, 1.5, Gas constant, R =, , = 132.728 kN/ m 2, Final pressure, p2 =, , m. R.T2 2.25 × 0.29693 × 473, =, V2, 1. 5, , = 210.672 kN/ m 2, Result, 1. Initial pressure, p1 =132.728 kN/ m 2, 2. Final pressure, p2 = 210.672 kN/ m 2, Unit – I , , P2.1
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Example 2.2, 0.5 m3 of air at 35 o C and 5 bar is heated at constant, volume until the final temperature is 185 o C . Find the final, pressure, work done and heat transferred. Take C p =1kJ/kg.K, and Cv =0.71kJ/kg.K, Given data, Co nstant volume heating, V2 = V1 =0.5 m 3, Initial temperature, t1 = 35 o C ; T1 = 35 + 273 = 308 K, Initial pressure, p1 = 5 bar = 5 × 105 N/ m 2, Final temperature, t2 = 185 o C ; T2 = 185 + 273 = 458 K, C p = 1.00 kJ/kg.K, , Cv = 0.71 kJ.kg.K, To find, 1. Final pressure, p2 2. Work done, W 3. Heat transfer, Q, Solution, 1. To find the final pressure, For a constant volume process,, , p2 =, , p1 p2, =, T1 T2, , p1 .T2 5 × 10 5 × 458, =, = 743506 N/ m 2 = 7.435 bar, T1, 308, , 2. To find the work done, In a constant volume process, work done, W = 0 ( Q dV = 0 ), 3. To find the heat transferred, Q, R = C p − Cv = 1 − 0.71 = 0.29 kJ/kg.K = 290 J/kg.K, By gas equation, p1 .V1 = m. R.T1, m=, , p1 .V1 5 × 10 5 × 0.5, =, = 2.7989 kg., R.T1, 290 × 308, , Heat transferred, Q = m.Cv (T2 − T1 ), = 2.7989 × 710 × (458 − 308), = 298082.85 J =298.083 kJ, Unit – I , , P2.2
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Result, 1. Final pressure, p2 = 7.435 bar, 2. Work done, W = 0, 3. Heat transfer, Q = 298.083 kJ, Example 2.3, 5 kg of a gas was heated from a temperature of 100 o C at, constant volume till its pressure become three times its original, pressure. For this process, calculate (1) the heat transfer, (2) change, in internal energy, (3) change in enthalpy and (4) change in entropy., Assume C p =1.00 kJ/kg.K and Cv =0.71kJ/kg.K, Given data, Mass of gas, m = 5 kg, Initial temperature, t1 = 100 o C ; T1 = 100 + 273 = 373 K, Constant volume heating, V2 = V1, Final pressure, p2 = 3 × original pressure = 3 p1, C p = 1.00 kJ/kg.K, , Cv = 0.71 kJ.kg.K, To find, 1. Heat transfer, Q, 2. Change in internal energy, ∆U, 3. Change in enthalpy, ∆H 4. Change in entropy, ∆S, Solution, 1. To find the heat transfer, Q, Heat transfer, Q = m.Cv (T2 − T1 ), , p1 p2, =, T1 T2, p, 3 p1, × 373 = 3 × 373 = 1119 K, Final temperature, T2 = 2 × T1 =, p1, p1, For a constant volume process,, , ∴ Q = 5 × 0.71 × (1119 − 373) =2648.3 kJ, 2. To find the change in internal energy, ∆U, We know that, Q = W + ∆U, or, ∆U = Q = 2648.3 kJ [ Q W = 0 ], 3. To find the change in enthalpy, ∆H, ∆H = m.C p (T2 − T1 ) = 5×1×(1119−373) = 3730 kJ., Unit – I , , P2.3
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4. To find change in entropy, ∆S, T , 1119 , ∆S = m.C v . ln 2 = 5 × 0.71 × ln, = 3.9 kJ/K, 373 , T1 , Results, 1. Heat transfer, Q = 2648.3 kJ, 2. Change in internal energy, ∆U = 2648.3 kJ, 3. Change in enthalpy, ∆H =3730 kJ, 4. Change in entropy, ∆S =3.9 kJ/Kn−m, Example 2.4, 0.35 m3 of air at 22 o C and under atmospheric pressure is, heated under constant volume to a temperature of 100 o C . Determine, (1) mass of air, (2) the final pressure, (3) the heat transfer, (4) the, change in internal energy, (5) the work done, (6) the change in, enthalpy and (7) the change in entropy. Assume C p = 1kJ/kg.K and, , Cv = 0.71kJ/kg.K, Given data, Initial volume, V1 = 0.35 m 3, Initial temperature, t1 = 22 o C ; T1 = 22 + 273 = 295 K, Initial pressure, p1 = Atm. Pressure = 101.325 kN/ m 2, Constant volume heating, V2 = V1 =0.35 m 3, Final temperature, t2 = 100 o C ; T2 = 100 + 273 = 373 K, C p = 1 kJ/kg.K, , Cv = 0.71 kJ.kg.K, To find, 1. Mass of air, m, , 2. Final pressure, p2, , 3. Heat transfer, Q, 4. Change in internal energy, ∆U, 5. Work done, 6. Change in enthalpy, ∆H, 7. Change in entropy, ∆S, Solution, 1. Mass of air, m, R = C p − Cv = 1 − 0.71 = 0.29 kJ/kg.K, By gas equation, p1 .V1 = m. R.T1, Unit – I , , P2.4
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m=, , p1 .V1 101.325 × 0.35, =, = 0.41454 kg., R.T1, 0.29 × 295, , 2. Final pressure, p2, For a constant volume process,, ∴ p2 =, , p1 p2, =, T1 T2, , p1 .T2 101.325 × 373, =, = 128.116 kN/ m 2, T1, 295, , 3. Heat transfer, Q, Q = m.Cv (T2 − T1 ) = 0.41454 × 0.71 × (373 − 295) = 22.957 kJ, 4. Change in internal energy, ∆U, ∆U = m.Cv (T2 − T1 ) =22.957 kJ, 5. Work done, W, During the constatn volume process, no work is done. ∴W= 0, 6. Change in enthalpy, ∆H, ∆H = m.Cv (T2 − T1 ) =0.41454 × 1 × (373 − 295) = 32.334 kJ/K, 7. Change in entropy, ∆S, T , 373 , ∆S = m.Cv . ln 2 = 0.41454 × 0.71 × ln, =0.069 kJ/K, T, 295 , 1, Result, 1. Mass of air, m = 0.41454 kg, 2. Final pressure, p2 =128.116 kN/ m 2, 3. Heat transfer, Q = 22.957 kJ, 4. Change in internal energy, ∆U = 22.957 kJ, 5. Work done, W = 0, 6. Change in enthalpy, ∆H = 32.334 kJ, 7. Change in entropy, ∆S = 0.069 kJ/K, Example 2.5, A gas having a volume of 0.28 m3 at a pressure of 700 kN/, , m 2 is expanded at constant pressure until its volume becomes 1.68 m3, . Determine the work done by the gas., Unit – I , , P2.5
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Given data, Initial volume, V1 = 0.28 m 3, Initial pressure, p1 = 700 kN/ m 2, Final volume, V2 = 1.68 m 3, To find, Work done, W, Solution, For a constant pressure process,, Work done, W = p.(V2 − V1 ) = 700 (168 − 0.28) = 980 kJ, Result, Work done, W =980 kJ., Example 2.6, A gas whose pressure, volume and temperature are 5 bar,, 0.23 m and 185 o C respectively has its state changed at constant, 3, , pressure until its temperature becomes 70 o C . Determine (1) the, work done, (2) the change in internal energy and (3) the heat, transferred during the process. R=290 J/kg.K; C p =1.005 kJ/kg.K, Given data, Initial pressure, p1 = 5 bar = 500 kN/ m 2, Initial volume, V1 = 0.28 m 3, Initial temperature, t1 = 185 o C ; T1 = 185 + 273 = 458 K, Final volume, V2 = V1 = 0.28 m 3, Final temperature, t2 = 70 o C ; T2 = 70 + 273 = 343 K, To find, 1. Work done, W 2. Change in internal energy, ∆U, 3. Heat transferred, Q, Solution, Specific heat , Cv = C p − R = 1.005 − 0.29 = 0.715 kJ/kg.K, By gas equation, p1 .V1 = m. R.T1, p .V, 500 × 0.23, m= 1 1 =, = 0.8658 kg., R.T1 0.29 × 458, Unit – I , , P2.6
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1. Work done, W = p.(V2 − V1 ) = m. R.(T2 − T1 ), = 0.8658 × 0.29 × (343 − 458) = −28.8744 kN−m, The negative sign indicates that work is done on the system., 2. Change in internal energy, ∆U = m.Cv .(T2 − T1 ), ∆U = 0.8658 × 0.715 × ( 343 − 458) = − 71.19 kJ, The negative sign indicates that there is loss of internal energy., , 3. Heat transferred, Q = W + ∆U, Q = −28.8744 − 71.19 = − 100.0644 kJ, The negative sign indicates that heat is rejected from the gas., Result, 1. Work done on the gas, W = 28.874 kN−m, 2. Change in internal energy, ∆U = −71.19 kJ, 3. Heat transferred, Q = 100.0644 kJ, Example 2.7, A gas has a density of 1.95 kg/ m3 at 1 bar and 18 o C . 0.8, kg of this gas is heated from 20 o C to 225 o C at constant pressure, by adding 180 kJ of heat. Calculate the specific heat at constant, pressure and specific heat at constant volume of the gas. Also, calculate the external work doe and change in internal energy, during the process., Given data, Density of the gas, ρ =, Pressure of gas, p =, Temperature of gas, t =, Mass of gas, m =, Initial temperature, t1 =, , 1.95 kg/ m 3, 1 bar = 100 kN/ m 2, 18 o C ; T= 18 + 273 = 291 K, 0.8 kg, 20 o C ; T1 = 20 + 273 = 293 K, , Final temperature, t2 = 225 o C ; T2 = 225 + 273 = 498 K, Heat added, Q = 180 kJ, To find, 1. Specific heats, C p and Cv, , 2. Workdone , W, , 3. Change in internal energy, ∆U, Unit – I , , P2.7
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Solution, By gas equation, p = ρ. R.T, p, 100, ∴R=, =0.176227 kJ/kg.K, =, ρ.T 1.95 × 291, Heat supplied, Q = m.Cp.(T2 − T1 ) = 1.09756 kJ/kg.K, C p − Cv = R; ∴ Cv = Cv − R, , =1.09756 − 0.176227 = 0.92133 kJ/kg.K, For a constant pressure process,, W = p.(V2 − V1 ) = m. R.(T2 − T1 ), = 0.8 × 0.176227 × 9498 − 293) = 28.9 kN−m, By first law of thermodynamics, Q = W + ∆U, ∴ ∆U = Q − W = 180 − 28.9 = 151.1 kJ, Result, 1. Specific heats, C p = 1.09756 lJ/kg.K and Cv = 0.92133 kJ/kg.K, 2. Workdone , W = 28.9 kJ, 3. Change in internal energy, ∆U =151.1 kJ, Example 2.8, 0.25 kg of air at a pressure of 1 bar occupies a volume of 0.3, m3 . If this air expands isothermally, to a volume of 0.9 m3 ,, determine (1) the initial temperature, (2) the final temperature, (3), external work done, (4) change in internal energy, (5) heat absorbed, by the air and (6) change in entropy. Assume R= 0.29kJ/kg.K, Given data, Mass of the air, m = 0.25 kg, Initial pressure , p1 = 1 bar = 100 kN/ m 2, Isothermal expansion : T2 = T1, Final volume, V2 = 0.9 m 3, Gas constant, R = 0.29 kJ/kg.K, To find, 1.Initial temperature, T1, 3. Work done, W, 5. Heat absorbed, Q, , 2. Final temperature, T2, 4. Change in internal energy, ∆U, 6. Change in entropy ∆S, , Unit – I , , P2.8
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Solution, 1. Initial temperature, T1, By gas equation, p1 .V1 = m. R.T1, p .V, 100 × 0.3, ∴Initial temperature, T1 = 1 1 =, = 419.79 K, m. R 0.25 × 0.29, T1 = 419.79 K (or) t1 = 419.79 − 273 = 140.79 o C, 2. Final temperature, T2, For isothermal process, T2 = T1 = 419.79 K, , V , 0. 9 , 3. Work done, W = p1 .V1 . ln 2 = 100 × 0.3 × ln, = 32.9583 kJ, 0.3 , V1 , 4. Change in internal energy, ∆U, 2, , For any process, ∆U = m.Cv . dT ., 1, , But, in isothermal process, dT = 0. ∴ ∆U = 0, There is no change in internal energy, 5. Heat absorbed by the air, Q, By first law of thermodynamics, Q = W + ∆U, Q = 32.9583 + 0 = 32.9583 kJ, 6. Change in entropy, V , 0. 9 , dS = m. R. ln 2 = 0.25 × 0.29 × ln, = 0.07965 kJ/K, V, 0.3 , 1, Results, 1.Initial temperature, T1 = 419.79 K ( t1 = 140.79 o C ), 2. Final temperature, T2 = 419.79 K ( t2 = 140.79 o C ), 3. Work done, W = 32.9583 kJ, 4. Change in internal energy, ∆U = 0, 5. Heat absorbed, Q =32.9583 kJ, 6. Change in entropy ∆S = 0.07965 kJ/K, Example 2.9, 1 kg of air is compressed from 1 bar and 30 o C to 7 bar, keeping the temperature constant. Determine the work done and, heat exchange during the process., Unit – I , , P2.9
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Given data, Mass, m = 1 kg, Initial pressure , p1 = 1 bar = 100 kN/ m 2, Initial temperature t1 : 30 o C ; T1 = 30 + 273 = 303 K, Final pressure, p2 = 7 bar = 700 kN/ m 2, Assume, gas constant, R = 0.287 kJ/kg.K, To find, 1. Work done , W, , 2. Heat transferred, Q, , Solution, 1. Work done, W, , V , p , 1, W = m. R.T1 . ln 2 = m. R.T1 . ln 1 = 1 × 0.287 × 303 × ln , 7, V1 , p2 , = −169.222 kJ, Negative sign indicates that work is done on the gas., 2. Heat transferred, Q, For isothermal process, Q = W = −169.222 kJ, Result, 1. Work done , W = −169.222 kJ, 2. Heat transferred, Q = −169.222 kJ, Example 2.10, A gas is compressed hyperbolically from initial conditions, of 80 kN/ m 2 and 0.007 m3 to a final pressure of 100 kN/ m 2 ., Determine the final volume of the gas and work done., Given data, Initial pressure, p1 = 80 kN/ m 2, Initial volume, V1 = 0.007 m 3, Final pressure, p2 = 100 kN/ m 2, To find, 1. Final volume, V2, , 2. Work done, W, , Unit – I P2.10
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Solution, 1. Final volume, V2, For an hyperbolic process, p1 .V1 = p2 .V2, p .V, 80 × 0.007, ∴ V2 = 1 1 =, = 0.0056 m 3, p2, 100, 2. Work done, W, , V , 0.0056 , W = p1 .V1 ln 1 = 80 × 0.007 × ln, = − 0.125 kJ, V, 0.007 , 2, Negative sign indicates that work is done on the gas., Result, 1. Final volume, V2 = 0.005 m 3 2. Work done, W = −0.125 kJ, Example 2.11, 0.9 kg of air at a pressure of 15 bar and 250 o C expanded, adiabatically and reversibly to a pressure of 1.5 bar. Determine the, work done. R = 0.287 kJ/kg.K; γ = 1.4, Given data, Mass of air, m = 0.9 kg, Initial pressure, p1 = 15 bar = 1500 kN/ m 2, Initial temperature, t1 = 250 o C ; T1 =250 + 273 = 523 K, Final pressure, p2 = 1.5 bar = 150 kN/ m 2, Gas constant, R = 0.287 kJ/kg.K, γ = 1.4, To find, Work done, W, Solution, , T p , In an adiabatic process, 2 = 2 , T1 p1 , p , ∴ T2 = T1 × 2 , p1 , , γ −1, γ, , 150 , = 523 × , , 1500 , , γ −1, γ, , 1 .4 − 1, 1 .4, , Unit – I P2.11, , = 270.886 K
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m. R.(T2 − T1 ), γ −1, 0.9 × 0.287 × 9523 − 270.886), W=, =162.8026 kJ., 1.4 − 1, , Work done, W =, , Result, Work done, W = 162.8026 kJ, Example 2.12, 1 kg of gas expands isentropically and its temperature is, observed to fall from 240 o C to 110 o C while its volume is doubled., The work done by the gas is 90 kJ in the process. Determine C p ,, , Cv and R for the gas., Given data, Mass of gas, m = 1 kg, Initial temperature, t1 = 240 o C ; T1 = 240 + 273 = 513 K, Final temperature, t2 = 110 o C ; T2 = 110 + 273 = 383 K, Final volume, V2 = 2 × V1, Work done, W = 90 kJ, To find, 1. Specific heat, C p, , 2. Specific heat, Cv, , 3. Gas constant,, , R, Solution, 1.Specific heat , Cv, In an adiabatic process, work done, W = m.Cv .(T2 − T1 ), W, 90, =, ∴ Cv =, = 0.6923 kJ/kg.K, m.(T1 − T2 ) 1(513 − 383), 2. Specific heat, C p, By adiabatic law,, γ −1, , T1 V2 , 513 2.V1 , , = , =, T2 V1 , 383 V1 , Taking log on both sides,, log (1.3394) = (γ−1) × log 2, 0.12691 = (γ−1) × 0.301, 0.12691 = 0.301 γ − 0.301, , γ −1, , Unit – I P2.12
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0.301 γ = 0.12691 + 0.301, 0.42791, γ=, = 1.4216, 0.301, Cp, We know that,, = γ C p = γ × Cv, Cv, = 1.4216 × 0.6923 = 0.98423 kJ/kg.K, 3. Gas constant, R = C p − Cv = 0.9892 − 0.6923 = 0.2918 kJ/kg.K, Result, 1. Specific heat, Cv = 0.9842 kJ/kg.K, 2. Specific heat, C p = 0.6923 kJ/kg.K, 3. Gas constant, R = 0.2918 kJ/kg.K, Example 2.13, 2 kg of gas at 8×105 N/ m 2 expands adiabatically till the, pressure falls to 4×105 N/ m 2 . During the process 1,20,000 N−m, of work is done by the system and the temperature falls from 377, o, C to 257 o C . Calculate the value of index of expansion and, characteristic gas constant., Given data, Mass of gas, m = 2 kg, Initial pressure, p1 = 8 × 105 N/ m 2 = 800 kN/ m 2, Final pressure, p2 = 4 × 105 N/ m 2 = 400 kN/ m 2, Work done, W = 1,20,000 N−m = 120 kN−m, Initial temperature, t1 = 377 o C ; T1 = 377 + 273 = 650 K, Final temperature, t2 = 257 o C ; T2 = 257 + 273 = 530 K, To find, 1. Index of expansion, γ 2. Characteristic gas constant, R, Solution, 1. Index of expansion, γ, γ −1, , p γ, T, 800 , In an adiabatic process, 1 , = 1 , , T2, 400 , p2 , Taking log on both sides,, Unit – I P2.13, , γ −1, γ, , =, , 650, 530
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γ −1, 800 , 650 , log , = log , , γ, 400, , , 530 , 800 , log , , γ −1, 400 = 0.2944, =, γ, 650 , log , , 530 , γ −1 = 0.2944 γ, γ − 0.2944 γ = 1, 0.7056 γ = 1, 1, γ=, = 1.4172, 0.7056, 2. Characteristic gas constant, Work done during adiabatic process, W =, ∴R=, , m. R.(T1 − T2 ), γ −1, , W (γ − 1), 120 × (1.4172 − 1), =, = 0.2086 kJ/kg.K, m.(T1 − T1 ), 2 × (650 − 530), , Result, 1. Index of expansion, γ =1.4172, 2. Characteristic gas constant, R = 0.2086 kJ/kg.K, Example 2.14, The initial volume of 0.18 kg of a certain gas was 0.15 m3, at a temperature of 15 o C and pressure of 1kg/c m 2 . After adiabatic, compression to 0.056 m3 , the pressure was found to be 4 kg/c m 2 ., Find (1) ratio of specific heats, (2) change in internal energy., Given data, Mass of gas, m = 0.18 kg, Initial volume, V1 = 0.15 m 3, Initial temperature, t1 = 15 o C ; T1 = 15 + 273 = 288 K, Initial pressure, p1 = 1 kg/c m 2 = 100 kN/ m 2, Final volume, V2 = 0.056 m 3, Final pressure, p2 = 4 kg/c m 2 = 400 kN/ m 2, To find, 1. Ratio of specific heats, γ 2. Change in internal energy, ∆U, Unit – I P2.14
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Solution, 1. Ratio of specific heat, γ, , p , 400 , log 2 , log , , p, 1=, 100 =1.407, In an adiabatic process, γ =, V , 0.15 , , log 1 log , 0.056 , V2 , 2. Change in internal energy, ∆U, m. R.(T2 − T1 ) p2 .V2 − p1 .V1, ∆U =, =, γ −1, γ −1, (400 × 0.056) − (100 × 0.15), ∆U =, = 18.182 kJ, 1.407 − 1, Result, 1. Ratio of specific heats, γ = 1.407, 2. Change in internal energy, ∆U = 18.182 kJ, Example 2.15, Air at 1 bar pressure and 40 o C is compressed to 1/10th of, the original volume isentropically. Determine the final pressure and, temperature and the work done on 1 m 3 of air. Assume R=287, J/kg.K. C p / Cv = 1.41, Given data, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial temperature, t1 = 40 o C ; T1 = 40 + 273 = 313 K, Final volume, V2 = 1/10th of initial volume, V1, V2, 1, V1, =, = 10, (or), V1 10, V2, 1, Initial volume, V1 = 1 m 3 ; ∴ V2 =, =0.1 m 3, 10, Gas constant, R = 287 J/kg.K, C p / Cv = γ = 1.41, To find, 1. Final pressure, p2 2. Final temperature, t2 3. Work done, W, Solution, 1. Fina pressure, p2, γ, , In isentropic process, p1 .V1 = p2 .V2, Unit – I P2.15, , γ
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γ, , V , ∴ p2 = p1 × 1 = 100 × (10)1.41 = 2570.396 kN/ m 2, V2 , 2. Final temperature, T2, V , T, In isentropic process, 1 = 1 , T2 V2 , , γ −1, , 1.41 −1, , V , = 313 × (10)0.41 = 804.534 K, ∴ T2 = T1 × 1 , V2 , t2 = 804.534 − 273 = 531.534 o C, 3. Work done, W, p .V − P2 .V2 (100 × 1) − (2570 .396 × 0.1), W= 1 1, =, 1.41 − 1, γ −1, = − 383.0234 kN−m, The negative sign indicates that work is done on the air., Results, 1. Final pressure, p2 = 2570.396 kN/ m 2, 2. Final temperature, T2 = 804.534 K; t2 = 531.534 o C, 3. Work done, W = 383.0234 kN−m, Example 2.16, 1 kg of air at 11 bar and 80 o C is expanded to 10 times the, original volume by (1) isothermal process and (2) isentropic, process. Determine the work done in each of the cases., R=287J/kg.K and γ=1.4, Given data, Initial pressure, p1 = 11 bar = 1100 kN/ m 2, Initial temperature, t1 = 80 o C ; T1 = 80 + 273 = 353 K, Final volume, V2 = 10 × original volume ( V1 ), V1, 1, =, V2 =10 V1 ;, V2 10, Gas constant, R = 287 J/kg.K = 0.287 kJ/kg.K, γ = 1.4, To find, 1. Work done in isothermal process, W, 2. Work done in isentropic process, W, Unit – I P2.16
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Solution, 1. Isothermal process, In isothermal prorcess,, , V , V , Work done, W = p1 .V1 . ln 2 = m. R.T1 . ln 2 , V1 , V1 , W = 1 × 0.287 × 353 × ln(10) = 233.2772 kN−m, 2. Isentropic process, p .V − P2 .V2 m. R.(T1 − T2 ), W= 1 1, =, γ −1, γ −1, , V , T, In isentropic process, 1 = 1 , T2 V2 , 1.41 −1, , γ −1, , 0.41, , V , 1 , = 353 × =140.532 K, ∴ T2 = T1 × 1 , V, 10 , 2, 1 × 0.287 × 9353 − 140.532), W=, = 152.4458 kN−m, 1.4 − 1, Result, 1) Work done in isothermal process, W = 233.2772 kN−m, 2) Work done in isentropic process, W = 152.4458 kN−m, Example 2.17, A perfect gas is compressed according to the law pV 1.25 = C, from an an initial pressure of 1 bar and volume 1 m3 to a final, volume of 0.5 m3 . Determine the final pressure and change in, entropy per kg of gas during the process. γ=1.4; R = 287 J/kg.K, Given data, Polytropic index, n = 1.25, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial volume, V1 = 1 m 3, Final volume, V2 = 0.5 m 3, Gas constant, R = 287 J/kg.K = 0.287 kJ/kg.K, γ = 1.4, To find, 1. Final pressure, p2, , 2. Change in entropy, ∆S, , Unit – I P2.17
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Solution, 1. Final pressure, p2, In polytropic prorcess, p1 .V1 = p2 .V2, n, , n, , n, , 1.25, , V , p2 V1 , = p2 = p1 × 1 , p1 V2 , V2 , 1.25, , 1 , p2 = 100 × , , 0.5 , , = 238.841 kN/ m 2, , 2. Change in entropy, ∆S, In a polytropic process, ∆S =, , V , γ−n, m. R. ln 2 , γ −1, V1 , , 1.4 − 1.25, 0. 5 , × 1 × 0.287 × ln, = −0.0746 kJ/kg.K, 1.4 − 1, 0.1 , Result, 1. Final pressure, p2 = 238.841 kN/ m 2, 2. Change in entropy, ∆S = −0.0746 kJ/kg.K, =, , Example 2.18, 0.25 kg of air at a pressure of 140 kN/ m 2 occupies 0.15, , m and from this condition, it is compressed to 1.4 mN/ m 2, according to the law p.V 1.25 = C . Determine (1) change in internal, 3, , energy (2) work done on or by the air (3) heat received or rejected, by the air (4) change in entropy. Assume C p =1.005 kJ/kg.K and, , Cv = 0.718 kJ/kg.K, Given data, Polytropic index, n = 1.25, Mass of the air, m = 0.25 kg, Initial pressure, p1 = 140 kN/ m 2, Initial volume, V1 = 0.15 m 3, Final prressure, p2 = 1.4 mN/ m 2 = 1400 kN/ m 2, C p = 1.005 kJ/kg.K, , Cv = 0.718 kJ/kg.K, Unit – I P2.18
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To find, 1. Chane in internal energy, ∆U, 3. Heat transfer, Q, , 2. Work done, W, 4. Change in entropy, dS, , Solution, R = C p − Cv = 1.005 − 0.718 = 0.287 kJ/kg.K; γ =, , Cp, Cv, , =1.4, , 1. Change in internal energy, ∆U, ∆U = m.Cv .(T2 − T1 ), By characteristic gas equation, p1V1 = m. R.T1, p .V, 140 × 0.15, ∴ T1 = 1 1 =, = 292.683 K, m. R 0.25 × 0.287, , p , T, In polytropic process, 2 = 2 , T1 p1 , , n −1, n, , n −1, , 0.25, , p n, 1400 1.25, = 292.683 × , ∴ T2 = T1 × 2 , = 463.871 K, , 140 , p1 , ∴ ∆U = 0.25 × 0.718 (463.871 − 292.683) = 30.728 kJ/kg.K, 2. Work done, W, m. R.(T1 − T2 ) 0.25 × 0.287 × (292.633 − 463.871), W=, =, =−49.130 kJ, 1.25 − 1, n −1, The negative sign indicates that work is done on the sytem., 2. Heat transfer, Q, 1.4 − 1.25, γ−n, × (−49.13) = −18.4237 kJ, Q=, ×W =, 1.4 − 1, γ −1, The negative sign indicates that heat is rejected from the system., 3. Change in entropy, ∆S, V , T , dS = m. R. ln 2 + m.Cv . ln 2 , V1 , T1 , n, , 1, , V , p n, p, In a polytropic process, 1 = 2 V2 = V1 × 1 , p2 V1 , p2 , 1, , 140 10, 3, ∴ V2 = 0.15 × , = 0.02377 m, 1400 , Unit – I P2.19
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0.02377 , 463.371 , dS = 0.25 × 0.287 × ln, + 0.25 × 0.712 × ln, , 0, ., 15, , , 292.683 , = − 0.1322 + 0.08197 = −0.05022 kJ/K, Result, 1. Chane in internal energy, ∆U = 30.728 kJ, 2. Work done, W = − 49.130 kJ, 3. Heat transfer, Q = − 18.423 kJ, 4. Change in entropy, ∆S = − 0.05022 kJ/K, Example 2.19, 2.5 kg of an ideal gas is expanded from a pressure of 7 bar, and volume of 1.5 m3 to a pressure of 1.4 bar and volume 4.5 m3 ., The change in internal energy is 500 kJ. Specific heat at constant, volume is 0.72 kJ/kg.K. Determine (1) gas constant, (2) index of, polytropic expansion, (3) work done, (4) initial and final, temperature., Given data, Mass of the air, m = 2.5 kg, Initial pressure, p1 = 7 bar = 700 kN/ m 2, Initial volume, V1 = 1.5 m 3, Final prressure, p2 = 1.4 bar = 140 kN/ m 2, Final volume, V2 = 4.5 m 3, Change in internal energy, ∆U = 500 kK, Cv = 0.72 kJ/kg.K, To find, 1. Gas constant, R 2. Index of polytropic expansion, n, 3. Work done, W 4. Initial and final temperature, T1 and T2, Solution, 1.Polytropic index, n, p , log 10 1 log 10 700 , p2 =, 140 = 1.465, In polytropic process, n =, V , 4. 5 , , log 10 2 log 10 , 1. 5 , V, 1, 2. Work done, W, p .V − p2 .V2 (700 × 1.5) − (140 × 4.5), W= 1 1, =, = 903.226 kJ, n −1, 1.465 − 1, Unit – I P2.20
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3. Initial and final temperature, T1 and T2, ∆U = m.Cv .(T1 − T2 ) =500, 500, 500, =, = 277.7778 K, T1 − T2 =, m.Cv 2.5 × 0.72, n −1, , ….. (1), , 1.465 −1, , T1 V2 , 4.5 , = , =, = 1.6667, , T2 V1 , 1.5 , ∴ T1 = 1.6667 T2, ….. (2), Substitute (2) in (1), 1.6667 T2 − T2 = 277.7778, 0.6667 T2 = 277.7778, 277.7778, T2 =, = 416.646 K ; t2 = 416.646 − 273 = 143.646 o C, 0.6667, T1 = 1.6667 T2 = 1.6667 × 416.646 = 694.424 K, Also,, , t1 = 694.424 − 273 = 421.424 o C, 4. Gas constant, R, , m. R.(T1 − T2 ), n −1, W .(n − 1), 903.226 × (1.465 − 1), =, = 0.6048 kJ/kg.K, ∴R =, m.(T1 − T2 ), 2.5 × (277.7778), Result, 1. Gas constant, R = 0.0648 kJ/kg.K, 2. Index of polytropic expansion, n = 1.465, 3. Work done, W = 903.226 kJ, 4. Initial temperature, T1 = 694.424 K ( t1 = 421.424 o C ), Work done, W =, , Final temperature, T2 = 416.646 K ( t2 = 143.646 o C ), Example 2.20, An ideal gas of molecular weight 30 and specific heat ratio, 1.38 is compressed according to the law p.V 1.25 = constant, from a, pressure of 1 bar abs and 15 o C to a pressure of 16 bar abs., Calculate the temperature of the gas at the end of compression, the, heat received or rejected and the work done on or by the gas during, the process. Assume the mass of the gas to be 1kg. use the, calculated value of Cv ., Unit – I P2.21
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Given data, Molecular weight, M = 30, Ration of specific heats, γ = 1.38, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial temperature, t1 = 15 o C , T1 = 15 + 273 = 288 K, Final prressure, p2 = 16 bar = 160 kN/ m 2, Mass of gas, m = 1 kg, To find, 1. Temperature at the end of compression, T2, 2. Heat transferred, Q, , 3. Work done, W, , Solution, 1. Temperature at the end of compression, T2, , p , T, In polytropic process, 2 = 2 , T1 p1 , p , ∴ T2 = T1 × 2 , p1 , , n −1, n, , n −1, n, , 1600 , = 288 × , , 100 , , 1.25 − 1, 1.25, , = 501.437 K, , T2 = 501.437 K ; t2 = 501.437 − 273 = 228.437 o C, 2. Work done, W, m. R(T1 − T2 ) 1 × 0.277 × (288 − 501.437), W=, =, = − 236.488 kN−m, 1.25 − 1, n −1, The negative sign indicates that work is done on the system., 3. Heat transferred, Q, Universal gas constant, Ru = 8.314 kJ/kg.K, R, 8.314, Characteristic gas constant, R = u =, = 0.277 kJ/kg.K, M, 30, Specific heat, Cv =, , R, 0.27, = 0.729, =, γ − 1 1.38 − 1, , γ−n, 1.38 − 1.25, ×W =, × 236.488, γ −1, 1.38 − 1, Q = − 80.904 kJ, The negative sign indicates that heat is rejected from the system., , Heat transferred, Q =, , Unit – I P2.22
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Result, 1.Temperature at the end of compression, T2 =501.437 K; t2 = 228.437 o C, 2. Heat transferred, Q = − 80.904 kJ, 3. Work done, W =− 236.488 kN−m, , PROBLEMS FOR PRACTICE, 1., , A certain gas occupies a a space of 0.3 m 3 at a pressure of 2, bar and temperature of 77 o C . It is heated at a constant, volume, until the pressure is 7 bar. Determine the (1) the, temperature at the end of the process (2) mass of the gas (3), change in internal energy and (4) change in enthalpy during, the process. Assume C p =1.005 kJ/kg.K; C p =0,712 kJ/kg.K;, and R=287J/kg.K, [Ans: t2 =952 o C , m=0.597 kg, ∆U =372 kJ, ∆H =525kJ], , 2., , 5 kg of gas is heated from a temperature of 100 o C at a, constant volume till its pressure becomes three times it, original pressure. Calculate (1) the heat transfer (2) change, in internal energy (3) change in enthalpy and (4) change in, entropy. Assume C p =1.00 kJ/kg.K and Cv =0.71 kJ/kg.K, [Ans: Q=2648 kJ, ∆U =2648 kJ, ∆H =3730kJ, ∆S =3.9kJ/K, , 3., , A gas has a density of 1.875 kg/ m 3 at 1 bar and 15 o C ., Calculate the characteristic gas constant R. When 0.9 kg of, this gas is heated from 15 o C to 250 o C at constant pressure,, the heat required is 176 kJ. Calculate the specific heat, capacity of the gas at constant pressure and specific heat, capacity of the gas at constant volume. Also calculate the, chang ein internal energy and external workdone during the, process. [Ans: R= 0.1852 k/kg.K; C p = 0.8274 kJ/kg.K ;, , Cv = 0.6422 kJ/kg.K; ∆U =135.8253 kJ and W=39.1747 kJ], 4., , A quantity of gas has a volume of 0.14 m 3 , pressure 1.5 bar, and a temperature 100 o C . If the gas is compressed at a, constant pressure, until its volume becomes 0.113 m 3 ,, determine (1) the temperature at the end compression (2), Unit – I P2.23
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work done in compressing the gas (3) decrease in internal, energy and (4) heat given out by the gas. Assume C p =1.005, kJ/lg.K; Cv =0.712 kJ/kg.K and R=285 J/kg.K, [Ans: t2 = 25.4 o C , W=4.2 kJ, ∆U =10.46 kJ, Q=14.77 kJ], 5., , A gas whose pressure, volume and temperature are 275 kN,, 0.09 m 3 , and 185 o C respectively has changed its state at, constant pressure until its temperature becomes 15 o C ., Calculate (1) the heat transferred (2) work done and (3) change, in entropy. Assume R=0.28 kJ/kg.K and C p =1.005 kJ/kg.K, [Ans: Q = −32.974 kJ, W = −9.1868 kJ, ∆S = 0.0899 kJ/K], , 6., , One kg of air at 1.5 bar and 30 o C undergoes a constatn pressure, process until the volume is trebled. Calculate (a) change in, internal energy, (b) change in enthalpy, (c) work done and (d), change in entropy. C p =1 kJ/kg.K and Cv =0.71 kJ/kg.K, [Ans: ∆U =430.26kJ, ∆H =606kJ, W=175.74 kN−m,, ∆S =1.0986 kJ/K], , 7., , A quantity of air has a volume of 0.4 m 3 at a pressure of 5 bar, and a temperature of 80 o C . It is expanded in cylinder at a, constant temperature to a pressure of 1 bar. Determine the, amount of work done by the air during expansion., [Ans: W=321.54 kJ], , 8., , 0.2 kg of air at a pressure of 1.1 bar and 15 o C is compressed, isothermally to a pressure of 5.5 bar. Calculate (a) the final volume, (b) heat rejected, (c) change in entropy and (d) change in internal, energy. Assume R=0.292 kJ/kg.K. [Ans: V2 =0.03058 m 3 ,, Q=27.069 kJ, ∆S = − 0.09399 kJ/K , ∆U =0], , 9., , 0.5 kg of air at a pressure of one bar occupies a volume of 0.4, m 3 . If this air expands isothermally to a volume of 0.8 m 3 ,, dinf (1) the initial temperature, (2) external work done and, (3) change in internal energy. Assume R=0.29 kJ/kg.K, [Ans: t1 =2.862 o C , W=27.726 kN−m, ∆U = 0], , 10. A gas is compressed hyperbolically from a pressure and, volume of 100 kN/ m 2 and 0.056 m 3 respectively to a volume, Unit – I P2.24
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of 0.007 m 3 . Determine the final pressure and the work done, on the gas., [Ans: p2 =800 kN/ m 2 , W=11.6449 kJ], 11. A quantity of air occupies a volume of 30 litres at a, temperature of 38 o C and a pressure of 104 kN/ m 2 . The, temperature of the air is raised by adiabatic compression until, the volume becomes 6 litres. Find the final temperature, the, external work done, the change of internal energy, the heat, transferred, the chang ein enthalpy and the chagne in entropy., Take R=0.29 kJ/kg.K and γ=1.4. [Ans: t2 =319.036 o C , W=, , −7.0485 kN−m, ∆U =7.0485 kJ, Q=0, ∆H =9.868 kJ, ∆S =0], 12. 1 kg of gas expands adiabatically and its temperature is, observed to fall from 240 o C to 100 o C while its volume increase, to 1.8 times the initial volume. The work done by the gas is 85, kJ in the process. Determine C p , Cv and R for the gas., [Ans: C p =0.934 kJ/kg.K, Cv =0.607 kJ/kg.K,, R=0.327kJ/kg.K], 13. 0.5 kg of a gas occupies 0.3 m at 20 C and 140 kN/ m 2 and, 3, , o, , after adiabatic compression to 0.15 m 3 , the pressure is 370 kN/, m 2 . Find the value of gas constant and the two specific heats., [Ans: R=0.287 kj?kg.K, C p =1.001 kJ/kg.K,, , Cv =0.714 kJ/kg.K], 14. 0.3 m 3 of a gas at 14 o C and 1 bar is compressed isentropically, to 7 bar. Calculate (a) mass of the gas, (b) the final temperature,, (c) work done. Assume C p =1 kJ/kg.K and Cv =0.715 kJ/kg.K, [Ans: m=0.36677 kg, t2 =226.73 o C , W=55.78 kJ], 15. A gas expands according to the law p.V 1.3 = C from a pressure, of 900 kN/ m 2 and a volume of 0.0025 m 3 to a pressure of 100, kN/ m 2 . How much heat was received or rejected by the gas, during this process? Determine also polytropic specific heat., Take γ=1.4 and Cv =0.718 kJ/kg.K, [Ans; Q=0.7458 kJ, Cn=0.239 kJ/kg.K], 16. 2 kg of gas is compressed from 150 kN/ m 2 and 16 o C to 750, kN/ m 2 according to the law p.V 1.3 = C . Determine (1) initial, Unit – I P2.25
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volume, (2) final temperature, (3) work done, (4) change in, internal energy, (5) heat transferred, (6) change in enthalpy, and (7) change in entropy. Assume R=0.280 kJ.kg.K and, [Ans: V1 =1.0789 m3 , t2 =145.985 o C ,, C p =1.0 kJ/kg.K, W=−242.638 kJ, ∆U = 187.178 kJ.kg.K, Q= − 55.46, kJ/kg.K, ∆H = 259.97 kJ, ∆S = − 0.1584 kJ/kg.K], 17. 0.675 kg of a gas at 14 bar and 280 o C is expanded to four times, the original volume according to the law p.V 1.32 = C . Determine, (1) the initial and final volume of the gas, (2) final pressure and, temperature fo the gas and (3) work done. Assume R=287 J/kg.K., [Ans: V1 =0.07652 m3 , V2 =0.30608 m 3 , p2 =224.6 kN/ m 2 ,, , t2 =81.87 o C , W=119.945 kN−m], 18. A perfect gas is compressed according to the law p.V 1.25 = C from, an initial pressure of 1 bar and volume 1 m 3 to a final volume of, 0.5 m 3 . Determine the finalpressure and the change of entropy, per kg of gas during the process. γ=1.4 and R=287 J/kg.k., [Ans: p2 =237.84 kN/ m 2 , ∆S = − 0.0746 kJ/kg.K], 19. A gas having a volume of 0.075 m 3 at 16 o C and 7 bar pressure, expands to 0.35 m 3 according to the law p.V 1.2 = C . Determine, (a) the final temperature, (b) work done, (c) heat transferred,, and (d) change in entropy. Assume γ=1.4 and R=0.2927 kJ/kg.K., [Ans; t2 = −60.63 o C , W=69.6 kJ, Q=34.8 kJ, ∆S =0.1399 kJ/K], , Unit – I P2.26
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Unit – II, Chapter 3., , THERMODYNAMIC AIR CYCLES, , 3.1 Introduction, A thermodynamic cycle consists of a series of, thermodynamic operations, which takes place in a certain order,, and the initial conditions are restored at the end of the process., These processes can be plotted on p−V diagram and T−S diagram., Each process is represented by its own curve and thus form a closed, figure., 3, , p, , 3, , 2, , 4, 4, , 2, T, , 1, , 1, S, , V, (a) p-V diagram, , (b) T-S diagram, , Fig.3.1 Thermodynamic cycle, , The area enclosed by the complete curve in the p−V, diagram gives the work done during the complete cycle. The net, heat transfer during the complete cycle is given by the are enclosed, by the complete curve in T−S diagram., 3.2 Air cycles and air standard efficiency, In air cycles, the air is used as the working fluid. The air in, an engine cylinder may be subjected to series of operations which, cause the air to return to its original state. This is called as air, cycle. The thermal efficiency obtained using air as working fluid is, known as air standard efficiency. This efficiency is the standard, efficiency for all cycles. It is used to compare the performance of, engines working on various cycles., , Unit – II , , 3.1
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3.3 Efficiencies of cycles, Thermal efficiency, During a cyclic process, a certain quantity of heat is taken, by the engine. A portion of this heat is converted into useful work, and the remaining heat is rejected. Therefore, work done during, the cycle is given by,, Work done, W = Heat supplied − Heat rejected, , W = Qs − Qw, , Efficiency of the cycle is given by,, , η=, , Output, Work done, Heat supplied- Heat rejected, =, =, Input Heat supplied, Heat supplied, , η=, , W, Q − Qr, = s, Qs, Qs, , This efficiency is the theoretical efficiency and is known as, thermal efficiency of the cycle. In this efficiency, the practical losses, which may occur during the running of an engine are not, considered. Hence, the actual thermal efficiency is always less than, the theoretical thermal efficiency., Relative efficiency, The relative efficiency is defined as the ratio of actual, thermal efficiency and theoretical thermal efficiency. It is also, called as efficiency ratio., Relative efficiency (or) Efficiency ratio,, Indicated thermal efficiency (or), Actual thermal efficiency, η rel =, Theoretica l (Ideal) theraml efficiency (or), Air standard efficiency, 3.4 Reversible and irreversible cycle, Reversible cycle, A thermodynamic process is said to be reversible, if the, system and surroundings are completely restored back to their, initial state when the process reversed., Unit – II , , 3.2
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2, , 2, , p, , p, 1, , 1, , V, , V, , (a) Reversible process, , (b) Irreversible process, , Fig.3.2 Reversible and irreversible process, , A reversible process occurs when the system passes, through a serious of equilibrium state. In a reversible process,, there should not be any loss of heat due to friction, radiation or, conduction, etc. A cycle will be reversible if all the processes, constituting the cycle are reversible. Thus in a reversible cycle, the, initial conditions are restored at the end of the cycle., Irreversible cycle, When the system and surroundings are not completely, restored by reversing the process, then the process is known as, irreversible process. In an irreversible process, there is a loss of, heat due to friction, radiation or conduction., • The main causes for the irreversibility are:, • Mechanical and fluid friction, • Unrestricted expansion, • Heat transfer with large temperature difference, A cycle will be irreversible, if any of the processes,, constituting the cycle, is irreversible. Thus in an irreversible cycle,, the initial conditions are not restored at the end of the cycle., 3.5 Thermodynamic reversibility, A process is thermodynamically reversible, if, • It passes through a number of equilibrium stages, • The change from one stage to successive stage is very slow, or quasi−static., • There is no frictional resistance, Unit – II , , 3.3
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Effects of thermodynamic reversibility on efficiency, • No other cycle can be more efficient than a reversible cycle, (Carnot cycle) when working between the same two fixed, temperature., • All reversible engines working between the same, temperature limit have the same efficiency., • Two reversible engines of different capacities working with, different mediums and operating between the same, temperature limits have the same efficiency., 3.6 Conditions for reversibility, The following are the conditions for reversibility of a cycle, • There should be no loss of energy during the cycle of, operation, • There should not be any free expansion process or, throttling process., • The pressure and temperature of the working substance, should be same as that of the surroundings., •, •, , The working substance must be a good conductor of heat., All the processes taking place in the cycle of operation must, be extremely slow., , •, , The working parts of the engine must be friction free., , 3.7 Assumptions in deriving air standard efficiency, • Air is the working fluid, • The gas in the engine cylinder is a perfect gas. It obeys the, gas laws., •, , All the compression and expansion processes are adiabatic., , •, •, , There is no friction inside the cylinder., Heat is supplied by bringing a hot body in contact with the, cylinder at certain points during the process. Similarly, heat is rejected by bringing a cold body in contact with the, cylinder at these points., , •, , The cycle is considered as a closed cycle., , •, , There is no chemical reaction in the engine cylinder., Unit – II , , 3.4
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3.8 Thermodynamic cycles, The following are the important thermodynamic cycles, 1) Carnot cycle [constant temperature cycle], 2) Otto cycle [constant volume cycle], 3) Diesel cycle, 4) Joule or Brayton cycle [constant pressure cycle], 5) Dual combustion cycle, 6) Rankine cycle, 7) Stirling cycle, 8) Ericsson cycle, 3.9 Carnot cycle [Constant temperature cycle], 2, , Qs, , Isothermal, 3, , W, , p, 1, , Adiabatic, 4, , Qr, , T2, , 2, , Isothermal, 3, Adiabatic, , Q, , T, T1, , 1, , 4, , S3=S4, S, (b) T-S diagram, S1 = S2, , V, (a) p-V diagram, , Fig.3.3 Carnot cycle, , This cycle was introduced by Nicolas Leonard Sadi Carnot., It consists of two isothermal processes and two isentropic, (reversible adiabatic) processes. The p−V and T−S diagrams of this, cycle are shown in the figure., Working of the cycle, At point 2, the clearance volume of the cylinder is occupied, by compressed air. A hot object is placed at the end of cylinder head, and the heat is supplied to the air at constant temperature ( T2 )., The air expands and forces the piston forward by doing work on the, piston. At point 3, the hot object is removed and the air expands, isentropically till the temperature falls to T1 at point 4. The piston, reverses and compression stroke starts. A cold object is placed at, the end of the cylinder head and the air is cooled at constant, temperature ( T1 ) up to point 1. At point 1, the cold object removed, and the air is compressed adiabatically (isentropically) till it, returns to its original state 2., Unit – II , , 3.5
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Heat supplied during the constant temperature process 2−3,, dQ , Qs = T2.dS Q dS =, T , , Heat rejected during the constant temperature process 4−1,, , Qr = T1.dS, Work done during the cycle,, , W = Qs − Qr = T2.dS− T1.dS, , W = (T2 − T1).dS, W (T2 − T1 ).dS, =, Qs, T2 .dS, (T − T1 ), T, = 2, = 1− 1, T2, T2, , Efficiency of the cycle, ηcarnot =, , ηcarnot, where,, , T1 = Minimum temperature of the cycle,, T2 = Maximum temperature of the cycle, , For the two given temperature limits, only Carnot cycle, gives the maximum possible efficiency. But, an engine working on, Carnot cycle is not possible because of the following reasons:, • For isothermal compression, the piston should move very, slowly and for isentropic compression, the piston should, move as fast as possible. This speed variation during the, same stroke of the piston is not possible., • It is not possible to provide a heat source which will supply, heat without change in temperature., • It is not possible to avoid friction between moving parts, completely., 3.10 Reversed Carnot cycle, The p−V and T−S diagram for a reversed Carnot cycle is, shown in the figure., Working of the cycle, 1−2 : Isentropic compression in a compressor, 2−3 : Isothermal heat rejection to a hot body ( T2 = T3 ), 3−4 : Isentropic expansion in expansion valve, 4−1 : Isothermal heat extraction from a cold body ( T1 = T4 ), Unit – II , , 3.6
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3, , n, , 4, , on, ssi, , W, , 2, , 3, , 2, re, mp, Co, , io, ans, Exp, , p, , Qr, , Qe, , Q, 1, , V, (a) p-V diagram, , T, 4, , 1, , S, (b) T-S diagram, , Fig.3.4 Reversed Carnot cycle, , Heat is extracted during the isothermal process 4−1., Heat extracted,, , Qe = T1.dS, , Heat is rejected during the isothermal process 2−3, , Qr = T2.dS, Work input, W = Qr − Qe = T2.dS− T1.dS= (T2 − T1).dS, Heat rejected,, , The main purpose of Carnot cycle is to extract heat from a, cold body and reject heat to a hot body. In this case, the, performance of cycle is called as Coefficient of Performance (COP)., In the case of refrigerator, we require a cooling effect. So, COP is, defined as the ratio of heat rejected to work input., Heat extracted Qe, T1 .dS, COPref =, =, =, Work input, W (T2 − T1 ).dS, ∴ COPref =, , T1, T2 − T1, , In the case of heat pump, heating effect is required. So,, COP is defined as the ratio of heat delivered to the work input., Heat rejected Qr, T2 .dS, COPhp =, =, =, Work input, W (T2 − T1 ).dS, ∴ COPhp =, , T2, T2 − T1, , 3.11 Otto cycle or Constant volume cycle, The first successful engine working on this cycle was, introduced by Nicholas A. Otto in 1876. Nowadays, many gas,, Unit – II , , 3.7
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petrol and many of the oil engines run on this cycle. Since, the heat, is supplied and rejected at a constant volume, it is also known as, constant volume cycle., 3, Qs, , 3, , Adiabatic, , 2, , S=C, , C, V=, , Vc, , 4, , 4, Qr, 1, , p, Vs, , V2=V3, , 2, T, , 1, , C, V=, , S1= S2, , S3=S4, S, (b) T-S diagram, , V1=V4, , V, (a) p-V diagram, , S=C, , Fig.3.5 Otto cycle, , The ideal p−V diagram and T−S diagram of this cycle are, shown in the figure. It consists of two reversible adiabatic, (isentropic) processes and two constant volume processes., Working of Otto cycle, 1−2 : Isentropic compression of air : Pressure and temperature, increases. Volume decreases. Entropy remains constant ( S1, , = S2 ), , 2−3 : Constant volume heating of air : Pressure, temperature and, entropy increases. Volume remains constant ( V1 = V2 ), 3−4 : isentropic expansion of air : Pressure and temperature, decreases. Volume increases. Entropy remains constant ( S3, , = S4 ), , 4−1 : Constant volume heat rejection from air: Pressure,, temperature and entropy decreases. Volume remains constant (, , V4 = V1 ), Heat is supplied at constant volume during the process 2−3 :, ∴ Heat supplied,, , Qs = m.Cv.(T3 − T2), , Heat is rejected at constant volume during the process 4−1 :, , Qr = m.Cv.(T4 − T1), Work done, W = Qs − Qr = m.Cv.(T3 − T2) − m.Cv.(T4 − T1), ∴ Heat rejected,, , Unit – II , , 3.8
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3.13 Diesel cycle, Diesel cycle was introduced by Rudolph Diesel. This cycle, is mostly used in diesel engines. It consists of two adiabatic, processes, one constant volume process and one constant pressure, process. The p−V diagram and T−S diagram for this cycle are, shown in the figure., 2, , p2 = p3, , Qs, , 3, Adiabatic, , 3, p=, , 4, Qr, 1, , p, Vc, , Vs, , 4, 2, T S=C, 1, , V1=V4, , V, (a) p-V diagram, , S=C, , C, , V=, , C, , S1 = S2, , S3=S4, S, (b) T-S diagram, , Fig. 3.7 Diesel cycle, , Working of the cycle, , = S2 ), 2−3:Air is heated at constant pressure ( p2 = p3 ). Heat supply is, stopped when temperature reaches to T3 at point 3. This point 3, is known as cut−off point. The volume at this point ( V3 ) is, , 1−2:Air is compressed isentropically ( S1, , known as cut−off volume., , = S4 ), 4−1:Air is cooled at constant volume ( V4 = V1 ), , 3−4:Air is expanded isentropically ( S3, , Heat is supplied during constant pressure process 2−3, Heat supplied, Qs = m.Cp.(T3 − T2 ), Heat is rejected during constant volume process 4−1, Heat rejected,, , Qr = m.Cv.(T4 − T1), , Work done during the cycle, W =, , Qs − Qr, , W = m.Cp.(T3 − T2 ) − m.Cv.(T4 − T1), Efficiency, η diesel =, , m.C p .(T3 − T2 ) − m.C v (T4 − T1 ), W, =, Qs, m.C p .(T3 − T2 ), , Unit – II 3.12
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3.14 Comparison of Otto cycle and Diesel cycle, Otto cycle, 1. It consists of two adiabatic and, two, constant, volume, processes., 2. Heat is supplied at constant, volume., 3. Efficiency of the cycle depends, on compression ratio (r) only., , Diesel cycle, It consists of two adiabatic,, one constant pressure and one, constant volume processes., Heat is supplied at constant, pressure., Efficiency of the cycle depends, on compression ratio (r) and, cut−off ratio ( ρ )., , 4. Compression ratio is equal to, expansion ratio., 5. For the same compression, ratio and same heat input, the, efficiency of Otto cycle is more, than that of Diesel cycle., 6. Compression ratio is less. It, varies from 5 to 8., , Compression ratio is not equal, to expansion ratio., For the same compression, ratio and same heat input, the, efficiency of Otto cycle is less, than that of Diesel cycle., Compression ratio is more. It, varies from 12 to 18., , 3.15 Actual p−V diagram of Otto cycle, d, , c, p, , e, b, , a, Vc, , Vs, V, , Fig.3.8 Actual p−V diagram of Otto cycle, , The actual p−V diagram of Otto cycle is shown in the figure., An engine working on Otto cycle consists of the following strokes:, 1. Suction stroke (a−b): The air−fuel mixture is sucked in to the, engine cylinder by the outward movement of the piston., 2. Compression stroke (b−c): The air− fuel mixture is compressed, by the piston during the return stroke. At the end of compression,, the mixture occupies clearance volume ( Vc ) only., Unit – II 3.14
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3. Heat addition at constant volume (c−d) : The air−fuel, mixture is ignited by an electric spark. This rises the pressure and, temperature of the mixture., 4. Expansion or working or power stroke (d−e) : The burnt, gases force the piston to move forward. As this expansion occurs at, high speed, it is considered as adiabatic (isentropic). This working, stroke gives sufficient energy to drive the piston through the other, three strokes. The excess energy is stored in the fly wheel., 5. Exhaust stroke (e−g) : The burnt gases are pushed out through, the exhaust valve by the inward movement of the piston. Thus the, cycle is completed., We know that the net work done during a cycle is, represented by the net area of the p−V diagram. The piston, requires energy to act as a pump during suction and exhaust, strokes. This energy can be supplied by the fly wheel. The energy, lost due to the area ‘eab’ is called as negative energy or pumping, energy and it is negative., ∴ Net work done = Area ‘bcde’ − Area ‘eab’, 3.16 Actual p−V diagram of Diesel cycle, c, , d, Adiabatic, , e, p, , a, Vc, , b, Vs, , V, Fig.3.9 Actual p−V diagram of Diesel cycle, , Unit – II 3.15
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The actual p−V diagram of the Diesel cycle is shown in the, figure. It consists of the following strokes., 1. Suction stroke (a−b) : During this stroke, air alone is drawn, into the engine cylinder by the outward movement of the piston., 2. Compression stroke (b−c) : The air is compressed adiabatically, by the inward movement of the piston. The air now occupies, clearance volume ( Vc ) only. The final pressure is above 40, atmosphere at the end of this stroke. The temperature of this high, pressure air is sufficient to ignite the fuel., 3. Heat addition at constant pressure (c−d) : The atomized fuel, is injected into the engine cylinder and it is immediately ignited by, the hot air in the cylinder. The burning of fuel takes place at, constant pressure. The fuel supply is cut−off when the piston, reaches ‘d’., 4. Working or expansion of power stroke (d−e) : The burnt, mixture forces the piston outward, thus doing work. This, expansion is adiabatic on account of high speed. This is the only, working stroke during the cycle. It gives sufficient power to drive, the engine and drive the piston through the other three strokes., 5. Exhaust stroke (e−a) : The burnt gases are pushed through the, exhaust valve by the inward movement of the piston. Thus the, cycle is completed., The net work done during a cycle is represented by the net, area of the p−V diagram. The piston requires energy to act as a, pump during suction and exhaust strokes. This energy can be, supplied by the fly wheel. The energy lost due to the area ‘eab’ is, called as negative energy or pumping energy and it is negative., ∴ Net work done = Area ‘bcde’ − Area ‘eab’, , Unit – II 3.16
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3.17 Comparison of theoretical an actual p−V diagram, , 1., , 2., 3., , 4., , Theoretical p−V, diagram, The pressure during the, suction, stroke, is, atmospheric pressure., Exhaust takes place at, atmospheric pressure., Net work done during the, cycle is given by the, enclosed are of the p−V, diagram., The, corners, of, the, theoretical p−V diagram, are sharp., , Actual p−V diagram, The pressure during the suction, stroke is less than atmospheric, pressure., Exhaust takes place above, atmospheric pressure., Net work done is obtained after, subtracting the energy lost due to, pumping during suction and, exhaust strokes., The corners of the theoretical p−V, diagram are rounded., , 3.18 Dual combustion cycle or semi−diesel cycle, 3, , Qs, , 4, Adiabatic, , Qs, 2, , 3, , W, , p, Vc, , Vs, , V2=V3, , V, (a) p-V diagram, , 5, Qr, 1, , C, p=, , C, V=, , 4, S=C, , 5, , 2, T, , V1=V5, , S=C, , 1, , V=C, , S1 = S2, , S4=S5, S, (b) T-S diagram, , Fig.3.10 Dual combustion cycle, , This cycle is a combination of Otto and Diesel cycles. In this, cycle, heat is absorbed partly at constant volume and partly at, constant pressure., Working of the cycle, , = S2 ), 2−3 : Air is heated at constant volume ( V2 = V3 ), 3−4 : Air is heated at constant pressure ( p3 = p4 ), 1−2 : Air is compressed isentropically ( S1, , Unit – II 3.17
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Referring the p−V diagram,, Area ‘abcda’ = Area ‘aefga’, ∴ pm =, , Workdoneduringthe cycle, ;, Strokevolume, , pm =, , W, Vs, , Mean effective pressure is determined with the help of, indicator diagram., Let,, , A − Area of indicator (p−V) diagram ( m 2 ), S − Spring scale or pressure scale (kN/ m 2 /m), L − Length of indicator diagram (m), , Then, mean effective pressure is given by,, , pm =, , Area of indicatordiagram× Springscale A × S, =, Lenghtof indicatordiagram, L, , If the mean effective pressure is based on the area of p−V, diagram or on indicated power, it is known as indicated mean, effective pressure ( pmi )., , pmi =, , Indicatedpower, No.of workingstrokesper second× Strokevolume, , pm =, , IP, n × Vs, , If the mean effective pressure is based on the brake power, of an engine, it is known as brake mean effective pressure ( pmb)., Brake power, BP, pmb =, =, n × Vs, n × Vs, , Unit – II 3.20
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REVIEW QUESTIONS, 1., , Define air standard efficiency., , 2., , What are the assumptions made in deriving the air−standard, efficiency?, , 3., , Draw the p−V and T−S diagrams of Carnot cycle., , 4., , Derive an expression for air−standard efficiency of Carnot cycle., , 5., , No engine can work on Carnot cycle. Why?, , 6., , Derive an expression for air−standard efficiency of reversed, Carnot cycle., , 7., , Explain the reversible and irreversible processes., , 8., , State the conditions of reversibility., , 9., , Draw the p−V and T−S diagram of the Otto cycle and indicate, the various processes., , 10. Derive an expression for the air−standard efficiency of Otto cycle., 11. What is the effect of compression ratio on efficiency of Otto cycle., 12. Draw the p−V and T−S diagram of Brayton’s (Joule’s) cycle., 13. Derive an expression for the air−standard efficiency of, Brayton cycle in terms of compression ratio., 14. Derive an expression for the air−standard efficiency of, Brayton cycle in terms of pressure ratio., 15. Draw the p−V and T−S diagrams of Diesel cycle and mention, the various processes., 16. Derive an expression for the air−standard efficiency of Diesel cycle., 17. What are the limitations of compression ratio/, 18. Explain the effect of cut−off ratio and compression ratio on, efficiency of Diesel cycle., 19. Compare Otto cycle and Diesel cycle., 20. Draw the p−V and T−S diagrams for Dual combustion cycle., 21. Draw the theoretical and actual p−V diagrams of Diesel cycle, and explain the reasons for differences., 22. What is meant by mean effective pressure., Unit – II 3.21
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SOLVED PROBLEMS, Example 3.1, While undergoing a Carnot cycle, the working fluid receives, heat at a temperature of 317 o C and reject heat at a temperature of, 22 o C . Find the theoretical efficiency of the cycle., Given data, Minimum temperature, t1 = 22 o C ; T1 =22 + 273 = 295 K, Maximum temperature, t2 = 317 o C ; T2 =317 + 273 = 590 K, To find, Efficiency of the cycle, η carnot, Solution, Efficiency of Carnot cycle,, T − T1 590 − 295, η carnot = 2, =, = 0.5 = 50%, T2, 590, Result, Efficiency of the cycle, η carnot = 50%, , Example 3.2, An ideal heat engine working on Carnot cycle converts 20, percentage of heat supplied into work. When the temperature of the, sink is reduced by 80 o C , its efficiency is doubled. Determine the, temperature of the source and sink., , Given data, Case I:, Efficiency of Carnot cycle, η carnot = 20% = 0.2, Case II:, Temperature of the sink, t1, = 80 o C ; T1 = 80 + 273 = 353 K, Efficiency of Carnot cycle, η carnot = 40% = 0.4, , To find, 1. Temperature of sink, T1, , Unit – II , ,, , 2. Temperature of source, T2, , P3.1
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Solution, Case I:, Efficiency of Carnot cycle,, T, η carnot = 1 − 1 = 0.2, T2, T1, = 1 − 0.2 = 0.8, T2, T1 = 0.8 T2, , ….. (1), , Case II:, , η carnot = 1 −, , T1, = 0.4, T2, , T1, = 1 − 0.4 = 0.6, T2, , T1 = 0.6 T2, 353 = 0.6 T2, 353, = 588.33 K ; t1 = 588.33 − 273 = 315.33 o C, T2 =, 0.6, Substitute the value of T2 in (1), , T1 = 0.8 T2 = 0.8 × 588.33 = 470.664 K ;, t1 = 470.664 − 273 = 197.664 o C, Result, 1. Temperature of sink, T1 = 470.664 K ( t1 =197.664 o C ), 2. Temperature of source, T2 = 588.33 K ( t2 = 315.33 o C ), , Example 3.3, An inventor claims a new engine that will develop 2.5 kW, for a heat addition of 300 kJ/min. The highest temperature of the, cycle is 1800 K and the lowest temperature is 600 K. Examine the, feasibility of the engine., , Given data, Power developed, P = 2.5 kW, ∴ Work done, W = 2.5 kJ/s, Heat supplied, Qs = 300 kJ/min =, , Unit – II , ,, , P3.2, , 300, = 5 kJ/s, 60
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Highest temperature, T2 = 1800 K, Lowest temperature, T1 = 600 K, To examine, The feasibility of the engine, Solution, For the two given temperature limits only Carnot cycle gives, maximum possible efficiency., Maximum efficiency,, T − T1 1800 − 600, η carnot = 2, =, = 0.6667 = 66.67%, T2, 1800, Efficiency of the engine,, ηengine = Work done = W = 2.5 = 0.5 = 50%, Heat supplied Qs, 5, Result, , ηcarnot > ηengine . Hence this is possible. Hence inventor’s claim, is correct., , Example 3.4, A Carnot engine working between 650K and 310K produces, 150 kJ of work. Find (i) thermal efficiency and (ii) heat added, during the process., , Given data, Maximum temperature, T2 = 650 K, Minimum temperature, t1 = 10 o C ; T1 = 10 + 273 = 283 K, To find, 1. Thermal efficiency, η carnot, , 2. Heat added, Qs, , Solution, Efficiency,, T2 − T1 650 − 310, =, = 0.5231 = 52.31%, T2, 650, Efficiency, η carnot = Work done = W, Heat added Qs, , η carnot =, , Unit – II , ,, , P3.3
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∴ Heat added, Qs =, , W = 150 = 286.752 kJ, ηcarnot 0.5231, , Result, 1. Thermal efficiency, η carnot = 52.31 %, 2. Heat added, Qs = 286.752 kJ, Example 3.5, An ideal Otto cycle takes in air at 27 o C . The adiabatic, expansion ratio for this engine is found to be 6. Find the air, standard efficiency of the engine., Given data, Lowest temperature, t1 = 27 o C ; T1 = 27 + 273 = 300 K, Expansion ratio, r = 6, To find, 1. Air−standard efficiency, ηotto, Solution, Efficiency,, , ηotto = 1 − 1, = 1 − 11.4 −1 = 0.512 = 51.2%, r γ −1, 6, Result, 1. Air−standard efficiency, ηotto = 51.2%, Example 3.6, Calculate the air−standard efficiency of an engine working, on Otto cycle, if the pressure at he beginning and the end of, compression are 103.5 kN/ m 2 and 827.5 kN/ m 2 respectively., Take γ=1.4., Given data, Pressure before compression, p1 = 103.5 kN/ m 2, Pressure after compression, p2 = 827.5 kN/ m 2, γ = 1.4, To find, 1. Air−standard efficiency, ηotto, , Unit – II , ,, , P3.4
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Solution, 1, , V p γ, Compression ratio, r = 1 = 2 , V2 p1 , 1, , 1 .4, r = 827.5 = 4.4145, 103, ., 5, , , , Air standard efficiency,, 1, ηotto = 1 − 1, = 1−, = 0.4479 = 44.79 %, r γ −1, 4.41451.4 −1, Result, 1. Air−standard efficiency, ηotto = 44.79%, Example 3.7, An engine working on ideal constant volume cycle has a, piston of 120mm diameter and 120mm stroke. The clearance, volume is 0.2 litres. If its relative efficiency is 40%, calculate the, actual thermal efficiency. Take γ=1.4, Given data, Piston diameter, d = 120 mm = 0.12 m, Stroke, l = 120 mm = 0.12 m, Clearance volume, Vc = 0.2 litres = 0.0002 m 3, Relative efficiency, ηrel = 40% = 0.4, γ = 1.4, To find, 1. Actual thermal efficiency, η actual, Solution, 2, 2, Stroke volume, Vs = π.d × l = π × (0.12) × 0.12 = 0.0013572 m 3, 4, 4, Vc + Vs 0.0002 + 0.0013572, = 7.786, Compression ratio, r =, =, Vc, 0.0002, Air standard efficiency,, 1, ηotto = 1 − 1, = 1−, = 0.56 = 56 %, 7.7861.4 −1, r γ −1, Relative efficiency, ηrel = Actual thermal efficiency, Air standard efficiency, ∴ ηactual = ηrel × ηotto = 0.4 × 0.56 = 0.224 = 22.4%, Result, 1. Actual thermal efficiency, η actual = 22.4%, , Unit – II , ,, , P3.5
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Example 3.8, An engine working on Otto cycle has a compression ratio of, 6. The temperature of air at the start of compression is 50 o C and, the maximum temperature is 1350 o C . Determine the work done per, cycle and efficiency of the cycle. Assume C p =1.005 kJ/kg.K,, , Cv =0.717 kJ/kg.K, Given data, Compression ratio, r = 6, Initial temperature, t1 = 50 o C ; T1 = 50+273 = 323 K, Maximum temperature, t3 = 1350 o C ; T3 =1350+273=1623K, , C p = 1.005 kJ/kg.K; = Cv = 0.717 kJ/kg.K, To find, 1. Efficiency, ηotto 2. Work done / cycle, W, Solution, , Cp 1.005, =, = 1.4017, Cv 0.717, 1, Efficiency, ηotto = 1 − 1, = 1 − 1.4017, −1 = 0.513 = 51.3 %, r γ −1, 6, Efficiency, ηotto = Work done (W), Heat supplied (Qs), ∴ Work done , W = ηotto × Q s, Heat supplied, Qs = m.Cv .(T3 − T2 ), γ=, , From the adiabatic compression process 1−2,, γ −1, , T2 V1 , =, = rγ −1 = 60.4017 = 2.0539, T1 V2 , ∴ T2 = 2.0539 × T1 = 2.0539 × 323 = 663.41 K, Qs = m.Cv .(T3 − T2 ) = 1 × 0.717 × (1623 − 663.41) = 688.026 kJ/kg, ∴ Work done, W = ηotto × Qs = 0.513 × 688.026 = 352.96 kJ/kg, Result, 1. Efficiency, ηotto = 51.3 %, 2. Work done / cycle, W = 352.96 kJ/kg, , Unit – II , ,, , P3.6
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Example 3.9, In an Otto cycle operation the compression ratio is changed, from 4.5 to 8.5. Determine the increase in air standard efficiency of, the cycle., Given data, Compression ratio, r1 = 4.5, Compression ratio, r2 = 8.5, To find, Increase in air standard efficiency, η2 − η1, Solution, 1 = 1− 1, = 0.4521 = 45.21 %, 4.51.4 −1, (r1)γ −1, Final efficiency, η2 = 1 − 1γ −1 = 1 − 11.4 −1 = 0.5752 = 57.52 %, 8 .5, (r2), , Initial efficiency, η1 = 1 −, , Increase in efficiency = η2 − η1 = 57.52 − 45.21 = 12.31%, Result, Increase in air standard efficiency, η2 − η1 = 12.31%, Example 3.10, In an ideal constant volume cycle, the pressure and, temperature of the air at the beginning of compression are 97 kN/, m 2 and 50 o C respectively. The ratio of compression is 5:1. The, heat supplied during the cycle is 970 kJ/kg of the working fluid., Determine: (1) maximum temperature of the cycle, (2) the thermal, efficiency of the cycle and (3) the work done during the cycle per kg, of working fluid. Assume γ=1.4 and Cv =0.717 kJ/kg.K, Given data, Pressure at the beginning, of compression, p1 = 97 kN/ m 2, Temp. at the beginning, of compression, t1 = 50 o C ; T1 =323K, , Compression ratio, r = 5, Heat supplied, Qs = 970 kJ/kg, γ = 1.4; = Cv =0.717 kJ/kg.K, Unit – II , , ,, , P3.7
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To find, 1. Maximum temperature of the cycle, T3 ,, 2. Thermal efficiency, ηotto, , 3. Work done, W, , Solution, From the isentropic process 1−2,, γ −1, , T2 V1 , =, = (r)γ −1 ;, T1 V2 , ∴ T2 = T1 .(r)γ −1 = 323 × (5)1.4 −1 = 614.88 K, From the constant volume process 2−3,, Qs = m.Cv.(T3 − T2), Qs, = 614.88 + 970 = 1967.74 K, ∴ T3 = T2 +, 1 × 0.717, m.Cv, T3 = 1967.74 K ; t3 = 1967.74 − 273 = 1694.74 o C, Thermal efficiency,, ηotto = 1 − 1γ −1 = 1 − 11.4 −1 = 0.4747 = 47.47 %, 5, (r2), Work done, W = ηotto × Qs = 0.4747 × 970 = 460.46 kJ/kg, , Result, 1.Maximum temperature of the cycle, T3 = 1967.74 K ( t3 =1694.7 o C ), 2. Thermal efficiency, ηotto = 47.47%, 3. Work done, W = 460.46 kJ/kg, Example 3.11, A cycle consists of two isentropic and two constant volume, process. The compression ratio is 6. The pressure and temperature, at the beginning of compression are 1 bar and 35 o C respectively., The maximum pressure of the cycle is 30 bar. If air is working fluid, and the cycle is theoretical cycle, determine the heat added and work, done per kg of air and the thermal efficiency of the cycle., , Given data, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial temperature, t1 = 35 o C ; T1 =308 K, Unit – II , ,, , P3.8
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Result, 1. Heat added / kg, Qs = 651.9586 kJ, 2. Work done / kg, W = 333.531 kJ, 3. Thermal efficiency, ηotto = 51.16 %, Example 3.12, The following data refer to a four cylinder petrol engine:, Total swept volume = 2000 c.c, Clearance volume = 60 c.c per cylinder, Maximum cycle temperature = 1400 o C, At the beginning of compression, the pressure is 10 5 N/ m 2, and the temperature is 24 o C .Calculate the air standard efficiency, and the mean effective pressure., Given data, Petrol engine, No. of cylinders, k, Total swept volume, Clearance volume, Vc, , −, =, =, =, , Otto cycle, 4, 2000 c.c = 0.002 m 3, 60 c.c / cylinder = 0.00006 m 3 /cylinder, , Max. temperature, t3 = 1400 o C ; T3 = 1673 K, Initial pressure, p1 = 10 5 N/ m2 =100 kN/ m 2, Min. temperature, t1 = 24 o C ; T1 =297 K, To find, 1. Air standard efficiency, ηotto, , 2. Mean effective pressure, pm, , Solution, Total swept volume = 0.002 m 3, Swept volume per cylinder, Vs = 0.002 = 0.0005 m 3, 4, Vc + Vs 0.00006 + 0.0005, = 9.3333, Compression ratio, r =, =, Vc, 0.00006, Air standard efficiency,, 1, ηotto = 1 − 1γ −1 = 1 −, = 0.5908 = 59.08 %, (r2), (9.3333)1.4 −1, From the adiabatic process 1−2,, , T2 V1 , =, T1 V2 , , γ −1, , = (r)γ −1, Unit – II P3.10, , ,
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∴ T2 = T1.(r)γ −1 = 297 × (9.3333)1.4 −1 = 725.72 K, From the adiabatic process 3−4, γ −1, , γ −1, 0.4, T4 V3 , = = 1 = 1 = 0.4092, T3 V4 , r, 9.3333 , ∴ T4 = 0.4092 × T3 = 0.4884 × 1673 = 684.67 K, , V1 = Vc + Vs = 0.00006 + 0.0005 = 0.00056 m 3, From characteristic gas equation, p1 .V1 = m. R.T1, p1.V1 100 × 0.00056, = 0.000657 kg, =, R.T1, 0.287 × 297, Stroke volume / kg, Vs = 0.0005 = 0.761 m 3 /kg, 0.000657, , ∴ Mass, m =, , Heat supplied at constant volume 2−3, Qs = m.Cv.(T3 − T2) = 1 × 0.718 × (1673 − 725.72) = 680.147 kJ/kg, Heat rejected at constant volume 4−1, Qr = m.Cv.(T4 − T1) = 1 × 0.718 × (684.67 − 297) = 278.347 kJ/kg, Work done during the cycle,, W = Qs − Qr = 680.147 − 278.347 = 401.8 kJ/kg, Mean effective pressure, pm = W = 401.8 = 527.99 kN/ m 2, Vs 0.761, , Result, 1. Air standard efficiency, ηotto = 59.08%, 2. Mean effective pressure, pm = 527.99 kN/ m 2, , Example 3.13, Calculate the air standard efficiency of diesel cycle having, compression ratio 18 and expansion ratio 10., Given data, Compression ratio, r = 18, Expansion ratio, r = 10, ρ, To find, Air standard efficiency, ηdiesel, Unit – II P3.11, ,
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Solution, Expansion ratio 18, =1.8, =, r, 10, Air standard efficiency of diesel cycle,, ργ − 1 , ηdlesel = 1 − 1 γ −1 , γ.(r) ρ − 1 , , Cut−off ratio, ρ =, , =1 −, , (1.8)1.4 − 1 , 1, = 0.6412 = 64.12 %, 1.4 − 1 , 1. 4 × ( 1. 8), (1.8 − 1) , , Result, Air standard efficiency, ηdiesel = 64.12 %, Example 3.14, Find the air standard efficiency of a diesel cycle engine if the cut, off is 6% of the stroke and the clearance is 1th of the stroke. Take γ=1.4., 13, Given data, Cut off = 6% of Vs, Clearance volume, Vc = 1 Vs, 13, To find, Air standard efficiency, ηdiesel, Solution, Cut−off volume, V3 = Vc + 6% Vs = Vc + 0.06Vs, Clearance volume, Vc = V2 = 1 Vs, 13, Compression ratio,, 1, Vs 1 + 1 14 , Vc + Vs 13 Vs + Vs, 13, = 13 = 14, r=, =, = , Vc, 1 V, 1 V, 1, 13 s, 13 s, 13, Cut−off ratio,, 1, Vs 1 + 0.06 , V, V, V + 0.06Vs 13 Vs + 0.06Vs, 13, , ρ= 3 = 3 = c, =, = , 1 V, V2 Vc, Vc, 1 V, 13 s, 13 s, = 13 × 1 + 0.06 = 1.78, 13, , Unit – II P3.12, ,
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ηdlesel = 1 −, =1 −, , γ, 1 ρ − 1, γ −1 , γ.(r) ρ − 1 , , (1.78 )1.4 − 1 , 1, = 0.6043 = 60.43 %, 1 .4 − 1 , 1.4 × (1.78 ), (1.78 − 1) , , Result, Air standard efficiency, ηdiesel = 60.43 %, Example 3.15, Estimate the air standard efficiency of a diesel engine, having cylinder diameter 250mm, stroke 400mm, clearance volume, 1.5 litre, fuel cut off at 5% of the stroke., Given data, Cylinder diameter, d = 250 mm = 0.25 m, Stroke, l = 400 mm = 0.4 m, Clearance volume, Vc = V2 = 1.25 litres = 0.00125 m 3, Cut off = 5% of Vs, To find, Air standard efficiency, ηdiesel, Solution, 2, 2, Stroke volume, Vs = π × d × l = π × 0.25 × 0.4 = 0.0196 m 3, 4, 4, Vc + Vs 0.00125 + 0.0196, Compression ratio, r =, = 16.68, =, Vc, 0.00125, , Cut−off volume, V3 = Vc + 6% Vs = Vc + 0.05Vs, = 0.0025 + (0.05 × 0.0196) = 0.00223 m 3, Cut−off ratio,, V, V, ρ = 3 = 3 = 0.00223 = 1.784, V2 Vc 0.00125, ∴ ηdlesel = 1 −, =1 −, , γ, 1 ρ − 1, γ −1 , γ.(r) ρ − 1 , , (1.784)1.4 − 1 , 1, = 0.6308 = 63.08 %, 1 .4 − 1 , 1.4 × (16.68), (1.784 − 1) , , Result, Air standard efficiency, ηdiesel = 63.08 %, <, , Unit – II P3.13, ,
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Example 3.16, In an ideal diesel cycle, the compression ratio is 14:1 and, expansion ratio is 8:1. The pressure and temperature at the beginning of, compression are 100 kJ/ m 2 and 45 o C respectively and the pressure at, the end of expansion is 219 kN/ m 2 . Determine the 1) maximum, temperature of the cycle 2) thermal efficiency of the cycle. Take γ=1.4., Given data, Compression ratio, r = 14, Expansion ratio = 8, Initial pressure, p1 = 100 kN/ m 2, Initial temperature, t1 = 45 o C ; T1 =318 K, Pressure at the end of expansion, p4 = 219 kN/ m 2, To find, 1. Maximum temperature, T3, , 2. Thermal efficiency, ηdiesel, , Solution, 1) Maximum temperature, T3, Expansion ratio = r = 8, ρ, ∴ Cut off ratio, ρ = r = 14 = 1.75, 8, 8, From the adiabatic compression process 1−2, γ −1, , T2 V1 , = = (r ) γ − 1, T1 V2 , ∴ T2 = T1.(r)γ −1 = 318 × (14)1.4 −1 = 913.85 K, From the constant pressure process 2−3, T3 V3, =, =ρ, T2 V2, ∴ T3 = ρ.T2 = 1.75 × 913.85 = 1599.237 K, , t3 = 1599.237 − 273 = 1326.237 o C, Thermal efficiency, ηdlesel = 1 −, =1 −, ,, , γ, 1 ρ − 1, γ −1 , γ.(r) ρ − 1 , , (1.75)1.4 − 1 , 1, = 0.605 = 60.5 %, 1 .4 − 1 , 1.4 × (14 ), (1.75 − 1) , Unit – II P3.14
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Result, 1. Maximum temperature, T3 = 1599.237 K ( t3 =1326.237 o C ), 2. Thermal efficiency, ηdiesel = 60.5%, Example 3.17, What will be the loss in ideal efficiency of a diesel engine, with the compression ratio of 14, if the fuel cut off is delayed from, 6% to 9% of the stroke?, Given data, Compression ratio, r = 14, Fuel cut off in case I = 6 % of Vs, Fuel cut off in case II = 9 % of Vs, To find, Loss in thermal efficiency, η1 − η2, Solution, Compression ratio, r =, , V1, = 14, V2, , ∴ V1 = 14 V2, , V2 + Vs = 14V2, ∴ Stroke volume, Vs = 14V2 − V2 = 13V2, 1. Fuel cut off is 6% of Vs, Fuel cut off = 0.06 × Vs = 0.06 × 13V2 = 0.78V2, Cut off volume, V3 = V2 + Fuel cut off = V2 + 0.78V2 = 1.78V2, V, 1.78V2, = 1.78, ∴ Cut off ratio, ρ = 3 =, V2, V2, γ, 1 ρ − 1, , , γ.(r) γ −1 ρ − 1 , (1.78)1.4 − 1 , (1.78 − 1) = 0.6043 = 60.43 %, , , , Thermal efficiency, η1 = 1 −, =1 −, , 1, 1.4 × (14)1.4 −1, , 2. Fuel cut off is 9% of Vs, Fuel cut off = 0.09 × Vs = 0.09 × 13V2 = 1.17V2, Cut off volume, V3 = V2 + Fuel cut off = V2 + 1.17V2 = 2.17V2, Unit – II P3.15, ,
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∴ Cut off ratio, ρ =, , V3 2.17V2, =, = 2.17, V2, V2, γ, 1 ρ − 1, , , γ.(r) γ −1 ρ − 1 , (2.17)1.4 − 1 , (2.17 − 1) = 0.5840 = 58.40 %, , , , Thermal efficiency, η2 = 1 −, , 1, 1.4 × (14)1.4 −1, Loss in thermal efficiency, η1 − η2 = 60.43 − 58.40 = 2.03 %, =1 −, , Result, Loss in thermal efficiency, η1 − η2 = 2.03 %, Example 3.18, Find the power output of a diesel engine working on a diesel, cycle with a compression ratio of 16 and an air flow rate of 0.25, kg/s. The initial condition of air is at 1 bar absolute and 27 o C, temperature. Heat added per cycle is 2500 kJ/kg. Assume C p =1.00, kJ/kg.K and Cv =0.714 kJ/kg.K, Given data, Compression ratio, r = 16, Mass flow rate, m = 0.25 kg/s, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial temperature, t1 = 27 o C ; T1 =27 + 273 = 300 K, Heat supplied, Qs = 2500 kJ/kg, C p = 1.00 kJ/kg.K, Cv = 0.714 kJ/kg.K, To find, Power output, P, Solution, , Cp, 1, =, = 1.4, Cv 0.74, From the adiabatic compression process 1−2, γ=, , γ −1, , T2 V1 , = = (r ) γ − 1, T1 V2 , ∴ T2 = T1.(r)γ −1 = 300 × (16)1.4 −1 = 909.43 K, Unit – II P3.16, ,
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Heat supplied, Qs = m.C p (T3 − T2 ), 2500 = 1 × 1 × ( T3 −909.43), , T3 = 2500 + 909.43 = 3409.43 K, From the constant pressure process 2−3, T3 V3, =, =ρ, T2 V2, T, 3409.43, ∴ ρ= 3 =, = 3.749, T2, 909.43, Air standard efficiency, ηdiesel = 1 −, =1 −, , γ, 1 ρ − 1, , γ −1 , γ.(r) ρ − 1 , , (3.749)1.4 − 1 , 1, = 0.5405 = 54.05 %, 1.4 × (16)1.4 −1 (3.749 − 1) , , Also, Air standard efficiency, ηdiesel =, , Work done (W ), Heat supplied (Qs ), , ∴ Work done, W = ηdiesel × Qs, = 0.5405 × 2500 = 1351.125 kJ/kg, Power = Work done × Mass flow rate, = 1351.25 × 0.25 = 337.813 kW, , Result, Power output, P = 337.813 kW, , Example 3.19, An air standard diesel cycle has a compression ratio of 18, and, the heat transferred to the working fluid per cycle is 1800 kJ/kg. At the, beginning of compression stroke the pressure is 1 bar and the, temperature is 300K. Calculate the temperature at each point in the, cycle. C p =1.005 kJ/kg.K; Cv =0.718 kJ/kg.K; R=0.287 kJ/kg.K, Given data, Compression ratio, r = 18, Heat transferred, Qs = 1800 kJ/kg, Initial pressure, p1 = 1 bar = 100 kN/ m 2, Initial temperature, T1 = 300 K, C p = 1.005 kJ/kg.K ; Cv = 0.718 kJ/kg.K, , R = 0.287 kJ/kg.K, Unit – II P3.17, ,
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PROBLEMS FOR PRACTICE, 1., , The temperature limits of a Carnot cycle using air as working, fluid between 650K and 310K produces 150 kJ of work. Find, a) thermal efficiency and b) heat added during the process., [Ans: ηCarnot =52.31%, Qs =286.75 kJ], , 2., , The temperature limits for a Carnot cycle using air as working, fluid are 500 o C and 10 o C . Calculate efficiency of the cycle and, the ratio of adiabatic expansion. Assume γ=1.4., [Ans: ηCarnot = 63.39%, r = 12.33], , 3., , An inventor claims that his engine develops 5 kW for a heat, addition of 435 kJ/min. The highest possible temperature of, the cycle is 200K and the lowest is 800K. Find out whether the, inventor is correct in his statement., [Ans: Incorrect, ηengine(68.96%) > ηcarnot(60%) ], , 4., , An ideal Carnot engine takes in air at 27 o C . The absolute, expansion ratio for this engine is found to be 6. Find the air, standard efficiency of the engine., [Ans: ηCarnot = 51.16%], , 5., , An engine has a bore of 90mm and stroke of 100mm. The, clearance volume is 82 c.c. If the engine works on Otto cycle,, determine the air standard efficiency of the engine., [Ans: ηOtto = 58.02%], , 6., , An engine working on Otto cycle uses air as working fluid. Its, compression ratio is raised from 5 to 6. Find the increase in, efficiency., [Ans: η1 − η2 = 3.69%], , 7., , An engine working on Otto cycle, has a cylinder diameter of, 150mm and stroke of 225mm. The clearance volume is 0.00125, m 3 . Find the air standard efficiency of this engine. Take γ=1.4, [Ans: ηOtto = 43.6%], , 8., , An engine working on Otto cycle has a piston diameter 100mm, and stroke 100mm. The clearance volume is 0.105 litres. If its, relative efficiency is 40%, calculate the actual thermal, efficiency. Take γ=1.4, [Ans: ηOtto = 23%], Unit – II P3.19, , ,
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9., , In an ideal constant volume cycle, the pressure and, temperature at the beginning of compression are 90 kN/ m 2, and 45 o C respectively. The ratio of compression is 5:1. The, heat supplied during the cycle is 900 kJ/kg of working fluid., Determine (a) the maximum temperature attained in the cycle,, (b) the thermal efficiency of the cycle and (c) the work done, during the cycle per kg of the working fluid. Assume γ=1.4 and, Cv =0.717 kJ/kg.K., [Ans: T3 =1860.6K, ηOtto =47.47%, W= 427.23 kJ/kg], , 10. An Otto cycle has a compression ratio of 6 and works between, the temperature limits 350K and 1800K. Determine the, maximum power developed when the working fluid air is used, at the rate of 0.4 kg/min. Also determine the temperature at, the end of compression. Take γ = 1.4, [Ans: P = 2.653 kW, T2 =716.7K], 11. A car possesses a four−stroke cylinder in−line diesel engine, with compression ratio 21:1 and expansion ratio 10.5:1. Find, the cut off ratio and air standard efficiency., [Ans: ρ = 2, ηdiesel = 65.36%], 12. A compression ignition engine working on diesel cycle has the, following data: Cylinder bore = 150mm, Stroke = 250mm,, Clearance volume = 400 c.c. The fuel injection takes place at, constant pressure for 5% of the stroke. Find the air standard, efficiency., [ ηdiesel =59.33%], 13. A diesel cycle has a compression ratio of 18 and cut off occurs, at 5% of stroke. Determine the air−standard efficiency of the, cycle., [Ans: ηdiesel =63.87%], 14. What will be the loss of thermal efficiency of a diesel engine, with a compression ratio of 14, if the fuel cut off is delayed from, 5% to 9% of the stroke., [Ans: η1 − η2 =2.75%], 15. A compression ignition engine has a stroke of 270mm and, cylinder of 165mm. The clearance volume is 0.000435 m3 and, the fuel injection takes place at constant pressure for 4.5% of, the stroke. Find the efficiency of the engine assuming it works, Unit – II P3.20, ,
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on diesel cycle. If the cut−off is increased to 7% with the, compression ratio unchanged, find the change in air−standard, efficiency., [Ans: η1 − η2 =1.81%], 16. In a diesel cycle, the pressure and temperature at the start of, compression are 100kN/ m2 and 60 o C respectively. The, maximum pressure attained in the cycle is 4.5 bar and the heat, received during the cycle is 600 kJ/kg of the working fluid., Determine (a) the temperature at the end of compression and, (b) the temperature at the end of combustion. Assume γ=1.4, and C p =1.003 kJ/kg.K, [Ans: T2 =511.72K, T3 =1110K], 17. The compression ratio and cut off ratio of a diesel is 14 and 22, respectively. Pressure and temperature at the beginning of, compression are 0.98 bar and 100 o C respectively. Obtain the, values of pressure and tem at all salient points of the cycle., [Ans: p1 =98kN/ m2 , p2 = p3 =3943kN/ m2 , p4 =295.5kN/,, , T1 =373K, T2 =1072K, T3 =2358K, T4 =1125K], 18. In an engine working on ideal diesel cycle, the pressure and, temperature at the beginning of compression stroke are, 98.5kN/ m2 and 60 o C . The maximum pressure attained, during the cycle is 4.5MN/ m2 and heat received during the, cycle is 580kJ/kg of air. Determine compression ratio and, efficiency of the cycle., [Ans: r = 15.33, ηdiesel =62.9%], , Unit – II P3.21, ,
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Unit – II, Chapter 4., , HEAT TRANSFER, , 4.1 Heat transfer, Heat is defined as the transfer of thermal energy across a, well-defined boundary around a thermodynamic system. Heat, transfer is a discipline of thermal engineering that concerns the, generation, use, conversion, and exchange of thermal energy and, heat between physical systems., 4.2 Modes of heat transfer, The fundamental modes of heat transfer are:, 1) Conduction or diffusion, 2) Convection, 3) Radiation, 1. Conduction, Conduction is the transfer of heat energy by microscopic, diffusion and collisions of particles within a body due to a, temperature gradient. The microscopically diffusing and colliding, objects include molecules, electrons, atoms, and phonons., Conduction takes place in solids, liquids and gases., Heat is transferred by conduction when adjacent atoms, vibrate against one another, or as electrons move from one atom, to another. Conduction is greater in solids because the network of, relatively close fixed spatial relationships between atoms helps to, transfer energy between them by vibration. Fluids are less, conductive. This is due to the large distance between atoms in a, gas. Conductivity of gases increases with temperature., Steady-state conduction : Steady state conduction is, the form of conduction that happens when the temperature, difference is constant. In steady state conduction, the amount of, heat entering any region of an object is equal to amount of heat, coming out . There is no change in the internal energy of the, Unit – II , , 4.1
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system during such process. Typical examples of steady state heat, transfer are :, • Cooling of an electric bulb by the surrounding atmosphere, • Heat flow from the hot to cold fluid in a heat exchanger, Unsteady state conduction : When the temperatures, are changing in an object with respect to time, the mode of heat, transfer is termed as unsteady state conduction. A change in, temperature indicates a change in internal energy of the system., Energy storage is thus a part of unsteady heat flow. Typical, examples of unsteady heat transfer are :, • Warm-up periods of furnaces, • Boilers and turbines, • Cooling of castings in a foundry., Transient conduction : It is a special kind of unsteady, process in which the system is subjected to cyclic variations in the, temperatue of its environment. The temperature at a particular, point of the system returns periodically to the same value. The, rate of heat flow and energy storage also undergo periodic, variations. Typical example include :, • Heating and cooling of the water of an I.C engine., 2. Convection, Convection is the transfer of thermal energy from one place, to another by the movement of fluids or gases. Convection is usually, the dominant form of heat transfer in liquids and gases. Convection, describes the combined effects of conduction and fluid flow or mass, exchange. Two types of convections are described below:, Natural of free convection : It is a type of heat, transfer, in which the fluid motion is not generated by any, external source but only by density differences in the fluid, occurring due to temperature gradients. In natural convection,, fluid surrounding a heat source receives heat, becomes less dense, and rises. The surrounding, cooler fluid then moves to replace it., This cooler fluid is then heated and the process continues,, Unit – II , , 4.2
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forming a convection current. This process transfers heat energy, from the bottom of the convection cell to top. The driving force for, natural convection is buoyancy, a result of differences in fluid, density. Because of this, the presence of a proper acceleration, such as arises from resistance to gravity, or an equivalent force,, is essential for natural convection., Examples of natural, convection include:, • The upward flow of air due to a fire or hot object, • The circulation of water in a pot that is heated from, below., • Cooling of billets in the atmosphers, Forced convection : It is a type of transfer in which fluid, motion is generated by an external source,like a pump, fan,, suction device, etc. It should be considered as one of the main, methods of useful heat transfer as significant amounts of heat, energy can be transported very efficiently. Forced convection is, often encountered by engineers designing or analyzing heat, exchangers, pipe flow, and flow over a plate at a different, temperature than the stream. Examples of forced convection, include:, • Flow of water in condenser tubes, • Fluid passing through the tubes of a heat exchanger, • Cooling of internal combustion engines, 3. Radiation, Thermal radiation is electromagnetic radiation generated, by the thermal motion of charged particles in matter. All matter, with a temperature greater than absolute zero emits thermal, radiation. The mechanism is that bodies with a temperature, above absolute zero have atoms or molecules with kinetic energies, which are changing, and these changes result in chargeacceleration and/or dipole oscillation of the charges that compose, the atoms. This motion of charges produces electromagnetic, radiation in the usual way. The main properties of thermal, radiation include:, Unit – II , , 4.3
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•, •, , •, , •, , Thermal radiation emitted by a body at any temperature, consists of a wide range of frequencies., The dominant frequency (or colour) range of the emitted, radiation shifts to higher frequencies as the temperature, of the emitter increases. For example, a red hot object, radiates mainly in the long wavelengths ., The total amount of radiation of all frequencies increases, steeply as the temperature rises; it grows as t4, where t is, the absolute temperature of the body., The rate of electromagnetic radiation emitted at a given, frequency is proportional to the amount of absorption, from the source. Thus, a surface that absorbs more red, light thermally radiates more red light., , Examples of thermal radiation include:, • The visible light from sun, • TheInfrared light emitted by an incandescent light bulb, • The Infrared radiation emitted by animals, 4.3 Heat transfer by conduction, 4.3.1 Fourier's Law, The law of heat conduction, also known as Fourier's law,, states that the time rate of heat transfer through a material is, proportional to the negative gradient in the temperature and to, the area, at right angles to that gradient, through which the heat, is flowing. Mathematically,, dt, dt, Q ∝ A, or Q = − k A, dx, dx, where, Q = Amount of heat flow in a unit time (kJ/s), A = Surface area of heat flow, taken at right angles to the, direction of flow ( m2 ), dt = Temperature difference on the two faces of the body (K), dx = Thickness of the body through which heat flows,, taken along the direction of flow (m), k = Constant of proportionality known as thermal, conductivity of the body (W/mK), Unit – II , , 4.4
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4.3.2 Thermal Conductivity, Thermal conductivity is defined as the quantity of heat, (Q) transmitted through a unit thickness (dt) in a direction, normal to a surface of unit area (A) due to a unit temperature, difference (dt) under steady state conditions and when the heat, transfer is dependent only on the temperature gradient., Mathematically,, Q. dx, Thermal conductivity, k =, A. dt, The unit of thermal conductivity is W/mK., Thermal conductivity for some common materials are, given below., Thermal Conductivity, Material, (W/m K), Air at oC, , 0.024, , Aluminum, , 205.0, , Brass, , 109.0, , Concrete, , 0.8, , Copper, , 385.0, , Glass, , 0.8, , Gold, , 310, , Ice, , 1.6, , Lead, , 34.7, , Silver, , 406.0, , Steel, , 50.2, , Wood, , 0.12-0.04, , 4.3.3 Heat conduction through plane wall, Let , Q=Heat flow rate through the wall, A=Area of the wall surface, δ =Thickness of the wall, dx=Thickness of elementary strip, dt=Temperature difference across the strip, t1 =Temperature at surface 1, t2 =Temperature at surface 2, Unit – II , , 4.5
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dt, , t1, , Q, , t2, x, , dx, , δ, , Fig.4.1 Heat conduction through plane wall, , The Fourier's equation is given by,, dt, Q=−k A, dx, Q. dx = −k. A dt, Integrating on both sides,, , Q, , δ, , t2, , 0, , t1, , dx = − k. A dt, , Q δ = k. A (t1 − t2), Q =, , k.A (t1 − t2), , δ, , The above expression for heat flow rate can be written as,, (t − t2), (t − t2), Q = 1, = 1, (δ / kA), Rt, where, Rt =, , δ = thermal resistance to heat flow., , kA, , 4.3.4 Heat conduction through composite wall, A composite wall refers to a wall of a several, heterogeneous layers. Walls of furnaces, boilers and heat, exchange devices consist of several layers. Consider a wall, composed of three layers as shown in the figure., Let, Q=Heat flow rate through the wall, A=Area of the wall surface, δ1, δ 2, δ3 =Thickness of the layers 1,2,3, Unit – II , , 4.6
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4.3.5 Heat conduction through a cylinder, Cylindrical metal tubes constitute an essential element of, power stations, oil refineries and most process industries. The, boilers have tubes in them, the condensers contain bank of, tubhes, the heat exchangers are tubulars and all these units are, connected by tubes., Q, r2, r1, , dr, dt, , t1, Temperature profile, , t2, , Fig.4.3 Heat conduction through cylinder, , Let, Q =Heat flow rate through the cylinder, A =Surface area of the cylinder, r1 =Inner radius of the cylinder, r2 =Outer radius of the cylinder, , l =Length of the cylinder, t1 =Temperature at inside surface, t2 =Temperature at outside surface, dr=Thickness of elementary strip, dt =Temperature difference across the strip, The Fourier's equation is given by,, dt, dt, Q = −k A, = −k(2πrl), dr, dr, , Q, dr, ., 2π kl, r, , = − dt, , Integrating on both sides,, , Q, 2πkl, , r2, , , r1, , dr, = −, r, , t2, , dt, t1, , Unit – II , , 4.8
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r, Q, loge 2 = (t1 - t2), 2πkl, r1, or, , Q = 2πkl, , (t1 − t2), (t − t2), = 1, loge r2 / r1, Rt, , where, Rt = Thermal resistence =, , loge r2 / r1, 2πkl, , 4.4 Heat transfer by convection, 4.4.1 Newton's law of cooling, The convective heat transfer between a surface and an adjacent, fluid is prescribed by Newton's law of cooling. It states that the, rate of heat loss of a body is proportional to the difference in, temperatures between the body and its surroundings., Mathematically, Q ∝ (ts - t f ) or Q = h A (ts - t f ), where, Q =Convective heat flow rate (Joules), A =Area exposed to heat transfer ( m2 ), h =Convective heat transfer co-efficient (W/ m2 K), tS =Surface temperature (K), t f =Fluid temperature (K), 4.4.2 Heat exchanger, Heat exchanger is a process equipment designed for the, effective transder of heat energy between two fluids.The purpose, may be either to remove heat from a fluid or to add heat to a fluid., Examples include:, • Boilers, superheaters and condensers of a power plant, • Automobile radiators and oil coolers of heat engines, • Condensers and evaporators in refrigeration units, • Water heaters and coolers, Based on the direction of flow of fluids, the heat, exchangers are classified into three types:, • Parallel flow heat exchangers, • Counter flow heat exchangers, • Cross flow heat exchangers, Unit – II , , 4.9
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Fluid B, , Fluid A, , Fluid A, , Fluid B, , Fig.4.4 Parallel flow heat exchanger, , Parallel flow heat exchangers : In parallel flow, arrangement, the fluid (hot and cold) enter the unit from the, same side, flow in the same direction and subsequently leave from, the same side. Obviously the flow of fluids is unidirectional and, parallel to each other., Fluid B, , Fluid A, , Fluid A, , Fluid B, , Fig.4.5 Counter flow heat exchanger, , Counter flow heat exchangers : In counter flow, arrangement, the fluid (hot and cold) enter the unit from opposite, ends, travel in opposite directions and subsequently leave from, opposite ends. Obviously the flow of fluid is opposite in direction, to each other. For a given surface area, the counter-flow, arrangement gives the maximum heat transfer rate and is, naturally preferred for the heating and cooling of fluids., Fluid B, , Fluid B, , Fluid A, , Fluid A, , Stream A unmixed, Stream B mixed, , Stream A unmixed, Stream B unmixed, , Fig.4.6 Cross flow heat exchanger, , Unit – II 4.10
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Cross flow heat exchangers : In the cross-flow, arrangement, the two fluids (hot and cold) are directed at right, angles to each other. Figure shows the common cross flow, arrangements. When mixing occurs, the temperature variations, are primarily in the flow direction. When unmixed, ther is, temperature gradient along the stream as well as in the direction, perpendicular to it. Apparently, temperaturess of the fluids, leaving the unit are not uniform for the unmixed streams. The, cross flow heat exchangers are commonly employed in air or gas, heating and cooling applications., 4.4.3 Overall heat transfer cofficient, A heat exchanger is ennentially a device in which energy, is transferred from one fluid to another across a good conducting, solid wall. Therefore the heat transfer is the combined effect of, conduction and covection. In this case an overall heat transfer, coefficient is used. Mathematically,, 1, U =, 1, δ, 1, +, +, hi, k, h0, where, Q =Overall heat transfer coefficient, hi =Convective heat transfer coefficient of inner surface, , h0 =Convective heat transfer coefficient of outer surface, k =Thermal conductivity of heat exchanger material, , δ =Thickness of the material, 4.4.4 Logrithmic mean temperature difference (LMTD), During heat exchange between two fluids, the temperature, of the fluids change in the direction of flow. In a parallel flow, system, the thermal head causing the flow of heat is maximum at, the inlet. However, the thermal head goes on decreasing along the, flow path and is minimum at the outlet. In a counter flow system,, both the fluids are in their coldest state at the exit. To calculate the, rate of heat transfer, an average value of the temperature, difference between the fluids has to be determined. It is known as, logrithmic mean temperature difference (LMTD)., Unit – II 4.11
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Mathematically, LMTD, tm =, , ∆t1 − ∆t2, ∆t, loge 1 , ∆t2 , , where, ∆t1 =Temperature difference between hot and cold fluid at entrance, ∆t2 =Temperature difference between hot and cold fluid at exit, 4.4.5 Properties of heat exchanger, 1. Capacity ratio (C), The product mc (mass x specific heat) of a fluid flowing in, a heat exchanger is termed as the capacity rate. It indicate the, capacity of the fluid to store energy at a given rate., Capacity rate of the hot fluid, Ch = mh ch, Capacity rate of the cold fluid, Cc = mc cc, The capicity ratio (C) is defined as the ratio of the, minimum to maximum capacity rate., m c, if mh ch > mc cc , then C = mc cc, h h ;, m c, if mh ch < mc cc , then C = mh ch, c c, 2. Heat exchanger effectiveness ( ∈ ):, The effectiveness of a heat exchanger is defined as the, ratio of the energy actually transferred to the maximum, theoretical energy transfer., Q, actual heat transfer, ∈= act =, Qmax, maximum possible heat transfer, 3. Number of transfer units (NTU), The number of transfer units (NTU) is a measure of the, size of heat exchanger. It is defined as:, UA, NTU =, mc cc when mh ch > mc cc, , NTU =, , UA, mh ch when mh ch < mc cc, , The denominator is always the smaller thermal capacity rate ., UA, UA, =, Therefore, NTU = (m c), C, min, min, Unit – II 4.12
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4.5 Heat transfer by radiation, 4.5.1 Properties of radiation, The important properties of thermal radiation are, described below:, 1. Emissivity ( ε ), The emissivity of a material (usually written ε or e) is the, relative ability of its surface to emit energy by radiation. It is the, ratio of energy radiated by a particular material to energy, radiated by a black body at the same temperature., , Energy emitted by the material, E, Mathematically, ε = E = Energy emitted by the block body, b, Emissivity is a dimensionless quantity. A true black body, would have an ε = 1 while any real object would have ε < 1., 2. Absorptivity ( α ):, It is the ratio of the radiation energy absorbed by the body, to the total incident radiation energy., , Q, Energy absorbed by the body, Mathematically, α = Qa =, Total incident energy, o, 3. Reflectivity ( ρ ):, It is the ratio of the radiation energy reflected from the, body surface to the total incident radiation energy., , Q, Energy reflected from the body, Mathematically, ρ = Qr =, Total incident energy, o, 4. Transmissivity ( τ ):, It is the ratio of the radiation energy transmitted through, the body surface to the total incident radiation energy., Mathematically,, , τ =, , Qt, Energy transmitted through the body, =, Qo, Total incident energy, , Unit – II 4.13
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5. Black body, A black body is an idealized physical body that absorbs, all incident electromagnetic radiation, regardless of frequency or, angle of incidence. The radiations are neither reflected from the, surface nor transmitted through it. For a black body α = 1 and, ρ = τ = 0 . In actual practice, there is no perfect black body, which will absorb all the incident radiations. Snow with 0.985, absorptivity is neary black to thermal radiations., 6. Gray body :, When a surface absorbs a certain percentage of impinging, radiations, the surface is called the gray body. The absorptivity of, a gray body is necessarily below unity. But it remains constant, over the entire range of temperature and wavelength of incident, radiation. For a gray body α, ρ and τ are uniform for all, wavelengths., 7. White body :, A body that reflects all the incident thermal radiations is, called an absoutely white body. For such bodies ρ = 1 and, α = τ = 0., 8. Transparent body :, A body that allows all the incident radiations to pass, through it is called transparent or diathermaneous. For such, bodies τ = 1 and α = ρ = 0 . Transmissivity varies with, wavelength of incident radiation. A material may be nontransparent for a cenrtain wavelength band and transparent for, another., 9. Opaque body :, An opaque body is one that transmits none of the, radiation that reaches it. But it reflects some of the radiation., Thus for an opque body, τ = 0 and α + ρ = 1., , Unit – II 4.14
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REVIEW QUESTIONS, 1., , Explain the various modes of heat transfer., , 2., , Define convection., , 3., , What are the properties of thermal radiation?, , 4., , State Fourier's Law., , 5., , Define thermal conductivity of a material., , 6., , Discuss about the heat conduction through plane wall and, composite wall., , 7., , Derive the expression for heat trnasfer through a cylinder., , 8., , List out the applications heat exchanger., , 9., , Explain the types of heat exchanger., , 10. Define LMTD., 11. What is overall heat transfer?, 12. Define: Capacity ratio, NTU., 13. What is effectiveness of a heat exchanger., 14. List out the properties of radiation., 15. Define : Emissivity, 16. Define : Absorbtivity, Reflectivity, Transmissivity, 17. Explain : Black body, opaque body, grey body, transparent, body., , Unit – II 4.15
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SOLVED PROBLEMS, Example 4.1, The interior of an oven is maintained at a temperature of, 850 C . The oven walls are 500 mm thick and are fabricated from, a meterial of thermal conductivity 0.3 W/mK. For an outside wall, temperature of 250 oC , workout the resistance to heat and the heat, flow per m2 of wall surface., o, , Given data, Temperature at surface 1, t1 = 850 oC, Temperature at surface 2, t2 = 250 oC, Thickness of the wall, δ = 500mm = 0.5 m, Thermal of conductivity, k = 0.3 W/mK, Area , A = 1 m2, To find, Heat flow rate, Q (W/ m2 ), Solution, , t1 − t2, 0 .5, =, = 1.667 K/W, kA, 0 .3 × 1, t − t2, 850 − 250, Heat flow rate, Q = 1, =, = 359.92 W/ m2, Rt, 1.667, Thermal resistance, Rt =, , Result, Heat flow rate, Q = 359.92 W/ m2, Example 4.2, A rod of 3cm diameter and 20cm length is maintained at, 100 oC at one end and 10 oC at the other end. These temperature, conditions are attained when there is heat flow rate of 6 watts. If, cylindrical surface of the rod is completely insulated, determine, the thermal condictivity of the rod material., Given data, Diameter of the rod, d = 3 cm = 0.03 m, Temperature at surface 1, t1 = 100 oC, Temperature at surface 2, t2 = 10 oC, Thickness of the wall, δ = 20cm = 0.2 m, Heat flow rate, Q = 6 Watts, Unit – II P4.1
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To find, Thermal of conductivity, k (W/mK), Solution, Area, A =, , π × d 2 = π × 0.032, , Heat flow rate, Q =, , 4, 4, k A (t1 − t2), , δ, , Qδ, 6 × 0.2, k =, =, =18.87 W/mK, A (t1 − t2), π × 0.032 , , × (100 − 10), 4, , , Result, Thermal of conductivity, k = 18.87 W/mK, Example 4.3, A storage chamber of interior dimensions 10m x 8m x 2.5m, high has its inside maintained at a temperature of -20 oC whilst, the outside is at 25 oC . The walls and ceiling of the chamber have, three layers made of, 60 mm thick board (k = 0.2 W/mK) on the inside, 90 mm thick insulation (k = 0.04 W/mK) at mid, 240 mm thick concrete (k = 1.8 W?mK) on the outside, Neglecting flow of heat through the floor, determine the rate at, which heat can flow towards inside of the chamber., Given data, Interior dimensions = 10m x 8m x 2.5m, Temperature difference, ∆t = 25-(-20) = 45 oC, Thickness of layer 1, δ1 = 60 mm = 0.06m, Thickness of layer 1, δ 2 = 90 mm = 0.09m, Thickness of layer 1, δ3 = 240 mm = 0.024m, Thermal conductivity of layer 1, k1 = 0.2 W/mK, Thermal conductivity of layer 1, k2 = 0.04 W/mK, Thermal conductivity of layer 1, k3 = 1.8 W/mK, To find, Heat flow rate, Q (Watts), Solution, Area, A = 2(10 × 2.5) + 2(8 × 2.5) + 2(10 × 8) = 170 m2, Unit – II P4.2
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δ3 , δ1 δ 2, k1 + k2 + k3 , 1 0.06, 0.09, 0.24 , =, +, +, = 0.01578 K/W, 170 0.2, 0.04, 1.8 , 45, ∆t, Heat flow rate, Q =, = 2851.7 W, =, Rt, 0.01578, Thermal resistance, Rt =, , 1, A, , Result, Heat flow rate, Q = 2851.7 W, Example 4.4, The walls of a house in cold region comprise three layers., 15 cm outer brick work (k = 0.75 W/mK), 1.25 cm inner wooden paneling (k = 0.2 W/mK), 7.5 cm intermediate layer of insulating material., The insulation layer is stated to offer resistance twice the thermal, resistance of brick work. If the inside and outside temperatures of, the composite wall are 20 oC and -15 oC respectively, determine the, rate of heat loss per unit area of the wall and the thermal, condictivity of the insulating material., Given data, Area, A = 1 m2, Temperature difference, ∆t = 20-(-15) = 35 oC, Thickness of brick work, δ1 = 15 cm = 0.15m, Thickness of wooden panel, δ 2 = 1.25 cm = 0.0125m, Thickness of insulation, δ3 = 7.5 cm = 0.075m, Thermal conductivity of brick work, k1 = 0.75 W/mK, Thermal conductivity of wooden panel, k2 = 0.2 W/mK, Thermal conductivity of insulation, k3 = 1.8 W/mK, To find, Thermal conductivity of insulating material, k3 (W/mK), Solution, Resistance of brick work, Rt1 =, , k1, 0.15, =, = 0.2 K / W, A δ1, 1 × 0.75, , Unit – II P4.3
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Resistance of wooden panel, Rt2 =, , k2, 0.0125, =, = 0.0625 K / W, A δ2, 1 × 0.2, , Resistance of insulating material, Rt3 = 2 × 0.2 = 0.4 K / W, Total resistance,, Rt = Rt1 + Rt2 + Rt3 = 0.2 + 0.0625 + 0.4 = 0.6625 K/W, 35, ∆t, Heat loss, Q =, =, = 52.83 W, Rt, 0.6625, Thermal conductivity of insulating material,, δ = 0.075 = 0.1875 W/mK, k3 =, A Rt, 1 × 0 .4, Result, Thermal conductivity of insulating material, k3 = 0.1875, W/mK, Example 4.5, A cylindrical cement tube of radii 0.05 cm and 1.0 cm has a wire, embedded into it along its axis. To maintain a steady temperature, difference of 120 oC between the inner and outer surfaces, a, current of 5 amprer is made ot flow in the wire. Determine the, amount of heat generated per meter length and the thermal, conductivity of cement. Take resistance of wire equal to 0.1 ohm, per cm of length., Given data, Inner radius, r1 = 0.05 cm = 0.0005 m, Outer radius, r2 = 1.0 cm = 0.01 m, Temperature difference, t1 − t2 = 120 oC, Current, I = 5 amp., Resistance of wire, R = 0.1 Ω per cm length, 10 Ω per m length, To find, (1) Heat generated, Q (W/m length), (2) Thermal conductivity of cement, k, Solution, Heat generated, Q = I 2R = 52 × 10 = 250 W/m length, Under steady state condition, the heat generated equal, the heat transfer throuhg the cylindrial element., Unit – II P4.4
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Unit – II, Chapter 5., , STEADY FLOW ENERGY, EQUATION AND APPLICATIONS, , 5.1 Introduction, In a steady flow system, the rates of flow of mass and energy, across the control surface are constant. In other words, at the steady, state of a system, any thermodynamic property will have a fixed, value at particular location, and will not change with time., Example: The flow system in boiler, steam condenser,, steam nozzles and air compressors are examples of steady flow, system., 5.2 Steady flow energy equation (SFEE), Q, System, boundary, , 1, , W, Z2, , 2, , Z1, Datum, , Fig.5.1 Steady flow system, , Consider a thermodynamic system as shown in the figure,, in which the rate of fluid flow is constant. Fluid enters the system, at point 1 and leaves the system at point 2., Assumptions made in the system analysis, 1. The rate of flow of mass and energy through the control, volume is constant., 2. Only potential, kinetic, internal and flow energies are, considered. Other forms of energies are neglected., 3. The rate of work and heat transfer between the system and, surroundings is constant., 4. The properties of fluid at any point remain constant at all times., Unit – II , , 5.1
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Let, p1, , =, , u1, C1, Z1, v1, h1, , =, , Internal energy at inlet, , =, , Velocity of flow at the inlet, , =, , Height above the datum at the inlet, , =, , Specific volume of the fluid at the inlet, , Q, W, , =, =, , Intensity of pressure at the inlet, , = Enthalpy of working fluid at the inlet, p2 , u2 , C2 , v2 , and h2 are the corresponding values at the outlet, Heat transfer during the flow through system, Work transfer during the flow through the system, , Total energy entering the system, = Potential energy + Kinetic energy + Internal energy, + Flow energy + Heat transfer, 2, C, = g.Z1 + 1 + u1 + p1 .v1 + Q, 2, Total energy leaving the system, = Potential energy + Kinetic energy + Internal energy, + Flow energy + Work transfer, C2, = g. Z2 + 2 + u2 + p2 .v2 + W, 2, By law of conservation of energy,, Energy entering the system = Energy leaving the system, C2, C 2, g . Z1 + 1 + u1 + p1 .v1 + Q = g . Z 2 + 2 + u2 + p2 .v2 + W, 2, 2, The above equation is called steady flow energy equation., We know that, u + p.v = h, ∴ The above equation can be written as,, 2, 2, C, C, g.Z1 + 1 + h1 + Q = g.Z2 + 2 + h2 + W, 2, 2, If ‘m’ is the mass flow rate of the fluid, the steady flow energy, equation may be written as,, , , , , C2, C 2, m g . Z1 + 1 + h1 + Q = m g . Z 2 + 2 + h2 + W , 2, 2, , , , , , Unit – II , , 5.2
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5.3 Application of steady flow energy equation, The application of steady flow energy equation includes the, following:, a) Steam generators ( Boilers) b) Steam condensers, c) Steam nozzles, d) Air compressors, e) Steam or gas turbines, f) Air heaters, etc., a) Steam generator or boiler, The main function of the boiler is to transfer the heat to the, steady flow system (water). The heat transfer takes place at, constant pressure., Steam out, , Water in, Control, Volume, Heat supply, , Fig. 5.2 Steam generator, , In a boiler,, i) No mechanical work is done. ∴ W=0, ii) Fluid velocity at the inlet and exit is small. ∴ There is no, change in kinetic energy, i.e. C1 = C2, iii) Potential energy between inlet and exit is also negligible, i.e., Z1 = Z 2, Applying steady flow energy equation to inlet and exit,, h1 + Q = h2, or, Heat transfer, Q = h2 − h1 J/kg., b) Steam condenser, The main function of the steam condenser is to transfer the, heat from the steady flow system (steam)., Unit – II , , 5.3
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In, 1, , Coolant, exit, , Coolant, in, , 2, , Condensate exit, , Fig. 5.3 Steam condenser, , In a condenser,, i) No work is done (W=0), ii) No change in kinetic energy ( C1 = C2 ), iii) No change in potential energy ( Z1 = Z 2 ), The steady flow energy equation will be reduced to, h1 + Q = h2, or, heat transfer, Q = h2 − h1 J/kg., c) Steam nozzles, , 2, , 1, , Entry, , Exit, , Control volume, Fig.5.4 Steam nozzle, , Nozzle is a device used for increasing the velocity of flowing, fluid at the cost of pressure drop. IN a nozzle,, i) No work is done (W=0), ii) No heat transfer takes place (Q=0), iii) No change in potential energy ( Z1 = Z 2 ), The steady flow energy equation can be written as,, 2, , 2, , C, C, C1 2, C 2, + h1 = 2 + h2 or, 2 − 1 = h1 − h2, 2, 2, 2, 2, , This equation shows that enthalpy of the system is, reduced. The energy lost is converted into the kinetic energy., C2 2 − C12 = 2(h1 − h2 ), or, final velocity, C2 = 2(h1 − h2 ) + C1, Unit – II , , 2, , 5.4
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If initial velocity is negligible,, C2 = 2(h1 − h2 ), For a perfect gas, h = C p .t, ∴ Final velocity can be written as, , C2 = 2 × C p × (T1 − T2 ) + C1, , 2, , The expansion of a fluid in the nozzle is treated as an, isentropic process., , p , T, ∴ 2 = 2 , T1 p2 , , γ −1, γ, , p , or, T2 = T1 . 2 , p2 , , γ −1, γ, , γ −1, , , γ, , , p, 2, , 2, ∴ C2 = 2 × C p × T1 − T1 + C1, p, 1 , , , , or, final velocity,, , C2 =, , γ −1, , , p2 γ , 2, , 2 × C p × T1 × 1 − , + C1, , p1 , , , , d) Air compressors, Air compressor is a machine used to produce high pressure, air. It takes in atmospheric air and compresses it to a high, pressure. Air compressors are generally classified as:, (i) Rotary compressors and (ii) Reciprocating compressors, (i) Rotary compressor : In rotary compressors, a rotor is used to, develop pressure. The rate of flow is very high in rotary, compressors. Hence this process is treated as reversible adiabatic, or isentropic. The steady flow energy equation is written as,, h1 = h2 + W, or, Work input, W = h1 − h2, (ii) Reciprocating compressor: In this type of compressors, a, piston with cylinder is used to produce high pressure. The rate of, flow is comparatively low and large area is in contact with the, surroundings. Hence heat transfer is not negligible. The steady, flow energy equation can be written as,, h1 + Q = h2 + W, or, Work input, W = Q + ( h1 − h2 ), Unit – II , , 5.5
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e) Steam turbine and gas turbines, A steam turbine is a device which converts the energy of, steam into mechanical work. Steam is expanded through a nozzle, and a certain amount of heat energy is converted into kinetic, energy. The steam with high velocity flows over curved blades and, its direction of motion is changed. This causes a change of, momentum and force thus developed drive the turbine shaft., In a gas turbine, air is compressed in a compressor. This, air is heated by the combustion of fuel. The hot products of, combustion is expanded in turbine blades, thus doing mechanical, work., In both the turbines, expansion of working fluid is treated, as reversible adiabatic (isentropic). Neglecting the changes in the, potential and kinetic energies, steady flow energy equation is, written as,, h1 = h2 + W, or, Work input, W = h1 − h2, 5.4 Non flow energy equation, In a closed system, there is no mass transfer across the, boundary. Hence the flow energy, kinetic energy and potential, energies are neglected. Therefore for a closed system, the energy, equation is written as,, u1 + Q = u2 + W, or, Q = W + u2 − u1, Q = W + ∆u, , REVIEW QUESTIONS, 1. What is meant by steady flow system?, 2. State and explain the steady flow energy equation., 3. List out the applications of steady flow energy equation., 4. Write down the steady flow energy equation for a steam nozzle., 5. Derive the steady flow energy equation for reciprocating air, compressor., , Unit – II , , 5.6
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Example 5.3, A boiler produces steam from water at 30 o C . The enthalpy, of steam is 2800 kJ/kg. Neglecting potential and kinetic energies,, calculate the heat transferred per kg. Take specific heat of water, as 4.19 kJ/kg.K, Given data, Feed water temperature, tw = 30 o C = 30+273 = 303 K, Enthalpy of steam, h2 = 2800 kJ/kg, Specific heat of water, Cpw = 4.19 kJ/kg.K, To find, 1. Heat transfer, Q (kJ/kg), Solution, Steady flow energy equation for a boiler, Q = h2 − h1, Enthalpy of feed water, h1 = C pw .tw = 4.19 × 30 = 125.7 kJ/kg., Heat transferred, Q = 2800 − 125.7 = 2674.3 kJ/kg, Result, 1. Heat transfer, Q = 2674.3 kJ/kg., , Example 5.4, In a steady flow system, a substance flows at the rate of 5, kg/s. At the entry the pressure, internal energy and specific volume, are 600 kN/ m 2 , 300 m/s, 2000 kJ/kg and 0.3 m3 /kg, respectively. At the exit the pressure, velocity, internal energy and, specific volume are 120kN/ m 2 , 150 m/s, 1400 kJ/kg and, 1.2, , m3 /kg respectively. During the flow through the system, the heat, lost to the surroundings is 32 kJ/kg. Determine the power of the, system in kW, stating whether it is from or to the system. Neglect, the change in potential energy., Given data, Mass rate of flow, m = 5 kg/s, Heat transferred, Q = −32 kJ/kg, , Unit – II P5.2
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At entry, p1 = 600 kN/ m 2, , At exit, p2 =120 kN/ m 2, , Internal energy, , C1 = 300 m/s, u1 = 2000 kJ/kg, , C2 = 150 m/s, u2 = 1400 kJ/kg, , Specific volume, , v1 = 0.3 m 3 /kg, , v2 = 1.2 m 3 /kg, , Pressure, Velocity, , To find, 1. Power of the system, P, Solution, Neglecting the potential energy, the steady flow energy, equation is given,, C1 2, C2 2, + u1 + p1 .v1 + Q =, + u2 + p2 .v2 + W, 2 × 1000, 2 × 2000, C 2 − C2 2 , ∴ Work done, W = 1, + (u1 − u2 ) + ( p1 .v1 − p2 .v2 ) + Q, 2 × 2000 , 300 2 − 150 2 , W=, + (2000 − 1400) + [(600 × 0.3) − (120 × 1.2)] − 32, 2 × 2000 , W = 33.75 + 600 + 36 − 32 = 637.75 kJ/kg., Since W is positive, work is done by the system., Power output, P = W × Mass rate = 637.75 × 5 = 3188.75 kW, Result, 1. Power of the system, P = 3188.75 kW, Example 5.5, Steam passes through a nozzle. At the entrance, the, enthalpy of steam is 3250 kJ/kg and the velocity is 70m/s. The, enthalpy of steam at the exit is 3100 kJ/kg. Assume there is no, heat losses, determine the velocity at the exit., Given data, Enthalpy of steam at entrance, h1 = 3250 kJ/kg, Velocity of steam at entrance, C1 = 60 m/s, Enthalpy of steam at the exit, h2 = 3100 kJ/kg, Unit – II P5.3
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To find, Velocity of steam at the exit, C2, Solution, Neglecting the potential energy and heat losses, the steady, flow energy equation is given by,, C1 2, = h1 − h2, 2 × 1000, , C2 = 2 × 1000 × (h1 − h2 ) + C1, , 2, , = 2 × 1000 × (3250 − 3100) + 60 2 = 551 m/s, Result, Velocity of steam at the exit, C2 = 551 m/s, Example 5.6, Steam enters a turbine at a velocity of 12 m/s and the, specific enthalpy is 2900 kJ/kg. At the exit of the turbine, the, steam velocity is 30 m/s and specific enthalpy is 2400 kJ/kg. The, heat lost during the flow through turbine is 20 kJ/kg and steam, flow rate is 85 kg/s. Determine the power output from the turbine., Given data, Velocity of steam at entrance, C1 = 12 m/s, Enthalpy of steam at entrance, h1 = 3250 kJ/kg, Velocity of steam at exit, C2 = 30 m/s, Enthalpy of steam at the exit, h2 = 2400 kJ/kg, Heat transferred, Q = − 20 kJ/kg, Mass flow rate, m = 85 kg/s, To find, Power output, P, Solution, The steady flow energy equation for steam turbine is given by,, C12, C2 2, + h1 + Q =, + h2 + W, 2 × 1000, 2 × 1000, C 2 − C2 2, ∴ Work output, W = 1, + (h1 − h2 ) + Q, 2 × 1000, Unit – II P5.4
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122 − 30 2, + (2900 − 2400) − 20, 2 × 2000, = − 0.378 + 500 − 20 = 479.622 kJ/kg, =, , Power output, P = W × Mass flow rate, = 85 × 479.622 = 40767.87 kW, Result, Power output, P = 40767.87 kW, Example 5.7, Air expands from 3 bar to 1 bar in a nozzle. The initial, velocity is 90 m/s and temperature is 150 o C . Estimate the final, velocity. C p = 1.005 kJ/kg.K, Given data, Initial pressure, p1 = 3 bar, Final pressure, p2 = 1 bar, Initial velocity, C1 = 90 m/s, Initial temperature, t1 = 150 o C ; T1 =150 + 273 = 423 K, To find, Final velocity, C2, Solution, The exit velocity in a nozzle is given by,, γ −1, , , γ, , , p, 2, , 2, C2 = 2 × 1000 × C p × T1 × 1 − + C1, p, 1 , 1 .4 − 1, , , 1 1 .4 , , C2 = 2 × 2000 × 1.005 × 423 1 − , + 90 2, 3, , , , , = 486.982 m/s, Result, Final velocity, C2 = 486.982 m/s, Unit – II P5.5
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Example 5.8, In a gas turbine, air flows at the rate of 5 kg/s. The velocity, and the enthalpy of air at the entrance are 200 m/s and 7000, kJ/kg. The exit velocity is 160 m/s and the enthalpy is, 5000, kJ/kg. As the air passes through the turbine, a heat loss of 40, kJ/kg occurs. Find the power developed by the turbine., Given data, Mass flow rate, m = 5 kg/s, Velocity at entrance, C1 = 200 m/s, Velocity at exit, C2 = 160 m/s, Enthalpy at exit, h2 = 5000 kJ/kg, Heat loss, Q = − 40 kJ/kg, To find, Power developed, P, Solution, The steady flow energy equation for gas turbine is given by,, C12, C2 2, + h1 + Q =, + h2 + W, 2 × 1000, 2 × 1000, C 2 − C2 2, + (h1 − h2 ) + Q, ∴ Work output, W = 1, 2 × 1000, 200 2 − 160 2, + (7000 − 5000) − 40, =, 2000, = 7.2 + 2000 − 40 = 1967.2 kJ/kg, Power output = W × Mass flow rate, = 1967.2 × 5 = 9836 kW, Result, Power developed, P = 9836 kW, , Unit – II P5.6
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PROBLEMS FOR PRACTICE, 1. A boiler produces steam from feed water at 30 o C . The, enthalpy of steam is 2700 kJ/kg. Neglecting the potential and, kinetic energies, calculate the heat transferred per kg. Take, specific heat capacity of water as 4.19 kJ/kg.K, [Ans: Q = 2574.3 kJ/kg], 2. A boiler produces 500 kg of steam per hour from feed water at, 40 o C . The enthalpy of steam is 2600 kJ/kg. Determine the rate, at which the heat is transferred. Neglect potential and kinetic, energies., [ Ans: Q=12,16,200 kJ/hr], 3. In a flow process, the internal energy of a gas decreases by 210, kJ/kg of gas and the increase in flow work is 4710 kN−m per 8 kg, of gas. Determine (a) the change in specific enthalpy and (b) the, change in total enthalpy. [Ans: ∆h =378.75 kJ, ∆H =3030 kJ], 4. In a steady flow system, a fluid flows at the rate of 6 kg/s. At, the entry, the pressure, velocity, internal energy and specific, volume are 600 kN/ m 2 , 250 m/s, 2000 kJ/kg and 0.33 m 3 /kg, respectively. At the exit, the pressure, velocity, internal energy, and specific volume are 120 kN/ m 2 , 100 m/s, 1500 kJ/kg, 1.25, m 3 /kg. The heat lost to the surroundings is 33 kJ/kg., Determine the power of the system. Neglect the change in, potential energy., [Ans: P = 3247.5 kW], 5. Steam passes through a nozzle. At the entrance, the enthalpy, of steam is 2940 kJ/kg and the velocity is 60 m/s. At the exit,, the enthalpy of steam is 2730 kJ/kg. Assuming the nozzle to be, horizontal and there is no losses, determine the exit velocity of, steam., [Ans: C2 = 650.8456 m/s], 6. In a steady flow of air through a nozzle, the enthalpy decreases, by 40 kJ between two sections. Assuming that there are no, other energy changes except the kinetic energy, determine the, increase in velocity at section 2, if the velocity at section 1 is, 100 m/s., [Ans: C2 − C1 = 200 m/s], , Unit – II P5.7
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7. Steam enters a turbine at a velocity of 12 m/s and specific, enthalpy is 2900 kJ/kg. At the exit of the turbine, the steam, velocity is 30m/s and the specific enthalpy is 2400 kJ/kg. The, heat lost during the flow through the turbine is 20 kJ/kg and, the steam flow rate is 85 kg/s. Determine the work output from, the turbine., [Ans: P = 40,767.87 kW], 8. Air expands from 4 bar to 1 bar in a nozzle. The initial velocity, is 100 m/s and the temperature is 100 o C . Determine the final, [Ans: C2 = 542.8087 m/s], velocity. C p =1.005 kJ/kg.K., , Unit – II P5.8
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Unit – IV, Chapter, , 8.NTERNAL COMBUSTION ENGINES, , 8.1 Introduction, A machine which converts heat energy into mechanical, work is known as heat engine. Internal combustion (I.C) engine is, a heat engine in which combustion of fuel takes place inside the, engine cylinder., Example: Petrol engine, diesel engine, gas engine, gas, turbine, etc., If the combustion takes place outside the working cylinder,, then it is called external combustion engine., Example: Steam engines and steam turbines., 8.2 Classification of I.C. engines, Internal combustion engines may be classified according to, 1. The type of fuel used, a) Petrol engines b) Diesel engines, c) Gas engines, 2. Number or strokes per cycle, a) Four stroke engine b) Two stroke engine, 3. Method of ignition, a) Spark Ignition (S.I) engines, b) Compression Ignition (C.I) engines, 4. Cycle of operation, a) Otto cycle engines, b) Diesel cycle engines, c) Dual combustion cycle engines, 5. Arrangement of cylinders, a) Vertical engines, b) Horizontal engines c) V-type, engines, d) In-line engines e) Radial engines, 6.Cooling system used, a) Air cooled engines, b) Water cooled engines, Unit – IV , , 8.1
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7. Method of fuel injection, a) Carburetor engines b) Air injection engines, c) Airless or solid injection engines, 5. Number of cylinders, a) Single cylinder engines, , b) Multi-cylinder engines, , 9. Method of governing, a) Quantity governed engine b) Quality governed engines, c) Hit and miss governed engines, 10. Location of valves, a) Overhead valve engines, , b) Side valve engines, , 11. Method of lubrication, a) Wet sump engines, , b) Dry sump engines, , 12. Speed of the engine, a) Low speed engines, c) High speed engine, , b) Medium speed engines, , 8.3 Major components of I.C. engines, The figure shows the constructional details of a four stroke, cycle petrol engine. The major components of the engine are, explained below., 1. Cylinder block, It is the main part of the engine. It contains cylinders. The, cylinders are accurately finished to accommodate piston. The, cylinder block houses piston, crank, connecting rod, cam shaft and, other engine parts. In water cooled engines, the cylinder block is, provided with water jackets for the circulation of cooling water., The materials used for making cylinder block are grey cast iron,, aluminium alloys, etc., 2. Cylinder head, The cylinder head is bolted to the top of cylinder block by, means of studs. The cylinder block contains inlet and exhaust, ports and valves. In the cylinder head, a spark plug is provided in, Unit – IV , , 8.2
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the case of petrol engine and a fuel injection nozzle is provided in, the case of diesel engine. A copper or asbestos gasket is provided, between the engine cylinder and cylinder head to make an air, tight joint. The materials used for making cylinder head are cast, iron, aluminium alloy, etc., Valve spring, Rocker arm, Inlet port, , Spark plug, Cylinder head, Exhaust port, Exhaust valve, , Inlet valve, , Cylinder, , Gudgeon pin, Piston rings, Piston, , Connecting rod, Push rod, , Cam, Crank pin, , Crank, Crank shaft, , Crank case, Crank web, , Fig.8.1 Four stroke cycle petrol engine, , 3. Cylinder liners, The liner is a sleeve, which is fitted into the cylinder bore., It provides wear resisting surface for the cylinder bores. Liners, may be of two types : (a) Dry liners and (b) Wet liners., (a) Dry liners: Dry liners have metal−to−contact with the, cylinder block. They are not directly in touch with cooling water., (b) Wet liners: These liners are surrounded or wetted by, cooling water. It provides wear resisting surface for the piston to, reciprocated. Also it acts as a seal for the water jacket., Liner material should withstand abrasive wear and, corrosive wear. Chromium plated mil steel tubes are used as, liners. Good quality cast iron liners produced by centrifugal, casting is also commonly used in engines., Unit – IV , , 8.3
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4. Crank case, It may be cast integral with the cylinder block. Some times,, it is cast separately and fitted to the cylinder block. It also serves, as a sump for the lubricating oil. The material used for crankcase, are cast iron, alumimium alloys or alloy steels., 5. Oil pan or oil sump, Oil sump is the bottom part of the engine. It contains, lubricating oil. A drain plug is provided in the oil sump to drain, out the oil. It is made of pressed steel sheet., 6. Piston, The piston is a cylindrical part. The main function of the, piston is to transmit the force exerted to the burning of fuel in the, cylinder to crank shaft through connecting rod. It also acts as a, movable gas tight seal to keep the gases inside the cylinder. It is, opened the bottom and closed at the top. The top of the piston is, called crown. The bottom portion is called skirt. Three grooves are, provided on the circumference of the piston to fit piston rings. The, piston may be made of cast iron, aluminium alloy or cast steel., 7. Piston rings, Compression rings and oil rings are inserted in the grooves, provided on the piston. Compression rings provide an effective, seal for the high pressure gases inside the cylinder. At least two, compression rings are provided in each piston., Oil rings wipe off the excess oil from the cylinder walls., This excess oil is returned to the oil sump through the oil holes., Generally piston rings are made of wear resistant materials such, as alloy cast iron, alloy steel, etc., 8.Connecting rod, It connects the piston and crank shaft. It transmits the, force exerted due to the burning of fuel during power stroke to the, crank shaft. The upper end of the connecting rod is fitted to the, piston by means of gudgeon pin. The lower end is called big end, and is connected to the crank., Unit – IV , , 8.4
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The connecting rods must withstand heavy thrusts. Hence, it must have more strength and rigidity. The connecting rod is, usually made with drop forged I–sections. The materials used for, connecting rod are plain carbon steel, aluminium alloys, nickel, alloy steels, etc., 9. Crank shaft, Crank shaft is the main rotating shaft of the engine. The, main function of crank shaft is to convert the reciprocating, motion of the piston into rotary motion with the help of, connecting rod. The crank shaft is held in position by the, bearings. This shaft contains one or more eccentric portions called, cranks. A crank is connected to the connecting rod by a crank pin., Front end of the crankshaft is provided with the following:, (a) A gear or sprocket: It drives the camshaft at half the, speed of the crankshaft., (b) Vibration damper: It protects the crankshaft from the, torsional vibrations set up during power strokes., (c) Pulley : The pulley is fitted with V−belts. It drives, engine fan, water pumps, dynamo, etc., The material of the crankshaft must be strong enough to, withstand heavy forces of the piston. They are made from carbon, steel, nickel–chromium alloy, heat treated alloy steels, etc., 10. Camshaft, Camshaft contains number of cams. It is used to convert, rotary motion into linear or straight line motion. It has number of, cams equal to the number of valves in an engine. An additional, cam is also provided to drive the fuel pump. A gear is provided in, the cam shaft to drive the distributor or oil pump. The opening, and closing of the engine valves are controlled by the cams, provided on the cam shaft. The camshaft rotates inside the main, bearings. It is driven by crankshaft through chain or gear drive., It is rotated at half the speed of the crankshaft., Unit – IV , , 8.5
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11. Valves, The inlet valve and the exhaust valve are the main valves, in an engine. The fresh charge enters into the cylinder through, inlet valve. The exhaust gases are forced out of the engine, cylinder through the exhaust valves. The opening and closing of, these valves are controlled by the cams provided in the camshaft., Inlet valve is made of plain nickel, nickel−chrome or, chrome−molybdenum. The exhaust valve is subjected to more, heat. Hence it should be capable of withstanding high, temperatures. Exhaust valves are made of silicon−chrome steel,, high speed steel, cobalt−chrome steel, nickel−chrome steel and, tungsten steel, etc., 12. Valve actuating mechanism, Valve actuating mechanism is provided in an engine to, open and close the valves. The following are the important, mechanisms:, (a) Side valve mechanism, (b) Overhead valve mechanism, (c) Overhead inlet and side exhaust valve mechanism, (a) Side valve mechanism: Side valve mechanism consists, of valves, valve spring, valve tappet, valve guide, camshaft and its, drive, etc. The cam mounted on the camshaft operates the valve, tappet. The valve tappet is pushed up. The valve tappet pushes the, valve from its seat against the spring force. Thus the valve is, opened. When the cam is not in action, the valve returns back to its, seat by the valve spring and spring retainer., (b) Overhead valve mechanism: This type of mechanism, is commonly used in i.c engines. The valves are located in the, cylinder head. This mechanism includes valves, valve springs,, valve tappet or valve lifter, push rod, rocker arm, etc. The cam, mounted on the camshaft moves the valve lifter. The valve lifter, pushes the push rod upwards. Push rod moves the rocker arm. The, rocker arm pushes the valve off its seat against the spring force., Thus the valve is opened. When the cam is not in action, the valve, returns back to its seat by the valve spring and spring retainers., Unit – IV , , 8.6
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(c) Overhead inlet and side exhaust valve mechanism:, In this system, the inlet valves are located in the cylinder head., They are operated by the overhead valve mechanism. The exhaust, valves are located in the cylinder block. They are operated by the, side valve mechanism principle., 8.4 Engine parts, materials and manufacturing methods, Sl., No., 1., , Part name, Cylinder, block, , 2., , Cylinder, head, , 3., , Cylinder, head, Crankcase, Piston, , 4., 5., , 6., , Piston, rings, , 7., , Connecting, rod, , 8., , Crankshaft, , 9., , Inlet valves, , 10., , Exhaust, valves, , Materials used, Grey cast iron,, aluminium alloys,, alloy steel, etc., Grey cast iron,, aluminium alloys,, alloy steel, etc., Cast iron, chromium, plated mild steel tubes, Alloy steels, cast iron, Cast iron, aluminium, alloy, nickel−chrome, alloy, cast steel, etc., Alloy cast iron containing, silicon and manganese,, alloy steels, etc., Plain carbon steel,, aluminium alloys, nickel, alloy steels, etc., Hot billet steel, carbon, steel, nickel−chrome,, other heat treated alloy, steels, etc., Plain nickel,, nickel−chrome, or chrome, molybdenum, Silicon chrome steel,, high speed steel,, cobalt−chrome steel,, nickel−chrome steel,, tungsten steel, etc., Unit – IV , , 8.7, , Manufacturing, method, Casting, , Casting or, forging, Centrifugal, casting, Forging, casting, Casting or, forging, Casting, , Drop forging, , Drop forging or, casting, , Forging, , Forging
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8.5 Four stroke cycle petrol engine or spark ignition engine, A four stroke petrol engine requires four strokes of piston to, complete one cycle of operation. In petrol engines, as the air fuel, mixture is ignited by a spark plug, it is also known as spark, ignition engines., , Inlet valve, , Cylinder, , Spark plug, , Piston, Connecting rod, Crank shaft, Crank, , 1. Suction stroke, , 2. Compression stroke, , Exhaust valve, , 3. Power stroke, , 4. Exhaust stroke, , Fig.8.2 Working of four stroke cycle petrol engine, , Construction, The engine contains a piston, which reciprocates inside the, cylinder. The piston is connected to the crank shaft by means of, connecting rod and crank. The inlet and outlet valves are, mounted on the cylinder head. A spark plug is also provided on, the cylinder head for spark ignition. The events in a four stroke, cycle petrol engine are as follows., 1. Suction or charging stroke, In this stroke, the required air–fuel mixture called as charge, is sucked into the engine cylinder. During this stroke, piston, Unit – IV , , 8.8
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moves down from Top Dead Centre (TDC) to Bottom Dead Centre, (BDC). Due to the downward movement of the piston, the, pressure inside the cylinder is reduced below the atmospheric, pressure. Now inlet valve opens and fresh air–fuel mixture is, sucked into the cylinder through the inlet valve. Exhaust valve, remains closed during this stroke., 2. Compression stroke, In this stroke, the air–fuel mixture inside the cylinder is, compressed. During this stroke, the piston moves up from BDC to, TDC and both the inlet and outlet valves are kept closed. Due to, the upward movement of the piston, the air–fuel mixture in the, cylinder is compressed. As a result of compression, the pressure, and temperature of air–fuel mixture increases considerably. This, completes one revolution of crank shaft., Ignition: When the piston reaches near TDC, an electric, spark is produced by the spark plug. The compressed air–fuel, mixture is ignited by the spark. It results in a sudden increase of, temperature and pressure of combustion products. However, the, volume remains constant during combustion., 3. Expansion stroke or working stroke or power stroke, In this stroke, the heat energy is converted into useful, mechanical work. The burning gases expand rapidly. A force is, exerted on the piston due to the high pressure and expansion of, hot gases. Now the piston is pushed downward from TDC to BDC., The movement of the piston is converted into rotary motion of the, crank shaft through connecting rod. Thus, the heat energy is, transformed into useful mechanical work. Both inlet and exhaust, valves are closed during this stroke., 4. Exhaust stroke, In this stroke, the burnt gases are exhausted from the, cylinder. Now the exhaust valve is opened. The piston moves, upward from BDC to TDC. This movement of piston pushes out, the burnt gases to the atmosphere through exhaust valve. During, Unit – IV , , 8.9
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this stroke, the inlet valve remains closed. This completes one, cycle and the cylinder is ready to suck air–fuel mixture to start a, new cycle., Thus in a four stroke engine, there is only one power stroke, and three idle strokes. The power stroke supplies the required, momentum to carry out all other strokes. Thus the engine is kept, running, 8.6 Two stroke cycle petrol engine, A two stroke engine require two strokes of piston to, complete one cycle of operation. It means that the suction,, compression, expansion and exhaust are completed in two strokes, of the piston. Thus, in a two stroke engine, there is one power, stroke for every revolution of crank shaft., Compressed charge, , Cylinder, , Spark plug, , Exhaust port, , Transfer port, , Inlet port, , Piston, Crank shaft, , Connecting rod, Crank case, a) Compression, , c) Expansion and, crank case compression, , b) Ingnition and, inductance, , d) Exhaust and, transference, , Fig.8.3 Working of two stroke cycle petrol engine, , Unit – IV 8.10
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Construction, The engine contains a piston, which reciprocates inside the, cylinder. The piston is connected to the crank shaft by means of, connecting rod and crank. In two stroke engines, ports are, provided in the cylinder walls instead of valves. There are three, ports namely inlet port, exhaust port and transfer port. The, closing and opening of the ports is obtained by the movement of, the piston. The crown of the piston is made with a shape so as to, deflect the air–fuel mixture upwards inside the cylinder. A spark, plug is also provided on the cylinder head for spark ignition. The, events in a two stroke cycle petrol engine are as follows., 1. First stroke (upward movement) of the piston, During the upward movement of the piston, compression,, ignition and inductance of air–fuel mixture take place., a) Compression, The piston moves from BDC to TDC. Both transfer and, exhaust ports are covered by the piston. Due to the upward, movement of the piston, the air–fuel mixture which is transferred, already into the engine cylinder is compressed by the piston. As a, result of compression, the pressure and temperature of air–fuel, mixture increases considerably., b) Ignition and inductance, When the piston reaches near TDC, an electric spark is, produced by the spark plug. The compressed air fuel mixture is, ignited by the spark. It results in a sudden increase of, temperature and pressure of combustion products., At the same time, the inlet port is uncovered by the piston., Now the fresh air–fuel mixture from the carburetor enters into, the crank case through the inlet port., 2. Second stroke (downward movement) of the piston, During the downward movement of the piston, expansion of, hot gases, crank case compression of air–fuel mixture, exhaust of, burnt gases and transference of air–fuel mixture take place., Unit – IV 8.11
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c) Expansion and crank case compression, The burning gases expand rapidly in the cylinder. A force is, exerted on the piston due to the high pressure and expansion of hot, gases. Now the piston is moved downward. The movement of the, piston is converted into rotary motion of crank shaft through, connecting rod. Thus, the heat energy is transformed into useful, mechanical work., When the piston moves downward, the air–fuel mixture, inside the crank case is partially compressed. This process is, known as crank case compression., d) Exhaust and transference, When the piston further moves downward, the exhaust port, is uncovered. The burnt gases escape to atmosphere through, exhaust port. The transfer port is also opened. The partially, compressed air fuel mixture inside the crank case enters into the, cylinder through transfer port. Due to the deflected shape in, crown of the piston, the air–fuel mixture is deflected upwards, inside the cylinder. Thus, the escape of fresh air–fuel mixture, along with exhaust gases is reduced., Again, when the piston moves up, the transfer port is first, closed and then the exhaust port is closed. The air–fuel mixture, inside the cylinder is compressed. When the piston reaches near, TDC, the inlet port is uncovered. Fresh air–fuel mixture enters, into the crank case. At the same time, the compressed air–fuel, mixture is ignited and the cycle is repeated., Application, The two stroke cycle petrol engines are generally used in, light vehicles such as scooters, motor cycles, three wheelers, etc., 8.7 Four stroke cycle diesel engine or, compression ignition (C.I) engine, A four stroke diesel engine requires four strokes of piston to, complete one cycle of operation. In diesel engines, as the air–fuel, mixture is ignited due to the temperature developed by the, compression, it also known as compression ignition engines., Unit – IV 8.12
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Inlet valve, , Cylinder, , Fuel injector, , Piston, Connecting rod, Crank shaft, Crank, , 1. Suction stroke, , 2. Compression stroke, , Exhaust valve, , 3. Power stroke, , 4. Exhaust stroke, , Fig.8.4 Four stroke diesel engine, , Construction, This engine contains a piston, which reciprocates inside the, cylinder. The piston is connected to the crank shaft by means of, connecting rod and crank. The inlet and outlet valves are, mounted on the cylinder head. A fuel injector is also provided on, the cylinder head for injection of fuel. The events in four stroke, cycle diesel engine are as follows., 1.Suction stroke, In this stroke, the air is sucked into the engine cylinder., During this stroke, piston moves down from Top Dead Centre, (TDC) to Bottom Dead Centre (BDC). Due to the downward, movement of the piston, the pressure inside the cylinder is, reduced below the atmospheric pressure. Now the air from the, atmosphere is sucked into the cylinder through the inlet valve, after filtering. The exhaust valve remains closed during this, stroke., Unit – IV 8.13
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2. Compression stroke, In this stroke, the air inside the cylinder is compressed., During this stroke, the piston moves up from BDC to TDC and, both the inlet and outlet valves are kept closed. Due to the, upward movement of the piston, the air in the cylinder is, compressed. As a result of compression, the pressure and, temperature of air increases considerably. This completes one, revolution of crank shaft., Ignition: When the piston reaches near TDC, the fuel is, injected into the cylinder in the form of fine spray with the help of, fuel injector. Due to high pressure and temperature of compressed, air, the air–fuel mixture is ignited. It results in a sudden increase, of temperature and pressure of combustion products. However,, the volume remains constant during combustion., 3. Expansion stroke or working stroke or power stroke, In this stroke, the heat energy is converted into useful, mechanical work. The burning gases expand rapidly. A force is, exerted on the piston due to the high pressure and expansion of, hot gases. Now the piston is pushed downward from TDC to BDC., The movement of the piston is converted into rotary motion of the, crank shaft through connecting rod. Thus, the heat energy is, transformed into useful mechanical work. Both inlet and exhaust, valves are closed during this stroke., 4. Exhaust stroke, In this stroke, the burnt gases are exhausted from the, cylinder. Now the exhaust valve is opened. The piston moves, upward from BDC to TDC. This movement of piston pushes out, the burnt gases to the atmosphere through exhaust valve. During, this stroke, the inlet valve remains closed. This completes one, cycle and the cylinder is ready to suck air to start a new cycle., Thus in a four stroke engine, there is only one power stroke, and three idle strokes. The power stroke supplies the required, momentum to carry out all other strokes. Thus the engine is kept, running, Unit – IV 8.14
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Application, The diesel engines are relatively slow speed engines and, they are generally used in heavy duty vehicles such as buses,, trucks, earth moving machines, etc., 8.8 Comparison of two stroke and four stroke engines, Two stroke engines, , Four stroke engines, , 1., , The power stroke is obtained, in one revolution of the crank, shaft, , One power stroke is obtained, in two revolutions of the crank, shaft., , 2., , Simple in design and lighter in Complicated design and, weight, heavier in weight, , 3., , The torque obtained is uniform The torque obtained is not, uniform, , 4., , Starting is easy, , Starting is not so easy, , 5., , Mechanical efficiency is high, , Mechanical efficiency is low, , 6., , Air cooling is generally used, , Water cooling is generally, used, , 7., , Fuel consumption is more, , Fuel consumption is less, , 8., , Consumption of lubricating oil, is more, , Consumption of lubricating oil, is less, , 9., , Some of the fuel may escape, with exhaust gases, , Fuel cannot escape with, exhaust gases, , 10. Thermal efficiency is less, , Thermal efficiency is more, , 11. Wear and tear of the moving, parts is more, , Wear and tear of the parts is, less, , 12. It gives more noise, , Noise is less, , 13. The initial cost and, maintenance cost are less, , The initial cost and, maintenance cost are more, , 14. These engines are generally, These engines are generally, used in light vehicles such as, used in heavy vehicles such as, motor cycles, scooters, mopeds, buses, lorries, trucks, etc., etc., Unit – IV 8.15
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8.9 Comparison of petrol (S.I) engine and diesel (C.I) engine, Petrol (S.I) engines, , Diesel (C.I) engines, , 1. It works on the basis of Otto It works on the basis of Diesel, cycle, cycle, 2. Air–fuel mixture is sucked into Air alone is sucked into the, the cylinder during suction cylinder during suction stroke, stroke, 3. Petrol engine operates with Diesel engine operates with, low pressure and temperature high, pressure, and, temperature, 4. Carburetor and spark plugs Fuel injection pump, are provided, injectors are provided, 5. Ignition of air–fuel mixture, takes place by an electric, spark produced by the spark, plug, , and, , Ignition of air–fuel mixture, takes place due to the heat, developed by the compression, of air, , 6. They are quantity governed They are quality governed, engines, engines, 7. Operating speed is more, , Operating speed is less, , 8. Weight per unit power is less, , Weight per unit power is more, , 9. Starting is easy, , Starting is not so easy, , 10. Initial cost and maintenance Initial cost and maintenance, cost are less, cost are more, 11. These engines produce less These engines produce more, noise, noise, 12. Thermal efficiency is less, 13. Fuel consumption, power is more, , Thermal efficiency is more, , per, , unit Fuel consumption per unit, power is less, , 14. The cost of petrol is more, , The cost of diesel is less than, petrol, , 15. Petrol engines are widely used Diesel engines are widely, in light vehicles, automobiles, used in heavy vehicles such as, airplanes, etc., buses, lorries, trucks, tractors,, etc., , Unit – IV 8.16
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8.10 Valve timing diagram (V.T.D), The valves in an i.c engines are assumed to open and close, at the dead centre positions of the piston. But, in actual practice, the valves operate some degrees before or after the dead centres, for the better utilization of charge. The ignition is also timed to, occur a little before the top dead centre. The timings of these, sequence of events are represented graphically in terms of crank, angle from dead centre position. This diagram is known as valve, timing diagram., 8.10.1 Valve timing diagram for four−stroke petrol engine, TDC, , Ig, , EVC, IVO - Inlet valve opens, IVC - Inlet valve closes, Ig, - Ignition, EVO - Exhaust valve opens, EVC - Exhaust valve closes, , 10°, , 25°, , IVO, 20°, , IVO - IVC : Suction, IVC - Ig, : Compression, Ig, - EVO : Expansion, EVO - EVC : Exhaust, IVO - EVC : Valve overlap, , 30°, , IVC, , 40°, , EVO, , BDC, , Fig.8.5 Valve timing diagram for four stroke petrol engine, , The valve timing diagram for four stroke petrol engine is, shown in the figure 8.5. The inlet valve opens10−30 o before the, top dead centre (TDC) position. The air−fuel mixture is sucked, into the engine cylinder till the inlet valve closes. The inlet valve, closes 30−40 o or even 60 o after the bottom dead centre (BDC), position. The air−fuel mixture is compressed till the spark occurs., The spark is produced 20−40 o before the TDC position. This gives, sufficient time for the fuel to burn. The pressure and temperature, increases considerably. The burning gases expand and force the, piston to do useful work., Unit – IV 8.17
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The burning gases expand till the exhaust valve opens. The, exhaust valve opens 30−60 o before the BDC position. The, exhaust gases are forced out of the cylinder till the exhaust valve, closes. The exhaust valve closes 8−20 o after the TDC position., Before it closes, again the inlet valve opens 10−30 o before the, TDC position. The period between the inlet valve opening and, exhaust valve closing is known as valve overlap period. During, this period both inlet and exhaust valves are open. The angle, between the inlet valve opening and exhaust valve closing is, known as angle of valve overlap., 8.10.2 Valve timing diagram for four−stroke diesel engine, TDC, , EVC, , FIB, , FIC, 10°, , IVO - Inlet valve opens, IVC - Inlet valve closes, FIB - Fuel injection begins, FIC - Fuel injection ceases, EVO - Exhaust valve opens, EVC - Exhaust valve closes, , 15°, , IVO, 18° 20°, , IVO, IVC, FIB, FIC, EVO, , 30°, , IVC, , - IVC, - FIB, - FIC, - EVO, - EVC, , : Suction, : Compression, : Fuel injection, : Expansion, : Exhaust, , 40°, , EVO, , BDC, , Fig.8.6 Valve timing diagram for four stroke diesel engine, , The actual valve timing diagram for four stroke diesel, engine is shown in figure 8.6. The inlet valve opens10−25 o before, the top dead centre (TDC) position. Fresh air is sucked into the, engine cylinder till the inlet valve closes. The inlet valve closes, 25−50 o after the bottom dead centre (BDC) position. The air is, compressed till the fuel is injected. The fuel injection starts, 5 o −10 o before TDC position in the compression stroke. The, air−fuel mixture burns. The pressure and temperature increases, considerably. The burning gases expand and force the piston to do, useful work., Unit – IV 8.18
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The burning gases expand till the exhaust valve opens. The, exhaust valve opens 30−50 o before the BDC position. The, exhaust gases are forced out of the cylinder till the exhaust valve, closes. The exhaust valve closes 10−15 o after the TDC position., Before exhaust valve closes, again the inlet valve opens 10−25 o, before the TDC position. The period between the inlet valve, opening and exhaust valve closing is known as valve overlap, period. During this period both inlet and exhaust valves are open., The angle between the inlet valve opening and exhaust valve, closing is known as angle of valve overlap., 8.11 Port timing diagram (P.T.D), Ports are provided in two−stroke cycle engines. The timings, of the sequence of events, such as opening and closing of ports,, ignition, etc. are represented graphically in terms of crank angle, from dead centre position. This diagram is known as port timing, diagram., 8.11.1 Port timing diagram of two−stroke petrol engine, Ig, , TDC, , IPO, 20°, , IPC, 100°, , EPC, TPC, , 120°, 140°, , EPO, TPO, , IPO - Inlet port opens, IPC - Inlet port closes, TPO - Transfer port opens, TPC - Transfer port closes, Ig, - Igniton, EPO - Exhaust port opens, EPC - Exhaust port closes, , IPO - IPC : Suction, IPO - TPO : Crank case compression, TPO - TPC : Tranfer, TPC - Ig, : Compression, Ig - TPO : Expansion, EPO - EPC : Exhaust, , BDC, , Fig.8.7 Port timing diagram for two stroke petrol engine, , The port timing diagram of a two−stroke petrol engine is, shown in figure 8.7. The inlet port is uncovered (opened) by the, piston 45−55 o before TDC position. The inlet port is covered, (closed) by the piston 45−55 o after TDC position., Unit – IV 8.19
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The exhaust port is uncovered by the piston 65−75 o before, the BDC position. The exhaust port is covered 65−75 o after the, BDC position. The transfer port is uncovered and covered by the, piston 55−65 o before and after BDC respectively. Ignition occurs, 15−25 o before the TDC. The sequence of events are given below:, IPO − IPC : Air−fuel mixture is sucked into the crankcase, IPC − TPO : Air−fuel mixture is partially compressed in the, crankcase (crankcase compression), TPO − TPC : Partially compressed air−fuel mixture, transferred into the engine cylinder., , is, , TPC − Ig : Air−fuel mixture is compressed in the cylinder., Ig : Air−fuel mixture is ignited by an electric spark., Air−fuel mixture burns., Ig − EPO : The burning gases expand and move the piston to, do work., EPO − EPC : Burnt gases are pushed out of the engine cylinder., 8.11.2 Port timing diagram of two−stroke diesel engine, TDC, , IPO, , IPC, FIB 10° 15, , °, , 100°, , EPC, EPC TPC, , 120°, , FIC, , PVO - Plate valve opens, PVC - Plate valve closes, TPO - Transfer port opens, TPC - Transfer port closes, FIB - Fuel injection begins, FIC - Fuel injection ceases, EPO - Exhaust port opens, EPC - Exhaust port closes, PVO - PVC : Air induction, PVC - TPO : Crank case compression, TPO - TPC : Air tranfer, EPO, TPC - FIB : Compression, TPO, FIB - FIC : Fuel injection, FIC - EPO: Expansion, EPO - EPC : Exhaust, , BDC, , Fig.8.8 Port timing diagram for two stroke diesel engine, , The port timing diagram of a two−stroke diesel engine is, shown in figure 8.8. The plate valve opens after the bottom dead, centre position. It remains open till the piston reaches top dead, centre. The fuel is injected into the cylinder 10−15 o before TDC, position. Fuel injection continues 15−20 o after TDC position., Unit – IV 8.20
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The exhaust port is uncovered and covered by the piston, 35−40 o before and after the BDC position respectively. The, transfer port is uncovered and covered by the piston 30−35 o, before and after BDC respectively. The sequence of events are, given below:, PVO − PVC : Air is induced into the crankcase through plate, valve, PVC − TPO : Air is partially compressed in the crankcase, (crankcase compression), TPO − TPC : Partially compressed air is transferred into the, engine cylinder through transfer port., TPC − FIB : Air is compressed in the cylinder. The pressure, and temperature of the air increases., :, Fuel, is injected into the hot compressed air. Fuel, FIB − FIC, mixes with hot air and burns., FIC − EPO : The burning gases expand and move the piston to, do work., EPO − EPC : Burnt gases are pushed out of the engine cylinder., 8.12 Fuel supply system in petrol (S.I)engines, The fuel supply system is used to supply required amount, of fuel to the engine. Generally, the following fuel supply systems, are used in petrol engines., a) Gravity feed system, b) Pressure feed system or pump feed system, Gravity feed system: In this system, the fuel storage tank, is kept at a higher level than the carburetor and the engine. The, fuel from the tank flows into the carburetor under gravitation, force. This system is employed in motor cycles, scooters, etc., , Storage tank, Fuel pump, , Fuel filter, , Carburettor, , To engine, cylinder, , Fig.8.9 Fuel supply system in petrol engine, , Unit – IV 8.21
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Pump feed system: In this system, a feed pump is used to, feed the fuel to the carburetor. It consists of a storage tank, fuel, pump, fuel filter and carburetor. The fuel tank is mounted away, form the engine to avoid fire risk. The fuel pump supplies fuel to, the carburetor through a fuel filter. The filter removes dirt, grit, and other foreign particles from the fuel. The carburetor supplies, required air–fuel mixture to the engine. This system is widely, employed in automobiles and air–crafts., 8.13 AC mechanical fuel pump, Fuel pump is used to supply fuel to the carburetor by, pumping. A commonly used diaphragm type AC mechanical fuel, pump is shown in the figure., Construction, Fuel from tank, Outlet valve, , Suction valve, , Fuel out to, carburettor, , Diaphragm, , Rocker arm, , Spring, Pivot, , Diaphragm lever, , Pull rod, , Fig.8.10 AC mechanical fuel pump, , This pump consists of a cam, rocker arm, pull rod, lever,, diaphragm, inlet and outlet check valves, return springs, etc. The, diaphragm is made of high grade cotton impregnated with, synthetic rubber. The valves are non–return check valves and, made of bakelite. A spring is provided under diaphragm to, maintain tension on diaphragm. The bottom of the diaphragm is, fitted with a pull rod. One end of the rocker arm is connected to, the pull rod and another end slides over the cam, the rocker arm, is pivoted at a point in the pump body., Unit – IV 8.22
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Working principle, When the cam rotates and pushes the rocker arm, the, rocker arm moves towards the body of the pump. This causes the, pull rod to pull down the diaphragm through the lever. The, pressure inside the chamber falls below the atmospheric pressure., As the result of this pressure difference, the fuel is sucked into, the main chamber through the suction valve from the fuel tank., During this movement, the outlet valve remains closed., When the cam is released from the rocker arm, the rocker, arm moves towards the cam shaft. This causes the diaphragm to, move upwards by the spring force. The pressure in the main, chamber is increased. Now the outlet valve opens and the fuel is, pumped out of the main chamber to the carburetor. During this, movement, the inlet valve remains closed. The suction and, pumping action is repeated continuously and the fuel is pumped, to the carburetor from the tank., 8.14 Carburetor, The process of atomizing and vaporizing the fuel and, mixing it with air is called carburetion. The device used for, carburetion is known as carburetor., Functions of a carburetor., 1) It atomizes and vaporizes the fuel., 2) It prepares a mixture of petrol and air in correct proportion., 3) It measures and supplies the proper quantity of air–fuel, mixture under all conditions of engine operations., 4) It maintains a small reserve of fuel at a constant head., 5) It provides easy starting of the engine in cold condition., 8.15 Simple carburetor, Construction, A simple carburetor is shown in the figure. It consists of a, float chamber and a mixing chamber. In the float chamber, a float, and a needle valve is provided to maintain a constant level of, petrol. The float chamber is vented to atmosphere., Unit – IV 8.23
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Air fuel mixture, to engine, , Fuel from, fuel pump, , Throttle valve, Fuel jet, , Vent to, atmosphere, , Venturi, Float, Float chamber, , Air, Choke valve, , Fig.8.11 Simple carburetor, , The mixing chamber contains a constricted section called, venturi. The fuel is supplied to the jet from the float chamber. The, jet keeps same level of petrol as the level in the float chamber., The mixing chamber has two butterfly type valve. One is called, choke valve and is used to allow air into the mixing chamber., Another one is called throttle valve and is used to allow air–fuel, mixture to the engine. The outlet of carburetor is fitted to the, inlet manifold of the engine to connect inlet valve., Working principle, When the fuel level goes down in the float chamber, the, float also goes down. This opens the needle valve and the fuel, enters into the float chamber. When the correct level is reached,, the float closes the needle valve. Now fuel supply is stopped., During suction stroke, vacuum is created inside the engine, cylinder. This causes pressure difference between the cylinder, and outside the carburetor. Due to this pressure difference,, atmospheric air flows into the carburetor. When the air passes, through the venturi, the velocity of air increases and pressure, reduces. Due to this pressure drop, the air sucks the fuel from the, jet. The fuel comes out in the form of fine particles and mixes, with air to form air–fuel mixture. This air–fuel mixture is, supplied to the engine through a throttle valve. The throttle valve, is operated by an accelerator to control the amount of air–fuel, mixture supplied to the engine., Unit – IV 8.24
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During cold stating, the choke valve is closed completely. This, causes a greater pressure difference and more fuel is sucked from, the jet. The engine can be started easily by this rich mixture., Disadvantages of simple carburetor, 1. It is suitable only for engines running at constant speed and load., 2. The working of a simple carburetor is affected by the charges in, atmospheric pressure and temperature. It gives rich mixture at, high altitude., 3. A week mixture supplied by the carburetor at low speeds may not, ignite properly., 4. It gives a weak mixture when the throttle valve is opened suddenly., 5. It does not have arrangement for providing rich mixture during, starting and warm up., 8.16 Solex carburettor, Air, 5, 10, 12, , 3, , 18, 13, , 6, , 14, , Acceleration, pump, , Spring, , 21, , 19, , 22, , 9, 20, 4, , 8, , Idling, circuit, , Bi-starter, , 7, 11, , 2, , 1, , 17, 16, , 15, , 1−Float, 2−Main jet, 3−Venturi, 4−Emulsion tube, 5−Air correction jet, 6−Spraying, nozzle, 7− Butterfly valve (Throttle valve), 8−Flat disc, 9−starter petrol jet, 10−Starter air, jet, 11−Cold starting passage, 12−Bi-starter lever, 13−Pilot jet, 14−Air bleed orifice,, 15−Volume control (idle) screw, 16−Idle port, 17−Slow speed opening, 18− Pump, injector, 19−Pump lever, 20−Pump jet, 21−Pump inlet valve, 22−Well, , Fig.8.12 Solex carburettor, , Unit – IV 8.25
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Solex carburettor is mostly used in modern automobiles., Fig.8.12 shows the schematic arrangement of solex carburettor., The drawbacks in simple carburettor can be overcome by using, this solex carburettor. The special feature of this carburetor is the, provision of bi−starter for cold starting. The working of the solex, carburettor is explained below., Normal running circuit :, Air, correction jet, Spraying nozzle, , Float, , Main jet, Emulsion tube, , Fig.8.13 Normal running circuit of solex carburettor, , The normal running circuit is shown in fig.8.13. During, normal running, the air is supplied through the choke tube into, the venturi. Petrol is supplied through the main jet. Lateral holes, are provided in the emulsion tube. The air passing through this, tube draws petrol through the lateral holes. The air−correction jet, mixes air and fuel in required ratio. The correct quantity of, air−fuel mixture is sprayed through spray nozzles. This mixture, is supplied to the engine through the throttle valve., Cold starting and warming circuit :, The cold starting circuit is shown in fig.8.14. A bi−starter or, progressive starter is provided for cold starting. The bi−starter has, a flat disc with different size holes. These holes connect the starter, petrol jet and starter air jet to the passage. This passage opens just, Unit – IV 8.26
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below the throttle valve. The bi−starter lever is operated from the, dashboard. This connects either bigger or smaller hole to the, passage. For starting, bigger holes are connected to the passage., The throttle valve is closed. The engine suction is applied to the, starting passage. More is drawn through the starter petrol jet and, less air is drawn through air−jet. Thus a rich mixture is supplied to, the engine through the starting passage., Choke valve, Breather, , Starter, petrol jet, , Float, Lever, , Float chamber, , Bistarter, Throttle valve, , Fig.8.14 Cold starting circuit of solex carburettor, , The starter lever is brought to an intermediate position after, the engine has started. This connects the smaller hole in the circuit, and the fuel supply becomes normal. The starter lever is brought to, off position when the engine reaches normal running condition., Idling and slow running circuit :, The idling and slow running circuit of solex carburettor is, shown in fig.8.15. During idling, the engine is kept running while, the throttle valve is almost closed. The engine suction is applied, at the idle port. Fuel is drawn from the pilot jet. Air is drawn, through the small air bleed orifice. The air and fuel mixes, together in the idling passage. This air−fuel mixture is supplied, to the engine through the idle port. A by pass orifice is provided, above the throttle valve. It ensures smooth transfer from the idle, circuit to the main circuit and avoids flat spot., Unit – IV 8.27
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Pilot jet, , Air bleed, orifice, , Slow speed opening, Volume control screw, Idle port, , Fig. 8.15 Idling and slow running circuit of solex carburettor, , During slow running, the throttle valve is partly opened., The suction at the idle port is not sufficient. The slow speed, opening supplies air−fuel mixture to the engine., Accelerating circuit:, Air, , Spring, Pump jet, , Diaphragm, , Injector, Float, chamber, Pump inlet valve, , Pump, lever, , Accelerator pedal, , Fig.8.16 Accelerating circuit, , Extra fuel is need during acceleration. The acceleration, pump injects extra fuel through the pump jet and the pump, Unit – IV 8.28
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injector. This pump is actuated by the accelerator through a lever., When accelerator is released, fuel is drawn into the pump, through a non−return valve. when the accelerator is pressed, down, the lever operates the pump diaphragm. The pump, diaphragm moves against the spring and pushes the fuel through, pump jet and the pump injector into the venturi., 8.17 Fuel supply system in diesel (C.I) engines, Leakage, Surplus fuel, return, , Feed pump, Fuel, tank, , Fuel filter, , Injection, pump, , Fuel injectors, , Fig.8.17 Fuel supply system in diesel engine, , In diesel engines, the pressure of compressed air reaches, 3500 to 4000 KN/m2 or even above at the end of compression stroke., So the fuel must be injected with a high injection pressure., The fuel injection system in a diesel engine is shown in the, figure. It consists of a fuel tank, fuel feed pump, fuel filter, fuel, injection pump and injectors. The fuel from the tank is pumped to, the fuel injection pump through the filters by fuel feed pump., From the fuel injection pump, the fuel is delivered to the injector., The injectors spray the fuel into the cylinder as fine particles., 8.18 Fuel injection pump, The main function of fuel injection pump is to deliver, correct quantity of fuel to the injector at high pressure. The most, widely used CAV fuel injection pump is shown in the figure., Construction, It consists of a plunger, which reciprocates inside a barrel., The plunger can be rotated by the rack and pinion arrangement. At, the same time, the plunger can be moved up and down in the, Unit – IV 8.29
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barrel by means of a cam. The plunger has a vertical rectangular, groove. This groove extends from the top to another helical groove., The fuel delivery non–return valve is seated in its seat by, spring force. Supply port and spill ports are provided in the, barrel. These ports are opened and closed by movement of the, plunger. The fuel passage is connected to the fuel injector., Working principle, When the plunger is at the bottom of its stroke, it uncovers, supply port and spill port. Diesel from feed pump is forced into, the barrel after filtration. When the cam pushes the plunger, the, inlet port and spill port are closed. The fuel above the plunger is, compressed. Due to the high pressure, the delivery valve opens, against the spring pressure. The fuel flows through the fuel, passage to the injectors., Fuel to injector, , spring, Barrel, Supply port, , Delivery valve, Spill port, Plunger, , Rack, , Tappet, , Cam, , Fig.8.18 CAV fuel injection pump, , When the plunger moves up further, the helical groove, connect the spill port to the top of the plunger. The remaining fuel, escapes through the spill port and the pressure is reduced, Unit – IV 8.30
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suddenly. This causes the delivery valve to come back to its seat, by the spring force. This cycle is repeated and the fuel is supplied, to the injector intermittently., The quantity of fuel delivered is controlled by the rotary, movement of the plunger. The plunger is rotated inside the barrel, by using the rack and pinion arrangement. This alters the effective, stroke of the plunger and the amount of fuel supply is varied., 8.19 Fuel injector (Atomizer), The function of a fuel injector is to spray the fuel into the, engine cylinder in the form of fine particles at the end of, compression stroke., , Leak off, , Adjusting screw, Spring, Spindle, , Fuel in, , Nozzle cap, Plunger, Nozzle valve, Nozzle, Fig.8.19 Fuel injector, , Pressure chamber, , Construction, The fuel injector consists of a nozzle. The nozzle is made with, the injector body. The nozzle valve is seated in its seat by spring, force. A piston is provided in between the spring and the valve. The, tension in the spring can be adjusted by an adjusting screw. A fuel, passage is provided in the body of the injector to allow the fuel from, the injection pump into pressure chamber. The fuel leaking from, the injector is collected and sent back to the tank through leak off, Unit – IV 8.31
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passage. The main types of nozzles used in diesel engine are single, hole nozzle, multiple hole nozzle and pintle nozzle., Working principle, The high pressure fuel from the fuel injection pump enters, into the injector through the fuel passage. Due to the high, pressure of the fuel, the nozzle valve is lifted off its seat against, the spring force. Now the fuel is injected into the engine cylinder, through the nozzle in the form of fine particles. When the fuel, pressure falls, the nozzle valve is brought back to its seat by the, spring force. This causes the nozzle hole to close. This process is, repeated and the fuel is sprayed into the cylinder at the end of, every compression stroke., 8.20 Injection nozzles, The function of the fuel injection nozzle is to inject the fuel, delivered by the pump into the combustion chamber. It also, atomizes the fuel. The following are the important types of, nozzles used in diesel engines., a) Single hole nozzles (flat and conical end), b) Multiple hole nozzles (short and long stem), c) Pintle nozzles, d) Delay type nozzles, a) Single hole nozzles: This type of nozzle contains a, single hole at the end of the nozzle. The fuel passes through this, hole into the combustion chamber. It is closed at the inner end by, the nozzle valve. Single hole nozzles are available with flat end, or conical end. In flat end nozzles, the hole is drilled at he centre, of the nozzle axis. In conical type, the hole is drilled at an angle, (4 o to 15 o ) to the nozzle axis. Single hole nozzles are used in, engines having turbulent chambers., b) Multiple hole (orifice) nozzle : This nozzle contains, many drilled holes. The number of holes may vary from 3 to 18. It, gives finely atomized spray of fuel with greater penetration. They, are used in open combustion chambers. They are of two types : (i), Nozzle with axial hole (ii) Long stem nozzles without axial hole., Unit – IV 8.32
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c) Pintle nozzles : The pintle nozzle consists of a plunger, (pintle) in the fuel passage. This plunger protrudes through the, mouth of the nozzle. The movement of the plunger controls the, fuel flow into the combustion chamber. The shape of the spindle is, made according to the required spray pattern. A hollow, cylindrical jet or a hollow cone shaped spray can be obtained by, suitably shaping the pintle., d) Delay type nozzles : This is also a type of pintle nozzle., The shape of shape of the pintle is change. The shape is in such a, way that it reduces the rate of fuel injection at the beginning of, the delivery and increases towards the end of delivery. It is used, in pre−combustion chambers to reduce the combustion noise when, the engine is idling., 8.21 Fuel filters, The petrol or diesel used in i.c engines may contain, impurities such as dust particles, abrasive particles, water, droplets, etc. Fuel filters are used to prevent these impurities, from going into the fuel pump and engine cylinder., 8.21.1 Fuel filter for petrol engines, Screw, , Body, , In, , Out, , Gasket, , Bowl, Strainer, , Srping, Retainer, , Fig.8.20 Ceramic fuel filter for petrol engine, , Different types of fuel filters are used in petrol engines., Nowadays, ceramic type filters are commonly used. If filters out, fine particles and water droplets from the fuel before it is, Unit – IV 8.33
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supplied to the carburettor. It consists of a glass bowl, ceramic, strainer, spring, body, etc. These filters are fitted in between the, fuel tank and fuel pump. The fuel is filtered when it passes, through the ceramic filter., The sediments and water are collected at the bottom of the, bowl. The filtered fuel enters the fuel pump through the outlet., The sediments settled at the bottom of the bowl may be cleaned, periodically by removing the glass bowl., 8.21.2 Fuel filter for diesel engines, There are two types of fuel filters are used in diesel engines., a) Primary filter or pre−filter, b) Secondary filter of micro−filter, a) Primary filter or pre−filter: The primary filter is, fitted in between fuel tank and the suction side of the fuel feed, pump. The element of this filter is made of nylon mesh or bronze, mesh. They filter longer size dust particles above 40 µ m., Outlet, , Inlet, Conical, diffuser, Sediment, chamber, , Drain plug, , Fig.8.21 Sedimentation type primary filter, , The fuel from the tank enters into the filter. It flows around, the conical diffuser funnel and moves downward to the sediment, chamber. The heavier impurities settle down at the bottom of the, sediment chamber. The filtered diesel moves up and passes, through the outlet. The impurities collected at the bottom of the, chamber can be drained off periodically., , Unit – IV 8.34
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b) Secondary filter or micro−filter : the micro−filter is, fitted in between the fuel feed pump and fuel injection pump. The, elements of this type of filter are made of cotton, cloth, various of, types of paper or cellulose. They filter out very minute particles, (up to 3 µ m) and water content in the fuel., , Inlet, , Outlet, , Canister, Filter, element, , Drain plug, , Fig.8.22 Secondary filter or micro filter, , The filtering element contains fine pores and it retains the, abrasive particles and other solid impurities except water, droplets. When water droplets pass through the element, an, electrostatic attraction is produced between them by the, resin−impregnated element. This causes the droplets to be pulled, towards each other and form larger droplets. These heavier, droplets settle down in the sediment chamber. The filtered fuel, passes through the filter outlet to the fuel injection pump., 8.22 Ignition system of I.C engines, In an internal combustion engine, the air–fuel mixture is, ignited at correct instant at the end of compression stroke. This, results in efficient and smooth running of the engine. The, following two methods are employed in internal combustion, engines., a) Compression ignition (for diesel engines), b) Spark ignition (for petrol engines), Unit – IV 8.35
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8.23 Compression ignition system, This system is used in heavy oil engines working on Diesel, cycle. In these engines, fresh air alone is sucked into the engine, cylinder during suction stroke. This air is compressed to a high, compression ratio ranging from 12 to 20. The pressure and, temperature of compressed air at the end of compression are, approximately 3500 KN/m2 and 600oC respectively. The, atomized fuel is injected into the engine cylinder at the end of, compression stroke., The temperature of compressed air is higher than the self, ignition temperature of fuel and hence fuel–air mixture is ignited., Since the ignition takes place due to high temperature of, compressed air, the diesel engines are known as compression, ignition engines., 8.24 Spark ignition system, This system is used in gas engines, petrol engines and light, oil engines working on Otto cycle. In this system the air–fuel, mixture is ignited by an electric spark produced by the spark, plug. The voltage required for producing an electric spark is about, 8000 Volts. So, the ignition system in a petrol engine has to step, up the battery voltage from 6 volts to 8000 volts. Moreover, it has, to provide a spark in each cylinder at the right time. The, following ignition methods are used in petrol (S.I) engines., 1) Coil ignition system, 2) Magneto ignition system, 3) Electronic ignition system, 8.24.1 Coil ignition system, Coil ignition system is generally employed in medium and, heavy duty petrol engines such as cars. The wiring diagram of a coil, ignition system for a four cylinder petrol engine is shown in the figure., Construction, This system consists of a battery, ignition switch, ignition, coil, condenser, contact breaker, distributor and spark plugs., Generally, 6V or 12V battery is used. The ignition coil consists of, Unit – IV 8.36
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primary and secondary windings. The primary winding consists of, thick wire with few number of turns. The secondary winding, consists of thin wire with several thousand turns., Ignition coil, , Secondary winding, , Ignition switch, , Distributor, , Rotor, 1 2, , Primary winding, , 3 4, , Contact breaker, , 1, , 2, , 3, , 4, , Cam, Battery, Condensor, , Spark plug, , Fig.8.23 Coil ignition system, , The circuit of this system can be divided into primary, circuit and secondary circuit. The primary circuit consists of, battery, primary winding, condenser and contact breaker. The, secondary circuit consists of secondary winding, distributor and, spark plugs. One end of the primary coil is connected to the, battery through ignition switch. Another end is connected to a, condenser and contact breaker., The contact breaker makes and breaks the primary circuit., The timing of the spark is controlled by the contact breakers. The, contact breaker is operated by a cam revolving at half crank, speed. The condenser is connected parallel with the contact, breaker. If prevents sparking across the gap between the contact, breaker points. It also causes a rapid break of the primary circuit,, giving higher voltage in the secondary circuit., The secondary coil is connected to the distributor. A carbon, contact carries the current to the revolving rotor arm. The rotor, distributes the high voltage to the respective spark plug at, regular intervals in the sequence of firing order of the engine. The, outer terminal of the spark plugs are earthed together and, connected to the body of the engine., Unit – IV 8.37
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Working principle, The ignition switch is switched on and the engine is, cranked. The cranking of the engine opens and closes the contact, breaker points through the cam., When the contact breaker points are closed, the current flows, from the battery to the contact breaker through the switch and, primary winding. This current sets up a magnetic field around the, primary winding of ignition coil. When the primary current is at, the highest peak, the contact breaker points are opened by the cam., The magnetic field set up in the primary winding is collapsed, suddenly. Due to this, a high voltage (about 10,000 Volts) is, generated in the secondary winding of the ignition coil. This high, voltage is supplied to the rotor of the distributor. The rotor, distributes the high voltage to various spark plugs in the sequence, of firing order of the engine. When this high voltage tries to escape, through the spark plug gap, an electric spark is produced. This, spark ignites the air–fuel mixture., 8.24.2 Magneto ignition system, Fixed armature, Secondary winding, N, , Distributor, , Rotor, S, , 1 2, , S, , Rotating magnet, assembly, , 3 4, , Contact breaker, , N, Primary winding, , 1, , 2, , 3, , 4, , Cam, Condensor, , Spark plug, , Fig.8.24 Magneto ignition system, , Magneto ignition system is generally employed in light duty, petrol engines such as motor cycles, scooters, etc. The wiring, diagram of magneto ignition system for a four cylinder petrol, engine is shown in the figure., , Unit – IV 8.38
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Construction, This system consists of a magneto, ignition switch, contact, breaker, condenser, distributors and spark plugs. The magneto, consists of a fixed armature and a rotating magnet assembly. The, armature contains primary and secondary windings. The magnet, is placed on the outer rim of the fly wheel., The circuit of this system can be divided into primary, circuit and secondary circuit. The primary circuit consists of, battery, primary winding, condenser and contact breaker. The, secondary circuit consists of secondary coil, distributor and spark, plug. The functions of all other parts are similar to the coil, ignition system., Working principle, The ignition switch is switched on and the engine is, cranked. The cranking of the engine opens and closes the contact, breaker points through the cam., When the contact breaker points are closed, the current flows, from the battery to the contact breaker through the switch and, primary winding. This current sets up a magnetic field around the, primary winding of ignition coil. When the primary current is at, the highest peak, the contact breaker points are opened by the cam., The magnetic field set up in the primary winding is collapsed, suddenly. Due to this, a high voltage (about 10,000 Volts) is, generated in the secondary winding of the ignition coil. This high, voltage is supplied to the rotor of the distributor. The rotor, distributes the high voltage to various spark plugs in the sequence, of firing order of the engine. When this high voltage tries to escape, through the spark plug gap, an electric spark is produced. This, spark ignites the air–fuel mixture., 8.24.3 Electronic ignition system, In this system, contact breaker points are replaced by the, magnetic pickup and reluctor. This eliminates the defects due to, contact breaker points. The schematic diagram of an electronic, ignition system is shown in the figure., Unit – IV 8.39
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Electronic control module, , Spark plug, Ignition switch, , Ignition coil, , Distributor, Magnetic pickup, , Vacuum advance, Battery, , Reluctor or armature, , Fig.8.25 Electronic ignition system, , Construction, This system consists of a battery, ignition switch, electronic, control module, magnetic pickup, reluctor or armature, ignition, coil, distributor and spark plugs. A number of metal tips are, provided on the reluctor according to the number of cylinders of, the engine. A permanent magnet is carried by the distributor. The, reluctor provides a path for the magnetic lines from this magnet., Working principle, The ignition switch is switched on and the engine is, cranked. When the reluctor passes the pick up coil, an electric, pulse is generated and magnetic field is induced. This causes a, magnetic field in the primary winding. This small current, triggers the electronic control module. This control module stops, the flow of battery current to the ignition coil. The magnetic field, set up in the primary winding is collapsed suddenly. Due to this,, a high voltage is generated in the secondary winding. This high, voltage is supplied to the distributor. The distributor distributes, the high voltage to various spark plugs in the sequence of firing, order of the engine. When this high voltage tries to escape, through the spark plug gap, an electric spark is produced. This, spark ignites the air–fuel mixture., , Unit – IV 8.40
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8.25 Governing of internal combustion engine, Governing is the process of regulating the fuel supply, according to the load condition so that to run the engine almost at, constant speed. The device used to regulate the fuel supply is, known as governor. The following are the different methods of, governing an internal combustion engines:, a) Quantity governing, b) Quality governing, c) Hit and miss governing, a) Quantity governing : In this method, the quantity of, air−fuel mixture entering into the engine cylinder is varied. The, composition of the air−fuel mixture remains constant when the, load varies. In petrol engines, the control is obtained by a throttle, valve in the carburettor. In automobiles, the throttle valve is, operated by the accelerator pedal through links., b) Quality governing : Quality governing is commonly, adopted in high speed diesel engines. In this type of governing,, the composition of air−fuel mixture is changed according to the, variation is load. The air flow rate remains constant. The, variation in fuel supply is obtained by the following methods:, (1) Adjusting the effective stroke of the fuel injection pump, (2) By−passing a part of the fuel to the reservoir, (3) Delaying the closure of the suction valve of the fuel, injection pump., c) Hit and miss governing : In this method, an explosion, is omitted when the speed of the engine increases above the, normal speed i.e. one power stroke is missed. The strength of the, air−fuel mixture is not varied. In gas engines, the centrifugal, governor closes the inlet valve. In diesel engines, the governor, makes the fuel pump out of action. No fuel will be supplied and, hence there will be no power stroke., 8.26 Cooling system of I.C engines, The temperature of the burning gases in the engine, cylinder is about 2000 to 2500oC. This heat is absorbed by the, Unit – IV 8.41
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engine components like cylinder head, cylinder walls, piston and, valves. The heating of these parts may cause the following., 1. The strength of piston and cylinder body may be reduced., 2. The engine valves may twist due to overheating., 3. The lubricating oil may decompose and become gummy. It also, gives carbon deposits., 4. It causes thermal stresses in the engine parts, which may lead, to their distortion., 5. Pre–ignition may occur due to the over heating of spark plug., 6. It reduces the volumetric efficiency of the engine., In order to avoid the effects of over heating, it is necessary to, provide a cooling system. However, a cooling system should remove, only about 30% of the heat generated by the combustion of fuel., The following two methods of cooling are generally used in, internal combustion engines., 1) Air cooling or direct cooling, 2) Water cooling or indirect cooling, 8.26.1 Air cooling, , Fins, Cylinder, , Fig.8.26 Air cooling, , Air cooling is used in small engines like, air craft engines,, engines used in scooters, motor cycles, etc. In air cooling system,, the air is allowed to circulate around the cylinder block and, cylinder head. Radiating fins are provided on the outer surface of, Unit – IV 8.42
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the cylinder block and cylinder head. This increases the heat, radiating surface. As more air comes in contact with the cylinder,, the heat is removed efficiently. The high velocity of air required, for cooling is obtained by the forward motion of the engine itself., Air circulating fan is provided in stationary engines., Advantages of air cooling, 1) The design of air cooling system is much simpler., 2) Lighter in weight., 3) The system is free from leakage and freezing of water., 4) Less space is sufficient., 5) This is very much useful in water scarcity areas., 6) The amount of cooling air required is less., Disadvantages, 1) This system is limited to small engines., 2) The engine parts are not uniformly cooled., 3) Air cooled engine produces loud noise., 4) Separate circulating fan is needed in stationary engines., 8.26.2 Water cooling, Water cooling is used in light and heavy vehicles such as, automobiles, buses, lorries, trucks, etc. In water cooling system,, the water is allowed to circulate around the cylinder block and, cylinder head to carry the heat. It consists of radiator, fan, water, pump, water jacket, thermostat valve, radiator shutters, etc., The water pump driven by the engine draws cold water, from the radiator. The cold water is forced to circulate through, water jackets of cylinder block and cylinder head. Thus, the, circulating water cools the engine parts by absorbing heat. Then, the water enters the radiator top and flows from top to bottom., While flowing through radiator tubes, the heat of water is, transmitted to the air drawn by a fan through radiator. Then the, water is cooled and recirculated., However, the temperature of the cooling water should not, fall below 75oC. A thermostat valve is provided between the, engine and the radiator top to control the temperature of the, Unit – IV 8.43
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cooling water. The valve remains open under normal operating, temperatures. When the temperature falls below normal, the, valve is closed to bypass the water. Thus cooling of water is, automatically ceased., Filler cap, , Thermostat valve, Impeller pump, , Water jacket, Cylinder, , Fan, , Piston, , Air, , Drain tap, , Fig.8.27 Water cooling system, , When the heat of the engine is more, the radiator shutters, are opened to allow more air to flow through the radiator. Thus,, the rate of cooling is increased. A drain tap is provided at the, bottom of the radiator to remove the water periodically., Advantages of water cooling, 1) More efficient cooling can be obtained. Hence, efficiency of the, engine is more., 2) Higher compression ratio is permitted., 3) The engine parts are uniformly cooled., 4) Water cooled engines can be installed anywhere in the vehicle., Disadvantages, 1) Water cooling system is heavier in weight., 2) More maintenance is required., 3) A slight leakage of water may stop the engine., 4) Freezing of water causes trouble during cold weather., 5) Water circulating pump and radiator pump consume power., Unit – IV 8.44
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8.27 Comparison of air cooling and water cooling system, Air cooling system, , Water cooling system, , 1. Engine design is simple, , Engine design is not much, simple, , 2. Weight is less, , Weight is more, , 3. No leakage problems, , Leakage of water may occurs, , 4. No risk of freezing of water in Freezing of water in cold, cold climates, climate may cause trouble, 5. Less maintenance is sufficient, , More maintenance is required, , 6. Requires less space, , More space is required, , 7. No danger occurs due to the A small damage in the cooling, damage in cooling system, system may cause danger, 8. The cost for cooling system is The cost for cooling system is, less, more, 9. Cooling is not so efficient, , Cooling is more efficient, , 10. Engine parts are not uniformly Engine parts are uniformly, heated, heated, 11. Higher compression ratio is Higher compression ratio is, not permitted, permitted, 12. It is used in light duty vehicles It is used in heavy duty, vehicles, 8.28 Lubrication of I.C engines, In an internal combustion engine, moving parts and, rotating parts rub against each other and frictional force is, developed. Due to this frictional force, heat is generated and the, engine parts wear quickly. Also, power is lost due to friction., Lubricants are introduced between moving and rotating parts to, reduce heat loss wear and tear., , Unit – IV 8.45
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8.28.1 Purposes of lubrication (or), Functions of lubrication in I.C engines, 1. To reduce friction between moving parts, 2. To reduce wear and tear of the moving parts, 3. To reduce the power loss due to friction, 4. To dissipate the heat generated from the moving parts, 5. To provide cushioning effect against the shocks of the engines, 6. To clean the moving parts by dissolving the impurities during, its circulation, 7. To provide an effective seal against high pressure gases in the, cylinder from leaking out, 8. To reduce the noise while the moving parts rub against each other, 9. To reduce the corrosion and erosion of moving parts, 8.28.2 Properties of lubricants, The important properties of lubricants are explained below:, 1) Viscosity : Viscosity is the resistance offered to the flow, of the lubricant. Viscosity of the oil decreases with the increase in, temperature. Viscosity of the lubricating oil should be within a, specified range even at the highest operating temperature of the, bearings. If too thick oil is used, it will lead to power loss, and, excessive wear and tear of the part. If the oil is too thin, it cannot, lubricate properly and lead to rapid wear of moving parts., 2) Oiliness: It is the property of an oil to spread and, adhere itself firmly with the bearing surfaces. Oiliness of the, lubricating oil should be high for better lubrication., 3) Flash and fire point : Flash point of an oil is the, minimum temperature at which it gives off enough vapour so that, a momentary flame is obtained when a flame is brought near the, oil surface. Fire point is the minimum temperature at which an, oil continuously burns. Fire point is always greater than the flash, point. The flash point of the lubricating oil should be higher than, the operating temperature of the bearing., Unit – IV 8.46
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4) Delegency : The lubricating oil should carry away small, impurities to keep the interior of the engine clean. This property, of lubricating oil is known as delegency., 5) Demulsibility (water separation) : The lubricating oil, should not form an emulsion when brought in contact with water., The property of resisting emulsification is known as, demulsibility., 6) Foaming : It is the condition in which small air bubbles, are present in the oil. This will reduce mass flow or oil and also, increase oxidation. Hence the lubricating oil should be free from, foaming problems., 7) Corrosiveness : The lubricating oil should prevent the, engine parts from corrosion and it should not contain sulphur., 8.29 Methods of lubrication, The following are the methods of lubrication systems are, used in internal combustion engines., 1. Petroil lubrication system or mist lubrication system., 2. Wet sump lubrication system, a) Gravity lubrication system, b) Splash lubrication system, c) Pressure lubrication system, d) Semi–pressure lubrication system, 3. Dry sump lubrication system, 8.29.1 Splash lubrication system, Connecting rod, Splashed oil, , Crank, Lubricating oil, , Scoop, , Fig.8.28 Splash lubrication system, , Unit – IV 8.47
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This lubrication system is employed in single cylinder, stationary internal combustion engines. In this system, the, lubricating oil is filled in the sump at the bottom of the crank, case. Scoops are attached to the big end of the connecting rod., When the piston is at the bottom of its stroke, the big end of, the connecting rod, crank pin and scoop dip into the oil. As the, crank shaft rotates, the scoop picks up the lubricating oil from the, sump and the surplus oil is splashed over the engine parts in the, form of droplets by centrifugal action. Splashed oil droplets settle, on the surface of the piston, cylinder walls, cam shaft and crank, shaft bearings, etc. and lubricate these parts. The splashed oil is, drained back to the sump after lubricating., 8.29.2 Pressure lubricating system (or) Forced lubrication system, Pressure, gauge, , Piston, To valve assembly, Gears, , Connecting rod, , Cam shaft, Crank, Oil gallery, , Filter, Oil sump, , Gear pump, Strainer, , Fig.8.29 Pressure lubrication system, , This system is used in light engines like automobile, engines. In this method, the lubricating oil is forced under, pressure for efficient lubrication. The line diagram of pressure, lubrication system is shown in the figure. It consists of oil sump,, oil pump, oil gallery, oil filter, pressure relief valve, pressure, gauges and oil dip stick., Unit – IV 8.48
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The oil pump (gear pump) is submerged in the lubricating, oil and it is driven by the cam shaft. The gear pump supplies oil, to the oil gallery with high pressure through filters. From the oil, gallery, the oil is distributed under pressure to the different parts, of the engine through oil tubes. Separate oil tubes carry oil for, lubricating big end bearings, timing gears, crank shaft, cam shaft,, valve assembly, etc. Oil from the gallery is supplied to the crank, shaft bearing. The oil flows to the connecting rod big end bearing, through the diagonally drilled hole in the crank shaft. A through, hole is drilled at the centre of the connecting rod. The oil from the, big end bearings flows to the small end bearings through the hole, and lubricate it., The excess oil is drained back into the oil sump. The, delivery pressure of the oil is controlled by relief valve. A pressure, gauge indicates the oil pressure in the system. An oil dip stick is, provided to measure the oil level in the sump., 8.30 Oil pumps, The function of an oil pump is to supply lubricating oil, under pressure to the bearings and the various parts of the, engine to be lubricated. It is located inside the crankcase and, submerged in the oil sump. It is driven by the camshaft through, a worm or spiral gear. The commonly used oil pumps are :, 1) Gear pump, 2) Rotor pump., 8.30.1 Gear pump, Gear pump consists of two spur gears meshed together. The, driver gear is driven by the camshaft. The two gears have a small, clearance with the casing. The gears are adjusted so that they, form a fluid−tight joint when they mesh together., Working principle, The gear pump is submerged in the oil in the crankcase., When the gears rotate, the oil is tapped between their teeth and, discharged through the outlet. The gears gradually build up, sufficient pressure to force the oil through the delivery pipe., Unit – IV 8.49
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Spging, , Adjusting screw, By-pass, , Ball valve, , Outlet, , Inlet, , Fig.8.30 Gear pump, , A pressure relief valve is provided to relieve excessive, pressure caused by high engine speeds. When the oil pressure, exceeds the specified limit, the ball in the relief valve is lifted off, its seat against the spring force and the oil is forced through the, by pass. An adjusting screw is provided to set the working, pressure of the pump., 8.30.2 Rotor pump, Outlet port, , Pump housing, , Inlet port, Oil, , Fig.8.31 Rotor pump, , Rotor pump consists of inner rotor, outer rotor, inlet and outlet, ports. The two rotors mesh internally. The inner rotor has a number, of lobes (teeth) one less than that on the outer rotor. The inner rotor, Unit – IV 8.50
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is driven by the camshaft. The inner rotor rotates the outer rotor., The two rotors are rotated with different axis of rotation. This causes, the variation in the space between these two rotors., Working principle, When the inner rotor is rotated through the camshaft, the, outer rotor also rotates. The oil enters the pump through the inlet, and is filled and squeezed between the tooth spaces. The rotation, of the rotor reduces the clearance between the teeth and oil is, discharged under pressure through the outlet. This pump is, compact than the gear pump. The operation is smooth and, noiseless. Hence rotor pumps are widely used in automobiles., 8.31 Oil filters, The lubricating oil may contains impurities such as sand, particles, metal particles, etc. These impurities cause rapid wear, of the engine bearings and other parts. So, the oil should be, filtered before supplying to the engine parts. The following two, types of filtering systems are commonly used:, 1) By−pass system, 2) Full flow system, 1) By−pass system : In this system, a small amount (10%), of oil passes through the filter and filtered out. The filtered oil, goes to the pump. The remaining oil goes to the engine parts, without being filtered. A very fine filtering element is used in this, system. This system is not widely used because it is not possible, to filter the entire oil content of the system., To engine, bearings, etc, Filter by-pass, valve, , Bearings, , Filter, , Filter, Main relief valve, Oil pump, , Sump, Strainer, (a) By-pass system, , (b) Full flow system, , Fig.8.32 Oil filtering systems, , Unit – IV 8.51
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2) Full flow system : In this system, the entire oil passes, through the filter before it is supplied to the engine part. A spring, loaded relief valve is provided in the filter in addition to the main, relief valve. when the filter is blocked, the by−pass valve diverts, the unfiltered oil to the engine parts. As the entire oil passes, through the filter, the faltering elements used are comparatively, coarser. The commonly used filtering elements are wire gauze,, cotton, felt, paper, plastic−impregnated paper, etc. This type of, system is widely used in automobiles nowadays., 8.32 Supercharging of internal combustion engines, Supercharging is forcing of air−fuel mixture or air alone, into the engine cylinder during the suction stroke with a pressure, greater than the atmospheric pressure. The density of air−fuel, mixture or air entered into the engine cylinder is increased by, supercharging. This is done by a compressor known as, supercharger or blower. As it increases the power output of the, engine, it is also called as boosting., 8.32.1 Effects or advantages of supercharging, 1) The power output of the engine is increased., 2) It provides better mixing of air and fuel., 3) It gives better vaporization of fuel and thus increases the, temperature of the charge., 4) It reduces the possibility of knocking in C.I engines., 5) Supercharged engines produces greater output even at high, altitude., 8.32.2 Applications of supercharging, 1) it is used in air crafts and stationary installations in, mountains to maintain a constant power output even at, higher altitudes., 2) It is used in locomotives and marine engine to reduce the size, of the engine so as to fit into a limited space., 3) Supercharging reduces weight of the engine and increases the, volumetric efficiency. Hence it is used in racing car engines., 4) It is used to increase the power of an existing engine when, more power is required., Unit – IV 8.52
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8.32.3 Turbo charger, Turbo charger is a device which utilizes the exhaust gas, energy (pressure) to drive a super charger through a turbine. Super, charger forces the air−fuel mixture or air alone into the engine, cylinder with a pressure greater than the atmospheric pressure., The turbo charger basically consists of a rotary compressor, and a gas turbine. Both the compressor and turbine are mounted, on the same shaft. The turbine rotates by utilizing the exhaust, energy of the engine. The compressor coupled with the turbine, also rotates. It forces the air or air−fuel mixture in to the engine, cylinder. This increases the thermal efficiency of the engine and, reduces the fuel consumption., , REVIEW QUESTIONS, , 1. How the I.C engines are classified?, 2. Draw a sketch of an I.C engine and indicates the main parts., 3. Write brief notes on the following: 1) Cylinder block, 2) Piston,, 3) Piston rings 4) Connecting rods 5) Crankshaft., 4. Write short notes on : 1) Cylinder head, 2) Valves, 3) Crankshaft,, and 4) connecting rod., 5. What are the different types of piston rings? Explain the, functions of these rings in an I.C engine., 6. Describe the working of a four−stroke cycle petrol engine with, neat sketches., 7. Explain the working of a four−stroke cycle diesel engine with, neat sketches., 8. Describe the working of a two−stroke cycle petrol engine with, neat sketches., 9. Explain the working of a two−stroke cycle diesel engine with, neat sketches., 10. Discuss the merits and demerits of 4−stroke cycle engines over, 2−stroke cycle engines., Unit – IV 8.53
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11. Differentiate between two stroke and four stroke engine., 12. Compare the petrol engine with diesel engine., 13. Explain the valve operating mechanism in an I.C engine., 14. Draw a typical valve timing diagram of a four stroke cycle, petrol engine and indicate the various processes., 15. Draw a typical valve timing diagram of a four stroke cycle, diesel engine and indicate the various processes., 16. Draw and explain a typical port timing diagram of a two stroke, cycle petrol engine., 17. Draw and explain a typical port timing diagram of a two stroke, cycle diesel engine., 18. Explain with sketches the working of A.C mechanical fuel pump., 19. What is the function of a carburettor? Draw a line diagram of a, simple carburettor and explain the working principle., 20. Sketch and name the parts of a solex carburettor., 21. With a neat sketch explain the working of starting and, accelerating circuits of solex carburettor., 22. Sketch a fuel feed pump used in diesel engine and explain its, working principle., 23. Describe the fuel injection pump of a diesel engine with a neat sketch., 24. Draw a neat sketch of an injector of a diesel engine and, explain its working., 25. Explain the different types of nozzles used for diesel engines., 26. What is need of fuel filters? Explain the working of a fuel filter, for a petrol engine., 27. Explain the working of a primary filter used in diesel engine., 28. Explain the working of a secondary or micro−filter used in, diesel engine., 29. Describe with a line diagram a coil ignition system and explain, the working principle., 30. Explain the working of the magneto ignition system used in S.I, engines with a line diagram., 31. With a line diagram, explain the working of electronic ignition, system., Unit – IV 8.54
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32. Briefly describe the various methods of governing employed in, I.C engines., 33. Explain the different types of cooling systems employed in I.C, engines., 34. Explain briefly the water cooling system used in I.C engines., 35. What are the desirable properties of a good lubricant?, 36. What are the functions of lubrication in I.C engines?, 37. Explain with a neat sketch the flash lubrication system., 38. Explain the pressure system of lubrication with a neat sketch., 39. What is the function of an oil pump? Explain the working of a, gear pump., 40. Explain the working of a rotor pump used in lubrication, system of I.C engines., 41. What is the purpose of oil filters? Briefly explain the various, types of oil filters., 42. What is meant by supercharging? What are the effects of, supercharging in an I.C engine?, 43. Briefly explain the turbo charger used in I.C engines., , Unit – IV 8.55
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Unit – IV, Chapter 7., , FUELS AND COMBUSTION OF FUELS, , 7.1 Introduction, A fuel may be defined as a substance which produces a, large amount of heat when burning with oxygen in the, atmospheric air. The amount of heat generated is known as, calorific value of the fuel., 7.2 Classification of fuels, The fuels may be classified as follows:, 1. Solid fuels, 2. Liquid fuels, 3. Gaseous fuels, Each of these fuels may be further subdivided into, (a) Natural fuels, (b) Prepared fuels, 1. Solid fuels, Natural solid fuels are readily available in nature., Examples of such fuels include wood, peat (mixture of water and, decayed vegetable matter), lignite, bituminous coal, anthracite, coal, etc., Wood consists of mainly carbon and hydrogen. It is, converted into coal when burnt in the absence of air. It is used in, places where large amount of waste wood is available. Peat is not, widely used because it has a large amount of water content., Lignite is an intermediate variety between bituminous coal and, peat. It contains more moisture and lower calorific value., Bituminous coal contains little moisture and higher carbon. This, is widely used for power generation in thermal power plants., Anthracite coal is the final stage of coal formation. It is, mostly used for domestic purposes because of its smokeless, combustion. It has high calorific value., The examples of prepared solid fuels include charcoal,, coke, briquetted coal, pulverized coal, etc., Unit – V , , 9.1
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Charcoal is prepared by the dry distillation of wood. It is a, good prepared solid fuel and is used for various metallurgical, processes. Coke is produced when coal is strongly heated, continuously for 42 to 48 hours in the absence of air in a closed, vessel. It has high carbon content and has a higher calorific value, than coal. Briquetted coal is produced form the finely ground coal, by moulding under pressure. Pulverized coal is obtained by, pulverizing the low grade fuels., 2. Liquid fuels, Liquid fuels are obtained from natural petroleum or crude, oil. The liquid fuels consist of hydrocarbons. They are mostly used, in internal combustion engines. The important liquid fuels are, gasoline (petrol), paraffin, diesel, oil, etc., Advantages of liquid fuels:, • Higher calorific value, • Lower storage space required, • Better control of consumption by using valves, • Better cleanliness and freedom from dust, • Easy handling and transportation, • Higher efficiency, Disadvantages, • Higher cost, • More risk of fire, • Requires costly containers for storage, 3. Gaseous fuels, The natural gas is found in or near the petroleum fields,, under the earth’s surface. It consists of methane ( CH4 ), ethane, ( C2 H6 ), carbon dioxide ( CO2 ) and carbon monoxide (CO)., The important prepared gases are coal gas, producer gas,, water gas, mond gas, blast furnace gas, coke over gas, etc. Coal gas, is obtained by the carbonization of coal. It is largely used for, domestic lighting and heating. Producer gas is obtained by the, partial combustion of coal, coke, anthracite coal or charcoal in a, Unit – V , , 9.2
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mixed air−steam blast. It is used in furnaces and for power, generation. Water gas is a mixture of hydrogen and carbon, monoxide. It is produced by passing steam over incandescent, coke. The water gas is used in furnaces and for welding., Mond gas is produced by passing air and a large amount, of steam over waste coal. It is used for power generation and, heating. Blast furnace gas is a by−product in the production of pig, iron in the blast furnace. It is used for power generation in gas, engines, foe steam raising in boilers and for preheating the blast, for furnace. Coke over gas is a by−product from coke oven. It is, obtained by the carbonization of bituminous coal. It is used for, industrial heating and power generation., Advantages of gaseous fuels, • The gaseous fuels undergo complete combustion with, minimum air supply., • They do not produce ash or smoke., • They are free from solid and liquid impurities., • The high temperature can be obtained at a moderate cost., • The supply of gaseous fuel can be accurately controlled., • Gas firing system is very clean., Disadvantages, • Gaseous fuels are easily inflammable, • The require large storage capacity, 7.3 Requirements of a good fuel, The following are the important requirements of a good fuel:, A good fuel should, • have high calorific value, • have a low ignition temperature, • not produce any harmful gases., • burn freely with high efficiency, once it is ignited., • produce less smoke and gases, • be economical, • be easily stored and transported, Unit – V , , 9.3
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7.4 Composition of fuels, Solid fuel consists of mainly carbon and hydrogen. They, also contain oxygen, nitrogen and mineral substances in various, compositions. The chemical composition of coal, coke, peat, etc can, be determined by proximate and ultimate analysis., a) Proximate analysis, Proximate analysis is carried out to determine the following in, the fuel:, (1) Moisture content, (2) Volatile matter, (3) Fixed carbon and, (4) Ash, (1) To determine moisture content : A known mass of fuel, ( mf ) is heated to about 120 o C to remove water by evaporation., Final mass of the fuel ( me ) is estimated. The difference between, the initial and final mass gives the amount of moisture content., Mass of moisture, mm = mf − me, The percentage of moisture in the fuel is calculated as:, m, mm % = m × 100, mf, (2) To determine the volatile matter : A known mass of, fuel ( mf ) is kept in an air tight crucible and heated for 7 minutes, at about 960 o C . The volatile matter along with moisture is, removed from the fuel. The final mass of the fuel ( me ) is, estimated. The mass of volatile matter is determined as follows:, Mass of volatile matter, mv = mf − me, The percentage of volatile matter in the fuel is calculated as:, m, mv % = v × 100, mf, (3) To determine the ash content : A known mass of, fuel ( mf ) is taken and is burnt at 730 o C in the presence of, oxygen. All combustible matter burns at this temperature leaving, the ash as residue. Final mass of the ash is ( mash ) is estimated., Unit – V , , 9.4
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The percentage of ash contents is calculated as:, m, mash % = ash × 100, mf, (4) To determine the fixed carbon : The fixed carbon is, determined as follows:, Mass of carbon, mc = mf − mm − mv − mash, The percentage of fixed carbon is determined as follows:, mc % = 100 − mm % − mv % − mash %, b) Ultimate analysis, The ultimate analysis is carried out to determine the, percentages of the ultimate constituents of the following:, (1) Carbon, (2) Hydrogen (3) Oxygen, (4) Sulphur, (5) Nitrogen, (6) Ash, (1) To determine carbon and hydrogen percentage :, A sample of fuel is burnt in a oxidised atmosphere. Carbon, dioxide and water vapours are produced by burning of fuel. The, masses of carbon dioxide are determined. The carbon dioxide is, absorbed in a solution of potassium hydroxide for estimating the, mass. Water vapours are collected in chloride tube., (2) To determine ash percentage : A known sample of, fuel is burnt at 730 o C in the in the presence of oxygen. All, combustible matter burns at this temperature leaving the ash as, residue. The mass of the ash is estimated., (3) To determine the nitrogen and sulphur percentage :, This is estimated by complex chemical analysis on different, samples of fuel., (4) To determine the oxygen percentage : The, percentage of oxygen is determined as follows:, Percentage of oxygen = 100 − Sum of percentages of remaining, constituents in the fuel., Unit – V , , 9.5
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7.5 Elements and compounds, An element is a basic substance which cannot be broken, to form another substance. The elements may be in solid, liquid or, gaseous state. A compound is a substance formed by chemically, combining two or more elements so that the properties of the, compound entirely differ from those of the individual elements., Example: Sodium chloride (Common salt) is a compound, which is formed by the chemical combination of two elements, sodium and chlorine., 7.6 Atom and molecule, An atom is the smallest particle of an element. It retains the, properties of that element. The mass of an atom varies from element, to element. Atomic mass or atomic weight is the relative measure of, an atom., A molecule is the smallest quantity of substance which, retains the properties of the compound. Molecular weight is the, relative measure of a molecule. The atomic weight and molecular, weight of some important substances are tabulated below:, Atomic weight, (Atomic mass), , Molecular, Weight, , 12, 1 (1.008), , 12, 2, , 14, , 28, , Oxygen, , C, H2, N2, O2, , 16, , 32, , Sulphur, , S, , 32, , 32, , Compounds, Carbon dioxide, , CO2, , −, , 12 + 32 = 44, , Carbon monoxide, Water vapour, , CO, H 2O, , −, −, , 12 + 16 = 28, 2 + 16 = 18, , Sulphur dioxide, , SO2, , −, , 32 + 32 = 64, , Methane, , CH4, , −, , 12 + 4 = 16, , Substance, Elements, Carbon, Hydrogen, Nitrogen, , Symbol, , Unit – V , , 9.6
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7.7 Combustion of solid fuels, The combustion process of fuels can be represented by, chemical equations., Combustion of carbon: A complete combustion of carbon takes, place when carbon in the fuel combines with oxygen in the air and, produces carbon dioxide. During this combustion, a large amount, of heat (33,822 kJ) is released., Carbon + Oxygen Carbon dioxide, O2, CO2, , C, +, Molecular weights :, , 12, , +, , Divided by 12 :, , 1, , +, , i.e., , 1 kg of C, , +, , 32, 8, 3, , 8, kg of O2, 3, , Thus, 1 kg of carbon requires, , , , , , 44, 11, 3, , 11, kg of CO2, 3, , 8, kg of oxygen for, 3, , 11, kg of carbon dioxide., 3, If the air supplied is not sufficient, carbon in the fuel is, burnt into carbon monoxide. This is an incomplete combustion., Carbon + Oxygen Carbon monoxide, O2, 2 CO, , 2C, +, , 32, (2 × 28), Molecular weights : (2 × 12) +, 4, 7, , Divided by 24 :, 1, +, 3, 3, 4, 7, kg of CO, i.e. 1 kg of C + 3 kg of O2 , 3, 4, Thus, 1 kg of carbon burning with, kg of oxygen, 3, 11, and produces, kg of carbon monoxide., 3, complete burning and produces, , Combustion of carbon monoxide: Carbon monoxide combines, with oxygen to form carbon dioxide., 2 CO + O2 2 CO2, Molecular weights : (2 × 28) + 32 (2 × 44), 4, 11, Divided by 56 :, 1, + 7 , 7, Unit – V , , 9.7
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11, kg of CO2, 7, 4, Thus, 1 kg of carbon monoxide requires, kg of, 7, 11, kg of carbon dioxide., oxygen to form, 7, i.e., , 1 kg of CO, , +, , 4, kg of O2, 7, , , , Combustion of hydrogen : During combustion, hydrogen in the, fuel combines with oxygen in the air and produces water vapour., Hydrogen + Oxygen Water vapour, 2 H2, O2, 2 H 2O, , +, , Molecular weights : (2 × 2) +, 32, (2 × 18), , 8, 9, Divided by 4 :, 1, +, 1, kg, of, H, 8, kg, of, O, 9, kg, of, H 2O, , i.e., +, 2, 2, Thus, 1 kg of hydrogen requires 8 kg of oxygen to, produce 9 kg of water vapour., Combustion of sulphur: During combustion, sulphur in the fuel, combines with oxygen in the air and produces sulphur dioxide., Sulphur + Oxygen Sulphur dioxide, O2, SO2, , S, +, Molecular weights :, Divided by 32 :, , 32, 1, , , 32, +, , 1, +, 1 kg of O2 , , 64, 2, 2 kg of SO2, , i.e. 1 kg of S +, Thus, 1 kg of sulphur requires 1 kg of oxygen to, produce 2 kg of sulphur dioxide., Combustion of methane : Methane burns with oxygen and, produces carbon dioxide and vapour., Methane + Oxygen Carbon dioxide + Water vapour, , CO2, CH4 + 2 O2, Molecular weights : 16 + (2 × 32) 44, 11, , Divided by 16 :, 1 +, 4, 4, 11, 9, i.e. 1 kg of CH4 +4 kg of O2 , kg of CO2 +, 4, 4, Unit – V , , 9.8, , + 2 H 2O, + (2 × 18), 9, +, 4, kg of H2O
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Thus, 1 kg of methane requires 4 kg of oxygen to, 11, 9, produce, kg of carbon dioxide and, kg of water vapour., 4, 4, Combustion of ethylene: Ethylene burns with oxygen and, produces carbon dioxide and water vapour., 2 CO2 + 2 H2O, C2 H4 + 3 O2, Molecular weights : 28 + (3× 32) (2 × 44) + (2 × 18), 24, 22, 9, , Divided by 28 :, 1, +, +, 7, 7, 7, 24, 22, 9, i.e. 1 kg of C2 H4 +, kg of O2 , kg of CO2 +, kg of H2O, 7, 7, 7, 24, Thus, 1 kg of ethylene requires, kg of oxygen to, 7, 22, 9, kg of carbon dioxide and, kg of water, produce, 7, 7, vapour., 7.8 Combustion of gaseous fuels, Gaseous fuels are generally expressed in terms of volume., Combustion of carbon monoxide gas :, 2 CO + O2 2 CO2, 3, 3, 2 m 3 of CO + 1 m of O2 2 m of CO2, 3, 3, 3, Divided by 2 : 1 m of CO + 0.5 m of O2 1 m of CO2, , Thus, 1 m 3 of carbon monoxide requires 0.5 m 3 of, oxygen to produce 1 m 3 of carbon dioxide., Combustion of hydrogen gas:, 2 H2 + O2 2 H2O, 2 m 3 of H2 + 1 m 3 of O2 2 m 3 of H2O, 3, 3, 3, Divided by 2 : 1 m of H2 + 0.5 m of O2 1 m of H2O, , Thus, 1 m 3 of hydrogen requires 0.5 m 3 of oxygen to, produce 1 m 3 of water vapour., , Unit – V , , 9.9
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Combustion of methane :, CH4 + 2 O2 CO2 + 2 H2O, 1 m 3 of CH4 + 2 m 3 of O2 1 m 3 of CO2 + 2 m 3 of H2O, Thus, 1 m 3 of methane requires 2 m 3 of oxygen to, produce 1 m 3 carbon dioxide and 2 m 3 of water vapour., Combustion of ethylene :, C2 H4 + 3 O2 2 CO2 + 2 H2O, 1 m 3 of C2 H4 + 3 m 3 of O2 2 m 3 of CO2 + 2 m 3 of H2O, Thus, 1 m 3 of ethylene requires 3 m 3 of oxygen to, produce 2 m 3 carbon dioxide and 2 m 3 of water vapour., 7.9 Atmospheric air, Oxygen is necessary for the combustion of any fuel. This, oxygen is obtained from atmospheric air. For all practical, purposes, the composition of air taken as follows:, Nitrogen = 77 % and Oxygen = 23 % (By weight or mass), Nitrogen = 79 % and Oxygen = 21 % (By volume), 7.10 Stoichiometric (minimum) air required for complete, combustion of solid and liquid fuels, Consider 1 kg of a fuel. Let the fuel contains, Carbon = C kg,, Hydrogen = H2 kg, Oxygen = O2 kg, and Sulphur = S kg., For the complete combustion of the fuel,, 8, 1 kg of carbon requires kg of O2, 3, ∴ C kg of carbon requires, , 8, × C kg of O2, 3, , 1 kg of hydrogen requires 8 kg of O2, ∴ H2 kg of hydrogen requires 8 × H 2 kg of O2, 1 kg of sulphur requires 1 kg of O2, ∴ S kg of sulphur requires S kg of O2, Unit – V 9.10
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8, C + 8 H2 + S kg, 3, Oxygen available in the fuel = O2 kg., Total Oxygen required =, , ∴ Net oxygen required for the combustion of the fuel, 8, = C + 8 H 2 + S − O2 kg, 3, Air required for 23 kg of oxygen = 100 kg, 8, Air required for C + 8 H 2 + S − O2 kg of oxygen, 3, , 100 8, mmin =, C + 8 H 2 + S − O2 kg., 23 3, , , 7.11 Excess air, The theoretical air supplied may not be sufficient for the, complete combustion of the fuel. In actual practice, the amount of, air supplied is always in excess than the theoretical air required., The amount of excess air supplied depends on the quantity of fuel, to be burnt, rate of combustion, method of firing, etc. Generally, 25% to 50% excess air used. If excess air is supplied,, Total air supplied = Theoretical (minimum) air + Excess air, mtotal = mmin + mex, Mass of excess air : The mass of excess air supplied can, be determined by the mass of unused oxygen available in the flue, gases., 100, kg of air is required for 1 kg of oxygen., 23, 100, ∴ Excess air supplied, mex =, × Mass of O2 in flue gases., 23, 7.12 Product of combustion, The products of combustion during the combustion of a, fuel can be estimated if the percentage of constituents of the fuel, are known., Unit – V 9.11
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Products of, combustion, , CO2, , Amount of products of, combustion, Calculated, Formula, mass, m(kg), 11, ×C, m1, 3, , H2O, , 9 H2, , m2, , SO2, , 2S, , m3, , O2, in excess air, N2, in total air, , 0.23 × mex, , m4, , 0.77 × mtot, , m5, m, , Percentage, by mass, m, × 100%, m, m1, × 100, m, m2, × 100, m, m3, × 100, m, m4, × 100, m, m5, × 100, m, 100 %, , 7.13 Theoretical (minimum) volume of air required for, complete combustion, Consider 1 m 3 of a gaseous fuel. Let it contains the, following constituents: Carbon monoxide = CO m 3 , Hydrogen = H2, m 3 , Methane = CH4 m 3 , Ethylene = C2 H4 m 3 , Oxygen = O2 m 3 ., , For the complete combustion of the gas,, 1 m 3 of carbon monoxide requires 0.5 m 3 of O2, ∴ CO m 3 of carbon monoxide requires 0.5 CO m 3 of O2, 1 m 3 of hydrogen requires 0.5 m 3 of O2, ∴ H2 m 3 of hydrogen requires 0.5 H 2 m 3 of O2, 1 m 3 of methane requires 2 m 3 of O2, ∴ CH4 m 3 of methane requires 2 CH 4 m 3 of O2, 1 m 3 of ethylene requires 3 m 3 of O2, ∴ C2 H4 m 3 of ethylene requires 3 C2 H 4 m 3 of O2, Total volume of O2 required = 0.5CO + 0.5 H2 + 3CH4 + 3C2 H4 m 3, , O2 available in the fuel = O2 m 3, ∴ Net volume of O2 required, = 0.5CO + 0.5 H 2 + 3CH 4 + 3C2 H 4 − O2 m 3, Unit – V 9.12
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Air required for 21 m 3 of oxygen = 100 m 3, Air required for [0.5CO + 0.5 H 2 + 3CH4 + 3C2 H 4 − O2 ] m 3 of oxygen, 100, =, [0.5CO + 0.5 H2 + 3CH4 + 3C2 H4 − O2 ], 21, ∴ Minimum volume of air required,, 100, [0.5CO + 0.5 H 2 + 3CH 4 + 3C2 H 4 − O2 ], Vair =, 21, 7.14 Volumetric and gravimetric analysis, The analysis of the composition of a mixture of gases, based on volume is known as volumetric analysis. This gives the, percentage by volume of each of the constituents of a gaseous fuel., The analysis of a mixture of gases based on mass or, weight is known as gravimetric analysis. This gives percentage by, mass or weight of each of the constituents of a gaseous fuel., 7.15 Conversion of volumetric to gravimetric (mass) analysis, The conversion of volumetric to gravimetric analysis, involves the following steps:, (1) The percentage volume V and the molecular weight M of each, constituent are noted., (2) The mass of each constituent is found out by multiplying the, volume and its molecular weight ( m = V × M ), (3) Total mass of the mixture of gases is found out by adding the, individual mass of each constituents. ( m = m1 + m2 + ... ), (4) The percentage by mass of each constituents is calculated by, dividing the mass of the constituent by total mass of the gas., m, % mass =, m, 7.16 Conversion of gravimetric (mass) analysis into, volumetric analysis, The conversion of gravimetric to volumetric analysis, involves the following steps:, (1) The percentage mass m and the molecular weight M of each, constituent are noted., (2) The volume of each constituent is found out by dividing the, m, mass of the constituent by its molecular weight ( V =, ), M, Unit – V 9.13
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(3) Total volume of the mixture of gases is found out by adding, the individual volume of each constituents. ( V = V1 + V2 + ... ), (4) The percentage by volume of each constituents is calculated, by dividing the volume of the constituent by total volume of, the gas., V, % volume =, V, 7.17 Mass of air actually supplied and air fuel ratio, The air−fuel ratio in volumetric analysis is given by,, N2, , , Air, mtot =, = 3.03C , kg / kg of fuel, CO, CO, Fuel, +, 2, , , where,, , C = Mass of carbon in 1 kg of fuel, N2 = Percentage by volume of N2 in flue gases, , CO2 = Percentage by volume of CO2 in flue gases, CO = Percentage by volume of CO in flue gases, The air−fuel ratio in gravimetric analysis is given by,, 100 × N 2 × C, Air, mtot =, kg / kg of fuel, =, Fuel 21CO2 + 33CO, where,, , C = Mass of carbon in 1 kg of fuel, N2 = Percentage by mass of N2 in flue gases, , CO2 = Percentage by mass of CO2 in flue gases, CO = Percentage by mass of CO in flue gases, 7.18 Calorific value (CV), The calorific value of a fuel is defined as the amount of, heat liberated by the complete combustion of unit quantity (1 kg, in case of solid and liquid fuel, and 1 m 3 in case of gaseous fuels), of a fuel. It is expressed as kJ/kg for liquid and solid fuels, and, kJ/ m 3 for gaseous fuels., (a) Gross or Higher Calorific Value (HCV), It is defined as the amount of heat obtained by the, complete combustion of unit mass (or unit volume) of a fuel, when, the products of combustion are cooled down to the temperature of, the air (15 o C ) supplied., , Unit – V 9.14
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Dulong's formula for HCV:, The higher calorific value (HCV) is given by the Dulong’s, formula, O , , HCV = 33800C + 144000 H 2 − 2 + 9270 S kJ/kg, 8 , , (b) Net or Lower Calorific Value (LCV), Lower calorific value is defined as the amount of heat, obtained by the complete combustion of unit mass (or unit, volume) of a fuel, when the heat absorbed by the product of, combustion is not recovered., The lower calorific value for solid and liquid fuels is given by:, LCV = [HCV − (9 H 2 + ms )]× 2466 kJ/kg, where, H2 = Mass of hydrogen (kg/kg of fuel), , ms = mass of moisture or steam (kg/kg of fuel), In the case of gaseous fuels,, m, , LCV = HCV − s × 2466 kJ/ m 3, V, s, , where, ms = amount of steam condensed (kg), , Vs = Volume of gas used at S.T.P ( m 3 ), 7.19 Water equivalent ( me ), The water equivalent of a substance (or an apparatus) is, defined as the quantity of water which requires the same amount, of heat as the substance to raise its temperature through 1 o C . It, is mathematically expressed as,, mc .C pc, me =, C pw, where, me = Water equivalent of apparatus, , mc = Mass of apparatus, C pc = Specific heat of apparatus material, C pw = Specific heat of water (4.19 kJ/kg.K), , Unit – V 9.15
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7.20 Experimental determination of calorific value, Calorimeters are used to determine the calorific value of any, fuel. The generally used calorimeters are :, 1) Bomb calorimeter (for solid and liquid fuels), 2) Junker’s calorimeter (for gaseous fuels), 3) Boy’s calorimeter (for gaseous fuels), 1) Bomb calorimeter, This calorimeter is used to determine the calorific value of, solid and liquid fuels., Thermometer, Non return valve, Release valve, Copper vessel, Water bath, Bomb, , Fuse wire, Crucible, , To battery, , Fig. 7.1 Bomb calorimeter, , Construction:, The schematic arrangement of bomb calorimeter is shown in, the figure. It consists of a bomb made of thick walled stainless steel., It is designed so as to withstand high pressure, heat and corrosion. A, crucible is provided inside the bomb to place the fuel. The oxygen, required to burn the fuel is supplied to the bomb through a, non−return valve. A pressure release valve is provided to discharge, the burnt gases. Two guide rods are provided at the bottom. The, crucible is supported by one of the guide rods. The guide rods are, connected with a fuse wire made of platinum of nichrome. The other, ends of fuse wire are connected to a battery. The fuel can be ignited, by the heat generated in the fuse wire when high voltage current is, passed through the wire. The bomb is placed in a water bath, containing a known quantity of water. A stirrer is also provided., Unit – V 9.16
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To compensate the loss of heat due to radiation, a cooling, correction is added to the observed temperature rise., Corrected temp. rise = Observed temp. rise + Cooling correction., (T2 − T1 ) c = (T1 − T1 ) + Tc, The corrected temperature rise is substituted in the above, expression to estimate the HCV of the fuel., The lower calorific value (LCV) is calculated from the, amount of moisture ( mm ) and hydrogen present in the fuel., , LCV = HCV − (9 H 2 + mm ) × 2466, 2) Junker’s gas calorimeter, This calorimeter is used to find out the calorific value of, gaseous fuels., Construction:, It consists of a cylindrical chamber with vertical tubes. A, burner is provided at the bottom of the chamber to burn the gas., The hot burnt gases pass to the top of the burner and then, descend through small metal tubes. These tubes are surrounded, by water. Water enters through bottom of the chamber and leaves, at the top. A pressure regulator is provided to control the, pressure of the gas entering the gas meter. A pressure gauge and, a thermometer are also provided., Thermometer(T), Water outlet, T, Cobmustion chamber, Over, flow, , T, Water, inlet, , Gas meter, , Pressure regulator, , Gas outlet, Condensate, , Fig.7.2 Junker’s gas calorimeter, , Unit – V 9.18
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Procedure:, The gaseous fuel is passed through the burner. The gas is, burnt under steady uniform conditions until the equilibrium, condition is reached. The volume of the gas (V) passing through, the calorimeter is measure by the gas meter. The temperature of, the gas (T) is noted. The pressure of the gas ( pg ) and the, barometric pressure ( pb ) are noted. The inlet temperature ( T1 ), and outlet temperature ( T2 ) of the cooling water are noted. The, mass of the cooling water ( mw ) is measured. The amount of, condensate ( ms ) is measured by collecting it in a small graduated, vessel., Let,, , ps, Ts, Vs, , =, , Standard pressure, , =, , Standard temperature, , = Volume of the gas burnt reduced to S.T.P, The calorific value of gaseous fuels is generally expression, kJ/ m 3 of the gas at S.T.P conditions., The volume of gas reduce to S.T.P conditions is calculated, using gas equation., p .V, p.V, p.V Ts, By gas equation, s s =, ; ∴ Vs =, ×, Ts, T, T, ps, Heat released by the fuel, Qf = HCV × Vs, Heat carried by the water, Qw = mw .C pw .(T2 − T1 ), Heat released by the fuel = Heat carried by the water, HCV × Vs = mw .C pw .(T2 − T1 ), ∴ HCV =, , mw .C pw .(T2 − T1 ), Vs, , The lower calorific value is determined from the amount, of condensate collected., ms, , , LCV = HCV − , , V, 2466, ×, s, , , Unit – V 9.19
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7.21 Orsat apparatus for flue gas analysis, The constituents of the flue gases are determined for, checking the combustion efficiency of boilers. Orsat apparatus is, used to carry out the flue gas analysis., Aspirator bottle, d, , c, , C, , b, , B, , a, , A, , Flasks, Eudiometer tube, , Fig.7.3 Orsat apparatus, , Construction:, Orsat apparatus consists of a graduate measuring glass, tube, called as eudiometer, and three flasks A, B and C. An, aspirator bottle containing water is connected to the bottom of the, eudiometer tube by means of rubber tube. It can be moved up and, down for producing a suction or pressure effect on the sample of, the flue gas., The flask A contains caustic soda (NaOH) and is used to, absorb carbon dioxide in the sample of the flue gas. The flask B, contains caustic soda and pyrogallic acid, which absorbs oxygen, from the sample of the flue gas. The flask C contains a solution of, cuprous chloride ( Cu 2Cl 2 ) in hydrochloric acid (HCl). It is used to, absorb carbon monoxide (CO) from the sample of the flue gas., Each of the three flasks has stop cock ‘a’, ‘b’, and ‘c’ respectively, and a three−way cock ‘d’ which can be opened to either, atmosphere or flue gas., Procedure :, The stop cocks (a, b and c) are closed. The three−way cock, is opened and 100 c m 3 of flue gas is sucked into the measuring, glass. Then the three−way cock is closed. The level of the, aspirator bottle is adjusted and the flue gas is forced in to flask A, Unit – V 9.20
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by opening the cock ‘a’. The flue gas is allowed to remain in the, flask A for sometime. The chemical in the flask A absorbs carbon, dioxide. Then the flue gas is sucked back into the glass tube by, moving the aspirator flask. The contraction in the volume of the, flue gas is noted by reading the level of water in the measuring, glass. This gives the amount of carbon dioxide present in the, sample of the flue gas., The procedure is repeated with the other two flasks B and, C to find the amount of oxygen and carbon monoxide present in, the sample of flue gas., , REVIEW QUESTIONS, 1., , How fuels are classified? List out the advantages and, disadvantages of each., , 2., , What are the requirements of good fuel?, , 3., , Explain the ultimate and proximate analysis of fuel?, , 4., , How the stoichiometric air requirement for a fuel is determined?, , 5., , What is meant by excess air? Explain the significance of, excess air., , 6., , Explain how the volumetric analysis of a gas is converted into, gravimetric analysis., , 7., , What is calorific value of a fuel? What is the difference, between HCV and LCV?, , 8., , Explain the procedure for finding the calorific value of a solid fuel., , 9., , Explain any one method of finding calorific value of a gaseous, fuel., , 10. Explain the procedure for the analysis of exhaust gases uses, Orsat apparatus, Unit – V 9.21
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SOLVED PROBLEMS, Example 9.1, Determine the stoichiometric air required for the complete, combustion of coal whose analysis by weight is given below:, Carbon=78%, Hydrogen=3%, Oxygen=3%, Sulphur=1%, Ash=10%, and Moisture=5%., Given data, Carbon, C = 78 % = 0.78 kg/kg of coal, Hydrogen, H2 = 3 % = 0.03 kg/kg of coal, Oxygen, O2 = 3 % = 0.03 kg/kg of coal, Sulphur, S = 1 % = 0.01 kg/kg of coal, Ash = 10% = 0.10 kg/kg of coal, Moisture = 5% = 0.05 kg/kg of coal, To find, Stoichiometric air required, mmin, Solution, Stoichiometric air required,, 100 8, mmin =, C + 8 H2 + S − O2 , 23 3, , =, , , 100 8, , × 0.78 + (8 × 0.03) + 0.01 − 0.03, 23 3, , , = 10 kg/kg of coal, , Result, Stoichiometric air required, mmin = 10 kg/kg of coal, Example 9.2, A coal has the following analysis by weight: Carbon=65%,, Hydrogen=7%, Oxygen=1%, Rest is ash. Determine the weight of, air required for the complete combustion of one kg of fuel., Given data, Carbon, C = 65 % = 0.65 kg/kg of coal, Hydrogen, H2 = 7 % = 0.07 kg/kg of coal, Oxygen, O2 = 1 % = 0.01 kg/kg of coal, To find, Weight of air required, Wair, Unit – V , , P9.1
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Solution, Mass of air required for complete combustion,, 100 8, mair =, C + 8 H2 + S − O2 , 23 3, , , 100 8, , =, × 0.65 + (8 × 0.07) + 0 − 0.01, , 23 3, , , = 9.928 kg/kg of coal, Weight of air required,, Wair = mair × g = 9.928 × 9.81 = 97.394 N/kg of coal, Result, Weight of air required, Wair = 97.394 N/kg of coal, Example 9.3, Steam boiler uses pulverised coal in the furnace. The, ultimate analysis of coal by weight as received is: Carbon=78%,, Hydrogen=3%, Oxygen=3%, Sulphur=1%, Ash=10%. Excess air, supplied is 30%. Calculate the weight of air to be supplied., Given data, Carbon, C = 78 % = 0.78 kg/kg of coal, Hydrogen, H2 = 3 % = 0.03 kg/kg of coal, Oxygen, O2 = 3 % = 0.03 kg/kg of coal, Sulphur, S = 1 % = 0.01 kg/kg of coal, Ash = 10% = 0.10 kg/kg of coal, Moisture = 5 % = 0.05 kg/kg of coal, To find, Weight of the air supplied, Wtot, Solution, Minimum air required,, 100 8, mmin =, C + 8 H2 + S − O2 , 23 3, , , 100 8, , =, × 0.78 + (8 × 0.03) + 0.01 − 0.03, 23 3, , , = 10 kg/kg of coal, Unit – V , , P9.2
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30, × 10 = 3 kg/kg of coal, 100, + mex = 10 + 3 = 13 kg/kg of coal, , Excess air supplied, mex = 30% =, Total air supplied, mtot = mmin, , Weight of air required,, Wtot = mtot × g = 13 × 9.81 = 127.53 N/kg of coal, Result, Weight of the air supplied, Wtot = 127.53 N/kg of fuel, Example 9.4, A fuel oil consists of the following percentage analysis by, mass: 82% C, 12% H2 , 2% O2 , 1% S and 3% N2 . Determine, stoichiometric mass of air required to completely burn the fuel and, also determine the products of combustion by mass as a, percentage., Given data, Carbon, C = 82 % = 0.82 kg/kg of oil, Hydrogen, H2 = 12 % = 0.12 kg/kg of oil, Oxygen, O2 = 2 % = 0.02 kg/kg of oil, Sulphur, S = 1 % = 0.01 kg/kg of oil, Nitrogen, N2 = 3% = 0.03 kg/kg of oil, To find, Stoichiometric air required, mmin, Products of combustion in percentage of mass, Solution, Stoichiometric air required,, 100 8, mmin =, C + 8 H2 + S − O2 , 23 3, , =, , , 100 8, , × 0.82 + (8 × 0.12) + 0.01 − 0.02, 23 3, , , , = 13.6377 kg/kg of oil, , Unit – V , , P9.3
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Amount of products of combustion, Products of, combustion, , Mass m (kg), , m, × 100%, m, , 11, × 0.82, 3, , 3.0067, , 20.54, , 9 × 0.12, , 1.0800, , 7.38, , 2 × 0.01, , 0.0200, , 0.14, , (0.77×13.6377), + 0.03, , 10.5310, , 71.94, , m =14.6377, , 100.00, , Formula, , Calculation, , H 2O, , 11, ×C, 3, 9 H2, , SO2, , 2S, , CO2, , N2, , 0.77 × mtot, , in air & fuel, , + N2 in fuel, , Percentage, by mass, , Result, Stoichiometric air required, mmin = 13.6377 kg/kg of oil, Products of combustion : CO2 =20.54%, H2O =7.38%, SO2 =0.14%,, and N2 =71.94%, Example 9.5, A fuel has a gravimetric analysis as follows: Carbon=86%,, Hydrogen=10%, Oxygen=1%, Sulphur=1%, Remaining is ash. If, 50% excess air supplied find (1) total amount of air supplied, (2) gravimetric analysis of exhaust gas., Given data, Carbon, C = 86 % = 0.86 kg/kg of fuel, Hydrogen, H2 = 10 % = 0.10 kg/kg of fuel, Oxygen, O2 = 1 % = 0.01 kg/kg of fuel, Sulphur, S = 1 % = 0.01 kg/kg of fuel, Ash = 100 − (86+10+1+1) = 2% = 0.02 kg/kg of fuel, To find, (1) Total amount of air supplied, mtot, (2) Gravimetric analysis of exhaust gases., Solution, Stoichiometric (minimum) air required,, 100 8, mmin =, C + 8 H2 + S − O2 , 23 3, , , 100 8, , =, × 0.86 + (8 × 0.1) + 0.01 − 0.01, 23 3, , , = 13.45 kg/kg of fuel., Unit – V , , P9.4
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Excess air supplied, mex = 50% =, , 50, × 13.45 = 6.725, 100, , Total air supplied, mtot = mmin + mex = 13.45 +6.7253 = 20.175 kg, Amount of products of combustion, Products of, combustion, , Formula, , Percentage, by mass, , Calculation Mass m (kg) m × 100%, m, , CO2, , 11, ×C, 3, , 11, × 0.86, 3, , 3.1533, , 14.906, , H 2O, , 9 H2, , 9 × 0.10, , 0.9000, , 4.254, , SO2, , 2S, , 2 × 0.01, , 0.0200, , 0.095, , O2, , 0.23 × mex, , 0.23 × 6.725, , 1.5468, , 7.312, , (0.77×20.175), , 15.5348, , 73.433, , m =2.1549, , 100.00, , in excess air, , N2, in total air, , 0.77 × mtot, , Result, (1) Total amount of air supplied, mtot = 20.175 kg/kg of fuel, (2) Gravimetric analysis of exhaust gases: CO2 =14.906%,, H2O =4.254%, SO2 =0.094%, O2 =7.312% and N2 =73.433%, Example 9.6, In a boiler trial, the analysis of the coal by weight, indicated as follows: 60% C, 4.5% H2 , 7.5% O2 , and remaining is, incombustible. The dry flue gas has the following composition by, volume: CO2 =9%, CO=1%, N2 =80%, O2 =10%. Determine (1) the, weight of air supplied per kg of coal burnt and (2) the percentage, of excess air., Given data, Analysis by weight: In 1 kg of coal,, Mass of carbon, C = 60 % = 0.60 kg, Mass of hydrogen, H2 = 4.5 % = 0.45 kg, Mass of oxygen, O2 = 7.5 % = 0.75 kg, Unit – V , , P9.5
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Dry flue gas composition by volume : In 1 m 3 of gas,, Volume of carbon dioxide, CO2 = 9 % = 0.09 m 3, Volume of carbon monoxide, CO = 1% = 0.01 m 3, Volume of nitrogen, N2 = 80% = 0.80 m 3, Volume of oxygen, O2 = 10% = 0.10 m 3, To find, (1) Weight of air supplied, Wtot, (2) Percentage of excess air, Solution, Actual air supplied is given by,, N2, 0.8, , , = 14.544 kg, mtot = 3.03 C , = (3.03 × 0.6) × , (0.09 + 0.01) , CO2 + CO , Weight of air required,, Wtot = mtot × g = 14.544 × 9.81 = 142.68 N, Stoichiometric (minimum) air required,, 100 8, mmin =, C + 8 H2 + S − O2 , 23 3, , , 100 8, , =, × 0.6 + (8 × 0.45) + 0 − 0.075, , 23 3, , , = 8.196 kg, Excess air supplied, mex = mtot − mmin = 14.544 − 8.196 = 6.348 kg, Percentage of excess air =, , mex, 6.348, × 100 =, × 100 = 77.46%, mmin, 8.196, , Result, (1) Weight of air supplied, Wtot = 142.68 kg, (2) Percentage of excess air = 77.46%, Example 9.7, A gas has the following percentage of composition by, volume: CO=10%, H2 =48%, CH4 =28%, O2 =3%, CO2 =2%, N2 =9%., Estimate the volume of air required for complete combustion of 1, m3 of fuel if 30% excess air is supplied., , Unit – V , , P9.6
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Given data, Volume of carbon monoxide, CO = 10 % = 0.10 m 3, Volume of hydrogen, H2 = 48 % = 0.48 m 3, Volume of methane, CH4 = 28 % = 0.28 m 3, Volume of oxygen, O2 = 3 % = 0.03 m 3, Volume of carbon dioxide, CO2 = 2 % = 0.02 m 3, Volume of nitrogen, N2 = 9 % = 0.09 m 3, To find, (1) Volume of air required, Vtot, Solution, Minimum volume of air required,, 100, Vair =, [0.5CO + 0.5 H2 + 2CH4 + 3C2 H4 − O2 ], 21, 100, =, [(0.5 × 0.1) + (0.5 × 0.48) + (2 × 0.28) + (3 × 0) − 0.03], 21, = 3.905 m 3 / m 3 of fuel, 30, Excess air supplied, Vex =, × 3.905 = 1.172 m 3, 100, Total volume, Vtot = Vair + Vex = 3.905 + 1.172 = 5.077 m 3 / m 3 of fuel., Result, Volume of air required, Vtot = 5.707 m 3, Example 9.8, A fuel oil has the following analysis by weight: C=85%,, H2 =12.5%, O2 =2%, and the residue=0.5%. The dry flue gas has, the following composition by volume: CO2 =9%, CO=1%, O2 =7.7%,, and N2 =82.23%. Determine the air−fuel ratio., Given data, Analysis of fuel oil by weight: In 1 kg of fuel,, Mass of carbon, C = 85 % = 0.85 kg, Mass of hydrogen, H2 = 12.5 % = 0.125 kg, Mass of oxygen, O2 = 2 % = 0.02 kg, Residue = 0.5% = 0.005 kg, Dry flue gas composition by volume : In 1 m 3 of gas,, Volume of carbon dioxide, CO2 = 9 % = 0.09 m 3, Unit – V , , P9.7
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Volume of carbon monoxide, CO = 1% = 0.01 m 3, Volume of nitrogen, N2 = 82.23% = 0.8223 m 3, Volume of oxygen, O2 = 7.7% = 0.077 m 3, To find, Air − fuel ratio, Solution, The fuel does not contain Nitrogen. Hence the mass of air actually, supplied or air−fuel ration is given by,, N2, Air, , , mtot =, = 3.03 C , , CO, +, CO, Fuel, 2, , 0.8223 , = (3.03 × 0.85) , = 21.18 kg/kg of fuel, 0.09 + 0.01 , Result, Air − fuel ratio = 21.18 : 1, Example 9.9, A flue gas has following composition by mass: C=80%,, H2 =10%, O2 =3%, S=2% and the rest is incombustible. Calculate, HCV and LCV., Given data, Carbon, C = 80 % = 0.80 kg/kg of fuel, Hydrogen, H2 = 10 % = 0.10 kg/kg of fuel, Oxygen, O2 = 3 % = 0.03 kg/kg of fuel, Sulphur, S = 2 % = 0.02 kg/kg of fuel, Incombustible = 100 − (80+10+3+2) = 5% = 0.05 kg/kg of fuel, To find, (1) Higher calorific value, HCV, (2) Lower calorific value, LCV, Solution, By Dulong’s formula,, , O, HCV = 33800 C + 144000 H2 − 2 + 9270 S, 8 , , 0.03 , = (33800 × 0.80) + 144000 0.01 −, + (9270 × 0.02), 8 , , = 41085.4 kJ/kg, Unit – V , , P9.8
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LCV = HCV − (9 H2 × 2466) = 41085.4 − (9 × 0.1 × 2466), = 38866 kJ/kg, Result, (1) Higher calorific value, HCV = 41085.4 kJ/kg, (2) Lower calorific value, LCV = 38866 kJ/kg, Example 9.10, Calculate the higher calorific value of a coal specimen, from the following data: Mass of coal burnt=1gm, Quantity of, water in calorimeter = 2500 gm, Increase in temperature of water =, 2.6 o C , Mass of bomb calorimeter and accessories = 3900 gm,, Mean specific heat of calorimeter= 0.39kJ/kg.K. If the fuel, contains 6% hydrogen, also calculate its lower calorific value., Given data, Mass of cola burnt, mf = 1 gm = 0.001 kg, Mass of water, mw = 2500 gm = 2.5 kg, Temperature rise, t2 − t1 = 2.6 o C ; T2 − T1 =2.6 K, Mass of calorimeter, mc = 3900 gm = 3.9 kg, Specific heat of calorimeter, C pc = 0.39 kJ/kg.K, Mass of H2 = 6 % = 0.06 kg/kg of fuel, To find, (1) Higher calorific value, HCV, (2) Lower calorific value, LCV, Solution, Water equivalent of bomb calorimeter,, mc .C pc 3.9 × 0.39, me =, =, = 0.363 kg, C pw, 4.19, Heat liberated by the coal, Qf = mf × HCV, Heat taken by water and calorimeter,, Qwc = (mw + me ).C pw .(T2 − T1 ), = (2.5 + 0.363) × 4.19 × 2.6 = 31.1895 kJ, Heat liberated by the coal, Qf, = Heat taken by water and calorimeter, mf × HCV = 31.1895, Unit – V , , P9.9
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∴Higher Calorific Value,, 31.1895 31.1895, HCV =, =, = 31189.5 kJ/kg, mf, 0.001, Lower calorific value, LCV = HCV − (9 H2 × 2466), = 31189.5 − (9 × 0.06 × 2466) = 29857.86 kJ/kg, Result, (1) Higher calorific value, HCV = 31189.5 kJ/kg, (2) Lower calorific value, LCV = 29857.86 kJ/kg, Example 9.11, The following particulars refer to an experiment for the, determination of calorific value of a sample coal containing carbon, 88% and hydrogen 4.2%. Mass of coal burnt=0.78gm, Quantity of, water in calorimeter = 1900 gm, Mass of fuse wire = 0.02gm of, calorific value 7000 J/kg, Water equivalent of calorimeter =, 1900gm, Observed temperature rise = 2.98 o C , Cooling correction =, 0.016 o C . Find the lower calorific value., Given data, Mass of carbon, C = 88% = 0.88 kg/kg of fuel, Mass of hydrogen, H2 = 4.2% = 0.042 kg/kg of fuel, Mass of fuel burnt, mf = 0.78 gm = 0.0078 kg, Mass of water, mw = 1900 gm = 2.5 kg, Mass of fuse wire, mfw = 0.02 gm = 0.00002 kg, Calorific value of fuse wire, C fw = 7000 J/kg = 7 kJ/kg, Observed temperature rise, t2 − t1 = 2.98 o C ; T2 − T1 =2.6 K, Cooling correction, tc = 0.016 o C, To find, Lower calorific value, LCV, Solution, Corrected temperature rise, = Observed temperature rise + Cooling correction, T2 − T1 = 2.98 + 0.016 = 2.996 o C, Heat liberated by the coal, Qf = mf × HCV, Unit – V P9.10
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Heat taken by water and calorimeter,, Qwc = (mw + me ).C pw .(T2 − T1 ), = (1.9 + 0.35) × 4.19 × 2.996 = 28.24479 kJ, Heat released by fuse wire, Qfw = mfw × C fw, = 0.00002 × 7 = 0.00014 kJ, Heat taken by water and calorimeter, Qwc = Heat released by fuel ( Qf ), + Heat released by fuse wire ( Qfw ), ∴ Heat released by fuel, Qf = Qwc − Qfw, = 28.24479 − 0.00014 = 28.24465 kJ, Heat released by fuel, Qf = mf × HCV, ∴ Higher calorific value, HCV =, , Qf, 28.24465, =, = 36211 kJ/kg, mf, 0.00078, , Lower calorific value, LCV = HCV − (9 H2 × 2466), = 36211 − (9 × 0.042 × 2466) = 35279 kJ/kg, Result, Lower calorific value, LCV = 35279 kJ/kg, Example 9.12, The following results were obtained in an experiment to, determine the calorific value of a gas using Junket’s gas calorimeter., Gas burnt = 0.06 m 3 , Pressure of gas = 75.29 cm of Hg, Temperature, of gas = 24 o C , Temperature of water at inlet = 30 o C , Temperature of, water at outlet = 42 o C , Mass of water passing through the calorimeter, = 18 kg, Steam condensed during the test = 0.05 kg. Determine the, higher and lower calorific values of gas at 15 o C and standard, barometric pressure of 76cm of Hg., Given data, Volume of gas burnt, V = 0.06 m 3, Pressure of gas, p = 75.29 cm of Hg, Temperature of gas, t = 24 o C ; T = 297 K, Inlet water temperature, t1 = 30 o C ; T1 = 303 K, Outlet water temperature, t2 = 42 o C ; T2 = 315 K, Unit – V P9.11
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Mass of water, mw = 18 kg, Standard pressure, ps = 76 cm of Hg, Standard temperature, ts = 15 o C ; Ts = 288 K, To find, (1) Higher calorific value, HCV, LCV, , (2) Lower calorific value,, , Solution, The volume of gas under S.T.P condition,, p.V Ts 75.29 × 0.06 288, Vs =, ×, =, ×, = 0.05764 m 3, T, ps, 297, 76, Higher calorific value is given by,, mw .C pw .(T2 − T1 ) 18 × 4.19 × (315 − 303), HCV =, =, = 15701.6, Vs, 0.05764, kJ/ m 3, Lower calorific value is given by,, m, , 0.05, LCV = HCV − s × 2466 = 15701 .6 − , × 2466 , V, 0, ., 05764, , , s, , , 3, = 13562.5 kJ/ m, Result, (1) Higher calorific value, HCV = 15701.6 kJ/ m 3, (2) Lower calorific value, LCV = 13562.5 kJ/ m 3, , PROBLEMS FOR PRACTICE, 1. Determine the mass of air required for the complete, combustion of 1 kg of coal whose analysis by weight is given, below:, Carbon=83%,, Hydrogen=5%,, Oxygen=2%,, Sulphur=0.2%, Nitrogen=9.8%., [Ans: mmin =11.284 kg], 2. The gravimetric analysis of coal is as follows: C=84%,, H2 =10%, O2 =4%, N2 =2%. Determine the theoretical air, required for complete combustion of 1 kg of coal., [Ans: mmin =13.04 kg], 3. Determine the mass of air required for the complete, combustion of 1 kg of coal whose analysis by weight is given, below:, Carbon=83%,, Hydrogen=5%,, Oxygen=2%,, Sulphur=0.2%, Nitrogen=9.8%., [Ans: mmin = 11.284kg], Unit – V P9.12
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4. Calculate the mass of the air to be supplied for the combustion, of a fuel containing 75% carbon, 8% hydrogen, 3% oxygen, if, 50% excess air is supplied. [Ans: mtot =17.022kg], 5. A coal has the following analysis: Carbon=65%, Hydrogen=7%,, Oxygen=1%, rest is ahs. Determine the weight of the air, required for the complete combustion of 1kg of coal., [Ans: Wmin =97.4 N/kg of fuel], 6. A fuel has the following composition: Carbon=90%,, Hydrogen=6%, Sulphur=2%, Oxygen=1%, ash=1%. Find the, theoretical quantity of air required to burn 1 kg of the above, fuel completely. What is the gravimetric composition of the, products of combustion. [ mmin =12.565kg, CO2 =24.35%,, H 2 O =3.98%, SO2 =0.29%, N 2 =71.38%], 7. The percentage composition by mass of a coal is given below:, C=89.3%, H2 =5%, O2 =4.2%, N2 =1.5%. If 30% excess air is, supplied, find the percentage composition of dry flue gases by, volume., [Ans: CO2 =19.86%, H 2 O =2.73%, O2 =4.99%, N 2 =72.42%], 8. Convert the given gravimetric analysis of the flue gas into, volumetric analysis: CO2 =10%, CO=1.5%, O2 =8.5% and N2 =80%, [Ans: CO2 =6.68%, CO =1.57%, O2 =7.8%, N 2 =83.95%], 9., , Convert the following volumetric analysis of a gas sample to, weight analysis: CO2 =14%, CO =1.2%, O2 =3.5%, N2 =81.3., [Ans: CO2 =20.28%, CO =1.1%, O2 =3.69%, N 2 =74.93%], , 10. A certain fuel has the following composition by weight: C=88%,, H2 =3.6%, O2 =4.8% and ash=3.6%. The flue gas analysis by, volume is CO2 =10.9%, CO =1%, O2 =7.1%, N2 =81%. Estimate, (1) the mass of air supplied per kg of fuel (2) the percentage of, excess air supplied., [Ans: mtot =18.15kg/kg of fuel, mex =61.33%], 11. The analysis of a sample of oil gives the following composition, by weight. C=72%, H2 =28%. Determine the lower calorific, value of the fuel using Dulong’s formula. Assume the, elements are in free state., [Ans: LCV=58441.7 kJ/kg], Unit – V P9.13
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12. A fuel has the following composition by mass: C=80%, H2 =10%,, , O2 =3%, S=2% and the rest is incombustible. Calculate HCV and, LCV., [Ans: HCV=41085.5 kJ/kg, LCV=38866 kJ/kg], 13. In an experiment to determine the higher calorific value of a, fuel, the following observations were made: Quantity of water, = 1500c m 3 , Increase in temperature of water=14.5 o C , Water, equivalent of calorimeter=250gm, Coal burnt=2.5gm., Calculate the higher calorific value. If the fuel contains 10%, of hydrogen per kg of fuel, also find the lower calorific value., [Ans: HCV=42528.5 kJ/kg, LCV=40309 kJ/kg], 14. The following refers to a calorific value test of a fuel by means, of a gas calorimeter. Volume of gas used=0.7 m 3 , Amount of, water heated = 25 kg, Rise in temperature of cooling water at, inlet and outlet = 14 o C , Amount of steam condensed=0.028 kg., Find the higher and lower calorific values per m 3 at S.T.P., Take the heat liberated in condensing water vapour and, cooling the condensate as 2470 kJ/kg. Assume C p for water as, 4.19 kJ/kg.K., , [Ans: HCV=2095 kJ/kg, LCV=1996 kJ/kg], , Unit – V P9.14
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Unit – V, Chapter 10., , PERFORMANCE OF I.C ENGINES, , 10.1 Testing of I.C engines, An internal combustion (i.c) engine should be tested to, check the performance under various operating conditions. These, tests are generally classified as:, a) Commercial tests, b) Thermodynamic tests., The following are the important reasons for conducting, tests in i.c engines., • To determine the power developed by the engine, • To determine the various efficiencies of the engine, • To determine the fuel consumption, • To ensure the design data, • To detect the sources of heat losses, • To prepare the heat balance sheet, a) Commercial tests: This test is conducted for the, following purposes:, • To determine the overload carrying capacity of the engine, • To determine the lubricating oil consumption, • To determine the amount of cooling water required, • To check the valve timings, • To check the stability of the engine at various load, condition, b) Thermodynamic tests: This test is conducted for, comparing the actual performance of the engine with the, theoretical performance., The following observations are made during this test:, Indicated power, brake power, fuel consumption, heat carried, away by the exhaust gases, heat carried away by the cooling, water, etc. The performance of the engine is determined through, various efficiencies and heat balance sheet., Unit – V , , 10.1
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10.2 Indicated power (IP), Indicated power is the actual power developed inside an, engine cylinder. It is measured with the help of indicated mean, effective pressure., Indicator diagram, 4, , 3, , + ve, , p, - ve, , 1, , 5, 6, 2, , V, , Fig.10.1 Indicator diagram, , An indicator diagram is the graphical representation of, pressure volume variations during the working cycle. It is drawn, by an instrument called indicator. The indicator diagram has a, positive loop and a negative loop. The work done during the cycle, is given by the area of positive loop. The pumping loss due to, suction and exhaust period is given by the area of the negative, loop. Therefore the net work done during the cycle is given by,, W= Are of positive loop − Area of negative loop., W = (Area 3−4−5−6−7−3) − (Area 1−2−7−1), The area of the indicator diagram is measured by a, planimeter or by counting the number of squares in the loops., Mean height of the indicator diagram is given by,, Area of indicator diagram, A, h=, =, Lenght of indicator diagram, L, Indicated mean effective pressure ( pmi ), It is the algebraic sum of the mean pressures acting on, the piston during each stroke in one complete working cycle. It is, measured from the actual indicator diagram., The indicated mean effective pressure is given by,, Unit – V , , 10.2
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pmi = Mean height of indicator diagram × Spring number, A, pmi = h × S =, ×S, L, Number of explosions: It is the number of power strokes or, working strokes of an engine in a given time. It is expressed by, the symbol ‘n’. The number of explosions per second is given by,, N, n=, for four stroke cycle engines, 2, n = N for two stroke cycle engines, where, N= Engine speed in r.p.s., The indicated power of an engine is given by,, IP = pmi .l.a. n. k, where, pmi = Indicated mean effective pressure, l = Length of stroke of the piston, a = Area of cross−section of piston or cylinder, n = No. of explosions or working strokes per second, k = No. of cylinder., 10.3 Brake power, Brake power is the useful power available at the, crankshaft. This power is less than the actual power (indicated, power) developed inside the engine., 10.4 Brake power measurement, Torque is needed for determining the brake power., Dynamometers are used to measure torque., Prony brake, dynamometers and rope brake dynamometers are widely used for, measuring brake power. Electrical dynamometers and hydraulic, dynamometers may also used to determine the brake power., a) Prony brake dynamometer : It consists of a brake, arm, wooden blocks and weights. The principle of working is the, conversion of power into heat by dry friction. The wooden blocks, are mounted on a flexible rope or band. The weights are attached, to brake arm. Spring loaded blots are provided to loosen or, tighten the wooden blocks for varying the friction. A brake is, provided to stop the engine., Unit – V , , 10.3
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Spring, Load bar, Flywheel, Tie bolt, Wooden block, , W, , Fig.10.2 Prony brake dynamometer, , The wooden blocks are pressed against the rotating, drums. The wooden blocks absorb the engine torque and dissipate, it as frictional resistance. The entire power is absorbed and, converted into heat., W = Net load on brake (kN), R= Effective radius of brake drum (m), N = Speed of the engine (r.p.s), Torque is given by, T = W . R, Brake power is given by, BP = 2. π. N .T = π. D. N .W ( Q 2 R = D ), Let,, , b) Rope brake dynamometer: It consists of a number of, ropes wound around a rotating drum. The drum is driven by the, crankshaft of the engine. One end of the rope is connected to a, spring balance. The other end is connected to the load. The power, is absorbed due to the friction between the rope and the drum., Let, D1 = Diameter of the brake drum (m), d1 = Diameter of the rope or thickness of the band, (m), W1 = Weights applied (kN), W2 = Spring balance reading (kN), N = Speed of the engine (r.p.s), , Unit – V , , 10.4
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Flywheel, Rope, Spring, balance, Load, Chain, , Fig.10.3 Rope brake dynamometer, , Effective diameter of brake drum, D = D1 + d1, D + d1, Effective radius of brake drum, R = 1, 2, Net brake load, W = W1 − W2, Brake torque , T = W . R, Brake power, BP = 2. π. N .T = π. D. N .W ( Q 2 R = D ), c) Brake power can also be calculated, if brake mean, effective pressure is given,, BP = pmb .l. a. n. k, where, pmb = Brake mean effective pressure, 10.5 Friction power (FP), Friction power is the power lost mainly due to the friction, in the moving parts. It also takes into account the follow losses:, • Mechanical losses in bearings, • Pumping loss due to suction and exhaust, • Air resistance flywheel, Frictional power is the difference between indicated, power and brake power., FP = IP − BP, , Unit – V , , 10.5
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10.6 Efficiencies of engines, a) Indicated thermal efficiency ( η I.T ), It is defined as the ratio of the actual power (indicated, power) developed to the heat energy supplied to the engine. It is, also called as actual thermal efficiency or thermal efficiency., , Indicated thermal efficiency =, η I .T =, , Indicated power, Heat supplied, , IP, IP × 3600, × 100 or η I . T =, × 100, FC × CV, FC × CV, , , 3600 , , where, IP = Indicated power (kW), FC = Fuel consumption (kg/hr or m 3 /hr), CV = Calorific value of the fuel (kJ/kg or kJ/ m 3 ), b) Brake thermal efficiency ( η B.T ), It is defined as the ratio of heat converted into useful, work (BP) to the heat energy supplied to the engine. Brake, thermal efficiency is called as overall efficiency. It varies from, 25% to 30%., Brake power, Brake thermal efficiency =, Heat supplied, BP, BP × 3600, η B. T =, × 100 or η B. T =, × 100, FC × CV, FC × CV, , , 3600 , where, BP = Indicated power (kW), c) Mechanical efficiency ( ηm ), It is defined as the ratio of brake power to indicated power, of an engine. Its value varies from 70% to 90%. It depends upon, the loading condition and design of the engine., , Mechanical efficiency =, ηm =, , Brake power, Indicated power, BP, IP, , Unit – V , , 10.6
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Mechanical efficiency is also given by,, Brake thermal efficiency, η, ηm =, = B.T, Indicated thermal efficiency η I .T, Brake mean effective pressure, p, or, ηm =, = mb, Indicated mean effective pressure pmi, d) Relative efficiency ( ηrel ), It is defined as the ratio of indicated (actual) thermal, efficiency to the theoretical (ideal) thermal efficiency. it value, varies from 60% to 80%. It is also called as efficiency ratio., Relative efficiency or efficiency ratio,, Indicated (actual) thermal efficiency, =, Ideal thermal efficiency (Air standard efficiency), η, ηrel = B. T, ηideal, e) Volumetric efficiency ( ηvol ), It is defined as the ratio of the actual volume (reduce to, N.T.P) of the charge admitted into the engine cylinder during the, suction stroke to the stroke volume of the engine., Volume of the charge admitted during suction, ηvol =, Stroke volume of the piston, Vn, ηvol =, Vs, p.V Tn, The volume reduced to N.T.P is given by, Vn =, ×, T, pn, 10.7 Specific fuel consumption (SFC), Specific fuel consumption is defined as the amount of fuel, consumption to obtain one brake power−hour of work. It is also, called as brake specific fuel consumption (BSFC), FC, BSFC =, BP, where, FC = mass of fuel consumed (kg/hr), BP = Brake power (kW), Unit – V , , 10.7
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Brake thermal efficiency is given by,, BP × 3600, 3600, η B.T =, =, FC × CV, BSFC × CV, 3600, or, BSFC =, η B. T × CV, , Q FC = BSFC , BP, , , Specific fuel consumption may also be given on indicated, power basis. It is known as indicated specific fuel consumption, (ISFC)., FC, 3600, ISFC =, =, IP, η I . T × CV, 10.8 Morse test, Morse test is used to find out the indicated power of a, multicylinder i.c. engine. No indicator is used is used in this, method., Procedure for Morse test:, 1) The engine is coupled with a brake and the brake power is, determined (BP), 2) The first cylinder is now cut−off. This is done by shorting, out the spark plug of the first cylinder in the case of petrol, engine, and stopping the fuel supply to the first cylinder, in the case of diesel engine. The speed of the engine, reduces considerably., 3) Load is reduced until the engine attains normal speed., 4) Under this condition, the brake power is determined, ( BP1 ), 5) Then the first cylinder is operated. The second cylinder is, cut−off., 6) The engine speed is restored to the normal speed and the, brake power is determined ( BP2 ), 7) The same procedure is repeated for each cylinder in, sequence. In each case, the brake power is determined, Let, IP1 , IP2 , IP3 ,and IP4 are the indicated power of the cylinders, 1, 2, 3, and 4 respectively., , Unit – V , , 10.8
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BP = Total brake power of all 4 cylinders, BP1 = Total brake power of cylinders with cylinder 1 cut−off., BP2 = Total brake power of cylinders with cylinder 2 cut−off., BP3 = Total brake power of cylinders with cylinder 3 cut−off., BP4 = Total brake power of cylinders with cylinder 4 cut−off., FP1 , FP2 , FP3 and FP4 are the frictional power of each cylinder., , If all cylinders are working,, BP = ( IP1 − FP1 ) + ( IP2 − FP2 ) + ( IP3 − FP3 ) + ( IP4 − FP4 ), BP = ( IP1 + IP2 + IP3 + IP4 ) − ( FP1 + FP2 + FP3 + FP4 ) ..(1), When the first cylinder is cut−off, the IP of the cylinder is cut off., But BP of the cylinder exists. Therefore brake power of the other, 3 cylinders is given by,, BP1 = ( IP2 + IP3 + IP4 ) − ( FP1 + FP2 + FP3 + FP4 ) ….. (2), Subtracting equation(2) from (1), we get, BP − BP1 = IP1 (Indicated power of first cylinder), Similarly,, BP − BP2 = IP2 (Indicated power of second cylinder), BP − BP3 = IP3 (Indicated power of third cylinder), BP − BP4 = IP4 (Indicated power of fourth cylinder), The indicated power of the engine is given by,, IP = IP1 + IP2 + IP3 + IP4, 10.9 Heat balance sheet, Heat energy is supplied in an internal combustion engine, by burning the fuel. A part of the heat supplied is converted into, useful mechanical work. It is available in the engine crankshaft., The remaining heat energy is lost. The main sources of heat, losses are as follows:, 1) Heat carried away by the cooling water ( Qw ), 2) Heat carried away by exhaust gases ( Qg ), 3) Heat lost due to radiation ( Qrad ), 4) Heat lost due to incomplete combustion ( Qic ), Unit – V , , 10.9
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Accounted heat loses : The heat losses which can be, determined from the observed values during the test are known, as accounted heat losses., Unaccounted heat losses : The heat losses which cannot be, determined accurately are known as unaccounted heat losses. The, unaccounted heat losses can be determined by subtracting the, accounted heat losses and heat converted into useful work from, the heat supplied to the engine., Procedure for preparing heat balance sheet., The procedure for preparing the heat balance sheet includes, the following:, 1) The engine is made to run on constant load at constant speed., 2) The brake power of the engine is estimated., 3) The amount of fuel consumption (FC)is determined., 4) The inlet and outlet temperatures ( T1 and T2 ) of the cooling, water are noted., 5) The amount of cooling water ( mw ) circulate is noted., 6) The temperature of exhaust gas is ( Tg ) is noted, 7) The amount of exhaust gases ( m g ) is noted., 8) The room temperature ( Ta ) is noted., The heat balance sheet can be prepared on hour basis, or, minute basis or second (kW) basis., 1) Heat supplied by the fuel ( Qs ), Qs = FC × CV, where, FC = Fuel consumption (kg/hr or m 3 /hr), CV = Calorific value of fuel (kJ/kg.K), 2) Heat equivalent of useful work (Brake power) ( QBP ):, QBP = BP × 3600, Percentage of heat used for brake power,, Q, QBP % = BP × 100, Qs, 3) Heat lost due to cooling water:, Qw = mw .C pw .(T2 − T1 ), Unit – V 10.10
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where, mw = Mass of cooling water, C pw = Specific heat capacity of cooling water (4.19 kJ/kg.K), , T1 = Inlet temperature of cooling water, T2 = Outlet temperature of cooling water, Percentage of heat lose due to cooling water:, Q, Qw % = w × 100, Qs, 4) Heat lost due to exhaust gases ( Q g ), , Q g = m g .C pg .(Tg − Ta ), where, m g = Mass of exhaust gases, C pg = Specific heat capacity of exhaust gases (1.005 kJ/kg.K), Tg = Temperature of exhaust gases, Ta = Ambient air temperature or room temperature, , The mass of the exhaust gases is the sum of the fuel, consumption and air consumption., m g = FC + ma, Percentage of heat lose due to cooling water:, Qg, Qg % =, × 100, Qs, 5) Unaccounted heat loss ( Qua ), The unaccounted heat losses may be calculated as follows:, Qua = Qs − [QBP + Qw + Qg + .......], Percentage of unaccounted heat losses, Q, Qua % = ua × 100, Qs, , Unit – V 10.11
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REVIEW QUESTIONS, 1., , What is the purpose of testing an i.c. engine? State the various, quantities that can be determined by thermodynamic tests., , 2., , What is mean effective pressure? How it is measured?, , 3., , Define: (a) Indicated power; (b) Brake power; (c) Specific fuel, consumption., , 4., , Distinguish between brake power, indicated power and friction, power., , 5., , List out the various methods of determining brake power of, an i.c.engine and explain any one method., , 6., , Define: Indicated thermal efficiency and brake thermal efficiency., , 7., , Define: Mechanical efficiency and relative efficiency., , 8., , What is the purpose of conducting a Morse test in an i.c. engine., , 10. Explain the Morse test for finding out the indicated power of a, multicylinder engine., 10. Why is the heat balance sheet prepared for an i.c. engine?, Discuss in detail the quantities to be determined in order to, complete the heat balance sheet?, 11. State the procedure to be adopted for preparing the heat, balance sheet during an i.c. engine test., , Unit – V 10.12
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SOLVED PROBLEMS, Example 9.1, A gas engine has piston diameter of 150mm, length of, stroke 400mm and mean effective pressure 5.5 bar. The engine, makes 120 explosions per minute. Determine the mechanical, efficiency of the engine, if the brake power is 5kW., Given data, Piston diameter, d = 150mm = 0.15m, Length of stroke, l = 3 % = 0.03 kg/kg of coal, Mean effective pressure, pmi = 5.5 bar = 550 kN/ m 2, 120, = 2 /s, No. of explosions, n = 120/min =, 60, Brake power, BP = 5 kW, To find, Mechanical efficiency, ηm, Solution, Area of cross section, a =, , π.d 2 π(0.5)2, =, =0.01767 m 2, 4, 4, , Let, k = 1, Indicated power, IP = pmi .l.a.n.k, = 550 × 0.4 × 0.01767 × 2 × 1 = 7.7754 kW, Mechanical efficiency,, BP, 5, ηm =, =, = 0.6431 = 64.31%, IP 7.7754, Result, Mechanical efficiency, ηm = 64.31%, Example 9.2, During a test on a four−stroke cycle oil engine, the, following data were obtained: Swept volume of the cylinder = 14, litres; Speed of the engine = 6.6 rev/s; Effective brake load = 77 kg;, Effective brake radius = 0.7 m; Indicated mean effective pressure =, 567 kN/ m 2 ; Determine the indicated and brake power and the, mechanical efficiency., Unit – V P10.1
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Given data, Swept volume, Vs = l.a = 14 litres = 0.014 m 3, Speed, N = 6.6 rev/s, Effective brake load, W = 77 kg, Effective brake radius, R = 0.7 m, Indicated mean effective pressure, pmi != 567 kN/ m 2, To find, (1) Indicated power, IP, (2) Brake power, BP, (3) Mechanical efficiency, ηm, Solution, 77 × 9.81, Effective brake load, W =, =0.75537 kN, 1000, N, ( Q Four stroke), No. of explosions, n =, 2, 6.6, n=, = 3.3 /s; Let, k = 1, 2, Indicated power, IP = pmi .l.a.n.k = pmi .Vs .n.k, = 567 × 0.014 × 3.3 × 1 = 26.1954 kW, Brake torque, T = W × R = 0.75537 × 0.7 = 0.528759 kN−m, Brake power, BP = 2.π. N .T = 2 × π × 6.6 × 0.528759 = 21.9271 kW, Mechanical efficiency,, BP 21.9271, ηm =, =, = 0.8371 = 83.71%, IP 26.1954, Result, (1) Indicated power, IP = 26.1954 kW, (2) Brake power, BP = 21.9271 kW, (3) Mechanical efficiency, ηm = 83.71 %, Example 9.3, In a test on diesel engine, the SFC was 260kg/kW−hr. The, calorific value of the fuel is 42000 kJ/kg. Determine the brake, thermal efficiency., Given data, SFC = 260 g/kW−hr = 0.26 kg/kW−hr, Calorific value, CV = 42000 kJ/kg, Unit – V P10.2
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To find, Brake thermal efficiency, η BT, Solution, Let, Brake power, BP = 1 kW., BP × 3600, 1 × 3600, Brake thermal efficiency, η BT =, =, FC × CV, 0.26 × 42000, = 0.3297 = 32.97 %, Result, Brake thermal efficiency, η BT = 32.97 %, Example 9.4, A six cylinder S.I engine works on 4 stroke cycle. The bore, of each cylinder is 70mm and the stroke 100 mm. The clearance, volume per cylinder is 67c.c. At a speed of 3300 rpm, the fuel, consumption is 18.5 kg/hr and the torque developed is 135 N−m., Calculate the brake thermal efficiency, if the calorific value of the, fuel is 45000 kJ/kg. Take γ=1.4 for air., Given data, No. of cylinders, k, Bore diameter, d, Stroke, l, Clearance volume, Vc, , =, =, =, =, , Speed, N =, Fuel consumption, FC, Torque, T, Calorific value of fuel, CV, γ, , =, =, =, =, , 6; Four stroke cycle, 70 mm = 0.07 m, 100 mm = 0.1 m, 67 c.c = 0.000067 m 3, 3300, 3300 rpm =, =55 rps, 60, 18.5 kg/hr, 135 N−m = 0.135 kN−m, 45000 kJ/kg, 1.4, , To find, Brake thermal efficiency, η BT, Solution, Brake power, BP = 2.π. N .T = 2 × π × 55 × 0.135 = 46.535 kW, BP × 3600, 1 × 3600, Brake thermal efficiency, η BT =, =, FC × CV, 0.26 × 42000, Unit – V P10.3
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=, , 46.653 × 3600, = 0.2017 = 20.17 %, 18.5 × 45000, , Result, Brake thermal efficiency, η BT = 20.17 %, Example 9.5, The following results were obtained during a test on a four, cylinder four stroke oil engine: Bore = 100mm; Stroke = 115mm;, Speed = 1650 rpm; Fuel used = 0.2 kg/min.; Calorific value of fuel, = 41900 kJ/kg; Net load on the brake drum = 390 N;, Circumference of the brake drum = 3.3m; Mechanical efficiency =, 80%. Determine (1) Brake thermal efficiency, (2) Indicated thermal, efficiency and (3) indicated mean effective pressure., Given data, No. of cylinders, k = 4; Four stroke cycle, Bore diameter, d = 100 mm = 0.1 m, Stroke, l = 115 mm = 0.115 m, 1650, =27.5 rps, Speed, N = 1650 rpm =, 60, 0.2 kg/min = 0.2×60=12, Fuel consumption, FC =, kg/hr, Calorific value of fuel, CV = 41900 kJ/kg, Net load on brake drum, W = 390 N = 0.39 kN, Circumference of brake drum,, = 3.3 m, π. D, Mechanical efficiency, ηm = 80 % = 0.8, To find, (1) Brake thermal efficiency, η BT, (2) Indicated thermal efficiency, η IT, (3) Indicated mean effective pressure, pmi, Solution, Brake power, BP = 2.π. N .T = π. D. N .W, = 3.3 × 27.5 × 0.39 = 35.3925 kW, Unit – V P10.4
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Brake thermal efficiency, η BT =, , BP × 3600, FC × CV, , 35.3925 × 3600, = 0.2534 = 25.34 %, 12 × 41900, Indicated thermal efficiency,, η, 0.2534, η IT = BT =, = 0.3168 = 31.68 %, ηm, 0.8, BP 35.3925, Indicated power, IP =, =, = 44.24 kW, ηm, 0.8, N 27.5, n=, =, ; IP = pmi .l.a.n.k, 2, 2, IP, 44.24, pmi =, =, =890.56 kN/ m 2, l.a.n.k, π.(0.1)2 27.5 , , 0.115 ×, ×, ×4, 4 2 , Result, (1) Brake thermal efficiency, η BT = 25.34 %, =, , (2) Indicated thermal efficiency, η IT = 31.68 %, (3) Indicated mean effective pressure, pmi =890.56 kN/ m 2, Example 9.6, An engine working on constant volume cycle is having, 150mm bore and 165mm stroke and clearance volume is 0.51 litre., The fuel used has a calorific value of 45350 kJ/kg and the, consumption is 0.334 kg/kW−hr. The mechanical efficiency of the, engine is 81%. Calculate (1) the indicated thermal efficiency and, (2) relative efficiency. Takeγ=1.4, Given data, Bore diameter, d, Stroke, l, Clearance volume, Vc, Fuel consumption, FC, Calorific value of fuel, CV, Mechanical efficiency, ηm, , =, =, =, =, =, =, , 150 mm = 0.15 m, 165 mm = 0.165 m, 0.51 litre = 0.00051 m 3, 0.334 kg/kW−hr, 45350 kJ/kg, 81 % = 0.81, , To find, (1) Indicated thermal efficiency, η IT, (2) Relative efficiency, ηrel, Unit – V P10.5
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Solution, Let, BP = 1 kW, Mechanical efficiency, ηm =, ∴ Indicated power, IP =, , BP, IP, , BP, 1, =, =1.2346 kW, ηm 0.81, , (1) Indicated thermal efficiency,, IP × 3600 1.2346 × 3600, η IT =, =, = 0.2934 = 29.34 %, 0.334 × 45350, FC × CV, (2) Relative efficiency, Relative efficiency, ηrel =, , η IT, ηideal, , Ideal thermal efficiency, ηideal = 1 −, , 1, r γ −1, , Vs + Vc, Vc, π, π, Stroke volume, Vs = × d 2 × l = × 0.152 × 0.165 = 2.9158 × 10 − 3 m 3, 4, 4, 2.9158 × 10 −3 + 0.00051, r=, = 6.72, 0.00051, 1, Ideal thermal efficiency, ηideal = 1 −, = 0.5332 = 53.32 %, 6.721.4 −1, 0.2934, Relative efficiency, ηrel =, = 0.5502 = 55.02 %, 0.5332, Result, (1) Indicated thermal efficiency, η IT = 29.34 %, Compression ratio, r =, , (2) Relative efficiency, ηrel = 55.02 %, Example 9.7, A six cylinder, four stroke engine had a bore to stroke ratio, of 360:500. During the trial, following results were obtained: Mean, area of indicator diagram = 0.00073 m 2 ; Length of indicator, diagram = 0.075 m; Spring number = 70,000 kN/ m 2 per m of, compression; Brake torque = 14 kN−m; Speed = 500 rpm; Fuel, consumption = 240 kg/hr. Calculate (1) Total indicated power, developed,, (2) Brake power, (3) Mechanical efficiency and, (4) Specific fuel consumption., Unit – V P10.6
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Given data, No. of cylinders, k, Bore diameter, d, Stroke, l, Area of indicator diagram, A, Length of indicator diagram, L, Spring number, S, Brake torque, T, , =, =, =, =, =, =, =, , 6; Four stroke, 360 mm = 0.36 m, 500 mm = 0.5 m, 0.00075 m 2, 0.075 m, 70000 kN/ m 2 /m, 14 kN−m, 500, rps, Speed. N = 500 rpm =, 60, Fuel consumption, FC = 240 kg/hr, , To find, (1) Indicated power, IP, (2) Brake power, BP, (3) Mechanical efficiency, ηm, (4) Specific fuel consumption, SFC, Solution, Indicated mean effective pressure, pmi =, , A, ×S, L, , 0.00075, × 70000 = 700 kN/ m 2, 0.075, π, π, Stroke volume, Vs = × d 2 × l = × 0.36 2 × 0.5 = 0.0509 m 3, 4, 4, N, 500, No. of explosions, n =, =, 2 60 × 2, Indicated power,, 500, IP = pmi .l.a.n.k = 700 × 0.0509 ×, × 6 = 890.75 kW, 60 × 2, 500, Brake power, BP = 2.π. N .T = 2 × π ×, × 14 = 733.04 kW, 2 × 60, BP, 733.04, Mechanical efficiency, ηm =, =, = 0.823 = 82.3 %, 890.75, IP, Specific fuel consumption,, FC, 240, BSFC =, =, =0.3274 kg/kW−hr., BP 733.04, Result, (1) Indicated power, IP = 890.75 kW, (2) Brake power, BP = 733.04 kW, Unit – V P10.7, =
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(3) Mechanical efficiency, ηm = 82.3%, (4) Specific fuel consumption, BSFC = 0.3274 kg/kW−hr., Example 9.8, The following results were obtained during a test on a, single cylinder four stroke cycle petrol engine: Brake power = 73.5, kW when running at 400 rpm; Brake mean effective pressure = 8.5, bar; Mechanical efficiency = 80%; Specific fuel consumption =, 0.346 kg/kW−hr; Calorific value of fuel = 44100 kJ/kg;, Compression ratio = 6:1. Determine (1) Bore and stroke (assuming, them to be equal), (2) Brake thermal efficiency, (3) Indicated, thermal efficiency, (4) Air standard efficiency and (5) Relative, efficiency., Given data, No. of cylinders, k = 1; Four stroke, Brake power, BP = 73.5 kW, 400, =6.667 rps, Speed. N = 400 rpm =, 60, Brake mean effective pressure, pmb = 8.5 bar = 850 kN/ m 2, Mechanical efficiency, ηm = 80 % = 0.8, Specific fuel consumption, SFC = 0.346 kg/kW−hr, Calorific value, CV = 44100 kJ/kg, Compression ratio, r = 6, To find, (1) Bore diameter (d); Stroke length (l), (2) Brake thermal efficiency, η BT, (3) Indicated thermal efficiency, η IT, (4) Air standard efficiency, ηideal, (5) Relative efficiency, ηrel, Solution, (1) Bore diameter and stroke length, N 6.667, n=, =, = 3.3335, 2, 2, Brake power, BP = pmb .l.a.n.k, Unit – V P10.8
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π, × d 2 × 3.3335 × 1, 4, π, 73.5 = 850 × l × × l 2 × 3.3335 × 1 (Q d = l), 4, 73.5 = 2225 .128 × l3, 1, 73.5, l3 =, =0.033032; l = (0.033032) 3 = 0.321 m, 2225 .128, Bore diameter, d = l = 0.321 m, , 73.5 = 850 × l ×, , (2) Brake thermal efficiency, η BT, FC, SFC =, ; FC = SFC × BP = 0.346 × 73.5 = 24.431 kg/hr, BP, BP × 3600, 73.5 × 3600, η BT =, =, = 0.2359 = 23.59%, FC × CV, 25.431 × 44100, (3) Indicated thermal efficiency, η IT, , η IT =, , η BT 0.2359, =, = 0.2949 = 29.49 %, ηm, 0.8, , (4) Air standard efficiency, ηideal, 1, 1, ηideal = 1 − γ −1 = 1 − 1.4 −1 = 0.5116 = 51.16%, r, 6, (5) Relative efficiency, ηrel, , ηrel =, , η IT, 0.2949, =, = 0.5764 = 57.64 %, ηideal 0.5116, , Result, (1) Bore diameter d = 321 mm; Stroke length l = 321 mm, (2) Brake thermal efficiency, η BT = 23.59 %, (3) Indicated thermal efficiency, η IT = 29.49 %, (4) Air standard efficiency, ηideal = 51.16 %, (5) Relative efficiency, ηrel = 57.64 %, Example 9.9, A four stroke cycle, four cylinder petrol engine has 625 mm, cylinder diameter and 950mm stroke. On test, it develops a torque, of 64 Nm, when running at 3000 rpm. The clearance volume of, each cylinder is 63.5 c.c. The brake thermal efficiency ratio based, on air standard cycle is 0.5 and calorific value of petrol is 44800, kJ/kg. Determine the fuel consumption in litres per hour and, brake mean effective pressure. Take γ=1.4 for air., Unit – V P10.9
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Given data, No. of cylinders, k, Diameter of cylinder, d, Stroke, l, Torque, T, , =, =, =, =, , 4; Four stroke, 625 mm = 0.0625 m, 950 mm = 0.095 m, 64 N−m, 3000, =50 rps, Speed. N = 3000 rpm =, 60, Clearance volume, Vc = 63.5 c.c= 0.0000635 m 3, Efficiency ratio, ηrel = 0.5, Calorific value, CV = 44800 kJ/kg, γ = 1.4, , To find, (1) Fuel consumption, FC, (2) Brake mean effective pressure, pmb, Solution, , π, π, × d 2 = 0.095 × (0.0625)2, 4, 4, = 0.000291456 m 3, V + Vs 0.0000635 + 0.000291456, Compression ratio, r = c, =, =5.59, Vc, 0.0000635, Stroke volume, Vs = l.a = l. ×, , Air standard efficiency,, 1, 1, ηideal = 1 − γ −1 = 1 −, = 0.4976 = 49.76%, r, 5.591.4 −1, Relative efficiency based on brake thermal efficiency is given by,, Brake thermal efficiency η B.T, ηrel =, =, Air standard efficiency, ηideal, Brake thermal efficiency, η B.T = ηrel × ηideal, = 0.5 × 0.4976 = 0.2488 = 24.88 %, Brake power, BP = 2.π. N .T = 2 × π × 50 × 0.064 = 20.106 kW, BP × 3600, BP × 3600, η B.T =, ; ∴ FC =, FC × CV, η B.T × CV, 20.106 × 3600, =, = 0.494 kg/hr, 0.2488 × 44800, N 50, No. of working strokes per second, n =, =, =25, 2, 2, Brake power, BP = pmb .l.a.n.k = pmb .Vs .n.k, Unit – V P10.10
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∴ Brake mean effective pressure,, BP, 20.106, pmb =, =, = 689.85 kN/ m 2, Vs .n.k 0.000291456 × 25 × 4, Result, (1) Fuel consumption, FC = 0.494 kg/hr, (2) Brake mean effective pressure, pmb = 689.85 kN/ m 2, Example 9.10, A six cylinder four stroke petrol engine has a compression, ratio of 4.5:1. The clearance volume of each cylinder is 100 c.c. The, engine consumes 10 kg of fuel per hour whose calorific value is, 42000 kJ/kg. The engine runs at 2400 rpm. Assume the relative, efficiency of 60%. Estimate the average indicated mean effective, pressure developed., Given data, No. of cylinders, k, Compression ratio, r, Clearance volume, Vc, Fuel consumption, FC, Calorific value, CV, , =, =, =, =, =, , 6; Four stroke, 4.5, 100 c.c = 0.0001 m 3, 10 kg/hr, 42000 kJ/kg, 2400, =40 rps, Engine speed. N = 2400 rpm =, 60, Relative efficiency, ηrel = 60 % = 0.6, To find, Indicated mean effective pressure, pmi, Solution, Air standard efficiency,, 1, 1, ηideal = 1 − γ −1 = 1 −, = 0.4521 = 45.21 %, 4.51.4 −1, r, Indicated thermal efficiency η I .T, Relative efficiency, ηrel =, =, Air standard efficiency, ηideal, η IT = ηideal × ηrel = 0.4521 × 0.6 = 0.2712 = 27.12 %, IP × 3600, Indicated thermal efficiency, η IT =, FC × CV, Unit – V P10.11
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η IT × FC × CV, 3600, 0.2712 × 10 × 42000, = 31.64 kW, =, 3600, N 40, No. of explosions, n =, =, = 20, 2, 2, V + Vc Vs, Compression ratio, r = s, =, + 1 (or), Vc, Vc, Vs = (r − 1) × Vc, ∴ Indicated power, IP =, , Stroke volume, Vs = (4.5 − 1) × 0.0001 = 0.00035 m 3, IP = pmi .l.a.n.k = pmi .Vs .n.k, IP, 31.64, ∴ pmi =, =, = 753.33 kN/ m 2, Vs .n.k 0.00035 × 20 × 6, Result, Indicated mean effective pressure, pmi = 753.33 kN/ m 2, Example 9.11, A four stroke cycle gas engine has a bore of 160mm and a, stroke of 240mm. The compression ratio is 4 to 1 and the effective, pressure is 350 kN/ m 2 . If the engine speed is 300 rpm and the, thermal efficiency is 30%, calculate the fuel consumption in m3, per IP hr. and the efficiency relative to the air standard cycle., Calorific value of gas is 17375 kJ/ m3 ., Given data, =, =, =, =, =, Speed, N =, Indicated thermal efficiency, η IT =, Calorific value, CV =, No. of cylinders, k, Diameter of bore, d, Stroke, l, Compression ratio, r, Mean effective pressure, pmi, , 1; Four stroke, 160 mm = 0.16 m, 240 mm = 0.24 m, 4, 350 kN/ m 2, 300 rpm = 5 rps, 30 % = 0.3, 17375 kJ/ m 3, , To find, (1) Specific fuel consumption on IP basis, ISFC, (2) Relative efficiency, ηrel, Unit – V P10.12
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Solution, , π 2 π, d = × (0.16)2 m 2, 4, 4, N 5, No. of explosions, n =, = = 2.5, 2 2, Indicated power, IP = pmi .l.a.n.k, π, = 350 × 0.24 × × (0.16)2 × 2.5 × 1 = 4.222 kW, 4, Specific fuel consumption on IP basis,, 3600, 3600, ISFC =, =, = 0.69065 kg/IP−hr., η IT × CV 0.3 × 17375, Area of cross section, a =, , Air standard efficiency,, 1, 1, ηideal = 1 − γ −1 = 1 − 1.4 −1 = 0.4257 = 42.57 %, r, 4., Indicated thermal efficiency η I .T, =, Relative efficiency, ηrel =, Air standard efficiency, ηideal, 0.3, ηrel =, = 0.7047 = 70.47 %, 0.4257, Result, (1) Specific fuel consumption on IP basis, ISFC = 0.69065 kg/IP−hr, (2) Relative efficiency, ηrel = 70.47 %, Example 9.12, The following results were obtained during a Morse test on, a four stroke petrol engine., BP with all cylinder working = 11.92 kW, BP with cylinder 1 cut off = 8.64 kW, BP with cylinder 2 cut off = 8.50 kW, BP with cylinder 3 cut off = 8.54 kW, BP with cylinder 4 cut off = 8.50 kW, Calculate the mechanical efficiency of the engine. If the, fuel consumption is 3.5 kg/hr, find the indicated thermal, efficiency. Calorific value of fuel is 42000 kJ/kg., Given data, Brake power with all cylinders working, BP, Brake power with cylinder 1 cut off, BP1, Brake power with cylinder 2 cut off, BP2, Brake power with cylinder 3 cut off, BP3, Unit – V P10.13, , =, =, =, =, , 11.92 kW, 8.46 kW, 8.50 kW, 8.54 kW
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Brake power with cylinder 4 cut off, BP4 = 8.50 kW, Fuel consumption, FC = 3.5 kg/hr, Calorific value, CV = 4200 kJ/kg, To find, (1) Mechanical efficiency, ηm, (2) Indicated thermal efficiency, η IT, Solution, Indicated power of first cylinder, IP1 = BP − BP1, = 11.92 − 8.46 = 3.46 kW, Indicated power of second cylinder, IP2 = BP − BP2, = 11.92 − 8.60 = 3.32 kW, Indicated power of third cylinder, IP3 = BP − BP3, = 11.92 − 8.54 = 3.38 kW, Indicated power of fourth cylinder, IP4 = BP − BP4, = 11.92 − 8.50 = 3.42 kW, Indicated power of engine, IP = IP1 + IP2 + IP3 + IP4, = 3.46 + 3.32 + 3.38 + 3.42 = 13.58 kW, BP 11.92, Mechanical efficiency, ηm =, =, = 0.8778 = 87.78 %, IP 13.58, Indicated thermal efficiency,, IP × 3600 13.58 × 3600, η IT =, =, = 0.3326 = 33.26 %, FC × CV, 3.5 × 42000, Result, (1) Mechanical efficiency, ηm = 87.78 %, (2) Indicated thermal efficiency, η IT = 33.26 %, Example 9.13, A four cylinder four stroke petrol engine has a bore of, 65mm and a stroke of 90mm. It is at full throttle and constant, speed. The fuel supply is fixed at 0.064 kg/min and the plugs of, the cylinders are short circuited without change in speed and the, brake torque is adjusted correspondingly. The brake power, measurements are as follows:, BP with all cylinder working = 12.25 kW, BP with cylinder 1 cut off = 8.84 kW, BP with cylinder 2 cut off = 8.73 kW, Unit – V P10.14
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BP with cylinder 3 cut off = 8.69 kW, BP with cylinder 4 cut off = 8.80 kW, Calculate the indicated power of the engine under these, conditions. Also calculate the indicated thermal efficiency of the, engine if the calorific value of the fuel is 44000 kJ/kg. Compare, this efficiency with air standard efficiency if the clearance volume, of one cylinder is 75 c m3 ., Given data, Diameter of bore, d = 65 mm = 0.065 m, Stroke length, l = 90 mm = 0.09 m, 0.064 kg/min =3.84, Fuel consumption, FC =, kg/hr., Brake power with all cylinders working,, = 12.25 kW, BP, Brake power with cylinder 1 cut off, BP1 = 8.86 kW, Brake power with cylinder 2 cut off, BP2 = 8.73 kW, Brake power with cylinder 3 cut off, BP3 = 8.69 kW, Brake power with cylinder 4 cut off,, = 8.80 kW, BP4, Clearance volume, Vc = 75c m 3 = 75 ×10 −6 m 3, To find, (1) Indicated power, IP, (2) Indicated thermal efficiency, η IT, (3) Air standard efficiency, ηideal, (4) Relative efficiency, ηrel, Solution, Indicated power of first cylinder, IP1 = BP − BP1, = 12.25 − 8.84 = 3.41 kW, Indicated power of second cylinder, IP2 = BP − BP2, = 12.25 − 8.73 = 3.52 kW, Indicated power of third cylinder, IP3 = BP − BP3, = 12.25 − 8.69 = 3.56 kW, Indicated power of fourth cylinder, IP4 = BP − BP4, = 12.25 − 8.80 = 3.45 kW, Unit – V P10.15
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Indicated power of engine, IP = IP1 + IP2 + IP3 + IP4, = 3.41 + 3.52 + 3.56 + 3.45 = 13.94 kW, IP × 3600, Indicated thermal efficiency, η IT =, FC × CV, 13.94 × 3600, = 0.2970 = 29.70 %, =, 3.84 × 44000, Stroke volume,, π, π, Vs = × d 2 × l = × (0.065)2 × 0.09 = 2.9865 × 10 −4 m 3, 4, 4, Vc + Vs 75 × 10 −6 + 2.9865 × 10 −4, Compression ratio, r =, =, =4.982, Vc, 75 × 10 − 6, Air standard efficiency,, 1, 1, ηideal = 1 − γ −1 = 1 −, = 0.4739 = 47.39 %, (4.982)1.4 −1, r, Indicated thermal efficiency η I .T, =, Relative efficiency, ηrel =, Air standard efficiency, ηideal, 0.2970, ηrel =, = 0.6267 = 62.67 %, 0.4739, Result, (1) Indicated power, IP = 13.94 kW, (2) Indicated thermal efficiency, η IT = 29.70 %, (3) Air standard efficiency, ηideal = 47.39 %, (4) Relative efficiency, ηrel = 62.67 %, Example 9.14, During a test on a four stroke cycle diesel engine the, following data and results were obtained: Mean height of the, indicator diagram = 12mm; Spring index = 27 kN/ m 2 /mm;, Swept volume of the cylinder = 14 litres; Speed of the engine = 396, rpm; Net load on the brake = 0.7554 kN; Radius of the brake drum, = 0.7 m; Fuel consumption = 7.2 kg/hr; Calorific value of fuel =, 44000 kJ/kg; Cooling water circulation = 540 kg/hr; Rise in, temperature of cooling water = 33 o C ; Specific heat of water = 4.18, kJ/kg.K; Energy to exhaust gas = 33.6 kJ/s. Determine (1), Mechanical efficiency, (2) the heat balance sheet expressed as kJ/s, as percentage of heat supplied to engine., , Unit – V P10.16
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Given data, Mean height of indicator diagram, h = 21 mm, Spring index, S = 27 kN/ m 2 /mm, Swept volume, Vs = l.a = 14 litres = 0.014 m 3, 396, =6.6rps, Speed, N = 396 rpm=, 60, Load on brake, W = 0.7554 kN, Radius of brake drum, R = 0.7 m, Fuel consumption, FC = 7.2 kg/hr., Calorific value, CV = 44000 kJ/kg, 540, kg/s, Cooling water circulation, mw = 540 kg/hr=, 3600, Rise in temperature of, 33 o C ; T1 − T2 = 33K, cooling water, ( t2 − t1 ) =, Specific heat of water, C pw = 4.18 kJ/kg.K, Energy to exhaust, Qg = 33.6 kJ/s, To find, (1) Mechanical efficiency, ηm, (2) Heat balance sheet expressed as kJ/s and as percentage of Qs, Solution, Mean effective pressure,, pmi = h × S = 21 × 27 = 567 kN/ m 2, No. of working strokes, n =, Indicated power, IP = pmi .l.a.n.k, , N 6.6, =, = 3. 3, 2, 2, , = 567 × 0.014 × 3.3 × 1 = 26.1954 kW, Brake torque, T = W × R = 0.7554 × 0.7 = 0.52878 kN−m, Brake power, BP = 2.π. N .T, = 2 × π × 6.6 × 0.52878 = 21.928 kW, BP, 21.928, Mechanical efficiency, ηm =, =, = 0.8371 = 83.71%, IP 26.1954, Heat balance sheet [kJ/s (kW) basis or second basis], (1) Heat supplied in fuel, FC, 7.2, Qs =, × CV =, × 44000 = 88 kJ/s, 3600, 3600, (2) Heat equivalent of brake power, QBP = BP = 29.928 kJ/s, Unit – V P10.17
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Percentage of heat converted into brake power,, Q, 12.928, QBP % = BP × 100 =, × 100 = 24.92 %, Qs, 88, (3) Heat carried away by cooling water, Qw, 540, Qw = m.C pw (T2 − T1 ) =, × 4.18 × 33 = 20.691 kW, 3600, Percentage of heat carried away by cooling water,, Q, 20.691, Qw % = w × 100 =, × 100 =23.51 %, Qs, 88, (4) Heat lost due to exhaust gases, Qg = 33.6 kJ/s, Percentage of heat lost duet to exhaust gases,, Qg, 33.6, Qg % =, × 100 =, × 100 = 38.18 %, Qs, 88, (5) Heat loss unaccounted, Qua, Qua = Qs − [QBP + Qw + Q g ], = 88 − [21.928 + 20.691 + 33.6] = 11.781 kJ/s, Percentage of heat loss unaccounted,, Q, 11.781, Qua % = ua × 100 =, × 100 =13.39 %, Qs, 88, Heat balance sheet [kJ/s (kW) basis or second basis], Sl., No, 1., , Heat, supplied, Heat, supplied in, fuel, , kJ/s, , %, , Heat spent, , kJ/s, , %, , 88, , 100, , −, , −, , −, , 21.928, , 24.92, , 20.691, , 23.51, , 33.600, , 38.18, , 11.781, , 13.39, , 88.000, , 100.00, , 2., , −, , −, , −, , 3., , −, , −, , −, , 4., , −, , −, , −, , 5., , Heat equivalent, of brake power, ( QBP ), Heat carried by, cooling water, ( Qw ), Heat lost due to, exhaust gases, ( Qg ), , −, , −, , −, , Heat loss, unaccounted, ( Qua ), , Total, , 88, , 100, , Total, , Unit – V P10.18
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Example 9.15, A single cylinder oil engine working on four stroke cycle, has a bore of 110mm and stroke of 130 mm runs at 600 rpm. The, mean effective pressure is 6 bar. It consumes 10 c.c of oil in 28, seconds. The diesel oil used is having calorific value of 42000, kJ/kg and the specific gravity is 0.85. The engine cooling water, enters at a temperature of 18 o C and leaves at 60 o C . The quantity, of cooling water circulated is 1.5 litres per minute. The brake, wheel diameter is 850 mm and rope diameter is 20mm. The net, load on the brake is 0.11 kN. The exhaust gas temperature is, 420 o C and its specific heat is 1.0 kJ/kg.K. The air fuel ratio is, 22:1 by weight. Room temperature is 30 o C . Determine (a), Indicated power, (b) Brake power, (c) Mechanical efficiency and (d), Indicated thermal efficiency. Also draw up the heat balance sheet, on hour basis., Given data, =, =, =, =, =, Fuel consumption, FC =, Calorific value, CV =, Specific gravity, s =, Inlet temperature of cooling water, t1 =, No. of cylinders, k, Diameter of bore, d, Stroke length, l, Speed, N, Mean effective pressure, pmi, , Outlet temperature of cooling water, t2 =, Mass of cooling water, mw =, Diameter of brake, D1 =, Rope diameter, d1 =, Net load, W =, Exhaust gas temperature, Tg =, , 1, 110 mm = 0.11 m, 130 mm = 0.13 m, 600 rpm = 10 rps, 6 bar = 600 kN/ m 2, 10 c.c in 28 seconds, 42000 kJ/kg, 0.85, 18 o C, 60 o C, 1.5 lit/min, 850 mm = 0.85 m, 20 mm =0.02 m, 0.11 kN, 420 o C, , Specific heat of exhaust gas, C pg = 1.0 kJ/kg.K, Air fuel ratio = 22:1, Room temperature, Ta = 30 o C, , Unit – V P10.19
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To find, (a) Indicated power, IP, (b) Brake power, BP, (c) Mechanical efficiency, ηm, (d) Indicated thermal efficiency, ηm, (e) Heat balance sheet on hour basis, Solution, (a) Indicated power, IP, Area of cross section, a =, , π 2 π, d = × (0.11)2 =, 4, 4, , 9.503 × 10 −3 m 2, , N 10, =, =5, 2, 2, Indicated power, IP = 600 × 0.13 × 9.503 × 10 −3 × 5 × 1, = 3.706 kW = 3.706 × 3600 = 13341.6 kJ/hr., No. of explosions, n =, , (b) Brake power, BP, , D + d 0.85 + 0.02, =, = 0.435 m, 2, 2, Brake torque, T = W. R, Brake power, BP = 2.π. N .T = 2.π. N .W . R, = 2 × π × 10 × 0.11 × 0.435 = 3.007 kW = 10825.2 kJ/hr, Effective radius, R =, , (c) Mechanical efficiency, ηm, BP 3.007, ηm =, =, = 0.814 = 81.4%, IP 3.706, (d) Indicated thermal efficiency, η IT, 10 × 10 −6, Fuel consumption / sec =, = 3.571 × 10 − 7 m 3 /sec, 26, Fuel consumption / hr = 3.571 × 10 −7 × 3600 = 1.286 × 10 −3 m 3 /hr., FC in litres / hr = 1.286 × 10 −3 × 1000 =1.286 litres /hr, FC in kg/hr = FC in litres / hr × Sp. Gravity, = 1.286 × 1.286 = 1.093 kg /hr, IP × 3600, 3.706 × 3600, η IT =, =, = 0.2906 = 29.06 %, FC × CV, 1.093 × 42000, Unit – V P10.20
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(e) Heat balance sheet, (1) Heat supplied in fuel, Qs = FC × CV = 1.093 × 42000 = 45906 kJ/hr, (2) Heat equivalent of brake power, QBP, QBP = BPE in kW × 3600 = 3.007 × 3600 = 10825.2 kJ/hr, Percentage of heat converted into brake power,, Q, 10825.2, QBP % = BP × 100 =, × 100 = 23.58 %, Qs, 45906, (3) Heat carried away by cooling water, Qw, Qw = m.C pw (T2 − T1 ) = 90 × 4.19 × (60 − 18) = 15838.2 kJ/hr, Percentage of heat carried away by cooling water,, Q, 15838 .2, × 100 = 34.50 %, Qw % = w × 100 =, 45906, Qs, (4) Heat lost due to exhaust gases, Qg, Mass of exhaust gas, m g = Mass of air ( ma ) + mass of fuel, ( mf ), Air fuel ratio = 22:1, Mass of air = 22 × Fuel consumption = 22 × 1.093 = 24.046 kg/hr, m g = ma + mf = 24.046 + 1.093 = 25.139 kg/hr, , Qg = m gC pg .(Tg − Ta ) = 25.139 × 1 × (420 − 30) = 9804.21 kg/hr, Percentage of heat lost duet to exhaust gases,, Qg, 9804.21, Qg % =, × 100 =, × 100 = 21.36 %, Qs, 45906, (5) Heat loss unaccounted, Qua, Qua = Qs − [QBP + Qw + Q g ], = 45906 − [10825.2 + 15838.2 + 9804.21] = 9438.39 kJ/hr, Percentage of heat loss unaccounted,, Q, 9438 .39, Qua % = ua × 100 =, × 100 =20.56 %, Qs, 45906, Unit – V P10.21
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Heat balance sheet [kJ/s (kW) basis or second basis], Sl., No, 1., , Heat, supplied, Heat, supplied, in fuel, , kJ/s, , %, , Heat spent, , kJ/s, , %, , 45906, , 100, , −, , −, , −, , 10825.2, , 23.58, , 15838.2, , 34.50, , 9804.2, , 21.36, , Heat, equivalent of, brake power, ( QBP ), Heat carried, by cooling, water ( Qw ), Heat lost due, to exhaust, gases ( Qg ), , 2., , −, , −, , −, , 3., , −, , −, , −, , 4., , −, , −, , −, , 5., , −, , −, , −, , Heat loss, unaccounted, ( Qua ), , 9438.4, , 20.56, , Total, , 45906, , 100, , Total, , 45906, , 100.00, , Example 9.16, In a trial of oil engine, the following data were obtained., Duration of trial = 30 min.; Speed =1750 rpm; Brake torque = 300, Nm; Fuel consumption = 9.35 kg; Calorific value of fuel = 42300, kJ/kg; Jacket cooling water circulation = 483 kg; Inlet and outlet, temperatures = 17 o C and 77 o C respectively. Air consumption =, 182 kg; Exhaust gas temperature = 486 o C ; Atmospheric, temperature = 17 o C . Calculate the brake power, indicated, thermal efficiency, specific fuel consumption on BP base, if, mechanical efficiency is 83%. Assume specific heat of exhaust, gases as 1.25 kJ/kg.K. Draw heat balance sheet on minute basis., Given data, Duration of trial = 30 min., Speed, N = 1759 rpm = 29.166 rps, Brake torque, T = 300 N−m = 0.3 kJ−m, Unit – V P10.22
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Fuel consumption, FC = 9.35 kg/30 min, = 0.3117 kg/min, Calorific value, CV = 42300 kJ/kg, Mass of cooling water, mw = 483 kg/30 min = 16.1 kg/min, Inlet temperature of cooling water,, t1 = 17 o C, Outlet temperature of cooling, water, t2 = 77 o C, Mass of air, ma = 182 kg/30 min = 6.067 kg/min, Exhaust gas temperature, t g = 486 o C, Atmospheric temperature, ta = 17 o C, Mechanical efficiency, ηm = 93 % = 0.83, Sp. Heat of exhaust gases, C pg = 1.25 kJ/kg.K, To find, (a) Brake power, BP, (b) Indicated thermal efficiency, ηm, (c) Specific fuel consumption, SFC, (e) Heat balance sheet on minute basis, Solution, (a) brake power, BP, Brake power, BP = 2.π. N .T = 2.π. × 29.166 × 0.3 = 54.977 kW, = 54.977 × 60 = 3298.67 kJ/min., (b) Indicated thermal efficiency, η IT, BP, BP, ηm =, IP =, IP, ηm, 3298 .67, ∴ IP =, = 3874.3 kJ/min, 0.83, IP, 3974.3, η IT =, =, = 0.3014 = 30.14 %, FC × CV, 0.3117 × 42300, (c) Specific fuel consumption, SFC, FC 0.3117 × 60, SFC =, =, =0.340 kg/kW.hr, BP, 54.977, , Unit – V P10.23
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(d) Heat balance sheet, (1) Heat supplied in fuel, Qs = FC × CV = 0.3117 × 42300 = 13184.91 kJ/hr, (2) Heat equivalent of brake power, QBP = BP = 3298.67 kJ/min, Percentage of heat converted into brake power,, Q, 3298.67, QBP % = BP × 100 =, × 100 = 25.02 %, Qs, 13184 .91, (3) Heat carried away by cooling water, Qw, Qw = m.C pw (T2 − T1 ) = 16.1 × 4.19 × (77 − 7) = 4047.54 kJ/min, Percentage of heat carried away by cooling water,, Q, 4047.54, × 100 = 30.70 %, Qw % = w × 100 =, 13184 .91, Qs, (4) Heat lost due to exhaust gases, Qg, Mass of exhaust gas, m g = Mass of air, ma + mass of fuel, mf, Air fuel ratio = 12:1, Mass of air = 12 × Fuel consumption = 12 × 0.3117 = 3.7404 kg/hr, m g = ma + mf = 3.7404 + 0;3117 = 4.0521 kg/min., , Qg = m gC pg .(Tg − Ta ) = 4.0521 × 1.25 × (200 − 17) = 926.918 kg/min, Percentage of heat lost duet to exhaust gases,, Qg, 926.918, Qg % =, × 100 =, × 100 = 7.03 %, Qs, 13184 .91, (5) Heat loss unaccounted, Qua, Qua = Qs − [QBP + Qw + Q g ], = 13184.91 − [3298.67 + 4047.54 + 926.918] = 49111.782, kJ/min., Percentage of heat loss unaccounted,, Q, 4911 .782, Qua % = ua × 100 =, × 100 = 37.25 %, Qs, 13184 .91, , Unit – V P10.24
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Heat balance sheet [kJ/s (kW) basis or second basis], Sl., No, 1., , Heat, supplied, Heat, supplied, in fuel, , kJ/s, , %, , Heat spent, , kJ/s, , %, , 13184.9, , 100, , −, , −, , −, , Heat, equivalent, of brake, power ( QBP ), Heat carried, by cooling, water ( Qw ), Heat lost, due to, exhaust, gases ( Qg ), , 3298.67, , 25.02, , 4047.54, , 30.7, , 926.918, , 7.03, , 2., , −, , −, , −, , 3., , −, , −, , −, , 4., , −, , −, , −, , 5., , −, , −, , −, , Heat loss, unaccounted, ( Qua ), , 4911.782, , 37.25, , Total, , 13184.9, , 100, , Total, , 13184.9, , 100.00, , PROBLEMS FOR PRACTICE, 1. A single cylinder internal combustion engine working on, 4−stroke cycle has a cylinder bore of 100mm and diameter and, piston stroke of 150mm. Speed of the engine is 400 rpm. If the, mean effective pressure is 650 kN/ m 2 , calculate the indicated, power developed by the engine. [Ans: IP=2.553 kW], 2. A gas engine has a ratio of diameter piston to stroke as 0.66, and average mean effective pressure is 450 kN/ m 2 . Calculate, the size of a four stroke cycle engine if it has a brake power of, 8kW at 250 rpm. Take mechanical efficiency as 75%., [Ans: l=321.6mm, d=212.3mm], Unit – V P10.25
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3. The brake power developed by an oil engine is 37.5 kW. The, mean effective pressure is 6 bar. No. of explosions are 80 per, minute. The mechanical efficiency is 80%. Determine the, dimensions of the cylinder assuming the ratio of stroke to, diameter as 2., [Ans: l=670mm], 4. A four cylinder, two stroke engine had a bore to stroke ratio of, 140:210 mm. during the trial, the following results were, obtained. Area of the indicator diagram = 700mm2, Length of, the indicator diagram = 70mm, Spring number =, 6 × 10 −5 kN/m m 2 /mm; Brake torque = 1kN−m; Speed = 430, rpm; Fuel consumption = 200 kg/hr. Calculate the total, indicated power, brake power, mechanical efficiency and, specific fuel consumption; [Ans: IP=103.72, BP=65.97,, ηm =63.61%, SFC = 3.032 kg/kW−hr], 5. The following observations were made during a trial on two, stroke engine: Stroke = 400mm; Bore = 250mm; Indicated, mean effective pressure = 6bar; Speed of the engine = 300, rpm; Effective brake load = 250 kg; Mean circumference of, brake drum = 3.6m; Fuel consumed = 4.35 kg per half an hour;, Calorific value of the fuel = 42000 kJ/kg. Determine the, indicated power, brake power, mechanical efficiency and, thermal efficiencies. [Ans: IP=58.9kW, BP=44.15kW,, ηm =74.95%, η IT =58%, ηrel =43.5%], 6. An engine working on constant volume cycle is 150mm bore, by 165 mm stroke and its clearance volume is 0.51 litre. The, fuel used has a calorific value as 45350 kJ/kg and the, consumption is 0.334 kg/kW−hr. The mechanical efficiency of, the engine is 81%. Calculate (1)Indicated thermal efficiency, and (2) Relative efficiency. Take γ=1.4, [Ans: η IT =29.35%, ηrel =55.045%], 7. The following data refer to a four cylinder petrol engine. At, the beginning of compression, the pressure and temperature, are 1 bar and 15 o C ; Maximum cycle temperature = 1300 o C ., Calculate the air standard efficiency and the mean effective, pressure., [ ηideal =51.16%, pmi =5.24 bar], 8. A four stroke cycle gas engine has a bore of 160mm and a, stroke of 240mm. The compression ratio is 4 to 1 and the, Unit – V P10.26
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engine speed is 300 rpm. The mean effective pressure is 350, kN/ m 2 . The thermal efficiency is 30%. Calorific value of the, gas is 17400 kJ/kg. Calculate the fuel consumption in, m 3 /IP−hr and the efficiency ratio., [Ans: ISFC=0.6897 m 3 /IP−hr, ηrel =70.48%], , 9. An engine working on constant volume cycle has a compression, ratio of 8. It uses petrol having a calorific value of 44000 kJ/kg. If, the brake thermal efficiency of the engine is 60% of the air, standard efficiency, determine the specific fuel consumption in, kg/kW−hr. Take γ = 1.4 [Ans: BSFC = 0.2415 kg/kW−hr], 10. The following results were obtained during a test on a single, cylinder, four stroke cycle gasoline engine: Brake power = 73.5, kW at a speed of 400 rpm; Brake mean effective pressure = 8.5, bar; Fuel consumption =0.346 kg/BP−hr of calorific value, 44100 kJ/kg; Compression ratio =6; Mechanical efficiency =, 0.8. Assume diameter of cylinder is equal to stroke of the, piston. Calculate (1) Bore and stroke (2) Brake thermal, efficiency (3) Indicated thermal efficiency (4) Air standard, efficiency and (5) Relative efficiency., [Ans: d=l=320.8 mm; η BT =23.59%, η IT =29.49%,, , ηideal =51.16%, ηrel =57.64%], 11. An ideal i.c engine works on four stroke Otto cycle. The, suction pressure is 1 bar. The pressure at the end of, compression is 2.75 bar. The total swept volume is 3000c.c., Clearance volume is 100c.c per cylinder. Find the mean, effective pressure. [Ans: pmi =6.713 bar], 12. A six cylinder four stroke cycle petrol engine is to be designed, to develop a brake power of 302 kW at 2500 rpm. The bore to, stroke ratio is 1:1.25. Mechanical efficiency is 83%. Indicated, mean effective pressure is 950 kN/ m 2 . Compression ratio of, the engine is 6.5:1. Relative efficiency is 55%. Calorific value, of petrol is 44630 kJ/kg. Determine (a) Bore and stroke (b), Fuel consumption in kg/hr and in kg/BP−hr., [Ans: d=146.16mm, l=182.7mm, FC=101.3 kg/hr,, BSFC=0.335kg/BP−hr], 13. The mean effective pressure was measured as 3 bar., Determine the power developed by a 4 cylinder four stroke, Unit – V P10.27
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cycle engine of 80mm bore and 90mm stroke, if the speed is, 2000 rpm. If the mechanical efficiency is 82% and the brake, thermal efficiency is 32.3%, determine the fuel consumption, per hour at this load. The calorific value of the fuel is 43500, kJ/kg. [Ans: IP=9.048kW, BP=7.419kW, FC=1.946 kg/hr], 14. In a test with a four cylinder, four stroke petrol engine, following results were obtained:, BP with all cylinders working = 23.54 kW, BP with No.1 cylinder cut−off = 15.88 kW, BP with No.2 cylinder cut−off = 16.40 kW, BP with No.3 cylinder cut−off = 16.55 kW, BP with No.4 cylinder cut−off = 16.92 kW, Estimate the indicated power of the engine and the, mechanical efficiency. [Ans: IP=28.41kW, ηm =82.86%], 15. A four cylinder petrol was tested at constant speed. The four, cylinder engine was supplied with fuel at 0.1 kg/min. and the, plugs were short circuited successively at constant speed for a, Morse test. The following data were obtained:, Power with all cylinders working = 14.86 kW, Power with cylinder 1 cut−off = 10.30 kW, Power with cylinder 2 cut−off = 10.44 kW, Power with cylinder 3 cut−off = 10.52 kW, Power with cylinder 4 cut−off = 10.37 kW, Calorific value of petrol was 41868 kJ/kg. Determine (1), Mechanical efficiency, (2) Brake thermal efficiency and, (3) Indicated thermal efficiency. [Ans: ηm =83.44%,, , η B.T =21.3%, η IT =25.52%], 16. In a trial of an oil engine the following data were obtained:, Duration of trial = 30min; Speed = 1750 rpm; Brake torque = 330, N−m; Fuel consumption = 9.35 kg of calorific value of 42300, kJ/kg. Jacket cooling water circulation = 483 kg; Inlet and outlet, temperatures = 17 o C and 77 o C respectively. Air consumption =, 182 kg; Exhaust gas temperature = 486 o C ; Atmospheric, temperature = 17 o C . Calculate the brake power, indicated, thermal efficiency, specific fuel consumption on BP basis, if, mechanical efficiency is 83%. Assume specific heat of exhaust, gases as 1.25 kJ/kg.K. Draw up an energy balance in kJ/min., Unit – V P10.28
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[Ans: BP=60.476 kW, η IT =33.16%, BSFC=0.309 kg/BP−hr,, Qs =12183.5 kJ/min (100%), QBP =3628.56 kJ/min (27.52%),, Qw =4047.54 kJ/min (30.70%), Qg =3739.28kJ/min (28.36%),, Qua =1768.12% kJ/min (13.42%)], , 17. The following observations were obtained during the trial on, two stroke cycle oil engine. Brake=200mm; Stroke=250mm;, Speed=350rpm; Effective brake drum diameter=1.2m; Net, brake load = 450N; Mean effective pressure = 2.8 bar; Oil, consumed = 3.6 kg/hr; Calorific value of oil = 41800 kJ/kg;, Mass of cooling water circulated=455 kg/hr; Rise in, temperature of cooling water = 28 o C ; Temperature of exhaust, gases = 320 o C ; Mass of exhaust gases = 2.68 kg/min; Room, temperature = 20 o C ; Specific heat of the gases = 2.68 kg/min;, Determine the mechanical efficiency, indicated thermal, efficiency and brake thermal efficiency. draw up the heat, balance sheet on minute basis and also express each item as, the percentage of heat supplied., [Ans: ηm =77.14%, η IT =30.69%, Qs =2508kJ/mon (100%),, QBP =593.76 kJ/min (23.67%), Qw =889.68 kJ/min (35.47%),, Qg =804.00 kJ/min (32.06%),, Qua =220.56% kJ/min, , (8.80%)], , Unit – V P10.29
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