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THERMODYNAMICS I INTRODUCTION Thermodynamics is the science dealing with the relations between the properties of a substance and the quantities (work and heat) which cause a change of state. It is concerned with the means necessary to convert heat energy from available sources such as chemical or nuclear piles into mechanical work. 1.0Thermodynamic Concepts The idea of a system plays an important part in thermodynamics. A system may be defined as a region in space containing a quantity of matter whose behavior is being investigated. For the purpose of thermodynamics, a system may be defined generally as a region in space containing a quantity of fluid. This quantity of matter or fluid may be separated from its surroundings by a boundary, which may be a physical boundary, such as the walls of a vessel, or some imaginary space enveloping the region. A boundary may therefore be defined as a physical or imaginary material or substance used to separate matter (fluid) from the surrounding. Surroundings may also be defined to those portions of matter external to the system, which are affected by changes occurring within the system. A system may contain matter, but when the same matter remains within the region throughout the process under investigation, it is called a CLOSED SYSTEM, and only work and heat cross the boundary. An open system, on the other hand, is a region in space defined by a boundary across which matter may flow in addition to work and heat. The fluid may flow into, out of, or through the region. Examples of open systems are; 1. A gas expanding from a container through a nozzle 2. Steam flowing through a turbine, and 3. Water entering a boiler and leaving as steam Examples of a closed system are; 1. A mixture of water and steam in a closed vessel 2. A gas expanding in a cylinder by displacing a piston From the last example, it will be seen that the boundary separating a closed system from its surroundings need not be fixed. It may expand or contract to accommodate any change of volume undergone by the fixed quantity of fluid. The processes undergone by the fluid in a closed system are described as NON-FLOW PROCESSES, whereas those undergone by the fluid in an open system are referred to as FLOW PROCESSES. 1 ISOLATED SYSTEMS It is a system in which no heat and work crosses the boundaries. Such a system can be described as being in thermodynamic equilibrium, which means pressure and temperature remains constant throughout the process. In this case, properties such as pressure and temperature must be uniform throughout the system. Boundary System Surrounding (a) boundary Outlet Inlet Turbine (Fluid in a turbine) Open System (b) Surrounding 2 Surrounding System Piston (Fluid in a cylinder) Boundary Closed system STEADY FLOW AND NON-FLOW ENERGY EQUATION From the concept of conservation of energy that energy can neither be created nor destroyed, the first law of thermodynamics is merely one statement of this principle covering all cyclic processes with a particular reference to thermal energy (heat) and mechanical energy (work). It is considered that the internal energy of a system undergoing a thermodynamic cycling is the same at the beginning and at the end of the cycling. The first law can therefore be stated as “when a system undergoes a thermodynamic cycle then the net heat supplied to the system is proportional to the net work done by the system. Σ Q ∝Σ W It has been proved by experiment that the constant of proportionality which is 4.18kJ = 1kilocaloric. Therefore, the value of the constant of proportionality is taken to be unity (1) ⇒ ΣQ = ΣW Sign convention All inputs to the system are taken to be positive and outputs are taken to be negative Heat supplied to the system = +ve Heat rejected to the system = -ve Work done on the system = +ve Work done by the system = -ve 3 From the law ΣQ = - ΣW ⇒ ΣQ + ΣW = 0 or ΣdQ + ΣdW = 0 ie if there is difference by heat and work NON-FLOW ENERGY EQUATION In a cycle where the final internal energy of the system is greater than the initial internal energy, it implies that the sum of the net heat supplied and the net work done has caused an increase in the internal energy of the system. ⇒ Net heat supplied + Network input = gain in internal energy ( or loss in internal energy) The internal energy of mass, m of a fluid is given as U, which is equivalent to mU where u is the specific internal energy. U = internal energy Consider a system whose internal energy has changed from state 1 to state 2 after it has gone through a complete cycle, then ⇒ ∑ dQ + ∑ dW = U2 – U1 For any non-flow process, heat and work can either be supplied or rejected and not both ⇒ ∑ dQ + ∑ dW = U2 – U1 ⇒ Q + W = U2 – U1 Q + W = mU2 – mU1 Q + W = m(U2 – U1) For a unit mass Q + W = U2 – U1 Non-flow energy equation E.g. 1. In a cylinder, the compressed air has a specific internal energy of 650 kJ/kg initially and after expansion, the specific internal energy becomes 380 kJ/kg. Calculate the heat flow to or from the cylinder if the work done by the air during the expansion is 150 kJkg-1. Solution U1 = 650 kJkg-1 U2 = 380 kJkg-1 W = -150 kJkg-1 4 But from the non-flow energy equation Q + W = U2 – U1 Q + W = 380kJkg-1 – 650 kJkg-1 Q + W = - 270kJkg-1 But W = - 150 ⇒ Q = -270kJkg-1 + 150kJkg-1 Q = - 120kJkg-1 ⇒ Heat was rejected by the system E.g. 2. In a cylinder, work done on the system is 480kJkg-1. The initial specific internal energy of the system is 320kJkg-1. Calculate the final specific internal energy when 400kJkg-1 of heat is released to the surrounding. Solution W = 480kJkg-1 U1 = 320kJkg-1 Q = 400kJkg-1 U2=? From Non flow energy equation Q + W = U2 – U1 - 400 + 480 = U2 – U1 ⇒ 80 = U2 – 320 U2 = 80 + 320 U2 = 400kJkg-1 Final specific internal energy is 400kJkg-1 ⇒ Air was compressed 5 STEADY-FLOW ENERGY EQUATION. Inlet (1) Q z1 Boundary W out Outlet 2 z2 Datum Steady-flow open system Inlet 1 Boundary P1 P1 l Section at inlet to the system 6 Consider unit mass of a fluid with internal energy U, moving with velocity c and at a height above a datum or reference level. The total energy of the fluid can be obtained as U + 2 + Zg Where 2 is the kinetic energy of unit mass of fluid and zg is the potential energy of unit mass of fluid ⇒ mU +m + mzg 2 ⇒ m( U + 2 + zg) but m = unit (1) = U + 2 + zg Let ṁ be the mass flow rate of the fluid flowing steadily. It is assumed that a steady rate of flow of heat Q̇ is supplied and the work input on the fluid is Ẇ. Also, an expenditure of energy is required to introduce the fluid across the boundary and out of the boundary taking an element of fluid length l Cross-sectional area of the inlet pipe A1͘ Then the energy required to push or introduce the element of fluid across the boundary is (P1A1) x l ⇒ P1 x A1l, but A1l = V1 ⇒ P1 x V1 ∴ The energy needed = P1V1 Where V1 = specific volume of the fluid at section 1, P1= pressure. ⇒ Energy = P1V1 Proofing Pressure = 1 ⇒ Force = P1A1 But Energy = Force x distance ( l) 7 Energy = F x l 2 From equation 1 F = P1A1 3 Put equation 3 into 2 ⇒ Energy = P1A1 x l = P1A1l⇒ V1 = A1l ⇒ Energy = P1V1 Energy to push in unit mass of fluid in the system Similarly, energy required at exit (out) to push unit mass of fluid across the boundary = P2V2 Now energy entering the system = m (U1 ++ zc1g)+ mP1V1 + Q̇ + Ẇ 2 Energy leaving the system m(U2 + +z2g)c+ mP2V2 2 Since there is steady flow of fluid into and out of the system, energy entering must be equal to energy leaving the system ⇒ ṁ(U1 + c + z1g) + ṁP1V1 + Q̇ + Ẇ = ṁ(U2 + c +z2g) + mP2V2 2 2 ⇒ ṁ(U1 + c + z1g + P1V1) + Q̇ + Ẇ = ṁ(U2 + c +z2g + P2V2) 2 2 In a flow process, the fluid always possesses a certain internal energy U and the term PV always occurs at inlet and outlet. Therefore, the sum of these terms can be replaced by the symbol (h). h is called specific enthalpy ⇒ h = U + PV The equation becomes ṁ( c + z1g + h1) + Q̇ + Ẇ = ṁ ( c + z2g + h2) 2 2 The steady flow equation 8 Changes in height are negligible in almost all problems in thermodynamics. Therefore, the height term in the equation can be omitted ṁ( c + h1) + Q̇ + Ẇ = ṁ ( c + h2) ⇒ 2 2 E.g.1 Gases flow through the turbine of a gas turbine unit at 15 kg/s and the turbine develops a power of 12,000kW. The velocities of the gases at inlet and outlet are 50 m/s and 120 m/s and the specific enthalpies of the gases at inlet and outlet are 1,400 kJ/kg and 400 kJ/kg respectively. Calculate the rate at which heat is released from the turbine. Solution From steady flow energy equation ṁ( c + h1) + Q̇ + Ẇ = ṁ ( c + h2) 2 2 For unit mass flow rate c 50⬚ K.E.at inlet = 2 = 2 = 1250 J/kg ⇒ K.E. at inlet = 1.250 kJ/kg c 120⬚ K.E. at outlet = 2 = 2 = 7200 = 7.2 kJ/kg Power (Ẇ) = - 12000 kW h1 = 1400 kJ/kg h2 = 400 kJ/kg m = 15 kg/s From the equation ⇒ 15 (1.25 + 1400) + Q̇ - 12000 = 15(7.24 + 400) 15 (1401.25) + Q̇ - 12000 = 15 (407.2) 21,018.75 + Q̇ - 12000 = 6,108 9018.75 + Q̇ = 6108 Q = - 2910.75 kW 9 Example 2 In the turbine of a gas turbine unit, the gases flow through the turbine at 17 kg/s and the power developed by the turbine is 4000kW. The specific enthalpies of the gases at inlet and outlet are 1200 kJ/kg and 360 kJ/kg respectively, and the velocities of the gases at inlet and outlet are 60 m/s and 150 m/s respectively. Calculate the rate at which its rejected from the turbine. Find also the area of the inlet pipe given that the specific volume of the gases at inlet is 0.5 m3/kg. [Hint: to find the inlet area, use ṁ = , A = ] WORK AND HEAT In thermodynamics, it is considered that, a change of state must be accompanied by the appearance of work and heat at the boundary. Work: Work is said to be done when a force moves an object through a distance. If part of the boundary of a system undergoes a displacement under the action of a pressure, then work is done. Work = Force (Pressure x Area) x the distance it moves in the direction of the force W = Fd = (PA) d SI unit is Joule (J) or Nm. Thus in thermodynamics, work can be defined as a form of energy which appears at the boundary when a system changes its state due to the movement of a part of the boundary under the action of a force. Similarly, heat can be defined as a form of energy that appears at the boundary when a system changes state due to the difference in temperature between the system and its surroundings. REVERSIBLE PROCESS A process is said to be reversible when a system changes state such that at any instant during the process, the state point can be located on a diagram. The fluid going through the process passes through a continuous series of equilibrium states. This can be represented by a line between the states. E.g.: 10 P 1 2 V It can therefore be said that when a fluid undergoes a reversible process, both the liquid and its surroundings can always be restored to their original state. For a reversible process 1. The process must be frictionless 2. Differences in pressure between the fluid and the surroundings must be infinitely small, i.e. the process must take place slowly 3. The difference in temperature between the fluid and the surroundings must also be infinitely small, i.e. the heat supplied or rejected to or from the fluid must be transferred infinitely slowly. IRREVERSIBLEPROCESS A fluid undergoing a process cannot in reality be kept in equilibrium in its intermediate states and therefore a continuous path cannot be traced on a diagram of thermodynamic properties. This process is therefore described as irreversible and is represented on a diagram by a dotted line. 11 P 1 2 V WORK AND REVERSIBILITY Consider an ideal frictionless fluid contained in a cylinder behind a piston. Assume that the pressure and temperature of the fluid are uniform and that there is no friction between the piston and cylinder walls. Let the cross-sectional area of the piston be (A), let the pressure of the fluid be (P) and let the pressure of the surrounding be (P + dP). The force exerted by the piston on the fluid is (PA) Let the piston move under the action of the force exerted, a distance (dl) to the left, then the work done on the fluid by the piston is given as Work = Force x distance ⇒ dW = -(PA) x dl ⇒ dW = -PdV i.e. Al = V wheredV is the small increase in volume. The negative sign is necessary because the volume is decreasing as a result of the movement of the piston to the left For a mass, m dW = -mPdV where V is the specific volume The work done on the fluid during any reversible process W is given by the area under the line of the process plotted on a property diagram. E.g. 12 p 1 2 V1 V2 v p 1 2 V1 V2 v dV On the P-Vdiagram, the work is given as = − !"# ⇒ m(shaded area) When P can be expressed in terms of V, then the integral $ !"# can be evaluated. 13 Example A unit mass of a fluid at a pressure of 4.5 bar and with a specific volume of 0.22 m3/kg contained in a cylinder behind a piston expands reversibly to a pressure of 0.8 bar according to a law P = c/V2, where c is a constant. Calculate the work done during the process. Data P1 = 4.5 bar = 4.5 x 105 N/m2 P2 = 0.8 bar = 0.8 x 105 N/m2 V1 = 0.22 m3/kg 4.5 bar P = # 0.8 bar V1 V2 Solution Using ! = # ⇒ ! = # ⇒ 4.5 x 105 = c0.22 ⇒ c = 4.5 x 105 x 0.0484 ⇒ c = 21780 Nm/kg Now ! = # / 21780 0 V = 'cP = ). 0.8 x 10- 14 ⇒ V2 = 0.52 m3/kg From = − $ !"# ⇒ W = −m $ 0 dV 3 30 53 ⇒ W = −mc $3 0 / 3 30 ⇒ W = −mc6− 1V7 3/ ⇒ W = −mc 8− 9 − ;< :0 : / ⇒ W = −mc 8 − < :/ : 0 But unit mass = 1 ⇒ c = -21780 Nm/kg V1 = 0.22 m3/kg V2 = 0.52 m3/kg ⇒ W = −21780 8 − < =. =.- ⇒ W = -21780[2.62] ⇒ = -57063.6 Nm/kg Proving but P = # W = −m $ PdV ⇒ W = −m $ 30 dV >0 53 ⇒ W= −mc $> 30 / > 53 = −mc $> 0 930 ; / > > >@0A/ = −mc $> 0 V ? ⇒$> 0 ? / / ? >0 ⇒ W = −mc 8 > < >/ 15 New formulaPolytropic process = constant volume > Work done = $> 0 PdV 1 / Now PVn = c from c = PV2 Or P = cV-n 2 Substituting equation 2 into equation 1 >0 W= cV ?B dV >/ >0 ⇒ W = c $> V ?B dV / ⇒ DV ?BC ]>>0/ ?BC ⇒ DV?BC − V?BC ] ?BC ⇒ DV?B V − V?B V ] butcV?B = P ?BC E0 30 ? E/ 3/ ⇒ from equation 2 ?BC Multiply top and bottom by -1 P V − P V Work done = n−1 Question 0.014m3 gas at a pressure of 2070 kN/m2 expands to a pressure of 207 kN/m2 according to the law PV1.35 = c. Determine the work done by the gas during the expansion. Data V1 = 0.014 m3 V2= ? P1 = 2070 kN/m2 P2 = 207 kn/m2 16 Law PV1.35 = c Solution PV1.35 = c ⇒ P1V11.35 = c ⇒ 2070 x (0.014)1.35 ⇒ 2070 x 0.0031 ⇒ 6.505 ∴ c = 6.505 To find V2⇒ P2V21.35 = c V.G- = cP /.JK H.-=- ⇒ V = ' =I ⇒ V = √0.0314 /.JK ⇒ V2 = 0.077 m3 > Now work done = $> 0 PdV / But P = 3/.JK from law > ⇒ W = − $> 0 3/.JK dV / > ⇒ W= − $> 0 cV ?.G- dV / > ⇒ W= −c $> 0 V ?.G- dV / > [email protected] 0 ⇒ −mc 8 ?=.G- < >/ ?P ⇒ DV ?=.G- ]>>0/ ?=.G- For the unit mass, m = 1 >0 ⇒ 8 =.G- 3O.JK < >/ ⇒ But c = 6.505 ?H.-=- ⇒ ?=.G- 8: O.JK − : O.JK < 0 / 17 Also V1 = 0.014m3 and V2 = 0.077m3 H.-=- ⇒ =.G- 8=.=IIO.JK − =.=Q O.JK < ⇒ 18.586 8=.Q=I − =.Q-< ⇒ 18.586D2.45 − 4.45] 18.586[-2] ⇒ W = -37.2 kJ Work done by the gas = 37.2kJ SOURCE Source in thermodynamics are areas of fixed high temperature from which the heat engine can draw heat. It has an infinite thermal capacity and any amount of heat can be drawn from it at constant temperature. Examples of sources are: 1. Gases produced by the continuous combustion of fuel. 2. Mass of continuous vapour SINK Sink in thermodynamics are areas of fixed lower temperature to which any amount of heat can be rejected. It should have an infinite thermal capacity and its temperature remains constant. WORKING SUBSTANCE Working substance is the fluid e.g. perfect gas, contained in a cylinder with non-conducting sides and conducting bottom. A working substance undergoes complete cyclic operation and the non-conducting substance ensures that the working substance undergo an adiabatic operation. An adiabatic process is one in which no heat is transferred to or from the fluid during the process. Adiabatic process could be reversible or irreversible. HEAT AND REVERSIBILITY During heat exchanges between the surroundings and a system, the part, which receives or delivers the heat, is called the heat reservoir. It may be a source or a sink of heat such that any quantity of heat which crosses a boundary during a process is insufficient to change its temperature e.g. source: gases produced by the continuous combustion of fuel, mass of condensing vapour. Sink e.g. a river, earth’s atmosphere. A transfer of energy by virtue of a 18 temperature difference can be carried out reversibly only if the temperature difference is infinitesimally small. Considering the non-flow energy equation for a reversible process dQ + dQ = dU butdW = -PdV ⇒ dQ – PdV = dU Constant volume process (isochoric process) In constant volume process, the boundaries of the system do not move because the working substance is contained in a rigid vessel and therefore no work is done on or by the system. It therefore implies that for a constant volume process Work done = 0 (zero) From the energy equation Q + W = U2 – U1 If W = 0, then Q = U2 – U1 For unit mass constant volume equation Hence all the heat supplied in a constant volume process goes to increase the internal energy of the system. For a small change in temperature i.e. infinitesimally change the process will be reversible, the equation becomes dQ = dU Constant pressure process (isobaric) or ( isopiestic) In a constant pressure process, in like constant volume process, the boundary must move against an external resistance as heat is supplied to the system. Hence, work is done by the system in overcoming the external force. 19 P P2 V2 P1 P2 V1 = V2 P1 V1 V V1 V2 Constant volume process Constant pressure process (Isochoric process) (Isobaric process) Work is done Work is not done For a unit mass, W = − $ PdV For reversible process ⇒ W = −P $ dV = -P(V2 – V1) From the energy equation Q + W = U2 – U1 We have Q = (U2 – U1) – W But W = -P(V2 – V1) ⇒ Q = (U2 – U1) – [-P(V2 – V1)] ⇒ Q = (U2 – U1) + P(V2 – V1) ⇒ Q = U2 – U1 + PV2 – PV1 ⇒ Q = U2 + PV2 – U1 – PV1 ⇒ grouping the terms ⇒ (U2 + PV2) – (U1 + PV1) ⇒ But h = U + PV ⇒ Q = h2 – h1 constant pressure equation Hence, all the heat supplied to a system equals the changes in enthalpy of the system. 20 POLYTROPIC PROCESS In practice many reversible processes for expansion and compression can be described appropriately by a relation or law of the form PVn = constant, where n is a constant called the index of expansion or compression, P and V are average values of pressure and specific volume of the system. For reversible process W= − PdV For a unit mass From PVn = c ⇒ P1V1n = P2V2n = PVn Now PVn = c ⇒P = cV S W = −c dV B V 3@TA/ ⇒ −c 8 < ?BC 3@TA/ 3/@T ⇒ W =c8 < = c 8 B? < B? 30/@T ?3//@T ⇒ W = c9 ; but c = PVn B? 30/@T ?3//@T ⇒ W = PV B 9 ; B? ⇒ But P1V1n = P2V2n = PVn E0 3T 0 30 /@T ?E 3T 3 /@T ⇒ W= 9 ; / / / B? E0 30 ?E/ 3/ ⇒ W= B? But from the non-flow energy equation Q + W = U2 – U1 P V − P V Q+ ). = U − U ⇒ n−1 21 CONSTANT TEMPERATURE PROCESS (ISOTHERMAL) When the quantities of heat and work are so proportioned during an expansion or compression that the temperature of the fluid remains constant. The process is said to be ISOTHERMAL. It is possible to show that for a reversible isothermal process, a certain relation exist between pressure, volume and temperature. Where temperature T, pressure P and volume V, T = F (PV) As temperature is clearly associated with heat, it might be easy to express heat in terms of temperature and other thermodynamic properties i.e. Q = TdU Where T is temperature and dU is some other property. Two suitable properties can be used to express the quantity of heat supplied to a system and can be written in the form Q = Tds for a reversible process Where T is the absolute thermodynamic temperature and s is the property enthropy. Integrating the expression when T is constant Q = T $ ds Q = TZs − s [ If P1V1 are known in other properties including internal energy. U can be obtained from tables and therefore work transfer can also be found from W = (U2 – U1) – Q FIRST LAW OF THERMODYNAMICS When a system is taken through a complete cycle i.e. the final state is equal in all respect to its initial state, the net amount of heat supplied to the system is proportional to the net amount of work done by the system on the surroundings. The net heat supplied can be represented by ΣQ, and the network input as ΣW, where the Σ represents the sum of a complete cycle. The intrinsic energy of the system remains unchanged and the first law thus be stated as: “when a system undergoes a thermodynamic cycle then, the net heat supplied to the system from its surroundings plus the network input to the system from its surroundings mut be equal to zero” 22 i.e. ΣQ + ΣW = 0 This is associated with the general principles of conservation of energy. COROLLARIES OF THE FIRST LAW COROLLARY 1: There exist a property of a closed system such that a change in its volume is equal to the sum of the net heat and work transfers during any change of state. PROOF: Assume that the converse of the proposition is true i.e. ∑Z$ Q + $ W[ depends upon the process, as well as upon the end states 1 and 2, and therefore cannot be equal to the change in a property. Consider two processes A and B by which a system can change from state 1 to state 2. The assumption is that in general ΣZ$ Q + $ W[A ≠ ΣZ$ Q + $ W[Ba In each case let the system return to its original state by a third process c x Property y Property For each of the complete cycles AC and BC `) Q + W. AC = ` ) Q + W. + ` ) Q + W. C 23 For the cycle BC we have `) Q + W. BC = ` ) Q + W. b + ` ) Q + W. C If the assumption described by the inequality in equation a is true, then `) Q + W. AC ≠ ` ) Q + W. BC But this contradicts the first law which implies that these quantities must be equal since they are both zero (0). Therefore the original proposition that ∑Z$ Q + $ W[is independent of the process must be true. If the property is denoted by Q, then the corollary can be expressed mathematically as : ` ) c+ . = d − d The property U is called the internal energy of the system. COROLLARY 2: The internal energy of a closed system remains unchanged if the system is isolated fron its surroundings. No formal proof of this statement is needed once the first one has been established (first corollary) Once the system is isolated i.e. Q = 0, W = 0 U2 – U1 = 0 This corollary is often called the law of conservation of energy. COROLLARY 3: A perpetual motion machine of the first kind is impossible. The perpetual motion machine was thought to be a machine that will continue to run forever once set in motion. Such a machine will be of no practical value and the presence of friction will make this impossible. What will be of immense value is a machine producing a continuous supply of work without absorbing energy from surroundings, such a machine called a perpetual motion machine of the first kind. 24 It is possible to divert a machine to deliver a limited quantity of work without requiring a source of energy in the surroundings. Such a device cannot do work continuously, however for this to be achieved, the machine must be capable of undergoing a succession of cyclic processes. But if net amount of heat is not supplied by the surroundings during a cycle, no net amount of work can be delivered by the system. Thus, the first law implies that a perpetual motion machine of the first kind is impossible. EXAMPLE The heat supplied to the steam in a boiler of a steam plant is 3200 kJ/kg and the heat rejected by the steam to the cooling water in the condenser is 2400 kJ/kg. The feed-pump work required to pump the condensate back into the boiler is 6kW. If the turbine develops 1100kW, what will be the steam flow rate? Solution Let me represent the steam flow rate ΣQ = 3200 – 2400 = 800 kJ/kg Using the steam flow rate ΣQ = 800me (kJ/kg x kg/s) = 800me kJ/s ∴ ΣQ = 800me kW ΣW = 6 – 1100 = -1094 kW From the first law of thermodynamics ⇒ ΣQ + ΣW = 0 800me + (-1094) = 0 800me – 1094 = 0 800me 1094 = 800 800 me = 1.3675 kg/s 25 SECOND LAW OF THERMODYNAMICS In the second law unlike the first law, some amount of heat must be rejected during a thermodynamic cycle and therefore the cycle efficiency is always less than unity. The second law can thus be stated as “It is impossible to construct a system which will operate in a cycle, extract heat from a reservoir, and do an equivalent amount of work on the surroundings” For the first law Q1 is equivalent to W1 i.e.|Q | = |W | but for the second law |Q | > |W | where Q1 is heat supplied and thus |Q | − |Q | = |W| for heat engine cycles. The second law implies that if a system is to undergo a cycle and produce work, it must operate between at least two reservoirs of different temperature however small this difference may be. 26 Hot reservoir Q1 W (ǀQ1ǀ – ǀQ2ǀ) Q2 Cold reservoir Heat engine cycle Hot reservoir Q2 W Q1 Cold reservoir Heat pump or refrigerator A machine which will produce work continuously while exchanging heat with only a single reservoir is known as a perpetual motion machine of the second kind. Such a machine will contradict the second law. An important consequence of the second law is that work is a more valuable form of energy transfer than heat. Heat can never be transferred continuously and completely into work whereas work can always be transferred continuously and completely into heat. 27 CONSEQUENCES OF THE SECOND LAW 1. If a system is taken through a cycle and thus a net work on the surroundings, it must be exchanging heat with at least two reservoirs at different temperatures. 2. If a system is taken through a cycle while exchanging heat with only one reservoir, the work transfer must either be zero or positive. 3. Since heat can never be converted continuously and completely into work whereas work can always be converted continuously and completely into heat, work is a more valuable form of energy transfer than heat. COROLLARY I (CLAUSINS STATEMENT OF THE SECOND LAW) “It is impossible to construct a system which will operate in a cycle and transfer heat from a cooler to the hotter body without work been done on the system by the surroundings”. PROOF OF THE FIRST COROLLARY Suppose that the converse of the proposition is true, the system could be represented by a heat pump for which |W| = 0. If it takes |Q| units of heat to pump from the cold reservoir, it must deliver |Q| units to the hot reservoir to satisfy the first law. A heat engine could also be operated between to reservoirs. Let it be such a size that it delivers units of heat to the cold reservoir whiles performing |W| units of work. Then the first law states that the engine must be supplied with (|Q|+ |W|) units of heat from the hot reservoir. Hot reservoir (ǀWǀ + ǀQǀ) = Q1 W Q Cold reservoir 28 The combined plat represent a heat engine extracting (|W| + |Q| − |Q| = |W|) units of heat from a reservoir and delivering an equivalent amount of work. This is impossible according to the second law. Consequently, the converse of the proposition cannot be true and therefore the original proposition must be true. P = Pressure (bar) t = Temperature g = dry saturated vapour f = saturated liquid Vg = specific volume of dry saturated vapour m3/kg Vf = specific volume of saturated liquid m3/kg Uf = specific internal energy of saturated liquid kJ/kg Ug = specific internal energy of dry saturated vapour (kJ) hf = specific enthalpy of saturated liquid (kJ/kg) hg = specific enthalpy of dry saturated vapour (kJ/kg) x = dryness fraction = mass of dry vapour in 1kg of a mixture of liquid and vapour Wetness fraction = mass of liquid in 1kg of the mixture ⇒ wetness fraction = (1- x) hfg = change in specific enthalpy from hf to hg From the energy equation, Q + W = U2 – U1 U2 is equivalent to Ug From state 1 to 2 U1 is equivalent to Uf ∴Q + W = U2 – U1 = Ug – Uf 1 Also, W = -PdV = -P(V2 – V1) From equation 1 Q = (Ug – Uf) – W ⇒ Q = (Ug – Uf) + P (V2 – V1) ⇒ Q = (Ug – Uf) + P (Vg – Vf) ⇒ Q = (Ug + PVg) – (Uf + PVf) 29 ⇒ But h = U + PV ⇒ Q = hg - hf i.e Q = h2 – h1 The heat required to change a saturated liquid to dry saturated vapour is called the specific enthalpy of vapourization, hfg For a wet vapour, the total volume of the mixture is given by the volume of liquid pressure plus the volume of dry vapour present. Therefore, the specific volume V is given by > ijP k ilmjl5C > ijP k 5 n >o j V= p pi Pqq k r p >o j For 1kg of wet vapour, there are x kg of dry vapour and (1 – x) kg of liquid. x is the dryness fraction 3s Z?t[C3u t ∴ V= but volume of liquid is usually negligible compared to the dry saturated vapour, hence for most practical problems V = xVg The enthalpy of a wet vapour is given by the sum of the enthalpy of the liquid and the enthalpy of the dry vapour. i.e. h = (1 – x)hf + xhg h = hf + x (hg–hf) ⇒ h = hf + xhfg = specific enthalpy Similarly: U = (1 – x) Uf + xUg ⇒ U = Uf + x(Ug – Uf) Specific internal energy Example 1 Calculate the specific volume, specific enthalpy and specific internal energy of wet steam at 18 bar and 0.9 dryness fraction. Solution i. V = xVg Vg can be obtained in tables x = 0.9 30 V = 0.9 (0.1104) = 0.0994m3/kg ii. h = hf + xhfg = 885 + 0.9(1912) = 2605.8 kJ/kg iii. U = Uf + x (Ug – Uf) = 883 + 0.9(2598 – 883) ⇒ U = 2426.5 kJ/kg Example 2 Calculate the dryness fraction, specific volume and specific internal energy of steam at 7 bar and specific enthalpy of 2600 kJ/kg Solution i. From h = hf + xhfg v? vw tvwx = vwx vwx v? vw x= vwx H==?HyI y=G x= = =HI =HI ∴ x = 0.92 ii. V = xVg = specific volume V = 0.92 (0.2728) V = 0.251 m3/kg iii. Specific internal energy U = Uf + x(Ug – Ug) = 696 + 0.92 (2573 – 696) = 696 + 0.92 (1877) = 696 + 1726.84 U = 2422.84 kJ/kg Example 3 Calculate the dryness fraction, specific internal energy, specific volume of steam at 15 bar and specific enthalpy of 2700 kJ/kg. What will be the heat rejected if the steam is cooled and expanded to a pressure of 8 bar with the same dryness fraction. Find the change in the specific internal energy and hence the work done on the system during the change of state. 31 Data P = 15 bar h = 2700 kJ/kg x =? U=? V=? Solution (i) h = hf + xhfg z? zs ⇒ x= zsu I==?{Q- ⇒ x= yQI {-- ⇒ yQI ∴ x i.e. the dryness fraction = 0.95 (ii) U = Uf – x(Ug-Uf) = 843 + (0.95)(2595 – 843) = 843 + 1664.4 U1 = 2507.4 kJ/kg (iii) V = xVg V = 0.95(0.1317) V = 0.125 m3/kg At pressure of 8 bar with the same dryness fraction ⇒ h1 = 2700 kJ/kg, x = 0.95 ⇒ h2 = hf + 0.95(2045) ⇒ = 721 + 1945.6 ⇒ h2 = 2666.6 kJ/kg Heat rejected i.e. Q (iv) Q = h 2 – h1 ⇒ = 2666.6 -2700 ⇒ = -33.4 32 ∴ Heat rejected = 33.4 kJ/kg (v) Change in specific internal energy ∆U = U1 – U2 but U2 = Uf + x(Ug – Uf) = 720 + 0.95(2577 – 720) = 720 + 1764.15 ∴ U2 = 2484.15 kJ/kg Now ∆U = U1 –U2, where U1 = 2507.4 ⇒ ∆U = 2507.4 – 2484 i.e. it moves from state 1 to state 2 ∴ ∆U = 23.25 kJ/kg (vi) The work done on the system is Q + W = ∆U W = ∆U – Q But Q = -33.4 ⇒ W = 23.25 + 33.4 W = 56.7 kJ/kg SUPERHEATED VAPOUR For the steam in the super heat region, temperature and pressure are independent propertis and when these values are given, other properties can be found. Vg, Ug, hg and Sg are given in the super heat tables. Example 1 Steam at 80 bar, 400℃ has an enthalpy of 3139 kJ/kg and specific volume of 3.428 x 10x-2 m3/kg. Find the specific internal energy. Solution h = 3139 kJ/kg 33 v = 3.428 x 10-2 m3/kg From h = U + PV U = h – PV {= } =K =3139 - (3.428 x 10-2) =J = 3139 – 8000(3.428 x 10-2) = 3139 – 27424 x 10-2 = 3139 – 274.24 U = 2864.76 kJ/kg Example 2 Steam at 110 bar has a specific volume of 0.0196 m3/kg. calculate the temperature, the specific enthalpy and the specific internal energy. Comparison h > hg (super heat) h < hg (wet) Solution P = 110 bar = 110 x 105 V = 0.0196 m3/kg Find out whether the steam is wet, dry saturated or super-heated t = 350℃ Degree of super heat = 350 – 318 = 32℃ h = U + PV but h = 2889 U = h – PV INTERPOLATION Question Find the temperature, specific volume, specific internal energy and specific enthalpy of dry saturated steam at 9.8 bar. Solution 34 y.{?y tg at 9.8 bar = (tg at 9 bar) + 9 =.y ; x (tg at 10 bar) y.{?y ⇒tg at 9.8 bar = 175.4 + 9 =?y ; x (179.9 – 175.4) = 175.4 + 0.9 x 4.5 = 175.4 + 4.05 tg at 9.8 bar = 179℃ or p (10 bar) y x a 175.4 t 179.9 b a = ~xb a = t – 175.4, x = 9.8 -9, y = 10 – 9 and b = 179.9 – 175.4 y.{?y ⇒ t – 175.4 = =?y x (179.9 – 175.4) y.{?y t = 175.4 + x 4.5 ⇒ t = 179℃ y.{?y (ii) Vg at 9.8 = Vg9bar + x (V10 – V9) =?y y.{?y (iii) hg at 9.8 = hg9bar + =?y x (h10 – h9) 35 ASSIGNMENT 0.05kg of steam is contained in a rigid vessel of volume 0.0076m3. What is the temperature of the steam? If the vessel is cooled, at what temperature will it just be dry saturated? Cooling is contained until the pressure in the vessel is 11 bar. calculate the final dryness fraction of the steam and the rejected between the initial and final states. Hint: V = mv ⇒ v = Vm v = specific volume, V = volume, m = mass OTTO CYCLE The Otto cycle is the ideal air standard cycle and forms the bases of spark-ignition and high compression ignition engines.the process may be imagined to occur in a cylinder filled with a reciprocating piston having a swept volume equal to m(V1 – V2) where m = the mass of the fluid in the cylinder. P PVγ = constant m(v1 – v2) 3 4 2 1 Swept volume V2 V V1 The process for the cycle is as follows: 36 1-2: Air is compressed as isentropically through a volume ratio V1/V2, known as compressed ratio rv, i.e. (rv = V1/V2) 2-3: A quantity of heat Q23 is added at constant volume until the air is in state 3. 3-4: The air is expanded isentropically to the original volume. 4-1: Heat Q41 is rejected at constant volume until the cycle is completed. The efficiency of the cycle is || 0J C/ η= 0J or 0J orη = 0J Assuming constant specific heat capacities for the air and considering unit mass of fluid, the heat transfers are Q23 = cV(T3 – T2) and Q41 = cV(T4 – T1) / η =1− 0J 3Z ?/ [ i.e. η=1− 3ZJ ?0 [ Z ?/ [ η=1− J ?0 ?/ ∴ η=1− J ?/ For a two isentropic process 1 to 2 and 3 to 4 0 J = = rVγ-1 / For a unit mass PV = RT P= 1 : γ Also PV = constant PVγ = c 2 Putting equation 1 into 2 Vγ = c : Dividing both sides by R But R = 1 (unity) 37 Vγ = : R = gas constant, thermal resistance :γ γ V =c : But from = Vγ-1 : TVγ-1 = c ⇒ T1V1γ-1 = T2V2γ-1 = c Divide through by T1V2γ-1 γ@/ 0 :/ ⇒ = γ@/ / :0 : γ? 0 = 9 /; : ⇒ / 0 # But rV = # 0 ∴ = rVγ-1 / ⇒ T2 = T1rVγ-1 ⇒ T3 = T4rVγ-1 ? / J? 0 ⇒ From η = 1 - ? / γ@/ ? : γ@/ : η= 1 - / ? / : γ@/ Z ? 0 [ η = 1- η=1- : γ@/ Example: Calculate the ideal area standard cycle efficiency based on the Otto cycle for a petrol engine with a cylinder bore of 50 mm, a stroke of 75 mm and a clearance volume of 21.3 cm3 Take γ = 1.4 :/ qr op > ijP C i B > ijP HINT: Compression ratio rV = = :0 i B > ijP 38 Data Diameter d = 50 mm = 50 x 10-3m Clearance volume = 21.3 x 106m3 Swept volume = stroke x area Stroke = 75 mm = 75 x 10-3 m Solution 0 Swept volume = Q xstroke G.Q ⇒ Q x 0.052 x 0.075 G.Q } =.==- ⇒ x 0.075 Q ⇒ 0.001964 x 0.075 = 0.0001473 m3 ∴ The swept volume = 0.0001473 m3 qr op > ijP C i B > ijP Now compression ratio (rV) = i B > ijP =.===QIGC.G }= = .G }= ⇒ Compression ratio (rV) = 0.0001473 m3 Using the equation η = 1 - : γ@/ But γ = 1.4 ⇒ η=1– Z=.===QIG[/.@/ η=1- =.===QIGO. =1- =.=yGG = 1 – 34.09 η = -33.09 39 DIESEL CYCLE In a diesel cycle the engine works on the idea of spontaneous ignition of fuel which is blasted into the cylinder and in this way the heat addition occurs at constant pressure instead of constant volume. P 3 2 PVγ = constant 4 1 V # From 1 – 2: Air is compressed isentropically through the compression ratio rV = # 2 – 3: Heat Q23 is added while the air expands at constant pressure to volume V3. At state 3 the heat supplied is cut-off and the volume ratioV3/V2 may conveniently be called cut-off ratio rc (i.e. rc = V3/V2) 3 – 4: The air is expanded isentropicaly to the original volume 4 – 1: Heat Q41 is rejected at constant volume until the cycle is completed 0J C / η= 0J Q23 = cP(T3 – T2) Q41 = cV(T1 – T4) / :Z / ? [ η= = 0J Z J ? 0 [ :Z ? /[ Z J ? 0[ ⇒ η=- ⇒ But cP/cV = γ ZQ − [ η =1− γZG − [ ⇒ 40 # # Also rV = # and rc = G# In terms of rc and rV γ ? : γ@/ Z?[ ⇒ η=1- 0 = rVγ-1 / 1 ⇒ T2 = T1rVγ-1 or T1 = T2).a rVγ−1 For the constant pressure process J = rc⇒ T3 = T2rc b 0 3J γ? 3J 3 rc γ? =9 ; =9 x 30 ; =9 ; γ? J 3 30 3 = 9 3; γ? ⇒ J ⇒ T4 = T39 3; γ? c Putting equations a, b and c into ?/ η=1- γZJ ?0 [ THE DUAL COMBUSTION CYCLE Modern oil engines, which is still called diesel engine, used solid injection of fuel, the fuel being injection by a spring-loaded injection. A cam driven from the engine crankshaft operates the fuel pump. The ideal cycle used as a basis for comparism is called the DUEL OR MIXED COMBUSTION CYCLE. 41 P P3= P4 3 4 γ PV = constant 2 5 1 V2 = V3 V1 V 1-2: isentropic compression 2-3: reversible constant volume heating 3-4: reversible constant pressure heating 4-5: isentropic expansion 5-1: reversible constant volume cooling The heat is supplied in two parts: 1. At constant volume 2. At constant pressure. Hence the names dual combustion cycle For the heat transfers, Heat added Q23 =cV (T3-T2) Q34 = cP(T4 – T3) Total heat added QT= Q23+Q34 ⇒ cV(T3-T2) + cP(T4-T3) Heat rejection = cV(T1-T5) Efficiency η = 42 C K/ K/ = ⇒1+ cVZT- − T [ η =1− cVZTG − T [ + cPZTQ − T [ V 1. compression ratio,rV = V VQ 2. Ratio of volumes rc = V - PG 3. Ratio of pressure rP = P E γ@/ η=1− 3γ@/ Z E?[C γ EZ ?[ ⇒ If rP = 1 P3 = P2 and the equation reduces to the thermal efficiency of the diesel cycle. DETONATION OR HEAVY KNOCK This is the instantaneous combustion of the remaining portion of petrol vapour and air mixture(Enet gas) as a result of the gas been subjected to temperatures and pressures above those normal for a combustion chamber. This always occurs after the mixture has been ignited in the usual way by spark plug. It is heard as a knocking sound in the chamber. PRE-IGNITION This a is a form of uncontrolled combustion which occurs towards the end of a compression stroke. In this case the mixture is ignited before spark occurs at the spark plug. Pre-ignition tries to make the engine reverse its direction of rotation and therefore results in abnormal stresses being imposed on the piston, connecting rods and bearings. Pre-ignition is generally caused by white hot spots or particles of metal or carbon instead of the spark igniting the mixture. It also causes detonation. POST-IGNITION This is another form of uncontrolled combustion, which occurs after the mixture is ignited. This will usually result in detonation. 43 REFERENCES (1) Applied Thermodynamics for Engineering Technology by T.D. Eastop and A. McConkey (2) Engineering Thermodynamics: Work and Heat Transfer by G.F.C Rogers and Y.R. Mayhew 44

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