C4.1) ds derivative is the slope of the constant pressure legs of the Brayton cycle on a T-s plane. For a en ideal gas(specific c,) the slope is positive and increases as T We can also plot the Brayton cycle in an h-s plane. This has advantages because chang in enthalpy directly show the work of the compressor and turbine and the heat added and reject The slope of the constant pressure legs in the h-s plane is(ah/as),=T Note that the similarity in the shapes of the cycles in T-s and h-s planes is true for ideal only. As we will see when we examine two-phase cycles, the shapes look quite different in two planes when the medium is not an ideal gas Plotting the cycle in T-s coordinates also allows another way to address the evaluation of the Brayton cycle efficiency which gives insight into the relations between Carnot cycle efficien and efficiency of other cycles. As shown in Figure C-2, we can break up the Brayton cycle into Qcy many small Carnot cycles. The"i "h "Carnot cycle has an efficiency of n,=1-low /Thigh where the indicated lower temperature is the heat rejection temperature for that elementary cycle and the higher temperature is the heat absorption temperature for that cycle. The upper and lower curves of the Brayton cycle, however, have constant pressure. All of the elementary Carnot cycles herefore have the same pressure ratio PIT PR= constant (the same for all the cycles) PT From the isentropic relations for an ideal gas, we know that pressure ratio, PR, and temperature ratio, TR, are related by: PRlY-I)Y=TR 1C-6
1C-6 ∂ ∂ T s T c p p = (C.4.1) The derivative is the slope of the constant pressure legs of the Brayton cycle on a T-s plane. For a given ideal gas (specific cp ) the slope is positive and increases as T. We can also plot the Brayton cycle in an h-s plane. This has advantages because changes in enthalpy directly show the work of the compressor and turbine and the heat added and rejected. The slope of the constant pressure legs in the h-s plane is ∂ ∂ hs T p ( ) = . Note that the similarity in the shapes of the cycles in T-s and h-s planes is true for ideal gases only. As we will see when we examine two-phase cycles, the shapes look quite different in these two planes when the medium is not an ideal gas. Plotting the cycle in T-s coordinates also allows another way to address the evaluation of the Brayton cycle efficiency which gives insight into the relations between Carnot cycle efficiency and efficiency of other cycles. As shown in Figure C-2, we can break up the Brayton cycle into many small Carnot cycles. The " " i th Carnot cycle has an efficiency of ηci lowi highi = − [ ] 1 ( ) T T , where the indicated lower temperature is the heat rejection temperature for that elementary cycle and the higher temperature is the heat absorption temperature for that cycle. The upper and lower curves of the Brayton cycle, however, have constant pressure. All of the elementary Carnot cycles therefore have the same pressure ratio: P T P T PR high low ( ) ( ) = = constant (the same for all the cycles). From the isentropic relations for an ideal gas, we know that pressure ratio, PR, and temperature ratio, TR, are related by : PR TR ( ) γ γ − = 1 /
Figure C-2: Ideal Brayton cycle as composed of many elementary Carnot cycles [Kerrebrock The temperature ratlos(lowi/high: of any elementary cycle"i""are therefore the same and each of the elementary cycles has the same thermal efficiency. We only need to find the temperature ratio across any one of the cycles to find what the efficiency is. We know that the temperature ratio of the first elementary cycle is the ratio of compressor exit temperature to engine entry(atmospheric for an aircraft engine)temperature, T,/To in Figure C-2. If the efficiency of all the elementary cycles has this value, the efficiency of the overall Brayton cycle(which is composed ofthe elementary cycles)must also have this value. Thus, as previously. bRayton =l- or ernt a benefit of this view of efficiency is that it allows us a way to comment on the efficiency of any thermodynamic cycle. Consider the cycle shown on the right, which operates between some maximum and minimum temperatures. We can break it up into small Carnot cycles and evaluate the efficiency of each. It can be seen that the efficiency of any of the small cycles drawn will be less than the efficiency of a Carnot cycle between T,mar and Tmin. This graphical argument shows that the efficiency of any other thermodynamic cycle operating between these maximum and minimum temperatures has an efficiency less than that of a Carnot cycle Arbitrary cycle operating between TT Maddy points If there is an ideal efficiency for all cycles, is there a maximum work or maximum power for all cycles? (MP 1C.7) IC-7
1C-7 Tmax Tmin T s Figure C-2: Ideal Brayton cycle as composed of many elementary Carnot cycles [Kerrebrock] The temperature ratios T T lowi highi ( ) of any elementary cycle “i” are therefore the same and each of the elementary cycles has the same thermal efficiency. We only need to find the temperature ratio across any one of the cycles to find what the efficiency is. We know that the temperature ratio of the first elementary cycle is the ratio of compressor exit temperature to engine entry (atmospheric for an aircraft engine) temperature, T T 2 0 / in Figure C-2. If the efficiency of all the elementary cycles has this value, the efficiency of the overall Brayton cycle (which is composed of the elementary cycles) must also have this value. Thus, as previously, η Brayton inlet compressor exit T T = − 1 . A benefit of this view of efficiency is that it allows us a way to comment on the efficiency of any thermodynamic cycle. Consider the cycle shown on the right, which operates between some maximum and minimum temperatures. We can break it up into small Carnot cycles and evaluate the efficiency of each. It can be seen that the efficiency of any of the small cycles drawn will be less than the efficiency of a Carnot cycle between Tmax and Tmin . This graphical argument shows that the efficiency of any other thermodynamic cycle operating between these maximum and minimum temperatures has an efficiency less than that of a Carnot cycle. Muddy points If there is an ideal efficiency for all cycles, is there a maximum work or maximum power for all cycles? (MP 1C.7) Arbitrary cycle operating between T T min, max
1. C5 Irreversibility, Entropy Changes, and"Lost work" Consider a system in contact with a heat reservoir during a reversible process. If there is heat O absorbed by the reservoir at temperature T, the change in entropy of the reservoir is different temperatures. To analyze these, we can visualize a sequence of heat reservoirs a AS=Q/T In general, reversible processes are accompanied by heat exchanges that occu different temperatures so that during any infinitesimal portion of the cycle there will not be any heat transferred over a finite temperature difference During any infinitesimal portion, heat dOrey will be transferred between the system and one of the reservoirs which is at T. If d@ey is absorbed by the system, the entropy change of the system is The entropy change of the reservoir is The total entropy change of system plus surroundings is dslotal= ds system ds reservoir=0 This is also true if there is a quantity of heat rejected by the system The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e the entropy of the system plus the entropy of the surroundings: ASo(a=0 We now carry out the same type of analysis for an irreversible process, which takes the system between the same specified states as in the reversible process. This is shown schematically at the right, with I and r denoting the irreversible and reversible processes In the irreversible process, the system receives heat do and does work di The change in internal energy for the irreversible process is du=do-dw(always true- first law) For the reversible pr du= tds-dW Irreversible and reversible state changes Because the state change is the same in the two processes (we specified that it was), the change in internal energy is the same Equating the changes in internal energy in the above two expressions yields dQactual-dWactual=TdS-dw IC-8
1C-8 1.C.5 Irreversibility, Entropy Changes, and “Lost Work” Consider a system in contact with a heat reservoir during a reversible process. If there is heat Q absorbed by the reservoir at temperature T, the change in entropy of the reservoir is ∆S QT = / . In general, reversible processes are accompanied by heat exchanges that occur at different temperatures. To analyze these, we can visualize a sequence of heat reservoirs at different temperatures so that during any infinitesimal portion of the cycle there will not be any heat transferred over a finite temperature difference. During any infinitesimal portion, heat dQrev will be transferred between the system and one of the reservoirs which is at T. If dQrev is absorbed by the system, the entropy change of the system is dS dQ T system rev = . The entropy change of the reservoir is dS dQ T reservoir rev = − . The total entropy change of system plus surroundings is dS dS dS total system reservoir =+ = 0. This is also true if there is a quantity of heat rejected by the system. The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e., the entropy of the system plus the entropy of the surroundings: ∆Stotal = 0. We now carry out the same type of analysis for an irreversible process, which takes the system between the same specified states as in the reversible process. This is shown schematically at the right, with I and R denoting the irreversible and reversible processes. In the irreversible process, the system receives heat dQ and does work dW. The change in internal energy for the irreversible process is dU dQ dW = − (Always true - first law). For the reversible process dU TdS dW = − rev . Because the state change is the same in the two processes (we specified that it was), the change in internal energy is the same. Equating the changes in internal energy in the above two expressions yields dQ dW TdS dW actual actual − =− rev . Irreversible and reversible state changes
The subscript "actual"refers to the actual process(which is irreversible). The entropy change associated with the state change is ds dActhal dWrey-dwactual (C.5.1) If the process is not reversible, we obtain less work(see IA W notes) than in a reversible process, dWtol dWv, So that for the irreversible process T There is no equality between the entropy change ds and the quantity d@/Tfor an irreversible process. The equality is only applicable for a reversible process The change in entropy for any process that leads to a transformation between an initial state"a and a final state“b” is therefore e deactual is the heat exchanged in the actual process. The equality only applies to a reversible process The difference dwrey-dWactual represents work we could have obtained, but did not. It is referred o as lost work and denoted by Wost. In terms of this quantity we can write ds= de (C.5.3) The content of Equation(C.5.)is that the entropy of a system can be altered in two ways: (i) through heat exchange and (ii) through irreversibilities. The lost work( dost in equation C5.3)is always greater than zero, so the only way to decrease the entropy of a system is through heat transfer To apply the second law we consider the total entropy change( system plus surroundings). If the surroundings are a reservoir at temperature t, with which the system exchanges heat reservoir surroundings= ctual The total entropy change is 1C-9
1C-9 The subscript “actual” refers to the actual process (which is irreversible). The entropy change associated with the state change is dS dQ T T =+ − actual [ ] dW dW rev actual 1 . (C.5.1) If the process is not reversible, we obtain less work (see IAW notes) than in a reversible process, dW dW actual < rev , so that for the irreversible process, dS dQ T actual > . (C.5.2) There is no equality between the entropy change dS and the quantity dQ/T for an irreversible process. The equality is only applicable for a reversible process. The change in entropy for any process that leads to a transformation between an initial state “a” and a final state “b” is therefore ∆SS S dQ T b a actual a b =−≥ ∫ where dQactual is the heat exchanged in the actual process. The equality only applies to a reversible process. The difference dW dW rev − actual represents work we could have obtained, but did not. It is referred to as lost work and denoted by Wlost . In terms of this quantity we can write, dS dQ T dW T actual lost = + . (C.5.3) The content of Equation (C.5.3) is that the entropy of a system can be altered in two ways: (i) through heat exchange and (ii) through irreversibilities. The lost work ( dWlost in Equation C.5.3) is always greater than zero, so the only way to decrease the entropy of a system is through heat transfer. To apply the second law we consider the total entropy change (system plus surroundings). If the surroundings are a reservoir at temperature T, with which the system exchanges heat, dS dS dQ T reservoir surroundings actual ( ) = = − . The total entropy change is dS dS dS dQ T dW T dQ T total system surroundings actual lost actual =+ = + −
dsylotal dw 0 The quantity(dOst /T)is the entropy generated due to irreversibility Yet another way to state the distinction we are making is (C.5.4) The lost work is also called dissipation and noted do. Using this notation, the infinitesimal entropy change of the system becomes or tds=dQ2+dφ Equation(C.5. 4) can also be written as a rate equation Either of equation(C5.4)or(C5.5)can be interpreted to mean that the entropy of the system, S, is affected by two factors: the flow of heat O and the appearance of additional entropy, denoted by dSGen. due to irreversibility. This additional entropy is zero when the process is reversible and always positive when the process is irreversible Thus, one can say that the system develops sources which create entropy during an irreversible process. The second law asserts that sinks of entropy are impossible in nature, which is a more graphic way of saying that dSGen and SGen are positive definite, or zero, for reversible processes the system, can be interpreted as a flux of entropy. The boundary is crossed by heat and the ratio of this heat flux to temperature can be defined as a flux of entropy. There are no strictions on the sign of this quantity, and we can say that this flux either contributes towards, or drains away, the system's entropy. during a reversible process, only this flux can affect the entropy of the system. This terminology suggests that we interpret entropy as a kind of weightless fluid, whose quantity is conserved (like that of matter)during a reversible process. During an irreversible process, r this fluid is not cor cannot disappear, but rather is created by sources throughout the system. While this interpretation should not be taken too literally, it provides an easy mode of expression and is in the same category of concepts such as those associated with the phrases"flux of This and the following paragraph are excerpted with minor modifications from A Course Thermodynamics, Volume 1, by J Kestin, Hemisphere Press(1979)
1C-10 dS dW T total lost = ≥ 0 . The quantity ( dW T lost / ) is the entropy generated due to irreversibility. Yet another way to state the distinction we are making is dS dS dS dS dS system from heat transfer generated due to irreversible processes =+ = + heat transfer Gen . (C.5.4) The lost work is also called dissipation and noted dφ. Using this notation, the infinitesimal entropy change of the system becomes: dS dS d T system = + heat transfer φ or TdS dQ d system = +r φ Equation (C.5.4) can also be written as a rate equation, dS dt SS S == + heat transfer Gen ˙˙ ˙ . (C.5.5) Either of equation (C.5.4) or (C.5.5) can be interpreted to mean that the entropy of the system, S, is affected by two factors: the flow of heat Q and the appearance of additional entropy, denoted by dSGen, due to irreversibility1 . This additional entropy is zero when the process is reversible and always positive when the process is irreversible. Thus, one can say that the system develops sources which create entropy during an irreversible process. The second law asserts that sinks of entropy are impossible in nature, which is a more graphic way of saying that dSGen and ˙ SGen are positive definite, or zero, for reversible processes. The term ˙ , ˙ S T dQ dt Q T heat transfer = or 1 , which is associated with heat transfer to the system, can be interpreted as a flux of entropy. The boundary is crossed by heat and the ratio of this heat flux to temperature can be defined as a flux of entropy. There are no restrictions on the sign of this quantity, and we can say that this flux either contributes towards, or drains away, the system's entropy. During a reversible process, only this flux can affect the entropy of the system. This terminology suggests that we interpret entropy as a kind of weightless fluid, whose quantity is conserved (like that of matter) during a reversible process. During an irreversible process, however, this fluid is not conserved; it cannot disappear, but rather is created by sources throughout the system. While this interpretation should not be taken too literally, it provides an easy mode of expression and is in the same category of concepts such as those associated with the phrases "flux of 1 This and the following paragraph are excerpted with minor modifications from A Course in Thermodynamics, Volume I, by J. Kestin, Hemisphere Press (1979)