1. C Applications of the Second Law N-Chapter6;VWB&S-8.1,8.2,8.5,8.6,8.7,8.8,9.6] 1. CI Limitations on the work that Can be supplied by a heat engine The second law enables us to make powerful and general statements concerning the maximum work that can be Q1 derived from any heat engine which operates in a cycle. To illustrate these ideas, we use a Carnot cycle which is shown schematically at e right. The engine operates between two heat Carnot cycle r reservoirs, exchanging heat QH with the high temperature reservoir at Ti and e, with the reservoir at TL. The entropy changes of the two reservoirs are AS,= Q Q The same heat exchanges apply to the system, but with opposite signs; the heat received from the high temperature source is positive, and conversely. Denoting the heat transferred to the engines by subscript“e”, The total entropy change during any operation of the engine is, ASL+△S Re servo at TH For a cyclic process, the third of these(As ) is zero, and thus(remembering that QH <0), atotal=As, AS,= QH 2r (C.1.1) For the engine we can write the first law as AUe=0(cyclic process)=QHe +OLe -We W Q1 Hence, using(C1.1) We=-QH-7i4Slotal eH\TH IC-1
1C-1 TH QH We QL TL Carnot cycle 1.C Applications of the Second Law [VN-Chapter 6; VWB&S-8.1, 8.2, 8.5, 8.6, 8.7, 8.8, 9.6] 1.C.1 Limitations on the Work that Can be Supplied by a Heat Engine The second law enables us to make powerful and general statements concerning the maximum work that can be derived from any heat engine which operates in a cycle. To illustrate these ideas, we use a Carnot cycle which is shown schematically at the right. The engine operates between two heat reservoirs, exchanging heat QH with the high temperature reservoir at TH and QLwith the reservoir at TL.. The entropy changes of the two reservoirs are: ∆S Q T H Q H H = L ; 0 The same heat exchanges apply to the system, but with opposite signs; the heat received from the high temperature source is positive, and conversely. Denoting the heat transferred to the engines by subscript “e”, Q QQ Q He =− =− H Le L ; . The total entropy change during any operation of the engine is, ∆ ∆ ∆∆ S S SS total H servoir at TH L servoir at TL e Engine =++ Re Re { { { For a cyclic process, the third of these ( ) ∆Se is zero, and thus (remembering that QH < 0), ∆ ∆∆ S SS Q T Q T total H L H H L L = + =+ (C.1.1) For the engine we can write the first law as ∆U Q QW e He Le = = +− e 0 (cyclic process) . Or, WQ Q e He Le = + = − − Q Q H L . Hence, using (C.1.1) W Q TS Q T T e HL total H L H =− − + ∆
(-Q T1△ The work of the engine can be expressed in terms of the heat received by the engine as we=2, The upper limit of work that can be done occurs during a reversible cycle, for which the total entropy change(ASa)is zero. In this situation Maximum work for an engine working between TH and TL: We=(CHe)\TH L Also, for a reversible cycle of the engine, 0 TH TL These constraints apply to all reversible heat engines operating between fixed temperatures. The thermal efficiency of the engine is Work done w Heat received Q The Carnot efficiency is thus the maximum efficiency that can occur in an engine working between two given temperatures We can approach this last point in another way. The engine work is given by We=-QH-TLAStota+QH (T/TH) total OH+QH(TL/TH)-We The total entropy change can be written in terms of the Carnot cycle efficiency and the ratio of the work done to the heat absorbed by the engine. The latter is the efficiency of any cycle we can devi total Q carnot- nAny other The second law says that the total entropy change is equal to or greater than zero. This means that the Carnot cycle efficiency is equal to or greater than the efficiency for any other cycle, with the equality only occurring if AS=0 1C-2
1C-2 = −( ) − Q − T T H T S L H L total 1 ∆ . The work of the engine can be expressed in terms of the heat received by the engine as W Q T T e H T S e L H L total = ( ) − 1 − ∆ . The upper limit of work that can be done occurs during a reversible cycle, for which the total entropy change ( ∆Stotal) is zero. In this situation: Maximum work for an engine working between T T H L and : W Q T T e He L H = ( ) − 1 Also, for a reversible cycle of the engine, Q T Q T H H L L + = 0. These constraints apply to all reversible heat engines operating between fixed temperatures. The thermal efficiency of the engine is η = = Work done Heat received W QHe =− = 1 T T L H ηCarnot . The Carnot efficiency is thus the maximum efficiency that can occur in an engine working between two given temperatures. We can approach this last point in another way. The engine work is given by W Q TS QTT e HL total =− − + ∆ HL H ( ) / or, TS Q QTT W L total ∆ =− + H HL H e ( ) / − The total entropy change can be written in terms of the Carnot cycle efficiency and the ratio of the work done to the heat absorbed by the engine. The latter is the efficiency of any cycle we can devise: ∆S Q T T T W Q Q T total He L L H e He He L Carnot Any other cycle = −− = − 1 η η . The second law says that the total entropy change is equal to or greater than zero. This means that the Carnot cycle efficiency is equal to or greater than the efficiency for any other cycle, with the equality only occurring if ∆Stotal = 0
Muddy b。in So, do we lose the capability to do work when we have an irreversible process and Why do we study cycles starting with the Carnot cycle? Is it because i is easier to work with?(MP 1C.2 1. C. 2 The Thermodynamic Temperature scale The considerations of Carnot cycles in this section have not mentioned the working medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium Because we derived the Carnot efficiency with an ideal gas as a medium, the temperature definition used in the ideal gas equation is not essential to the thermodynamic arguments. More specifically, we can define a thermodynamic temperature scale that is independent of the working medium. To see this, consider the situation shown below in Figure C-1, which has three reversible cycles. There is a high temperature heat reservoir at T, and a low temperature heat reservoir at T For any two temperatures T, T,, the ratio of the magnitudes of the heat absorbed and rejected in a Carnot cycle has the same value for all systems JT2 Q Q Figure C-1: Arrangement of heat engines to demonstrate the thermodynamic temperature scale We choose the cycles so Q i is the same for A and C. also Q3 is the same for B and c. for a Carnot cycle QL FITL TH; n is only a fune Q2/Q2=F(7,2 AT e/03=F(T,T) But 1C-3
1C-3 Muddy points So, do we lose the capability to do work when we have an irreversible process and entropy increases? (MP 1C.1) Why do we study cycles starting with the Carnot cycle? Is it because I is easier to work with? (MP 1C.2) 1.C.2 The Thermodynamic Temperature Scale The considerations of Carnot cycles in this section have not mentioned the working medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium. Because we derived the Carnot efficiency with an ideal gas as a medium, the temperature definition used in the ideal gas equation is not essential to the thermodynamic arguments. More specifically, we can define a thermodynamic temperature scale that is independent of the working medium. To see this, consider the situation shown below in Figure C-1, which has three reversible cycles. There is a high temperature heat reservoir at T3 and a low temperature heat reservoir at T1. For any two temperatures T T1 2 , , the ratio of the magnitudes of the heat absorbed and rejected in a Carnot cycle has the same value for all systems. WA WC A B C Q1 Q3 Q3 Q1 Q2 T2 T1 T3 Q2 WB Figure C-1: Arrangement of heat engines to demonstrate the thermodynamic temperature scale We choose the cycles soQ1 is the same for A and C. Also Q3 is the same for B and C. For a Carnot cycle η =+ = 1 ( ) Q Q FT T L H L H , ; η is only a function of temperature. Also Q Q FT T 1 2 12 = ( ) , Q Q FT T 2 3 23 = ( ) , Q Q FT T 1 3 13 = ( ) , . But
ee 22 23 2 Hence F(71,Ty)=F(T,T2)×F(T2,r) Not a function Cannot be a function of 72 We thus conclude that F(T, T)has the form f(T)/f(T2),and similarly F(T,T)=f(T)/f(T). The ratio of the heat exchanged is therefore g-n(,)- f(T3 In general OH f(TH er f( so that the ratio of the heat exchanged is a function of the temperature. We could choose any function that is monotonic, and one choice is the simplest: f(T)=T. This is the thermodynamic scale of temperature, QH/Q=TH/T. The temperature defined in this manner is the same as that for the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent 1. C3 Representation of Thermodynamic Processes in T-s coordinates. It is often useful to plot the thermodynamic state transitions and the cycles in terms of temperature(or enthalpy)and entropy, T, S, rather than P, The maximum temperature is often the constraint on the process and the enthalpy changes show the work done or heat received directly, so that plotting in terms of these variables provides insight into the process. A Carnot cycle is shown below in these coordinates, in which it is a rectangle, with two horizontal, constant temperature legs. The other two legs are reversible and adiabatic, hence isentropic (ds= dQ /T=0), and therefore vertical in T-s coordinates Isothermal Adiabatic 几L Carnot cycle in T,s coordinates If the cycle is traversed clockwise, the heat added is 1C-4
1C-4 Q Q Q Q Q Q 1 3 1 2 2 3 = . Hence FT T FT T FT T T T 1 3 2 12 23 2 ( ) , ,, = ( ) × ( ) Not a function of Cannot be a function of 1 24 341 2 444 3 444 . We thus conclude that FT T1 2 ( ) , has the form fT fT () () 1 2 / , and similarly FT T f T f T 23 2 3 ( ) , / = ( ) ( ). The ratio of the heat exchanged is therefore Q Q FT T f T f T 1 3 1 3 1 3 = ( ) = ( ) ( ) , . In general, Q Q f T f T H L H L = ( ) ( ) , so that the ratio of the heat exchanged is a function of the temperature. We could choose any function that is monotonic, and one choice is the simplest: fT T ( ) = . This is the thermodynamic scale of temperature, QQ TT H L HL = . The temperature defined in this manner is the same as that for the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent 1.C.3 Representation of Thermodynamic Processes in T-s coordinates. It is often useful to plot the thermodynamic state transitions and the cycles in terms of temperature (or enthalpy) and entropy, T,S, rather than P,V. The maximum temperature is often the constraint on the process and the enthalpy changes show the work done or heat received directly, so that plotting in terms of these variables provides insight into the process. A Carnot cycle is shown below in these coordinates, in which it is a rectangle, with two horizontal, constant temperature legs. The other two legs are reversible and adiabatic, hence isentropic ( dS dQ T = rev / = 0), and therefore vertical in T-s coordinates. T TH TL Isothermal Adiabatic s Carnot cycle in T,s coordinates If the cycle is traversed clockwise, the heat added is a c b d
Heat added: QH=fTaS=TH(S-Sa)=THAS The heat rejected (from c to d) has magnitude eL=TAS The work done by the cycle can be found using the first law for a reversible process dU=do-dw Tas-dw This form is only true for a reversible process) We can integrate this last expression around the closed path traced out by the cycle dU=∮Tas-fcW However du is an exact differential and its integral around a closed contour is zero: 0=∮TdS-∮dW The work done by the cycle, which is represented by the term fdw, is equal to fTds,the area enclosed by the closed contour in the T-s plane. This area represents the difference between the heat absorbed (fTds at the high temperature)and the heat rejected (fTds at the low temperature) Finding the work done through evaluation of fTdsis an alternative to computation of the work in a reversible cycle from fPdv. Finally, although we have carried out the discussion in terms of the entropy, S, all of the arguments carry over to the specific entropy, S; the work of the reversible cycle per unit mass is given by fTds Muddy points How does one interpret h-s diagrams? (MP 1C.3) Is it always oK to"switch"T-s and h-s diagram?(MP 1C. 4) What is the best way to become comfortable with T-s diagrams?(MP 1C.5 What is a reversible adiabat physically?(MP 1C.6) 1. C 4 Brayton Cycle in T-s Coordinates The Brayton cycle has two reversible adiabatic (i.e, isentropic)legs and two reversible constant pressure heat exchange legs. The former are vertical, but we need to define the shape of the latter. For an ideal gas, changes in specific enthalpy are related to changes in temperature by dh=cpdr, so the shape of the cycle in an h-s plane is the same as in a T-s plane, with a scale factor of c, between the two. This suggests that a place to start is with the combined first and second law, which relates changes in enthalpy, entropy, and pressure dh= tds dp P On constant pressure curves dP=0 and dh= Tds. The quantity desired is the derivative of temperature, T,with respect to entropy, s, at constant pressure: (dT/as),. From the combined first and second law and the relation between dh and dt. this is 1C-5
1C-5 Heat added: QH TdS T S S T S a b = ∫ = − Hb a H ( ) = ∆ . The heat rejected (from c to d) has magnitude Q TS L L = ∆ . The work done by the cycle can be found using the first law for a reversible process: dU dQ dW = − . = − TdS dW (This form is only true for a reversible process). We can integrate this last expression around the closed path traced out by the cycle: ∫ dU TdS dW = − ∫ ∫ However dU is an exact differential and its integral around a closed contour is zero: 0 = − ∫ TdS dW∫ . The work done by the cycle, which is represented by the term ∫ dW , is equal to ∫ Tds, the area enclosed by the closed contour in the T-S plane. This area represents the difference between the heat absorbed (∫ TdS at the high temperature) and the heat rejected (∫ TdS at the low temperature). Finding the work done through evaluation of ∫ TdSis an alternative to computation of the work in a reversible cycle from ∫ PdV. Finally, although we have carried out the discussion in terms of the entropy, S, all of the arguments carry over to the specific entropy, s; the work of the reversible cycle per unit mass is given by Tds. ∫ Muddy points How does one interpret h-s diagrams? (MP 1C.3) Is it always OK to "switch" T-s and h-s diagram? (MP 1C.4) What is the best way to become comfortable with T-s diagrams? (MP 1C.5) What is a reversible adiabat physically? (MP 1C.6) 1.C.4 Brayton Cycle in T-s Coordinates The Brayton cycle has two reversible adiabatic (i.e., isentropic) legs and two reversible, constant pressure heat exchange legs. The former are vertical, but we need to define the shape of the latter. For an ideal gas, changes in specific enthalpy are related to changes in temperature by dh c dT = p , so the shape of the cycle in an h-s plane is the same as in a T-s plane, with a scale factor of cp between the two. This suggests that a place to start is with the combined first and second law, which relates changes in enthalpy, entropy, and pressure: dh Tds dp = + ρ . On constant pressure curves dP=0 and dh Tds = . The quantity desired is the derivative of temperature, T, with respect to entropy, s, at constant pressure: ∂ ∂ T s p ( ) . From the combined first and second law, and the relation between dh and dT, this is
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 . (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)
