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 = < H ; 0 ∆S Q T L Q L L = > 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