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-51C-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