Thermodynamics

508 | ThermodynamicsFuelCombustionchamberq in23HeatexchangerCompressorTurbinew net231FreshairExhaustgases4CompressorTurbinew netFIGURE 9–29An open-cycle gas-turbine engine.1Heatexchanger4q outFIGURE 9–30A closed-cycle gas-turbine engine.T21q inP = const.P = const.34q out3-4 Isentropic expansion (in a turbine)4-1 Constant-pressure heat rejectionThe T-s and P-v diagrams of an ideal Brayton cycle are shown in Fig. 9–31.Notice that all four processes of the Brayton cycle are executed in steadyflowdevices; thus, they should be analyzed as steady-flow processes. Whenthe changes in kinetic and potential energies are neglected, the energy balancefor a steady-flow process can be expressed, on a unit–mass basis, as1q in q out 2 1w in w out 2 h exit h inlet(9–15)Therefore, heat transfers to and from the working fluid are(a) T-s diagramsq in h 3 h 2 c p 1T 3 T 2 2(9–16a)andPq out h 4 h 1 c p 1T 4 T 1 2(9–16b)2q in3Then the thermal efficiency of the ideal Brayton cycle under the cold-airstandardassumptions becomess = const.h th,Brayton w netq in 1 q outq 1 c p 1T 4 T 1 2in c p 1T 3 T 2 2 1 T 1 1T 4 >T 1 12T 2 1T 3 >T 2 12s = const.1 4q out(b) P-v diagramFIGURE 9–31T-s and P-v diagrams for the idealBrayton cycle.vProcesses 1-2 and 3-4 are isentropic, and P 2 P 3 and P 4 P 1 . Thus,T 2 a P 1k12>k2b a P 1k12>k3b T 3T 1 P 1 P 4 T 4Substituting these equations into the thermal efficiency relation and simplifyinggiveh th,Brayton 1 1r 1k12>kp(9–17)