Thermodynamics

576 | ThermodynamicsThus,andq in 1h 9 h 8 2 11 y2 1h 11 h 10 2Discussion This problem was worked out in Example 10–4 for the same pressureand temperature limits with reheat but without the regeneration process.A comparison of the two results reveals that the thermal efficiency of the cyclehas increased from 45.0 to 49.2 percent as a result of regeneration.The thermal efficiency of this cycle could also be determined fromwhere 13583.1 1089.82 kJ>kg 11 0.1766213674.9 3155.02 kJ>kg 2921.4 kJ>kgq out 11 y z21h 13 h 1 2 11 0.1766 0.13062 12335.7 191.812 kJ>kg 1485.3 kJ>kgh th 1 q outq in1485.3 kJ>kg 1 0.492 or 49.2%2921.4 kJ>kgh th w netq in w turb,out w pump,inq inw turb,out 1h 9 h 10 2 11 y21h 11 h 12 2 11 y z2 1h 12 h 13 2w pump,in 11 y z2w pump I,in 11 y2w pump II,in 1y2w pump III,inAlso, if we assume that the feedwater leaves the closed FWH as a saturatedliquid at 15 MPa (and thus at T 5 342°C and h 5 1610.3 kJ/kg), itcan be shown that the thermal efficiency would be 50.6.10–7 ■ SECOND-LAW ANALYSISOF VAPOR POWER CYCLESThe ideal Carnot cycle is a totally reversible cycle, and thus it does notinvolve any irreversibilities. The ideal Rankine cycles (simple, reheat, orregenerative), however, are only internally reversible, and they may involveirreversibilities external to the system, such as heat transfer through a finitetemperature difference. A second-law analysis of these cycles reveals wherethe largest irreversibilities occur and what their magnitudes are.Relations for exergy and exergy destruction for steady-flow systems aredeveloped in Chap. 8. The exergy destruction for a steady-flow system canbe expressed, in the rate form, asX # dest T 0 S # gen T 0 1S # out S # in2 T 0 a aoutm # s Q# outT b,out ainm # s Q# inT b,inb1kW2or on a unit mass basis for a one-inlet, one-exit, steady-flow device as(10–18)x dest T 0 s gen T 0 a s e s i q outT b,out q inT b,inb1kJ>kg2(10–19)