576 | Thermodynamics
Thus,andDiscussion This problem was worked out in Example 10–4 for the same pres-
sure and temperature limits with reheat but without the regeneration process.
A comparison of the two results reveals that the thermal efficiency of the cycle
has increased from 45.0 to 49.2 percent as a result of regeneration.
The thermal efficiency of this cycle could also be determined fromwhereAlso, if we assume that the feedwater leaves the closed FWH as a satu-
rated liquid at 15 MPa (and thus at T 5 342°C and h 5 1610.3 kJ/kg), it
can be shown that the thermal efficiency would be 50.6.10–7 ■ SECOND-LAW ANALYSIS
OF VAPOR POWER CYCLESThe ideal Carnot cycle is a totally reversible cycle,and thus it does not
involve any irreversibilities. The ideal Rankine cycles (simple, reheat, or
regenerative), however, are only internally reversible,and they may involve
irreversibilities external to the system, such as heat transfer through a finite
temperature difference. A second-law analysis of these cycles reveals where
the largest irreversibilities occur and what their magnitudes are.
Relations for exergy and exergy destruction for steady-flow systems are
developed in Chap. 8. The exergy destruction for a steady-flow system can
be expressed, in the rate form, as(10–18)or on a unit mass basis for a one-inlet, one-exit, steady-flow device asxdestT 0 sgenT 0 asesi (10–19)qout
Tb,outqin
Tb,inb¬¬ 1 kJ>kg 2
X#
destT 0 S#
genT 01 S#
outS#
in^2 T 0 aa
outm#sQ#
out
Tb,outa
inm#sQ#
in
Tb,inb¬¬ 1 kW 2
wpump,in 11 yz 2 wpump I,in 11 y 2 wpump II,in 1 y 2 wpump III,inwturb,out 1 h 9 h 102 11 y 21 h 11 h 122 11 yz 21 h 12 h 132hthwnet
qinwturb,outwpump,in
qinhth 1 qout
qin 1 1485.3 kJ>kg
2921.4 kJ>kg0.492 or 49.2%1485.3 kJ>kg 11 0.17660.1306 21 2335.7191.81 2 kJ>kgqout 11 yz 21 h 13 h 122921.4 kJ>kg 1 3583.11089.8 2 kJ>kg 11 0.1766 21 3674.93155.0 2 kJ>kgqin 1 h 9 h 82 11 y 21 h 11 h 102