576 | Thermodynamics
Thus,
and
Discussion 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 from
where
Also, 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 CYCLES
The 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 as
xdestT 0 sgenT 0 asesi (10–19)
qout
Tb,out
qin
Tb,in
b¬¬ 1 kJ>kg 2
X
#
destT 0 S
#
genT 01 S
#
outS
#
in^2 T 0 aa
out
m#s
Q
#
out
Tb,out
a
in
m#s
Q
#
in
Tb,in
b¬¬ 1 kW 2
wpump,in 11 yz 2 wpump I,in 11 y 2 wpump II,in 1 y 2 wpump III,in
wturb,out 1 h 9 h 102 11 y 21 h 11 h 122 11 yz 21 h 12 h 132
hth
wnet
qin
wturb,outwpump,in
qin
hth 1
qout
qin
1
1485.3 kJ>kg
2921.4 kJ>kg
0.492 or 49.2%
1485.3 kJ>kg
11 0.17660.1306 21 2335.7191.81 2 kJ>kg
qout 11 yz 21 h 13 h 12
2921.4 kJ>kg
1 3583.11089.8 2 kJ>kg 11 0.1766 21 3674.93155.0 2 kJ>kg
qin 1 h 9 h 82 11 y 21 h 11 h 102