State 2: 250 °C, 3976.2 kPa
Saturated vapor enthalpy h g
À Á
2
¼ 2801:0 kJ/kg
Saturated vapor entropy s g
À Á
2
¼ 6:0721 kJ/kg Á K
The isentropic condition of 2 ! 3 determines s 3 ¼ s 2 ¼ 6:0721 kJ/kg Á K.
Note from Table 5.2: s g 20 kPaÞ ¼ s f þ s fg ¼ 0:8320 þ 7:0752 ¼ 7:9073
À
Therefore,
s 3 ¼ s f þ x 3 s fg ; i:e:; 6:0721 ¼ 0:8320 þ x 3 Â 7:0752 yields;
Quality at state 3 is 0.74063.
It follows
State 3: 60.06 °C, 20 kPa (see Table 5.2)
x 3 ¼ 0:74063
h 3 ¼ h f þ x 3 Â h fg ¼ 251:42 þ 0:74063 Â 2357:5 ¼ 1997:45 kJ/kg
(see Table A-5 for the above enthalpy values)
Use the entropy value at state 1 and the same procedure of determining
state 3, i.e.,
2:7933 ¼ 0:8320 þ x 4 Â 7:0752
to find the properties at state 4.
State 4: 60.06 °C, 20 kPa (see Table 5.2)
x 4 ¼ 0:27721
h 4 ¼ h f þ x 4 Â h fg ¼ 251:42 þ 0:27721 Â 2608:9 ¼ 904:937 kJ/kg
Answers to (b), (c), and (a): Heat and work are related to enthalpy changes
for steady-flow processes:
(b) Heat rejected = h 3 À h 4 ¼ 1092:5 kJ/kg each cycle
(c) Net work output = h 2 À h 3
ð
ÞÀ h 1 À h 4
ð
Þ¼622:787 kJ=kg
(a) g th ¼
NetWorkOutput
h 2 Àh 1
¼
622:787
1715:3 ¼ 0:363
Which of course checks with the Kelvin formula
g th ¼ 1 À
T C
T H
¼ 1 À
60:06 þ 273:15
250 þ 273:15
¼ 0:363
114
5 Entropy and the Entropy Principle