Modern Engineering Thermodynamics

468 CHAPTER 13: Vapor and Gas Power Cycles
EXAMPLE 13.5 (Continued )
The thermodynamic states of the steam at the four monitoring stations shown in the equipment schematic are
Station 1
p 1 = 100: psia
T 1 = 500:°F
h 1 = 1279:1 Btu/lbm
s 1 = 1:7087 Btu/ðlbm . RÞ
Station 3
p 3 = 1:00 psia
x 3 = 0:00
Station 2s
Station 4s
p 4s = p 4 = 100: psia
s 4s = s 3 = 0:1326 Btu/ðlbm . RÞ
h 3 = h f = 69:7 Btu/lbm h 4s = h 3 + v 3 ðp 4 – p 3 Þ
s 3 = s f = 0:1326 Btu/ðlbm . RÞ
v 3 = v f ð1:00 psiaÞ = 0:01614 ft 3 =lbm
where we have calculated the following items:
= 69:7 + ð0:01614Þð100: – 1:00Þð144/778:16Þ
= 70:0 Btu/lbm
x 2s = s 2 – s f 2
/sfg2 = s 1 – s f 2
/sfg 2
= ð1:7087 – 0:1326Þ/1:8455 = 0:8540
h 2s = h f 2 + x 2s h fg 2 = 69:7 + ð0:8540Þð1036:0Þ
= 954:4Btu/lbm
a. The degree of superheat at the outlet of the boiler is determined from Table C.2a in Thermodynamic Tables to accompany
Modern Engineering Thermodynamics and Eq. (13.10) as
Degree of superheat = 500: – T sat ð100: psiaÞ
= 500: – 327:8 = 172°F
b. Here, the highest temperature in the cycle is T H = 500. + 459.67. = 960. R, and the lowest temperature in the cycle is
T L = T sat (1.00 psia) = 101.7 + 459.67 = 561.4 R. Then, Eq. (7.16) gives the Carnot cycle thermal efficiency as
ðη T Þ Carnot = 1 − T L
= 1 − 561:4 = 0:415 = 41:5%
T H 960:
c. Equation (13.9b) is used to determine the isentropic Rankine cycle thermal efficiency as
ðη T Þ isentropic
Rankine
= h 1 − h 2s − v 3 ðp 4 − p 3 Þ 1279:1 − 954:4 − ð0:01614Þð100: − 1Þð144/778:16Þ
=
h 1 − h 3 − v 3 ðp 4 − p 3 Þ 1279:1 − 69:7 − ð0:01614Þð100: − 1Þð144/778:16Þ
= 0:268 = 26:8%
Exercises
13. Determine the degree of superheat in Example 13.5 if the boiler outlet pressure were 120. psia instead of 100. psia.
Assume all the other variables remain unchanged. Answer: Degree of superheat = 159°F.
14. Find the equivalent Carnot cycle thermal efficiency of the engine in Example 13.5 if the condenser pressure were
2.00 psia instead of 1.00 psia. Assume all the other variables remain unchanged. Answer: (η T ) Carnot = 39.0%.
15. Determine the actual Rankine cycle thermal efficiency of the system described in Example 13.5 if the prime mover and
boiler feed pump isentropic efficiencies were 0.600 and 0.700, respectively. Assume all the other variables remain
unchanged. Answer: (η T ) Rankine = 16.1%.
Initially, the purpose of superheating the vapor was simply to eliminate or at least reduce the amount of
moisture in the engine’s low-pressure stages. However, at the higher boiler pressures and temperatures of
the 20th century, the effect of superheating the steam can add as many as 5 percentage points to the isentropic
Rankine cycle thermal efficiency.