Modern Engineering Thermodynamics

13.17 Aircraft Gas Turbine Engines 501
Similarly, the turbine’s (prime mover) isentropic efficiency is given by
ðη s Þ pm =
where, using the constant specific heats, we obtain
Then,
3a. The Brayton cold ASC thermal efficiency is given by
ð _W pm Þ actual
ð _W pm Þ isentropic
= T 1 − T 2
T 1 − T 2s
T 2s = T 1 ðp 2s /p 1 Þ ðk−1Þ/k = ð2060Þð28:0/190:
Þ ð1:40−1Þ/1:40 = 1190 R = 730:°F
ðη T Þ
Brayton
cold ASC
ðη s Þ pm
=
2060 − 1350
= 0:816 = 81:6%
2060 − 1190
ðconstant
specific heatsÞ
= T 1 − T 2s − ðT 4s − T 3 Þ
T 1 − T 4s
2060 − 1190 − 1100 − 520:
= ð
Þ = 0:302 = 30:2%
2060 − 1100
but the actual thermal efficiency of the engine, based on constant specific heats and the data provided in the schematic, is
ðη T Þ Brayton
ðactual, constant
specific heatsÞ
= T 1 − T 2 − ðT 4 − T 3 Þ
T 1 − T 4
2060 − 1350 − 1175 − 520:
= ð
Þ = 0:062 = 6:2%
2060 − 1175
2b. We can easily take into account the temperature-dependent specific heats by using Table C.16a in Thermodynamic Tables
to accompany Modern Engineering Thermodynamics. For the compressor, p r4 = ðp 4s /p 3 Þ = ð1:2147Þð200:/14:7Þ = 16:5 and,
by interpolation in Table C.16a, we find that T 4s = 1084 R = 624°F. Then,
Similarly, for the turbine,
ðη s Þ c
ðvariable
specific heatsÞ
= T 4s − T 3
=
1084 − 520:
= 0:861 = 86:1%
T 4 − T 3 1175 − 520:
p r2 = p r1 ðp 2s /p 1 Þ = ð196:16Þð28:0/190:Þ = 28:9
and, by interpolation in Table C.16a, we find that T 2s = 1261 R = 801°F. Then,
3b. Finally, the Brayton hot ASC can be easily determined from
where, from Table C.16a,
Then,
ðη s Þ pm
= T 1 − T 2 2060 − 1350
= = 0:889 = 88:9%
T 1 − T 2s 2060 − 1261
ðvariable
specific heatsÞ
ðη T Þ Brayton
= h 1 − h 2s − ðh 4s − h 3 Þ
h 1 − h 4s
hot ASC
h 3 = 124 Btu/lbm ðat 520:RÞ
h 4s = 262 Btu/lbm ðby interpolation at 1084 RÞ
h 1 = 521 Btu/lbm ðat 2060 RÞ
h 2s = 307 Btu/lbm ðby interpolation at 1261 RÞ
521 − 307 − ð262 − 124Þ
ðη T Þ Brayton
= = 0:293 = 29:3%
521 − 262
hot ASC
(Continued )