Page 418 - Advanced thermodynamics for engineers

408 CHAPTER 17 GAS TURBINES

The critical pressure ratio, from Eqn (17.74),is
p 04 1 1
¼ ¼ n o ¼ 1:918
4
p c 1 k 1 k=ðk 1Þ 1 0:333

1 1 0:95 2:333
h j k þ 1
Since p 04 /p a > p 04 /p c the nozzle is choked.
2 2 958:6
T 5 ¼ T c ¼ T 04 ¼ ¼ 821:8K
k þ 1 2:333
1 2:501

p 5 ¼ p c ¼ p 04 ¼ ¼ 1:304 bar
p 04 p c 1:918
p c 100 1:304 3
5
r ¼ ¼ ¼ 0:5529 kg m
RT c 0:287 821:8
1=2 1=2
V 5 ¼ðkRT c Þ ¼ð1:333 0:287 821:8 1000Þ ¼ 560:7m=s:
A 5 1 1 2
¼ ¼ ¼ 0:003226 m s kg:
_ m r V 5 0:5529 560:7
5
The specific thrust is
A 5
F s ¼ðV 5 V a Þþ p c p a
_ m

¼ 560:7 270 þ 0:003226 1:304 0:5404 10 5

¼ 290:7 þ 246:3 ¼ 537:0Ns=kg:
Temperature rise in engine is
T 03 T 02 ¼ 1200 5647:7 ¼ 635:3K:
Hence, using

_ m f Q 0
p
DT ¼
ð _ m a þ _ m f Þc p g
Q 0
p
¼
ðε þ 1Þc p g
gives an air–fuel ratio,
Q 0
p
ε ¼ 1 ¼ 59:38:
DT
c p g
This results in a sfc of
1 3600
¼ ¼ 0:1129 kg=hN:
εF s 59:38 537