Page 423 - Advanced thermodynamics for engineers

17.3 AIRCRAFT GAS TURBINES 413

and hence
1=2 1=2
V 8 ¼ 2c p T 02 T 8 ¼ð2 1:005 1000 42:8Þ ¼ 293 m=s:
Since the bypass ratio (b) is 3.0,
_ m t 115 3:0
_ m b ¼ ¼ ¼ 86:25 kg=s
b þ 1 4:0
F b ¼ 86:25 293 ¼ 25; 300 N:

Considering the work requirement of the HP rotor,
c pa 1:005 396:3
T 04 T 05 ¼ ðT 03 T 02 Þ¼ ¼ 350:3K
h c pg 0:99 1:147
m
and for the LP rotor
4:0 1:005 49:7
c p a
T 05 T 06 ¼ðb þ 1Þ ðT 02 T 01 Þ¼ ¼ 176 K:
m
h c p g 0:99 1:147
Hence
T 05 ¼ T 04 ðT 04 T 05 Þ¼ 1300 350:3 ¼ 949:7K
T 06 ¼ T 05 ðT 05 T 06 Þ¼ 949:7 176 ¼ 773:7K
p 06 may then be found as follows.

n=ðn 1Þ
1=0:225
p 04 T 04 1300
¼ ¼ ¼ 4:02
p 05 T 05 949:7
n=ðn 1Þ
1=0:225
p 05 T 05 949:7
¼ ¼ ¼ 2:48
p 06 T 06 773:7
p 04 ¼ p 03 Dp cc ¼ 19:0 1:0 1:25 ¼ 17:75 bar
p 04 17:75
p 06 ¼ ¼ ¼ 1:78 bar
ðp 04 =p 05 Þðp 05 =p 06 Þ 4:02 2:48
Thus the hot nozzle pressure ratio is
p 06
¼ 1:78
p a
while the critical pressure ratio is
p 06 1
¼ ¼ 1:914:
p c 1 0:333 4

1
0:95 2:333
This nozzle is also unchoked, and hence p 7 ¼ p a .
" # " #
ðk 1Þ=k 1=4
1 1
T 06 T 7 ¼ h T 06 1 ¼ 0:95 773:7 1 ¼ 98:5K
j
p 06 p a 1:78