TITLE.PM5

686 ENGINEERING THERMODYNAMICS

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T (K)

1148

300

3

4 ′

4

1

2

2 ′

4bar

1bar

s

Fig. 13.57

∴ T 2 = 300 × 1.486 = 445.8 K

ηcompressor =

TT

TT

21

21

−

′−

or 0.8 =
445 8 300
2 300

. −
T′−

or T 2 ′ =
445 8 300
08

.

.

−

+ 300 = 482.2 K

Now, heat supplied by the fuel = heat taken by the burning gases

0.9 × mf × C = ()()mm c TTa+××−′f p 32

∴ C =

mm

m

a f

f

F +

H

G

I

K

J ×

cT T m

m

p a cT T

f

() p

.

()

.

32 32

09

1

09

−′

=+

F

H

G

I

K

J×

−′

or 42000 =
m
m

a

f

+

F

HG

I

KJ

(^1) × 1.
.9
00 1148 482.27
0
()−
= 739.78
m
m
a
f

• F
  HG
  I
  KJ
  1
  ∴ m
  m
  a
  f
  =^42000
  739 78.

• 1 = 55.77 say 56

∴ A/F ratio = 56 : 1. (Ans.)
Example 13.39. A gas turbine unit receives air at 1 bar and 300 K and compresses it
adiabatically to 6.2 bar. The compressor efficiency is 88%. The fuel has a heating valve of 44186
kJ/kg and the fuel-air ratio is 0.017 kJ/kg of air. The turbine internal efficiency is 90%. Calculate
the work of turbine and compressor per kg of air compressed and thermal efficiency.
For products of combustion, cp = 1.147 kJ/kg K and γ = 1.333. (UPSC, 1992)
Solution. Given : p 1 (= p 4 ) = 1 bar, T 1 = 300 K ; p 2 (= p 3 ) = 6.2 bar ; ηcompressor = 88% ;
C = 44186 kJ/kg ; Fuel-air ratio = 0.017 kJ/kg of air,
ηturbine = 90% ; cp = 1.147 kJ/kg K ; γ = 1.333.