TITLE.PM5

688 ENGINEERING THERMODYNAMICS

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\M-therm\Th13-6.pm5

Example 13.40. The air enters the compressor of an open cycle constant pressure gas
turbine at a pressure of 1 bar and temperature of 20°C. The pressure of the air after compression
is 4 bar. The isentropic efficiencies of compressor and turbine are 80% and 85% respectively. The
air-fuel ratio used is 90 : 1. If flow rate of air is 3.0 kg/s, find :
(i)Power developed.
(ii)Thermal efficiency of the cycle.
Assume cp = 1.0 kJ/kg K and γ = 1.4 for air and gases.
Calorific value of fuel = 41800 kJ/kg.
Solution. Given :p 1 = 1 bar ; T 1 = 20 + 273 = 293 K
p 2 = 4 bar ; ηcompressor = 80% ; ηturbine = 85%
Air-fuel ratio = 90 : 1 ; Air flow rate, ma = 3.0 kg/s
(i)Power developed, P :
Refer to Fig. 13.59 (b)

T

T

2

1

=

p

p

2

1

(^1) 14 1
4 14
1
F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.
.
= 1.486
∴ T 2 = (20 + 273) × 1.486 = 435.4 K
ηcompressor =
TT
TT
21
21
−
′−
0.8 =
435 4 293
2 293

. −
T′−

∴ T 2 ′ =

435 4 293

08

.

.

−

+ 293 = 471 K

T(K)

293

(b)

1

2 2 ′

4 ′

4

3

s

p = 1 bar^1

2 ′ (^3) p = 4 bar 2
T Work
1 4 ′
C
C.C.
T = 20 + 273
= 293 K
1
(a)
Fig. 13.59
Heat supplied by fuel = Heat taken by burning gases
mf × C = (ma + mf) cp(T 3 – T 2 ′)
(where ma = mass of air, mf = mass of fuel)