GAS POWER CYCLES 693

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\M-therm\Th13-6.pm5

Example 13.44. In a gas turbine the compressor is driven by the high pressure turbine.
The exhaust from the high pressure turbine goes to a free low pressure turbine which runs the
load. The air flow rate is 20 kg/s and the minimum and maximum temperatures are respectively
300 K and 1000 K. The compressor pressure ratio is 4. Calculate the pressure ratio of the low
pressure turbine and the temperature of exhaust gases from the unit. The compressor and turbine
are isentropic. Cp of air and exhaust gases = 1 kJ/kg K and γ = 1.4. (GATE, 1995)

Solution. Given : m&a = 20 kg/s ; T 1 = 300 K ; T 3 = 1000 K,

p

p

2

1

= 4 ; cp = 1 kJ/kg K ; γ = 1.4.

Pressure ratio of low pressure turbine, p

p

4

5

:
Since the compressor is driven by high pres-
sure turbine,

∴

T

T

p

p

2

1

2

1

(^10)
=F 4)^1
HG
I
KJ

−γ
γ
(
.4
.4 = 1.486
or T 2 = 300 × 1.486 = 445.8 K
Also, mc&ap(T 2 – T 1 ) = m&acp (T 3 – T 4 )
(neglecting mass of fuel)
or T 2 – T 1 = T 3 – T 4
445.8 – 300 = 1000 – T 4 ,orT 4 = 854.2 K
For process 3-4 :
T
T
p
p
3
4
3
4
1
=F
HG
I
KJ
−γ
γ
or p
p
T
T
3
4
3
4
1
0
=F
HG
I
KJ
.4
.4
or
p
p
3
4
1000 35
854 2
=FHG IKJ
.
.
= 1.736
Now, p
p
p
p
p
p
3
4
3
5
5
4
=× = 4 ×
p
p
5
4
Q p
p
p
p
3
5
2
1
== 4
F
HG
I
KJ
∴
p
p
p
p
5
4
3
4
1
4
1
4

F
HG
I
KJ
= × 1.736 = 0.434
Hence pressure ratio of low pressure turbine =
p
p
4
5
1
0 434

. = 2.3. (Ans.)
Temperature of the exhaust from the unit, T 5 :

T

T

p

p

4

5

4

5

(^111)
=F 23 1
HG
I
KJ

γ− −
γ
(.)
.4
.4 = 1.269
∴ T 5 ==
T 4
1 269
854 2

. 1 269

.

.

= 673 K. (Ans.)

+Example 13.45. Air is drawn in a gas turbine unit at 15°C and 1.01 bar and pressure

ratio is 7 : 1. The compressor is driven by the H.P. turbine and L.P. turbine drives a separate

T(K)

s

1000

300

3

4

5

2

1

Fig. 13.62