682 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-6.pm5

T

s

2

1

3

Compressor 4

Turbine

p = 1 bar 1

p = 4 bar^2

(ii)Power required :

T

T

4

3

= p

p

1

2

(^1) 136 1
1 136
5
F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.

. = 0.653

∴ T 4 = 1000 × 0.653 = 653 K
Power developed per kg of gas per second
= cp (T 3 – T 4 )
= 1.0425 (1000 – 653) = 361.7 kW. (Ans.)
Example 13.34. An isentropic air turbine is used to supply 0.1 kg/s of air at 0.1 MN/m^2
and at 285 K to a cabin. The pressure at inlet to the turbine is 0.4 MN/m^2. Determine the
temperature at turbine inlet and the power developed by the turbine. Assume cp= 1.0 kJ/kg K.
(GATE, 1999)
Solution. Given : m&a = 0.1 kg/s ; p 1 =
0.1 MN/m^2 = 1 bar, T 4 = 285 K ; p 2 = 0.4 MN/m^2 = 4 bar ;
cp = 1.0 kJ/kg K.
Temperature at turbine inlet, T 3 :

T

T

p

p

3

4

2

1

(^1) 14 1
4 14
1

F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.
.
= 1.486
∴ T 3 = 285 × 1.486 = 423.5 K. (Ans.)
Power developed, P :
P = mc T T&a p() 34 −
= 0.1 × 1.0 (423.5 – 285)
= 13.85 kW. (Ans.)
Example 13.35. Consider an air standard cycle in which the air enters the compressor at
1.0 bar and 20°C. The pressure of air leaving the compressor is 3.5 bar and the temperature at
turbine inlet is 600°C. Determine per kg of air :
(i)Efficiency of the cycle,(ii)Heat supplied to air,
(iii)Work available at the shaft,(iv)Heat rejected in the cooler, and
(v)Temperature of air leaving the turbine.
For air γ = 1.4 and cp = 1.005 kJ/kg K.
Solution. Refer Fig. 13.52.
Pressure of air entering the compressor, p 1 = 1.0 bar
Temperature at the inlet of compressor, T 1 = 20 + 273 = 293 K
Pressure of air leaving the compressor, p 2 = 3.5 bar
Temperature of air at turbine inlet, T 3 = 600 + 273 = 873 K
(i)Efficiency of the cycle, ηcycle :
ηcycle = 1 –^11
()rp
γ
γ
− = 1 –
1
35
14 1
(.)^14
.
.
− = 0.30 or 30%. (Ans.)^
Q r p
p p

F
HG
I
KJ
2
1
35
10

. 35
.
.

Fig. 13.54