692 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-6.pm5

ηcompressor = 85%, ηturbine = 80%, ηcombustion = 85%.

For air and gases : cp = 1.024 kJ/kg K, γ = 1.4

Power developed by the plant, P = 1065 kW.

(i)The quantity of air circulation, ma :

For isentropic compression 1-2,

T

T

p

p

2

1

2

1

(^1) 14 1
5 14
1

F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.

. = 1.584

∴ T 2 = 293 × 1.584 = 464 K

Now, ηcompressor =

TT

TT

21

21

−

′−

i.e. 0.85 =

464 293

2 293

−

T′−

∴ T 2 ′ = 464 293

0.85

− + 293 = 494 K

For isentropic expansion process 3-4,

T

T

p

p

4

3

4

3

(^111)
1 1
49
= 0 635
F
HG
I
KJ
=FHG IKJ =
γ− −
γ
.
.
.4
.4
∴ T 4 = 953 × 0.635 = 605 K
Now, ηturbine =
TT
TT
34
34
−′
−
0.8 =
953
953 605
−′ 4
−
T
∴ T 4 ′ = 953 – 0.8(953 – 605) = 674.6 K
Wcompressor = cp (T 2 ′ – T 1 ) = 1.024 (494 – 293) = 205.8 kJ/kg
Wturbine = cp (T 3 – T 4 ′) = 1.024 (953 – 674.6) = 285.1 kJ/kg.
∴ Wnet = Wturbine – Wcompressor = 285.1 – 205.8 = 79.3 kJ/kg of air
If the mass of air flowing is ma kg/s, the power developed by the plant is given by
P = ma × Wnet kW
1065 = ma × 79.3
∴ ma =
1065
79.3
= 13.43 kg.
i.e., Quantity of air circulation = 13.43 kg. (Ans.)
(ii)Heat supplied per kg of air circulation :
Actual heat supplied per kg of air circulation

−′
= −
cT Tp() 32 024 953 494)(
ηcombustion 0

1. .85
   = 552.9 kJ/kg.
   (iii)Thermal efficiency of the cycle, ηthermal :
   ηthermal = Work output
   Heat supplied
   79.3
   552.9
   = = 0.1434 or 14.34%. (Ans.)