TITLE.PM5

GAS POWER CYCLES 687

dharm
\M-therm\Th13-6.pm5

For isentropic compression process 1-2 :

T

T

p

p

2

1

2

1

(^1) 14 1
62 14
1
=F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ.
.
.
= 1.684
∴ T 2 = 300 × 1.684 = 505.2 K
Now, ηcompressor = TT
TT
21
21
−
′−
0.88 =
505 2 300
2 300

. −
T′−

T 2 ′ =

505 2 300

088

. 300
.

F − +

HG

I

KJ

= 533.2 K

Heat supplied = (ma + mf) × cp(T 3 – T 2 ′) = mf × C

or^1 +

F

HG

I

KJ

m

m

f

a

× cp(T 3 – T 2 ′) =

m

m

f

a

× C

or (1 + 0.017) × 1.005(T 3 – 533.2) = 0.017 × 44186

∴ T 3 =

0.017 44186

(1 0.017) 1.005

×

+× + 533.2 = 1268 K

For isentropic expression process 3-4 :

T

T

p

p

4

3

4

3

(^1) 1 333 1
1 1 333
62
=F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.
.
.
= 0.634
∴ T 4 = 1268 × 0.634 = 803.9 K (Q γg=1 333. ......Given)
Now, ηturbine =
TT
TT
34
34
−′
−
0.9 =
1268
1268 803 9
−′ 4
−
T
.
∴ T 4 ′ = 1268 – 0.9(1268 – 803.9) = 850.3 K
Wcompresssor = cp(T 2 ′ – T 1 ) = 1.005(533.2 – 300) = 234.4 kJ/kg
Wturbine = cpg(T 3 – T 4 ′) = 1.147(1268 – 850.3) = 479.1 kJ/kg
Net work = Wturbine – Wcompressor
= 479.1 – 234.4 = 244.7 kJ/kg
Heat supplied per kg of air = 0.017 × 44186 = 751.2 kJ/kg
∴ Thermal efficiency, ηth. = Net work
Heat supplied
= 244.7
751.2
= 0.3257 or 32.57%. (Ans.)
T(K)
s
1268
(^3001)
2
2 ́
3
4 ́
4
6.2 bar
1 bar
Fig. 13.58