GAS POWER CYCLES 645

dharm
\M-therm\Th13-3.pm5

∴ T 2 = T 1 × 2.408 = 303 × 2.408 = 729.6 K

For the constant volume process 2-3,

T

p

T

p

3

3

2

2

=

∴ T 3 = T 2. p

p

3

2

= 729.6 ×

60

21.67

= 2020 K

Also,

ρ−

−

(^1) =
1
4
r 100 or 0.04
∴
ρ−
−
1
91 = 0.04 or ρ = 1.32
For the constant pressure process 3-4,
V
T
V
T
4
4
3
3
= or T
T
V
V
4
3
4
3
= = ρ
∴ T 4 = T 3 × ρ = 2020 × 1.32 = 2666.4 K
Also expansion ratio,
V
V
V
V
V
V
V
V
V
V
5 r
4
5
2
2
4
1
2
3
4
=×=×=
ρ
[and]QVV VV 51 ==2 3
For adiabatic process 4-5,
T
T
V
Vr
5
4
4
5
(^11)

F
HG
I
KJ
=F
HG
I
KJ
γ− γ−
ρ
∴ T 5 = T 4 × ρ
γ
r
F
HG
I
KJ
− 1
= 2666.4 ×
132
9

. 14 1.
F
HG

I

KJ

−

= 1237 K

Also p 4 V 4 γ = p 5 V 5 γ

p 5 = p 4. V

V

4

5

F

HG

I

KJ

γ

= 60 ×

r

ρ

F γ

HG

I

KJ = 60 ×

132

9

F.^14.

HG

I

KJ = 4.08 bar

Heat supplied, Qs = cv(T 3 – T 2 ) + cp(T 4 – T 3 )

= 0.71 (2020 – 729.6) + 1.0 (2666.4 – 2020) = 1562.58 kJ/kg

Heat rejected, Qr = cv (T 5 – T 1 )

= 0.71 (1237 – 303) = 663.14 kJ/kg

ηair standard-

1562.58

=QQ− = −

Q

sr

s

1562.85 663.14

= 0.5756 or 57.56%. (Ans.)

(ii)Power developed by the engine, P :

Mass of air in the cycle is given by

m =

pV

RT

11

1

= 1 10 0 0165

287 303

××^5

×

. = 0.0189 kg

∴ Work done per cycle = m(Qs – Qr)

= 0.0189 (1562.58 – 663.14) = 16.999 kJ

Power developed = Work done per cycle × no. of cycles per second

= 16.999 × 3 = 50.99 say 51 kW. (Ans.)