TITLE.PM5

GAS POWER CYCLES 609

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\M-therm\Th13-1.pm5

Solution. Refer Fig. 13.4.

Maximum pressure, p 1 = 18 bar

Maximum temperature, T 1 = (T 2 ) = 410 + 273 = 683 K

Ratio of isentropic (or adiabatic) compression,

V

V

4

1

= 6

Ratio of isothermal expansion, V

V

2

1

= 1.5.

Volume of the air at the beginning of isothermal expansion, V 1 = 0.18 m^3.

(i)Temperatures and pressures at the main points in the cycle :

For the isentropic process 4-1 :

T

T

1

4

=

V

V

4

1

F^1

HG

I

KJ

−γ

= (6)1.4 – 1 = (6)0.4 = 2.05

∴ T 4 = T^1

205.

=^683

052.

= 333.2 K = T 3

Fig. 13.4

Also, p

p

1

4

= V

V

4

1

F

HG

I

KJ

γ

= (6)1.4 = 12.29

∴ p 4 =

p 1

12.29

=

18

12.29

= 1.46 bar

For the isothermal process 1-2 :

p 1 V 1 = p 2 V 2

p 2 =

pV

V

11

2

=^18

1. 5
   = 12 bar
   For isentropic process 2-3, we have :
   p 2 V 2 γ = p 3 V 3 γ