TITLE.PM5

622 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-2.pm5

(i)Compression ratio.

(ii)Thermal efficiency of the cycle.

(iii)Work done.

Take γ for air = 1.4.

Solution. Refer Fig. 13.11.

Total volume

VS

V(m )^3

p (bar)

3

Adiabatic

Adiabatic 4

1

2

VC

T (K)

2223

311

2

1

3

4

s(kJ/kg K)

Constant volume

Constant volume

p 2

p 1

Fig. 13.11

Initial temperature, T 1 = 38 + 273 = 311 K

Maximum temperature, T 3 = 1950 + 273 = 2223 K.

(i)Compression ratio, r :

For adiabatic compression 1-2,

p 1 V 1 γ = p 2 V 2 γ

or

V

V

1

2

F

HG

I

KJ

γ

=

p

p

2

1

But

p

p

2

1

= 15 ...(given)

∴ ()rγ = 15 Q r

V

V

=

L

N

M

O

Q

(^1) P
2
or (r)1.4 = 15
or r = ()15
1

1. (^4) = (15)0.714 = 6.9
   Hence compression ratio = 6.9. (Ans.)
   (ii)Thermal efficiency :
   Thermal efficiency, ηth = 1 –
   1
   ()rγ−^1 = 1 –
   1
   (.) 69 14 1. −
   = 0.538 or 53.8%. (Ans.)