GAS POWER CYCLES 617

dharm
\M-therm\Th13-2.pm5

=

8 297 1 36

47

()()2. 0.

0.

−
× = 1.334 bar
Hence mean effective pressure = 1.334 bar. (Ans.)
Example 13.9. The minimum pressure and temperature in an Otto cycle are 100 kPa and
27°C. The amount of heat added to the air per cycle is 1500 kJ/kg.
(i) Determine the pressures and temperatures at all points of the air standard Otto cycle.
(ii)Also calculate the specific work and thermal efficiency of the cycle for a compression
ratio of 8 : 1.
Take for air : cv = 0.72 kJ/kg K, and γ = 1.4. (GATE, 1998)
Solution. Refer Fig. 13.7. Given : p 1 = 100 kPa = 10^5 N/m^2 or 1 bar ;
T 1 = 27 + 273 = 300 K ; Heat added = 1500 kJ/kg ;
r = 8 : 1 ; cv = 0.72 kJ/kg ; γ = 1.4.
Consider 1 kg of air.
p

v

3

4

1

2

Compression

pv = Cg

Expansion

Fig. 13.7

(i)Pressures and temperatures at all points :

Adiabatic compression process 1-2 :

T

T

2

1

=

v

v

(^1) r
2
1
F (^1811) 2 297
HG
I
KJ
== =
−
− −
γ
()γ ().4.
∴ T 2 = 300 × 2.297 = 689.1 K. (Ans.)
Also pv 11 γγ=pv2 2
or
p
p
v
v
2
1
1
2
=F 81 18 379
HG
I
KJ

γ
().4.
∴ p 2 = 1 × 18.379 = 18.379 bar. (Ans.)
Constant volume process 2-3 :
Heat added during the process,
cv (T 3 – T 2 ) = 1500