TITLE.PM5

GAS POWER CYCLES 637

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\M-therm\Th13-3.pm5

(iii)Mean effective pressure, pm :

Mean effective pressure of Diesel cycle is given by

pm = pr r

r

1

111

11

()[( ) ( )

()()

γγγρ ργ

γ

−− −

−−

−

=

115 4121 15 12 1

41151

×−^14 − −1 14^14

−−

()[ ( )()− ( )]

()()

..1. 2. 2..

1.

=

44.31 568 338 863

014

[]1. 0. 1.

.4

−×

× = 7.424 bar. (Ans.)

(iv)Power of the engine, P :

Work done per cycle = pmVs =

7.424 10 0 00942

10

5

3

××.
= 6.99 kJ/cycle
Work done per second = Work done per cycle × no. of cycles per second
= 6.99 × 380/60 = 44.27 kJ/s = 44.27 kW
Hence power of the engine = 44.27 kW. (Ans.)
Example 13.22. The volume ratios of compression and expansion for a diesel engine as
measured from an indicator diagram are 15.3 and 7.5 respectively. The pressure and tempera-
ture at the beginning of the compression are 1 bar and 27°C.
Assuming an ideal engine, determine the mean effective pressure, the ratio of maximum
pressure to mean effective pressure and cycle efficiency.
Also find the fuel consumption per kWh if the indicated thermal efficiency is 0.5 of ideal
efficiency, mechanical efficiency is 0.8 and the calorific value of oil 42000 kJ/kg.
Assume for air : cp = 1.005 kJ/kg K ; cv = 0.718 kJ/kg K, γ = 1.4. (U.P.S.C., 1996)

Solution. Refer Fig. 13.18. Given :

V

V

1

2

= 15.3 ;

V

V

4

3

= 7.5

p 1 = 1 bar ; T 1 = 27 + 273 = 300 K ; ηth(I) = 0.5 × ηair-standard ; ηmech. = 0.8 ; C = 42000 kJ/kg.

The cycle is shwon in Fig. 13.18, the subscripts denote the respective points in the cycle.

Fig. 13.18. Diesel cycle.