Power Plant Engineering

260 POWER PLANT ENGINEERING

Also, V 3 – V 2 = 0.05Vs = 0.05(V 1 – V 2 )

or, V 3 – V 2 = 0.05(13V 2 – V 2 ),^1

2

V

V

= 13

or, V 3 – V 2 = 0.06V 2

3

2

V

V

= 1.6

ηair standard = 1 – 1

1

()γ−





γr

()^1

(1)

ργ−

ρ−



= 1 – 1.4 1

1

1.4(14) −







(1.61.4 1)

(1.6 1)

−



−

= 0.615 = 61.5%

ηrelative = thermal

air standard

η

η

0.65 = thermal

0.615

η

ηthermal = 0.4

But, ηthermal =

I.P.

(C)mf×

0.4 =

I.P.

(^1441800)
3600



I.P. = 65 KW
ηmech =
B.P.
I.P.
0.76 =
B.P.
65
B.P. = 49.4 kW
Mean effective pressure can be calculated based on I.P. or B.P. of the engine
I.P. = mi.
(. LAN .10)
6
np k
, pmi = indicated mean effective pressure
65 =
2 0.25 (0.2)^2 300 1 10
4
6
pmi
π
×× × ×××
pmi = 8.27 bar
and brake mean effective pressure (pmb) = 0.76 × 8.27 = 6.28 bar.
V = V –VS1 2
1
4
2 3
0.05 VS
P
V
Fig. 8.18