Power Plant Engineering

DIESEL POWER PLANT 257

Given that;

Pm = 4.12 bar

L = 0.27 m

A =

4

π

× (0.165)^2 = 0.214 m^4

n = 264

I.P. =

(4.12 10^5 0.27 0.0214 264)

(60 1000 2)

×× × ×

××

= 5.24 kW

Now, indicated thermal efficiency

=

(Heat equivalent of I.P. per hour)

(Heat in fuel per hour)

=

(5.24 3600)

(1.076 39150)

×

×

= 44.78%

Now air standard efficiency = 1 –

r^1 k

k

−





(1)

(1)

ρ−k



ρ−

If clearance volume is taken as unity, then,

r = 14.3,

ρ = 1 + {(4.23 × 13.3)100} = 1.56

k = 1.4

Efficiency = 1 –

(14.3)0.4

1.4

−





(1.561.4 1)

(1.56 1)

−

−



= 62%

Now relative efficiency

=

Indicated Thermal efficiency

Air standard efficiency

=

0.4478

0.62

= 72.23%.

Example 3. A six-cylinder two-stroke cycle marine diesel engine with 100 mm bore and 120 mm
stroke delivers 200 B.H.P. at 2000. R.P.M. and uses 100 kg of fuel per hour. If I.H.P. is 240, determine
the following:

(a) Torque,

(b) Mechanical efficiency,

(c) Indicated specific fuel consumption.