Power Plant Engineering

258 POWER PLANT ENGINEERING

Solution. (a) BHP =

2NT

4500

Π

Where, T = torque

N = RPM

200 =

(2 2000 T)

4500

π× ×

T = 71.7 kg.m.

(b) ηm = Mechanical efficiency =

BHP

IHP

=

200

240

= 0.83

(c) Indicated specific fuel consumption = W/I.H.P.

Where, W = Amount of fuel used per hour

Indicated specific fuel consumption =

100
240
= 2.41 kg/IHP hour.
Example 4. A diesel engine develops 200 H.P. to over come friction and delivers 1000 BHP. Air
consumption is 90 kg per minute. The air fuel ratio is.15 to 1. Find the following:

(a) IHP, (b) Mechanical efficiency, (c) Specific fuel consumption.

Solution. (a) BHP = 1000

FHP = 200

IHP = BHP + FHP = 1000 + 200 = 1200

(b) ηm = Mechanical efficiency =

BHP

IHP

=

1000

1200

= 0.83 = 83%.

(c) K = Air fuel ratio = 15

W = Air consumed per hour

= 90 × 60 = 5400 kg per hour

S = Amount of fuel consumed =

W

K

=

5400

15

= 360 kg per hour.

Specific fuel consumption =

S

IHP

=

360

1200

= 0.3 kg/IHP hr.

Example 5. The brake thermal efficiency of a diesel engine is 30 percent. If the air to fuel ratio

by weight is 20 and the calorific value of the fuel used is 41800 kJ/kg, what brake mean effective

pressure may be expected at S.T.P. conditions?