Power Plant Engineering

Example 2. A four-stroke diesel engine has a piston diameter of 16.5 cm and a stroke of 27 cm.
The compression ratio is 14.3, the cut-off 4.23% of the stroke and the mean effective pressure 4.12 bar.
The engine speed is 264 rev/min and the fuel consumption is 1.076 kg of oil per hour, having a calorific
value of 39150 kJ/kg. Calculate the relative efficiency of the engine.
Solution. I.P. = 
P LA
(60 1000
2)
m
n
×
×
, for two stroke engine

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DIESEL POWER PLANT
257
Given that;
P
m
= 4.12 bar
L = 0.27 m
A = 
4
π
× (0.165)
2
= 0.214 m
4
n = 264
I.P. = 
5
(4.12 10
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 – 
1
k
r
k
−






(
1)
(
1)
k


ρ −




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 – 
0.4
(14.3)
1.4
−










1.4
(1.56
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.

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258
POWER PLANT ENGINEERING
Solution. (a) BHP = 
2 NT
4500
Π
Where,
T = torque
N = RPM
200 = 
(2
2000
T)
4500
π ×
×
T = 71.7 kg.m.
(b) 
η
m
= 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) 
η
m
= 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.