Question
Download Solution PDFThe frictional power of a single cylinder four-stroke engine is estimated to be 25 kW. If the mechanical efficiency is 75%, then the value of indicated power (in kW) will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Friction Power:
It is the power loss in overcoming the friction between piston and cylinder walls, between the crankshaft and camshaft and their bearings, etc.
It is the difference between Indicated power and Brake power.
F.P = I.P - B.P
∴ B.P = I.P - F.P
Mechanical Efficiency:
It is the ratio of brake power to indicated power.
\({η _m} = \frac{{B.P}}{{I.P}} = \frac{{{η _{bth}}}}{{{η _{ith}}}}\)
Calculation:
Given:
ηm = 75% = 0.75, FP = 25 kW
Mechanical Efficiency:
\({η _m} = \frac{{B.P}}{{I.P}} \)
\({η _m} = \frac{{I.P\;-\;F.P}}{{I.P}} \)
\(0.75 = \frac{{I.P\;-\;25}}{{I.P}} \)
I.P - 0.75I.P = 25
0.25I.P = 25
I.P = 100 kW
Last updated on May 28, 2025
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