Question
Download Solution PDFA single cylinder four stroke engine operating at 80% of mechanical efficiency develops a brake power of 60 kW. The indicated power and the power lost due to friction respectively are
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\(Indicated\;power\;\left( {I.P} \right) = \frac{{Brake\;power\;\left( {BP} \right)}}{{{\eta _{mech}}}}\)
Frictional power (F.P) = Indicated power (I.P) – Brake power (B.P)
Calculation:
Given:
Brake power = 60 kW, ηmech = 80 % = 0.8
Now,
\(I.P = \frac{{BP}}{{{\eta _{mech}}}}\)
\(I.P. = \frac{{60}}{{0.8}}\)
∴ Indicated power (I.P.) = 75 kW
Now,
F.P = I.P – B.P = 15 kW
∴ Power lost due to friction is 15 kWLast updated on Jun 23, 2025
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