SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 152 of 200

Marks: +1, -0

An engine at full load delivers 200 kW brake power. It requires 25 kW to rotate it without fuel at the same speed. The mechanical efficiency at half load is:

80 %
50 %
25 %
66.7 %

Validate

Solution:

Brake power (BP) = 200 kW, Half load = 100 kW Friction Power (FP) = 25 kW
Mechanical efficiency at half load

={BP}/{BP+FP}

Mechanical efficiency at half load

=100/125

Mechanical efficiency at half load = 0.8 ⇒ 80 %

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