Question

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:

1. 80 %
2. 50 %
3. 25 %
4. 66.7 %

Answer 1

Correct Answer - Option 1 : 80 %

Concept:

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Calculation:

Given: 

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

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Mechanical efficiency at half load \( = \frac{{100}}{{125 }}\)

Mechanical efficiency at half load = 0.8 ⇒ 80 %