Question Bank • QB_EXP_FP_PWR_007 MCQ • 1 Mark

Section 3: Farm Power

Engine Power: Indicated, Brake & Friction Power

Subtopic: Mechanical efficiency

Question Statement:

A tractor diesel engine develops a brake power of $42.5\text{ kW}$ while its frictional power losses are measured as $6.5\text{ kW}$. The mechanical efficiency of the engine is:

Options:

  • (A) $86.7\%$
  • (B) $80.7\%$
  • (C) $91.7\%$
  • (D) $73.7\%$

Answer Key:

A

Step-by-Step Solution & Explanation

Indicated power is the sum of brake power and friction power: $\text{IP} = \text{BP} + \text{FP} = 42.5 + 6.5 = 49.0\text{ kW}$ Mechanical efficiency is given by: $\eta_m = \frac{\text{BP}}{\text{IP}} \times 100 = \frac{42.5}{49.0} \times 100 = 86.7\%$
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