Allied & ICAR Exam Bank • ALLIED_0246 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A tractor engine develops $36\text{ kW}$ brake power. If friction power is $9\text{ kW}$, the mechanical efficiency in percentage is:

Answer Key:

80

Step-by-Step Solution & Explanation

Indicated power $IP = BP + FP = 36 + 9 = 45\text{ kW}$. Mechanical efficiency $\eta_m = (BP / IP) \times 100 = (36 / 45) \times 100 = 80.0\%$.
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