Allied & ICAR Exam Bank • ALLIED_0870 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses

Question Statement:

A tractor engine delivers torque $T = 250\text{ N}\cdot\text{m}$ at rated speed $N = 2100\text{ rpm}$ ($\omega = \frac{2 \pi \times 2100}{60} = 219.91\text{ rad/s}$). Brake power $P_b = T \cdot \omega$ in kW is (rounded off to one decimal place):

Answer Key:

55.0

Step-by-Step Solution & Explanation

$P_b = 250 \times 219.911 = 54,978\text{ W} \approx 55.0\text{ kW}$.
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