Question Bank • QB_FP_152 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

During a brake test of a tractor engine using a rope brake dynamometer, the dead load on the brake is $450\text{ N}$, the spring balance reading is $50\text{ N}$, and the engine speed is $1800\text{ rpm}$. The effective brake drum radius is $0.40\text{ m}$. Calculate the brake power developed in $\text{kW}$ (round off to 2 decimal places).

Answer Key:

30.16

Step-by-Step Solution & Explanation

Net braking load: $W_{\text{net}} = W - S = 450 - 50 = 400\text{ N}$ Braking torque: $T = W_{\text{net}} \times R = 400\text{ N} \times 0.40\text{ m} = 160\text{ N}\cdot\text{m}$ Brake power: $BP = \frac{2\pi N T}{60000} = \frac{2\pi \times 1800 \times 160}{60000} = \frac{1809557}{60000} \approx 30.16\text{ kW}$ Acceptable range: $29.8 - 30.5\text{ kW}$.
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