Question Bank • QB_FP_118 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

During a rope brake dynamometer test on an engine running at $1200\text{ rpm}$, the dead weight on the tight side is $450\text{ N}$ and the spring balance reading on the slack side is $50\text{ N}$. The effective brake drum diameter (including rope diameter) is $0.80\text{ m}$. Calculate the brake power developed in $\text{kW}$.

Answer Key:

20.11

Step-by-Step Solution & Explanation

Net brake load $W - S = 450 - 50 = 400\text{ N}$. Brake drum radius $R = \frac{0.80}{2} = 0.40\text{ m}$. Brake torque: $T = (W - S) \times R = 400 \times 0.40 = 160\text{ N}\cdot\text{m}$ Brake power: $BP = \frac{2 \pi N T}{60000} = \frac{2 \pi \times 1200 \times 160}{60000} = \frac{1206371}{60000} \approx 20.106\text{ kW}$
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