Question Bank • QB_FP_115 NAT • 2 Marks

Section 3: Farm Power

Farm Power

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

Question Statement:

A diesel engine running at $2200\text{ rpm}$ develops a brake torque of $150\text{ N}\cdot\text{m}$. If the brake specific fuel consumption ($bsfc$) is $240\text{ g/(kW}\cdot\text{h)}$, calculate the fuel consumption rate of the engine in $\text{kg/h}$.

Answer Key:

8.29

Step-by-Step Solution & Explanation

Brake power ($BP$): $BP = \frac{2 \pi N T}{60000} = \frac{2 \pi \times 2200 \times 150}{60000} = \frac{2073451}{60000} = 34.557\text{ kW}$ Fuel consumption rate $m_f$: $m_f = \frac{BP \times bsfc}{1000} = \frac{34.557 \times 240}{1000} = 8.294\text{ kg/h}$
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