Question Bank • QB_FP_154 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 diesel engine develops $30\text{ kW}$ brake power with a brake specific fuel consumption (BSFC) of $240\text{ g/(kW}\cdot\text{h)}$. If the calorific value of diesel is $42\text{ MJ/kg}$, calculate the brake thermal efficiency of the engine as a percentage (round off to 1 decimal place).

Answer Key:

35.7

Step-by-Step Solution & Explanation

Brake thermal efficiency: $\eta_{\text{bth}} = \frac{3600}{\text{BSFC (in kg/kW}\cdot\text{h)} \times \text{CV (in MJ/kg)}}$ $\text{BSFC} = 240\text{ g/kW}\cdot\text{h} = 0.240\text{ kg/kW}\cdot\text{h}$ $\eta_{\text{bth}} = \frac{3600}{0.240 \times 42000} = \frac{3600}{10080} = 0.3571 \implies 35.71\% \approx 35.7\%$ Acceptable range: $35.0 - 36.5\%$.
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