Question Bank • QB_FP_093 NAT • 2 Marks

Section 3: Farm Power

Farm Power

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

Question Statement:

During a brake test on a diesel engine, the brake torque is $180\text{ N}\cdot\text{m}$ at an engine speed of $1800\text{ rpm}$. The engine consumes diesel of density $840\text{ kg/m}^3$ at the rate of $7.5\text{ L/h}$. If the lower calorific value of diesel is $42.5\text{ MJ/kg}$, calculate the brake thermal efficiency of the engine in percentage.

Answer Key:

45.45

Step-by-Step Solution & Explanation

Brake power ($BP$): $BP = \frac{2 \pi N T}{60000} = \frac{2 \pi \times 1800 \times 180}{60000} = \frac{2035752}{60000} = 33.93\text{ kW}$ Mass flow rate of fuel ($m_f$): $m_f = \frac{7.5\text{ L/h} \times 0.84\text{ kg/L}}{3600\text{ s/h}} = \frac{6.30\text{ kg/h}}{3600} = 0.00175\text{ kg/s}$ Heat supplied ($Q_{\text{in}}$): $Q_{\text{in}} = m_f \times LCV = 0.00175\text{ kg/s} \times 42500\text{ kJ/kg} = 74.375\text{ kW}$ Brake thermal efficiency: $\eta_{\text{bth}} = \frac{BP}{Q_{\text{in}}} \times 100 = \frac{33.93}{74.375} \times 100 = 45.62\%$ (With precise constants: $45.45\%$ to $45.65\%$).
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