Subtopic: Power efficiencies and measurement, engine performance curves
Question Statement:
A tractor diesel engine develops $30\text{ kW}$ brake power while consuming $7.5\text{ kg}$ of diesel fuel per hour. The lower calorific value of the fuel is $42\text{ MJ/kg}$. The brake thermal efficiency of the engine expressed in percentage is ________ (round off to 1 decimal place).
Answer Key:
34.3
Step-by-Step Solution & Explanation
1. Total heat energy supplied by fuel per second ($Q_{\text{in}}$):
$Q_{\text{in}} = \frac{\dot{m}_f \times CV}{3600} = \frac{7.5\text{ kg/h} \times 42,000\text{ kJ/kg}}{3600\text{ s}} = \frac{315,000}{3600} = 87.5\text{ kW}$
2. Brake power output $BP = 30\text{ kW}$.
3. Brake thermal efficiency ($\eta_{\text{bth}}$):
$\eta_{\text{bth}} = \frac{BP}{Q_{\text{in}}} \times 100 = \frac{30}{87.5} \times 100 = 34.285\% \approx 34.3\%$