Question Bank • QB_FP_ADV_005 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A 4-cylinder, 4-stroke tractor diesel engine has a cylinder bore of $100\text{ mm}$ and stroke length of $120\text{ mm}$. During a dynamometer test at $2000\text{ rpm}$, the engine produces a brake torque of $180\text{ N}\cdot\text{m}$ while consuming fuel at a rate of $8.5\text{ kg/h}$. The calorific value of the diesel fuel is $42.5\text{ MJ/kg}$. The brake thermal efficiency of the engine is ________ $\%$ (round off to two decimal places).

Answer Key:

37.58

Step-by-Step Solution & Explanation

1. Brake Power ($BP$): $BP = \frac{2 \pi N T}{60000} = \frac{2 \times \pi \times 2000 \times 180}{60000} = \frac{2261946.7}{60000} = 37.699\text{ kW}$ 2. Total thermal energy input rate ($Q_{\text{in}}$): $Q_{\text{in}} = \dot{m}_f \times CV = \left( \frac{8.5}{3600}\text{ kg/s} \right) \times 42500\text{ kJ/kg} = 2.3611 \times 10^{-3} \times 42500 = 100.347\text{ kW}$ 3. Brake Thermal Efficiency ($\eta_{\text{bth}}$): $\eta_{\text{bth}} = \frac{BP}{Q_{\text{in}}} \times 100 = \frac{37.699}{100.347} \times 100 = 37.568\% \approx 37.58\%$
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