Allied & ICAR Exam Bank • ALLIED_0252 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 engine has fuel consumption rate $8.4\text{ kg/h}$. Fuel calorific value is $42000\text{ kJ/kg}$. Brake power is $35\text{ kW}$. Brake thermal efficiency in percentage is (rounded off to one decimal place):

Answer Key:

35.7

Step-by-Step Solution & Explanation

Heat input rate $= (8.4 / 3600) \times 42000 = 0.002333 \times 42000 = 98.0\text{ kW}$. $\eta_{bt} = (35 / 98) \times 100 \approx 35.71\%$.
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