Allied & ICAR Exam Bank • ALLIED_0243 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A 4-stroke 4-cylinder diesel engine with bore $100\text{ mm}$ and stroke $120\text{ mm}$ runs at $2000\text{ rpm}$. Mean effective pressure is $0.6\text{ MPa}$. Indicated power in kW is (rounded off to two decimal places):

Answer Key:

37.70

Step-by-Step Solution & Explanation

$A = \pi(0.1)^2 / 4 = 0.007854\text{ m}^2$. Stroke $L = 0.12\text{ m}$. For 4-stroke, $n = 2000 / 2 = 1000\text{ rpm}$. Cylinders $k = 4$. $P_m = 600\text{ kPa}$. $IP = (600 \times 0.12 \times 0.007854 \times 1000 \times 4)/60 = 37.70\text{ kW}$.
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