Allied & ICAR Exam Bank • ALLIED_1115 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Mean Effective Pressure

Question Statement:

A 4-stroke single cylinder engine develops an indicated power of 12 kW at 1800 rpm. If the swept volume is 1.2 litres ($0.0012\text{ m}^3$), the indicated mean effective pressure ($P_{imep}$) in bar is:

Answer Key:

6.67

Step-by-Step Solution & Explanation

For a 4-stroke engine, power strokes per second $n = \frac{N}{2 \times 60} = \frac{1800}{120} = 15\text{ s}^{-1}$. $IP = P_{imep} \times V_s \times n \implies 12,000\text{ W} = P_{imep} \times 0.0012 \times 15 = P_{imep} \times 0.018 \implies P_{imep} = \frac{12,000}{0.018} = 666,667\text{ Pa} = 6.67\text{ bar}$.
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