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Question

# A single cylinder 4-stroke engine using gas (heating value 11500 kJ/m3 ) as a fuel has an air/fuel p ratio of 10 : 1. The indicated power is 22.727 kW while running at 5500 rpm. The volumetric efficiency is 0.75 and indicated thermal efficiency is 0.3 What is the stroke volume?

A
1264 cc
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B
1916 cc
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C
1437 cc
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D
1686 cc
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Solution

## The correct option is B 1916 ccIndicated thermal efficiency, ηith=IPVf×CV Where, mf is the volume flow rate of fuel and CV is the calorific value in kJ/m3. So, ηith=22.727Vf×11500 [where ηith=0.3] Vf=6.5876×10−5m3/s Air fuel ratio = AF=10 So. volume flow rate of air, Va=10×6.5876×10−3 Va=0.06587 m3/s Volume flow rate of air per cycle, Va=0.06587×2×60N=0.06587×2×605500 =1.4371×10(−3)m3/cycle Volumetric efficiency, ηy=Va)Vs Where Vs is the swept volume, Vs=Va0.75=1.4371×10−30.75=1.9165×10(−3)m3 = 1916.13 cc

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