Question

A single cylinder 4-stroke engine using gas (heating value 11500 $kJ/m^3$ ) 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
• B. 1916 cc
• C. 1437 cc
• D. 1686 cc

Answer 1

The correct option is B 1916 cc

Indicated thermal efficiency,

${\eta }_{ith}=\frac{IP}{V_f\times CV}$

Where, $m_f$ is the volume flow rate of fuel and CV is the calorific value in $kJ/m^3$.

So, ${\eta }_{ith}=\frac{22.727}{V_f\times 11500}$

[where ${\eta }_{ith}=0.3$]

$V_f=6.5876\times {10}^{-5}{m}^3/s$

Air fuel ratio = $\frac{A}{F}=10$

So. volume flow rate of air,

$V_a=10\times 6.5876\times {10}^{-3}$

$V_a=0.06587m^3/s$

Volume flow rate of air per cycle,

$V_a=\frac{0.06587\times 2\times 60}{N}=\frac{0.06587\times 2\times 60}{5500}$

$=1.4371\times {10}^{(}-3)m^3/cycle$

Volumetric efficiency,

${\eta }_y=\frac{V_a)}{V_s}$

Where $V_s$ is the swept volume,

$V_s=\frac{V_a}{0.75}=\frac{1.4371\times {10}^{-3}}{0.75}=1.9165\times {10}^{(}-3)m^3$

= 1916.13 cc