GATE 2026 AG – Question 51
A four-cylinder four-stroke engine has 100 mm bore,120 mm stroke,1800 rpm mean speed and 0.9 MPa IMEP. Energy fluctuation/indicated work per crankshaft revolution is .30. Ignore other losses. For allowable speed variation of ±1% about the mean, find flywheel mass moment of inertia in $\mathrm{kg\,m^2}$,to two decimals. Use π=3.14.
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Correct answer: 0.70 to 0.74
Explanation
Swept volume per cylinder is $V_s=(3.14/4)(.100)^2(.120)=.000942$ m³. Four-stroke operation gives four cylinder cycles per two revolutions, so indicated work per revolution is $W_r=4p_mV_s/2=1695.6$ J. Energy fluctuation is $.30W_r=508.68$ J. The coefficient of speed fluctuation is $(1.01-.99)=.02$,and $\omega=2(3.14)(1800)/60=188.4$ rad/s. Using $\Delta E=I\omega^2C_s$ gives $I=.71656$,rounded **0.72 kg·m²**.