Answer to Question #260642 in Mechanical Engineering for Nilesh

Question

A vertical diesel engine running at 350 rpm develops 600 kW and has 4 impulses per

revolution. If the fluctuation of energy is 25 per cent of the work done during each

impulse, estimate the cross-sectional area of the rim of the flywheel required to keep the

speed within 2 rpm of the mean speed when the mean peripheral speed of the rim is 1350

m/min. Cast iron has a density of 7.2 Mg/m3

.

Answer 1

Mean angular velocity, $\omega=350rpm=36.6519rad/s=2199.115rad/min$

$C_s=\frac{2*max\ deviation}{mean\ velocity}$

$=\frac{2(2)}{350}=0.011429$

$r=\frac{mean\ peripheral\ velocity}{\omega}$

$=\frac{1350}{2199.115}=0.61388$

$m=\rho 2πr*A$

$=7200*2π*0.61388*A=27771.2769A$

$I=mr^2$

$=27771.2769A*0.61388=10465.568A$

$E_m=0.5I\omega^2$

$=0.5*10465.568A*36.6519^2=7029522A$

$i=\frac{350*4}{60}=23.333$

$∆E=\frac{P*0.25}{i}$

$=\frac{600000*0.25}{23.333}=6428.6J$

$∆E=2E_mC_s$

$6428.6=2*7029522A*0.011429$

$A=\frac{6428.6}{160680.813}$

=0.0400085m²

$\approx 40000mm^2$