Answer to Question #260643 in Mechanical Engineering for Nilesh

Question

A shaft fitted with a flywheel rotates at 250 rpm and drives a machine the resisting torque of

which varies in a cyclic manner over a period of three revolutions. The torque rises from 675 Nm to

2700 Nm in a uniform manner during 0.5 revolution and remains constant for 1 revolution, the cycle

being then repeated. If the driving torque applied to the shaft is constant and the flywheel has a

mass of 450 kg and a radius of gyration of 0.6 m, find the power necessary to drive the machine and

the percentage fluctuation of speed.

Answer 1

Power of the motor

P=Mean torque*Angular speed

$=T_m*\frac{2πN}{60}$

$=1875*\frac{2π*250}{60}$

$=49.0625W$

Maximum fluctuation energy is given by $∆E=\frac{1}{2}*\frac{π}{2}*1125+(2π*1125)+\frac{1}{2}*\frac{π}{2}*1125$

$=1125(\frac{π}{4}+\frac{π}{4}+2π)=1125(\frac{π}{2}+2π)$

$=\frac{5π}{2}*1125=8831.25$

$∆E=mk^2\omega^2C_s$

$8831.25=450(0.6)^2*(26.166)^2*C_s$

$8831.25=110914.8481C_s$

$C_s=0.079622$

Hence percentage fluctuation of speed =7.9622%