Question

A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine. The torque of machine varies in a cyclic manner over a period of 3 revolutions. The torque rises from 750 N-m to 3000 N-m uniformly during 1/2 revolution and remains constant for the following revolution. It then falls uniformly to 750 N-m during the next 1/2 revolution and remains constant for one revolution, the cycle being repeated thereafter. Determine the power required to drive the machine and percentage fluctuation in speed, if the driving torque applied to the shaft is constant and the mass of the flywheel is 500 kg with radius of gyration of 600 mm.

Answer 1

Given : N = 250 r.p.m or ω = 2π x 250/60 = 26.2 rad/s; m = 500 kg; k = 600 mm = 0.6 m

The turning moment diagram for the complete cycle is

we know that that the torque reuired for one complete cycle

 = Area of figure OABCDEF

= Area OAEF + Area ABG + Area BCHG + AreaCDH

= OF x OA + 1/2 x AG x BG + Gh x CH + 1/2 x HD x CH

If T_{mean is the torque in N-m, then torque required for one complete cycle}

= T_mean x 6 πN-m   ....(ii)

From equations (i) and (ii)

= T_mean  = 11 250π/6 π = 1875 N-m

Power required to drive the machine

We know that power required to drive the the machine,

P == T_mean x ω = 1875 x 26.2 = 49125 W

Coefficient of fluctuation of speed

Let C_s = Coefficient of  fluctuation of speed.

First of all, let us find the values of LM and NP. From similar triangles ABG and BLM.

_{Now, from similar triangles CHD and CNP,}

From Fig. we find that 

BN = CN = 3000 - 1875 = 1125 N-m

Since the area above the mean torque line represents the maximum fluctuation of energy. therefore, maximum fluctuation of energy,

We know that maximum fluctuation of energy (Δ E),