# Rotational Torque

A permanent magnet generator creates $4.25$ MW at 15 rpm. How much rotational torque is required to turn the shaft at 15 rpm when the generator load is at $4.25$ MW?

Answer

Let us suppose for simplicity that the magnet generator is a cylinder and its radius is R = 1m. This will not affect the generalization of the result.

The speed of a point on the exterior of the magnet cylinder is

$V = \omega*R$, where omega is the pulsation of the movement and

$\omega = 2*\pi*frequency = 2*\pi*F$

$frequency = 15rpm =15/60 =0.25$ Hz

Hence $v = 2*\pi*F*R = 2*\pi*0.25 = 1.57$ rad/sec

The power generated by the applied force must equal at least the power generated.

$P = F*v$, where F is the applied force $F = P/v = 4.25*10^6/1.57 = 2.70*10^6$ N

The rotational torque of this force is simply

$M = F*R = 45092.8*1=2.70*10^6 (N*m)$

One can see from here that the choosing of a 1 m cylinder radius does not affect the result.