Low speed permanent magnet snychronous generator

[ENRG]

Homework Statement

A permanent magnet synchronous generator (as used in the wind industry) develops 5MW of electricity at 12 RPM; What amount of mechanical torque would be required to turn this generator at 12 RPM.? These generators are said to be 95% efficient

Homework Equations

Torque is part of the basic specification of an engine the power of an engine is expressed as its torque multiplied by its rotational speed of the axis usually known as rpm (revolutions per minute)

So
Given P = 5 MW = 5000000 W.
Revolutions per minute (rpm) = 12

Power,
P = Torque * rpm.

The Attempt at a Solution

So Torque = P / rpm
Torque = 5000000/12
Torque = 416,616

 

[Valkarie]

.67 NmBut since the generator is 95% efficient, the mechanical torque required to turn the generator at 12 rpm would be 416,616.67/0.95 = 437,295.38 Nm

 

[Mene]

.7 Nm

To turn this generator at 12 RPM, a mechanical torque of approximately 416,616.7 Nm would be required. This is because torque is directly proportional to power and inversely proportional to rotational speed. In this case, the power output of the generator is 5MW and the rotational speed is 12 RPM, so the torque required would be 5,000,000/12 = 416,616.7 Nm.

It is important to note that this calculation assumes 100% efficiency. Since the generator is said to be 95% efficient, the actual torque required may be slightly higher. However, this value still provides a good estimate for the required torque. Additionally, the actual torque required may vary depending on the specific design and operating conditions of the generator.