Induction motor highest real power condition

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Edit: Misread the problem the first time.

The answer lays it out for you - as you progress forward from the startup, the power delivered to the motor is function of R * i^2 * (1-s)/s. Resistance is a factor, but it is constant (determined by the construction of the motor), so we ignore that for the purpose of the problem.

As speed progresses forward from startup, i^2 starts to decrease, but (1-s)/s increases fast enough to overcome that (slip is decreasing), and the power rises.

After the peak, the i^2 starts to dominate, and the power begins to fall, going all the way to 0 at synchronous speed.

The peak ACTIVE power doesn't happen at breakdown torque (close to 0 rpm here) - the current may be at its highest, but the delivered power is mostly being lost in the rotor (the P_rotor,losses = I^2*R) and never gets delivered to the load. Makes sense when you think about it - what happens when you overload an unprotected motor? It overheats.

Try thinking about it from the other side, with mechanical torque and power: P = Torque * Rotational Speed * Some Constant.

At breakdown torque, increasing the speed a little bit barely changes the torque, and so the power increases with speed. Approaching synchronous speed, the torque is approaching 0, and therefore power falls towards 0 too. So there is always some "catch-point" in between where the power hits a maximum - where the torque begins falling fast enough to counter the increase in speed.

Also, I'd take it easy with the copyrighted material you post online. The publishers of these study materials are extremely generous with how much they allow to be posted to these forums, in the name of freeing people up to ask questions - but you should really try to keep it to the minimum necessary to ask your question.
 
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