Asynchronous motors - the heat of the rotor

[fawk3s]

Main Question or Discussion Point

So I was reading about asynchronous motors, and it seems the motor works better when the resistance of the rotor is smaller. This is fairly obvious, because the smaller the resistance, the bigger the inductive currents and therefore more mechanical force is applied on the rotor, thus its doing more mechanical work (leaving out the frequency of the current at the moment for simplification).
Now as I started thinking about it in more depth, I stumbled upon this question:

If the resistance of the rotor of one motor is say 2x smaller than the resistance of the rotor of another motor, and therefore the first motor (with the smaller-resistance-rotor) is able to do say 2x more mechanical work in a time unit, does the rotor of the first motor also do more work via heating than the second?
Because of P=V*I=I^2*R

Thanks in advance,
fawk3s

 

[rcgldr]

In addition to "copper" (related to resitance) losses you also have "iron" losses (related to magnetic effects) in motors:

http://www.engineeringtoolbox.com/electrical-motor-efficiency-d_655.html

 

[Phrak]

To good approximation you can lump all the losses due to source resistance, lead resistance, winding resistance and core loose together as a single resistance in series with the load. The load is an effective variable resistance. This should make your comparative analysis of two motors fairly simple.

 

[rcgldr]

When considering the total losses to heat, a 80% efficient motor will have double the losses to heat of a 90% efficient motor. Assuming the same temperature limit and rate of heat dissipation, the 90% efficient motor can operate at twice the power of the 80% efficient motor.