So I was looking at a couple of the formulas for three phase motors, and here is what I came across. What do you guys think my logic:

1. RPM = (120 x Freq.) / Poles
2. Torque = (5252 x Power) / rpm

so replacing rpm from 1, we have,

Torque = ((5252 x Power) x Poles) / ( 120 x Freq.)

and in this last one, the Power and the Frequency, are changing, but we know them because:

1. For all practical purposes, the mechanical frequency of the motor = input current frequency

2. If we take into account efficiency (~88%), then we can compute power as V x I x eff.

3. We have updated values of motor inputs like Voltage, Current, Efficiency & Frequency from our PWM and H-Bridge circuit.

 

Anonymous... see Control.com Thread http://www.control.com/thread/1026212338!

Regards, Phil Corso

 

Anonymous... further to my earlier comment(s):

Your formula resullts in a torque-vs-speed curve that is a straight line between zero and operating speed. Its slope would be Ts/Ns, where Ts and Ns represent rated torque and speed. Additionally, at speeds above rated, your formula would still be a straight-line having the same positive slope. Unfortunately, such is not the case for an induction motor!

Actually, the shape of the classic torque-vs-speed curve is often shaped like that of a "sway-back horse" and not a straight line! That is, torque is related to slip (the difference between supply frequency and stator power (minus its losses) transferred across the air-gap to the rotor (minus its losses!}

I hesitate producing the formula in this post because some forum members could become annoyed. So, rather than spoil their upcoming Holidays, I suggest you search the Control.Com archives. If the search proves unfruitful, contact me!

Regards, Phil Corso