I am not a electrical expert so others will need to check my assumptions concerning motor operation with a VFD. However, the relationship between HP and Torque is as follows:
1) HP = (T x RPM) / 5252 And 2) T = (HP x 5252) / RPM
From 2) we see that 3 HP at 600 RPM requires 26.26 ft-lbs of torque ((3 x 5252) / 600 = 26.26).
From 1) we can can plug in the 26.26 ft-lbs of torque to determine what the motor HP would need to be at rated RPM. We can also see it would be better to start with a 1750 RPM motor rather than a 3600 RPM motor. We see that a 8.75 HP, 1750 RPM motor would be required to produce 26.26 ft-lbs of torque at 600 RPM ((26.26 x 1750) / 5252) = 8.75 HP).
Note the above assumes that a 3 phase motor will make full rated torque a 600 RPM when operated with a VFD. Since the motor is producing 3 HP it should draw similar wattage to that of a 3 HP motor running at rated speed. I say similar since there will be some loss in the VFD etc. The motor cannot produce its rated power (HP) at 600 RPM since it is not running at rated RPM, see equation 1).
Those who are more knowledgeable in electric motor operation please add any corrections as required.
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