Article 430.250

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Its not a calculation as much as its based on NEMA motor data. Its more conservative than nameplate data in case you change out a motor, the idea is you don't have to increase the conductor and disconnects.
But I really don't know where the table data actually comes from, its been in the NEC for a long time.
 
James@CHA said:
How are the full-load currents calculated. I have not been able to find the formula used.
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Well, first off, there is no need to calculate them, because they are given right there in the table. As for where the numbers in the table come from, it is my understanding that these are "worst case" scenarios for motors which are available on the market.

The actual full load amperage of a given HP motor will depend on its PF and efficiency. Some manufacturers may take the time to build a very high power factor, highly efficient motor, and this motor will have a relatively low FLA. Another manufacturer may build the same horsepower motor, but not do what it takes for high efficiency and high power factor. This motor will draw a relatively high current. Obviously the first motor will cost more than the second motor.

Now, at an actual installation of this motor, let's say the original motor was a high PF, high eff, 25 HP motor. If the wiring and equipment was installed based on this "low" FLA, and then years later the motor went bad, and it was replaced with a cheap low PF low eff 25 HP motor, the conductors and OCP may not be sufficient, even though they are both 25 HP motors.*

*I reserve the right to be totally and completely wrong about the above. None of this is substantiated by fact, this is just my thoughts on why the tables exist.

Edit: Darn it, Tom Baker, I hate it when someone beats me to the punch.
 
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The reason i ask is that i was trying to calculate FLA on a 32HP (elevator)motor. This doesnt sound like a standard motor. The elevator manufacturer is reluctant to give the FLA until the "product has been specified".

The FLA for a 30HP, 3-Phase, 480V Induction motor is 40A

But if i use I*V*1.732*.85(PF)/746+HP I come up with 28.7A??
 
James@CHA said:
The reason i ask is that i was trying to calculate FLA on a 32HP (elevator)motor. This doesnt sound like a standard motor. The elevator manufacturer is reluctant to give the FLA until the "product has been specified".

The FLA for a 30HP, 3-Phase, 480V Induction motor is 40A

But if i use I*V*1.732*.85(PF)/746+HP I come up with 28.7A??
Click to expand...

Don't forget the effciency of the motor.

The motor HP is the output of the motor, using the 746W/HP conversion factor would then give you output watts. The difference between input watts and output watts is the efficiency of the motor.

I believe no motor built to NEMA MG-1 standards is "allowed" to have FLA as high as those found in the NEC table.
 
The tables listed in 430.250 must be one of those things you take at face value and don't ask how they are achieved...
 
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