Calculating Motor kVA

[jrt5054]

One thing that has always confused me is the correct way to calculate the motor kW for a load study.

I have seen two different methods and they produce wildly different answers. I was hoping to get some input on the discrepancy and some advice on the best method. And I am only talking about 3 phase induction motors in this post.

The first method I've used is to take the Amps value from NEC Table 430.250 and convert it to kVA using V*I*1.73

If we take a 7.5hp motor, from the table we get 11A. 480V*11A*1.73 we get 9.13kVA

The other method I've seen is to use the conversion 1hp = 0.746kW. Using the same example as before we would get 7.5hp = 5.595kW. Now I understand that this is kW and not kVA and that the difference is the power factor. However, to get from 5.595kW to 9.13kVA you would need a power factor of about 0.6 which seemed excessively low.

Which method should be used?

Thanks!

 

[steve66]

I usually use the first. The second method ignores any inefficiency of the motor.

 

[Besoeker3]

kW and kVA are different measures. For a 100 kW motor is a measure of power. The kVA is different depending on the power factor.

 

[Besoeker3]

Just a little addendum motors have different power factors. For example a 4-pole motor is typically a lower factor than say, an 8-pole motor.

 

[jim dungar]

Motors are rated in output power, either kW or HP.
Your load study needs input power.

I would use the amps from the NEC table method.

 

[tom baker]

And the amps from the NEC tables is a conservative number, to account for motor changes

 

[Besoeker3]

And the amps from the NEC tables is a conservative number, to account for motor changes

Yes, I should have used that for USA. We just use actual calcs.

 

[winnie]

The amps from the NEC table will give you a conservative number.

Use the amps from the motor nameplate to get a reasonably accurate number.

For a real number you need to actually know the mechanical loading of the motor, or actually measure the current drawn by the motor.

Hp * 0.746 simply converts the rated mechanical output of the motor into kW, ignoring the actual loading, the efficiency of the motor, or the power factor. If you guess at power factor and efficiency, then Hp * 0.746 / PF / Eff is a way to convert those guesses into a kVA guess.

-Jon

 

[Shaneyj]

If using motor nameplate is one supposed to use terminal voltage or nameplate voltage?


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[Besoeker3]

The amps from the NEC table will give you a conservative number.

Use the amps from the motor nameplate to get a reasonably accurate number.

For a real number you need to actually know the mechanical loading of the motor, or actually measure the current drawn by the motor.

Hp * 0.746 simply converts the rated mechanical output of the motor into kW, ignoring the actual loading, the efficiency of the motor, or the power factor. If you guess at power factor and efficiency, then Hp * 0.746 / PF / Eff is a way to convert those guesses into a kVA guess.

-Jon

FWIW we don't use HP there days - not since fifty years ago. Yes, I that old!

 

[retirede]

If using motor nameplate is one supposed to use terminal voltage or nameplate voltage?


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If you’re using nameplate amps, use nameplate voltage.

 

[JoeStillman]

If you’re using nameplate amps, use nameplate voltage.

Likewise table 430.250. The correct column to use is the 460V column. so it's 11 x 460 x 1.732 = 8,764 VA. If you use 746W/HP to calculate the power factor, you're answer will be too low because it doesn't account for heat losses. Heat losses are in real power (Watts).

 

[nhee2]

I used to calculate using NEC values (and still do for feeder conductor/branch circuit conductor as required in 430) but for facility load calcs i use HP, factoring in assumptions for motor efficiciency and pf to get KVA.