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I was doing some research for a customer who is installing a hot tub and I just wanted to be sure we were not going to overload his existing electrical system. (I originally wanted to do a load calc which included a number for his existing AC unit.) I found this quote below from an old thread labeled "5 Ton AC Load" on Mike Holt's forum:

""For "guestimating" I have typically used 1.5KVA/ton, and hasn't failed me yet.

The only thing I will caution folks of is to watch your units, e.g. kW does not equal KVA when motors are involved.""


Can someone elaborate on how KW does not equal KVA when motors are involved?
 

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RIP 1959-2015
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I was doing some research for a customer who is installing a hot tub and I just wanted to be sure we were not going to overload his existing electrical system. (I originally wanted to do a load calc which included a number for his existing AC unit.) I found this quote below from an old thread labeled "5 Ton AC Load" on Mike Holt's forum:

""For "guestimating" I have typically used 1.5KVA/ton, and hasn't failed me yet.

The only thing I will caution folks of is to watch your units, e.g. kW does not equal KVA when motors are involved.""


Can someone elaborate on how KW does not equal KVA when motors are involved?
It has to do with the power factor which should also be on the name plate, such as 90% pf

I may be wrong but an incandescent lamp runs on 100% pf so you can use KW in that case..
 

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Bilge Rat
motors and controls.........
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KVA is volt-amps. This relates to wire size and overcurrent protection.

KW is watts. This relates to actual work being done.

In nearly all cases, KVA will be larger than KW.

KVA X Power Factor = KW.

When sizing wire and breakers for motors, KW doesn't matter much, KVA is what is used. Size the wire and breaker based on the nameplate amps and let the KW fall where it will.
 

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VA is the worst-case measurement of the amount of current a device will draw, but not an accurate measure of how much power in watts it actually uses (or in the case of a motor how much power is available at the shaft-end).

For example, a motor could draw 12kVA, but have a power-factor of 0.85 so the available shaft end power would only be 120V*10A*0.85pF=1.02kW.

So, if you used that kW rating for your load calculation, you would wrongly assume that it only drew 8.5A and your circuit would be sized improperly. That's why we use VA in service calculation.

That said, I don't believe I've ever seen anything except for some synchronous motors labeled in kVA. They're always based on shaft-end power which is either gonna be IEC kilowatts or NEMA horesepower.
 
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