Motor Current

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pcboiler

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What's the formula to determine motor current when horsepower and voltage are known? I'm trying to solve a problem where the answer uses I = 600*HP/V, but I can't for the life of me find this equation anywhere else. Any ideas?

Thanks
 
180206-2407 EST

pcboiler:

Quite often a fully loaded motor might have a rough estimate of of 1000 VA input for 1 HP of mechanical output. This covers power factor and motor efficiency. At 100% efficiency 1 HP = 746 W.

Your equation is more like 600 VA = 1 HP. This is not possible as a motor On the other hand if you supplied 1 HP to a generator and it could produce 600 VA output, then that is possible. That would be 80% efficiency.

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In my opinion, V it is line-to-neutral in a three phase system then from HP to W the factor is
1/.746*1000/3 and the product between power factor and efficiency [average]=0.745:
I=HP/.746*1000/3/.745/V=599.8*HP/V
 
My guess is that this formula was for guesstimating starting current on a motor; 600% is a usual value to use for that. But simply dividing HP by voltage doesn’t get you to FLA, and the percent sign may have been lost in the fog of time. So it might have originally been 600% of kVA/V, or even 600% of kW/V, which would be an over-estimated way to be sure you were covered, because by leaving out power factor, you are going to end up with a higher value and if using this to quickly estimate the size of a service or generator, it’s a safer bet. It’s nothing that is officially sanctioned anywhere though, it sounds more like a “rule of thumb” issue no matter what.
 
Julius Right said:
In my opinion, V it is line-to-neutral in a three phase system then from HP to W the factor is
1/.746*1000/3 and the product between power factor and efficiency [average]=0.745:
I=HP/.746*1000/3/.745/V=599.8*HP/V
Click to expand...

Oh that’s clever. But it would only apply to single phase motors connected L-N in a 3 phase Y system. Given that’s the way most of the rest of the world is configured; 400Y230V (nominal), that would make sense.

But it doesn’t work here in North America unless it’s a 208Y120 service and a 120V motor. (Technically it could apply to a 480Y277 service, but 277V motors are rarely used).
 
I use

i = (HP X 746)/(sqrt3 x 0.95 x 0.8 x V)
i = 565 x HP/V
0.95 for line V vs motor V ie 480 vs 460
0.8 ~ pf x eff

the likely used a smaller number for pf x eff
 
Ingenieur said:
I use

i = (HP X 746)/(sqrt3 x 0.95 x 0.8 x V)
i = 565 x HP/V
0.95 for line V vs motor V ie 480 vs 460
0.8 ~ pf x eff

the likely used a smaller number for pf x eff
Click to expand...

Yes, that is the way. JuliusRight divided HP by 746 instead of multiplying it to find watts.
:) And still more hilarious jraef objection to it.:D
 
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