Motor HP Conversion Formula

EFFICIENCY= POUT/PIN= 746/PIN

PIN = 921 W

PIN = V I COS (ANGLE)

COS (ANGLE) = PF =0.86

CURRENT= 921/V* PF

Your formula is correct.

I am not a practising electrical engineer and while my first impression was that your formula was correct, however it occurs to me that you may be double counting.

As you might know, apparent power is V xI but real power is V x I x PF where V = Volt, I = Current and PF = Power Factor. Efficiency is "power out"' over "power in" expressed as a percentage. It follows then that if the "power in" figure was apparent power (and I would think this was the most common used definition), then the efficiency term would already include an allowance for "PF" , and including it again would be double counting.

The formula is correct and an allowance for power factor is needed, as shown in the formula. Efficiency does not take power factor into account.

So just to confirm then (because I don't believe it).

You are saying that power in is always VI X Power Factor i.e. "real power" and that it is never expressed as "apparent power", and that if one saw the term "850VA" on the case of a 1 approximately HP output motor that such a VA is always real power

Since you are a guest - what is your authority.