That's the inverse of efficiency, plus a unit conversion constant... but I think you know that. It looks like you're calculating an "electrical power to produce mechanical power" factor, in watts/horsepower. As for actual efficiency:So here goes, someone throws a motor on the bench and ask what can it do? All we have is nameplate data that reads:
Operating Voltage = 48 vdc
Load Torque = 5.237 ft-lbs
Load Speed = 4150 Rpm
Load Current = 88 amps
Peak HP = 10
Continuous HP = 4.14
First thing it tells me I would be using 48 volts x 88 amps = 4224 watts. From that if I divide 4224 watts / 4.14 HP = 1020 w/hp efficiency...
input: 48 volts x 88 amps = 4224 watts (as above)
output: 4.14 hp = 4.14 x 746 = 3088 watts
efficiency = output / input = 3088 W / 4224 W = 73%
That's not good, but I don't know if any better can be expected.
The "Load Torque" and "Load Speed" do correspond to the listed "Continuous HP".