Motor Power Calculator

Motor Power and Full Load Current

This calculator does two jobs that come up constantly on the shop floor. It converts a motor rating between kilowatts (kW) and horsepower (HP), and it works out the full load current the motor draws from the supply. Knowing the current is the first step when you size cables, select a starter, or set an overload relay.

kW to HP Conversion

Motor ratings are quoted in kW in most of the world and in HP on a lot of older or imported equipment. The conversion is straightforward.

1 HP = 0.7457 kW (metric horsepower)
kW = HP x 0.7457
HP = kW / 0.7457

A quick note on standards. The metric (or mechanical) horsepower used here is 0.7457 kW. Some references round this to 0.746 kW, and the imperial horsepower is very close to that figure, so you may see a difference in the second decimal place depending on which value a supplier uses.

Full Load Current Formulas

For a three phase motor:

I = (kW x 1000) / (1.732 x V x PF x Eff)

For a single phase motor:

I = (kW x 1000) / (V x PF x Eff)

Here I is the full load current in amperes, V is the supply voltage, PF is the power factor (a value between 0.5 and 1), and Eff is the efficiency expressed as a decimal (for example 0.9 for 90 percent). The 1.732 in the three phase formula is the square root of 3, which comes from the line to line relationship in a balanced three phase system.

Worked Example

Take a 10 HP three phase motor on a 415 V supply, with a power factor of 0.85 and an efficiency of 90 percent. First convert to kW: 10 x 0.7457 = 7.457 kW. Then apply the three phase formula:

I = (7.457 x 1000) / (1.732 x 415 x 0.85 x 0.9)
I = 7457 / 549.9
I = 13.56 A (approximately)

So this motor draws roughly 13.6 A at full load. If the same 7.457 kW were run as a single phase motor at 230 V with the same power factor and efficiency, the current would be about 42 A, which shows why heavier motors stay on three phase supplies.

Practical Notes

• Use the motor nameplate current for sizing protection when it is available. Treat the calculated value as an estimate.
• Starting current can be six to eight times the full load current, which is why star delta or soft starters are used on larger motors.
• A poor power factor raises the current for the same shaft power, so power factor correction can reduce losses and demand charges.
• Efficiency drops at part load, so a motor that is oversized for its duty often runs less efficiently than a correctly matched one.