Three-phase power calculator

How do you calculate three-phase power?

kW = √3 × line-to-line voltage × current × power factor ÷ 1,000, the formula the U.S. Department of Energy uses for motor loads. Apparent power is kVA = √3 × V × I ÷ 1,000.

Example

The DOE’s motor example: 469.7 V, 37 A and a power factor of 0.763 give 1.732 × 469.7 × 37 × 0.763 ÷ 1,000 = 22.97 kW. The DOE sheet prints it as 22.9 kW.

What is the power factor?

Real power divided by apparent power (kW ÷ kVA). Motors often run below 1; resistive heaters are close to 1.

How do I find the current for a load?

Choose “Current from power”: amps = kW × 1,000 ÷ (√3 × V × PF). A 50 kW load at 480 V and PF 0.9 draws about 66.8 A.

Which voltage should I enter?

The line-to-line voltage, such as 208, 240, 480 or 600 V, and the average current of the three phases.

Why is the factor √3 in the formula?

A balanced three-phase load takes the same current in each of three phases, and the voltage from line to line is √3 (1.732) times the voltage from line to neutral. Adding the three phases, 3 × (V ÷ √3) × I × PF, gives √3 × V × I × PF when V is the line-to-line voltage. On a 480 V system each line is about 277 V from the neutral.

Does a low power factor waste power?

Not in itself. The Department of Energy's motor fact sheet explains that a low power factor means excess current: the reactive energy is stored in the motor's windings and returned to the line every cycle. The supply wiring still has to carry that extra current, and that cost is why capacitors are used to correct it.

What the formula assumes

The same expression works for a wye or a delta connection when the load is balanced, and the kVA figure simply drops the power factor. If the phases are not equal, the Department of Energy sheet uses the average of the three line-to-line voltages and of the three currents, which is what to enter here. It gives the power a load takes in, not the mechanical power a motor delivers, which is lower by the motor's efficiency.

Worked example

The Department of Energy's fact sheet measures a 40 hp motor at 467, 473 and 469 V, 36, 38 and 37 A, and power factors 0.75, 0.78 and 0.76. The averages are 469.7 V, 37 A and 0.763, so P = √3 × 469.7 × 37 × 0.763 ÷ 1,000 = 22.97 kW; the sheet prints 22.9. The apparent power is 30.1 kVA. The same 22.97 kW at a power factor of 1 would need only 28.2 A, so the low power factor adds 31.1% to the current. For a 50 kW load at 480 V and PF 0.9, the current is 66.8 A; at 208 V the same load draws 154.2 A, and single-phase at 480 V it would draw 115.7 A.

Common mistakes

Entering the voltage from a line to neutral instead of line to line is the commonest: 277 V in place of 480 V gives 28.8 kW for the same current instead of 50.0 kW. Treating kW as kVA is the next; they differ by the power factor. A motor's horsepower rating is output, not input: the sheet multiplies by 0.7457 kW per horsepower and divides by the full-load efficiency. Last, the result applies to the load as measured; for sizing wire and breakers, have a licensed electrician confirm it against the National Electrical Code, which has its own tables for motors.

Sources: U.S. Department of Energy, Motor Challenge Program, “Determining Electric Motor Load and Efficiency,” fact sheet DOE/GO-10097-517 (equations 1 and 2, the 40 hp example, glossary of power factor); the √3 relation is the standard line-to-line geometry. Related: watts to amps calculator, breaker size calculator, wire size calculator, electricity cost calculator. How we check our calculators.