sparkytools

Third year Three-phase · tool

Three-phase power calculator

Enter the line-to-line voltage, the power factor and any one of current, kW or kVA. The rest — and the kVAR — are worked out with the √3 formula, and the working is shown with your numbers.

Supply
I know the…
  • Current per line—
  • Apparent power—
  • Real power—
  • Reactive power—
  • Phase angle (lagging)—
Show the working

    Arithmetic on the numbers you entered, assuming a balanced load. It sizes nothing — conductors and protection come from your code book.

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    How it works

    Three quantities describe the power in an AC circuit. kVA (apparent power) is volts times amps — what the wires and the transformer have to carry. kW (real power) is the part that does work: heat, light, a turning shaft. kVAR (reactive power) is the part that sloshes back and forth building magnetic fields in motors and transformers. They form a right triangle, and the power factor is kW ÷ kVA.

    • kVA = √3 × V × I ÷ 1000three-phase, line values
    • kVA = V × I ÷ 1000single-phase
    • kW = kVA × PFreal power
    • kVAR = √(kVA² − kW²)reactive power
    • I = kW × 1000 ÷ (√3 × V × PF)three-phase current from kW

    Worked: the default numbers

    1. 480 V three-phase, 100 A per line, power factor 0.85.
    2. kVA = 1.732 × 480 × 100 ÷ 1000 = 83.1 kVA.
    3. kW = 83.1 × 0.85 = 70.7 kW.
    4. kVAR = √(83.1² − 70.7²) = 43.8 kVAR. The angle whose cosine is 0.85 is 31.8°.

    Motor loads are where the power factor matters: the same 70.7 kW at a power factor of 1.0 would need only 85 A. The extra 15 A is the reactive current, and it heats the conductors just as much as the useful current does — the subject of the power factor correction tool.

    Questions people ask

    Why is there a √3 in the three-phase power formula?

    Because the line voltage of a three-phase system is √3 times the voltage across one phase of a wye, and the three phases each carry their share. Written in line values — the ones you can actually measure — the total comes out as √3 × V line × I line. It is the same power as three single-phase circuits, just expressed in the numbers on your meter.

    Should I use line-to-line or line-to-neutral voltage?

    Line-to-line (208, 480, 600…) in the √3 formula, with the current in one line. If you only have line-to-neutral figures you can use 3 × V phase × I instead — it gives the same answer on a balanced load.

    What if the three line currents are different?

    The formula assumes a balanced load. For a modest imbalance, using the average of the three currents gets you close. For a badly unbalanced load, work each phase separately (V line-to-neutral × I × PF for each) and add them.

    Where do I get the power factor?

    From the equipment (a motor nameplate or data sheet gives it at full load), from a power-quality meter, or from the utility bill for a whole building. Resistive heat and incandescent lighting are 1.0. If you do not know it, the kVA and amps are still right — only the kW and kVAR need the PF.

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