TP

Engineering

Three-Phase Power Calculator

Calculate balanced three-phase apparent, real, reactive, useful, and loss power from line-to-line voltage, line current, power factor, and load efficiency. Extend real input power across operating hours to audit energy use and cost.

Apparent power (kVA)-
Real input power (kW)-
Reactive power (kvar)-
Useful load output (kW)-
Energy over entered hours (kWh)-
Energy cost over entered hours-
Modeled conversion loss (kW)-
Line current per real kW (A/kW)-

Decision view

Three-phase power triangle and real-power conversion

Three-phase power triangle and real-power conversionThe electrical triangle separates kVA, kW, and kvar while the output bar reconciles useful power and modeled loss.
Exact scenario comparisonPower factor changes while all other entered assumptions remain constant.
Power factorApparent power (kVA)Real input power (kW)Reactive power (kvar)Useful load output (kW)Energy over entered hours (kWh)Energy cost over entered hoursModeled conversion loss (kW)Line current per real kW (A/kW)

How to use Three-Phase Power Calculator

  1. Enter balanced line voltage and line current.
  2. Enter power factor as a decimal and load efficiency as a percentage.
  3. Choose operating hours and an energy rate.
  4. Read the power triangle before interpreting useful output and period cost.

Calculator guide

Understanding Three-Phase Power Calculator

Balanced three-phase power has three related but different magnitudes: apparent power sets electrical capacity, real power performs work, and reactive power sustains alternating fields. This calculator keeps those quantities, load efficiency, energy, and cost on their correct units.

Triangle first S, P, and Q must reconcile geometrically.
PF is not efficiency They act at different stages.
Energy uses kW Period kWh comes from real input power.
Units stay separate kVA, kW, and kvar are not interchangeable.

Calculation method

How the calculation works

Apply the square-root-of-three balanced three-phase relationship to line voltage and current, then separate apparent, real, reactive, useful, and loss power. Multiply line voltage and line current by the square root of three for apparent power, apply power factor for real input power, recover reactive power with the power triangle, and apply efficiency only to real power.

Detailed calculation process

Build the three-phase power triangle before applying load efficiency

The defaults use 400 V line voltage, 125 A line current, power factor 0.88, and 92% load efficiency.

General formula: S = sqrt(3) V_L I_L / 1000; P = S PF; Q = sqrt(S^2 - P^2); P_out = P eta; E = P h; C = E r The square-root-of-three factor converts balanced line quantities to total three-phase apparent power. Power factor projects S onto the real-power axis; efficiency then separates useful output from conversion loss.

What each symbol means

V_L Line-to-line RMS voltage, measured in volts (V).
I_L Line RMS current, measured in amperes (A).
S Three-phase apparent power, measured in kilovolt-amperes (kVA).
PF Power factor P/S, unitless from 0 to 1.
P / Q Real power in kW and reactive power in kvar.
eta / h / r Load efficiency, operating hours, and energy rate in currency per kWh.

Worked substitution with the default inputs

1. Calculate apparent power: S = 1.7320508 x 400 V x 125 A / 1000 = 86.6025 kVA Dividing by 1,000 converts volt-amperes to kilovolt-amperes.
2. Resolve real power: P = 86.6025 kVA x 0.88 = 76.2102 kW Power factor converts the apparent-power magnitude to the real-power component.
3. Recover reactive power: Q = sqrt(86.6025^2 - 76.2102^2) = 41.1339 kvar S is the hypotenuse of the right power triangle with legs P and Q.
4. Apply load efficiency: P_out = 76.2102 x 0.92 = 70.1134 kW; loss = 76.2102 - 70.1134 = 6.0968 kW Efficiency applies to real input power, not directly to kVA or kvar.
5. Reconcile energy and cost: E = 76.2102 x 720 = 54,871.37 kWh; C = 54,871.37 x 0.14 = 7,681.99 The period cost uses electrical real input energy.

The default system carries 86.603 kVA, consumes 76.210 kW, exchanges 41.134 kvar, delivers 70.113 kW, and loses 6.097 kW in the modeled load.

Power anatomy

Separate electrical capacity from useful work

The diagram combines the power triangle with a real-power output split.

Capacity S sets total electrical loading.
Work component P is the real input.
Field component Q completes the power triangle.
Conversion Efficiency splits P into useful output and loss.

Worked situations

Practical examples

  • A 400 V, 125 A balanced load carries 86.603 kVA.
  • At PF 0.88, real input power is 76.210 kW.
  • At 92% efficiency, modeled conversion loss is 6.097 kW.

Better inputs

Useful tips

  • Confirm whether source data uses line-to-line or phase voltage.
  • Do not treat kvar as consumed kWh.
  • Use measured demand intervals for tariff analysis.

Before relying on the result

Limitations and common mistakes

  • The model assumes a balanced sinusoidal three-phase system.
  • Harmonics, imbalance, starting current, conductor temperature, and protection are excluded.
  • Tariff demand rules and site electrical design require separate review.

Reference

Key terms

Apparent power
RMS voltage-current capacity measured in kVA.
Real power
Average power converted to work or heat, measured in kW.
Reactive power
Oscillating field power measured in kvar.

Important note

Calculated from the entered values using the displayed engineering relationship. Confirm design values, load cases, safety factors, standards, and field conditions with a qualified professional.

Frequently asked questions

Why is sqrt(3) used?

Balanced three-phase line quantities combine across three phases to produce the square-root-of-three relationship.

Is power factor the same as efficiency?

No. Power factor relates kW to kVA; efficiency relates useful output to real input kW.

Does reactive power create energy cost?

It does not directly create kWh, though it can affect capacity and utility charges.

Can I use phase-to-neutral voltage?

Not in this line-to-line formula unless you first convert the voltage convention.