OpenSpecEngineering

Finds the line voltage, line current and the real, reactive and apparent power of a balanced three-phase load. Pick a nominal system voltage and how the load is connected, then give the current in each phase and the power factor. Electricians and designers use it to check a load against its feeder and to size transformers, generators and breakers.

Equations

Wye: V_L = √3·V_φ, I_L = I_φ     Delta: V_L = V_φ, I_L = √3·I_φ

S = √3·V_L·I_L     P = S·cos φ     Q = S·sin φ

V_φ and I_φ are the voltage across and the current through one phase of the load. V_L and I_L are the line-to-line voltage and the current in each supply conductor [1]. S is the apparent power in kVA, P the real power in kW and Q the reactive power in kVAR. cos φ is the power factor, the cosine of the angle between each phase’s voltage and current.

The system list holds the nominal voltages of ANSI C84.1 [2] and IEC 60038 [3]. A wye load takes the line-to-neutral voltage, such as 277 V on a 480Y/277 V system. On a 3-wire system with no neutral, such as 480 V, the page uses V_L/√3. A delta load takes the line-to-line voltage. The row under the powers shows which phase voltage was used. For any other supply, choose Custom and enter the phase voltage.

Limits and common mistakes

• The load must be balanced, with the same current and power factor in all three phases. For an unbalanced load, add up the phases one at a time.
• Enter the phase current, not the line current. For a delta load the line current is √3 times larger.
• The nominal pair is rounded: 277 V × √3 is 479.8 V, not 480 V, so V_L may differ slightly from the system name.
• The power factor is the load’s displacement power factor. Harmonic currents from drives and rectifiers add to S but are not included.

Worked example

A heating and ventilating unit on a 480Y/277 V system has its loads wired in wye. Each phase draws 42 A at a power factor of 0.85. What does the unit take from the supply?

1. Choose Nominal system voltage 480Y/277 V and Winding Connection Wye (star).
2. Enter Phase current (I_φ) 42 A and Power factor (cos φ) 0.85. Press Solve. The Solution shows: Line voltage (V_L): 479.778 V
   Line current (I_L): 42 A
   Real power (P): 29.6667 kW
   Reactive power (Q): 18.3858 kVAR
   Apparent power (S): 34.902 kVA
   Phase voltage used, V_φ (line to neutral of 480Y/277 V): 277 V

Check: each phase takes 277 V × 42 A = 11.634 kVA, so three take 34.902 kVA. Then P = 34.902 × 0.85 = 29.667 kW, and Q = √(34.902² − 29.667²) = 18.39 kVAR.