What does power factor measure?

B. The fraction of the apparent power (kVA) that is real power (kW)

A. How efficient the motor is at turning electricity into shaft power

C. How balanced the three phases are

D. The ratio of line voltage to phase voltage

Troisième année Lesson 3 of 13 · about 14 min

Three-phase power and power factor

What you’ll learn

The three kinds of AC power — kW, kVA and kVAR — and the triangle that ties them together.
The three-phase power formula and where the √3 in it comes from.
How to work current from power and power from current, at 208, 480 and 600 V.
What power factor is (and is not), and why a low one costs current you pay for in copper.

Before this: Wye and delta · Second year: AC fundamentals

Three-phase power is worked out with one formula — P = √3 × V line × I line × PF — and everything else in this lesson is about what the pieces mean. The voltage and current are the ones you can put a meter on. The power factor says how much of that current does useful work. The √3 is the same one you met in wye and delta.

Three kinds of power

In a DC circuit, volts times amps is watts and that is the end of it. In an AC circuit with motors and transformers on it, the current does not rise and fall exactly in step with the voltage — it lags behind. That splits the power into three:

kW — real powerdoes the work: heat, light, a turning shaft. What the energy meter bills.
kVAR — reactive powerbuilds and collapses magnetic fields every cycle. Does no work, but still flows in the wires.
kVA — apparent powervolts × amps. What the conductors, breakers and transformer have to carry.

The power triangle. kW along the bottom, kVAR up the side, kVA on the slope. Power factor is the bottom over the slope — the cosine of the angle between them. A PF of 1 flattens the triangle: all the kVA is kW.

The formula, both ways

kVA = √3 × V × I ÷ 1000V line-to-line, I per line
kW = √3 × V × I × PF ÷ 1000real power
I = kW × 1000 ÷ (√3 × V × PF)current from real power
I = kVA × 1000 ÷ (√3 × V)current from apparent power

Where does the √3 come from? Three phases each carry Vphase × I × PF, so the total is 3 × Vphase × I × PF. On a wye, Vphase = Vline ÷ √3, and 3 ÷ √3 = √3. So 3 × (Vline ÷ √3) = √3 × Vline. The same result comes out for a delta, where the √3 sits on the current instead. Either way, in line values — the ones you measure — it is √3.

Worked: a motor on 480 V

Clamp reads 40 A on each line of a 480 V three-phase motor; the nameplate power factor is 0.86.
kVA = 1,732 × 480 × 40 ÷ 1000 = 33,3 kVA.
kW = 33,3 × 0,86 = 28,6 kW.
kVAR = √(33.3² − 28.6²) = 17.0 kVAR.
Sanity check: kW is always smaller than kVA unless PF is exactly 1.

Worked: one heater, three supply voltages

A 50 kW resistive duct heater (PF = 1.0).
On 600 V (common in Canada): I = 50,000 ÷ (1.732 × 600) = 48.1 A.
On 480 V (common in the US): I = 50,000 ÷ (1.732 × 480) = 60.1 A.
On 208 V: I = 50,000 ÷ (1.732 × 208) = 138.8 A.
Same useful heat, nearly three times the current at 208 V as at 600 V. That is why big loads run at the highest voltage available — the conductors and gear get smaller.

What a poor power factor costs

Take 20 kW of load at 480 V. At PF 1.0 it draws 20,000 ÷ (1.732 × 480) = 24.1 A. At PF 0.7 the same 20 kW needs 20 ÷ 0.7 = 28.6 kVA, and the current rises to 34.4 A — 43% more current for no more useful work. Every one of those amps heats the conductors (I²R), drops voltage and takes capacity from the transformer.

Induction motors are the usual cause, and they are worst lightly loaded: a motor idling on a conveyor still needs its full magnetising current, while its kW has dropped away. Capacitors supply that reactive current locally so the supply does not have to — see the power factor correction calculator.

Sur le chantier

Tips

Clamp all three lines and use the average if they differ a little. Badly different currents mean an unbalance problem to find, not a number to average.
Heaters and incandescent lamps are PF 1.0. Motors, transformers and fluorescent ballasts are not.
A power-quality meter reads kW, kVA and PF directly — handy for checking your arithmetic.
Remember the approximate current for 1 kVA at each voltage: about 1.2 A at 480 V, 0.96 A at 600 V, 2.8 A at 208 V.

Common mistakes

Using line-to-neutral voltage in the √3 formula (or line-to-line in the 3 × formula).
Multiplying by 3 instead of √3.
Treating kVA as kW and undersizing for a motor-heavy load.
Confusing power factor with efficiency — a motor’s current from horsepower needs both.
Entering PF as a percentage (85) instead of a fraction (0.85).

Check yourself

0 right of 0 answered 8 questions · pick one answer each

1.A 480 V three-phase motor draws 40 A per line at a power factor of 0.86. What is its apparent power?
2.Same motor: 33.3 kVA at PF 0.86. What real power is it taking?
3.A 50 kW resistive heater on a 600 V three-phase supply. Current per line?
4.The same 50 kW heater is moved to a 208 V three-phase supply. The line current…
5.Two loads each take 20 kW on the same 480 V three-phase feeder. Load X has PF 1.0, load Y has PF 0.7. Which draws more current?
6.What does power factor measure?
7.In the power triangle, kW = 30 and kVAR = 40. What are the kVA and the power factor?
8.You only have line-to-neutral voltage (277 V) and line current (50 A) on a balanced wye load at PF 1. Total power?

Answers

B. 33.3 kVA — kVA = √3 × V × I ÷ 1000 = 1.732 × 480 × 40 ÷ 1000 = 33.3 kVA. 28.6 is the kW (33.3 × 0.86). 19.2 forgot the √3; 57.6 used 3 instead of √3.
A. 28.6 kW — kW = kVA × PF = 33.3 × 0.86 = 28.6 kW. 17.0 is the kVAR, the reactive part.
B. 48.1 A — I = W ÷ (√3 × V × PF) = 50,000 ÷ (1.732 × 600 × 1.0) = 48.1 A. 83.3 A forgot the √3 (single-phase formula).
B. rises to about 139 A — Same power at lower voltage means more current: 50,000 ÷ (1.732 × 208) = 138.8 A. Current scales with 600 ÷ 208 ≈ 2.9 times.
B. Y, about 1.4 times as much — Current depends on kVA, and kVA = kW ÷ PF. Y needs 20 ÷ 0.7 = 28.6 kVA against X’s 20 kVA — 1.43 times the current for the same useful power.
B. The fraction of the apparent power (kVA) that is real power (kW) — PF = kW ÷ kVA — the cosine of the angle between voltage and current. It is not efficiency: a motor can be 95% efficient and still have a PF of 0.8.
B. 50 kVA, 0.6 — kVA = √(30² + 40²) = √2,500 = 50 kVA (a 3-4-5 triangle). PF = kW ÷ kVA = 30 ÷ 50 = 0.6.
B. 41.6 kW — Using phase values: P = 3 × Vphase × I × PF = 3 × 277 × 50 = 41,550 W. Check with line values: 1.732 × 480 × 50 = 41,570 W — the same, to rounding.

Questions fréquentes

What is the three-phase power formula?

Power (watts) = √3 × line voltage × line current × power factor. Leave off the power factor and you have volt-amperes (VA). Divide by 1,000 for kW or kVA. The √3 (1.732) is there because the formula uses line values; written with phase values it is 3 × Vphase × Iphase × PF, which gives the same answer.

Is power factor the same as efficiency?

No. Efficiency is shaft power out ÷ electrical power in — how much is lost as heat. Power factor is real power ÷ apparent power — how much of the current does useful work at all. A motor can be 95% efficient with a power factor of 0.8; the reactive current it draws is not lost as heat in the motor, it just circulates, heating the supply conductors on the way.

Why does the utility care about my power factor?

Because they have to supply the kVA — the full current — even though you only pay for kW. Low power factor means bigger transformers and conductors for the same useful power, so many utilities bill large customers for demand in kVA or add a power-factor charge. Correcting it is the subject of the power factor correction tool.

Voir aussi

Three-phase power calculator — this arithmetic, with your numbers
Calculateur de correction du facteur de puissance
Power factor (glossary)
Single & three phase formula sheet — printable
Wye and delta: line vs phase
Rusty on three-phase math?

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