Power factor, kW and kVA explained

Power factor (PF) is active power divided by apparent power: PF = kW ÷ kVA. It is the fraction of the apparent power that is active power. At a power factor of 1, every amp delivers energy. At 0.8, the same kW needs 25% more current (1 ÷ 0.8 = 1.25).

Read the three-phase power guide first if you need the formulas for kW and kVA.

kW, kVA and kvar

When voltage and current are sinusoidal, the three powers form a right-angled triangle: S² = P² + Q². A load of 80 kW and 60 kvar draws 100 kVA, so its power factor is 0.8:

Power factor is not efficiency. It tells you how much current a load draws for its kW. It does not tell you how much of that kW the load loses as heat. An induction motor draws an almost constant magnetising current, so at part load its reactive power stays nearly the same while its active power falls. The US Department of Energy motor fact sheet shows power factor falling steeply as load drops, while the efficiency of larger motors stays close to rated over a much wider range. An oversized motor can be efficient and still have a poor power factor.

Current and copper losses

For a balanced three-phase load, I = P ÷ (√3 × V_LL × PF). At 400 V line to line, 80 kW at a power factor of 0.80 draws 144 A per line. At 0.95 it draws 122 A.

Copper loss in cables and transformer windings is I²R, so it scales with (1 ÷ PF)². Correction from 0.80 to 0.95 multiplies those losses by (0.80 ÷ 0.95)² = 0.71, a 29% reduction. The released current is also capacity. A 1,000 kVA transformer at a power factor of 0.80 can supply 800 kW. At 0.95 it can supply 950 kW.

The reduction applies only upstream of the point where the capacitors connect. A central bank at the main switchboard relieves the transformer and the supply. It does not relieve the sub-main cables to the motors.

Reactive energy charges

In Ireland, ESB Networks applies a low power factor surcharge to its distribution use of system charges. The surcharge applies when the metered kvarh in a billing period is more than one third of the metered kWh. It is charged only on the kvarh above that third, and it does not apply to sites with a maximum export capacity above zero. One third of the kWh is tan φ = 0.333, which is a power factor of 0.949. In Great Britain, the common distribution charging methodology uses a power factor of 0.95, which is reactive energy above 33% of active energy. It calculates the excess for each half hour. Other contracts charge for maximum demand in kVA, and there the demand charge rises directly as the power factor falls.

A worked example: a site uses 100,000 kWh in a month at an average power factor of 0.80. Its reactive energy is 100,000 × tan(cos⁻¹ 0.80) = 75,000 kvarh. The allowance is 33,333 kvarh, so 41,667 kvarh are chargeable. At an average of 0.95, the reactive energy is 32,868 kvarh, and a monthly test charges nothing. A half-hourly test can still charge that site for the periods when its power factor drops, for example at night when the base load is mostly transformers and idle motors. Multiply the chargeable kvarh by the rate in your tariff to get the value of correction.

True and displacement power factor

Meters report one or both of two power factors. Displacement power factor (cos φ) comes from the phase angle between the fundamental (50 or 60 Hz) voltage and current. True power factor is P ÷ S with RMS values, so it includes harmonic current as well. The true RMS guide explains RMS values on distorted circuits.

With a sinusoidal voltage, true power factor = displacement power factor × 1 ÷ √(1 + THD_I²), where THD_I is the total harmonic distortion of the current. A six-pulse drive with no DC link choke can draw current with THD_I of 85% or more. With a displacement power factor of 0.97, its true power factor is 0.97 ÷ √(1 + 0.85²) = 0.74. A DC link choke brings THD_I down to about 30%, and the true power factor rises to 0.93. The THD and TDD calculator gives THD and TDD from harmonic magnitudes.

Capacitors correct displacement, not distortion. On a site with high THD_I, a plain capacitor bank gives little improvement in true power factor, and it can resonate with the supply.

Unbalanced three-phase loads

Total kW and total kvar add across the phases. Total kVA does not have one definition. Arithmetic apparent power adds the three phase values: S_A = S_a + S_b + S_c. Vector apparent power uses the totals: S_V = √(P² + Q²). When the phases have different power factors, the two give different answers.

The same site reads 0.90 on one meter and 0.949 on another. The vector value is exactly on the Irish one-third threshold. A tariff meter derives power factor from total kWh and kvarh, which is the vector form. A monitor that sums per-phase kVA gives the arithmetic form. IEEE 1459 recommends a third definition, effective apparent power S_e, which also counts the extra current that unbalance causes and gives the lowest power factor of the three. Find out which definition your meter uses before you compare it with a bill.

The same rule applies when you combine loads. Add kW and kvar, then calculate the power factor. A 100 kW load at 0.70 and a 20 kW load at 1.00 give 120 kW and 102 kvar, which is a power factor of 0.76. The mean of the two power factors, 0.85, is wrong.

Sizing power factor correction

The reactive power that a capacitor bank must supply to move a load from one power factor to another is Q_c = P × (tan φ₁ − tan φ₂). For 300 kW at 0.80, corrected to 0.95, Q_c = 300 × (0.750 − 0.329) = 126 kvar:

Size a real bank from measured demand over the whole operating cycle, not from one reading. A fixed 126 kvar bank on that site at night, with 60 kW at 0.80 (45 kvar), leaves 81 kvar leading and a power factor of about 0.6 leading. Leading reactive power raises the voltage. An automatic bank avoids this. Its controller reads the power factor at the incomer and switches steps in and out to follow the load. Switch the bank out when a standby generator carries the site, because most generators can absorb only a little leading reactive power before their voltage regulation fails.

Check for resonance before you add capacitors to a site with drives. The bank and the supply inductance resonate at harmonic order h₀ ≈ √(S_sc ÷ Q_c), where S_sc is the short-circuit power at the point of connection in kVA and Q_c is the bank rating in kvar. Behind a 1,000 kVA transformer with 6% impedance, S_sc is about 16,700 kVA before upstream impedance. With 300 kvar connected, h₀ is 7.5. At 600 kvar it is 5.3. As steps switch in, the resonance moves down through the 7th harmonic toward the 5th, the two largest harmonic currents from a six-pulse drive. Symptoms of resonance are blown capacitor fuses, swollen or hot capacitor cans, and voltage THD that rises when a step switches in.

The usual fix is a detuned bank, with a reactor in series with each capacitor step. A 7% reactor tunes each step to 50 ÷ √0.07 = 189 Hz on a 50 Hz supply, below the 5th harmonic at 250 Hz. A 14% reactor tunes to 134 Hz, for sites with a large 3rd harmonic. On sites where distortion is the main problem, an active harmonic filter or active front end drives treat the cause instead: they draw near-sinusoidal current, and an active filter can supply reactive power as well.

Read power factor on a monitoring system

Calculate an average power factor from energy counters. True average power factor is kWh ÷ kVAh for the same period. The form kWh ÷ √(kWh² + kvarh²) gives a different answer. Reactive energy meters to IEC 62053-24 measure the fundamental component only, so this form gives an average displacement power factor. On a distorted circuit it reads higher than the true value. If you use it, take the lagging (import) kvarh register, not a net of the leading and lagging registers. Never average power factor readings. An average of readings gives a lightly loaded night interval the same weight as a fully loaded day interval.

Keep kW and kVA beside the ratio. A 2 kW circuit at a power factor of 0.5 draws 3.5 kvar and hardly changes the site total. The same ratio on a 500 kW incomer means 866 kvar.

Treat power factor below about 5% of rated current as indicative only. At that load, the current is mostly magnetising or standby current, and current transformer phase error grows. IEC 61869-2 allows a class 0.5S current transformer 90 minutes (1.5°) of phase displacement at 1% of rated current, against 30 minutes at 20% and above. At a true power factor of 0.5 (60°), a 1.5° error reads as 0.48 or 0.52 and moves kW by about 4.5%.

Know the sign convention. Under the IEC convention, the sign of the power factor follows the direction of active power: positive when the site imports and negative when it exports. An L or C label then marks lagging or leading. Meters set to an IEEE convention sign the power factor by load type instead, and manufacturers do not agree on whether lagging or leading is negative. A negative value can also mean a reversed current sensor. The negative readings guide shows how to tell these cases apart.

To find the source of a low site power factor, start with the incomer profile and find the periods when the power factor drops. Then compare the kvar or kVA of each circuit over the same periods. A ZEM electricity monitor reports kWh, kVAh, kW, kVA and power factor for each phase and in total, so the true average power factor of a circuit for any period is the kWh difference divided by the kVAh difference. The ZEM does not report kvar. √(kVA² − kW²) gives the non-active power, which includes distortion and does not show whether the circuit leads or lags.