Power factor correction calculator: size your capacitor bank in kvar and stop paying reactive power charges

Calculator

System Active power P (kW) Current cos φ Target cos φ Voltage (V) Frequency

Capacitor bank rating 27.66 kvar

Factor k: 0.553
Line current: 96.23 A → 75.97 A (−21.1 %)
Cable losses: −37.7 %
Bank current: 39.93 A
Capacitance per capacitor: 183.4 µF

In three-phase, the voltage is the line-to-line voltage and the capacitance is that of each of the three capacitors connected in delta.

What you gain by compensating

No more surcharge. Many utilities charge for reactive energy when the power factor falls below a limit, typically 0.9 to 0.95. In Spain, for supplies above 15 kW, it is billed when reactive energy exceeds 33 % of active energy, which is exactly a cos φ of 0.95.
Less current. Less current flows for the same useful power: cables, protective devices and the transformer have more headroom and can take more load.
Lower losses. Cable losses go with the square of the current: cutting the current by 20 % reduces them by more than 35 %.
Lower voltage drop on long lines, especially with large motors.

Formulas

The reactive power the capacitor bank has to supply is:

Q_c = P × (tan φ₁ − tan φ₂) = P × k

P is the active power in kW; φ₁ and φ₂, the angles of the current and target power factor; and tan φ = √(1 − cos² φ) / cos φ. The capacitance of each capacitor, with ω = 2π × f:

System	Capacitance per capacitor	Bank current
Three-phase, three capacitors in delta	`C = Q_c / (3 × ω × U²)`	`I_c = Q_c / (√3 × U)`
Single-phase	`C = Q_c / (ω × U²)`	`I_c = Q_c / U`

The line current drops in proportion to the power factors, I₂ = I₁ × cos φ₁ / cos φ₂, and cable losses with the square of that ratio.

Worked example

Data: a workshop drawing 50 kW at 400 V three-phase with cos φ 0.75; the target is 0.95.
Factor k: tan φ₁ = 0.882 and tan φ₂ = 0.329, so k = 0.553.
Bank: Q_c = 50 × 0.553 = 27.66 kvar; in practice, a standard 30 kvar bank.
Result: the current drops from 96.23 A to 75.97 A (21 % less) and cable losses by 37.7 %.

Table of k factors

Multiply the active power in kW by the factor in the table to get the kvar needed:

Factor k from the current cos φ (rows) to the target (columns)

Current cos φ	0.90	0.92	0.95	0.98	1.00
0.50	1.25	1.31	1.40	1.53	1.73
0.55	1.03	1.09	1.19	1.32	1.52
0.60	0.85	0.91	1.00	1.13	1.33
0.65	0.68	0.74	0.84	0.97	1.17
0.70	0.54	0.59	0.69	0.82	1.02
0.75	0.40	0.46	0.55	0.68	0.88
0.80	0.27	0.32	0.42	0.55	0.75
0.85	0.14	0.19	0.29	0.42	0.62
0.90	–	0.06	0.16	0.28	0.48

Common questions

What is reactive power?

The energy that motors, transformers and reactors need to build their magnetic field. It does no useful work, but it flows through cables and the transformer and loads them just like active power.

Up to what cos φ should I compensate?

The usual target is 0.95, the limit above which many utilities stop charging the surcharge. Going beyond 0.98 costs a lot for little gain, and reaching 1 or overcompensating can cause overvoltages at light load.

Fixed or automatic bank?

Fixed for a constant load, such as a motor that always runs or a transformer's no-load consumption. Automatic, with steps switching in and out with demand, when the load changes during the day.

What if there are harmonics?

Capacitors can resonate with harmonics from power electronics and become overloaded. In such installations detuned banks (with blocking reactors) are used: ask the manufacturer.

How do I find my current cos φ?

From the electricity bill (active and reactive energy for the period: cos φ = kWh / √(kWh² + kvarh²)), from the power analyser in the main switchboard or, for a motor, from its nameplate.

Stop paying for reactive power