Power Factor Correction Capacitor Calculator (kvar, µF)

Formulas

• Qc (kvar) = P × (tan φ1 − tan φ2) = K × P, tan φ = √(1 − cos²φ) ÷ cos φ
• Motor: P = rated output ÷ efficiency (ABB Motor guide: Q = K × P ÷ η)
• Per capacitor: C = Q_each × 1000 ÷ (2π × f × U²) [F], Q_each = Qc ÷ 3 (three-phase)
• Delta: U = line voltage, star: U = line voltage ÷ √3 → star needs 3× the µF for the same kvar
• Capacitor line current = Qc × 1000 ÷ (√3 × V)

ABB's Motor guide gives C = 3.2 × 10⁶ × Q ÷ U² at 50 Hz. 3.2 × 10⁶ is 10⁹ ÷ (2π × 50) ≈ 3.18 × 10⁶ rounded; with Q in kvar and U in volts, C comes out in µF. This calculator uses C = Q ÷ (2π f U²) without rounding, so it also works at 60 Hz.

K table (tan φ1 − tan φ2)

These values are computed from the formula and agree with ABB Motor guide Table 4.7 within ±0.001. Multiply the output (or real power) kW by K for the kvar needed.

Directly at a motor (ABB Motor guide §4.10)

• Capacitors are usually connected in parallel with the motor or a group of motors. Over-compensation can make an induction motor self-excite and run as a generator, so it is normal practice not to compensate for more than the motor's no-load current.
• Capacitors must not be connected in parallel with single phases of the winding; that can make star-delta starting difficult or impossible.
• On a two-speed motor with separate windings and compensation on both, the capacitors should not remain in circuit on the unused winding. Such capacitors can in some cases increase winding heating and vibration.

Choosing the target

• The closer the target is to 1.0, the faster the required kvar grows, and the easier it is to end up leading (over-compensated) when the load drops. Plants with varying load often use automatic correction that switches several steps.
• Utility power factor requirements and tariffs differ by contract, so check your supply agreement. This page does not quote them because we have not checked the original terms.

FAQ

How many kvar raise a 100 kW load from 0.75 to 0.95?

K = tan(acos 0.75) − tan(acos 0.95) ≈ 0.882 − 0.329 = 0.553, so Qc ≈ 55.3 kvar. At 400 V 50 Hz in delta, each capacitor is about 18.4 kvar, about 367 µF.

How do I convert kvar to µF?

With the voltage U across one capacitor and the frequency f: C (µF) = kvar × 10⁹ ÷ (2π × f × U²). The same capacitor gives 1.2 times the kvar at 60 Hz as at 50 Hz.

Delta or star?

For the same kvar, star needs three times the capacitance per capacitor but sees 1/√3 of the voltage. Three-phase capacitor units are sold by rated voltage and kvar, so check the unit's rated voltage against the line voltage.

Should I correct to exactly 1.0?

It turns leading when the load drops, and at a motor it risks self-excitation, so targets are usually lower. See «Choosing the target» above.

To convert kVA, kW, kvar and amps, use the Three-Phase Power Calculator. For a motor's rated current, use the Motor Current Calculator. For one capacitor's reactance and current, use the Reactance & LC Resonance Calculator.

Sources

• ABB, Low voltage motors — Motor guide (2019): §4.10 Power factor, «Reactive-power compensation», Table 4.7 Phase compensation
• ABB, Technical guide No. 7 — Dimensioning of a drive system (3AFE64362569 Rev D): eq. 3.15