How the calculation works
Power factor is the ratio of active power (kW, which does useful work) to apparent power (kVA, which the supply has to carry). Inductive loads such as motors and transformers draw reactive power (kvar) to build their magnetic fields. Capacitors supply that reactive power locally, so the network no longer has to carry it.
The current drawn from the supply falls in proportion to the power factor improvement, and resistive losses in cables and transformers fall with the square of the current:
Loss reduction = 1 − (PF1 ÷ PF2)²
The capacitance needed in each phase depends on how the capacitors are connected, with V the line voltage and ω = 2πf:
Star: C = Qc ÷ (ω · V²)
Worked example
A 400 V, 50 Hz plant draws 400 kW at a power factor of 0.75 and wants to reach 0.95:
- tan(arccos 0.75) = 0.882 and tan(arccos 0.95) = 0.329, a difference of 0.553.
- Reactive power to add: 400 × 0.553 = 221 kvar. The next standard bank size is 225 kvar; many plants would choose 250 kvar for margin and load growth.
- Supply current falls from 770 A to 608 A, and apparent power from 533 kVA to 421 kVA.
- Losses in the cables and transformer fall by about 38%.
- For a delta-connected bank, the capacitance is about 1,470 µF per phase.
kvar multiplier table
Multiply your kW by the value where your existing power factor row meets your target column.
| Existing PF | 0.90 | 0.92 | 0.95 | 0.98 | 1.00 |
|---|---|---|---|---|---|
| 0.60 | 0.849 | 0.907 | 1.005 | 1.130 | 1.333 |
| 0.65 | 0.685 | 0.743 | 0.840 | 0.966 | 1.169 |
| 0.70 | 0.536 | 0.594 | 0.692 | 0.817 | 1.020 |
| 0.75 | 0.398 | 0.456 | 0.553 | 0.679 | 0.882 |
| 0.80 | 0.266 | 0.324 | 0.421 | 0.547 | 0.750 |
| 0.85 | 0.135 | 0.194 | 0.291 | 0.417 | 0.620 |
| 0.88 | 0.055 | 0.114 | 0.211 | 0.337 | 0.540 |
| 0.90 | – | 0.058 | 0.156 | 0.281 | 0.484 |
Practical points before you buy a bank
- Harmonics. With variable speed drives, rectifiers or UPS systems on site, plain capacitors can resonate with the network and overload. Measure harmonic distortion first; if it is significant, specify a detuned bank with series reactors.
- Varying load. If the load changes through the day, use an automatic bank with a controller and several steps. A fixed bank sized for full load will overcompensate at light load and push the voltage up.
- Correcting individual motors. A capacitor connected directly at a motor's terminals should not exceed about 90% of the motor's no-load magnetising kvar, to avoid self-excitation when the motor is switched off.
- Check your tariff. Savings come from avoiding reactive energy penalties, reducing kVA demand charges, freeing transformer and cable capacity, and lowering losses. Which of these applies depends on how your utility bills you.
Questions
What target power factor should I aim for?
Most plants aim for 0.95 to 0.98. Reaching 1.00 needs much more kvar for little extra benefit, and risks overcompensation at light load.
Why is the standard bank size larger than the calculated kvar?
Capacitor banks are sold in standard ratings, so the calculator rounds up to the next common size. Exact steps vary by manufacturer. Many engineers also add margin for load growth and capacitor ageing.
Can I use this for medium voltage?
Yes, the kvar calculation is the same. Select kV for the voltage. Medium voltage banks are usually star-connected, and their protection and switching need specific design work.