Work out the capacitive kvar that raises a lagging power factor to a target, and how much the line current falls.
Capacitive reactive power to add
42.13kvar
From power factor 0.8 to 0.95 on 100 kW.
Check the values to see a result.
Choose the bank and its detuning with a harmonic and resonance check.
Qc = P × (tan φ1 − tan φ2), with φ = cos⁻¹(PF)
Raising 100 kW from a power factor of 0.8 to 0.95 needs 42.13 kvar of capacitance, and cuts the line current by 15.79%.
Tip: Size correction from measured kW and power factor over a representative period, not from a single reading or the transformer rating.
Power factor correction adds capacitive kvar to cancel part of the lagging kvar that motors and transformers draw. The capacitance needed is the real power times the difference between the tangents of the two phase angles. Raising a 300 kW load from a power factor of 0.8 to 0.95 takes 126.4 kvar and cuts the line current by about 16%.
Qc = P × (tan φ1 − tan φ2)
P is the real power in kW, φ1 the phase angle now and φ2 the angle at the target, each found as cos−1 of the power factor. The real power does not change; only the reactive part does.
Qc = P × k
k = tan φ1 − tan φ2 depends only on the two power factors, so it can be read from a table. From 0.8 to 0.95, k = 0.421: every kW needs 0.421 kvar.
I2 = I1 × PF1 ÷ PF2
For the same real power, current falls in proportion to the rise in power factor. That frees capacity in cables and transformers and reduces their losses.
Correcting to exactly 1.0 is rarely worth it. The last few hundredths need a lot of kvar, and a bank sized for full load can push the power factor leading when the load is light. Most sites aim for 0.95 to 0.98, or whatever the tariff sets.
tan(cos−1 0.8) = 0.75 and tan(cos−1 0.95) = 0.329, so the bank needs 300 × (0.75 − 0.329) = 126.4 kvar. At 400 V the line current falls from 541.3 A to 455.8 A.
Capacitive reactive power to add 126.4 kvar Open in the calculator
From 0.82 to 0.95 the plant needs 73.86 kvar. A 75 kvar bank, the next size up in 25 kvar steps, brings the power factor to 0.9516.
Capacitive reactive power to add 73.86 kvar Open in the calculator
From 0.75 to 0.98 a 150 kW load needs 101.8 kvar, and the line current falls by 23.47%. The first half of the kvar does most of the work: the move from 0.95 to 0.98 alone accounts for about 19 kvar.
Capacitive reactive power to add 101.8 kvar Open in the calculator
Multiply the real power in kW by the figure for your present and target power factors to get the kvar to add.
| Power factor now | To 0.90 (kvar/kW) | To 0.95 (kvar/kW) | To 0.98 (kvar/kW) | To 1.00 (kvar/kW) |
|---|---|---|---|---|
| 0.60 | 0.849 | 1.005 | 1.130 | 1.333 |
| 0.65 | 0.685 | 0.840 | 0.966 | 1.169 |
| 0.70 | 0.536 | 0.692 | 0.817 | 1.020 |
| 0.72 | 0.480 | 0.635 | 0.761 | 0.964 |
| 0.74 | 0.425 | 0.580 | 0.706 | 0.909 |
| 0.76 | 0.371 | 0.526 | 0.652 | 0.855 |
| 0.78 | 0.318 | 0.474 | 0.599 | 0.802 |
| 0.80 | 0.266 | 0.421 | 0.547 | 0.750 |
| 0.82 | 0.214 | 0.369 | 0.495 | 0.698 |
| 0.84 | 0.162 | 0.317 | 0.443 | 0.646 |
| 0.86 | 0.109 | 0.265 | 0.390 | 0.593 |
| 0.88 | 0.055 | 0.211 | 0.337 | 0.540 |
Multiply the real power in kW by tan φ1 − tan φ2, or by the multiplier in the table above. The answer is the capacitive kvar at the supply voltage.
The level your tariff rewards, commonly 0.95 or above. Going to 1.0 costs more kvar for little gain and risks a leading power factor at light load.
Only if harmonics are low. A meter’s total power factor includes distortion, which capacitors cannot correct and which can cause resonance with them. Where total harmonic distortion is significant, use displacement power factor and a detuned bank.
It reduces reactive energy and kVA demand charges where the tariff has them, and slightly reduces losses in cables and transformers. It does not reduce the kWh the load itself uses.
Size correction from a representative period, not one reading. A ZEM records power factor and demand continuously, before and after the bank goes in.
