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Electrical Engineering DISCUSSION
Our workshop draws about 100 kW on a 400 V, 50 Hz three-phase supply at a power factor of 0.75 lagging, mostly induction motors. The utility charges a penalty below 0.9, so I want to correct to 0.95.
How many kvar of capacitors does that take, what does it do to the line current, and is there a reason not to correct all the way to 1.0?
The capacitors must supply the difference in reactive power between the two power factors: Q_c = P × (tan φ1 - tan φ2). For pf 0.75, tan φ1 = 0.882; for pf 0.95, tan φ2 = 0.329. So Q_c = 100 × (0.882 - 0.329) = 55.3 kvar.
As a check with the power triangle: before correction Q = 88.2 kvar and S = 133.3 kVA. After it, Q = 32.9 kvar and S = 105.3 kVA. The real power stays at 100 kW throughout.
Line current is I = P / (√3 × V × pf). Before: 100,000 / (1.732 × 400 × 0.75) = 192 A. After: 100,000 / (1.732 × 400 × 0.95) = 152 A, a drop of about 21 percent. Loss in the supply cable and transformer goes with current squared, (152 / 192)² = 0.62, so those losses fall by about 38 percent.
The reduction only applies upstream of the point where the capacitors are connected. The cables between the bank and each motor still carry the motor's full reactive current, which is why large motors are sometimes corrected individually.
If you want the capacitance itself: for a delta-connected bank each capacitor sees the line voltage, so C per phase = Q_c / (3 × 2π × f × V²) = 55,300 / (3 × 314 × 400²), about 367 µF. A star-connected bank would need three times that value, because each capacitor sees only 230 V, which is one reason low-voltage banks are normally connected in delta.
In practice you buy by kvar at a rated voltage, and the output scales with voltage squared. A unit rated 440 V and used on 400 V delivers (400 / 440)² = 83 percent of its nameplate kvar, so allow for that when choosing sizes.
Correcting to unity is rarely worth it. Going from 0.95 to 1.0 needs another 32.9 kvar, 60 percent more capacitors, for a further current reduction of only 5 percent. The bigger problem is light load: a fixed 88 kvar bank left connected when most motors are off makes the installation leading, which raises the voltage and can itself be penalised.
That is why a load that varies is corrected with an automatic bank, for example six 10 kvar steps switched by a power factor controller. For capacitors fitted directly at a motor's terminals, the usual guidance is to stay below the motor's no-load magnetising kvar to avoid self-excitation when the motor is switched off.
Check for harmonics before ordering. Capacitors and the supply inductance form a parallel resonant circuit at a harmonic order of about sqrt(S_sc / Q_c). With a 5 MVA short-circuit level and 55 kvar that is sqrt(5000 / 55), about 9.5, which lies between the 7th and 11th harmonics, both of which six-pulse drives produce. If a significant share of the load is variable-speed drives or rectifiers, use detuned banks: a 7 percent series reactor moves the resonance to 50 / sqrt(0.07) = 189 Hz, safely below the 5th harmonic at 250 Hz.
Capacitors stay charged after disconnection. The bank needs discharge resistors and capacitor-duty switchgear, and installation is work for a qualified electrician.
