Power-factor correction on a VFD-heavy 415 V bus
A fabrication shop has a maximum demand of 800 kW at 0.72 lagging power factor on a 415 V, 50 Hz bus. The utility tariff penalises power factor below 0.90 and the plant wants to target 0.95. The proposed solution is a 500 kVAr fixed capacitor bank connected at the main LV bus. The bus is fed by a 1600 kVA transformer with 6% impedance, and about 35% of the load is supplied through six-pulse VFDs. The electrical contractor has sized the capacitor cable at 400 A and the switch-fuse at 630 A. The plant has experienced unexplained capacitor fuse failures at another site.
1. Calculate the reactive power needed to raise the power factor from 0.72 to 0.95. Is 500 kVAr enough?
Read the worked answer
For 0.72, tan phi1 = sqrt(1 - 0.72²)/0.72 = 0.694/0.72 = 0.964. For 0.95, tan phi2 = sqrt(1 - 0.95²)/0.95 = 0.312/0.95 = 0.329. Qc = P(tan phi1 - tan phi2) = 800(0.964 - 0.329) = 508 kVAr. A 500 kVAr bank leaves Q = 800 x 0.964 - 500 = 271 kVAr; tan phi = 271/800 = 0.339; pf = 0.947, just under the 0.95 target. So 500 kVAr is marginally short; 525-550 kVAr would be needed, preferably split into steps rather than one fixed bank. The calculation uses maximum demand; at lower load the fixed bank would overcorrect and cause leading power factor.
2. Check the parallel resonance risk. What does this mean for the proposed plain capacitor bank?
Read the worked answer
Transformer short-circuit capacity = 1600/0.06 = 26.7 MVA. Resonance order h is approximately sqrt(Ssc/Qc) = sqrt(26.7/0.5) = sqrt(53.4) = 7.3. The parallel resonance is near the 7th harmonic, and six-pulse VFDs produce 5th and 7th harmonics. If the 7th harmonic current is amplified, capacitor current and voltage distortion rise, which explains fuse failures and capacitor overheating. A plain 500 kVAr bank is therefore risky on this bus. Check with a harmonic study. Use a detuned capacitor bank with a 7% series reactor tuned to about 189 Hz (below the 5th harmonic), which shifts the parallel resonance below the 5th; verify that the shifted resonance does not coincide with the 3rd. Alternatively install an active filter or increase the number of pulses. Do not energise a plain bank without checking the harmonic spectrum.
3. Check the contractor's cable and fuse ratings and state the switching requirements.
Read the worked answer
Capacitor rated current = Q/(sqrt(3) x V) = 500/(1.732 x 0.415) = 695 A. The 400 A cable is undersized. For capacitor circuits, cable continuous rating should be at least 1.35 times rated current: 1.35 x 695 = 938 A, so use a cable rated about 1000 A (multiple runs or a larger conductor). The 630 A switch-fuse is also below the capacitor current and must be replaced with a capacitor-duty device rated above 695 A, with high-breaking capacity for the bus fault level. Switching: use capacitor-duty contactors or breakers with damping reactors to limit inrush; provide discharge resistors so the bank discharges to below 50 V within the specified time; use steps (e.g. 100 + 150 + 250 kVAr) with a controller to avoid overcorrection at light load; interlock with the VFDs and verify protection settings.