Capacitor Bank Failure on a VFD-Loaded Bus
A 500 kVA, 415 V industrial consumer has a steady load of 300 kW at 0.78 PF lagging and is billed on kVAh. Two 75 kW 6-pulse VFDs and a 100 kVAr fixed capacitor bank are connected at the 415 V bus. The capacitor bank has blown fuses twice in six months and its case is hot. Measurements: voltage THD 6.5%, current THD 28%, 5th-harmonic current 85 A, capacitor current 160 A at 415 V. The utility requires PF ≥ 0.95. The plant proposes adding a second 100 kVAr fixed bank at the same bus. The 500 kVA transformer has 5% impedance. Capacitors are rated 415 V, 50 Hz.
1. Calculate the net reactive power needed to reach 0.95 PF. If a 7% detuned reactor is used, what capacitor voltage rating and approximate installed kVAr would you specify? Explain the steps.
Read the worked answer
P = 300 kW. tanφ1 = tan(cos⁻¹0.78) = 0.802; tanφ2 = tan(cos⁻¹0.95) = 0.329. Q_net = 300 × (0.802 − 0.329) = 142 kVAr. A 7% reactor raises the capacitor voltage to U/(1−p) = 415/0.93 ≈ 446 V. A 415 V capacitor would be overstressed, so specify at least 440 V, preferably 480 V, capacitors. The reactor also changes the net output; as a first pass order about 150 kVAr at the raised rating and confirm with the manufacturer's detuned-bank table. Split into steps of 25, 50, 50, 25 kVAr with an automatic PF controller, target 0.95–0.97, and set the CT at the PCC. Check voltage rise: ΔU ≈ QX/U; with X ≈ 0.0172 Ω, 150 kVAr gives only about 6 V (1.5%), so overvoltage is not a problem.
2. Why did the existing 100 kVAr bank fail? Calculate the parallel resonance order with the 500 kVA transformer and assess whether adding a second 100 kVAr fixed bank is safe.
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Ssc = 500 kVA / 0.05 = 10,000 kVA. Parallel resonance h ≈ √(Ssc/Qc) = √(10,000/100) = 10. So the bank is near the 10th harmonic, not the 5th; but the 5th-harmonic current is 85 A and current THD is 28%, so the capacitor carries harmonic current in addition to fundamental. Its rated current at 415 V is 100,000/(√3×415) = 139 A; measured 160 A is about 15% over, plus dielectric heating from voltage distortion. That is enough to blow fuses and overheat. Adding another 100 kVAr moves h to √(10,000/200) = 7.1, still above the 5th, but it raises the capacitor current and makes the bank a larger harmonic sink. Without detuning, the second bank is not safe. Also check capacitor voltage rating: 415 V units on a bus with 6.5% THD can see repeated peaks above rating.
3. Recommend a complete compensation scheme: location, stepping, detuning, switching and protection. What would you not do?
Read the worked answer
Install one automatic detuned bank at the main 415 V PCC, not two fixed banks and not at the VFD terminals. Use 7% reactors tuned to about 189 Hz (below the 5th), 440/480 V capacitors, and 25/50/50/25 kVAr steps controlled by a PF controller with a CT on the incoming side. Switch each step with a capacitor-duty contactor and HRC fuses; add overcurrent and overvoltage protection, and a thermostat/temperature trip. Target 0.95 lagging, avoid leading at light load, and interlock the bank with the main breaker. Do not connect a fixed bank directly to the VFD bus; do not use 415 V capacitors with a 7% reactor; do not size the bank only from the kVAh bill without checking harmonics. Re-measure THD after commissioning and, if the 5th/7th remain high, consider a tuned filter instead of plain detuning.