VFD Harmonic Resonance and Power-Factor Correction
A 1,600 kVA, 11/0.433 kV distribution transformer feeds an automobile parts plant. Monthly maximum demand is 1,050 kW at 0.82 power factor, giving about 1,280 kVA. A 400 kvar automatic capacitor bank was added at the main bus, but it trips on overcurrent during the shift when two 250 kW six-pulse VFDs ramp. Measured voltage THD is 9.5% and current THD on the VFD feeder is 31%. The transformer has a nameplate impedance of 6.5% and no K-factor rating. The utility charges a power-factor penalty below 0.90 and a demand charge of Rs 500/kVA. The plant is considering a larger capacitor bank, a transformer upgrade to 2,000 kVA, or an active harmonic filter. Production is three shifts, and VFD load is expected to grow by 30%.
1. Explain the trip and calculate the parallel-resonance order with the 400 kvar bank. Which proposed fix is technically wrong?
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Transformer short-circuit capacity = 1,600/0.065 = 24,615 kVA = 24.6 MVA. Parallel resonance order h ≈ sqrt(Ssc/Qc) = sqrt(24,615/400) = 7.8, close to the 7th harmonic, which six-pulse VFDs produce. The bank is acting as a harmonic sink and amplifier; overcurrent is from harmonic current, not fundamental kvar. A larger bank is worse: at 600 kvar, h ≈ 6.4, still near the 5th/7th and possibly lower. Plain capacitors must not be added. Use a detuned reactor bank rated for harmonic duty, or an active filter, or both. The capacitor bank should be designed for 400 kvar at 440 V with reactor and protected against overvoltage. A harmonic study is needed before switching.
2. Calculate the kVA and demand-charge saving from correcting PF to 0.95 at 1,050 kW, and assess the 2,000 kVA transformer upgrade.
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At 0.82, kVA = 1,050/0.82 = 1,280; at 0.95, kVA = 1,050/0.95 = 1,105; reduction = 175 kVA. At Rs 500/kVA, saving ≈ Rs 87,500/month, or Rs 10.5 lakh/yr, if billing follows kVA. Required reactive compensation = 1,050 × (tan acos 0.82 − tan acos 0.95) = 1,050 × (0.698 − 0.329) = 387 kvar, so 400 kvar is correct in size but must be detuned. The transformer is at 80% loading now; after PF correction it is 69%. A 2,000 kVA upgrade is not justified by fundamental loading and does not fix harmonics. Harmonic losses and K-factor may require derating, but first filter and correct PF. The demand-charge saving helps pay for the filter, not a transformer.
3. Specify the retrofit and M&V, including the 30% VFD growth.
Read the worked answer
Install a 400 kvar detuned capacitor bank with 7% reactor, or an active filter sized for at least the existing 31% current THD plus 30% growth. If VFDs total 500 kW, harmonic current is substantial; an active filter of roughly 150–200 A may be needed, but final size must come from a harmonic study. Keep the bank behind a circuit breaker with harmonic protection; do not use standard contactor switching. For M&V, use a power-quality logger for one week before/after at the transformer and VFD feeder; record THDv/THDi, PF, kW, kVA and capacitor current. Verify THDv below 5% and PF above 0.95. Track demand charge and penalty for 12 months, and recheck resonance after the VFD growth is installed.