Capacitor Bank Sizing and a Hidden Resonance
An 11 kV, 50 Hz radial feeder supplies a small industrial load at the end of an 8 km overhead line. The line series impedance is (0.20 + j0.30) Ω per km. The load is 3.0 MVA at 0.80 power factor lagging, and the sending-end voltage is held at 11.0 kV. The utility requires the load-bus power factor to be raised to at least 0.95 lagging, and the plant owner also wants to reduce the feeder voltage drop. A 1.0 MVAr capacitor bank is available. The 11 kV bus has a short-circuit level of 20 MVA, and a 400 kW six-pulse variable-frequency drive is already connected nearby. The engineer must decide whether to install the available bank directly at the load bus, and whether any additional precaution is needed. The plant has experienced unexplained capacitor failures at another site, so the owner is sensitive to harmonic problems.
1. Calculate the capacitor bank size required to raise the load power factor from 0.80 to 0.95 lagging, and compare it with the available 1.0 MVAr bank. What power factor would the available bank actually achieve?
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The active power is P = 3.0 × 0.80 = 2.4 MW. At 0.80 lagging, Q1 = P tan(cos⁻¹ 0.80) = 2.4 × 0.75 = 1.80 MVAr. At 0.95 lagging, Q2 = 2.4 × tan(cos⁻¹ 0.95) = 2.4 × 0.3287 = 0.789 MVAr. The required compensation is Qc = 1.80 − 0.789 = 1.011 MVAr.
A 1.0 MVAr bank leaves Q = 1.80 − 1.00 = 0.80 MVAr. The apparent power becomes S = √(2.4² + 0.8²) = 2.530 MVA, so the power factor is 2.4/2.530 = 0.949 lagging. That is marginally below the 0.95 requirement. The available bank is close but not sufficient if the limit is strict; a 1.05 MVAr or 1.1 MVAr bank would be needed, or the load must be reduced slightly.
2. Estimate the line-to-line voltage drop before and after installing the 1.0 MVAr bank. Use the nominal 11 kV load voltage and state your approximation.
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The line impedance is R = 0.20 × 8 = 1.6 Ω and X = 0.30 × 8 = 2.4 Ω. Using the nominal 11 kV for the current estimate:
Before compensation: I = 3.0 MVA/(√3 × 11 kV) = 157.5 A, cos φ = 0.80, sin φ = 0.60. The per-phase drop is I(R cos φ + X sin φ) = 157.5 × (1.6 × 0.8 + 2.4 × 0.6) = 157.5 × 2.72 = 428 V. The line-to-line drop is √3 × 428 ≈ 742 V, or about 6.7% of 11 kV.
After the 1.0 MVAr bank: S = 2.530 MVA, I = 2.530 MVA/(√3 × 11 kV) = 132.8 A, cos φ = 0.9487, sin φ = 0.3162. The per-phase drop is 132.8 × (1.6 × 0.9487 + 2.4 × 0.3162) = 132.8 × 2.277 = 302 V. The line-to-line drop is √3 × 302 ≈ 524 V, or about 4.8%.
The approximation treats the load voltage as 11 kV; the actual load voltage is slightly lower, so the currents and drops are slightly higher. The relative improvement is still clear: the bank reduces the drop by roughly 220 V line-to-line.
3. Assess the harmonic resonance risk of the 1.0 MVAr bank at this bus. Compute the parallel resonant order and recommend a practical mitigation.
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For a star-connected bank, the capacitive reactance per phase is X_C = V²/Qc = (11 kV)²/1.0 MVAr = 121 Ω. The supply reactance is X_S = V²/S_sc = (11 kV)²/20 MVA = 6.05 Ω. The parallel resonant harmonic order is h_res = √(X_C/X_S) = √(121/6.05) ≈ √20 ≈ 4.47.
A six-pulse drive produces strong 5th and 7th harmonic currents. A resonance near the 4.5th harmonic will amplify the 5th harmonic, causing high capacitor current, voltage distortion, and capacitor overheating or failure. This is consistent with the failures seen at the other site.
A practical mitigation is to add a detuned series reactor, typically 7% of the capacitor reactance, which shifts the resonance below the 4th harmonic and prevents amplification of the 5th. The reactor raises the capacitor voltage, so the bank voltage rating must be checked. Alternatives include relocating the bank, using a tuned filter, or adding an active filter, but the bank should not be installed alone without a harmonic study.