Capacitor Bank Sizing and Harmonic Resonance at a 480 V Plant
A 480 V, three-phase, 60 Hz plant is served by a 1000 kVA transformer with 5.75% impedance. The load is 600 kVA at 0.72 power factor lagging, made up mostly of across-the-line induction motors plus two six-pulse VFDs on the largest pumps. The utility applies a demand charge and a penalty below 0.90 pf, so maintenance proposes a fixed 300 kvar capacitor bank switched with the main breaker. Before approving, you must confirm the kvar needed to reach 0.95, decide the bank's voltage rating, and decide whether a plain bank or a detuned (reactor-tuned) bank belongs in this service. Treat the transformer as the only significant source impedance and neglect resistance and cable capacitance.
1. Determine the capacitor kvar required to raise the plant power factor from 0.72 to 0.95 lagging, and the resulting reduction in line current.
Read the worked answer
Real power (constant): P = 600 × 0.72 = 432 kW
Reactive power:
- Before: Q₁ = √(600² − 432²) = 416.4 kvar
- After: Q₂ = 432 × tan(cos⁻¹ 0.95) = 432 × 0.3287 = 142.0 kvar
Required bank: Q_c = 416.4 − 142.0 = 274 kvar (specify 275 kvar, or accept 300 kvar and note the plant runs slightly leading at light load: 432 kW with Q = 416.4 − 300 = 116.4 kvar gives pf = 0.966 leading).
Currents (I = kVA / (√3 × 0.480 kV)):
- Before: 600 / 0.8314 = 721.7 A
- After (275 kvar): S = √(432² + 142²) = 454.7 kVA → 546.9 A
Reduction ≈ 24%, which also cuts I²R losses in the transformer and service by about 42%. The savings must be checked against the resonance risk below before the bank is ordered.
2. Using the same 1000 kVA, 5.75% transformer, assess whether a 300 kvar bank creates a harmful parallel resonance with the supply, and state the conclusion you would put in the approval memo.
Read the worked answer
Short-circuit capacity of the source: S_sc = 1000 kVA / 0.0575 = 17,391 kVA
Parallel resonant order: h_r = √(S_sc / Q_c) = √(17,391 / 300) = √57.97 = 7.6
A resonance at h ≈ 7.6 sits directly between the 7th (420 Hz) and the 11th, and effectively right on top of the 7th once capacitor tolerance (±5–10%) and transformer impedance tolerance are considered. The two six-pulse VFDs inject continuous 5th and 7th harmonic current, so the bank would amplify 7th-harmonic voltage and current - the classic failure mode: bulged cans, blown fuses, overheated reactors, and distorted bus voltage.
Memo conclusion: do not install a plain (detuned-free) bank of this size. Specify a detuned bank tuned below the 5th harmonic so the resonance is moved well below any characteristic harmonic, and confirm the result with a harmonic study that includes the VFD spectrum and any future load growth. Also re-run h_r if the bank size changes - a smaller plain bank would push h_r higher (e.g., 150 kvar gives h_r ≈ 10.8, still risky).
3. For a detuned bank using a 7% series reactor, determine the capacitor voltage rating and nameplate kvar needed to deliver roughly 275 kvar at the 480 V bus, and explain why the nameplate differs from the delivered kvar.
Read the worked answer
Tuning: with p = 0.07, the series-resonant order is h = 1/√0.07 = 3.78 (227 Hz at 60 Hz), safely below the 5th, so the branch is capacitive at 300–420 Hz and no longer resonates with the source.
Voltage across the capacitors: the reactor raises the capacitor terminal voltage above bus voltage by approximately 1/(1 − p): V_c = 480 × 1/(1 − 0.07) = 480 × 1.075 = 516 V → specify 525 V rated units (never 480 V units).
Nameplate vs. delivered kvar: kvar output scales with V², so a unit rated at 525 V delivers only (480/525)² = 0.836 of nameplate at 480 V: Q_nameplate = 275 / 0.836 = 329 kvar → specify a 330 kvar at 525 V bank.
Checks to write into the spec:
- Reactor and capacitor continuous current rating must include harmonic loading (typically 1.35× fundamental for a detuned bank on VFD-rich bus).
- The 7% reactor also filters part of the 5th harmonic, which is a bonus but must not be counted as the primary harmonic solution.
- Short-circuit withstand of the reactor and the bank's switching device must be verified against the 17.4 MVA source.
If the 480 V nameplate figure is used directly, the plant ends up roughly 16% short of the target kvar and misses the 0.95 pf threshold.