Power-Factor Correction on a Bus Fed by a Six-Pulse Drive
A 480 V, three-phase, 60 Hz feeder supplies a 75 kW (100 hp) induction motor driving a slurry pump. At full load the motor efficiency is 93% and its power factor is 0.86. The feeder is 120 m of 3/0 AWG copper in PVC conduit, with conductor resistance at operating temperature of 0.21 Ω/km per phase and reactance of 0.08 Ω/km. The utility applies a power-factor penalty below 0.90, and the plant intends to correct to 0.95 using a capacitor bank switched at the motor terminals and left energized whenever the motor runs. Two months ago a six-pulse VFD was added to an identical motor on the same 500 kVA, 5.75%-impedance transformer; the main MCC already carries a 300 kvar capacitor bank. Since the VFD was commissioned, the capacitor fuses have blown twice and the electrician measured about 18% fifth-harmonic current on the bus. Your capacitor calculation gives roughly 21 kvar for the new bank.
1. Calculate the motor's full-load line current and the feeder voltage drop, and state whether the voltage drop is acceptable for a branch circuit plus feeder.
Read the worked answer
Input power: P_in = 75 kW / 0.93 = 80.65 kW Apparent power: S = P/pf = 80.65 / 0.86 = 93.78 kVA Line current: I = S/(√3 × V) = 93,780/(1.732 × 480) = 93,780/831.4 = 112.8 A
Feeder impedance (per phase, 0.120 km): R = 0.21 × 0.120 = 0.0252 Ω X = 0.08 × 0.120 = 0.0096 Ω sinφ = √(1 − 0.86²) = 0.511
Voltage drop (line-to-line, three-phase, approximate but adequate for a short feeder): VD = √3 × I × (R cosφ + X sinφ) = 1.732 × 112.8 × (0.0252 × 0.86 + 0.0096 × 0.511) = 1.732 × 112.8 × (0.02167 + 0.00491) = 1.732 × 112.8 × 0.02658 = 5.2 V, or 1.1% of 480 V. Receiving-end voltage ≈ 474.8 V.
Acceptability: a common design guide is ≤3% on a branch circuit and ≤5% combined feeder plus branch. At 1.1% this feeder is well inside the limit, so voltage drop is not the design driver here; conductor ampacity, terminations, and motor starting are. Note the calculation uses the operating power factor - correction to 0.95 will reduce the drop slightly (to roughly 4.5 V).
2. Determine the capacitor bank size needed to raise the power factor from 0.86 to 0.95, and quantify the resulting current reduction.
Read the worked answer
The capacitor supplies only reactive power; real power stays at 80.65 kW.
Q_c = P (tanφ₁ − tanφ₂) φ₁ = cos⁻¹ 0.86 = 30.68°, tanφ₁ = 0.5933 φ₂ = cos⁻¹ 0.95 = 18.19°, tanφ₂ = 0.3287 Q_c = 80.65 × (0.5933 − 0.3287) = 80.65 × 0.2646 = 21.3 kvar
At the corrected power factor: S₂ = 80.65 / 0.95 = 84.9 kVA I₂ = 84,900/(1.732 × 480) = 102.1 A
Current reduction = 112.8 − 102.1 = 10.7 A, about 9.5%. Because I²R losses scale with current squared, feeder and transformer copper losses fall by roughly 1 − (102.1/112.8)² ≈ 18% - usually a larger dollar benefit than the penalty avoidance on a feeder this short.
Practical checks before installing: (1) 21.3 kvar at 480 V is a nominal rating - capacitors must be rated for the actual sustained voltage, which rises with harmonics; (2) do not leave a fixed bank energized at light load or the motor may become leading and self-excite; switch it with the motor or use a automatic regulator; (3) confirm the utility's metering point, since correcting at the motor does not correct reactive burden elsewhere on the bus.
3. Using the transformer and existing-bank data, assess the resonance risk and recommend an installation approach. Justify why simply replacing fuses is wrong.
Read the worked answer
Estimate the parallel resonant order of the bus:
Short-circuit capacity of the transformer: S_sc = 500 kVA / 0.0575 = 8,696 kVA Total connected capacitance = 300 kvar existing + 21 kvar proposed ≈ 321 kvar Resonant order: h = √(S_sc/Q_c) = √(8,696/321) = √27.1 ≈ 5.2
Six-pulse drives generate characteristic harmonics at 5th, 7th, 11th, 13th. The bus resonance sits essentially on the 5th harmonic, and the measured 18% fifth-harmonic current confirms a strong source. The capacitor bank presents a low impedance at the resonant frequency, so 5th-harmonic current circulates through it and through the supply, producing high capacitor current and voltage distortion. That is why fuses blow: they are being asked to interrupt harmonic current the bank was never sized for, not a fault.
Replacing fuses is wrong because the fuse is a symptom, not the cause. A larger fuse will let the capacitor fail catastrophically or overheat the transformer; the resonance remains.
Recommended approach:
- Perform a harmonic study (or at minimum measure voltage and current THD and impedance vs. frequency) before energizing any new bank.
- Do not install a plain capacitor bank on a bus with significant 6-pulse load. Use detuned reactors tuned to about 4.2× fundamental (below the 5th) so the bank looks inductive at the 5th, or use an active filter.
- Add line reactors or a DC-link choke on the VFD to reduce harmonic current at the source.
- Relocate capacitors upstream of the drive load if possible, and specify capacitors at 1.1–1.2× nominal voltage rating for harmonic service.
- Recheck capacitor switching: vacuum contactors and adequate discharge resistors.