Capacitor bank resonance on a 33 kV industrial bus with 6-pulse drives
An industrial plant is fed from a 33 kV bus with a three-phase short-circuit level of 250 MVA. The load is 8 MVA at 0.85 pf lagging, and a group of 6-pulse thyristor drives on the same bus injects 5th-harmonic current that already produces 3.2% voltage distortion. To escape the utility's power-factor penalty below 0.90, the plant proposes a single 10 MVAr capacitor bank switched by one vacuum contactor at the 33 kV bus, with 33 kV rated units. At night the plant load falls to about 2 MW at 0.85 pf. The design engineer argues that no series reactor is needed because 'the parallel resonance works out at the 5th harmonic, which is a coincidence the load damping will take care of', and the capacitor supplier has quoted the cheapest option: no reactor, one step.
1. Calculate the parallel resonance order and the steady-state voltage rise at the bus, and judge whether the engineer's argument is acceptable.
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The resonant order for a capacitor bank in parallel with the source impedance is h = sqrt(Ssc/Qc) = sqrt(250/10) = 5.0. That is not a coincidence to be tolerated: it is exactly the dominant characteristic harmonic of a 6-pulse drive, where the 5th-harmonic current is typically 20-25% of the fundamental. At resonance the parallel circuit presents a high impedance to the 5th harmonic, so the small existing 5th current is amplified into a large 5th voltage and a large circulating current through the capacitors - the exact set-up for capacitor fuse blowing, dielectric failure and drive overvoltage trips. The engineer's argument reverses the risk: damping is lowest at light load, when the load's resistive component disappears, so the amplification is worst at night.
The steady-state voltage rise is approximately Qc/Ssc = 10/250 = 4%, i.e. about 1.3 kV on a 33 kV base, added to whatever the tap position already gives. That is significant on its own, and it is compounded at light load (next part).
Conclusion: the bank as proposed is not acceptable; a 5.0 resonance with 6-pulse drives present must be detuned.
2. What happens at night with the single 10 MVAr step, and what are the consequences for the 33 kV and 11 kV voltages?
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At night the load is 2 MW at 0.85 pf, i.e. about 1.24 MVAr lagging. With the full 10 MVAr bank in service the net reactive export is 10 - 1.24 = 8.76 MVAr, so the plant runs at a heavily leading power factor and the bus voltage rises by roughly 8.76/250 = 3.5% from the capacitor alone on top of the 4% already calculated at higher load.
The consequences are: (a) the 33 kV bus and the 11 kV busbars fed through the plant transformer sit above the upper statutory limit, forcing tap changes that then conflict with the daytime voltage; (b) the plant transformer may self-excite if it is disconnected while the bank stays connected to the tertiary or the bus, giving a dangerous overvoltage; (c) the resonance point shifts as the effective source impedance changes with the number of incoming cables in service, so the 5.0 order is not fixed and may move onto the 5th or 7th at some switching states; (d) at light load the circuit Q is highest, so the 5th-harmonic voltage amplification is at its worst exactly when the drives are running and the plant is quiet.
A single-step bank cannot follow this load profile. The bank must be split into steps controlled by a power-factor controller with a target of 0.95-0.98 lagging, and must be blocked from making the plant leading.
3. Recommend the corrected design with numbers, including the reactor, the capacitor voltage rating and the switching scheme.
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Split the bank into steps: for a load of 8 MW at 0.85 pf (4.96 MVAr) and a target of 0.98 lagging (about 1.6 MVAr), the required compensation is roughly 3.4 MVAr at full load, so 2 x 2 MVAr or 4 x 1 MVAr steps give adequate resolution without over-compensating at night. Never energise more steps than the load requires.
Add a 7% series reactor in each step. With Xl = 0.07 Xc the branch's series resonance is at sqrt(1/0.07) = 3.78, i.e. 189 Hz on a 50 Hz system. Above that frequency the branch is inductive, so it cannot parallel-resonate with the source inductance at the 5th (250 Hz) or 7th (350 Hz), and it presents a low impedance that absorbs those currents instead of amplifying them.
The reactor raises the voltage across the capacitor units: Vc = V/(1 - 0.07) = 33/0.93 = 35.5 kV, so 33 kV units are under-rated. Specify units rated at least 36 kV, or use a 12% reactor with correspondingly higher-rated units. Check the capacitor current: 10 MVAr at 33 kV is 175 A, and each step's contactor must be rated for the capacitor-switching inrush with the reactor damping it. Finally, obtain the actual harmonic spectrum and recalculate h = sqrt(Ssc/Qc) for the maximum and minimum source configurations, and confirm the resulting 5th and 7th voltage distortion stays within the 5% planning level after the bank is in service.