Power Factor Penalty at a 33/11 kV Construction Township Substation
A 33/11 kV township substation serving a construction colony has a recorded maximum demand of 4.60 MVA at 0.78 lagging, measured on 30-minute integration, on a 5 MVA, 33/11 kV transformer. The utility tariff penalises any monthly average power factor below 0.90, and the regulator's supply code requires 0.95 or better at the point of connection. The connected load is dominated by about 900 kW of VFD-driven borewell pumps, plus a large base of fluorescent and LED lighting. On the 11 kV bus there is an existing 1 MVAr capacitor bank in five 200 kVAr steps; three units have already failed and blown fuses, and spares are not available. The substation engineer proposes buying a new 1 MVAr fixed bank switched by a single 11 kV contactor and asks whether it will clear the penalty and whether it is safe to run with the drives.
1. Calculate the reactive compensation needed to reach 0.95 and determine whether the proposed 1 MVAr bank is sufficient.
Read the worked answer
Real power: P = 4.60 × 0.78 = 3.588 MW.
Existing reactive power: with cos φ₁ = 0.78, tan φ₁ = 0.8023, so Q₁ = 3.588 × 0.8023 = 2.879 MVAr.
Target: with cos φ₂ = 0.95, tan φ₂ = 0.3287, so Q₂ = 3.588 × 0.3287 = 1.179 MVAr.
Compensation required: 2.879 − 1.179 = 1.70 MVAr.
Check the proposal: adding 1 MVAr leaves 1.879 MVAr, so new kVA = √(3.588² + 1.879²) = √(12.874 + 3.531) = 4.05 MVA, giving a power factor of 3.588/4.05 = 0.886. That is still below the 0.90 penalty threshold - the bank would cost money and the penalty would continue.
Sizing: order about 1.8 MVAr, split as 2 × 900 kVAr or 4 × 450 kVAr. After compensation the transformer load is 3.588/0.95 = 3.78 MVA, i.e. 75.5% loading on the 5 MVA unit, which is comfortable. Note that the healthy 2 × 200 kVAr of the old bank could be reused, but units that have already burst are usually at end of life; patching them is false economy.
2. Explain the technical risks of a single fixed 1 MVAr step on this bus and recommend a switching arrangement.
Read the worked answer
Risk 1 - light-load overvoltage and leading power factor. A fixed bank on a colony load that drops at night will export reactive power, raise the 11 kV bus voltage, and can take the supply into a leading power factor, which the tariff may penalise as well. Compensation must follow the load, so it needs an APFC relay with 4–6 switched steps and a minimum-step lockout.
Risk 2 - harmonic resonance. The VFDs inject 5th and 7th harmonics. Parallel resonance occurs near h = √(Ssc/Qc). If the 11 kV bus fault level is around 200 MVA and 1.8 MVAr is connected, h ≈ √(200/1.8) ≈ 10.5 - close to the 9th–11th range - and with a stiffer bus it can land right on the 11th. Capacitor current then rises, fuses blow and units fail exactly as the existing bank has. The remedy is a detuned reactor in series with each step, typically 7% impedance tuned to about 189 Hz, placing the resonance below the 5th harmonic. Capacitors should be rated for the harmonic voltage and current (typically 440 V units on a 415 V system, or 11 kV units derated), not standard units.
Risk 3 - switching transients. One large step means a single large inrush and a voltage dip each time it closes. Smaller steps with contactors rated for capacitor duty, and a short delay between steps, avoid this.
Recommend a 1.8 MVAr bank in five or six detuned steps with an APFC relay sensing on the 11 kV side, plus a power quality survey before ordering.