Overloaded 500 kVA Transformer and Power-Factor Correction
A cooperative industrial estate is supplied by a 500 kVA, 11 kV/415 V, Dyn11 distribution transformer with an off-circuit tap changer on the nominal tap. Summer measurements taken at the transformer secondary during the peak shift show a line current of 740 A, a power factor of 0.75 lagging (active power about 400 kW), a top-oil temperature of 86 °C and an ambient of 43 °C. The largest consumer runs a 200 kW compressor and several induction motors and refuses to shed load. The manufacturer's data sheet states that the winding hot-spot rise limit is reached at rated load and that sustained operation above nameplate kVA shortens insulation life. The estate proposes installing a capacitor bank at the transformer secondary. The tariff penalises a power factor below 0.90, and the estate wants to reach 0.95 lagging with a margin. Assume the measured active load stays constant at 400 kW.
1. Show by calculation that the transformer is overloaded, and explain why tap changing or a larger oil cooler is not the correct first remedy.
Read the worked answer
Loading check
S = √3 × V × I = 1.732 × 415 × 740 = 531.8 kVA
531.8 / 500 × 100 = 106.4 % of nameplate kVA - the transformer is running about 6 % above rating continuously, at 43 °C ambient, with a top-oil temperature of 86 °C (a 43 K top-oil rise). Active power = 531.8 × 0.75 ≈ 399 kW, consistent with the measured 400 kW, so the figures cross-check.
Why not taps or cooling alone
- The off-circuit tap changer only alters the turns ratio and therefore the secondary voltage; it does not reduce kVA, current or winding loss.
- Forcing extra cooling lowers the winding temperature but the current still exceeds the nameplate value, so the copper loss (I²R) and the mechanical and thermal duty on bushings, cables and the tap changer remain above design.
Insulation ageing roughly doubles for every 6–8 K of sustained hot-spot temperature rise above the rated value, so the overload must be removed at source, not masked.
2. Calculate the capacitor kVAr needed to raise the power factor to 0.95 lagging, choose a practical bank from 50 kVAr steps, and compute the resulting line current and percentage reduction.
Read the worked answer
Reactive power required
Qc = P (tan φ1 − tan φ2)
- cos φ1 = 0.75 → tan φ1 = 0.8819
- cos φ2 = 0.95 → tan φ2 = 0.3287
Qc = 400 × (0.8819 − 0.3287) = 221 kVAr
Practical selection: a 200 kVAr bank in four 50 kVAr steps is preferred to a single 250 kVAr unit, because the controller can then switch steps in and out as load varies (see the over-correction risk below).
With 200 kVAr connected: Q2 = (400 × 0.8819) − 200 ≈ 153 kVAr; S = √(400² + 153²) = 428 kVA, power factor = 400/428 = 0.935 lagging. That is short of the 0.95 target, so a fifth 50 kVAr step (250 kVAr total) is required to reach Q2 ≈ 100 kVAr, S = √(400² + 100²) = 412 kVA, power factor 0.970.
Current
I = 412 000 / (1.732 × 415) = 573 A, against 740 A before correction - a reduction of (740 − 573)/740 = 22.6 %.
The transformer now runs at 412/500 = 82 % of rating, which also removes the thermal problem.
3. The estate proposes one fixed 250 kVAr bank switched with the main breaker. Explain why an automatic, stepped bank is safer, supporting your answer with a light-load calculation and one other technical caution.
Read the worked answer
Light-load calculation
At night the estate's load falls to, say, 100 kW. Its original reactive demand is 100 × 0.8819 = 88 kVAr. A fixed 250 kVAr bank would leave a net reactive of 88 − 250 = −162 kVAr (leading):
S = √(100² + 162²) = 190 kVA, power factor = 100/190 = 0.53 leading
A leading power factor of this size causes voltage rise at the secondary, can trip the utility's reverse-reactive protection, attracts the same tariff penalty and adds unnecessary capacitor losses and switching transients.
Why stepped, automatic control
A controller with a sensing CT switches 50 kVAr steps so that the residual is always small and lagging at any load from 100 kW to 400 kW. Steps should be rated for capacitor duty (contactor with damping resistors or a capacitor-duty contactor) and the bank must be discharged through resistors before re-energisation.
Second caution: if the estate's drives include VFDs or other harmonic sources, the capacitor bank can resonate with the transformer and supply impedance, amplifying a harmonic and destroying capacitors. Check the harmonic spectrum and, if present, use detuned reactors or a filtered bank.