100 kVA transformer at 112% load with one hot phase and a new tubewell request
A 100 kVA, 11 kV/433 V, three-phase ONAN distribution transformer with a declared 55 °C temperature rise supplies a mix of domestic and small industrial consumers. On a summer afternoon, the measured secondary current is 150 A on two phases and 170 A on the third, the power factor is 0.82, and the top-oil temperature is 92 °C. The ambient in the transformer yard is 45 °C. Before summer the loading was around 60%. The lineman proposes to disconnect a few consumers and also wants to release a new 15 kW tubewell connection from the same unit, arguing that the transformer 'still has 40 kVA spare because it is 100 kVA'.
1. Calculate the present loading in kVA and kW and decide whether the transformer is overloaded.
Read the worked answer
Using the measured current and the secondary line voltage:
S = √3 × V × I = 1.732 × 433 V × 150 A ≈ 112,500 VA, i.e. 112.5 kVA
That is about 112% of the 100 kVA rating. The real-power output at 0.82 pf is roughly 112.5 × 0.82 ≈ 92 kW.
The third phase at 170 A is worse: if all three phases carried 170 A, the load would be about 1.732 × 433 × 170 ≈ 127 kVA. So the unit is not only overloaded, it is also badly unbalanced, which causes a neutral current, extra losses and local heating in one limb. The lineman's arithmetic - '100 kVA minus something equals 40 kVA spare' - is wrong: at 112.5 kVA there is no spare capacity at all; the transformer is already beyond rating.
2. Assess the significance of 92 °C top-oil temperature and the unbalanced current, and list the checks and immediate actions.
Read the worked answer
A declared 55 K rise on a 45 °C ambient means the design permits a top-oil temperature of roughly 100 °C at rated load, so 92 °C on its own is not proof of a fault - but combined with 112% loading and unbalance it shows the unit is running at its thermal limit. The winding hot spot sits well above oil temperature, and insulation ageing roughly doubles for every additional 6–8 K, so sustained operation here shortens life.
Checks: verify the temperature gauge against a second instrument, check oil level, leaks, silica gel and oil BDV, clean radiator fins and confirm cooling, check for loose or hot connections, and measure current on all phases and the neutral.
Immediate actions: reduce load, redistribute single-phase consumers across phases to bring the 170 A phase down, and record oil temperature against load through the day. Continued operation at this level without action risks a failure that takes out the whole feeder.
3. Evaluate the new 15 kW tubewell request and state what should happen before any capacity is released.
Read the worked answer
At 0.82 pf, 15 kW is 15 / 0.82 ≈ 18.3 kVA. Adding it to the present 112.5 kVA gives about 130.8 kVA, or 131% of rating - before allowing for the motor's starting current and for further unbalance. The request must be refused or deferred as things stand.
What should happen first:
- Power-factor correction: capacitors bringing 0.82 up to about 0.95 would cut the existing demand from 112.5 kVA to roughly 92.2 / 0.95 ≈ 97 kVA, which releases real capacity more cheaply than adding transformer size.
- Rebalancing: spread single-phase domestic loads so the phases are within a few amperes of each other and the neutral current falls.
- Load management: stagger the tubewell and other motor loads; check the starting method and the voltage dip.
- Augmentation: if demand is genuinely growing, plan a larger transformer or a separate feeder - but decide this from measured, balanced, power-factor-corrected data, not from nameplate arithmetic.