All-Day Efficiency and Loading Decision for a 630 kVA Plant Transformer
A dairy plant runs a 630 kVA, 11 kV/433 V ONAN distribution transformer. Its test report gives a no-load loss of 1.35 kW and a full-load copper loss of 7.2 kW at 75 °C. The logged daily cycle is: 6 hours at 480 kW at 0.80 pf, 10 hours at 200 kW at 0.80 pf, and 8 hours with the process shut down but the transformer still energised. The plant is proposing to replace this unit with a 500 kVA transformer of similar per-unit losses 'to stop running at part load'. On a summer afternoon the yard ambient is 42 °C and the top-oil thermometer reads 78 °C. Two forced-cooling fans are mounted but have never been wired. Tariff is ₹7.10 per kWh and the plant runs 330 days a year.
1. Compute the all-day efficiency of the existing transformer and the annual cost of its losses.
Read the worked answer
Energy delivered:
- 480 kW × 6 h = 2880 kWh
- 200 kW × 10 h = 2000 kWh
- Total = 4880 kWh/day
Iron loss (constant while energised): 1.35 × 24 = 32.4 kWh
Copper loss = 7.2 × (load kVA / 630)² × hours:
- Peak: 480/0.8 = 600 kVA → ratio 0.952 → 7.2 × 0.907 × 6 = 39.2 kWh
- Day: 200/0.8 = 250 kVA → ratio 0.397 → 7.2 × 0.157 × 10 = 11.3 kWh
- Total copper = 50.5 kWh
All-day efficiency = 4880 / (4880 + 32.4 + 50.5) = 4880 / 4962.9 = 98.33 %
Loss cost = 82.9 kWh/day × 330 × ₹7.10 ≈ ₹1.94 lakh per year.
Note that the 8 no-load hours alone burn 10.8 kWh/day - about 13 % of all losses - which is where any energy-conservation argument should start, not at part-load efficiency.
2. At what loading does maximum efficiency occur, and does the proposed 500 kVA replacement make engineering sense?
Read the worked answer
Maximum-efficiency loading: efficiency peaks when copper loss equals iron loss, i.e. at load fraction k = √(P₀/P_k) = √(1.35/7.2) = 0.433, or 273 kVA (≈ 43 % loading).
The daily average demand is (2880 + 2000)/24 = 203 kW ≈ 254 kVA - very close to the maximum-efficiency point. So for 18 of 24 hours the existing unit is already close to optimum; the claim that it is 'inefficiently part-loaded' is largely false.
The 500 kVA proposal fails on a simpler ground: the 6-hour peak is 600 kVA, which is 120 % of 500 kVA. The replacement would be overloaded for a quarter of every day, with accelerated insulation ageing and a real risk of tripping the plant on a hot afternoon. A transformer must first be able to carry the peak; efficiency is a secondary optimisation.
Correct actions: keep the 630 kVA unit; consider a smaller dedicated transformer or switched LV supply for the night load only if the no-load loss is genuinely significant compared with the tariff; and evaluate the replacement on total owning cost (no-load loss capitalised at ₹/W), not on nameplate kVA.
3. Given 42 °C ambient, 78 °C top oil and two unconnected fans, what is your loading recommendation and what would you verify?
Read the worked answer
Temperature-rise check: top-oil rise = 78 − 42 = 36 K. For an ONAN unit the permissible top-oil rise is typically 50–55 K, so the transformer is not in distress - it is simply running hot because the ambient is hot, not because of an internal fault.
But the margin is thin. At 600 kVA the unit is at 95 % loading, and IEC 60076-7 requires the permissible loading to be reduced when the ambient exceeds 40 °C unless the load cycle has been light beforehand. Two independent measures are justified:
- Wire and commission the fans. ONAF typically raises the sustainable rating by 20–25 %, which converts a 95 % loaded unit at 42 °C into a comfortably loaded one and slows insulation ageing (roughly halving thermal life for every additional 6–8 K of hot-spot temperature).
- Verify before relying on the thermometer: confirm the top-oil indicator is calibrated, check that the radiators are clean and unobstructed, confirm the oil level and test the oil (BDV and acidity) if the unit has run above 75 °C for long periods.
Do not simply raise the trip setting of the winding-temperature alarm to stop nuisance alarms; that removes the only protection the asset has.