How the calculation works
A transformer has two kinds of loss. No-load (iron) losses are present whenever it is energised, whatever the load. Load (copper) losses rise with the square of the load current:
Annual losses (kWh) = P0 × hours energised + Pk × (S ÷ Sn)² × hours on load
Efficiency is highest at the load where copper losses equal iron losses:
Worked example
An older 1,000 kVA distribution transformer has 1.7 kW no-load loss and 10.5 kW load loss. It is energised all year at an average 60% load, and power costs 0.12 per kWh.
- Losses: 1.7 + 10.5 × 0.6² = 5.48 kW, or 48,000 kWh a year, costing about 5,760.
- A modern low-loss unit with 0.693 kW and 7.6 kW losses would lose 3.43 kW, or 30,040 kWh a year.
- Saving: about 17,970 kWh and 2,160 a year.
- Best efficiency for the old unit is at 40% load, where its 1.7 kW of iron loss equals its copper loss.
The loss values in this example are illustrative. Always use the figures from your transformer's test report or the supplier's offer.
Practical points
- Lightly loaded transformers that stay energised all year are dominated by no-load losses. Where two transformers share a lightly loaded site, switching one off outside production hours can save more than any replacement.
- When buying a transformer, compare offers on total cost of ownership: purchase price plus the capitalised cost of its no-load and load losses over its life, not on purchase price alone.
- Harmonic currents increase load losses beyond this calculation, especially eddy current losses in the windings.
Questions
How do I find a transformer's no-load and load losses?
They are measured in the factory acceptance test and given in the test report. Some nameplates show them. For a new purchase, suppliers quote guaranteed values in their offer.
At what load is a transformer most efficient?
Where load losses equal no-load losses, at a load of √(P0 ÷ Pk) of rating. For many distribution transformers this is between 30% and 50% load.
Should I use average or RMS load?
Load losses depend on the square of current, so for a varying load the RMS value is more accurate. Using the plain average underestimates load losses.