Use the no-load and rated-current load losses from the transformer test report. Move the loading point to see the squared-current effect and the equal-loss point.
The two loss components are equal at 32.74% load. At fixed positive power factor, this is the model's maximum-efficiency point.
Total transformer loss
4,465W
50% of 1,000 kVA at power factor 0.9.
Check the values to see a result.
Constant voltage, frequency and reference-temperature losses. Harmonics, cooling auxiliaries and changing temperature are not modelled. Use separate periods for a varying load.
Loss = no-load loss + rated load loss × (load / 100)²
No-load loss of 1,340 W plus load loss of 3,125 W gives 4,465 W in total.
Tip: For a changing load, calculate each interval and sum its loss energy. Squaring the average load usually understates losses.
An energised transformer consumes power even when its secondary supplies no load. Its no-load loss remains approximately constant at rated voltage and frequency, while its load-dependent loss rises approximately with current squared.
Use the loss figures from the transformer test report at its stated reference temperature. The interactive curve shows how a loading change affects both total loss and efficiency. It does not authorise overload or certify regulatory efficiency.
P₀ = manufacturer no-load test loss
This loss remains while the transformer is energised, even at zero output. It is not multiplied by the loading percentage. De-energised periods are excluded from the entered energised hours.
Pk = rated load loss × load fraction²
Enter the load-dependent loss measured at rated current, not the combined full-load loss. The model holds its reference temperature constant and does not estimate harmonic or auxiliary cooling losses.
Efficiency = output kW / (output kW + loss kW)
Output real power depends on kVA loading and power factor. The same current loading causes the same modelled loss at different power factors, but the useful real-power output changes.
Loss energy = loss W × hours / 1000
For a variable load, calculate each period and sum. Half the time at zero and half at full load produces more load loss than a steady 50% load. Squaring an average hides this difference.
Schneider Electric’s calculation example uses 1,340 W no-load loss and 12,500 W rated load loss. At 50% loading the load-dependent part is 3,125 W, giving 4,465 W total. At power factor 0.9, output is 450 kW and efficiency is approximately 99.018%. Over 8,760 unchanged hours, losses total 39,113.4 kWh.
Total transformer loss 4,465 W Open in the calculator
Load-dependent loss falls to zero but the 1,340 W no-load loss remains. Over 8,760 hours that is 11,738.4 kWh. This is not the loss of a transformer that has been disconnected and de-energised.
Total transformer loss 1,340 W Open in the calculator
No. Subtract the no-load component first if the report provides only the combined figure. Otherwise the tool counts no-load loss twice.
With fixed power factor and these constant loss coefficients, differentiation gives a maximum when the two loss components are equal. Real temperature, harmonics and auxiliary loads can move the operating optimum.
Only if you accept a constant-load approximation. For a variable profile, sum the losses of its separate time intervals. The mean of current squared is generally greater than the square of mean current.
No. Watts of loss are a heat source, but temperature rise also depends on cooling, enclosure, ambient conditions and thermal time constants. Use manufacturer thermal data for that assessment.