How to use this calculator
- Enter or choose rating.
- Enter or choose load.
- Enter or choose load power factor.
- Enter or choose no-load (core) loss.
- Enter or choose full-load copper (load) loss.
- The efficiency updates as you type. Use Copy result or Share to keep it.
Method
Output = kVA × x × pf; Copper loss = Pcu,FL × x²; η = Output ÷ (Output + Pcore + Pcu,FL × x²) × 100; Maximum efficiency at x = √(Pcore ÷ Pcu,FL)
- x
- load as a fraction of rating
- pf
- load power factor
- Pcore
- no-load (iron) loss
- Pcu,FL
- copper loss at rated load
Worked example
- Output: 100 kVA × 1 × 0.8 = 80 kW.
- Losses: 1,000 W core + 1,500 W copper = 2,500 W.
- Efficiency: 80 ÷ (80 + 2.5) = 96.97%. Maximum efficiency at √(1000 ÷ 1500) = 81.6% load.
At full load and 0.8 power factor the transformer is 96.97% efficient; its efficiency peaks at about 82% load, where copper loss equals core loss.
What the result means
Because core loss never stops, a lightly loaded transformer can be surprisingly inefficient over a year. For energy over time, weight the losses by how many hours the transformer spends at each load (all-day efficiency).
Assumptions and limits
- Copper loss scales with the square of current; core loss is constant at rated voltage.
- Stray and dielectric losses are included in the two test figures.
- Temperature correction of copper loss and harmonic loads are not modelled.
Questions
Why does power factor change efficiency?
Losses depend on current (kVA), but output is measured in kW. A lower power factor means less useful output for the same losses.
When is a transformer most efficient?
When the variable copper loss equals the fixed core loss, at a load of √(core ÷ full-load copper loss).