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Transformers & PF

Transformer Loading & Efficiency Calculator

Check how heavily a transformer is loaded and what that costs in losses. Enter the nameplate rating, voltages and test-report losses, then the actual load as kVA, kW with PF, or a measured secondary current. You get percentage loading, full-load and actual currents on both windings, copper loss scaled with load squared, and efficiency.

Last reviewed

ηmax050100 %Pcu ∝ x²Pfe
  • Efficiency
  • Copper loss
  • Core loss
Core loss is present whenever the transformer is energised; copper loss grows with the square of the load. Efficiency peaks at the load where the two are equal (qualitative shape).

For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.

Inputs

Transformer

Nameplate

kVA
V
V
W

From test report or datasheet

W

Copper loss at full load

Actual load

kVA

How to use this calculator

  1. Choose single- or three-phase and enter the rated kVA and both voltages (line-to-line for three-phase).
  2. Enter the no-load loss P0 and the load loss Pk at rated current from the test report or datasheet.
  3. Enter the actual load as kVA, as kW with power factor, or as a measured current on the secondary side.
  4. Read the loading, losses and efficiency. The tool also gives the load at which efficiency peaks.

How it works: the formula

x = S_load / S_rated (per-unit loading) I_FL = S_rated / (√3 × V_LL) (each winding, three-phase) I_FL = S_rated / V (single-phase) P_cu = P_k × x² P_loss = P_0 + P_cu η = P_out / (P_out + P_0 + P_cu), P_out = S_load × PF η is maximum at x = √(P_0 / P_k)

Load (copper) loss is I²R loss in the windings. Current is proportional to load, so this loss scales with the square of loading. No-load (core) loss depends on voltage and frequency and stays almost constant whenever the transformer is energised. Efficiency is highest where the two losses are equal.

Variables and units

SymbolMeaningUnit
S_ratedNameplate ratingkVA
S_loadActual loadkVA
xLoading as a fraction of rating—
P_0No-load (core, iron) lossW
P_kLoad (copper) loss at rated currentW
ηEfficiency%

Worked example

A 1000 kVA, 11 kV / 415 V transformer with P0 = 1700 W and Pk = 10 500 W, carrying 750 kVA at PF 0.9:

x = 750 / 1000 = 0.75 → 75 % I_HV = 1 000 000 / (1.73205 × 11 000) = 52.49 A I_LV = 1 000 000 / (1.73205 × 415) = 1391.2 A P_cu = 10 500 × 0.75² = 5906 W P_loss = 1700 + 5906 = 7606 W η = 675 000 / (675 000 + 7606) = 98.89 % Peak efficiency at √(1700 / 10 500) = 40.2 % load

Assumptions

  • Losses entered are at reference temperature and rated voltage.
  • Currents ignore magnetising current and voltage regulation under load.
  • Cooling-fan or pump losses are not included in the efficiency.

Limitations

  • Overload capability depends on ambient temperature, the load cycle and the cooling class. IEC 60076-7 (oil-immersed) and IEC 60076-12 (dry-type) are the loading guides, together with the manufacturer's limits.
  • Harmonic currents increase eddy and stray losses beyond the x² scaling. Transformers feeding non-linear loads may need a K-factor rating or derating.
  • Efficiency at a single operating point says nothing about energy losses over a varying load cycle. Weight the result by hours at each load for that.

References

Standards and texts that cover this topic. Use the edition adopted in your jurisdiction.

  • IEC 60076-1, Power transformers — Part 1: General (definitions of no-load loss, load loss and routine tests).
  • IEC 60076-7, Loading guide for mineral-oil-immersed power transformers; IEC 60076-12 for dry-type transformers.
  • IEEE Std C57.12.00, General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers.
  • IEEE Std C57.91, Guide for Loading Mineral-Oil-Immersed Transformers.

Frequently asked questions

Where do I find P0 and Pk?

On the factory test report, or in the maker's datasheet as "no-load losses" and "load losses" (sometimes "short-circuit losses" at 75 °C or 120 °C reference). Regional ecodesign or efficiency rules may also set maximum values for your size.

Why is efficiency highest below full load?

Core loss is fixed and copper loss grows with load squared. Efficiency peaks where they are equal, which for many distribution transformers is somewhere around 30–50 % load.

Can I run a transformer above 100 %?

Sometimes, for limited periods. It depends on ambient temperature, the preceding load and insulation ageing. That judgement belongs to the loading guides and the manufacturer, not to a simple calculation.

Which current should I use for cable sizing?

Use the full-load current of the relevant winding for conductors rated to the transformer, together with the protection rules in your code. The actual current tells you how much headroom is left.

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How these formulas were checked: methodology and testing.