Calculation
η = β·SN·cosφ / [β·SN·cosφ + P0 + β²·PkN]
Core loss remains approximately constant while copper loss grows with the square of load fraction.
Busch 2006, Abschnitt 8.3.3.3 Betriebsverhalten
Transformer partial-load efficiency
Core loss remains approximately constant while copper loss grows with the square of load fraction.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- Efficiency η (0 to 1) — Ratio of output real power to useful power plus modelled losses.
- Rated apparent power SN in kVA — Nameplate apparent-power rating; enter its numerical kVA value through this power field.
- Load fraction β — Current or apparent load divided by rated value; 0.75 means 75% load.
- Power factor cos φ — Displacement factor of the connected load at the operating point.
- No-load loss P0 — Real power measured at rated voltage in the no-load test, approximately core loss.
- Rated copper loss PkN — Real power measured in the short-circuit test at rated current; scale by β squared at partial load.
Example
10 kVA, β=0.75, power factor 0.8, P0=120 W and PkN=300 W give P2=6.0 kW and η≈0.954 or 95.4%.
Assumptions and limits
Constant rated voltage, load-independent P0 and constant winding resistance; stray loss, temperature change and harmonics are excluded.
Understand Transformer partial-load efficiency
This calculator connects no-load and short-circuit tests to an actual operating point, showing why a lightly loaded transformer may be inefficient despite small absolute loss.
What does this quantity describe?
At constant rated voltage, P0 is treated as nearly constant core loss. Rated-current copper loss PkN falls to β²·PkN at load fraction β. Useful output is β·SN·power factor.
Formula and variables
η = β·SN·cosφ / [β·SN·cosφ + P0 + β²·PkN]
P2 = β·SN·cosφPcu = β²·PkNη = P2/(P2+P0+Pcu)
| Symbol / input | Meaning |
|---|---|
| Efficiency η (0 to 1) | Ratio of output real power to useful power plus modelled losses. |
| Rated apparent power SN in kVA | Nameplate apparent-power rating; enter its numerical kVA value through this power field. |
| Load fraction β | Current or apparent load divided by rated value; 0.75 means 75% load. |
| Power factor cos φ | Displacement factor of the connected load at the operating point. |
| No-load loss P0 | Real power measured at rated voltage in the no-load test, approximately core loss. |
| Rated copper loss PkN | Real power measured in the short-circuit test at rated current; scale by β squared at partial load. |
Choose the inputs correctly
SN comes from the nameplate. β is actual current or apparent load divided by rated value. Power factor belongs to the load. P0 comes from the no-load test and PkN from the rated-current short-circuit test.
How to use the calculator
Ensure all data refer to the same transformer, tap and temperature condition. Obtain β and power factor at the actual operating point, then compare several load points rather than rated load only.
Worked example
10 kVA at β=0.75 and power factor 0.8 delivers 6.0 kW. With 120 W core loss and 0.75²·300 W=168.75 W copper loss, η=6000/6288.75≈95.4%.
Understand the result and units
At low load P0 dominates; at high load copper loss grows quadratically. Maximum efficiency occurs approximately when the two loss components are equal.
Enter SN numerically in kVA; in the real-power balance 1 kVA contributes 1 kW per unit power factor. P0 and PkN may be entered in watts.
Typical applications
Loading comparisons, choosing several small versus one large transformer, and evaluating teaching experiments.
Assumptions, limits and common mistakes
Constant voltage and equivalent losses. Temperature-dependent winding resistance, stray/harmonic loss and thermal transients are excluded.
Common mistake: Do not use apparent power directly as useful real power; power factor is essential. Scale PkN with β squared, not linearly.
Frequently asked questions
What is “Transformer partial-load efficiency” used for?
Loading comparisons, choosing several small versus one large transformer, and evaluating teaching experiments.
Where do the input values come from?
SN comes from the nameplate. β is actual current or apparent load divided by rated value. Power factor belongs to the load. P0 comes from the no-load test and PkN from the rated-current short-circuit test.
What does the result not cover?
Constant voltage and equivalent losses. Temperature-dependent winding resistance, stray/harmonic loss and thermal transients are excluded.