Transformer Efficiency Calculator
Transformer Efficiency Calculator
Efficiency and voltage regulation for a mains transformer from the figures it is actually specified by — no-load loss, load loss and impedance voltage — with the load fraction at maximum efficiency, the leading-power-factor case where regulation goes negative, and all-day efficiency from a duty schedule.
Efficiency, regulation and all-day efficiency
a 100 kVA transformer with 130 W of no-load loss, 1,250 W of load loss and 4.0% impedance, at 0.6 load and 0.90 power factor lagging
Two losses, and a phasor
maximum at x = √(P₀ ÷ Pk), where the two losses are equal
%R = Pk ÷ S × 100 %X = √(%Z² − %R²)
regulation = |1 + x(cos φ − j sin φ)(%R + j%X)/100| − 1, as a percentage
all-day η = Σ(x·S·cos φ·h) ÷ [ Σ(x·S·cos φ·h) + Σ(x²·Pk·h) + P₀·henergised ]
- P0
- no-load loss: hysteresis and eddy currents in the core, drawn every hour the transformer is energised and almost independent of load
- Pk
- load loss at FULL load: I²R in the windings plus stray loss. It scales with the square of the load fraction
- sin phi
- taken NEGATIVE for a leading power factor. That one sign is what makes regulation go negative
- %Z
- the nameplate impedance voltage: the percentage of rated voltage that drives rated current into a short circuit
Worked example
a 100 kVA transformer with 130 W of no-load loss, 1,250 W of load loss and 4.0% impedance, at 0.6 load and 0.90 power factor lagging
At 0.6 load it delivers 54 kW and loses 130 W in the core plus 0.6² × 1,250 = 450 W in the copper — 580 W in all
So η = 54,000 ÷ (54,000 + 580) = 98.937%
Its best point is where copper loss equals core loss, at x = √(130 ÷ 1,250) = 0.3225 — only 32.2 kVA, where it reaches 99.112%
The load loss makes %R = 1,250 ÷ 100,000 = 1.25%, so %X = √(4.0² − 1.25²) = 3.800% and regulation at this load is 1.683% — the usual approximation gives 1.669%. At the same power factor LEADING it would be -0.290%: the voltage rises on load
Over a day of 8 h at 0.8, 8 h at 0.5 and 8 h at 0.2, it delivers 1,080 kWh and wastes 9.30 kWh in copper plus 3.12 kWh in iron — an all-day efficiency of 98.863%
Chapman’s solved problem 2-6, worked through this page
| Quantity | From the test data | Chapman’s answer |
|---|---|---|
| Equivalent resistance referred to the secondary | 0.1258 Ω | 0.126 Ω |
| Equivalent reactance referred to the secondary | 0.4750 Ω | 0.476 Ω |
| %R and %X on a 1,000 VA, 115 V base | 0.952% and 3.592% | — |
| Copper loss at rated current | 9.515 W | — |
| Voltage regulation at rated load, 0.8 PF lagging | 2.94% | 2.96% |
| Efficiency at rated load, 0.8 PF lagging | 98.35% | 98.3% |
IEC 60076-5 recognised minimum short-circuit impedance
| Rated power | Minimum %Z |
|---|---|
| 25 to 630 kVA | 4.0% |
| 631 to 1,250 kVA | 5.0% |
| 1,251 to 2,500 kVA | 6.0% |
| 2,501 to 6,300 kVA | 7.0% |
| 6,301 to 25,000 kVA | 8.0% |
| 25,001 to 40,000 kVA | 10.0% |
| 40,001 to 63,000 kVA | 11.0% |
| 63,001 to 100,000 kVA | 12.5% |
One loss that never stops, and one that only shows up on load
A transformer is specified by two measurements, and everything on this page comes out of them. The open-circuit test energises one winding at rated voltage with the other open: almost no current flows, so almost none of the reading is copper loss, and what the wattmeter shows is the core loss. The short-circuit test shorts one winding and raises the voltage on the other until rated current flows: the voltage needed is a few per cent of rated — that is the impedance voltage — the flux is correspondingly tiny so there is almost no core loss, and what the wattmeter shows is the copper loss. IEC 60076-1 defines the two in clauses 11.5 and 11.4.
The consequence is the whole of transformer economics. Core loss is there every hour the unit is energised, whatever it is doing. Copper loss goes as the square of the load, so at a third of rated load it is a ninth of its full-load value. Efficiency therefore peaks not at full load but where the two are equal, at x = √(P₀/Pk) — for the defaults here, about a third of rated. A distribution transformer sized for the peak half-hour of a summer afternoon spends most of its life below that, which is why efficiency standards weight no-load loss so heavily, and why all-day efficiency — energy out over energy in across a real duty cycle — is the number that decides what a transformer costs to own.
Regulation is a different question from efficiency and it does not follow it. The load current flows through the winding resistance and the leakage reactance, and the drop across them is what the secondary loses. The textbook shortcut is x(%R·cos φ + %X·sin φ), which is the projection of that drop onto the terminal voltage and is very good at ordinary power factors. This page computes the exact phasor magnitude as well, and prints the difference, because the shortcut loses accuracy as the power factor falls and the drop stops being nearly parallel to the voltage.
Leading power factor is the case worth understanding. Put a capacitive load on and sin φ changes sign: the reactive part of the drop now opposes the source rather than adding to it, and past the point where it overcomes the resistive part the secondary voltage is HIGHER on load than off it. Regulation goes negative. This is not a textbook curiosity — it is what happens on a feeder at night when the power-factor correction capacitors are still connected and the motors are not, and it is why regulation is always quoted at a stated power factor.
What this page leaves out: temperature. Load loss is quoted at a reference temperature, usually 75 °C, and a hot winding loses more; the core loss varies a little with voltage and frequency; and how much overload a transformer will take is a thermal question for the IEC 60076-7 loading guide, not an arithmetic one. For the currents see the transformer kVA and current calculator, for the turns the turns ratio calculator, and for the load side the power factor calculator and the voltage drop calculator.
Frequently asked questions
At what load is a transformer most efficient?
Where the copper loss equals the core loss, which is at a load fraction of √(no-load loss ÷ full-load copper loss). For a typical distribution transformer that is between a third and a half of rated — not at full load, which is where most people assume it must be.
What is all-day efficiency and why is it different?
It is energy out divided by energy in over a whole day, rather than power out over power in at one instant. It is lower than the instantaneous figure because the core loss runs for all 24 hours while the load does not, and it is the number that decides the transformer’s running cost. Two transformers with identical peak efficiency can differ substantially on all-day efficiency if one has more iron loss than the other.
Why can voltage regulation be negative?
Because a leading (capacitive) load’s current leads the voltage, so its drop across the leakage reactance subtracts from the source instead of adding to it. Once that outweighs the resistive drop the secondary voltage is higher loaded than unloaded. It is a real effect on lightly loaded feeders with power-factor correction connected.
How do I get %R and %X from the test data?
%R is the full-load copper loss divided by the rating — a 1,250 W load loss on 100 kVA is 1.25%. %X is what is left of the nameplate impedance: √(%Z² − %R²). That is why this page asks for the load loss and the impedance voltage rather than for %R and %X directly; those two are what appears on a nameplate.
Does power factor change a transformer’s losses?
No. The losses depend on the current, and the current at a given load fraction is the same whatever the phase angle. What changes is the useful power that current delivers, so the efficiency falls with the power factor even though nothing about the transformer has changed — and the regulation gets worse, because more of the drop is now in phase with the voltage.
Is the short-circuit test loss the same as the full-load copper loss?
Only if the test was done at rated current. It usually is, but not always — Chapman’s own worked problem reproduced on this page has a short-circuit test at twice rated current, and using its 38.1 W directly instead of scaling it by the square of the current ratio turns a 98.3% answer into 95.0%.
Related calculators
References
- IEC 60076-1:2011, Power transformers — Part 1: General. Clause 11.4, measurement of short-circuit impedance and load loss; clause 11.5, measurement of no-load loss and current; and the definition of rated power in 3.4.6 as the apparent power which, with the rated voltage, determines the rated current.
- IEC 60076-5:2006, Power transformers — Part 5: Ability to withstand short circuit, Table 1. The recognised minimum short-circuit impedances reproduced above — 4.0% for 25 to 630 kVA, rising to 12.5% at 63 to 100 MVA.
- Chapman S J. Electric Machinery Fundamentals, solved problem 2-6. The 1,000 VA 230/115 V transformer whose open- and short-circuit tests this page reproduces end to end: R_eq,s = 0.126 Ω, X_eq,s = 0.476 Ω, 2.96% regulation and 98.3% efficiency at rated load and 0.8 PF lagging.
- United for Efficiency, Model Regulation Guidelines and Model Procurement Specification for Distribution Transformers, Annex B, citing IEC 60076-20. The Tier 2 maxima used as defaults here for a 100 kVA liquid-immersed unit: 130 W no-load loss and 1,250 W load loss.