Transformer Efficiency and Voltage Regulation Calculator, Maximum and All-Day Efficiency
From the losses and the impedance get the transformer efficiency, the point of maximum efficiency, the voltage regulation and the all-day efficiency.
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How this calculator works
Enter the rating in kVA, the iron loss (open-circuit test), the copper loss at full load (short-circuit test) and the percentage impedance. Then enter the load as a fraction of full load and its power factor.
The calculator finds the efficiency at that load, the load at which the copper loss equals the iron loss (maximum efficiency) and the maximum efficiency. It splits the impedance into resistance and reactance and gives the regulation exactly, from the phasors.
For the all-day efficiency enter how many hours the transformer carries the load. For the rest of the day it is energised with no load, so the iron loss continues.
Formulas used
- \eta = \frac{x S\cos\phi}{x S\cos\phi + P_i + x^2P_{cu}}
- x_m = \sqrt{P_i/P_{cu}}
- R = \frac{P_{cu}}{S},\quad X = \sqrt{Z^2 - R^2}
- \text{VR} \approx x(R\cos\phi + X\sin\phi)
Worked example
With the default numbers in the calculator above:
- Output power at this load: 300.00 kW
- Copper loss at this load (it varies with the square of the load): 3.38 kW
- Efficiency at this load: 98.08 %
- Load for maximum efficiency (copper loss equals iron loss): 64.55 % of full load, which is 322.75 kVA
- Maximum efficiency at this power factor: 98.10 %
- Resistance and reactance as a fraction of the rating: R = 1.20 % and X = 4.34 %
- Voltage regulation (exact, from the phasors): 2.69 % (the common approximation x(R cos + X sin) gives 2.67 %)
- All-day (energy) efficiency over 24 hours: 97.28 % (3600.00 kWh out, 100.50 kWh lost)
Change the numbers in the calculator to see the full working, every step, for your own case.
Common questions
At what load is a transformer most efficient?
When the copper loss equals the iron loss, which is at a load fraction equal to the square root of the iron loss over the full-load copper loss.
Why is all-day efficiency lower?
A distribution transformer is energised all day but often lightly loaded, so the fixed iron loss is a large share of the energy it handles.
Can regulation be negative?
Yes, with a leading power factor (a capacitive load) the secondary voltage can be higher on load than on no load.
Where do I get the losses?
From the transformer test certificate or data sheet, listed as no-load loss (iron) and load loss (copper).
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Got a question about this? Ask an engineering question and get a plain-language answer with references.
Further reading: Any electrical machines text (Fitzgerald, Kingsley and Umans; Theraja, A Textbook of Electrical Technology); AS 60076 Power transformers.
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