At 50% loading, a transformer does not necessarily produce half its full-load losses. In the simplified model used here, its load-dependent component falls to one quarter, while no-load loss remains. At 80% loading, the load-dependent component is 64% of its rated value. That difference can reverse the result when comparing two designs.
The following figures are hypothetical, not product specifications or a claim that either loading level is optimal. They demonstrate how to use the transformer loss comparison calculator and how to document a result someone else can reproduce.
Use two loss inputs, not one efficiency headline
Let P0 mean no-load loss in kW, Pk mean the load-dependent loss measured at rated current in kW, and x mean the fraction of rated current. The screening formula is P_loss = P0 + Pk × x². Schneider Electric describes this relationship in its partial-load heat-loss explanation. The approximation holds the relevant test conditions fixed; it is not a thermal simulation.
For this exercise, both options have the same hypothetical 1,250 kVA rating and operating voltage. Their loss figures are assumed comparable at the same reference conditions. At 50% current loading, each carries approximately 625 kVA; at 80%, approximately 1,000 kVA. We compare equal delivered apparent load, not two differently sized transformers at the same percentage.
| Assumed input | Option A | Option B |
|---|---|---|
| No-load loss P0 | 1.10 kW | 1.70 kW |
| Rated load loss Pk | 9.00 kW | 7.00 kW |
| Energized annual hours | 8,760 | 8,760 |
| Energy-only tariff | 0.15 currency/kWh | 0.15 currency/kWh |
At 50% load, Option A loses less
Enter 50 into the calculator’s load-percentage field, not 0.5. Internally, the model converts it to a fraction before squaring.
- Option A: 1.10 + 9.00 × 0.50² = 3.35 kW.
- Option B: 1.70 + 7.00 × 0.50² = 3.45 kW.
- B minus A: 0.10 kW more loss.
If this operating point remained unchanged for all 8,760 hours, A would lose 29,346 kWh and B would lose 30,222 kWh. B therefore consumes 876 additional kWh for losses, costing 131.40 additional currency units at the assumed tariff. In the calculator’s savings convention, B versus A shows a negative saving of −131.40. The minus sign is meaningful, not an error.
At 80% load, the comparison reverses
Change only the loading field to 80%. Leave the ratings, loss inputs, hours and tariff unchanged.
- Option A: 1.10 + 9.00 × 0.80² = 6.86 kW.
- Option B: 1.70 + 7.00 × 0.80² = 6.18 kW.
- A minus B: 0.68 kW less loss with B.
The annual loss-energy values are now 60,093.60 kWh for A and 54,136.80 kWh for B. Their difference is 5,956.80 kWh, or 893.52 currency units. B’s higher fixed loss is outweighed by its lower load-dependent loss at this operating point. This is an operating-energy result, not a purchase recommendation.
Find the crossover without guessing a best loading band
Set the two modelled loss expressions equal: 1.10 + 9x² = 1.70 + 7x². Rearranging gives 2x² = 0.60, so x = √0.30 = 0.5477, or about 54.77%. Below that loading, A has lower modelled losses; above it, B has lower losses. This crossover belongs to these two hypothetical input sets. It is neither a universal efficiency optimum nor an acceptable loading limit.
A varying load needs separate time blocks
Suppose A operates for 4,380 hours at 50% and another 4,380 hours at 80%. Its annual losses are 3.35 × 4,380 + 6.86 × 4,380 = 44,719.80 kWh. The average loading is 65%, but using a constant 65% produces (1.10 + 9 × 0.65²) × 8,760 = 42,945.90 kWh. That shortcut understates this scenario by 1,773.90 kWh.
The reason is visible in the arithmetic: the average of 0.50² and 0.80² is 0.445, whereas 0.65² is 0.4225. For a multi-period profile, run each period separately and sum its loss energy. The website’s worksheet uses one constant operating point per run; it does not ingest an hourly load series.
What to record before comparing real equipment
Ask for the exact product configuration and separate loss figures. Schneider Electric’s dry-type transformer loss-data guidance notes that current loss information may need to be requested from product support. Do not mistake total full-load loss for the Pk input: adding P0 to that total would count the no-load component twice.
- Record ratings, voltage, frequency, reference temperature and the source document date.
- Keep energized hours separate from hours with useful output; an energized idle period still belongs in this model.
- For unequal ratings, establish the same served load and calculate each option’s loading fraction separately.
- Identify omitted temperature changes, harmonics and auxiliary cooling; verify their treatment with the manufacturer.
Energy-only savings exclude purchase price, financing, maintenance and demand charges. Use this comparison to make assumptions explicit, then combine it with the transformer sizing framework and project-specific engineering review. A precise annual number is useful only when the loading history and loss inputs justify it.
Sources & further reading
- Schneider Electric: Liquid-filled transformer heat loss at percentage loads ↗
- Schneider Electric: Obtaining low-voltage dry-type transformer loss data ↗