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Drawing No. EH–EE–022 // Electrical Engineering

Transformer Cooling Calculator

Reviewed August 2026

Estimate transformer cooling airflow, oil flow, water flow, heat-transfer area, fan and pump power, cooling margin and simplified temperature rise for AN, AF, ONAN, ONAF, OFAF and OFWF systems.

Enter heat, not transformer output: A 1,000 kVA transformer does not require 1,000 kW of cooling. The cooling system removes the transformer’s electrical losses — usually a much smaller number obtained from test data, manufacturer data or the linked transformer calculator.

Start with the physical question

How many kilowatts of loss must be removed, which cooling class is used, and how much temperature increase is acceptable in the cooling medium?

Step 1Enter total heat loss at the operating load.
Step 2Select dry-type air cooling or liquid-immersed cooling class.
Step 3Choose acceptable air, oil or water temperature rise.
Step 4Compare required flow and area with the installed system.
Simple mode provides a quick heat-balance check. Advanced mode exposes coolant properties, pressure losses, auxiliary power and thermal assumptions.

Simple cooling inputs

1. Heat source and cooling class
Include no-load loss plus load-dependent loss at the actual operating load.
2. Coolant temperature rise
3. Installed cooling equipment
Total surface area available to reject heat to ambient or coolant.
A rough screening value used only for the hot-spot estimate.

Simple cooling results

Check that losses and temperature rises are positive.
Heat to remove
Required air flow
Equivalent / required oil flow
Required water flow
Required cooling area
Installed flow margin
Estimated hot-spot temperature
Simple mode assumes 80% of total transformer loss is winding-related heat when no loss split is available.
Approx. auxiliary power
Simple assumptions where applicable: fan 250 Pa / 60%; oil pump 80 kPa / 65%; water pump 120 kPa / 70%.
Cooling class interpretation
Cooling diagram.
Cooling arrangement for the selected class. The drawing is schematic and not to scale.
Hot coolant or heat leaving the active partCooled return, cooling air or cooling waterForced-circulation equipmentTank, radiator or heat exchanger
Temperature chart.
Temperature comparison. Bars 1–4 are ambient, bulk coolant, winding hot spot and entered target.
Bars 1–2: ambient and bulk coolantBar 3: estimated winding hot spotBar 4: entered target temperature
Installed-system screenResult
Air-flow margin
Oil-flow margin
Water-flow margin
Cooling-area margin
Entered coolant transport temperature rises

Cooling class explained

The modern IEC-style liquid cooling code uses four letters. Dry-type cooling is commonly described with two letters.

CodePlain-language meaningWhat moves the heat?
AN / AADry transformer, natural airNatural convection through winding ducts and enclosure openings
AF / FADry transformer, forced airFans increase airflow through the windings or enclosure
ONAN / OAOil natural, air naturalBuoyancy circulates oil; tank or radiators reject heat by natural air convection
ONAF / OA-FAOil natural, air forcedOil circulation remains natural; fans move air over radiators
OFAF / FOAOil forced, air forcedOil pumps and radiator fans are both active
OFWF / FOWOil forced, water forcedOil pump sends hot oil through an oil-to-water heat exchanger

What changes between air and liquid cooling?

Air cooling

Air is simple and readily available, but its density and heat capacity are low. Removing several kilowatts with a small air-temperature rise can require a large airflow and carefully designed ventilation paths.

Oil or ester cooling

A liquid carries much more heat per unit volume and also provides insulation. Natural circulation can be effective, while pumps are used when higher loading or controlled flow is required.

Water-cooled exchanger

Water has high heat capacity and can make the external cooler compact, but the system adds pumps, water quality control, leak barriers, maintenance and consequences of loss of cooling.

Worked example

A transformer dissipating 8 kW with a permitted 15 K air rise requires approximately:

Q_air ≈ 8 / (1.165 × 1.006 × 15) = 0.456 m³/s

For oil with density 860 kg/m³, heat capacity 1.9 kJ/kg·K and a 10 K rise, the equivalent heat-transport flow is about:

Q_oil ≈ 8 / (860 × 1.9 × 10) = 0.00049 m³/s = 0.49 L/s

Natural circulation does not use a pump, but the fluid still must circulate internally at a rate sufficient to transport the heat.

Background

Engineering background, equations and assumptions used by this calculator.

Frequently Asked Questions

Practical questions about assumptions, inputs and limitations.

References