About the kW to GPM Calculator
This tool estimates the water flow rate needed to carry a given thermal load using a measured or design temperature difference. It is useful for HVAC designers, hydronic system troubleshooters, and anyone converting heat transfer in kilowatts into gallons per minute.
How to use this calculator
- Enter the thermal power in kW.
- Enter the temperature difference across the equipment.
- Select whether the temperature difference is in °F or °C.
- Click Calculate GPM to view the estimated flow.
- Use Reset to restore the default 10 kW and 10 °F inputs.
How it works
The calculator converts thermal power from kilowatts to BTU/hr using 1 kW = 3412.142 BTU/hr. It then applies the common water heat transfer relationship: BTU/hr = 500 × GPM × ΔT°F.
Solving for flow gives GPM = kW × 3412.142 ÷ (500 × ΔT°F). If the temperature difference is entered in Celsius, the calculator first converts it to Fahrenheit by multiplying by 1.8.
The result assumes clean water near typical HVAC operating temperatures, where the constant 500 is a practical approximation based on water density, specific heat, and minutes per hour. Glycol mixtures, oils, seawater, or unusual temperatures may require a different fluid constant.
Example calculation
For 10 kW of heat transfer with a 10 °F temperature difference, the heat rate is 10 × 3412.142 = 34,121 BTU/hr. The flow is 34,121 ÷ (500 × 10) = 6.82 GPM. The same result is about 25.82 L/min.
Frequently asked questions
What does ΔT mean in this calculation?
ΔT is the temperature difference between the supply and return water, or the rise/drop across the equipment.
Can I enter Celsius for the temperature difference?
Yes. Select °C, and the calculator converts the difference to °F using ΔT°F = ΔT°C × 1.8 before calculating GPM.
Why is the constant 500 used?
It is a standard HVAC approximation for water that combines water density, specific heat, and 60 minutes per hour.
Does this work for glycol systems?
Not exactly. Glycol changes the fluid density and specific heat, so the 500 constant may not be accurate for glycol mixtures.
What happens if the temperature difference is smaller?
For the same kW load, a smaller ΔT requires a higher flow rate because less heat is carried per gallon of water.