Hydronic Heat Transfer (Water)

Also known as 500 rule · GPM delta T formula · BTU per hour water

Q˙=ρwcwV˙ΔT\dot{Q} = \rho_w c_w \dot{V} \, \Delta T

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Every hydronic system is a delivery truck: the water is the truck, the temperature drop is how much cargo it unloaded, and the flow rate is how many trips per minute it makes. The physics is nothing more than ṁcΔT — mass flow times specific heat times temperature change — which in volumetric terms becomes ρ·c·V̇·ΔT. The solver runs it in SI with ρ = 998.3 kg/m³ and c = 4186.8 J/(kg·K), then converts, so you can type gpm and °F and read watts, or type kilowatts and read litres per second.

North American technicians know this as BTU/hr = 500 × GPM × ΔT, and the 500 is not magic: 60 min/hr × 8.33 lb/gal × 1.00 BTU/(lb·°F) = 499.8, rounded to 500 in every textbook since the 1930s. Try it — 20 gpm at a 20 °F drop gives 500 × 20 × 20 = 200,000 BTU/hr, and this page returns 58.6 kW, the same number in SI clothes. The classic field trap is measuring ΔT across the boiler when the load is downstream of a bypass or a primary/secondary tee: you get the boiler's ΔT, not the load's, and your calculated capacity is fiction. The second trap is glycol — a 40 % propylene mix knocks roughly 10–15 % off the 500, so use the glycol page instead of this one.

Hydronic Heat Transfer (Water)
Q˙=ρwcwV˙ΔT\dot{Q} = \rho_w c_w \dot{V} \, \Delta T
Where
  • Q˙\dot{Q}= Heat transfer rate
  • V˙\dot{V}= Water flow rate
  • ΔT\Delta T= Supply-to-return ΔT