A worked calculation of the heat delivered into domestic hot water from flow and temperature rise, and what it means for boiler choice.
A common calculation is the heat going into domestic hot water from the flow rate and the temperature rise.
Formula: power (kW) = flow rate (litres per minute) × temperature rise (°C) × 4.2 ÷ 60. The 4.2 is the specific heat of water in kJ per kg per °C, and 1 litre of water is taken as 1 kg.
- Inputs. A draw-off of 10 litres per minute, with the water raised by 30 °C from inlet to outlet.
- Energy per minute. 10 kg × 30 °C × 4.2 kJ per kg per °C = 1,260 kJ every minute.
- Convert to power. There are 60 seconds in a minute, so 1,260 ÷ 60 = 21 kJ per second, which is 21 kW.
- Interpret. That 21 kW is the useful heat delivered into the water. At steady operation the gross fuel input must also cover heat losses, so it is higher than the useful output. A data plate may state net input, which excludes the latent heat recovered by condensation, so check its basis before comparing figures.
- Compare with demand. Check both the space-heating and hot-water duties. Whether they operate together or with hot-water priority depends on the system; checking both does not always mean adding their peak ratings together.
- Apply it to a combi. At a constant useful output of 21 kW and a 30 °C rise, the total available flow is about 10 litres per minute. Two outlets must share that capacity. A modulating combi may increase its output as demand rises, up to its limit; only within that limit can it maintain both the requested flow and temperature rise. There is no large stored reserve to cover demand beyond its capacity.
Use the method to sanity-check whether a given flow and temperature rise are realistic for an appliance.
Heating circuits
The same relationship sizes a heating circuit, usually worked in kilograms per second: output (kW) = mass flow (kg/s) × 4.2 × temperature drop (K). A circuit carrying 0.3 kg/s that cools by 15 K between flow and return delivers 0.3 × 4.2 × 15 = 18.9 kW. Rearranged, mass flow = output ÷ (4.2 × drop): an 18 kW load designed for a 20 K drop needs 18 ÷ 84 = 0.21 kg/s, about 13 litres per minute. Halving the design drop doubles the flow the pump and pipes must carry.