Radiator Sizing:

This is an important step in determining the thermal requirement needed to heat a home. For this, a correct calculation must be made based on the area of the home and its thermal coefficient, which is determined by the volume of air that needs to be heated, the number and size of windows, and the level of insulation. This way, the appropriate size of the radiators is established. Of course, the type of boiler to be installed will also be considered, knowing that a condensing boiler requires a sufficiently large radiant surface for thermal discharge to occur and to operate in condensing mode.

Boiler Sizing:

It is very important to know the power of the heating boiler that we want to install, so that it is sufficient for the area we want to heat. To calculate the power of the boiler, we will perform the product between: the surface area of the rooms, their height, the caloric coefficient, and the correction factor. The caloric coefficient depends on the thermal insulation of the house and the surface area of the windows, while the last factor represents the index for converting power from calories to watts.

The caloric coefficient of a home is determined by the thermal insulation level of the house, as well as the surface area of the windows that serve it. Usually, this coefficient has a value between 30 and 70 kcal/m³, with the note that it is lower when the house is better insulated. If there are too many windows or if the house is not properly insulated, the coefficient will be higher, significantly affecting the thermal comfort. To perform a quick/estimated calculation, if the house is insulated, you can use an average value of 50 kcal/m³.

Correction Factor:

This step involves converting calories into watts. This happens because caloric power is measured in calories, while thermal power is measured in watts. The calculation of the correction factor is made through this conversion from calories to watts, using an index (a ratio between the two units of measurement) whose value is 1.163.

Let’s take the example of a 100 m² home with a height of 2.5 meters, giving a volume of 250 m³. Therefore, we multiply these 250 m³ by an average caloric coefficient of 50 kcal/m³ for an insulated home and the correction factor:

250 m³ X 50 kcal/m³ X 1.163 W/mcal = 14,537 W (approximately 14.5 kW)

For this calculation, given the multiple variables involved in calculating the caloric coefficient, we recommend an error margin of 20%, resulting in a requirement of 17.4 kW.

Of course, this calculation is an estimate, and an exact calculation can be made for each individual home by a specialist.