How to calculate the number of radiator sections

When modernizing the heating system, in addition to replacing the pipes, radiators are also changed. And today they are made of different materials, shapes and sizes. What is not less important, they have different heat output: the amount of heat that can transfer to the air. And this is necessarily taken into account when making the calculation of sections of radiators.

The room will be warm if the amount of heat that goes out will be compensated. Therefore, the calculations are based on the heat loss of the premises (they depend on the climatic zone, the material of the walls, insulation, window area, etc.). The second parameter is the thermal capacity of one section. This is the amount of heat it can give out at maximum system parameters (90°C at the inlet and 70°C at the outlet). This characteristic is necessarily indicated in the passport, often present on the packaging.

One important point: when making calculations on your own, take into account that most manufacturers indicate the maximum figure that they received under ideal conditions. Therefore, any rounding should be made to the larger side. In the case of low-temperature heating (coolant inlet temperature below 85°C), look for the heat output for the corresponding parameters or make a recalculation (described below).

Calculation by area

This is the simplest method that allows you to roughly estimate the number of sections needed to heat the room. On the basis of many calculations, norms for the average heating power of one square of area are derived. To take into account the climatic peculiarities of the region, SNiP prescribed two norms:

• for the regions of the middle strip of Russia it is necessary from 60 W to 100 W;
• for areas above 60°, the norm of heating per square meter is 150-200 W.

Why is such a large range given in the norms? In order to be able to take into account the materials of the walls and the degree of insulation. For houses made of concrete the maximum values are taken, for brick houses the average values can be used. For insulated houses – the minimum. Another important detail: these norms are calculated for an average ceiling height – not higher than 2.7 meters.

Knowing the area of the room, multiply its heat input rate, the most suitable for your conditions. You get the total heat loss of the room. In the technical data for the selected radiator model, find the heat output of one section. Divide the total heat loss by the capacity and get their number. It is not difficult, but to make it clearer, let’s give an example.

Example of calculating the number of sections of radiators on the area of the room

Corner room 16^m2, in the middle strip, in a brick house. Install will be batteries with a thermal capacity of 140 W.

For a brick house we take heat losses in the middle of the range. Since the room is corner, it is better to take a larger value. Let it be 95 W. Then it turns out that to heat the room requires 16^{m2 *} 95 W = 1520 W.

Now we calculate the number of radiators for heating this room: 1520 W / 140 W = 10,86 pcs. Rounded off, we get 11 pcs. So many sections of radiators will need to be installed.

Calculation of heating batteries on the area is simple, but far from ideal: the height of the ceilings is not taken into account at all. With non-standard height use another method: by volume.

Counting batteries by volume

There are in SNiP norms and for heating one cubic meter of premises. They are given for different types of buildings:

• for brick for 1 ^m3 requires 34 W of heat;
• for panel buildings – 41 W

This calculation of radiator sections is similar to the previous one, only now you need not the area, but the volume and the norms are taken different. The volume is multiplied by the norm, the resulting figure is divided by the power of one section of the radiator (aluminum, bimetallic or cast iron).

Example of calculation by volume

For an example, let’s calculate how many sections are needed in a room^{with an} area of 16^{m2 and a} ceiling height of 3 meters. The building is made of brick. Radiators will take the same power: 140 W:

• Find the volume. 16^m2 * 3 meters = 48 ^m3
• Calculate the required amount of heat (the norm for brick buildings is 34 W). 48 ^m3 * 34 W = 1632 W.
• Determine how many sections are needed. 1632 W / 140 W = 11,66 pcs. Rounded off, we get 12 pcs.

Now you know two ways of how to calculate the number of radiators per room.

Read more about calculating the room area and volume here.

Heat output of one section

Today, the range of radiators is large. With the external similarity of most, the thermal performance can differ significantly. They depend on the material from which they are made, on the size, wall thickness, internal cross-section and on how well the design is thought out.

Therefore, to say exactly how many kW in 1 section of aluminum (cast-iron bimetallic) radiator, can be said only in relation to each model. This data is specified by the manufacturer. After all, there is a significant difference in size: some of them are high and narrow, others are low and deep. The power of a section of the same height of the same manufacturer, but different models, may differ by 15-25 W (see in the table below STYLE 500 and STYLE PLUS 500) . Even more tangible differences can be from one manufacturer to another.

Nevertheless, for a preliminary assessment of how many sections of batteries are needed for space heating, derived average values of thermal capacity for each type of radiator. They can be used for approximate calculations (data for batteries with an inter-axial distance of 50 cm are given):

• Bimetallic – one section emits 185 W (0.185 kW).
• Aluminum – 190 W (0.19 kW).
• Cast iron – 120 W (0.120 kW).

Exactly how many kW in one section of the radiator bimetallic, aluminum or cast iron you can, when you choose a model and determine the dimensions. Very large can be a difference in cast iron radiators. They are there with thin or thick walls, because of which their heat output varies significantly. Above are the average values for batteries of the usual shape (accordion) and close to it. In radiators in the “retro” style, the thermal capacity is many times lower.

Based on these values and average norms in SNiP derived the average number of radiator sections per 1^m2:

• bimetallic section will heat 1.8^m2;
• aluminum – 1.9-2.0^m2;
• cast iron – 1,4-1,5^m2;

How to calculate the number of sections of the radiator on these data? It’s even simpler. If you know the area of the room, divide it by the coefficient. For example, the room is 16^m2, for its heating will approximately need:

• bimetallic 16^m2 / 1,8^m2 = 8,88 pcs, rounded off – 9 pcs.
• aluminum 16^m2 / 2^m2 = 8 pcs.
• cast iron 16^m2 / 1,4^m2 = 11,4 pcs, rounded off – 12 pcs.

These calculations are only approximate. According to them you will be able to roughly estimate the cost of purchasing radiators. Precisely calculate the number of radiators per room you can choose the model, and then still recalculating the number depending on what the temperature of the coolant in your system.

Calculation of radiator sections depending on real conditions

Once again we draw your attention to the fact that the heat output of one section of the radiator is specified for ideal conditions. This is the amount of heat the radiator will produce if the inlet temperature of the coolant is +90°C, the outlet temperature is +70°C and the room is maintained at +20°C. That is, the system temperature head (also called “system delta”) will be 70°C. What should you do if your system has an inlet temperature above +70°C or if you need a room temperature of +23°C? Recalculate the declared capacity.

To do this, you need to calculate the temperature head of your heating system. For example, you have +70°C at the supply, +60°C at the output, and you need a room temperature of +23°C. Find the delta of your system: this is the arithmetic mean of the inlet and outlet temperatures minus the room temperature.

For our case we get: (70°C+ 60°C)/2 – 23°C = 42°C. The delta for these conditions is 42°C. Then we find this value in the conversion table (below) and multiply the declared capacity by this coefficient. This gives you the power that this section will be able to deliver for your conditions.

When recalculating, proceed in the following order. In the columns colored in blue, find the line with a delta of 42°C. It corresponds to a coefficient of 0.51. Now we calculate the heat output of 1 radiator section for our case. For example, the declared power of 185 W, applying the found coefficient, we get: 185 W * 0,51 = 94,35 W. Almost twice less. This is the power and should be substituted when you calculate the sections of radiators. Only taking into account the individual parameters in the room will be warm.