When upgrading the heating system, in addition to changing pipes, the radiators are also changed. And today they are from different materials, different shapes and sizes. What is not less important, they have different heat transfer: the amount of heat that can transmit air. And this is taken into account when calculating radiator sections.
The room will be warm, if the amount of heat that goes away, will be compensated. Therefore, in calculations, heat losses of premises are taken as a basis (they depend on the climatic zone, wall material, insulation, window area, etc.). The second parameter is the thermal power of one section. This is the amount of heat that it can give at the maximum system parameters (90 ° C at the inlet and 70 ° C at the output). This characteristic is necessarily indicated in the passport, often present on the package.
One important point: when carrying out calculations yourself, note that most manufacturers indicate the maximum figure they received under ideal conditions. Therefore, any rounding up in the big party. In the case of low-temperature heating (inlet temperature below 85 ° C), they seek thermal power for the relevant parameters or make a recalculation (described below).
This is the simplest technique, allowing approximately to estimate the number of sections necessary for heating the room. Based on many calculations, the norms for the average heating power of one square of the area are derived. In order to take into account the climatic features of the region, two norms were prescribed in the SNiPe:
- for the regions of the central Russia it is necessary from 60 W to 100 W;
- for areas above 60 °, the heating rate per square meter is 150-200 watts.
Why are the norms given such a large range? In order to be able to take into account the wall materials and the degree of insulation. For houses made of concrete take the maximum values, for brick it is possible to use medium. For insulated houses - the minimum. Another important detail: these norms are calculated for an average ceiling height - not more than 2.7 meters.
Knowing the area of the room, multiply its rate of heat consumption, 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 thermal power of one section. The total heat loss is divided by the power, get their number. It's easy, but to make it clearer, let's give an example.
Example of calculating the number of radiator sections by area of a room
Corner room 16 m2, in the middle lane, in a brick house. Install will be batteries with a thermal power of 140 watts.
For a brick house we take heat loss in the middle of the range. Since the room is angular, it is better to take a larger value. Let it be 95 watts. Then it turns out that for heating the room it takes 16 m2 * 95 W = 1520 watts.
Now consider the number: 1520 W / 140 W = 10.86 pcs. Round off, it turns out 11 pcs. So many sections of radiators will need to be installed.
Calculation of heating batteries for the area is simple, but far from ideal: the height of ceilings is not taken into account completely. At non-standard altitude, use a different technique: by volume.
Count the batteries by volume
There are norms in SNiPe for heating one cubic meter of premises. They are given for different types of buildings:
- for brick on 1 m3 Requires 34 watts of heat;
- for panel - 41 W
This calculation of the radiator sections is similar to the previous one, only now we need not the area, but the volume and norms are taken by others. 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, we calculate how many sections per 16 m2 and a ceiling height of 3 meters. The building is built of brick. Radiators we take the same power: 140 W:
- We find the volume. 16 m2 * 3 m = 48 m3
- We consider the necessary amount of heat (the norm for brick buildings is 34 W). 48 m3 * 34 W = 1632 watts.
- Determine how many sections you need. 1632 W / 140 W = 11.66 pcs. Round off, we get 12 pcs.
Now you know how to calculate the number of radiators per room.
Heat transfer of one section
Today the range of radiators is big. With the external similarity of the majority, the thermal performance can vary significantly. They depend on the material from which they are made, on the size, wall thickness, internal section and how well the structure is designed.
Therefore, to say exactly, how many kW in 1 section of aluminum (cast iron bimetallic) radiator can be said only with reference to each model. This data is indicated by the manufacturer. After all, there is a significant difference in size: some are tall and narrow, others are low and deep. The power of a section of the same height of the same manufacturer, but different models, can differ by 15-25 W (see the table below STYLE 500 and STYLE PLUS 500). Even more tangible differences can be found in different manufacturers.
Nevertheless, for preliminary estimation of how many sections of batteries are needed for space heating, the values of thermal power for each type of radiators have been eliminated. They can be used for approximate calculations (data are given for batteries with an axial distance of 50 cm):
- Bimetallic - one section allocates 185 W (0.185 kW).
- The aluminum is 190 W (0.19 kW).
- Cast iron - 120 W (0.120 kW).
More precisely how many kW in one section of the radiator bimetallic, aluminum or cast iron you can when you choose a model and determine the size. Very big there can be a difference in cast-iron batteries. They are with thin or thick walls, because of what their thermal power significantly changes. Above are the average values for batteries of the usual form (accordion) and those close to it. At radiators in the style of "retro" thermal power is lower at times.
Based on these values and average norms in SNiPe, the average number of sections of the radiator was deduced by 1 m2:
- bimetallic section heats up 1,8 m2;
- aluminum - 1,9-2,0 m2;
- cast-iron - 1,4-1,5 m2;
How to calculate the number of radiator sections from these data? It's still easier. If you know the area of a room, divide it by a factor. For example, a room of 16 m2, for its heating is approximately necessary:
- bimetallic 16 m2 / 1.8 m2 = 8.88 pcs, round off - 9 pcs.
- aluminum 16 m2 / 2 m2 = 8 pcs.
- cast-iron 16 m2 / 1.4 m2 = 11.4 pieces, round off - 12 pcs.
These calculations are only approximate. On them you will be able to roughly estimate the cost of purchasing heating appliances. Precisely calculate the number of radiators per room, you can select the model, and then recalculate the number depending on the temperature of the coolant in your system.
Calculation of radiator sections depending on actual conditions
Once again, we draw your attention to the fact that the thermal power of one section of the battery is indicated for ideal conditions. The battery will give out so much heat if its coolant has a temperature of + 90 ° C at the inlet, at the outlet + 70 ° C, in the room it is maintained at + 20 ° C. That is, the temperature head of the system (also called the "delta system") will be 70 ° C. What to do if your system is above + 70 ° C at the entrance to the case? or is the room temperature + 23 ° C necessary? Recalculate the claimed capacity.
To do this, calculate the temperature head of your heating system. For example, at the feed you have + 70 ° C, at the outlet of 60 ° C, and in the room you need a temperature of + 23 ° C. Find the delta of your system: this is the average arithmetic temperature at the inlet and outlet, minus the room temperature.
For our case, we get: (70 ° C + 60 ° C) / 2 - 23 ° C = 42 ° C. Delta for such conditions is 42 ° C. Next we find this value in the conversion table (located below) and multiply the declared power by this coefficient. We learn the power that this section can give for your conditions.
We find in the columns, colored in blue, a line with a delta of 42 ° C. It corresponds to a coefficient of 0.51. Now calculate the thermal power of the 1 section of the radiator 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 half that. That's the power and you need to substitute when doing the calculation of radiator sections. Only taking into account the individual parameters in the room will be warm.