Contents
When modernizing the heating system, in addition to replacing pipes, radiators are also changed. And today they are made of different materials, different shapes and sizes. Just as important, they have different heat transfer: the amount of heat that can be transferred to the air. And this must be taken into account when calculating the sections of radiators.
The room will be warm if the amount of heat that is lost is compensated. Therefore, in the calculations, the heat losses of the premises are taken as a basis (they depend on the climatic zone, on the material of the walls, insulation, window area, etc.). The second parameter is the thermal power of one section. This is the amount of heat that it can produce at the maximum system parameters (90°C inlet and 70°C outlet). This characteristic must be indicated in the passport, often present on the packaging.
One important point: when doing the calculations yourself, keep in mind that most manufacturers indicate the maximum figure that they received under ideal conditions. Therefore, make any rounding up. In the case of low-temperature heating (heat carrier temperature at the inlet below 85°C), search for the heat output for the relevant parameters or make a recalculation (described below).
Calculation by area
This is the simplest technique that allows you to roughly estimate the number of sections needed to heat a room. Based on many calculations, the norms for the average heating power of one square of the area were derived. To take into account the climatic features of the region, two norms were prescribed in SNiP:
- for the regions of 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 is there such a large range 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 u2,7buXNUMXbare taken, for brick houses, you can use average values. For insulated houses – the minimum. Another important detail: these standards are calculated for the average ceiling height – no higher than XNUMX meters.
Knowing the area of uXNUMXbuXNUMXbthe room, multiply its rate of heat consumption, the most suitable for your conditions. 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 power, you get their number. It’s not difficult, but to make it clearer, let’s give an example.
An example of calculating the number of radiator sections according to the area of the room
Corner room 16 m2, in the middle lane, in a brick house. Batteries with a thermal power of 140 watts will be installed.
For a brick house, we take heat losses 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 16 mXNUMX is required to heat the room2 * 95 W = 1520 W.
Now we count the number of radiators for heating this room: 1520 W / 140 W = 10,86 pcs. We round up, it turns out 11 pieces. How many sections of radiators will need to be installed.
The calculation of heating batteries per area is simple, but far from ideal: the height of the ceilings is not taken into account at all. With a non-standard height, a different technique is used: by volume.
We count batteries by volume
There are norms in SNiP for heating one cubic meter of premises. They are given for different types of buildings:
- for brick 1 m3 34 watts of heat required;
- for panel – 41 W
This calculation of radiator sections is similar to the previous one, only now we do not need an area, but we take other volumes and norms. We multiply the volume by the norm, divide the resulting figure by the power of one section of the radiator (aluminum, bimetallic or cast iron).
Volume calculation example
For example, let’s calculate how many sections you need in a room with an area of 16 m2 and a ceiling height of 3 meters. The building is made of bricks. Let’s take radiators of the same power: 140 W:
- Finding volume. 16 m2 * 3 m = 48 m3
- We consider the required amount of heat (the norm for brick buildings is 34 W). 48 m3 * 34 W = 1632 W.
- Determine how many sections you need. 1632W / 140W = 11,66pcs Rounding up, we get 12 pcs.
Now you know two ways to calculate the number of radiators per room.
Read more about calculating room area and volume here.
Heat dissipation of one section
Today, the range of radiators is large. With the external similarity of the majority, thermal performance can vary significantly. They depend on the material from which they are made, on the dimensions, wall thickness, internal section and on how well the design is thought out.
Therefore, to say exactly how many kW in 1 section of an aluminum (cast-iron bimetallic) radiator can only be said in relation to each model. This information is provided 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 sections of the same height of the same manufacturer, but different models, may differ by 15-25 W (see the table below STYLE 500 and STYLE PLUS 500) . Even more tangible differences can be between different manufacturers.
However, for a preliminary assessment of how many sections of batteries are needed for space heating, we deduced the average values of thermal power for each type of radiator. They can be used for approximate calculations (data are given for batteries with a center distance of 50 cm):
- Bimetallic – one section emits 185 W (0,185 kW).
- Aluminum – 190 W (0,19 kW).
- Cast iron – 120 W (0,120 kW).
More precisely, how many kW in one section of a bimetallic, aluminum or cast iron radiator you can when you choose a model and decide on the dimensions. The difference in cast iron batteries can be very large. They are available with thin or thick walls, which significantly changes their thermal power. Above are the average values for batteries of the usual shape (accordion) and those close to it. Radiators in the style of “retro” thermal power is lower at times.
Based on these values and the average norms in SNiP, they deduced 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 radiator sections from this data? Still easier. If you know the area of the room, divide it by a factor. For example, a room of 16 m2, for its heating you will need approximately:
- bimetallic 16 m2 / 1,8 m2 u8,88d 9 pcs, rounding up – XNUMX pcs.
- aluminum 16 m2 / 2 m2 = 8 pcs.
- cast iron 16 m2 / 1,4 m2 u11,4d 12 pcs, rounding up – XNUMX pcs.
These calculations are only approximate. According to them, you can roughly estimate the cost of purchasing heating appliances. You can accurately calculate the number of radiators per room by choosing a model, and then recalculating 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, + 70 ° C at the outlet, while + 20 ° C is maintained indoors. That is, the temperature head of the system (also called the “delta system”) will be 70 ° C. What to do if your system does not get above +70°C at the inlet? or is a room temperature of +23°C required? Recalculate the declared power.
To do this, you need to calculate the temperature head of your heating system. For example, at the supply you have +70°C, at the outlet +60°C, and in the room you need a temperature of +23°C. We find the delta of your system: this is the arithmetic average of the inlet and outlet temperatures, minus the room temperature.
For our case, it turns out: (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 teach the power that this section can give out for your conditions.
When recalculating, we act in the following order. We find in the columns, tinted in blue, a line with a delta of 42°C. It corresponds to a coefficient of 0,51. Now we calculate the thermal power of 1 section of the radiator for our case. For example, the declared power is 185 W, applying the found coefficient, we get: 185 W * 0,51 = 94,35 W. Almost two times less. It is this power that needs to be substituted when making the calculation of radiator sections. Only taking into account individual parameters, the room will be warm.