How to count sections of a heating battery. How many sections do you need a radiator. Calculation. Calculation by room volume
Exist . To heat 1 m2 of a room to a comfortable temperature (+20 °C), the heater must emit 100 W of heat. This number must be used.
You need to do the following:
- Define thermal power one edge of the battery. Often it is equal to 180 watts.
- Calculate or measure the temperature of the coolant in the heating system. If the temperature of the water entering the heater is tin. \u003d 100 ° С and leaving it is tout. \u003d 80 ° C, then the number 100 is divided by 180. The result is 0.55. It is 0.55 sections that should be used for 1 sq. m.
- If the measured indicators are lower, then the ΔT indicator is calculated (in the above case it is 70 ° C). To do this, use the formula ΔT = (tin. + tout.) / 2 - tk, where tk is the desired temperature. Standard tk is 20 °C. Let tin. = 60 °С, and tout. \u003d 40 ° С, then ΔT \u003d (60 + 40) / 2 - 20 \u003d 30 ° С.
- Find a special plate in which a certain value of ΔT corresponds to a correction factor. For some radiators at ΔT = 30 ° C, it is 0.4. These labels should be asked from the manufacturers.
- Multiply the thermal power of one fin by 0.4. 180 * 0.4 = 72 watts. That is how much heat can be transferred by one section from a coolant heated to 60 ° C.
- Divide the norm by 72. Total 100/72 = 1.389 sections are needed to heat 1 m2.
This method has the following disadvantages:
- Norm 100 W is calculated for rooms whose height is less than 3 m. If the height is greater, then a correction factor must be used.
- Not taken into account heat loss through windows, doors and walls if the room is corner.
- The heat loss caused by a certain way of installing the heater is not taken into account.
Read also: What are the dimensions of aluminum radiators
Correct calculation
It provides multiplying the area of the room by the norm 100, adjusting the result depending on the characteristics of the room and dividing the final figure by the power of one rib (it is advisable to use the adjusted power).
The product of the area and the norm equal to 100 W is corrected as follows:
- For each window, 0.2 kW is added to it.
- For each door, 0.1 kW is added to it.
- For a corner room, the final figure is multiplied by 1.3. If the corner room is located in a private house, then the coefficient is 1.5.
- For a room with a height greater than 3 m, apply the coefficients 1.05 (height 3 m), 1.1 (height 3.5 m), 1.15 (4 m), 1.2 (4.5 m).
It is also necessary to take into account the way the heater is placed, which also leads to heat loss. These losses are:
- 3-4% – in case of installation of the heating device under wide window sill or shelf;
- 7% if the radiator is installed in a niche;
- 5-7% if it is near open wall, but the screen partially covers it;
- 20-25% - in case of complete screen closure.
Example of calculating the number of sections
It is planned to put the battery in a room of 20 square meters. m. The room is corner, has two windows and one door. The height is 2.7 m. The radiator will be placed under the windowsill (correction factor - 1.04). The boiler supplies heat carrier with a temperature of 60 °C. At the outlet of the heater, the water will have a temperature of 40 °C.
Before purchasing and installing sectional radiators (usually bimetallic and aluminum), most people have a question about how to calculate heating radiators according to the area of \u200b\u200bthe room.
In this case, it would be most correct to produce But it uses a huge number of coefficients, and as a result, something underestimated or, conversely, overestimated may come out. In this regard, many use simplified options. Let's consider them in more detail.
Main settings
Please note that the correct operation of the heating system, as well as its efficiency, largely depends on its type. However, there are other parameters that affect this indicator in one way or another. These options include:
- Boiler power.
- Number of heaters.
- The power of the circulation pump.
Conducted calculations
Depending on which of the above parameters will be subject to detailed study, an appropriate calculation is made. For example, determining the required power of a pump or a gas boiler.
In addition, it is often necessary to calculate heating appliances. In the process of this calculation, it is also necessary to calculate the buildings. This is due to the fact that, having made a calculation, for example, of the required number of radiators, one can easily make a mistake when choosing a pump. A similar situation occurs when the pump cannot cope with supplying the required amount of coolant to all radiators.
Enlarged calculation
The calculation of heating radiators by area can be called the most democratic way. In the regions of the Urals and Siberia, the figure is 100-120 W, in central Russia - 50-100 W. A standard heater (eight sections, the center distance of one section is 50 cm) has a heat output equal to 120-150 W. Bimetallic radiators have a slightly higher power - about 200 watts. If a we are talking about a standard coolant, then for a room of 18-20 m 2 with a height of 2.5-2.7 m, two cast-iron appliances of 8 sections will be required.
What determines the number of devices
Calculation of heating radiators by area
Given the above factors, you can perform the calculation. So, 100 W will be required for 1 m 2, that is, to heat a room of 20 m 2, 2000 W will be required. One cast-iron radiator of 8 sections is capable of delivering 120 watts. We divide 2000 by 120 and we get 17 sections. As mentioned earlier, this parameter is very enlarged.
Calculation of heating radiators of a private house with its own heater is carried out according to the maximum parameters. Thus, 2000 is divided by 150 and we get 14 sections. We will need such a number of sections to heat a room of 20 m 2.
Formula for accurate calculation
There is a rather complicated formula by which you can make an accurate calculation of the power of a heating radiator:
Q t \u003d 100 W / m 2 × S (rooms) m 2 × q1 × q2 × q3 × q4 × q5 × q6 × q7, where
q1 - type of glazing: conventional glazing - 1.27; double glazing - 1; triple - 0.85.
q2 - wall insulation: poor - 1.27; wall in 2 bricks - 1; modern - 0.85.
q3 - the ratio of the areas of window openings to the floor: 40% - 1.2; 30% - 1.1; 20% - 0.9; 10% - 0.8.
q4 - outdoor temperature (minimum): -35°C - 1.5; -25°C - 1.3; -20°C - 1.1; -15° C - 0.9; -10C° - 0.7.
q5 - number of outer walls: four - 1.4; three - 1.3; angular (two) - 1.2; one - 1.1.
q6 - type of premises located above the calculated one: cold attic - 1; heated attic - 0.9; heated residential - 0.8.
q7 - height of premises: 4.5m - 1.2; 4m - 1.15; 3.5m - 1.1; 3m - 1.05; 2.5m - 1.3.
Example
Let's calculate the heating radiators by area:
A room of 25 m 2 with two double-leaf window openings with triple glazing, 3 m high, enclosing structures of 2 bricks, located above the room cold attic. Minimum air temperature in winter period time - +20°C.
Q t \u003d 100W / m 2 × 25 m 2 × 0.85 × 1 × 0.8 (12%) × 1.1 × 1.2 × 1 × 1.05
The result is 2356.20 watts. We divide this number by So, for our premises we need 16 sections.
Calculation of heating radiators by area for a private country house
If the rule applies for apartments - 100 W per 1 m 2 of the room, then this calculation will not work for a private house.
For the first floor, the power is 110-120 W, for the second and subsequent floors - 80-90 W. In this regard, multi-storey buildings are much more economical.
The calculation of the power of heating radiators by area in a private house is carried out according to the following formula:
N=S×100/P
In a private house, it is recommended to take sections with a small margin, this does not mean that it will make you hot, just the wider the heater, the lower the temperature must be supplied to the radiator. Accordingly, the lower the temperature of the coolant, the longer it will serve heating system generally.
It is very difficult to take into account all the factors that have any effect on the heat transfer of the heating device. In this case, it is very important to correctly calculate heat loss, which depend on the size of the window and doorways, window leaf. However, the examples discussed above make it possible to determine the required number of radiator sections as accurately as possible and at the same time ensure a comfortable temperature regime in the room.
When designing heating systems, a mandatory measure is the calculation of the power of heating devices. The result obtained to a greater extent influences the choice of one or another equipment - heating radiators and heating boilers (if the project is carried out for private houses that are not connected to central systems heating).
The most popular at the moment are batteries made in the form of interconnected sections. In this article, we will talk about how to calculate the number of radiator sections.
Methods for calculating the number of battery sections
In order to calculate the number of sections of heating radiators, you can use three main methods. The first two are quite light, but they give only an approximate result that is suitable for typical multi-storey buildings. This includes the calculation of sections of radiators according to the area of \u200b\u200bthe room or according to its volume. Those. in this case, it is enough to find out the desired parameter (area or volume) of the room and insert it into the appropriate formula for calculation.
The third method involves the use for calculations of many different coefficients that determine the heat loss of the room. This includes the size and type of windows, floor, type of wall insulation, ceiling height and other criteria that affect heat loss. Heat loss can also occur for various reasons related to mistakes and shortcomings in the construction of a house. For example, there is a cavity inside the walls, the insulation layer has cracks, marriage in building material etc. Thus, the search for all causes of heat leakage is one of the prerequisites for performing an accurate calculation. For this, thermal imagers are used, which display on the monitor the places of heat leakage from the room.
All this is done in order to select such a power of radiators that compensates for the total value of heat loss. Let's consider each method of calculating battery sections separately and give a good example for each of them.
Calculation of the number of radiator sections according to the area of \u200b\u200bthe room
This method is the simplest. To obtain the result, you will need to multiply the area of \u200b\u200bthe room by the value of the radiator power required for heating 1 sq.m. This value is given in SNiP, and it is:
- 60-100W for the average climatic zone of Russia (Moscow);
- 120-200W for areas located to the north.
Calculation of radiator sections according to the average power parameter is carried out by multiplying it by the value of the room area. So, 20 sq.m. will require for heating: 20 * 60 (100) = 1200 (2000) W
Further, the resulting number must be divided by the power value of one section of the radiator. To find out what area 1 section of the radiator is designed for, just open the equipment data sheet. Let's assume that the power of the section is 200W, and the total power required for heating is 1600W (we take the arithmetic mean). It remains only to clarify how many radiator sections are needed per 1 m2. To do this, we divide the value of the required power for heating by the power of one section: 1600/200 = 8
Result: for heating a room of 20 sq. m. you will need an 8-section radiator (provided that the power of one section is 200W).
The calculation of sections of heating radiators according to the area of \u200b\u200bthe room gives only an approximate result. In order not to be mistaken with the number of sections, it is best to make calculations, provided that for heating 1 sq.m. 100W power required.
This, as a result, will increase the overall cost of installing the heating system, and therefore such a calculation is not always appropriate, especially when limited budget. A more accurate, but still the same, approximate result will give the following method.
The method of this calculation is similar to the previous one, except that now from the SNiP you will need to find out the power value for heating not 1 sq.m., but a cubic meter of the room. According to SNiP, this is:
- 41W for heating the premises of panel-type buildings; 34W for brick houses.
As an example, let's take the same room with an area of 20 square meters. m., and set the conditional height of the ceiling - 2.9 m. In this case, the volume will be equal to: 20 * 2.9 \u003d 58 cubic meters
Of this: 58*41 =2378W for panel house 58*34=1972W for brick house
Let's divide the obtained results by the value of the power of one section. Total: 2378/200 = 11.89 (panel house) 1972/200 = 9.86 (brick house)
If rounded up to a larger number, then to heat a room of 20 square meters. m. panel will need 12-section, and for a brick house 10-section radiators. And this figure is also approximate. To with high precision calculate how many battery sections you need for space heating, you need to use more in a complicated way, which will be discussed next.
To carry out an accurate calculation, special coefficients are introduced into the general formula, which can, how to increase (the increase factor) the value minimum power radiator for heating the room, and lower it (lowering factor).
In fact, there are many factors that affect the power value, but we will use the most those that are easy to calculate and easy to operate with. The coefficient depends on the values of the following room parameters:
- Ceiling height:
- With a height of 2.5m, the coefficient is 1;
- At 3m - 1.05;
- At 3.5m - 1.1;
- At 4m - 1.15.
- Type of window glazing in the room:
- Simple double glass - the coefficient is 1.27;
- Double-glazed window of 2 glasses - 1;
- Triple double-glazed window - 0.87.
- The percentage of the window area of the total area of the room (for ease of determination, you can divide the window area by the area of the room and then multiply by 100):
- If the calculation result is 50%, a coefficient of 1.2 is taken;
- 40-50% – 1,1;
- 30-40% – 1;
- 20-30% – 0,9;
- 10-20% – 0,8.
- Wall insulation:
- Low level of thermal insulation - the coefficient is 1.27;
- Good thermal insulation (laying in two bricks or insulation 15-20 cm) - 1.0;
- Increased thermal insulation (wall thickness from 50 cm or insulation from 20 cm) - 0.85.
- The average value of the minimum temperature in winter that can last a week:
- -35 degrees - 1.5;
- -25 – 1,3;
- -20 – 1,1;
- -15 – 0,9;
- -10 – 0,7.
- Number of outer (end) walls:
- 1 end wall - 1.1;
- 2 walls - 1.2;
- 3 walls - 1.3.
- Type of room above the heated room:
- Unheated attic - 1;
- Heated attic - 0.9;
- Heated living quarters - 0.85.
From this it is clear that if the coefficient is higher than one, then it is considered increasing, if it is lower, it is decreasing. If its value is one, then it does not affect the result in any way. To make a calculation, it is necessary to multiply each of the coefficients by the value of the area of \u200b\u200bthe room and the average specific heat loss per 1 sq.m., which is (according to SNiP) 100W.
Thus, we have the formula: Q_T= γ*S*K_1*…*K_7, where
- Q_T is the required power of all radiators for space heating; γ is the average value of heat loss per 1 sq.m., i.e. 100W; S- total area premises; K_1…K_7 are the coefficients influencing the amount of heat losses.
- Room area - 18 sq.m.;
- Ceiling height - 3m;
- Window with conventional double glass;
- The window area is 3 sq.m., i.e. 3/18*100 = 16.6%;
- Thermal insulation - double brick;
- The minimum temperature outside for a week in a row is -20 degrees;
- One end (external) wall;
- The room above is a heated living room.
Now let's replace the alphabetic values with numeric ones and get: Q_T= 100*18*1.05*1.27*0.8*1*1.3*1.1*0.85≈2334 W
It remains to divide the result by the power value of one section of the radiator. Let's say that it is equal to 160W: 2334/160 \u003d 14.5
Those. for heating a room with an area of 18 sq.m. and given heat loss coefficients, a radiator with 15 sections is required (rounded up).
There is another simple way to calculate the sections of radiators, focusing on the material of their manufacture. In fact, this method does not give an exact result, but it helps to estimate the approximate number of battery sections that will need to be used in the room.
Heating batteries are usually divided into 3 types depending on the material of their manufacture. These are bimetallic, which use metal and plastic (usually as an outer coating), cast iron and aluminum radiators heating. The calculation of the number of battery sections made of a particular material is the same in all cases. Here it is enough to use the average value of the power that one section of the radiator can produce, and the value of the area that this section is able to warm up:
- For aluminum batteries- this is 180W and 1.8 square meters. m;
- Bimetallic - 185W and 2 sq.m.;
- Cast iron - 145W and 1.5 sq.m.
Using a simple calculator, the calculation of the number of sections of heating radiators can be done by dividing the area of \u200b\u200bthe room by the value of the area that one section of the radiator from the metal of interest to us can heat. Let's take a room of 18 square meters. m. Then we get:
- 18 / 1.8 = 10 sections (aluminum);
- 18/2 = 9 (bimetal);
- 18 / 1.5 \u003d 12 (cast iron).
The area that one section of the radiator is capable of heating is not always indicated. Usually manufacturers indicate its power. In this case, you will need to calculate the total power required to heat the room, using any of the above methods. If we take the calculation by area and the power required to warm up 1 sq.m., in 80W (according to SNiP), then we get: 20*80=1800/180=10 sections (aluminum); 20*80=1800/185=9.7 sections (bimetal); 20*80=1800/145=12.4 sections (cast iron);
Rounding the decimal numbers to one of the sides, we get approximately the same result, as in the case of calculations by area.
It is important to understand that calculating the number of sections for the metal of the radiator is the most inaccurate method. It can help you decide on the choice in favor of a particular battery, and nothing else.
And finally advice. Almost every manufacturer heating equipment or an online store on its website places a special calculator for calculating the number of sections of heating radiators. It is enough to enter the required parameters into it, and the program will output the desired result. But, if you do not trust the robot, then the calculations, as you can see, are quite easy to do on your own, even on a piece of paper.
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At first glance, it is easy to calculate how many radiator sections to install in a particular room. How more room- the more sections the radiator should consist of. But in practice, how warm it will be in a particular room depends on more than a dozen factors. Given them, it is possible to calculate the required amount of heat from radiators much more accurately.
General information
The heat output of one section of the radiator is indicated in technical specifications products from any manufacturer. The number of radiators in a room usually corresponds to the number of windows. Radiators are most often located under the windows. Their dimensions depend on the area free wall between window and floor. It should be borne in mind that the radiator must be lowered from the window sill by at least 10 cm. And between the floor and the bottom line of the radiator, the distance must be at least 6 cm. These parameters determine the height of the device.
The heat output of one section of a cast-iron radiator is 140 watts, more modern metal ones - from 170 and above.
You can calculate the number of sections of heating radiators , leaving the area of \u200b\u200bthe room or its volume.
According to the norms, it is considered that 100 watts of thermal energy is needed to heat one square meter of a room. If we proceed from the volume, then the amount of heat per 1 cubic meter will be at least 41 watts.
But none of these methods will be accurate if you do not take into account the characteristics of a particular room, the number and size of windows, the material of the walls, and much more. Therefore, when calculating the radiator sections according to the standard formula, we will add the coefficients created by one or another condition.
Room area - calculation of the number of sections of heating radiators
Such a calculation is usually applied to premises located in standard panel residential buildings with a ceiling height of up to 2.6 meters.
The area of the room is multiplied by 100 (the amount of heat for 1m2) and divided by the heat output of one section of the radiator indicated by the manufacturer. For example: the area of the room is 22 m2, the heat transfer of one section of the radiator is 170 watts.
22X100/170=12.9
This room needs 13 radiator sections.
If one section of the radiator has 190 watts of heat transfer, then we get 22X100 / 180 \u003d 11.57, that is, we can limit ourselves to 12 sections.
You need to add 20% to the calculations if the room has a balcony or is located at the end of the house. A battery installed in a niche will reduce heat transfer by another 15%. But in the kitchen it will be 10-15% warmer.
We make calculations according to the volume of the room
For a panel house with a standard ceiling height, as already mentioned above, the heat calculation is based on the need for 41 watts per 1m3. But if the house is new, brick, double-glazed windows are installed in it, and the outer walls are insulated, then 34 watts per 1 m3 are already needed.
The formula for calculating the number of radiator sections looks like this: the volume (area multiplied by the height of the ceiling) is multiplied by 41 or 34 (depending on the type of house) and divided by the heat transfer of one section of the radiator indicated in the manufacturer's passport.
For example:
The area of the room is 18 m2, the ceiling height is 2.6 m. The house is a typical panel building. The heat output of one section of the radiator is 170 watts.
18X2.6X41 / 170 \u003d 11.2. So, we need 11 radiator sections. This is provided that the room is not corner and it does not have a balcony, otherwise it is better to install 12 sections.
Calculate as accurately as possible
And here is the formula by which you can calculate the number of radiator sections as accurately as possible :
The area of the room multiplied by 100 watts and by the coefficients q1, q2, q3, q4, q5, q6, q7 and divided by the heat transfer of one section of the radiator.
More about these ratios:
q1 - type of glazing : with triple glazing, the coefficient will be 0.85, with double glazing - 1 and with conventional glazing - 1.27.
Thanks to the correct calculations of the number of sections for a bimetallic radiator can be created indoors comfortable temperature regardless of the weather outside.
And also you can wisely cut costs for heating for the benefit of your wallet, but without sacrificing comfort.
If you want to use natural resources wisely, do not want to freeze in the cold season and do not want to overpay for heating, then replace the batteries with more energy efficient ones. And before replacing or buying new radiators, you need to calculate how many sections it should have.
How to calculate the heat transfer of a bimetallic radiator and one section
The power of a bimetallic radiator is related to its capacity and size. The less carrier in the battery, the more efficient and economical it is. Reason - small amount of water which heats up faster, so much less electricity is spent.
Photo 1. Bimetal 500/80 bimetallic radiator, heat output - 2280 W, manufacturer - "Konner".
Calculation of the number of sections
For each room, its own calculation of the required number of sections is made. To do this, a number of factors are taken into account: product model, heat transfer level and room area.
Methods for assessing heat transfer by room dimensions
In order to correctly calculate and select the desired model in terms of area and size, first find out how many sections are required for heating 1 sq. m. The easiest way to calculate by the area of the room.
By area per square meter
The calculation formula is:
- N = S/P x 100.
- N- the number of sections.
- S- area of the room.
- P- kW in each section.
For example, for a room with an area (3x4) 12 sq. m. you need to do the following calculations: 12 sq. mx100/200W = 6 (12 m2x100/200W).
So this room needs 6 sections, but it is important to consider that these calculations are approximate. There are factors that can affect the increase in the number of sections. This is the presence of an uninsulated balcony, two outer walls and cold bridges that make the radiator work less efficient.
For more accurate readings it is also important to consider the height of the ceiling, window locations, connection method, insulation quality external walls and their presence.
Heat dissipation of bimetallic heating radiators directly depends on several parameters, which, brought together, will show how many sections are required to fit a certain area.
As the practice of using bimetal in apartments with central heating shows, it is correct calculated power allows you to efficiently heat the room and significantly save on payment utilities.
Attention! The disadvantage of calculation by area is that the indicators are obtained approximate.
To have an accurate idea of how many sections should be in bimetallic radiator, use other formulas. For example, by volume.
By volume
Based on the center distance, the radiator volumes can vary:
- 200 mm - 0.1-0.16 l.;
- 350 mm - 0.17-0.2 l.;
- 500 mm - 0.2-0.3 l.
It turns out if in the design 10 sections and center distance 200 mm, then the volume of water is from 1 to 1.6 liters.
For 10 with center distance 350 mm the volume of water is from 1.7 to 2 liters. If you take 10 pieces with center distance 500 mm, then the volume of water will be 2-3 liters. The most popular bimetal options are models with 8, 10, 12, 14 sections.
You can also do volume calculations. . For 1 sq. m requires 41 watts. Calculate the parameters based on the following formula:
- V \u003d length * width * height (in meters) \u003d volume in cubic meters. m.
As a result, you can find out the heat dissipation of the battery.
- P=V*41= number in watts.
Correction factors
Actual heat transfer may differ from those stated in the passport. They are affected by operating conditions. Therefore, be aware of the correction factors B1 and B2.
| Radiator type | Radiator height, mm | B1 | B2 | |
| When installed at outer wall | When installed near external glazing | |||
| 10 | 300 | 1,005 | 1,04 | 1,1 |
| 10 | 500 | 1,01 | ||
| 11,2 | 300 | 1,02 | ||
| 11,2 | 500 | 1,027 | 1,03 | 1,08 |
| 21 | 300 | 1,035 | 1,02 | 1,06 |
| 500 | 1,05 | |||
| 22 | 300 | 1,08 | - | 1,04 |
| 500 | 1,09 | |||
| 33 | 300 | 1,15 | 1,01 | 1,02 |
| 500 | 1,2 | |||
Multiply the number obtained during the calculation by the coefficient:
- north and corner rooms 1,3;
- areas with severe frosts 1,6;
- boxes and screens (you can add 20%, if niche - 7% );
- 100 for window heat dissipation in the room increases, 200 for the door.
Useful video
Check out the video for the different methods for calculating the number of radiator sections.