In order to make the calculation of bimetallic radiators, sufficient for repair in an apartment or house, you will not need serious knowledge of accurate measurements. Heating in apartments is almost always implemented using sectional batteries. This is due to the fact that the heating center works with high pressure. For example, most often the working pressure of steel radiators is 10. atm. Aluminum or bimetallic batteries are able to withstand from 40. atm. At the same time, for each room, or the radiator, you can determine the number of necessary sections to drop the area even with different heat lines.

What is the calculation of sections in the battery

The organization of heating at home or apartment is one of the most expensive tasks during construction or repair. From the number of sections in the battery, not only the room temperature in the cold period is hung, but also the total repairs' costs. Too large radiator may be ineffective, do not warmly completely or not work as it should.

Each room has a different area, thermal losses, nuances and features of the furniture. From where the heating batteries will be located, their efficiency also depends. The main task is to compensate for the heat loss of the building, evenly warm all the premises, provide comfortable conditions for the use of radiators. What's better? One battery for 12 sections, or 2 batteries of 6? Calculate the number of sections can be having a plan, a calculator and a few minutes of its time.

Calculation of the number of sections based on the area

Focusing on the square when choosing a battery, you must make a correction to the height of the ceilings. The average is 2.5-2.8 m. For heating a square meter of living space, it will take about 100 W energy according to the construction standards. Naturally, the heating of the house from the cold brick and the insulated foam block will require different power of thermal instruments. The same can be said about the energy efficiency of the house, the presence of double-glazed windows, good ventilation, roof insulation or floor.

Example of calculation:

Room 30 sq.m., having two windows, and the height of the ceilings is 2.4 m. Calculate the number of sections is needed for several heat devices.

30 x 100 W \u003d 3000 W Energy will be needed on average for heating of this room.

Aluminum and bimetallic radiators have different power. Moreover, there are several typical sizes. The most common mid-sieve distance for sectional heating batteries - 500 mm, also exists 800 mm, 350 mm, or even 200 mm. In order to correctly calculate the number of sections in the radiator, it is necessary to pre-clarify the seller the thermal power from a particular manufacturer. Some firms mark their products based on the standard configuration in 10 sections, some indicate the power of the piece of each element.

The average power is in the range of 140-170 W. It is worth being attentive, since this parameter is determined based on the temperature of the coolant of 60 degrees. If you plan to use a low-temperature heating system, for example, through a heat accumulator, the number of sections will be needed greater than when heating is directly from the boiler.

Total: 3000 W / 150 \u003d 20 sections.

Considering that there are two windows in the room, as a result of our calculations, the installation of two radiators for 10 sections in each will be the best option.

What gives this number

We could calculate the ratio to any side, for example - 8 and 12, 6 and 14. Why is it better to install a 2 radiator for 10 sections? The fact is that the radiators from the manufacturer come in packs with a set of 10 sections. It guarantees you what exactly the manufacturer collected all the elements. Almost all radiators before admission to the implementation are to be tested. This is usually due to the discharge of pressure. In some cases, even the use of special fluids is allowed. There are also ways to process the radiator from the inside to increase its service life. It can be a spraying of paint, varnish, special anti-corrosion compositions.

The assembly of sections is made due to the connection between the nipples and sealing of the paronite gasket. Sometimes the laying is to sit on glue, sometimes on silicone, sometimes it works only at the expense of its plane. Having broken the gasket should not be reused, and immediately need to be replaced. As a result, the situation is obtained - you need a radiator, the length of which consists of 12 sections. The store should take the factory packaging in 10 sections, unscrew 2 sections from another battery, twist two nipples and put them on the gaskets. As a result, you will receive 12 sections, but also the place of manual assembly, to which more than a year warranty will not get. At the same time, the manufacturers provide a guarantee from 5 to 25 years to the factory battery assembly.

Question The second - what shop does with the remaining 8 sections? What kind of gaskets are used, nipple? What properties used sealant?

Installing 2 batteries for 10 sections is oriented under the windows. Aluminum and bimetallic radiators create sufficient convection to organize a thermal veil in front of the source of energy loss. This will save and save and make your home warmer.

Calculate the number of sections in the battery is simple enough, but it is worth remembering that the parameters of your heating system will change over time. This can affect equipment wear, garbage deposition in pipelines or inside radiators. Do not forget also about very cold winter, which can happen every 7-10 years. Given the service life of the heating system, it will not be superfluous in 20-30%.

If you plan to hide the battery behind the screen or dense curtains - it costs a 10% increase in the radiator power. The same applies to premises with high ceilings, the more internal volume - the greater the thermal power of the radiator will need.

No need to strive to calculate up to 1 units. Your boiler will not be able to give more of its rated power, and adjustment even with the wrong calculation, will be carried out due to the temperature of the coolant. Competently design the heating system is important that it is comfortable. Cranes installed, thermostatic valves should allow the volume of the coolant passing through the thermal device.

The heating system includes many different elements. All of them are important for normal functioning, including radiators. Today, various batteries are used for heating private houses and apartments (this is how the people are called radiators). They can be made of cast iron, aluminum or be bimetallic. But in order to be warm in the house, it is important to correctly calculate the number of sections in the radiator. It is about this that will be discussed in this article. And specifically, an approximate calculation of the number of sections of the bimetallic radiator will be given.

Simple calculation method when replacing old batteries

If you decide to replace the old cast-iron heating radiator, you can use an easy way and make the calculation of the required number of battery sections. For this you need consider some factors. Namely:

• the heat transfer in bimetallic and cast iron radiators is slightly different. If the first this value is 200 W per section, then the second is 180 W.
• how big battery warped. If her work satisfied you, then it is good. If not, you can increase the number of sections.
• after a certain time, the heating radiator will warm a little worse. This is due to the clogging of the inner cavities of the device.

As a rule, when replacing the cast-iron heating radiator, the bimetallic number of battery sections do not change. Of course, if the work of the old battery suited you. If the heat is missing, then you can increase the number of sections.

Calculation based on the size of the room

Another thing is when the installation of the heating system is produced in a new house. In this case, it is not possible to rely on the previous experience of exploitation of heating radiators. Here more accurate calculation is requiredBased on the size of the room.

Such calculations can be done based on:

There are a number of sanitary standards, according to which there should be a certain power of heating devices for each square meter of the room. These standards can be easily found via the Internet. So, for the middle strip of our country, the power per square meter must be at least 100 W. Based on this, it is easy to make the necessary calculations.

For example, if you take room Square in 12 square meters (three to four), the power of the heating devices should be 1200 W (12 sq.m. * 100 W). We divide this value to the power of one section of the bimetallic radiator (200 W at a temperature of the coolant 90 degrees) we obtain 6 sections.

To obtain more accurate calculations, you can use a method that relies on the volume of heated room. In this case, the data is also taken from sanitary standards. So, for the middle strip on one cubic meter it is necessary to have 41 W power of the heating devices.

If you take the same area as in the previous example, then with a ceiling height of 2.7 meters we obtain the volume of the entire room 32.4 cubic meters (20 sq.m. * 2.7 meters). Then the power of radiators should be 32.4 * 41 \u003d 1328.4 W. If you divide the thermal power of one bimetallic section, then we get 6.64. It means that it is desirable to install the 7-section radiator for heating.

As can be seen using the method of calculating the volume of the room, you can obtain more accurate data on the number of sections of bimetallic (and any other) heating radiator. But in this case, the presence of windows in the room and some other factors are not taken into account. To clarify, it is necessary to use correction coefficients.

Determine the correction coefficients

By making the calculation of the required number of sections of the bimetallic radiator, it is not enough to know the area or the volume of the room. Many factors are important: the condition of the walls, the presence in the neighborhood of unheated premises, the temperature of the coolant supplied (from this will depend on the thermal power of each section), etc.

To in the room, it was, the heat should be taken into account and some correction coefficients. Namely:

Another correction coefficient refers to private homes. In such buildings there is a cold attic room, and all the walls come out. So, the power of the heating devices should be greater. Thus, for private houses, when calculating the number of sections of the bimetallic radiator, a correction coefficient of 1.5 is applied.

The calculation of the required number of sections on a bimetallic radiator depends on many factors. This is the size of the room, and the availability of windows, and much more. For example, if the walls of the private house are insulated well, then there will be little heat loss. So, the radiators can be installed with a smaller length and power. Also number of sections May depend on the people themselves who live in the dwelling. If they like a lot of heat, then the heating devices are installed more powerful.

At the stage of preparation for capital repairs and in the process of planning a new home, there is a need to calculate the number of heating radiator sections. The results of such calculations allow you to know the number of batteries, which would be enough to ensure an apartment or at home enough warmth even in the coldest weather.

The procedure for calculating may vary depending on the set of factors. Check out the instructions for quick situations, calculating for non-standard rooms, as well as with the procedure for performing the most detailed and accurate calculations, taking into account all sorts of significant room characteristics.

The heat transfer indicators, the form of the battery and the material of its manufacture - these indicators do not take into account these indicators.

Important! Do not fulfill the calculation at once for the whole house or apartment. Spend a little more time and make calculations for each room separately. Only so you can get the most reliable information. At the same time, in the process of calculating the number of battery sections for heating the corner room, it is necessary to add 20% to the final result. You need to throw the same stock from above, if there are no effectiveness in the work of heating, it is not enough for high-quality warming up.

Let's start learning from considering the most frequently used method of calculation. It is unlikely to be considered the most accurate, but according to the ease of execution, it definitely breaks forward.

In accordance with this "universal" method for heating 1 m2 of the area of \u200b\u200bthe room, you need 100 W batteries. In this case, the calculation is limited to one simple formula:

K \u003d s / u * 100

In this formula:

For example, consider the procedure for calculating the required number of battery for room with dimensions of 4x3.5 m. The area of \u200b\u200bthis room is 14 m2. The manufacturer claims that each section of the battery released by them gives 160 W power.

We substitute the values \u200b\u200bin the above formula and we get that 8.75 sections of the radiator are needed for heating our room. Correct, of course, in the most side, i.e. To 9. If the angular room, add a 20% stock, round again, and we get 11 sections. If problems are observed in the work of the heating system, add another 20% to the initially calculated value. It will turn out about 2. That is, in the amount for heating a 14-meter corner room in the conditions of unstable operation of the heating system, 13 batteries will be needed.

Approximate calculation for standard rooms

Very simple calculation option. It is based on the fact that the size of the heating batteries of serial production is practically no different. If the room's height is 250 cm (standard value for most residential premises), then one section of the radiator will be able to heat 1.8 m2 of space.

The area of \u200b\u200bthe room is 14 m2. To calculate, it is enough to divide the value of the area to the previously mentioned 1.8 m2. The result is 7.8. Round up to 8.

Thus, to warm up the 14-meter room with a 2.5-meter ceiling you need to buy a battery for 8 sections.

Important! Do not use this method when calculating a low-power unit (up to 60 W). The error will be too big.

Calculation for non-standard rooms

This calculation option is suitable for non-standard rooms with too low or too high ceilings. The basis of the calculation is the approval in accordance with which 1 m3 of residential space is needed for 31 m3 of the battery power. That is, the calculations are performed according to a single formula that has this kind:

A \u003d bx 41,

• A - the desired number of heating battery sections;
• B - the size of the room. It is calculated as the product of the length of the room on its width and height.

For example, we consider room with a length of 4 m, a width of 3.5 m and a height of 3 m. Its volume will be 42 m3.

The general need of this room in thermal energy will calculate, multiplying its volume to the 41 W mentioned earlier. The result is 1722 W. For example, we take the battery, each section of which issues 160 W thermal power. The required number of sections will calculate, dividing the total need for thermal power to the power value of each section. It turns out 10.8. As usual, rounded to the nearest more integer, i.e. up to 11

Important! If you bought batteries that are not divided into sections, divide the overall need for heat to the power of the whole battery (indicated in the concomitant technical documentation). So you will learn the right amount of heating.

The maximum accurate calculation option

From the above calculations, we saw that none of them is perfectly accurate, because Even for the same premises, the results albeit a bit, but still differ.

If you need the maximum calculation accuracy, use the following method. It takes into account many coefficients capable of affecting the effectiveness of heating and other significant indicators.

In general, the calculated formula has the following form:

T \u003d 100 W / m 2 * A * B * C * D * E * F * G * S,

• where T is the total amount of heat required for heating the room under consideration;
• S - Square of the heated room.

The remaining coefficients need more detailed study. So, coefficient A takes into account the characteristics of the glazing of the room.

The values \u200b\u200bare as follows:

• 1.27 for rooms, whose windows are glazed with just two glasses;
• 1.0 - for premises with windows equipped with double double-glazed windows;
• 0.85 - If windows have a triple glass.

The coefficient in takes into account the characteristics of the insulation of the walls of the room.

The dependence is as follows:

• if the insulation is low-efficient, the coefficient is accepted equal to 1.27;
• with good insulation (for example, if the walls are posted in 2 bricks or are purposefully insulated with a high-quality heat insulator), a coefficient is 1.0;
• with a high level of insulation - 0.85.

The C ratio indicates the ratio of the total area of \u200b\u200bwindow openings and the floor surface in the room.

Dependence looks like this:

• with a ratio of equal to 50% coefficient C is taken as 1.2;
• if the ratio is 40%, the coefficient is equal to 1.1;
• with a ratio of equal to 30%, the coefficient value is reduced to 1.0;
• in the case of a less percentage ratio, coefficients are 0.9 (for 20%) and 0.8 (for 10%).

The D coefficient indicates the average temperature in the coldest period of the year..

Dependence looks like this:

• if the temperature is -35 and below, the coefficient is taken equal to 1.5;
• at temperatures up to -25 degrees, a value of 1.3 is used;
• if the temperature does not fall below -20 degrees, the calculation is carried out with a coefficient of 1.1;
• residents of regions in which the temperature does not fall below -15, the coefficient 0.9 should be used;
• if the temperature in the winter does not fall below -10, consider with a ratio of 0.7.

The coefficient E indicates the number of external walls.

If the outer wall is one, use the 1.1 coefficient. With two walls, increase it to 1.2; at three to 1.3; If the external walls 4, use a coefficient equal to 1.4.

Factor F takes into account the features of the above room. The dependence is as follows:

• if there is no heated attic room above, the coefficient is taken equal to 1.0;
• if the attic heated is 0.9;
• if the neighbor above is heated living room, the coefficient can be reduced to 0.8.

And the last coefficient of formula - G - takes into account the height of the room.

The procedure is as follows:

• in rooms with ceilings with a height of 2.5 m, the calculation is carried out using a coefficient of 1.0;
• if the room has a 3-meter ceiling, the coefficient is increased to 1.05;
• with a ceiling height of 3.5 m, consider with a 1.1 coefficient;
• rooms with a 4-meter ceiling are calculated with a 1.15 coefficient;
• when calculating the number of battery sections for heating the room with a height of 4.5 m, increase the coefficient up to 1.2.

This calculation takes into account almost all existing nuances and allows you to determine the necessary number of sections of the heating unit with the smallest error. Finally, you will only have to divide the calculated indicator on the heat transfer of the same battery section (specify in the supplied passport) and, of course, to round the number found to the nearest integer value towards the increase.

The number of bimetallic heating radiator can be calculated in two ways:

• one provides for the use of the room area;
• the second is to apply the volume of the room in which the battery will occur.

The first is advisable to apply when ceiling height is not more than 3 m. If the walls have a greater height, the second way becomes more reliable. Both methods are concluded in the calculation of the amount of heat required to create the optimum temperature in the room. The calculation is carried out in different ways:

• the first method is to multiply the area on the figure of 100 W (it is a normative thermal power by 1 m2);
• the second in the multiplication of the room for 41 watts.

Both methods have one common feature: the resulting digit is adjusted by correction coefficients that show the effect of room features on heat loss or its savings.

Factors affecting heat loss

1. View of glazing windows. Most of all warmly goes through windows with ordinary glass (correction factor 1.27). For a double and triple glass package, an indicator 1 and 0.85, respectively.
2. The magnitude of the windows. To determine the influence of this factor, the ratio of windows area to a similar indoor indicator is recognized. If it is 10 part, that is, it is 10% of the floor area, then k \u003d 0.8. With a further increase in the ratio by 10% K increases by 0.1. When the windows area is half of the same floor indicator, k \u003d 1.2.
3. Heat insulation. With low heat insulation, heat loss is 127% (correction coefficient K \u003d 1.27), with an average and high thermal insulation - 100 and 85%, respectively (K is 1 and 0.85).
4. Temperature on the street. What it is lower, the higher the heat loss becomes. At the same time, for the temperature of -10 ° C k \u003d 0.7. With a further decrease in the temperature of 5 degrees, the coefficient grows by 0.2. If outside the window -25 ° C, then K is 1.3.
5. Number of external walls. With one outer wall, the heat loss is small, therefore K is 1.1. If there are two and three outer walls, the coefficient is 1.2 and 1.3, respectively.
6. Type of room upstairs. If the same heated room is at the top, the heat loss is very small (k \u003d 0.8). In the presence of a heated attic K is 0.9. If the attic does not heal, then k \u003d 1.

See also: What radiators are better: bimetallic or aluminum

Calculation of the number of sections depending on the area

Q \u003d s * 100 * k1 * k2 * k3 * k4 * k5 * k6 / p,

• S - room area,
• k1 - heat loss coefficient caused by type of glazing,
• k2 is a digit that depends on the ratio of windows and premises,
• k3 is the thermal insulation coefficient,
• k4 is the temperature coefficient outside the window,
• k5 is an indicator of heat loss through a certain number of external walls,
• k6 - coefficient demonstrating the influence of the level of thermal insulation of the room, which is above the room,
• P is the thermal power of the same sector (you need to specify in W, so kW is translated into W).

Example: Let it be room with dimensions 4x3 m (ii s \u003d 12 m2). It has one outer wall, a window with a double glass and an area of \u200b\u200b3.6 m2. It is located under the heated room. The thermal insulation of the walls is medium, and it is often -25 ° C outside the window. In such a room, it is planned to establish bimetallic batteries with a heat transfer of 0.2 kW.

Since the indicators S and P are known, it remains to determine the value of the coefficients and calculate the amount of Röbembers. In this case, the coefficients are as follows:

• k1 \u003d 1,
• k2 \u003d 1, (3.6 / 12 * 100 \u003d 30%),
• k3 \u003d 1,
• k4 \u003d 1.3,
• k5 \u003d 1,1,
• k6 \u003d 0.8.

So, Q \u003d 12 * 100 * 1 * 1 * 1 * 1.3 * 1.1 * 0.8 / 200 \u003d 6.86 sectors. Since rounding is on the most side, in a room of 12 m2, you need to put the heating radiator with 7 sections. The final digit is still to increase by 30-40% because the thermal power of the sector (in this case is 0.2 kW) is defined for Δt \u003d 70 ° C, that is, for the heating system, in which the average temperature of the coolant is 90 ° C (100 at the entrance to the heating battery and 90 at the output). This is provided that the room should be 20 ° C.

See also: Choose batteries for heating an apartment or house

Individual heating systems do not have such a heated coolant, so the heating battery with 7 sections will not have enough kW. Given this, it is necessary to increase its number of ribs. To know how much you need to add, it is necessary to determine the heat transfer of one segment of the heating radiator with a smaller Δt.

For this use formula PC \u003d k * f * ΔtWhere:

• PC - thermal power of one segment of heating radiator,
• K is a heat transfer coefficient
• F is an area of \u200b\u200bheating surface (K and F are often indicated in tables compiled by manufacturers),
• ΔT is a temperature pressure (it is measured in ° C).

• tVC is the temperature of hot water at the entrance,
• tB is the temperature of the heated water at the outlet,
• tWN is the desired air temperature in the room.

Determining the number of sections per 1 m2

Some household owners often want to know how many sections need 1 square meters. m. knowing this indicator, you can calculate their total quantity, multiplying it on the area.

For various heating radiators, the number of sections per 1 m2 is different. This is explained by different thermal power. The number of battery sectors affect the features of the room.

Calculate the number of sections per quarter. m is possible by the above formula. However, it does not need to use the area of \u200b\u200bthe room. If you take into account the described condition without taking into account S, Q will be 100 * 1 * 1 * 1 * 1,3 * 1.1 * 0.8 / 200 \u003d 0.572 sections / sq. m. Next to define the total figure you need 0.572 to multiply by 12.

When designing heating systems, a mandatory measure is to carry out the calculations of the power of heating devices. The result obtained to a greater extent affects the selection of this or that equipment - heating radiators and heating boilers (if the project is performed for private houses that are not connected to central heating systems).

The largest popularity is currently using batteries made in the form of interconnected sections. This article we are talking 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, it is possible to use three main ways. The first two are slightly light, but they give only an approximate result that is suitable for typical rooms of multi-storey houses. This includes the calculation of the sections of radiators on the area of \u200b\u200bthe room or by its volume. Those. In this case, it is enough to know the desired parameter (area or volume) of the room and paste it into the appropriate formula for calculating.

The third method involves the use of a plurality of different coefficients that determine the heat loss of the room. This includes dimensions and type of windows, floor, type of wall insulation, ceilings height and other criteria affecting heat loss. Heat loss can also occur for various reasons associated with errors and shortcomings during the construction of the 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 mandatory conditions for performing an accurate calculation. To do this, thermal imagers are used, displaying heat leakage from the room on the monitor.

All this is done in order to pick up such power of radiators, which compensates for the total heat loss value. Consider each method of calculating the battery sections separately and give a visual example for each of them.

Calculation of the number of radiator sections on the area of \u200b\u200bthe room

This method is the most simple. To obtain the result, it will be necessary to multiply the area of \u200b\u200bthe room to the value of the radiator power required for heating 1kv.m. This value is given in SNiP, and it is:

• 60-100W for the average climatic zone of Russia (Moscow);
• 120-200W for regions located north.

The calculation of the radiators sections according to the averaged power parameter is carried out by multiplying it to the value of the room area. So, 20 sq.m. We will require for heating: 20 * 60 (100) \u003d 1200 (2000) W

Next, the resulting number must be divided into the power value of one section of the radiator. To find out what area 1 section of the radiator is calculated, it is enough to open the equipment to the equipment. Suppose that the power of the section is 200W, and the total capacity required for heating is 1600W (we take the arithmetic average). It remains only to clarify how much the radiator sections are 1 m2. To do this, we divide the value of the required power for heating to the power of one section: 1600/200 \u003d 8

Result: For heating room with an area of \u200b\u200b20 square meters. m. A 8-section radiator will need (provided that the power of one section is 200W).

The calculation of the sections of the heating radiators by the value of the area of \u200b\u200bthe room gives only an approximate result. In order not to make a mistake with the number of sections, it is best to produce calculations, provided that for heating 1 sq.m. Power required 100W.

This, as a result, will increase the total cost of installing the heating system, and therefore the holding of such a calculation is not always appropriate, especially with a limited budget. More accurate, but, all the same, the approximate result will give the following method.

The method of this calculation is similar to the previous one, except that now it will be necessary to find out the power value for heating not 1 sq.m., and the cubic meter of the room. According to SNiP - this is:

41W for heating the premises of the panel type buildings; 34W for brick houses.

As an example, take the same room at 20 square meters. m., and set the conditional ceiling height - 2.9m. In this case, the volume will be equal to: 20 * 2.9 \u003d 58 cubic meters

From this: 58 * 41 \u003d 2378 W for the panel house 58 * 34 \u003d 1972 W for a brick house

We divide the results obtained to the power value of the same section. TOTAL: 2378/200 \u003d 11.89 (panel house) 1972/200 \u003d 9.86 (brick house)

If rounded to a larger number, then for heating the room at 20 square meters. The m. Panel will need 12-section, and for a brick house 10-section radiators. And this figure is also approximate. In order to calculate with high accuracy, how many battery sections are needed to heating the premises, it is necessary to take advantage of a more complex way, which will be discussed below.

To carry out accurate calculation, special coefficients are introduced into the general formula, which can, how to increase (increase the increase), the value of the minimum power of the radiator for heating the room and reduce it (lowering coefficient).

In fact, the factors affecting the power value, the set, but we will use the most those that are easy to calculate and with which it is easy to operate. The coefficient depends on the values \u200b\u200bof the following parameters of the room:

1. Ceiling height:
   • At a height of 2.5 m, the coefficient is 1;
   • At 3m - 1.05;
   • At 3,5m - 1.1;
   • At 4m - 1.15.
2. Type of glazing windows indoors:
   • Simple double glass - the coefficient is 1.27;
   • Double glass of 2 glass - 1;
   • Triple glass - 0.87.
3. The percentage of the area of \u200b\u200bthe window from the total area of \u200b\u200bthe room (for ease of definition, you can divide the area of \u200b\u200bthe window to the area of \u200b\u200bthe room and multiply then 100):
   • If the result of calculations is 50%, the coefficient 1.2 is taken;
   • 40-50% – 1,1;
   • 30-40% – 1;
   • 20-30% – 0,9;
   • 10-20% – 0,8.
4. Wall insulation:
   • Low heat insulation - the coefficient is 1.27;
   • Good thermal insulation (masonry in two bricks or insulation 15-20cm) - 1.0;
   • Increased thermal insulation (wall thickness from 50cm or insulation from 20cm) - 0.85.
5. The average value of the minimum temperature in winter, which can hold a week:
   • -35 degrees - 1.5;
   • -25 – 1,3;
   • -20 – 1,1;
   • -15 – 0,9;
   • -10 – 0,7.
6. The number of external (end) walls:
   • 1 end wall - 1.1;
   • 2 walls - 1.2;
   • 3 Walls - 1.3.
7. Type of room above the heated room:
   • Unheated attic - 1;
   • Heated attic - 0.9;
   • Heated residential premises - 0.85.

From here it is clear that if the coefficient is above one, it is considered to be increasing, if lower is lower. If the unit is worth it, then it does not affect the result. To make a calculation, it is necessary to multiply each of the coefficients to the value of the room and the averaged specific value of thermal losses per 1 sq. M., which is (according to SNiP) 100W.

Thus, we have the formula: q_t \u003d γ * s * k_1 * ... * k_7, where

• Q_t - the required power of all radiators for heating the room;
γ - the average temperature of the heat loss per 1 sq.m., i.e. 100W; S is the total area of \u200b\u200bthe room; K_1 ... k_7 - coefficients affecting the magnitude of thermal losses.

• Room area - 18 sq.m.;
• Ceiling height - 3m;
• Window with conventional double glass;
• The area of \u200b\u200bthe window is 3 sq.m., i.e. 3/18 * 100 \u003d 16.6%;
• Thermal insulation - double brick;
• The minimum temperature on the street during the week in a row -20 degrees;
• One end (outer) wall;
• Premises from above - heated living room.

Now replace the letter values \u200b\u200bto numeric and get: Q_t \u003d 100 * 18 * 1.05 * 1,27 * 0.8 * 1 * 1,3 * 1.1 * 0.85≈2334 W

It remains to split the result on the power value of one radiator section. Suppose that on equal to 160W: 2334/160 \u003d 14.5

Those. For heating room with an area of \u200b\u200b18 sq.m. And the reduced coefficients of thermal losses will require a radiator with 15 sections (rounded to the most side).

There is another simple way to calculate the radiators sections, focusing on the material of their manufacture. In fact, this method does not give an accurate result, but it helps to estimate an approximate number of batteries sections that will need to use indoors.

The heating batteries are made to divide on 3 types depending on the material of their manufacture. These are bimetallic, in which metal and plastic are used (usually as an external coating), cast-iron and aluminum heating radiators. Calculation of the number of battery sections made from a particular material are the same in all cases. It is enough to use the averaged power value that one section of the radiator can give, and the area of \u200b\u200bthe 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 amount of heating radiators sections can be made by separation of the area of \u200b\u200bthe room to the value of the area, which is able to warm up one section of the radiator from the metal of interest to us. Take the room at 18 square meters. m. Then we get:

• 18 / 1.8 \u003d 10 sections (aluminum);
• 18/2 \u003d 9 (Bimetal);
• 18/15 \u003d 12 (cast iron).

The area that is able to warm up one radiator section is not always indicated. Usually manufacturers indicate its power. In this case, it will be necessary to calculate the overall power required for heating the room, any of the above methods. If you take the calculation of the area and the power required to warm up 1 sq.m., in 80W (according to SNiP), then we obtain: 20 * 80 \u003d 1800/180 \u003d 10 sections (aluminum); 20 * 80 \u003d 1800/185 \u003d 9.7 sections (bimetal); 20 * 80 \u003d 1800/145 \u003d 12.4 sections (cast iron);

Radiated decimal numbers in one of the parties, we get about the same result, as in the case of calculations in the area.

It is important to understand that the calculation of the number of sections for the metal of the manufacture of the radiator is the most inaccurate method. It can help determine the choice of one or another battery, and with anything else.

And finally, the Council. Almost every manufacturer of 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 in it, and the program will display the desired result at the output. But, if you do not trust the robot, then calculations, as you can see, it is easy to produce and independently even on a sheet of paper.

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