How to calculate the number of radiator sections per room: 3 calculation schemes of varying complexity

Watts and sections

To calculate the number of heating radiator sections, you need to know two values:

  • The amount of heat that is lost through the building envelope and that we need to compensate;
  • Heat flow from one section.

Dividing the first value by three, we get the desired one - the number of sections.

About power

In calculations for batteries of different types, it is customary to operate with the following values ​​of thermal power per section:

  • Cast iron radiator - 160 watts;

  • Bimetallic - 180 watts;

  • Aluminum - 200 watts.

As always, the devil is in the details.

In addition to the standard size of radiators (500 mm along the axes of the collectors), there are also low batteries designed to be installed under windowsills of non-standard heights and to create a thermal curtain in front of panoramic windows. With a center-to-center distance along the collectors of 350 mm, the heat flux per section decreases by 1.5 times (say, for an aluminum radiator - 130 watts), with 200 mm - 2 times (for aluminum - 90-100 watts).

In addition, the actual heat dissipation is greatly influenced by:

  1. Coolant temperature (read - the surface temperature of the heater);
  2. Indoor temperature.

Typically, manufacturers indicate heat flux for a difference between these temperatures of 70 degrees (say, 90 / 20C). However, the real parameters of the heating system are often far from the maximum allowable 90-95C in it: in the central heating system, the supply temperature reaches 90C only at the peak of frost, and in the autonomous circuit, the typical temperature of the coolant is completely equal to 70C for the supply and 50C for the return pipeline.

Reducing the temperature delta by half (for example, from 90/20 to 60/25 degrees) will reduce the section power by exactly half. An aluminum radiator will give no more than 100 watts of heat per section, a cast-iron one - no more than 80 watts.

Calculation schemes

Method 1: by area

The simplest calculation scheme takes into account only the area of ​​the room. According to the norms of half a century ago, 100 watts of heat should fall on one square meter of the room.

Knowing the thermal power of the section, it is easy to find out how many radiators are needed for 1m2. With a power of 200 watts per section, it is capable of heating 2 m2 of area; 1 square of the room corresponds to half of the section.

Let us, as an example, calculate the heating of a 4x5 meter room for MC-140 cast iron radiators (rated power 140 watts per section) at a coolant temperature of 70C and a room temperature of 22C.

  1. The delta of temperatures between environments is 70-22 = 48C;
  2. The ratio of this delta to the standard, for which the declared power of 140 watts is 48/70 = 0.686. This means that the real power in the given conditions will be equal to 140x0.686 = 96 watts per section;
  3. The area of ​​the room is 4x5 = 20 m2. Estimated heat demand - 20x100 = 2000 W;
  4. The total number of sections is 2000/96 = 21 (rounded to the nearest whole value).

Such a scheme is extremely simple (especially if you use the nominal value of the heat flux), but it does not take into account a number of additional factors that affect the room's heat demand.

Here is a partial list:

  • Rooms can vary in ceiling heights. The higher the overlap, the greater the volume to be heated;

Increasing the ceiling height increases the temperature variation at and below the ceiling. In order to get the cherished +20 on the floor, it is enough to warm up the air under a 2.5-meter-high ceiling to + 25C, and in a room 4 meters high under the ceiling, everything will be +30. The rise in temperature increases the loss of thermal energy through the ceiling.

  • In general, more heat is lost through windows and doors than through main walls;

The rule is not universal. For example, a triple-glazed unit with two energy-saving glasses in terms of thermal conductivity corresponds to a 70-centimeter brick wall. A double glazing unit with one i-glass allows 20% more heat to pass through, while its price is 70% lower.

  • The location of an apartment in an apartment building also affects heat loss. Corner and end rooms with common walls with the street will be clearly colder than those located in the center of the building;

  • Finally, the climatic zone has a very strong effect on heat loss. In Yalta and Yakutsk (the average January temperature is +4 and -39, respectively), the number of radiator sections per 1 m2 will be predictably different.

Method 2: by volume for standard insulation

Here are instructions for buildings that meet the requirements of SNiP 23-02-2003, which normalizes the thermal protection of buildings:

  • We calculate the volume of the room;
  • We take 40 watts of heat per cubic meter;
  • For corner and end rooms, multiply the result by a factor of 1.2;
  • For each window we add 100 W to the result, for each door leading to the street - 200;

  • The resulting value is multiplied by the regional coefficient. It can be taken from the table below.
Average January temperature Coefficient
0 0,7
-10 1
-20 1,3
-30 1,6
-40 2

Let's find out how much heat is needed for our 4x5 meter room, specifying a number of conditions:

  • The height of the ceiling in it is 3 meters;
  • The room is a corner room with two windows;
  • It is located in the city of Komsomolsk-on-Amur (average January temperature -25C).

Let's get started.

  1. The volume of the room is 4x5x3 = 60 m3;
  2. The base value of the heat demand is 60x40 = 2400 W;
  3. Since the room is angular, multiply the result by 1.2. 2400x1.2 = 2880;
  4. Two windows add another 200 watts. 2880 + 200 = 3080;
  5. Taking into account the climatic zone, we use a regional coefficient of 1.5. 3080x1.5 = 4620 watts, which corresponds to 23 sections of aluminum radiators operating at rated power.

Now we will be curious and calculate how many radiator sections are needed per 1 m2. 23/20 = 1.15. Obviously, the calculation of the heat load according to the old SNiP (100 watts per square, or a section of 2 m2) will be too optimistic for our conditions.

Method 3: by volume for non-standard insulation

How to calculate the number of batteries per room in a building that does not meet the requirements of SNiP 23-02-2003 (for example, in a Soviet-built panel house or in a modern "passive" house with extremely effective insulation)?

Heat demand is estimated by the formula Q = V * Dt * k / 860, where:

  • Q is the required value in kilowatts;
  • V - heated volume;
  • Dt is the temperature difference between the room and the street;
  • k - coefficient determined by the quality of insulation.

The temperature difference is calculated between the sanitary standard for a dwelling (18-22C, depending on the climatic zone and the location of the room inside the building) and the temperature of the coldest five-day period of the year.

The insulation coefficient can be taken from another table:

As an example, we will again disassemble our room in Komsomolsk-on-Amur, once again clarifying the input data:

  • The temperature of the coldest five-day period for this climatic zone is -31C;

The absolute minimum is lower and is -44C. However, extreme cold does not last long and is not included in the calculations.

  • The walls of the house are brick, half a meter thick (two bricks). The glazing of the windows is triple.

So:

  1. The volume of the room has already been calculated by us earlier. It is equal to 60 m3;
  2. The sanitary norm for a corner room and a region with a minimum of winter temperatures below -31C - +22, which, combined with the temperature of the coldest five-day period, gives us Dt = (22 - -31) = 53;
  3. We take the coefficient of insulation equal to 1.2;

  1. The heat demand will be 60x53x1.2 / 860 = 4.43 KW, or 22 sections of 200 watts each. The result is approximately equal to that obtained in the previous calculation due to the fact that the insulation of the house and windows meets the requirements of the SNiP regulating the thermal protection of buildings.

Useful little things

The real heat transfer of heating radiators is influenced by a number of additional factors, which should also be taken into account in the calculations:

  • With one-sided lateral connection, the power of all sections corresponds to the nominal only when their number is no more than 7-10. The far edge of the longer battery will be much colder than the liners;

The problem is solved by diagonal connection. In this case, all sections will be evenly heated, regardless of their number.

  • In most newly built houses, the heating supply and return pipes are located in the basement, which means that the risers are bridged in pairs on the upper floor. The return riser radiator will always be colder than the supply radiator;
  • Various screens and niches, again, reduce the heat transfer of the heating system, and the difference with the nominal heat output can reach 50%;

  • Throttle valves on the inlet limit the flow of water through the radiator, even when fully open. The drop in thermal power is determined by the choke configuration and is usually 10-15%. The exception is full bore ball and plug valves;

  • Radiators with side one-sided connection in the central heating system are gradually silting up. As the siltation proceeds, the temperature of the outermost sections will drop.

To combat dirt, the battery is periodically flushed through a flush valve installed in the lower collector of the outermost section. The hose connected to it is sent to the sewer, after which a certain amount of coolant is discharged through it.

Conclusion

As you can see, simple heating calculation schemes do not always give an accurate result. The video in this article will help you learn more about the calculation methods. Feel free to share your own experience in the comments. Good luck, comrades!