Creating comfortable living conditions orwork is a paramount task of construction. A significant part of the territory of our country is in the northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always topical. With the growth of tariffs for energy, the reduction in energy consumption for heating comes to the fore.

Climatic characteristics

The choice of the construction of walls and roof depends beforeonly from the climatic conditions of the construction area. For their definition it is necessary to refer to SP131.13330.2012 "Construction climatology". The following values ​​are used in the calculations:

  • The temperature of the coldest five-day period is 0.92, denoted by Tn;
  • the average temperature is denoted by Thoth;
  • duration, is denoted by ZOT.

For the example for Murmansk, the values ​​have the following meanings:

  • Тн = -30 degrees;
  • That = -3.4 degrees;
  • ZOT = 275 days.

In addition, it is necessary to set the design temperature inside the TV room, it is determined in accordance with GOST 30494-2011. For housing, you can take TV = 20 degrees.

To perform the heat engineering calculation of the enclosing structures, the GSOP (degree-day of the heating period) is calculated in advance:
GSOP = (TV - TOT) x ZOT.
In our example, GSOP = (20 - (-3.4)) x 275 = 6435.

heat engineering calculation of enclosing structures

Main factors

For the right choice of materials protectingIt is necessary to determine what thermal characteristics they should possess. The ability of a substance to conduct heat is characterized by its thermal conductivity, denoted by the Greek letter l (lambda) and measured in W / (mx deg.). The ability of the structure to retain heat is characterized by its resistance to heat transfer R and is equal to the ratio of thickness to thermal conductivity: R = d / l.

In case the structure consists of several layers, the resistance is calculated for each layer and then summed.

Resistance to heat transfer is the mainan indicator of the external structure. Its value should exceed the normative value. Performing the heat engineering calculation of the building envelope, we must determine the economically justified composition of the walls and roof.

heat engineering calculation of the building envelope

Thermal conductivity values

The quality of thermal insulation is determined in the firstturn thermal conductivity. Each certified material undergoes laboratory tests, as a result of which this value is determined for the operating conditions "A" or "B". For our country most of the regions correspond to the operating conditions of "B". Performing the heat engineering calculation of the enclosing structures of the house, this value should be used. Thermal conductivity values ​​are indicated on the label or in the material passport, but if they are not available, reference values ​​from the Code of Practice can be used. Values ​​for the most popular materials are given below:

  • Masonry of ordinary bricks - 0.81 W (m х deg.).
  • Masonry made of silicate brick - 0,87 W (m х deg.).
  • Gas and foam concrete (density 800) - 0.37 W (mx degree).
  • The wood of coniferous species is 0.18 W (mx degree).
  • Extruded polystyrene foam - 0.032 W (mx degree).
  • Mineral wool plates (density 180) - 0.048 W (mx degree).

The normative value of the resistance to heat transfer

The calculated value of the resistance to heat transfer is notmust be less than the base value. The base value is determined according to Table 3 SP50.13330.2012 "Thermal Protection of Buildings". The table defines the coefficients for calculating the basic values ​​of the resistance to heat transfer of all enclosing structures and building types. Continuing the begun teplotehnichesky calculation of enclosing designs, the example of calculation can be presented as follows:

  • Rsten = 0.00035x6435 + 1.4 = 3.65 (mx deg / W).
  • Рпокр = 0,0005х6435 + 2,2 = 5,41 (m х degrees / W).
  • Рчерд = 0,00045х6435 + 1,9 = 4,79 (m х degrees / W).
  • Rocka = 0.00005x6435 + 0.3 = 0.62 (mx deg / W).

Thermotechnical calculation of external enclosingconstruction is carried out for all structures that close the "warm" circuit - the floor on the ground or the floor of the underground, the external walls (including windows and doors), the combined covering or the overlap of an unheated attic. Also, the calculation should be performed for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.

formula for heat engineering calculation of enclosing structures

Thermal engineering of walls

Most of the walls and ceilings are multilayered and heterogeneous in their construction. Thermotechnical calculation of the enclosing structures of the multilayer structure is as follows:
R = d1 / l1 + d2 / l2 + dn / ln,
where n is the parameters of the n-th layer.

If we consider a brick plastered wall, we get the following construction:

  • outer layer of plaster 3 cm thick, thermal conductivity 0.93 W (mx degree);
  • brickwork made of solid clay bricks 64 cm, thermal conductivity 0.81 W (mx degree);
  • the inner layer of plaster thickness of 3 cm, thermal conductivity 0.93 W (m × deg.).

The formula for the heat engineering calculation of the enclosing structures is as follows:

R = 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 = 0.85 (mx deg / W).

The obtained value is much smallerdetermined earlier the basic value of the resistance to heat transfer in the walls of a residential building in Murmansk 3.65 (mx deg / W). The wall does not meet regulatory requirements and needs to be insulated. We use mineral wool boards 150 mm thick and heat conductivity 0.048 W (mx degree) to warm the walls.

Having selected the thermal insulation system, it is necessary to perform the verification heat engineering calculation of the enclosing structures. The calculation example is given below:

R = 0.15 / 0.048 + 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 = 3.97 (mx deg / W).

The resulting calculated value is greater than the base value - 3.65 (mx deg / W), the insulated wall meets the requirements of the norms.

Calculation of overlaps and overlapping coatings is carried out in a similar way.

heat engineering calculation of external enclosing structure

Thermal calculation of floors in contact with the ground

Often in private homes or public buildingsthe floors of the first floors are made on the ground. Resistance to heat transfer of such floors is not standardized, but at least the construction of floors should not allow the fall of dew. Calculation of structures in contact with the ground is carried out as follows: the floors are divided into strips (zones) 2 meters wide, starting from the outer boundary. Such zones are allocated up to three, the remaining area belongs to the fourth zone. If the floor construction does not provide an effective insulation, the resistance to the heat transfer of the zones is adopted as follows:

  • 1 zone - 2.1 (m х degrees / W);
  • 2 zone - 4,3 (m х degrees / W);
  • 3 zone - 8.6 (m х degrees / W);
  • 4 zone - 14.3 (m х degrees / W).

It is easy to see that the further the floor areais located from the outer wall, the higher its resistance to heat transfer. Therefore, often limited to the insulation of the perimeter of the floor. In this case, resistance to the heat transfer of the insulated structure is added to the heat transfer resistance of the zone.
Calculation of the resistance to the heat transfer of the floorit is necessary to include in the general heat engineering calculation of the enclosing structures. An example of calculation of floors on the ground will be considered below. We take a floor area of ​​10 x 10, equal to 100 sq. M.

  • The area of ​​1 zone will be 64 square meters.
  • Area 2 zone is 32 square meters.
  • The area of ​​the 3rd zone will be 4 square meters.

Average value of resistance to heat transfer of the floor on the ground:
Рпола = 100 / (64 / 2,1 + 32 / 4,3 + 4 / 8,6) = 2,6 (m х degrees / W).

Having completed the insulation of the perimeter of the floor with a polystyrene plate with a thickness of 5 cm, a 1 meter wide strip, we obtain the average value of the resistance to heat transfer:

Рпола = 100 / (32 / 2,1 + 32 / (2,1 + 0,05 / 0,032) + 32 / 4,3 + 4 / 8,6) = 4,09 (m х degrees / W).

It is important to note that in this way not only the floors are calculated, but also the structures of the walls in contact with the ground (walls of the recessed floor, warm basement).

example of calculating cn

Thermal engineering of doors

A slightly different base value is calculatedresistance to the heat transfer of the entrance doors. To calculate it, you will first need to calculate the resistance to heat transfer of the wall according to the sanitary and hygienic criterion (non-falling dew):
Рст = (Тв - Тн) / (ДТн х ав).

Here DTN - the difference in temperature between the inner surface of the wall and the temperature of the air in the room, is determined by the Code of Standards and for housing is 4.0.
av is the heat transfer coefficient of the inner surface of the wall, according to the joint venture it is 8.7.
The base value of the doors is taken equal to 0.6 x Pst.

For the chosen design of the door, it is required to perform a verification heat engineering calculation of the enclosing structures. Example of calculating the entrance door:

Рдв = 0,6 х (20 - (-30)) / (4 х 8,7) = 0,86 (m х grad / W).

This is the calculated value that corresponds to the door, insulated with a mineral wool slab 5 cm thick. Its resistance to heat transfer is R = 0.05 / 0.048 = 1.04 (m × deg / W), which is greater than the calculated value.

Comprehensive requirements

Calculations of walls, ceilings or coatings are carried outto check the elemental requirements of the standards. The set of rules also establishes a complete requirement, characterizing the quality of insulation of all enclosing structures in general. This value is called the "specific heat-shielding characteristic". Without its verification, no heat engineering calculation of the enclosing structures is complete. The calculation example for the SP is given below.

DesignationAreaRA / R
Walls833.6522.73
Coating1005.4118.48
Overlapping the basement1004.7920.87
Window150.6224.19
Doors20.82.5
Amount88.77

Kob = 88.77 / 250 = 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for a house measuring 10 x 10 x 2.5 m. Resistance to heat transfer is equal to the base values.

The standardized value is determined in accordance with the JV, depending on the heated volume of the house.

In addition to the complex requirement, a heat engineering calculation of the enclosing structures is also used to compile an energy passport, an example of issuing a passport is given in the attachment to SP50.13330.2012.

 heat engineering calculation of building envelope

Coefficient of homogeneity

All the above calculations are applicable tohomogeneous structures. That in practice is very rare. In order to take into account the inhomogeneities that reduce the resistance to heat transfer, a correction coefficient of thermal engineering uniformity, r, is introduced. It takes into account the change in the resistance to heat transfer introduced by window and door openings, external corners, non-uniform inclusions (for example, crosspieces, beams, reinforcing belts), cold bridges,

Calculation of this coefficient is rather complicated,so in a simplified form, you can use the approximate values ​​from the reference literature. For example, for brickwork - 0,9, three-layer panels - 0,7.

 heat engineering calculation of enclosing structures calculation example

Effective insulation

Choosing a house insulation system, it's easy to see,that it is practically impossible to meet modern requirements for thermal protection without using an effective insulation. So, if you use traditional clay brick, you will need a masonry a few meters thick, which is economically impractical. At the same time, the low thermal conductivity of modern heaters based on expanded polystyrene or rock wool allows us to limit ourselves to thicknesses of 10-20 cm.

For example, in order to achieve a baseline heat transfer resistance of 3.65 (m × deg / W), you will need:

  • a brick wall 3 m thick;
  • masonry of foam concrete blocks of 1.4 m;
  • mineral wool insulation 0,18 m.
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