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Thermal calculation of a brick wall example. Methodology for thermal engineering calculation of an external wall

Thermal engineering calculations make it possible to determine the minimum thickness of enclosing structures to ensure that there are no cases of overheating or freezing during the operation of the structure.

Enclosing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, efficiency and architectural design, must first of all meet thermal engineering standards. Enclosing elements are selected depending on constructive solution, climatological characteristics of the development area, physical properties, humidity and temperature conditions in the building, as well as in accordance with the requirements of resistance to heat transfer, air permeability and vapor permeability.

What is the point of the calculation?

  1. If, when calculating the cost of a future building, only the strength characteristics are taken into account, then, naturally, the cost will be less. However, this is a visible saving: subsequently, significantly more money will be spent on heating the room.
  2. Properly selected materials will create an optimal microclimate in the room.
  3. When planning a heating system, a thermal engineering calculation is also required. For the system to be cost-effective and efficient, it is necessary to have an understanding of the real capabilities of the building.

Thermal requirements

It is important that external structures meet the following thermal requirements:

  • They had sufficient heat-shielding properties. In other words, it is impossible to allow overheating of premises in the summer, and excessive heat loss in the winter.
  • The difference in air temperatures between the internal elements of fences and premises should not be higher than the standard value. Otherwise, excessive cooling of the human body may occur by radiating heat onto these surfaces and condensation of moisture from the internal air flow on the enclosing structures.
  • In the event of a change in heat flow, temperature fluctuations inside the room should be minimal. This property is called heat resistance.
  • It is important that the airtightness of the fences does not cause strong cooling of the premises and does not impair the heat-insulating properties of the structures.
  • Fences must have normal humidity conditions. Since overmoistening of fences increases heat loss, causes dampness in the room, and reduces the durability of structures.

In order for structures to meet the above requirements, thermal engineering calculations are performed, and heat resistance, vapor permeability, air permeability and moisture transfer are calculated according to the requirements of regulatory documentation.

Thermal qualities

The thermal characteristics of external structural elements of buildings depend on:

  • Humidity conditions of structural elements.
  • The temperature of internal structures, which ensures that there is no condensation on them.
  • Constant humidity and temperature in the rooms, both in cold and in warm time of the year.
  • The amount of heat that is lost by a building during the winter.

So, based on all of the above, thermal engineering calculation of structures is considered an important stage in the design process of buildings and structures, both civil and industrial. Design begins with the choice of structures - their thickness and sequence of layers.

Problems of thermal engineering calculations

So, the thermal engineering calculation of enclosing structural elements is carried out with the aim of:

  1. Compliance of structures with modern requirements for thermal protection of buildings and structures.
  2. Providing a comfortable microclimate in the interior.
  3. Ensuring optimal thermal protection of fences.

Basic parameters for calculation

To determine the heat consumption for heating, as well as to make a thermal engineering calculation of the building, it is necessary to take into account many parameters depending on the following characteristics:

  • Purpose and type of building.
  • Geographical location of the building.
  • Orientation of walls according to cardinal directions.
  • Dimensions of structures (volume, area, number of floors).
  • Type and sizes of windows and doors.
  • Characteristics of the heating system.
  • The number of people in the building at the same time.
  • Material of walls, floors and ceilings of the last floor.
  • Availability of hot water supply system.
  • Type of ventilation systems.
  • Other design features buildings.

Thermal engineering calculation: program

To date, many programs have been developed to make this calculation. As a rule, the calculation is carried out on the basis of the methodology set out in the regulatory and technical documentation.

These programs allow you to calculate the following:

  • Thermal resistance.
  • Heat loss through structures (ceiling, floor, door and window openings, and walls).
  • The amount of heat required to heat the infiltrating air.
  • Selection of sectional (bimetallic, cast iron, aluminum) radiators.
  • Selection of panel steel radiators.

Thermal engineering calculation: example calculation for external walls

For the calculation, it is necessary to determine the following basic parameters:

  • t in = 20°C is the temperature of the air flow inside the building, which is taken for calculating fences based on the minimum values ​​of the most optimal temperature of the corresponding building and structure. It is accepted in accordance with GOST 30494-96.

  • According to the requirements of GOST 30494-96, the humidity in the room should be 60%, as a result the room will be provided with normal humidity conditions.
  • In accordance with Appendix B of SNiP 02/23/2003, the humidity zone is dry, which means that the operating conditions for the fences are A.
  • t n = -34 °C is the temperature of the external air flow in the winter, which is accepted according to SNiP based on the coldest five-day period, which has a probability of 0.92.
  • Z ot.per = 220 days - this is the duration of the heating period, which is accepted according to SNiP, while the average daily temperature environment≤ 8 °C.
  • T from.trans. = -5.9 °C is the ambient temperature (average) during the heating period, which is accepted according to SNiP, with a daily ambient temperature ≤ 8 °C.

Initial data

In this case, a thermal technical calculation of the wall will be carried out in order to determine the optimal thickness of the panels and the thermal insulation material for them. Sandwich panels will be used as external walls (TU 5284-001-48263176-2003).

Comfortable conditions

Let's look at how thermal engineering calculations are performed outer wall. First, you should calculate the required heat transfer resistance, focusing on comfortable and sanitary conditions:

R 0 tr = (n × (t in - t n)): (Δt n × α in), where

n = 1 is a coefficient that depends on the position of the external structural elements in relation to the outside air. It should be taken according to SNiP data 02/23/2003 from Table 6.

Δt n = 4.5 °C is the standardized temperature difference between the internal surface of the structure and the internal air. Accepted according to SNiP data from Table 5.

α in = 8.7 W/m 2 °C is the heat transfer of internal enclosing structures. The data is taken from table 5, according to SNiP.

We substitute the data into the formula and get:

R 0 tr = (1 × (20 - (-34)) : (4.5 × 8.7) = 1.379 m 2 °C/W.

Energy saving conditions

When performing a thermal engineering calculation of a wall, based on energy saving conditions, it is necessary to calculate the required heat transfer resistance of structures. It is determined by GSOP (heating period degree-day, °C) using the following formula:

GSOP = (t in - t from.trans.) × Z from.trans., where

t in is the temperature of the air flow inside the building, °C.

Z from lane and t from.per. is the duration (days) and temperature (°C) of a period with an average daily air temperature ≤ 8 °C.

Thus:

GSOP = (20 - (-5.9)) ×220 = 5698.

Based on energy saving conditions, we determine R 0 tr by interpolation according to SNiP from Table 4:

R 0 tr = 2.4 + (3.0 - 2.4) × (5698 - 4000)) / (6000 - 4000)) = 2.909 (m 2 °C/W)

R 0 = 1/ α in + R 1 + 1/ α n, where

d is the thickness of the thermal insulation, m.

l = 0.042 W/m°C is the thermal conductivity of the mineral wool board.

α n = 23 W/m 2 °C is the heat transfer of external structural elements, accepted according to SNiP.

R0 = 1/8.7 + d/0.042+1/23 = 0.158 + d/0.042.

Insulation thickness

Thickness thermal insulation material is determined based on the fact that R 0 = R 0 tr, while R 0 tr is taken under energy saving conditions, thus:

2.909 = 0.158 + d/0.042, whence d = 0.116 m.

We select the brand of sandwich panels from the catalog with the optimal thickness of the thermal insulation material: DP 120, while the total thickness of the panel should be 120 mm. Thermal engineering calculations of the building as a whole are carried out in a similar way.

The need to perform a calculation

Designed based on thermotechnical calculation If done correctly, enclosing structures can reduce heating costs, the cost of which regularly increases. In addition, saving heat is considered an important environmental task, because it is directly related to reducing fuel consumption, which leads to a reduction in environmental impact. negative factors on the environment.

In addition, it is worth remembering that improperly performed thermal insulation can lead to waterlogging of structures, which will result in the formation of mold on the surface of the walls. The formation of mold, in turn, will lead to spoilage interior decoration(peeling of wallpaper and paint, destruction of the plaster layer). In particularly advanced cases, radical intervention may be necessary.

Often construction companies in their activities they strive to use modern technologies and materials. Only a specialist can understand the need to use a particular material, both separately and in combination with others. It is the thermal engineering calculation that will help determine the most optimal solutions that will ensure the durability of structural elements and minimal financial costs.

If you are planning to build
small brick cottage, then you will certainly have questions: “Which
thickness should the wall be?”, “Do you need insulation?”, “Which side should you put it on?”
insulation? etc. and so on.

In this article we will try in
understand this and answer all your questions.

Thermal calculation
enclosing structure is needed, first of all, in order to find out which
thickness should be your exterior wall.

First, you need to decide how much
floors will be in your building and depending on this the calculation is made
enclosing structures according to bearing capacity(not in this article).

According to this calculation we determine
the number of bricks in your building's masonry.

For example, it turned out 2 clay
bricks without voids, brick length 250 mm,
mortar thickness 10 mm, total 510 mm (brick density 0.67
It will be useful to us later). You decided to cover the outer surface
facing tiles, thickness 1 cm (be sure to find out when purchasing
density), and the inner surface is ordinary plaster, layer thickness 1.5
cm, also do not forget to find out its density. A total of 535mm.

In order for the building not to
collapsed, this is certainly enough, but unfortunately in most cities
Russian winters are cold and therefore such walls will freeze. And so as not
The walls were frozen, we needed another layer of insulation.

The thickness of the insulation layer is calculated
in the following way:

1. You need to download SNiP on the Internet
II 3-79* —
“Construction Heat Engineering” and SNiP 23-01-99 - “Construction Climatology”.

2. Open SNiP construction
climatology and find your city in table 1*, and look at the value at the intersection
column “Air temperature of the coldest five-day period, °C, security
0.98" and lines with your city. For the city of Penza, for example, t n = -32 o C.

3. Estimated indoor air temperature
take

t in = 20 o C.

Heat transfer coefficient for internal wallsa in = 8.7 W/m 2˚С

Heat transfer coefficient for external walls in winter conditionsa n = 23W/m2·˚С

Standard temperature difference between internal temperature
air and the temperature of the inner surface of the enclosing structuresΔ tn = 4 o C.

4. Next
We determine the required heat transfer resistance using the formula #G0 (1a) from building heating engineering
GSOP = (t in - t from.trans.) z from.trans. , GSOP=(20+4.5)·207=507.15 (for the city
Penza).

Using formula (1) we calculate:

(where sigma is the direct thickness
material, and lambda density. Itook it as insulation
polyurethane foampanels with a density of 0.025)

We take the insulation thickness to be 0.054 m.

Hence the wall thickness will be:

d = d 1 + d 2 + d 3 + d 4 =

0,01+0,51+0,054+0,015=0,589
m.

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A long time ago, buildings and structures were built without thinking about what thermal conductivity qualities the enclosing structures had. In other words, the walls were simply made thick. And if you have ever happened to be in old merchant houses, then you might have noticed that the outer walls of these houses are made of ceramic bricks, the thickness of which is about 1.5 meters. This thickness brick wall provided and still provides a completely comfortable stay for people in these houses, even in the most severe frosts.

Nowadays everything has changed. And now it is not economically profitable to make the walls so thick. Therefore, materials have been invented that can reduce it. Some of them: insulation and gas silicate blocks. Thanks to these materials, for example, the thickness of brickwork can be reduced to 250 mm.

Now walls and ceilings are most often made of 2 or 3 layers, one layer of which is a material with good thermal insulation properties. And in order to determine the optimal thickness of this material, a thermal engineering calculation is carried out and the dew point is determined.

You can find out how to calculate the dew point on the next page. Thermal engineering calculations will also be considered here using an example.

Required regulatory documents

For the calculation, you will need two SNiPs, one joint venture, one GOST and one manual:

  • SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings." Updated edition from 2012.
  • SNiP 23-01-99* (SP 131.13330.2012). "Building climatology". Updated edition from 2012.
  • SP 23-101-2004. "Design of thermal protection of buildings".
  • GOST 30494-96 (replaced by GOST 30494-2011 since 2011). "Residential and public buildings. Indoor microclimate parameters".
  • Benefit. E.G. Malyavin "Heat loss of a building. Reference manual".

Calculated parameters

In the process of performing thermal engineering calculations, the following is determined:

  • thermal characteristics building materials enclosing structures;
  • reduced heat transfer resistance;
  • compliance of this reduced resistance with the standard value.

Example. Thermal engineering calculation of a three-layer wall without an air gap

Initial data

1. Local climate and indoor microclimate

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

The calculated relative humidity of the internal air under the condition of no condensation on the internal surfaces of external fences is equal to - 55% (SNiP 23-02-2003 clause 4.3. Table 1 for normal humidity conditions).

The optimal air temperature in a living room during the cold season is t int = 20°C (GOST 30494-96 table 1).

Estimated outside air temperature t ext, determined by the temperature of the coldest five-day period with a probability of 0.92 = -31°C (SNiP 23-01-99 table 1 column 5);

The duration of the heating period with an average daily outside air temperature of 8°C is equal to z ht = 215 days (SNiP 23-01-99 table 1 column 11);

Average outside air temperature for the heating period t ht = -4.1°C (SNiP 23-01-99 table 1 column 12).

2. Wall design

The wall consists of the following layers:

  • Decorative brick (besser) 90 mm thick;
  • insulation (mineral wool board), in the figure its thickness is indicated by the sign “X”, since it will be found during the calculation process;
  • sand-lime brick 250 mm thick;
  • plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but it exists.

3. Thermophysical characteristics of materials

The values ​​of the material characteristics are summarized in the table.


Note (*): These characteristics can also be found from manufacturers of thermal insulation materials.

Calculation

4. Determination of insulation thickness

To calculate the thickness of the thermal insulation layer, it is necessary to determine the heat transfer resistance of the enclosing structure based on the requirements of sanitary standards and energy saving.

4.1. Determination of the thermal protection standard based on energy saving conditions

Determination of degree-days of the heating period according to clause 5.3 of SNiP 02/23/2003:

D d = ( t int - t ht) z ht = (20 + 4.1)215 = 5182°C×day

Note: degree days are also designated GSOP.

The standard value of the reduced heat transfer resistance should be taken no less than the standardized values ​​determined according to SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:

R req = a×D d + b = 0.00035 × 5182 + 1.4 = 3.214m2 × °C/W,

where: Dd is the degree-day of the heating period in Nizhny Novgorod,

a and b - coefficients accepted according to table 4 (if SNiP 23-02-2003) or according to table 3 (if SP 50.13330.2012) for the walls of a residential building (column 3).

4.1. Determination of thermal protection standards based on sanitation conditions

In our case, it is considered as an example, since this indicator is calculated for industrial buildings with excess sensible heat of more than 23 W/m3 and buildings intended for seasonal operation (autumn or spring), as well as buildings with an estimated internal air temperature of 12 °C and below is the heat transfer resistance of enclosing structures (with the exception of translucent ones).

Determination of the standard (maximum permissible) resistance to heat transfer according to sanitation conditions (formula 3 SNiP 02/23/2003):

where: n = 1 - coefficient adopted according to Table 6 for the outer wall;

t int = 20°С - value from the original data;

t ext = -31°С - value from the original data;

Δt n = 4°С - the normalized temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure, taken according to Table 5 in this case for the external walls of residential buildings;

α int = 8.7 W/(m 2 ×°C) - heat transfer coefficient of the internal surface of the enclosing structure, taken according to Table 7 for external walls.

4.3. Thermal protection standard

From the above calculations, for the required heat transfer resistance we select R req from the energy saving condition and now denote it R tr0 = 3.214 m 2 × °C/W .

5. Determination of insulation thickness

For each layer of a given wall, it is necessary to calculate the thermal resistance using the formula:

where: δi - layer thickness, mm;

λ i is the calculated thermal conductivity coefficient of the layer material W/(m × °C).

1 layer (decorative brick): R 1 = 0.09/0.96 = 0.094 m2 × °C/W .

Layer 3 (sand-lime brick): R 3 = 0.25/0.87 = 0.287 m2 × °C/W .

4th layer (plaster): R 4 = 0.02/0.87 = 0.023 m2 × °C/W .

Determination of the minimum permissible (required) thermal resistance of a heat-insulating material (formula 5.6 by E.G. Malyavin “Heat loss of a building. Reference manual”):

where: R int = 1/α int = 1/8.7 - heat transfer resistance on the inner surface;

R ext = 1/α ext = 1/23 - heat transfer resistance on the outer surface, α ext is taken according to table 14 for external walls;

ΣR i = 0.094 + 0.287 + 0.023 - the sum of the thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the thermal conductivity coefficients of materials adopted in column A or B (columns 8 and 9 of table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 °C /W

The thickness of the insulation is equal to (formula 5.7):

where: λ ut - coefficient of thermal conductivity of the insulation material, W/(m °C).

Determination of the thermal resistance of the wall from the condition that the total thickness of the insulation will be 250 mm (formula 5.8):

where: ΣR t,i is the sum of the thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 °C/W.

From the obtained result we can conclude that

R 0 = 3.503 m 2 × °C/W> R tr0 = 3.214m 2 × °C/W→ therefore, the thickness of the insulation is selected Right.

Effect of air gap

In the case when three-layer masonry is used as insulation mineral wool, glass wool or other slab insulation, it is necessary to install a ventilated air layer between the outer masonry and the insulation. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to dry the insulation, which becomes wet from condensation.

This air gap is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:

a) the layers of the structure located between the air gap and the outer surface (in our case, this is decorative brick (besser)) are not taken into account in the thermal engineering calculation;

b) on the surface of the structure facing the layer ventilated by outside air, the heat transfer coefficient α ext = 10.8 W/(m°C) should be taken.

Note: influence air gap taken into account, for example, in the thermal calculation of plastic double-glazed windows.

Creating comfortable living conditions or labor activity is the primary task of construction. A significant part of the territory of our country is located in northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always important. With rising energy tariffs, reducing energy consumption for heating comes to the fore.

Climatic characteristics

The choice of wall and roof design depends primarily on the climatic conditions of the construction area. To determine them, you need to refer to SP131.13330.2012 “Building climatology”. The following quantities are used in the calculations:

  • the temperature of the coldest five-day period with a probability of 0.92 is designated Tn;
  • average temperature, designated Thot;
  • duration, denoted by ZOT.

Using the example for Murmansk, the values ​​have the following values:

  • Tn=-30 degrees;
  • Tot=-3.4 degrees;
  • ZOT=275 days.

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

To perform a thermal engineering calculation of enclosing structures, first calculate the GSOP value (degree-day of the heating period):
GSOP = (Tv - Tot) x ZOT.
In our example, GSOP = (20 - (-3.4)) x 275 = 6435.

Basic indicators

For the right choice materials of enclosing structures, it is necessary to determine what thermal characteristics they must have. 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/(m x deg.). The ability of a 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.

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

Heat transfer resistance is the main indicator of the external structure. Its value must exceed the standard value. When performing thermal engineering calculations of the building envelope, we must determine the economically justified composition of the walls and roof.

Thermal conductivity values

The quality of thermal insulation is determined primarily by thermal conductivity. Each certified material undergoes laboratory tests, as a result of which this value is determined for operating conditions “A” or “B”. For our country, most regions correspond to operating conditions “B”. When performing thermal engineering calculations of the building envelope, this value should be used. Thermal conductivity values ​​are indicated on the label or in the material passport, but if they are not available, you can use reference values ​​from the Code of Practice. Values ​​for the most popular materials are given below:

  • Masonry made of ordinary brick - 0.81 W (m x deg.).
  • Sand-lime brickwork - 0.87 W (m x deg.).
  • Gas and foam concrete (density 800) - 0.37 W (m x deg.).
  • Wood coniferous species- 0.18 W (m x deg.).
  • Extruded polystyrene foam - 0.032 W (m x deg.).
  • Mineral wool slabs (density 180) - 0.048 W (m x deg.).

Standard value of heat transfer resistance

The calculated value of heat transfer resistance should not be less than the base value. The basic value is determined according to Table 3 SP50.13330.2012 “buildings”. The table defines the coefficients for calculating the basic values ​​of heat transfer resistance of all enclosing structures and types of buildings. Continuing the started thermal engineering calculation of enclosing structures, an example of the calculation can be presented as follows:

  • Rsten = 0.00035x6435 + 1.4 = 3.65 (m x deg/W).
  • Rpokr = 0.0005x6435 + 2.2 = 5.41 (m x deg/W).
  • Rcherd = 0.00045x6435 + 1.9 = 4.79 (m x deg/W).
  • Rokna = 0.00005x6435 + 0.3 = x deg/W).

Thermal engineering calculations of the external enclosing structure are performed for all structures that close the “warm” circuit - the floor on the ground or the ceiling of a technical underground, external walls (including windows and doors), a combined covering or the ceiling of an unheated attic. Also, the calculation must be performed for internal structures if the temperature difference in adjacent rooms is more than 8 degrees.

Thermal calculation of walls

Most walls and ceilings are multi-layered and heterogeneous in their design. Thermal engineering calculation of enclosing structures of a multilayer structure is as follows:
R= d1/l1 +d2/l2 +dn/ln,
where n are the parameters of the nth layer.

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

  • outer layer of plaster 3 cm thick, thermal conductivity 0.93 W (m x deg.);
  • masonry made of solid clay brick 64 cm, thermal conductivity 0.81 W (m x deg.);
  • the inner layer of plaster is 3 cm thick, thermal conductivity 0.93 W (m x deg.).

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

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

The obtained value is significantly less than the previously determined base value of the heat transfer resistance of the walls of a residential building in Murmansk 3.65 (m x deg/W). The wall does not meet regulatory requirements and needs insulation. To insulate the wall we use a thickness of 150 mm and a thermal conductivity of 0.048 W (m x deg.).

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

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

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

The calculation of floors and combined coverings is carried out similarly.

Thermal engineering calculation of floors in contact with the ground

Often in private homes or public buildings they are carried out on the ground. The heat transfer resistance of such floors is not standardized, but at a minimum the design of the floors should not allow dew to occur. The 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 border. There are up to three such zones; the remaining area belongs to the fourth zone. If the floor design does not provide effective insulation, then the heat transfer resistance of the zones is assumed to be as follows:

  • 1 zone - 2.1 (m x deg/W);
  • Zone 2 - 4.3 (m x deg/W);
  • Zone 3 - 8.6 (m x deg/W);
  • Zone 4 - 14.3 (m x deg/W).

It is easy to notice that the further the floor area is from the external wall, the higher its resistance to heat transfer. Therefore, they are often limited to insulating the perimeter of the floor. In this case, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone.
The calculation of the heat transfer resistance of the floor must be included in the general thermal engineering calculation of the enclosing structures. We will consider an example of calculating floors on the ground below. Let's take a floor area of ​​10 x 10 equal to 100 square meters.

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

Average value of resistance to heat transfer of the floor over the ground:
Rpol = 100 / (64/2.1 + 32/4.3 + 4/8.6) = 2.6 (m x deg/W).

Having insulated the perimeter of the floor with an expanded polystyrene board 5 cm thick, a strip 1 meter wide, we obtain the average value of heat transfer resistance:

Rpol = 100 / (32/2.1 + 32/(2.1+0.05/0.032) + 32/4.3 + 4/8.6) = 4.09 (m x deg/W).

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

Thermal calculation of doors

The basic value of heat transfer resistance is calculated slightly differently entrance doors. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary and hygienic criterion (no dew):
Rst = (Tv - Tn)/(DTn x av).

Here DTn is the temperature difference between the inner surface of the wall and the air temperature in the room, determined according to the Code of Rules and for housing is 4.0.
ab is the heat transfer coefficient of the inner surface of the wall, according to SP is 8.7.
The basic value of doors is taken equal to 0.6xРst.

For the selected door design, it is necessary to perform a verification thermal engineering calculation of the enclosing structures. An example of calculating an entrance door:

Rdv = 0.6 x (20-(-30))/(4 x 8.7) = 0.86 (m x deg/W).

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

Comprehensive Requirements

Calculations of walls, floors or coverings are performed to verify the element-by-element requirements of the standards. The set of rules also establishes a comprehensive requirement characterizing the quality of insulation of all enclosing structures as a whole. This value is called “specific thermal protection characteristic”. Not a single thermal engineering calculation of enclosing structures can be done without checking it. An example of calculation for a joint venture is given below.

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 with dimensions of 10 x 10 x 2.5 m. Heat transfer resistances are equal to the basic values.

The normalized value is determined in accordance with the SP depending on the heated volume of the house.

In addition to the complex requirement, to draw up an energy passport, a thermal engineering calculation of the enclosing structures is also performed; an example of how to prepare a passport is given in the appendix to SP50.13330.2012.

Uniformity coefficient

All the above calculations are applicable for homogeneous structures. Which in practice is quite rare. To take into account inhomogeneities that reduce heat transfer resistance, a correction factor for thermal homogeneity - r - is introduced. It takes into account the change in heat transfer resistance introduced by window and door openings, external corners, heterogeneous inclusions (for example, lintels, beams, reinforcing belts), etc.

The calculation of this coefficient is quite complicated, so in a simplified form you can use approximate values ​​from reference literature. For example, for brickwork - 0.9, three-layer panels - 0.7.

Effective insulation

When choosing a home insulation system, it is easy to make sure that you can meet modern thermal protection requirements without using effective insulation almost impossible. So, if you use traditional clay bricks, you will need masonry several meters thick, which is not economically feasible. At the same time, the low thermal conductivity of modern insulation based on polystyrene foam or stone wool allows you to limit yourself to thicknesses of 10-20 cm.

For example, to achieve a basic heat transfer resistance value of 3.65 (m x deg/W), you will need:

  • brick wall 3 m thick;
  • masonry made of foam concrete blocks 1.4 m;
  • mineral wool insulation 0.18 m.