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 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 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 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.

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

The optimal air temperature in the living room is cold period year t int = 20°С (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);

Duration heating season 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 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/m 3 and buildings intended for seasonal use (autumn or spring), as well as buildings with a design internal air temperature of 12 ° C and a lower 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 - normalized temperature difference between the internal air temperature and the temperature inner surface enclosing structure, is 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 m 2 × °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 thermal insulation 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 external masonry and 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: the influence of the air gap is taken into account, for example, in the thermal engineering calculations of plastic double-glazed windows.

Example of thermal engineering calculation of enclosing structures

1. Initial data

Technical task. Due to the unsatisfactory heat and humidity conditions of the building, it is necessary to insulate its walls and mansard roof. For this purpose, perform calculations of thermal resistance, heat resistance, air and vapor permeability of the building envelope, assessing the possibility of moisture condensation in the thickness of the fences. Establish the required thickness of the thermal insulation layer, the need to use wind and vapor barriers, and the order of arrangement of layers in the structure. Develop design solution, meeting the requirements of SNiP 23-02-2003 “Thermal protection of buildings” for enclosing structures. Calculations should be performed in accordance with the set of rules for design and construction SP 23-101-2004 “Design of thermal protection of buildings”.

General characteristics of the building. A two-story residential building with an attic is located in the village. Sviritsa, Leningrad region. The total area of ​​external enclosing structures is 585.4 m2; total wall area 342.5 m2; total window area 51.2 m2; roof area – 386 m2; basement height - 2.4 m.

The structural design of the building includes load-bearing walls, reinforced concrete floors made of hollow-core panels, 220 mm thick and a concrete foundation. The external walls are made of brickwork and plastered inside and outside with mortar with a layer of about 2 cm.

The roof of the building has a truss structure with a steel seam roof, made over lathing with a pitch of 250 mm. The 100 mm thick insulation is made of mineral wool slabs laid between the rafters

The building has stationary electric-thermal storage heating. The basement has a technical purpose.

Climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, the calculated average temperature of the internal air is taken equal to

t int= 20 °C.

According to SNiP 01/23/99 we accept:

1) the estimated temperature of the outside air during the cold period of the year for the conditions of the village. Sviritsa, Leningrad region

t ext= -29 °C;

2) duration of the heating period

z ht= 228 days;

3) average outside air temperature during the heating period

t ht= -2.9 °C.

Heat transfer coefficients. The values ​​of the heat transfer coefficient of the internal surface of the fences are taken as follows: for walls, floors and smooth ceilings α int= 8.7 W/(m 2 ·ºС).

The values ​​of the heat transfer coefficient of the outer surface of the fences are taken as follows: for walls and coverings α ext=23; attic floors α ext=12 W/(m 2 ·ºС);

Standardized heat transfer resistance. Degree-days of the heating season G d are determined by formula (1)

G d= 5221 °C day.

Because the value G d differs from table values, standard value R req determined by formula (2).

According to SNiP 02/23/2003, for the obtained degree-day value, the normalized heat transfer resistance is R req, m 2 °C/W, is:

For external walls 3.23;

Coverings and overlaps over driveways 4.81;

Fencing over unheated undergrounds and basements 4.25;

Windows and balcony doors 0,54.

2. Thermal engineering calculation of external walls

2.1. Resistance of external walls to heat transfer

Exterior walls made of hollow ceramic bricks and have a thickness of 510 mm. The walls are plastered on the inside with lime-cement mortar 20 mm thick, and on the outside with cement mortar of the same thickness.

The characteristics of these materials - density γ 0, thermal conductivity coefficient in the dry state  0 and vapor permeability coefficient μ - are taken according to the table. Clause 9 of the application. In this case, in the calculations we use the thermal conductivity coefficients of materials  W for operating conditions B, (for wet operating conditions), which are obtained from formula (2.5). We have:

For lime-cement mortar

γ 0 = 1700 kg/m 3,

 W=0.52(1+0.168·4)=0.87 W/(m·°С),

μ=0.098 mg/(m h Pa);

For brickwork made of hollow ceramic bricks on cement-sand mortar

γ 0 = 1400 kg/m 3,

 W=0.41(1+0.207·2)=0.58 W/(m·°С),

μ=0.16 mg/(m h Pa);

For cement mortar

γ 0 = 1800 kg/m 3,

 W=0.58(1+0.151·4)=0.93 W/(m·°С),

μ=0.09 mg/(m h Pa).

The heat transfer resistance of a wall without insulation is equal to

R o = 1/8.7 + 0.02/0.87 + 0.51/0.58 + 0.02/0.93 + 1/23 = 1.08 m 2 °C/W.

In the presence of window openings forming wall slopes, the coefficient of thermal uniformity of brick walls with a thickness of 510 mm is accepted r = 0,74.

Then the reduced heat transfer resistance of the building walls, determined by formula (2.7), is equal to

R r o =0.74 1.08 = 0.80 m 2 °C/W.

The obtained value is much lower than the standard value of heat transfer resistance, so a device is needed external thermal insulation and subsequent plastering with protective and decorative compositions plaster mortar reinforced with fiberglass mesh.

To allow the thermal insulation to dry out, the covering plaster layer must be vapor-permeable, i.e. porous with low density. We select a porous cement-perlite mortar that has the following characteristics:

γ 0 = 400 kg/m 3,

 0 = 0.09 W/(m °C),

 W=0.09(1+0.067·10)=0.15 W/(m·°С),

 = 0.53 mg/(m h Pa).

Total heat transfer resistance of added layers of thermal insulation R t and plaster lining R w should be no less

R t + R w = 3.23/0.74-1.08 = 3.28 m 2 °C/W.

Preliminarily (with subsequent clarification) we accept the thickness of the plaster lining as 10 mm, then its resistance to heat transfer is equal to

R w =0.01/0.15=0.067 m 2 °C/W.

When used for thermal insulation of mineral wool boards produced by JSC "Mineral Wool" brand Facade Butts  0 =145 kg/m 3,  0 =0.033,  W =0.045 W/(m °C) the thickness of the thermal insulation layer will be

δ=0.045·(3.28-0.067)=0.145 m.

Rockwool slabs are available in thicknesses from 40 to 160 mm in 10 mm increments. We accept a standard thermal insulation thickness of 150 mm. Thus, the slabs will be laid in one layer.

Checking compliance with energy saving requirements. The design diagram of the wall is shown in Fig. 1. Characteristics of the layers of the wall and the total resistance of the wall to heat transfer without taking into account the vapor barrier are given in table. 2.1.

Table 2.1

Characteristics of wall layers andtotal wall resistance to heat transfer

	Layer material	Density γ 0, kg/m 3	Thickness δ, m	Calculated thermal conductivity coefficient λ W, W/(m K)	Design heat transfer resistance R, m 2 °C)/W
	Interior plaster (lime-cement mortar)
	Masonry made of hollow ceramic bricks
	External plaster ( cement mortar)
	Mineral wool insulation FACADE BATTS
	Protective and decorative plaster (cement-perlite mortar)

The heat transfer resistance of the building walls after insulation will be:

R o = 1/8.7+4.32+1/23=4.48 m 2 °C/W.

Taking into account the coefficient of thermal uniformity of external walls ( r= 0.74) we obtain the reduced resistance to heat transfer

R o r= 4.48 0.74 = 3.32 m 2 °C/W.

Received value R o r= 3.32 exceeds standard R req=3.23, since the actual thickness of the heat-insulating boards is greater than the calculated one. This situation meets the first requirement of SNiP 23-02-2003 for the thermal resistance of the wall - R o ≥ R req .

Verification of compliance with requirements forsanitary, hygienic and comfortable indoor conditions. Calculated difference between the internal air temperature and the internal wall surface temperature Δ t 0 is

Δ t 0 =n(t int – t ext)/(R o r ·α int)=1.0(20+29)/(3.32·8.7)=1.7 ºС.

According to SNiP 02/23/2003, for the external walls of residential buildings, a temperature difference of no more than 4.0 ºС is allowed. Thus, the second condition (Δ t 0 ≤Δ t n) done.

P
let's check the third condition ( τ int >t grew up), i.e. Is it possible for moisture to condense on the inner surface of the wall at the design outdoor temperature? t ext= -29 °C. Inner surface temperature τ int enclosing structure (without heat-conducting inclusion) is determined by the formula

τ int = t int –Δ t 0 =20–1.7=18.3 °C.

Indoor water vapor pressure e int equal to

Creation comfortable conditions for living or working 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 comfortable temperature in buildings is always relevant. 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 values ​​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.).
• Masonry from sand-lime brick- 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 doesn't satisfy 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).

Received calculated value more than the basic one - 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 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 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 comprehensive requirement for drawing up energy passport They also carry out thermal engineering calculations of enclosing structures; an example of obtaining 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, we introduce correction factor thermal homogeneity - r. It takes into account the change in heat transfer resistance introduced by window and doorways, 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 see that it is almost impossible to meet modern thermal protection requirements without using effective insulation. So, if you use traditional clay brick, a masonry several meters thick will be required, which is not economically feasible. However, low thermal conductivity modern insulation materials based on expanded polystyrene 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.

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).

By this calculation we define
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). Outer surface You decided to cover
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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