Heat engineering calculations. Thermal engineering calculation of the outer wall (calculation example)

13.06.2019

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

At present, 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: heaters and gas silicate blocks. Thanks to these materials, for example, the thickness brickwork can be reduced up 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 calculation is carried out and the dew point is determined.

How the calculation is made to determine the dew point, you can find on the next page. Here, the heat engineering calculation will be considered using an example.

Required regulatory documents

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

SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition from 2012.
SNiP 23-01-99* (SP 131.13330.2012). "Construction 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 the building. Reference guide".

Calculated parameters

In the process of performing a heat engineering calculation, the following are 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. The climate of the area and the microclimate of the room

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

The calculated relative humidity of the indoor air from the condition of no condensation on the inner surfaces of the outer fences is - 55% (SNiP 23-02-2003 p.4.3. Table 1 for normal humidity conditions).

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

Estimated outdoor temperature text, determined by the temperature of the coldest five-day period with a security of 0.92 = -31 ° С (SNiP 23-01-99 table 1 column 5);

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

The average outdoor temperature during the heating period t ht = -4.1 ° C (SNiP 23-01-99 table. 1 column 12).

2. Wall construction

The wall consists of the following layers:

Brick decorative (besser) 90 mm thick;
insulation (mineral wool board), in the figure its thickness is indicated by the sign "X", since it will be found in the calculation process;
silicate brick 250 mm thick;
plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but there is.

3. Thermophysical characteristics of materials

The values of the characteristics of the materials are summarized in the table.

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

Payment

4. Determining the thickness of the insulation

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

4.1. Determination of the norm of thermal protection according to the condition of energy saving

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

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

Note: also degree-days have the designation - GSOP.

The normative value of the reduced resistance to heat transfer should be taken not less than the normalized values determined by SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:

R req \u003d a × D d + b \u003d 0.00035 × 5182 + 1.4 \u003d 3.214m 2 × °С/W,

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

a and b - coefficients taken 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 the norm of thermal protection according to the condition of sanitation

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 operation(in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° C and below the reduced heat transfer resistance of enclosing structures (with the exception of translucent ones).

Determination of the normative (maximum allowable) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):

where: n \u003d 1 - coefficient taken from table 6 for outer wall;

t int = 20°C - value from the initial data;

t ext \u003d -31 ° С - value from the initial data;

Δt n \u003d 4 ° С - normalized temperature difference between the temperature of the indoor air and the temperature of the inner surface of the building envelope, is taken according to table 5 in this case for the outer walls of residential buildings;

α int \u003d 8.7 W / (m 2 × ° С) - heat transfer coefficient of the inner surface of the building envelope, taken according to table 7 for external walls.

4.3. Thermal protection rate

From the above calculations for the required heat transfer resistance, we choose R req from the condition of energy saving and denote it now R tr0 \u003d 3.214 m 2 × °С/W .

5. Determining the thickness of the insulation

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 - calculated coefficient of thermal conductivity of the layer material W/(m × °С).

1 layer (decorative brick): R 1 = 0.09 / 0.96 = 0.094 m 2 × °С/W .

3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .

4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .

Determination of the minimum allowable (required) thermal resistance thermal insulation material(formula 5.6 E.G. Malyavin "Heat loss of the building. Reference manual"):

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

R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer 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 thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the coefficients of thermal conductivity of materials taken 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 ° С /W

The thickness of the insulation is (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 - the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ·°С / W.

From the result obtained, it can be concluded that

R 0 \u003d 3.503m 2 × °С/W> R tr0 = 3.214m 2 × °С/W→ therefore, the thickness of the insulation is selected right.

Influence of the air gap

In the case when mineral wool, glass wool or other slab insulation is used as a heater in a three-layer masonry, it is necessary to install an air ventilated 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 drain the insulation, which gets wet from condensate.

This air layer 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) structural layers located between the air gap and outer surface(in our case, this is a decorative brick (besser)), they are not taken into account in the heat engineering calculation;

b) on the surface of the structure facing the layer ventilated by the 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 heat engineering calculation of plastic double-glazed windows.

A long time ago, buildings and structures were built without thinking about what heat-conducting qualities the enclosing structures have. In other words, the walls were simply made thick. And if you 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. Such a thickness of the brick wall provided and still provides quite a comfortable stay for people in these houses even in the most severe frosts.

How the calculation is made to determine the dew point, you can find on the next page. Here, the heat engineering calculation will be considered using an example.

Required regulatory documents

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

SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition from 2012.
SNiP 23-01-99* (SP 131.13330.2012). "Construction 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 the building. Reference guide".

Calculated parameters

In the process of performing a heat engineering calculation, the following are determined:

thermal characteristics of building materials of 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. The climate of the area and the microclimate of the room

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

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

Estimated outdoor temperature text, determined by the temperature of the coldest five-day period with a security of 0.92 = -31 ° С (SNiP 23-01-99 table 1 column 5);

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

The average outdoor temperature during the heating period t ht = -4.1 ° C (SNiP 23-01-99 table. 1 column 12).

2. Wall construction

The wall consists of the following layers:

Brick decorative (besser) 90 mm thick;
insulation (mineral wool board), in the figure its thickness is indicated by the sign "X", since it will be found in the calculation process;
silicate brick 250 mm thick;
plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but there is.

3. Thermophysical characteristics of materials

The values of the characteristics of the materials are summarized in the table.

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

Payment

4. Determining the thickness of the insulation

To calculate the thickness of the heat-insulating 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 norm of thermal protection according to the condition of energy saving

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

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

Note: also degree-days have the designation - GSOP.

R req \u003d a × D d + b \u003d 0.00035 × 5182 + 1.4 \u003d 3.214m 2 × °С/W,

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

a and b - coefficients taken 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 the norm of thermal protection according to the condition of sanitation

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 operation (in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° С and below the given resistance to heat transfer of enclosing structures (with the exception of translucent ones).

Determination of the normative (maximum allowable) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):

where: n \u003d 1 - coefficient adopted according to table 6 for the outer wall;

t int = 20°C - value from the initial data;

t ext \u003d -31 ° С - value from the initial data;

α int \u003d 8.7 W / (m 2 × ° С) - heat transfer coefficient of the inner surface of the building envelope, taken according to table 7 for external walls.

4.3. Thermal protection rate

From the above calculations for the required heat transfer resistance, we choose R req from the condition of energy saving and denote it now R tr0 \u003d 3.214 m 2 × °С/W .

5. Determining the thickness of the insulation

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 - calculated coefficient of thermal conductivity of the layer material W/(m × °С).

1 layer (decorative brick): R 1 = 0.09 / 0.96 = 0.094 m 2 × °С/W .

3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .

4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .

Determination of the minimum allowable (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 - resistance to heat transfer on the inner surface;

R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;

The thickness of the insulation is (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 - the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ·°С / W.

From the result obtained, it can be concluded that

R 0 \u003d 3.503m 2 × °С/W> R tr0 = 3.214m 2 × °С/W→ therefore, the thickness of the insulation is selected right.

Influence of the air gap

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

b) on the surface of the structure facing the layer ventilated by the 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 heat engineering calculation of plastic double-glazed windows.

When and determining the need additional insulation at home, it is important to know the heat loss of its structures, in particular. An online wall thermal conductivity calculator will help you make calculations quickly and accurately.

In contact with

Why do you need a calculation

The thermal conductivity of this element of the building is the property of the structure to conduct heat through a unit of its area with a temperature difference between inside and outside the room of 1 deg. WITH.

The heat engineering calculation of enclosing structures performed by the service mentioned above is necessary for the following purposes:

to select heating equipment and the type of system that allows not only to compensate for heat loss, but also to create a comfortable temperature inside residential premises;
to determine the need for additional insulation of the building;
when designing and constructing a new building to select a wall material that provides the least heat loss in certain climatic conditions;
to create indoor comfortable temperature not only during the heating period, but also in summer in hot weather.

Attention! Performing independent thermotechnical calculations wall structures, use the methods and data described in such regulatory documents as SNiP II 03 79 "Construction Heat Engineering" and SNiP 23-02-2003 "Thermal Protection of Buildings".

What does thermal conductivity depend on?

Heat transfer depends on factors such as:

The material from which the building is built various materials differ in their ability to conduct heat. Yes, concrete different kinds bricks contribute to a large loss of heat. On the contrary, galvanized logs, beams, foam and gas blocks, with a smaller thickness, have lower thermal conductivity, which ensures the preservation of heat inside the room and much lower costs for insulation and heating of the building.
Wall thickness - the larger this value, the less heat transfer occurs through its thickness.
Humidity of the material - the greater the moisture content of the raw material from which the structure is erected, the more it conducts heat and the faster it collapses.
The presence of air pores in the material - air-filled pores prevent accelerated heat loss. If these pores are filled with moisture, heat loss increases.
The presence of additional insulation - lined with a layer of insulation outside or inside the wall in terms of heat loss, have values \u200b\u200bmany times less than non-insulated ones.

In construction, along with the thermal conductivity of walls, such a characteristic as thermal resistance (R) has become widespread. It is calculated taking into account the following indicators:

coefficient of thermal conductivity of the wall material (λ) (W/m×0С);
construction thickness (h), (m);
the presence of a heater;
moisture content of the material (%).

The lower the thermal resistance value, the more the wall is subject to heat loss.

The thermotechnical calculation of enclosing structures according to this characteristic is performed according to the following formula:

R=h/λ; (m2×0С/W)

Example of thermal resistance calculation:

Initial data:

the load-bearing wall is made of dry pine timber 30 cm (0.3 m) thick;
thermal conductivity coefficient is 0.09 W/m×0С;
result calculation.

Thus, the thermal resistance of such a wall will be:

R=0.3/0.09=3.3 m2×0С/W

The values obtained as a result of the calculation are compared with the normative ones in accordance with SNiP II 03 79. At the same time, such an indicator as the degree-day of the period during which the heating season continues is taken into account.

If the obtained value is equal to or greater than the standard value, then the material and thickness of the wall structures are selected correctly. Otherwise, the building should be insulated to achieve the standard value.

In the presence of a heater, its thermal resistance is calculated separately and summarized with the same value of the main wall material. Also, if the material of the wall structure has high humidity, apply the appropriate coefficient of thermal conductivity.

For a more accurate calculation of the thermal resistance of this design, similar values \u200b\u200bof windows and doors facing the street are added to the result obtained.

Valid values

When performing a heat engineering calculation of the outer wall, the region in which the house will be located is also taken into account:

For southern regions with warm winters and small temperature differences, walls of small thickness can be erected from materials with an average degree of thermal conductivity - ceramic and clay fired single and double, and of high density. The thickness of the walls for such regions can be no more than 20 cm.
At the same time, for the northern regions, it is more expedient and cost-effective to build enclosing wall structures of medium and large thickness from materials with high thermal resistance - logs, gas and foam concrete of medium density. For such conditions, wall structures up to 50–60 cm thick are erected.
For regions with temperate climate and alternating temperature regime in winter they are suitable with high and medium thermal resistance - gas and foam concrete, timber, medium diameter. Under such conditions, the thickness of wall enclosing structures, taking into account heaters, is no more than 40–45 cm.

Important! The thermal resistance of wall structures is most accurately calculated by the heat loss calculator, which takes into account the region where the house is located.

Heat transfer of various materials

One of the main factors affecting the thermal conductivity of the wall is the building material from which it is built. This dependence is explained by its structure. So, materials with a low density have the lowest thermal conductivity, in which the particles are arranged quite loosely and there is a large number of pores and voids filled with air. These include various types of wood, light porous concrete - foam, gas, slag concrete, as well as hollow silicate bricks.

Materials with high thermal conductivity and low thermal resistance include various types of heavy concrete, monolithic silicate brick. This feature is explained by the fact that the particles in them are located very close to each other, without voids and pores. This contributes to faster heat transfer in the thickness of the wall and a large heat loss.

Table. Thermal conductivity coefficients of building materials (SNiP II 03 79)

Calculation of a sandwich structure

The thermotechnical calculation of the outer wall, consisting of several layers, is carried out as follows:

according to the formula described above, the value of thermal resistance of each of the layers of the "wall cake" is calculated;
the values of this characteristic of all layers are added together, obtaining the total thermal resistance of the wall multilayer structure.

Based on this technique, it is possible to calculate the thickness. To do this, it is necessary to multiply the thermal resistance missing to the norm by the thermal conductivity coefficient of the insulation - as a result, the thickness of the insulation layer will be obtained.

With the help of the TeReMOK program, the thermotechnical calculation is performed automatically. In order for the wall thermal conductivity calculator to perform calculations, it is necessary to enter the following initial data into it:

type of building - residential, industrial;
wall material;
construction thickness;
region;
required temperature and humidity inside the building;
presence, type and thickness of insulation.

Useful video: how to independently calculate the heat loss in the house

Thus, the thermotechnical calculation of enclosing structures is very important both for a house under construction and for a building that has already been built for a long time. In the first case, the correct heat calculation will save on heating, in the second case, it will help to choose the insulation that is optimal in terms of thickness and composition.

In modern conditions, people are increasingly thinking about the rational use of resources. Electricity, water, materials. To save all this in the world came for a long time and everyone understands how to do it. But the main amount in the bills for payment is heating, and not everyone understands how to reduce the expense for this item.

What is thermal engineering calculation?

Thermal engineering calculation is performed in order to select the thickness and material of the building envelope and bring the building in line with thermal protection standards. The main regulatory document regulating the ability of a structure to resist heat transfer is SNiP 23-02-2003 "Thermal protection of buildings".

The main indicator of the enclosing surface in terms of thermal protection was the reduced resistance to heat transfer. This is a value that takes into account the heat-shielding characteristics of all layers of the structure, taking into account cold bridges.

A detailed and competent heat engineering calculation is quite laborious. When building private houses, the owners try to take into account the strength characteristics of materials, often forgetting about heat conservation. This can lead to rather disastrous consequences.

Why is the calculation performed?

Before starting construction, the customer can choose whether he will take into account the thermal characteristics or ensure only the strength and stability of the structures.

The cost of insulation will definitely increase the estimate for the construction of the building, but will reduce the cost of further operation. Individual houses are built for decades, perhaps they will serve the next generations. During this time, the cost of effective insulation pay off several times over.

What does the owner get correct execution calculations:

Savings on space heating. Heat loss buildings are reduced, respectively, the number of radiator sections will decrease with a classical heating system and the power of the system warm floors. Depending on the heating method, the owner's costs for electricity, gas or hot water become smaller;
Savings on repairs. At proper insulation a comfortable microclimate is created in the room, condensation does not form on the walls, and microorganisms dangerous to humans do not appear. The presence of a fungus or mold on the surface requires repair, and a simple cosmetic one will not bring any results and the problem will arise again;
Security for residents. Here, as well as in the previous paragraph, we are talking about dampness, mold and fungus, which can cause various diseases in people who are constantly in the room;
respect for environment. There is a shortage of resources on the planet, so reducing the consumption of electricity or blue fuel has a positive effect on the ecological situation.

Normative documents for performing the calculation

The reduced resistance and its compliance with the normalized value is the main goal of the calculation. But for its implementation, you will need to know the thermal conductivity of the materials of the wall, roof or ceiling. Thermal conductivity is a value that characterizes the ability of a product to conduct heat through itself. The lower it is, the better.

During the calculation of heat engineering, they rely on the following documents:

SP 50.13330.2012 "Thermal protection of buildings". The document was reissued on the basis of SNiP 23-02-2003. The main standard for calculation;
SP 131.13330.2012 "Construction climatology". New edition of SNiP 23-01-99*. This document allows you to define climatic conditions the locality in which the object is located;
SP 23-101-2004 "Design of thermal protection of buildings" in more detail than the first document in the list, reveals the topic;
GOST 30494-96 (replaced by GOST 30494-2011 since 2011) Residential and public buildings;
Manual for students of construction universities E.G. Malyavin “Heat loss of the building. Reference manual".

Thermal engineering calculation is not complicated. It can be performed by a person without special education according to the template. The main thing is to approach the issue very carefully.

An example of calculating a three-layer wall without an air gap

Let's take a closer look at an example of a heat engineering calculation. First you need to decide on the source data. As a rule, you choose the materials for the construction of the walls yourself. We will calculate the thickness of the insulation layer based on the materials of the wall.

Initial data

The data is individual for each construction object and depends on the location of the object.

1. Climate and microclimate

Construction area: Vologda.
Purpose of the object: residential.
Relative air humidity for a room with a normal humidity regime is 55% (item 4.3. Table 1).
The temperature inside the residential premises tint is set by regulatory documents (Table 1) and is equal to 20 degrees Celsius.

text is the estimated outside air temperature. It is set by the temperature of the coldest five days of the year. The value can be found in, table 1, column 5. For a given area, the value is -32ᵒС.

zht = 231 days - the number of days in the period when additional space heating is needed, that is, the average daily temperature outside is less than 8ᵒС. The value is looked up in the same table as the previous one, but in column 11.

tht = -4.1ᵒС – average outside air temperature during the heating period. The value is in column 12.

2. Wall materials

All layers should be taken into account (even a layer of plaster, if any). This will allow the most accurate calculation of the design.

In this embodiment, consider a wall consisting of the following materials:

a layer of plaster, 2 centimeters;
an inner verst made of ordinary solid ceramic brick with a thickness of 38 centimeters;
a layer of Rockwool mineral wool insulation, the thickness of which is selected by calculation;
outer verst of front ceramic brick, 12 centimeters thick.

3. Thermal conductivity of adopted materials

All properties of materials must be presented in the passport from the manufacturer. Many companies represent full information about products on their websites. The characteristics of the selected materials are summarized in a table for convenience.

Calculation of the thickness of the insulation for the wall

1. Energy saving condition

Calculation of the value of degree-days of the heating period (GSOP) is carried out according to the formula:

Dd = (tint - tht) zht.

All letter designations presented in the formula are deciphered in the source data.

Dd \u003d (20-(-4.1)) * 231 \u003d 5567.1 ᵒС * day.

The normative resistance to heat transfer is found by the formula:

The coefficients a and b are taken according to table 4, column 3.

For initial data a=0.00045, b=1.9.

Rreq = 0.00045*5567.1+1.9=3.348 m2*ᵒС/W.

2. Calculation of the norm of thermal protection based on the conditions of sanitation

This indicator is not calculated for residential buildings and is given as an example. The calculation is carried out with an excess of sensible heat exceeding 23 W / m3, or the operation of the building in spring and autumn. Also, calculations are required at a design temperature of less than 12ᵒС indoors. Formula 3 is used:

The coefficient n is taken according to table 6 of the SP "Thermal protection of buildings", αint according to table 7, Δtn according to the fifth table.

Rreq = 1*(20+31)4*8.7 = 1.47 m2*ᵒС/W.

Of the two values obtained in the first and second paragraph, the largest is selected, and further calculation is carried out on it. In this case, Rreq = 3.348 m2*ᵒС/W.

3. Determination of the thickness of the insulation

The heat transfer resistance for each layer is obtained by the formula:

where δ is the layer thickness, λ is its thermal conductivity.

a) plaster R pcs \u003d 0.02 / 0.87 \u003d 0.023 m2 * ᵒС / W;
b) ordinary brick R row.brick. \u003d 0.38 / 0.48 \u003d 0.79 m2 * ᵒС / W;
c) facing brick Rut = 0.12 / 0.48 = 0.25 m2 * ᵒС / W.

The minimum heat transfer resistance of the entire structure is determined by the formula (, formula 5.6):

Rint = 1/αint = 1/8.7 = 0.115 m2*ᵒС/W;
Rext = 1/αext = 1/23 = 0.043 m2*ᵒС/W;
∑Ri = 0.023+0.79+0.25 = 1.063 m2*ᵒC/W, i.e. the sum of the numbers obtained in point 3;

R_tr ^ ut \u003d 3.348 - (0.115 + 0.043 + 1.063) \u003d 2.127 m2 * ᵒС / W.

The thickness of the insulation is determined by the formula (formula 5.7):

δ_tr^ut \u003d 0.038 * 2.127 \u003d 0.081 m.

The value found is the minimum. The insulation layer is taken not less than this value. In this calculation, we finally accept the thickness of the mineral wool insulation as 10 centimeters, so that you do not have to cut the purchased material.

For calculations of heat losses of the building, which are performed for design heating systems, it is necessary to find the actual value of the resistance to heat transfer with the found thickness of the insulation.

Rо = Rint+Rext+∑Ri = 1/8.7 + 1/23 + 0.023 + 0.79 + 0.1/0.038 + 0.25 = 3.85 m2*ᵒС/W > 3.348 m2*ᵒС/W.

The condition is met.

Influence of the air gap on the heat-shielding characteristics

When constructing a wall protected slab insulation a ventilated layer is possible. It allows you to remove condensate from the material and prevent it from getting wet. The minimum thickness of the gap is 1 centimeter. This space is not closed and has direct communication with the outside air.

In the presence of an air-ventilated layer, the calculation takes into account only those layers that are located before it from the side of warm air. For example, a wall pie consists of plaster, internal masonry, insulation, an air gap and external masonry. Only plaster, internal masonry and insulation are taken into account. The outer layer of masonry goes after the ventilation gap, therefore it is not taken into account. In this case, the outer masonry performs only an aesthetic function and protects the insulation from external influences.

Important: when considering structures where the airspace is closed, it is taken into account in the calculation. For example, in the case of window fillings. The air between the panes plays the role of an effective insulation.

Teremok program

To perform the calculation using a personal computer, specialists often use the program for thermal calculation "Teremok". It exists online and as an application for operating systems.

The program performs calculations based on all necessary normative documents. Working with the application is extremely simple. It allows you to work in two modes:

calculation of the required layer of insulation;
verification of an already thought-out design.

The database contains all the necessary characteristics for the settlements of our country, you just need to select the one you need. It is also necessary to choose the type of construction: external wall, mansard roof, ceiling over a cold basement or attic.

When you press the continue button, a new window appears that allows you to "assemble" the structure. Many materials are available in the program memory. They are divided into three groups for ease of search: structural, heat-insulating and heat-insulating-structural. You only need to set the layer thickness, the program will indicate the thermal conductivity itself.

Without necessary materials you can add them yourself, knowing the thermal conductivity.

Before making calculations, you must select the type of calculation above the plate with the wall structure. Depending on this, the program will give either the thickness of the insulation, or report on the compliance of the enclosing structure with the standards. After the calculations are completed, you can generate a report in text format.

"Teremok" is very convenient to use and even a person without a technical education is able to deal with it. For specialists, it significantly reduces the time for calculations and preparation of a report in electronic form.

The main advantage of the program is the fact that it is able to calculate the thickness of the insulation not only of the outer wall, but of any structure. Each of the calculations has its own characteristics, and it is quite difficult for a non-professional to understand all of them. To build a private house, it is enough to master this application, and you do not have to delve into all the difficulties. Calculation and verification of all enclosing surfaces will take no more than 10 minutes.

Thermal engineering calculation online (calculator overview)

Thermal engineering calculation can be done on the Internet online. Not bad, as in my opinion, is the service: rascheta.net. Let's take a quick look at how to work with it.

By going to the site online calculator, the first step is to choose the standards by which the calculation will be made. I choose the 2012 rulebook as it is a newer document.

Next, you need to specify the region in which the object will be built. If your city is not available, choose the nearest one. Big city. After that, we indicate the type of buildings and premises. Most likely you will calculate a residential building, but you can choose public, administrative, industrial and others. And the last thing you need to choose is the type of enclosing structure (walls, ceilings, coatings).

We leave the calculated average temperature, relative humidity and thermal uniformity coefficient the same if you do not know how to change them.

In the calculation options, set all two checkboxes except the first one.

In the table, we indicate the wall cake starting from the outside - we select the material and its thickness. On this, in fact, the whole calculation is completed. Below the table is the result of the calculation. If any of the conditions is not met, we change the thickness of the material or the material itself until the data complies with regulatory documents.

If you want to see the calculation algorithm, then click on the "Report" button at the bottom of the site page.

Now, in times of ever-rising energy prices, high-quality insulation has become one of the priorities in the construction of new and repair of already built houses. The cost of work associated with improving the energy efficiency of the house almost always pays off within a few years. The main thing during their implementation is not to make mistakes that will nullify all efforts at best, and at worst, they will also harm.

The modern building materials market is simply littered with all kinds of heaters. Unfortunately, the manufacturers, or rather, the sellers are doing everything so that we, ordinary developers, choose their material and give our money to them. And this leads to the fact that in various sources of information (especially on the Internet) there are many erroneous and misleading recommendations and advice. Get tangled up in them common man pretty easy.

In fairness, it must be said that modern heaters are really quite effective. But in order to use their properties one hundred percent, firstly, the installation must be correct, corresponding to the manufacturer's instructions, and, secondly, the use of insulation must always be appropriate and expedient in each specific case. So how do you do the right thing effective insulation Houses? Let's try to understand this issue in more detail ...

home insulation mistakes

There are three main mistakes that developers most often make:

incorrect selection of materials and their sequence for the "pie" of the building envelope (walls, floors, roofs ...);
inappropriate, chosen "at random" thickness of the insulation layer;
not correct installation with non-compliance with the technology for each specific type of insulation.

The consequences of these mistakes can be very sad. This is the deterioration of the microclimate in the house with an increase in humidity and constant fogging of windows in the cold season, and the appearance of condensate in places where it is not permissible, and the appearance of an unpleasantly smelling fungus with gradual decay of the interior decoration or building envelope.

The choice of insulation method

The most important rule to follow at all times is: insulate the house from the outside, not from the inside! The meaning of this important recommendation clearly shown in the following figure:

The blue-red line in the figure shows the change in temperature in the thickness of the "pie" of the wall. It clearly shows that if the insulation is made from the inside, in the cold season the wall will freeze through.

Here is an example of such a case, by the way, based on very real events. lives good man in a corner apartment of a multi-storey building panel house and in winter, especially in windy weather, it freezes. Then he decides to insulate the cold wall. And since his apartment is on the fifth floor, it’s impossible to think of anything better than to insulate from the inside. At the same time, one Saturday afternoon, he watches a program on TV about repairs and sees how, in a similar apartment, the walls are also insulated from the inside with the help of mats from mineral wool.

And everything there seemed to be shown correctly and beautifully: they put up a frame, laid a heater, covered it with a vapor barrier film and sheathed it with drywall. But they just didn’t explain that they used mineral wool, not because it is the most suitable material for wall insulation from the inside, but because the sponsor of today's release is a major manufacturer of mineral wool insulation.

And so our good man decides to repeat it. It does everything the same as on TV, and the apartment immediately becomes noticeably warmer. Only his joy from this does not last long. After a while, he begins to feel that some foreign smell has appeared in the room and the air seems to have become heavier. And a few days later, dark damp spots began to appear on the drywall at the bottom of the wall. It's good that the wallpaper did not have time to paste. So what happened?

And what happened was panel wall, closed from internal heat layer of insulation, quickly froze. Water vapor, which is contained in the air and, due to the difference in partial pressures, always tends from the inside of a warm room to the outside, began to enter the insulation, despite the vapor barrier, through poorly glued or not glued joints at all, through holes from stapler brackets and drywall fastening screws. Upon contact of vapors with a frozen wall, condensation began to fall on it. The insulation began to dampen and accumulate more and more moisture, which led to an unpleasant musty smell and the appearance of a fungus. In addition, wet mineral wool quickly loses its heat-saving properties.

The question arises - what then should a person do in this situation? Well, for starters, you still need to try to find an opportunity to make insulation from the outside. Fortunately, now there are more and more organizations involved in such work, regardless of height. Of course, their prices will seem very high to many - 1000 ÷ 1500 rubles per 1 m² on a turnkey basis. But this is only at first glance. If we fully calculate all the costs for internal insulation (insulation, its lining, putties, primers, new painting or new wallpaper plus wages for employees), then in the end the difference with external insulation becomes irrelevant and of course it is better to prefer it.

Another thing is if it is not possible to obtain permission for external insulation (for example, the house has some architectural features). In this extreme case, if you have already decided to insulate the walls from the inside, use heaters with minimal (almost zero) vapor permeability, such as foam glass, extruded polystyrene foam.

Foam glass is more eco-friendly material but unfortunately also more expensive. So if 1 m³ of extruded polystyrene foam costs about 5,000 rubles, then 1 m³ of foam glass costs about 25,000 rubles, i.e. five times more expensive.

Technology details internal insulation walls will be discussed in a separate article. Now we note only the moment that when installing the insulation, it is necessary to exclude the violation of its integrity to the maximum. So, for example, it is better to glue EPPS to the wall and completely abandon the dowels (as in the figure), or reduce their number to a minimum. As a finish, the insulation is covered with gypsum plaster mixtures, or also pasted over with sheets of drywall without any frames and without any self-tapping screws.

How to determine the required thickness of insulation?

With the fact that it is better to insulate a house from the outside than from the inside, we have more or less figured it out. Now the next question is how much insulation should be laid in each case? It will depend on the following parameters:

what are the climatic conditions in the region;
what is the required microclimate in the room;
what materials make up the "pie" of the building envelope.

A little about how to use it:

Calculation of insulation of the walls of the house

Let's say the "pie" of our wall consists of a layer of drywall - 10 mm ( interior decoration), gas silicate block D-600 - 300 mm, mineral wool insulation - ? mm and siding.

We enter the initial data into the program in accordance with the following screenshot:

So point by point:

1) Perform the calculation according to:- we leave a dot in front of "SP 50.13330.2012 and SP 131.13330.2012", as we see these norms are more recent.

2) Locality:- choose "Moscow" or any other that is on the list and is closer to you.

3) Type of buildings and premises- install "Residential."

4) Type of enclosing structure- select "External walls with a ventilated facade." , as our walls are sheathed on the outside with siding.

5) Estimated average temperature and relative humidity of indoor air are determined automatically, we do not touch them.

6) Coefficient of thermal homogeneity "r"- its value is selected by clicking on the question mark. We are looking for what suits us in the tables that appear. If nothing fits, we accept the value “r” from the instructions of the Moscow State Expertise (indicated at the top of the page above the tables). For our example, we took the value r=0.85 for walls with window openings.

This coefficient is not available in most similar online programs for thermal calculation. Its introduction makes the calculation more accurate, since it characterizes the heterogeneity of wall materials. For example, when calculating brickwork, this coefficient takes into account the presence of mortar joints, the thermal conductivity of which is much greater than that of the brick itself.

7) Calculation options:- check the boxes next to the items "Calculation of vapor permeability resistance" and "Calculation of dew point".

8) We enter into the table the materials that make up our “pie” of the wall. Please note - it is fundamentally important to make them in order from the outer layer to the inner.

Note: If the wall has an outer layer of material separated by a layer of ventilated air (in our example, this is siding), this layer is not included in the calculation. It is already taken into account when choosing the type of enclosing structure.

So, we entered the following materials into the table - KNAUF mineral wool insulation, gas silicate with a density of 600 kg / m³ and lime-sand plaster. In this case, the values of the coefficients of thermal conductivity (λ) and vapor permeability (μ) automatically appear.

The thicknesses of the layers of gas silicate and plaster are known to us initially, we enter them into the table in millimeters. And we select the desired thickness of the insulation until the inscription “ R 0 pr >R 0 norms (... > ...) the design meets the requirements for heat transfer.«

In our example, the condition begins to be fulfilled when the thickness of the mineral wool is 88 mm. We round this value up to 100 mm, since it is this thickness that is commercially available.

Also under the table we see inscriptions that say that moisture accumulation in the heater is impossible and condensation is not possible. This indicates the correct choice of insulation scheme and the thickness of the insulation layer.

By the way, this calculation allows us to see what was said in the first part of this article, namely, why it is better not to insulate the walls from the inside. Let's swap the layers, i.e. put a heater inside the room. See what happens in the following screenshot:

It can be seen that although the design still meets the requirements for heat transfer, the vapor permeability conditions are no longer met and condensation is possible, as indicated under the material plate. The consequences of this have been discussed above.

Another advantage of this online program is that by clicking on the " Report» at the bottom of the page, you can get the entire heat engineering calculation carried out in the form of formulas and equations with the substitution of all values. Someone might be interested in this.

Calculation of attic floor insulation

An example of a heat engineering calculation attic floor shown in the following screenshot:

This shows that in this example, the required thickness of mineral wool for attic insulation is at least 160 mm. Cover - by wooden beams, "pie" make up - insulation, pine boards 25 mm thick, fiberboard - 5 mm, air gap - 50 mm and drywall filing - 10 mm. The air gap is present in the calculation due to the presence of a frame for drywall.

Calculation of basement floor insulation

An example of a heat engineering calculation for a basement floor is shown in the following screenshot:

In this example, when the basement is a monolithic reinforced concrete 200 mm thick and the house has an unheated underground, the minimum required thickness of insulation with extruded polystyrene foam is about 120 mm.

Thus, the implementation of thermal engineering calculation allows you to correctly arrange the "pie" of the building envelope, select the required thickness of each layer and, in the end, perform effective insulation of the house. After that, the main thing is to make a high-quality and correct installation of insulation. Their choice is now very large and each has its own characteristics in working with them. This will certainly be discussed in other articles on our site devoted to the topic of home insulation.

We would love to see your comments on this topic!