Wooden houses will certainly never lose their relevance and will not go away from the peak of popularity. The warm, pleasant structure of high-quality wood, beneficial for human health, cannot be compared with either stone or mortars, and especially not with any polymers. Nevertheless, the thermal insulation qualities of wood, although quite high, are still not enough to ensure the most comfortable microclimate in the house, and it is necessary to resort to additional insulation of the walls.
Insulating wooden walls is a very delicate matter, since it is necessary to ensure that the thermal insulation layer is sufficient, but not excessive. In addition, much depends on the type of external and internal decoration of the walls, if any is provided. In a word, it is impossible to do without thermal engineering calculations. And in this matter, a calculator for calculating wall insulation should serve well wooden house.
Everyone associates home with comfort, warmth and coziness. Heat in the house is created using quality system heating, but insulation of a house or apartment remains an important factor, because often, especially in houses old building, the state of wall insulation leaves much to be desired or is absent altogether.
For insulation there is specialized material– insulation, which is mounted on external walls, on ceilings or floors.
This is usually not done indoors (on the inside of the walls). This is due to many factors, including the unprofitability of this activity.
An important indicator remains the thickness of the heat-insulating material itself, which is specially calculated for the required heating volumes, area and temperature outside the window.
Why is it so important to calculate correctly?
IN modern world thermal insulation is necessary not only for greater comfort, but also for savings. The cost of heating is constantly rising, which is hitting the pocket harder and harder, and the task of insulation is also to save money by retaining heat.
Properly selected thickness of both wall, floor or ceiling insulation allows you to reduce costs communal payments several times.
In winter, heat is retained indoors much longer, and in summer, on the contrary, it retains excess heat from the street.
Many people think that the thicker the slab, the warmer insulating material– the more savings. But this is far from true: in summer it will be cooler, and in winter it will be much hotter, but the wall structure may be subject to deformation and destruction. A smaller thickness can lead to an additional increase in energy consumption.
Insulation of the house structure (ceiling, walls, floor) is a necessary part during repair or construction (both in a residential building and in buildings intended for people to work). Selection quality materials for thermal insulation – important point in this matter, but much more important is the competent selection of the thickness of the material. Factors such as the durability of the structure and specifications during direct operation of the building.
There must be air ducts between the first and second floors, and a chimney at the top.
If you compare the thermal conductivity of different raw materials, you can see that a mineral wool slab conducts it better than a structure made of expanded clay concrete blocks.
Why is thermal insulation needed?
Many people do not fully understand how the thickness of insulation affects the durability and technical characteristics of a structure. In simple terms, thermal insulation allows you to save on costs utilities , because heat loss is reduced by almost a third, and in some cases by half.
It remains important by-effect thermal insulation, which is sound insulation. This is especially important for apartment buildings in urban areas, where sounds from the street can cause unnecessary discomfort. Panel houses also have extremely low sound insulation.
If we are talking about personal construction with your own hands, for example, your own mansion or a country home, then thermal insulation materials make it possible to reduce construction costs by replacing materials for building walls.
So, using thick polystyrene or mineral wool slabs (within 10 cm wide), it is possible to replace brick walls with them. The load on these walls must be small, so this method is suitable for one-story buildings, building verandas or houses for guests.
Requirements for thermal insulation materials
Eat a large number of requirements for thermal insulation materials, which are allocated depending on the operational load for a new building, weather conditions, material capabilities, etc.
One of the main and important characteristics of thermal insulation is considered technical feasibility conduct and retain heat. It depends on the various factors, such as: the structure and porosity of the material, its density, as well as the level of moisture and humidity absorption.
Based on thermal conductivity, there are three classes of thermal conductivity:
- A– low thermal conductivity and heat saving (0.06 W/sq. m);
- B– average thermal conductivity and heat saving (0.06 – 0.115 W/sq. m);
- IN– high thermal conductivity and heat saving (0.115 – 0.175 W/sq. m).
To guarantee high-quality thermal insulation of the facade (end), be it a high-rise building or a private small mansion, the thermal insulation must be quite durable and strong in order to be able to withstand the weight of the final finish.
As a result, it is necessary to carefully select the material, based on what the wall will be covered with at the stage of external finishing. Tiles, for example, weigh quite a lot, so a solid base is needed, but wallpaper (and also cork covering) will adhere perfectly in almost all cases, but applying such a coating outdoors is highly not recommended.
In addition to the fact that thermal insulation must be as vapor-tight as possible, it should not absorb moisture. This material should not ignite or burn, and also support combustion (should die out after ignition), release harmful and toxic substances, and should not be subject to deformation due to temperature changes.
Insulation methods
Reducing heat loss depends on the correct selection of material, as well as its location on the building. There are several methods for insulating walls, which differ in their properties, having both advantages and disadvantages.
There are the following methods for wall insulation:
- Wall. Is common brick partition with SniPov thickness from 40 cm.
- Multilayer insulation. It consists of wall cladding on both sides. This is done only at the time of construction of the structure, otherwise it will be necessary to dismantle part of the wall.
- External insulation. The most common method is done by insulating outside walls, after which a layer is applied finishing. One of the disadvantages of this method is the need for additional hydro- and vapor barrier.
What are the dimensions of the material?
If thermal insulation material very thin, cold and damp seep through the wall, but excessive thickness is also useless.
The following are considered standard material dimensions:
- 75 mm;
- 150 mm;
- 60 mm;
- 200 mm;
- 70 mm;
- 80 mm;
- 50 mm;
- 15 mm.
If the layer of thermal insulation material is at least a couple of centimeters less than required, the walls will begin to let in the cold and become damp.
For example, the dew point, which is located outside the structure, will shift slightly inside the wall due to the fact that the thermal insulation material will not be able to hold it. As a result, condensation will begin to appear on the wall plane, it will slowly become damp, collapse, and mold and mildew will appear.
A very thick layer of thermal insulation will lead to unjustified costs. Any good owner wants to build not just high-quality and reliable home, but also save as much as possible, and a thick layer of insulation costs good money. Also, with a large thickness of thermal insulation, the natural ventilation from inside the walls, as a result of which the inside of the building becomes very stuffy and uncomfortable. In addition, if insulation is carried out on the inside of the wall, a thick layer of material will take up a very large amount of free space, reducing the square footage of the room both visually and physically.
That is why it is important to be able to calculate the thickness of thermal insulation.
Another very significant point is that the determination of the thickness of the heat insulator depends directly on the raw material from which the wall is made. Based on this information, we can conclude about thermal conductivity and thermal properties this part of the structure. Such data makes it possible to qualify heat transfer at any square meter area. An absolute list of these materials is specified in SNiP No2-3-79. The density of insulation varies, but is usually used from 0.6 – 1000 kg/m3.
IN modern construction Foam blocks are often used, which are subject to certain thermal insulation requirements:
- GSOP – 6000;
- resistance in heat transfer and thermal transfer of walls - over 3.5 C/sq. m/W;
- resistance in heat transfer and thermal transfer of ceilings - over 6C/sq. m/W.
If you intend to lay a number of layers of thermal insulation, the heat transfer resistance characteristics are calculated as the sum of all layers. In this case, it is necessary to take into account the thermal conductivity and properties of the material from which the walls are made.
Calculation charts and calculators
In order to perform a thermal engineering calculation of a heat insulator, it is necessary to take into account several points that are quite difficult to understand for an inexperienced builder. The most necessary indicator is the characteristics of the wall and climatic features areas where construction is taking place, as well as their relationship. Once you have decided on the technology for performing the work and have chosen required material, you should start making calculations.
Necessary advice: to insulate the first floor in a private or apartment building It is recommended to choose the same material from the same manufacturer from the same batch.
It is imperative to insulate pipelines and other highways on the street side that lead inside the home. These are some of the most potentially dangerous places the occurrence of huge local heat loss and penetration of cold through them (up to 30% of the heat is lost).
When you have decided on the technology for performing the work and selected the appropriate material, you can begin to make calculations.
Comfortable living in the house involves creating conditions for maintaining optimal temperature air especially in winter. When building a house, it is very important to choose the right insulation and calculate its thickness. Any construction material whether it is brick, concrete or foam block, it has its own thermal conductivity and thermal resistance. Thermal conductivity refers to the ability of a building material to conduct heat. This value is determined in laboratory conditions, and the obtained data is provided by the manufacturer on the packaging or in special tables. Thermal resistance is the reciprocal value of thermal conductivity. The material that conducts heat well, therefore, has low heat resistance.
For the construction and insulation of a house, a material with low thermal conductivity and high resistance is chosen. To determine the thermal resistance of a building material, it is enough to know its thickness and thermal conductivity coefficient.
Calculation of wall insulation thickness
Let's imagine that the house has walls made of foam concrete with a density of 300 (0.3 m), the thermal conductivity coefficient of the material is 0.29. Divide 0.3 by 0.29 and get 1.03.
How to calculate the thickness of insulation for walls to ensure comfortable living in the house? To do this, you need to know the minimum value of thermal resistance in the city or region where the building to be insulated is located. Next, you need to subtract the resulting 1.03 from this value, as a result you will know the heat resistance that the insulation should have.
If the walls consist of several materials, their thermal resistance values should be summed up.
The thickness of the wall insulation is calculated taking into account the heat transfer resistance of the material used (R). To find this parameter, you should apply the standards of “Thermal protection of buildings” SP50.13330.2012. The value of GOSP (degree days heating season) is calculated by the formula:
In this case, t B reflects the temperature inside the room. According to established standards it should vary between +20-22°C. Average air temperature – t from, number of days of the heating period in a calendar year – z from. These values are given in the “Construction Climatology” SNiP 23-01-99. Special attention should be given to the duration and air temperature in the period when the average daily t≤ 8 0 C.
After the thermal resistance has been determined, you should find out what the thickness of the insulation of the ceiling, walls, floor, and roof of the house should be.
Each material of the “multilayer cake” structure has its own thermal resistance R and is calculated using the formula:
RTR = R 1 + R 2 + R 3 … R n,
Where n refers to the number of layers, and the thermal resistance of a certain material is equal to the ratio of its thickness (δ s) to thermal conductivity (λ S).
R = δS/λS
Thickness of wall insulation made of aerated concrete and brick
For example, in the construction of the structure, aerated concrete D600 with a thickness of 30 cm is used, basalt wool with a density of 80-125 kg/m 3 acts as thermal insulation, and hollow brick with a density of 1000 kg/m 3, 12 cm thick, is used as a finishing layer. The thermal conductivity coefficients given above materials are indicated in the certificates, they can also be seen in SP50.13330.2012 in Appendix C. So the thermal conductivity of concrete was 0.26 W/m* 0 C, insulation - 0.045 W/m* 0 C, brick - 0.52 W/m* 0 C. We determine R for each of the materials used.
Knowing the thickness of aerated concrete, we find its thermal resistance R Г = δ SH /λ SH = 0.3/0.26 = 1.15 m 2 * 0 C/W, the thermal resistance of brick - R К = δ SК /λ SК = 0.12/ 0.52 = 0.23 m 2 * 0 C/V. Knowing that the wall consists of 3 layers
R TP = R G + R U + R K,
find the thermal resistance of the insulation
R U = R TR - R G - R K.
Let's imagine that construction takes place in a region where RTR (22 0 C) is 3.45 m 2 * 0 C/W. We calculate R У = 3.45 - 1.15 – 0.23 = 2.07 m 2 * 0 C/W.
Now we know what resistance basalt wool should have. The thickness of the insulation for walls will be determined by the formula:
δ S = R Y x λ SU = 2.07 x 0.045 = 0.09 m or 9 cm.
If we imagine that RTR (18 0 C) = 3.15 m 2 * 0 C/W, then R У = 1.77 m 2 * 0 C/W, and δ S = 0.08 m or 8 cm.
Roof insulation thickness
This parameter is calculated by analogy with determining the thickness of the insulation of the walls of a house. For thermal insulation of attic rooms, it is better to use a material with a thermal conductivity of 0.04 W/m°C. For attics, the thickness of the peat insulating layer is not very important.
Most often, highly effective rolled, mat or slab thermal insulation is used to insulate roof slopes, and backfill materials are used for attic roofs.
The thickness of the insulation for the ceiling is calculated using the above algorithm. The temperature in the house depends on how well the parameters of the insulating material are determined. winter time. Experienced builders advise increasing the thickness of the roof insulation to 50% of the design thickness. If loose or crushable materials are used, they must be loosened from time to time.
Insulation thickness in a frame house
Glass wool, stone wool, ecowool, and bulk materials can act as thermal insulation. Calculation of insulation thickness in frame house simpler, because its design provides for the presence of insulation itself and external and external upholstery, usually made of plywood and practically not affecting the degree of thermal protection.
For example, the inner part of the wall is plywood 6 mm thick, the outer part is OSB board 9 mm thick, stone wool acts as insulation. The construction of the house is taking place in Moscow.
The average thermal resistance of the walls of a house in Moscow and the region should be R = 3.20 m 2 * 0 C/W. The thermal conductivity of insulation is presented in special tables or in the product certificate. For stone wool it is λ ut = 0.045 W/m* 0 C.
Insulation thickness for frame house determined by the formula:
δ ut = R x λ ut = 3.20 x 0.045 = 0.14 m.
Stone wool slabs are available in thicknesses of 10 cm and 5 cm. in this case It will be necessary to lay mineral wool in two layers.
Thickness of floor insulation on the ground
Before you begin calculations, you should know at what depth the floor of the room is located relative to ground level. You should also have an idea of the average ground temperature in winter at this depth. The data can be taken from the table.
First you need to determine the GSOP, then calculate the heat transfer resistance, determine the thickness of the floor layers (for example, reinforced concrete, cement strainer on insulation, flooring). Next, we determine the resistance of each layer by dividing the thickness by the thermal conductivity coefficient and summing the resulting values. Thus, we will find out the thermal resistance of all layers of the floor, except for the insulation. To find this indicator, from the standard thermal resistance we subtract the total thermal resistance of the floor layers, with the exception of the thermal conductivity coefficient of the insulating material. The thickness of the floor insulation is calculated by multiplying the minimum thermal resistance of the insulation by the thermal conductivity coefficient of the selected insulating material.
Everyone who builds own house, wants it to be warm. This can be achieved in several ways: build thick walls, make good insulation or heat the house well.
In practice, all these methods are used together, but from an economic point of view, insulating the house, or rather increasing the thickness of the insulation, has a higher priority.
Our calculation will consist of two main stages:
- Finding the heat transfer resistance of the walls, which is necessary for further calculations.
- Selection of the required insulation thickness depending on the design and material of the walls.
First, let's look at short video, in which the expert explains in detail why you need to install insulation in external walls brick house and what type of insulation to use.
Heat transfer resistance of walls
To find this parameter we use SP 50.13330.2012 " Thermal protection buildings" which can be downloaded on our website (link).
In paragraph 5 “Thermal protection of buildings” several formulas are presented that will help us calculate the thickness of the insulation and walls. In order to do this, there is a parameter called heat transfer resistance and denoted by the letter R. It depends on the required indoor temperature and the climatic conditions of a given city or area.
In general, it is calculated by the formula R TP = a x GSOP + b.
According to table 3, the values of coefficients a and b for the walls of residential buildings are 0.00035 and 1.4, respectively.
All that remains is to find the value of the GSOP. It stands for degree-day of the heating period. You'll have to tinker a little with this value.
Formula for calculation GSOP = (t B —t OT) xz FROM.
In this formula, t B is the temperature that should be inside the room. According to the norms, it is 20-22 0 C.
The values of the parameters t OT and z OT mean the average outside air temperature and the number of days of the heating period in a year. You can find them in SNiP 23-01-99 “Building climatology”. (link).
If you look at this SNiP, you will see a large table at the very beginning, where climatic parameters are given for each city or region.
We will be interested in the column that says “Duration and average air temperature of the period with an average daily air temperature ≤ 8 0 C.”
Example of calculation of parameter RTR
In order to make everything more clear, let's calculate the heat transfer resistance of the walls (RTR) for a house built in Kazan.
For this we have two formulas:
RTR = a x GSOP + b,
GSOP = (t B -t OT) x z OT
First, let's calculate the GSOP. To do this, we are looking for the city of Kazan in the right column of SNiP 01/23/99.
We find from the table that the average temperature t OT = - 5.2 0 C, and the duration z OT = 215 days/year.
Now you need to decide what indoor air temperature is comfortable for you. As was written above, t B = 20-22 0 C is considered optimal. If you like cooler or warmer temperatures, then when calculating the GPS for the value of t B may be different.
So, let’s calculate the GSOP for temperatures t B = 18 0 C and t B = 22 0 C.
GSOP 18 = (18 0 C-(-5.2 0 C) x 215 days/year = 4988.
GSOP 22 = (22 0 С-(-5.2 0 С) x 215 days/year = 5848
Now let's find the resistance to heat transfer. As we already know, the coefficients a and b for the walls of residential buildings, according to Table 3 from SP 50.13330.2012, are equal to 0.00035 and 1.4.
RTR (18 0 C) = 0.00035 x 4988 + 1.4 = 3.15 m 2 * 0 C/W, for 18 0 C indoors.
RTR (22 0 C) = 0.00035 x 5848 + 1.4 = 3.45 m 2 * 0 C/W, for 22 0 C.
The wall together with the insulation must have such resistance in order for there to be minimal heat loss in the house.
So, we have received the necessary initial data. Now let's move on to the second stage, to determining the thickness of the insulation.
Calculation of insulation thickness
Each material included in a multilayer wall pie has its own thermal resistance R. So, our task is to ensure that the sum of all the resistances of the materials included in the wall structure is equal to the thermal resistance R TR, which we calculated in the previous chapter, i.e. e.:
RTR = R 1 + R 2 + R 3 ...Rn, where n is the number of layers.
The thermal resistance of an individual material, R, is equal to the ratio of layer thickness (δ s) to thermal conductivity (λ S).
R = δS/λS
Examples of calculating the thickness of insulation for walls made of brick and aerated concrete
Example 1. The wall is made of aerated concrete blocks D600 30 cm thick, insulated on the outside with stone wool with a density of 80-125 kg/m 3 , and lined on the outside with ceramic hollow bricks with a density of 1000 kg/m 3 . Construction was carried out in Kazan.
To further determine the thickness of the insulation, we will need the values of the thermal conductivity of the materials λ S. This data must be present in the certificate for the materials.
If for some reason they are not there, then you can look at them in Appendix C to SP 50.13330.2012, which we used earlier.
λ SH = 0.14 W/m* 0 C - thermal conductivity of aerated concrete;
λ SК = 0.52 W/m* 0 C – thermal conductivity of brick.
R Г = δ SG /λ SG = 0.3/0.14 = 2.14 m 2 * 0 C/W - thermal resistance of aerated concrete;
R K = δ SK /λ SK = 0.12/0.52 = 0.23 m 2 * 0 C/B - thermal resistance of the brick.
Because our wall consists of three layers, then the equation will be correct:
R TP = R G + R U + R K,
then R Y = R TP - R G - R K
In the previous chapter we found the value of RTR (22 0 C) for the city of Kazan. We use it for our calculations.
R У = 3.45 - 2.14 - 0.23 = 1.08 m 2 * 0 C/W.
Thus, we found what thermal resistance the insulation should have. To find insulation thickness Let's use the formula:
δ S = R Y x λ SU = 1.08 x 0.045 = 0.05 m.
We found that for the given conditions, insulation with a thickness of 5 cm is sufficient.
If we take the value R TP (18 0 C) = 3.15 m 2 * 0 C/W, we get:
R У = 3.15 - 2.14 - 0.23 = 0.78 m 2 * 0 C/W.
δ S = R У x λ SU = 0.78 x 0.045 = 0.035 m
As you can see, the thickness of the insulation has changed by only one and a half centimeters.
Example 2. Let's consider an example when, instead of aerated concrete blocks, sand-lime brick with a density of 1800 kg/m 3 is laid. The thickness of the masonry is 38 cm.
By analogy with previous calculations, we find the thermal conductivity values from the table:
λ SК1 = 0.87 W/m* 0 C - thermal conductivity sand-lime brick density 1800 kg/m3;
λ SU = 0.045 W/m* 0 C – thermal conductivity of insulation;
λ SК2 = 0.52 W/m* 0 C – thermal conductivity of brick with a density of 1000 kg/m3.
R K1 = δ SK1 /λ SK1 = 0.38/0.87 = 0.44 m 2 * 0 C/W - thermal resistance of brick 1800 kg/m 3;
R K2 = δ SK2 /λ SK2 = 0.12/0.52 = 0.23 m2 * 0 C/V - thermal resistance of brick 1000 kg/m3.
We find the thermal resistance of the insulation:
R У = 3.45 – 0.44 – 0.23 = 2.78 m 2 * 0 C/W.
Now we calculate the thickness of the insulation:
δ S = R Y x λ SU = 2.78 x 0.045 = 0.12 m.
Those. For these conditions, an insulation thickness of 12 cm is sufficient.
Example 3. As clear example speaking about the importance of insulation, let’s consider a wall consisting only of D600 aerated concrete.
Knowing the thermal conductivity of aerated concrete blocks, λ SH = 0.14 W/m* 0 C, we can immediately calculate the required wall thickness because the wall is uniform.
δ S = R TP x λ SH = 3.45 x 0.14 = 0.5 m
We receive, in order to comply with all SNiP standards, we must lay out a wall 0.5 m thick.
In this case, you can go in two ways: make the wall immediately the required thickness or build a thinner wall and additionally insulate it.
The first option seems to us more reliable and less expensive, because there is no work on installing insulation. The second option is more suitable for already built houses.
All these examples show how the thickness of insulation depends on the material of the walls. By analogy with them, you can make calculations for any type of material.
Video “Wall insulation”
In conclusion, we suggest you watch a couple of videos that will be useful when choosing the thickness of insulation for the walls of a house built from foam concrete and aerated concrete.
A warm house is the dream of every owner; to achieve this goal, thick walls are built, heating is installed, and high-quality thermal insulation is installed. In order for insulation to be rational, it is necessary to choose the right material and correctly calculate its thickness.
The size of the insulation layer depends on the thermal resistance of the material. This indicator is the reciprocal of thermal conductivity. Each material - wood, metal, brick, foam plastic or mineral wool - has a certain ability to transmit thermal energy. The thermal conductivity coefficient is calculated during laboratory tests, and is indicated on the packaging for consumers.
If the material is purchased without labeling, you can find a summary table of indicators on the Internet.
The thermal resistance of a material ® is a constant value; it is defined as the ratio of the temperature difference at the edges of the insulation to the force of the heat flow passing through the material. Formula for calculating the coefficient: R=d/k, where d is the thickness of the material, k is the thermal conductivity. The higher the value obtained, the more effective the thermal insulation.
Why is it important to correctly calculate insulation indicators?
Thermal insulation is installed to reduce energy loss through the walls, floor and roof of a home. Insufficient insulation thickness will cause the dew point to move inside the building. This means the appearance of condensation, dampness and fungus on the walls of the house. An excess layer of thermal insulation does not significantly change temperature indicators, but requires significant financial costs, and is therefore irrational. This disrupts air circulation and natural ventilation between the rooms of the house and the atmosphere. To save money while ensuring optimal conditions residence requires an accurate calculation of the thickness of the insulation.
Calculation of the thermal insulation layer: formulas and examples
To be able to accurately calculate the amount of insulation, it is necessary to find the heat transfer resistance coefficient of all materials in a wall or other area of the house. It depends on the climatic indicators of the area, therefore it is calculated individually using the formula:
GSOP=(tv-tot)xzot
tв - indoor temperature indicator, usually 18-22ºC;
tot - average temperature value;
zot - duration heating season, day.
Values for calculation can be found in SNiP 01/23/99.
When calculating the thermal resistance of a structure, it is necessary to add up the indicators of each layer: R = R1 + R2 + R3, etc. Based on the average indicators for partial and multi-storey buildings Approximate coefficient values have been determined:
- walls - at least 3.5;
- ceiling - from 6.
The thickness of the insulation depends on the building material and its size; the lower the thermal resistance of the wall or roof, the larger the insulation layer should be.
Example: a wall made of sand-lime brick 0.5 m thick, which is insulated with foam plastic.
Rst.=0.5/0.7=0.71 - thermal resistance of the wall
R- Rst.=3.5-0.71=2.79 - value for foam plastic
For foam plastic, thermal conductivity k=0.038
d=2.79×0.038=0.10 m - foam boards 10 cm thick will be required
Using this algorithm, it is easy to calculate the optimal amount of thermal insulation for all areas of the house except the floor. When making calculations regarding base insulation, you must refer to the soil temperature table in your region of residence. It is from this that the data is taken to calculate the GSOP, and then the resistance of each layer and the required value of insulation are calculated.
Popular ways to insulate a home
Thermal insulation of a building can be done during the construction stage or after its completion. Among the popular methods:
- Monolithic wall of significant thickness (at least 40 cm) made of ceramic bricks or wood.
- The construction of enclosing structures by well masonry is the creation of a cavity for insulation between two parts of the wall.
- Installation of external thermal insulation in the form of a multilayer structure made of insulation, lathing, moisture-proof film and decorative finishing.
Using ready-made formulas, you can calculate the optimal thickness of insulation without the help of a specialist. When calculating, the number should be rounded up; a small margin of the thermal insulation layer will be useful for temporary temperature drops below the average.
