What thickness of insulation is better to choose. How thick should the insulation be, comparison of thermal conductivity of materials

To create in winter comfort at home, must be maintained indoors optimal temperature. This is not difficult if the owner has taken care of insulation in advance.

However, simply laying insulating material not enough. For effective thermal insulation, it is necessary that the insulation layer be of a certain thickness.

At first glance, there are no difficulties here. It is enough to lay more thermal insulation - and the warmth in the house is ensured. However, any insulation has a certain weight, to which is added the weight of the structure holding it. And all this weight is fixed on the wall, creating additional load.

If the additional load exceeds the strength limits of the wall, the thermal insulation will fall off along with pieces of the wall. But even when the wall strength is sufficient, excessive thermal insulation does not lead to additional fuel savings.

In this case, heat loss during ventilation or through ventilation comes to the fore, and this is achieved with the help of thermal insulation. cannot be eliminated. But the costs of laying excess insulation material can be significant. On the other hand, reducing the thickness of thermal insulation below a certain limit is also unprofitable - heat losses and heating costs increase.

In a building materials store, you can ask the seller to calculate the required thickness and total amount of insulation. This is done using special computer programs . But we must take into account that the store employees are interested in sales maximum quantity building materials, therefore they can significantly overestimate the figures. How to find the golden mean?

The issue of thermal insulation of buildings is dealt with by applied science of thermal engineering. In accordance with its recommendations, the Code of Rules SP 50.13330.2012 was created, which is included in SNiP 02/23/2003 and regulates the thermal protection of buildings.

IN SNiP 23-01-99(Building climatology) provides initial climatological data for localities and regions Russian Federation.

These documents serve as landmarks for calculations required thickness and total quantity thermal insulation materials. Having made such calculations, the home owner receives the necessary information for purchasing and starting work.

Thermal protection of buildings according to the Code of Practice must meet such requirements:

The thermal resistance of enclosing structures should not be lower than the values specified in the document.
The specific heat-shielding characteristic of a house should not exceed the specified norm.
The temperature of the inner surface of the enclosing structures should not fall below the minimum permissible value.

Of these three parameters the most important are thermal resistance and minimum internal temperature. They will serve as key quantities in the calculations.

Thermal resistance R TP is called the reciprocal of thermal conductivity. Its dimension is m 2 °C/W. The internal temperature of wall surfaces for residential premises is normalized in the range of 20–22°C.

The initial value for calculations is degree-days of the heating season(abbreviated as GSOP). The dimension of this parameter is °C day/year. GSOP is calculated using the following formula:

GSOP=(t B –t OT)·z OT ,

where t B – internal temperature (+22°C), t OT – average temperature air outside heating season, z ot – number of days of the heating period in a year when average daily temperature no higher than +8°C.

Moscow will serve as an example. For the capital of the Russian Federation, the duration of the heating period is 214 days/year, and the average outside temperature for this period t OT = –3.1°C (see Table 1, Building Climatology). We substitute the values into the formula and get:

GSOP = [(22 – (–3.1)] · 214 = 5371.4 degree days.

We are looking for the value of heat transfer resistance corresponding to this number of degree days (see Table 3, Code of Rules). The result is a number that differs from the round table values, and the table contains only round values. For other cases a formula with coefficients a and b is provided:

R TP = a · GSOP + b

We substitute the values into it and get:

R TP = 0.00035 · 5371.4 + 1.4 = 3.27999 m² °C/W.

However, the resulting value is total thermal resistance of the wall and insulation:

R TP = R CT + R y .

In the above Code of Practice, it is recommended to consider the thermal resistance of building materials taking into account operating conditions. According to the climate humidity map (Building Climatology), Moscow is located in the zone of normal humidity. Table 2 of the Code of Rules recommends taking into account the thermal conductivity of materials for these conditions in rooms with normal humidity (most rooms) under the letter B.

Let’s assume that you need to insulate walls made of solid clay bricks with a mortar of cement and sand 0.51 m thick (two bricks). The thermal conductivity coefficient of such masonry is 0.81 W/m °C. Thermal resistance of materials is determined by the relation:

where P is the thickness of the material, m, k is the thermal conductivity coefficient, W/m °C. Substituting the values, we get:

R CT = 0.51 / 0.81 = 0.6296 m² °C/W.

Thermal resistance of thermal insulation equal to the difference between the total resistance and the wall resistance:

R y = R TP – R CT = 3.27999 – 0.6296 = 2.65039 m² °C/W.

It remains to determine the thickness the insulation itself. We will use slabs made of stone wool density 50 kg/m³. Its thermal conductivity coefficient under the specified conditions is 0.045 W/m °C. To get the thickness, multiply its thermal resistance by the thermal conductivity coefficient:

P y = R y · k = 2.65039 · 0.045 = 0.11927 m or approximately 12 cm.

This calculation is suitable for insulating walls under plaster.

Stone wool, like porous material, outside on brickwork usually laid, covering it with a vapor-permeable membrane, and then a ventilated façade is installed.

Air constantly passes through the air gap of this facade from bottom to top. At the same time, it not only carries away steam from the stone wool layer, but also leads to loss some amount of thermal energy.

For ventilated facades large sizes on multi-storey buildings Heating engineers have developed formulas for calculating these heat losses. They allow you to calculate the thickness of an additional layer of insulation to compensate for these losses. However, the calculation mechanism is very complex and requires taking into account many quantities: the speed of air flow in the layer, its height, flow inhomogeneities, etc.

Doing such complex calculations for a one-story country house doesn't make sense. Experienced specialists advise when installing a ventilated facade increase the calculated thermal insulation thickness by approximately 30%. In our example it will be:

P = P y · 1.3 = 0.11927 · 1.3 = 0.1550 m or approximately 15 cm.

That is, to insulate a house in Moscow with solid brick masonry on a mortar of cement and sand with an external wall thickness of 0.51 cm, you will need to lay three layers of slabs 50 mm thick basalt wool, and then install a ventilated façade.

Calculation roof insulation thickness

Calculation of thermal insulation thickness when laying under the roof also has its own characteristics. Under a pitched or gable roof, insulation is installed according to the same principle as on a wall with a ventilated facade.

Air penetrates under the roof from below and, passing through the air gap above the insulation, exits through the cracks under the ridge. This also creates additional heat loss, which must be taken into account when calculating the thickness of thermal insulation.

Calculate the thickness of the insulation for roofing is much easier than for walls. After all, the roof itself has practically no thermal resistance, and under the insulation on a pitched or gable roof there is no solid thick structural material. This means that only the thermal resistance of the insulation needs to be taken into account.

When calculating, we will proceed from the same value of GSOP = 5371.4 degree-days and will use the same heat transfer resistance formula RTP = a · GPS + b. However, we take the resistance values in column 5 for attic floors. The coefficients a and b are different there: a = 0.00045; b = 1.9. Substituting these values into the formula, we get:

R У = 0.00045 · 5371.4 + 1.9 = 4.3171 m² °C/W.

Insulation thickness We count in the same way as for walls:

P Y = R Y · k = 4.3171 · 0.045 = 0.19427 m or approximately 20 cm.

In other words, for insulation pitched or gable roof at home in Moscow you will need four layers of basalt wool slabs 50 mm thick.

Thickness calculation insulation materials when laying on walls, you can do it yourself, taking into account the current data building codes and rules. Calculation of thermal insulation thickness for the roof is practically no different from the calculation for the walls, but in this case it is necessary to use the thermal resistance values from another column of the table.

How to calculate the thickness of insulation for walls using a table of thermal conductivity of materials, look at the video:

For any home, a comfortable and warm atmosphere is important, which will make your stay pleasant and convenient. The correct microclimate will allow you to get rid of many troubles, including dampness, heat loss, and too high heating costs. To avoid such negative aspects, it is necessary to select the correct type and thickness of insulation.

When choosing insulation, such parameters as the region of residence, the purpose of the room, as well as the material from which the house is built are important.

Today construction market offers numerous options for insulation, which differ not only in size and thickness, but also in the type of raw materials for production, operational characteristics. When choosing a heat insulator, it is necessary not only to clarify the thickness, but also to determine for which wall material it will be optimal. You should pay attention to the climatic region and wind loads. For example, the value for the thickness of the insulation will indicate for which specific room the insulator is selected. For a living room this will be one indicator, but for an attic or basement it will be completely different.

Parameters for insulation

Insulation materials are selected based not only on thickness, but also on other indicators. What thickness to take depends on the following:

climatic region for the construction site;
main wall material;
purpose of the room, its level above the ground;
manufacturing material.

Manufacturers offer various options. Many claim that aerated concrete or expanded clay concrete is an excellent option for construction warm home, here you can save on insulation. But is this really so? It is necessary to compare thermal conductivity coefficients. In order for the thickness to be selected correctly, it is necessary to take into account that all insulation materials differ in their characteristics, their thermal conductivity indicators will be different.

As comparative data you can take:

Expanded polystyrene thermal insulators with a thermal conductivity coefficient of 0.039 W/m*°C with a thickness of 0.12 m.
Mineral wool (basalt wool, stone) with data of 0.041 W/m*°C and 0.13 m.
Iron concrete walls with data of 1.7 W/m*°C and 5.33 m.
Solid sand-lime brick with data of 0.76 W/m*°C and 2.38 m.
Hollow (holey) brick with data of 0.5 W/m*°C and 1.57 m.
Glued laminated timber with values of 0.16 W/m*°C and 0.5 m.
Expanded clay concrete (warm concrete) with values of 0.47 W/m*°C and 1.48 m.
Gas silicate blocks with data of 0.15 W/m*°C and 0.47 m.
Foam concrete blocks with a thermal conductivity coefficient of 0.3 W/m*°C at 0.94 m.
Slag concrete with data of 0.6 W/m*°C and 1.8 m.

Based on the listed data, you can see that the thickness of the wall to ensure a normal and comfortable microclimate is from one and a half meters. But that's too much. It is best to make the wall thinner, but use a layer of mineral wool or expanded polystyrene with a thickness of only 12-13 cm. This will be much more economical.

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Comparative characteristics

Today, not only comfort and savings, but also the availability of free space in the house and on the site depend on what material you choose for insulation. Too thick brick walls take up a lot of space, it can be used more efficiently.

Comparison of thermal conductivity coefficients:

Expanded polystyrene sides PSB-S-25 with a value of 0.042 W/m*°C and a required thickness of 124 mm.
Rockwool mineral wool for facade insulation: thermal conductivity coefficient - 0.046 W/m*°C, required thickness -135 mm.
Glued laminated timber from spruce or pine with indicators of 500 kg/m³ according to GOST 8486: thermal conductivity coefficient - 0.18 W/m*°C, required thickness - 530 mm.
Special warm ceramic blocks with a layer of thermal insulating adhesive: thermal conductivity coefficient -0.17 W/m*°C, required thickness - 575 mm.
Aerated concrete blocks 600 kg/m³: thermal conductivity coefficient - 0.29 W/m*°C, required thickness - 981 mm.
Sand-lime brick according to GOST 379: thermal conductivity coefficient - 0.87 W/m*°C, required thickness - 2560 mm.

According to the data presented, it is clear that mineral wool, expanded polystyrene, and ordinary timber are the leaders among other materials.

Using them as insulation makes it possible to build brick or concrete walls of smaller thickness. If the house is being built in a warm region, then 10 cm of insulation is sufficient. For colder regions, 12-13 cm is already required, but taking into account the material from which the main wall of the house is made.

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Example of insulation calculation

The choice of thickness for a heat insulator must begin with the fact that the material is selected for its intended purpose. specific premises and by temperature zone. All zones that are used for calculations can be found in special reference books. Among the frequently used 4 are:

1 zone: from 3501 degree days;
Zone 2: 3001-3501 degree days;
Zone 3: 2501-3000 degree days;
Zone 4: up to 2500 degree days.

The following calculation options can be cited as an example:

Minimum valid values for thermal resistance are represented by 4 zones of 2.8; 2.5; 2,2 and 2.
Floors, coverings for unheated, unused attics: 4.95; 4.5; 3.9; 3.3.
Cold cellars, ground floors: 3,5; 3,3; 3; 2,5.
Floors for unheated plinths and basements located at ground level: 2.8; 2.6; 2.2; 2.
Floors for basements located below ground level: 3.7; 3.45; 3; 2.7.
Balcony structures, display cases and panoramic windows, walls around them, translucent special facades, verandas, covered terraces: 0.6; 0.56; 0.55; 0.5.
Front doors for apartment buildings, hallways for large public buildings: 0,44; 0,41; 0,39; 0,32.
Entrance premises, corridors, hallways, halls for private low-rise buildings: 0,6; 0,56; 0,54; 0,45.
Entrance halls and halls for premises located above the ground floor level: 0.25; 0.25; 0.25; 0.25.

Using this indicator, you can calculate the thickness of the heat insulator of any structure. For example, the walls of a house are made of sand-lime brick 51 cm. The insulation is carried out using 10 cm thick foam plastic boards. To determine whether the planned thickness of the insulation is suitable, you simply need to calculate the coefficient for the thermal resistance of the foam plastic and the wall, after which the resulting values are added and compared with those presented above.

For walls 51 cm, the following data is obtained:

The thermal conductivity coefficient of sand-lime brick is 0.87.
The thickness of the wall 51 must be divided by 0.87 to obtain the thermal resistance of the brick equal to 0.58.
With polystyrene foam they do things differently. Its thickness is divided by the thermal conductivity coefficient of this material 0.043, the result is 2.32.
Now we need to add up the obtained values, the result is 2.88. This indicator must be compared with those given above. If the received data for external walls made of sand-lime brick coincide with those required for a specific region (climatic zone), then 10 cm of foam plastic will be quite sufficient.

It must be remembered that if the insulation is used for colder areas, then its thickness should be 12-14 cm to create comfortable living conditions in the house.

To choose the right thermal insulation material, you need to carefully determine its parameters. The influence is exerted by the climatic zone in which the house is being built, what material its walls are made of, and for what part of the structure the heat insulator is used. It is important to immediately pay attention to the features of use certain type insulation. Usually purchased mineral wool or polystyrene foam, but their characteristics are different, so it is necessary to calculate separately for each material.

Proper thermal insulation for the walls of an apartment or house consists not only in choosing a certain type of thermal insulation material, but also in calculating its thickness.

Insufficient insulation will affect not only the temperature in the room, but will also cause the dew point to transfer to the inner surface of the wall. Condensation that appears will lead to increased humidity, mold and rot on the walls.

On the other hand, excessive thermal insulation, although it eliminates these problems, is not economically profitable. Even a significant excess of the thickness of the insulation layer over the calculated one will bring only a slight increase in the thermal protection indicator of the entire structure.

Calculation of thermal insulation thickness

In construction, there is such a thing as thermal resistance - this is an indicator that determines the ability of a material or structure to resist the transfer of heat from the room to the external environment.

The thermal resistance coefficient is a constant value, derived empirically based on the climatic characteristics of the region. It is individual for each region of Russia. The data is regulated by SNIP 23-01-99 “Building climatology”. The table shows some indicators by region:

The thermal resistance of a wall consists of the resistance to heat transfer of all layers of homogeneous materials, this includes load-bearing structures and insulation.

The thickness of the insulation will be calculated using the formula:

R reg =δ/k, where
R reg – average thermal resistance for the region;
δ – thickness of the insulation layer;
k – coefficient of thermal conductivity of thermal insulation W/m 2 ×ºС.

The calculation of wall insulation must take into account the thickness and material of the load-bearing external walls to which it will be attached.

Data on the thermal conductivity coefficient of some building materials and the most common types of modern insulation materials are given in the table.

Let's calculate the minimum required thickness of the most popular polystyrene foam insulation for Yakutsk - R reg = 4.9 m 2 ×ºС/W. If the house is built of sand-lime brick in two rows.

We determine the real thermal resistance of a wall with a thickness of two bricks δ brick = 0.51 m, k = 0.81 W/m 2 ×ºС, and substitute it into the formula.

R brick = δ/k = 0.51/0.81 = 0.62 m 2 ×ºС/W

We subtract the calculated value from the constant for the Yakutsk region. The value that the polystyrene foam should cover will be obtained.

R = R reg - R brick = 4.9 – 0.62 = 4.34 m 2 ×ºС/W This is the desired indicator that needs to be covered.

δ = R foam plastic × k = 4.34 × 0.035 = 0.1519 (m),

From the calculations it is clear that for a house built in Yakutia from double silicate brick, a layer of polystyrene foam thermal insulation 152 mm thick is required. Taking into account the thickness of the air gaps inside the wall (between the walls), we take the working thickness of polystyrene foam to be 150 mm.

Insulation for walls used indoors

The main requirements, in addition to low thermal conductivity, that apply to thermal insulation materials used indoors:

small thickness of the insulating structure to save usable space;
environmentally friendly - the material should not emit any harmful substances.

Several types of insulation meet these parameters, each of which has its own installation technology.

Foil insulation

Of the entire range of foil materials, thermal insulation based on polyethylene foam is most suitable for insulating walls from the inside.

Manufacturers produce many brands: Folgoizol, Alufom, Ecofol, Armaflex, Jermaflex, Penofol, Izolon, Isoflex. Thermal insulation of the room occurs on a double principle. Infrared radiation is reflected by the aluminum layer back into the room, and foamed polyethylene with a thickness of 2 to 10 mm prevents the cold from penetrating.

Installation is carried out with the reflective side facing the room. The joints of the panels are glued with aluminum tape. Main feature devices for such insulation is the presence of a gap of 10-20 mm between the foil and inside finishing decorative materials.

Some time after installing thin foiled polyethylene foam on the wall, it may sag and lose some of its effectiveness. In order to prevent this, installation is carried out with glue over the entire surface area (on concrete or brick bases), more frequent fastening of thermal insulation to wooden wall staples from a construction stapler or the use of reinforced material.

One of modern materials, which can be used to insulate walls even at the construction stage is ecowool. This is an environmentally friendly material, which consists of 80% cellulose fibers with active additives:

Borax – preventing combustion;
Boric acid – providing protection against fungi, rot, rodents and insects.

Ecowool is installed using special sprayers into the inter-wall space. The spraying process can be seen in more detail here:

Thermal insulation applied on the outside of the wall

Materials of this type have additional requirements related to resistance to negative influence external environment:

Low moisture absorption;
Frost resistance – the ability to withstand multiple freeze-thaw cycles without destruction;
UV resistance;
Strength.

Expanded polystyrene

It is the most common material for insulating facades. However, its installation is quite labor-intensive. In addition, when calculating polystyrene foam insulation, it is necessary to add cost additional materials and performing work on intermediate strengthening and finishing decorative finishing facade.

Brick wall;
Special assembly adhesive for insulation;
Expanded polystyrene;
Special plastic dowels"umbrella";
Fiberglass mounting mesh;
Glue for mesh;
Primer that increases the adhesion of plaster;
Decorative plaster.

Liquid thermal insulation for walls is a new and progressive thermal insulation material, which is not yet very common, but is rapidly gaining popularity.

It consists of ceramic and silicone porous microspheres based on a polymer acrylic adhesive composition. The main advantage of this material is the versatility of its use; it can be applied to any wall: concrete, brick, wood.

Application is easy to do with your own hands, with a brush or using a regular sprayer.

Having selected the necessary thermal insulation material and calculated its thickness, it is also necessary to follow the installation technology. Otherwise, the thermal insulation properties of the material will be significantly reduced.

Before purchasing insulation materials for your home, it is worth calculating the thickness of the insulation. No amount of recommendations or experience from neighbors will help determine how much protection your home specifically needs. The reason is that the effectiveness of thermal insulation is influenced both by the climate in a particular region and by the characteristics of the frame or roof of the house itself. The main goal of such calculations is to determine the required layer of insulation, which will allow, at minimal cost, to provide reliable protection against heat loss through the building envelope.

How to do this?

Any online calculator program will help simplify the task for inexperienced builders. These are easy to find on construction portals or on the official websites of manufacturers of thermal insulation materials. Or you can try to do all the calculations yourself. In any case, you need to know the requirements for thermal protection of buildings in your climatic region. They are in SNiP 02/23/2003 and on the Internet in the form of summary tables, which provide data on all major cities Russia.

For example, let’s take the data for Moscow and the region – 3.14 m2 °C/W. This is the resistance that all layers of the main structure, air and insulating layers, as well as exterior decoration. We will build on the given figure, not forgetting that we are talking about the minimum acceptable indicator.

Here, the thermal engineering calculation of the required insulation thickness begins with an analysis of the selected building material and capacity of load-bearing walls:

Concrete has the highest heat transfer coefficient - 1.5-1.6 W/m °C.
Brick has a relatively low thermal conductivity of 0.56 W/m °C, but in masonry this figure actually doubles and is already 1.2.
Good performance for cellular concrete and gas blocks is about 0.2-0.3 W/m °C.
Wood (depending on the selected species) – 0.10-0.18 W/m°C.

However, these figures in themselves only give an idea of thermal insulation characteristics different materials. For calculations, it is also necessary to take into account the thickness of the structure. Dividing it by the heat transfer coefficient, we get the resistance of real walls.

Let's take a standard 30 cm thick aerated concrete masonry: R = 0.3 m ÷ 0.2 W/m °C = 1.5 m °C/W.

We arm ourselves with a calculator and find that for thermal protection of the walls of such a house built in Moscow, it is not enough: 3.14-1.5 = 1.64 m ° C / W.

Now you can choose insulation for walls by considering several materials with different thermal conductivity values, but giving the same effect due to thickness:

Mineral wool (0.04 W/m °C) – 1.64x0.04 = 0.0656 m or 66 mm.
Foam plastic (0.05 W/m °C) – 1.64x0.05 = 0.082 m (82 mm).
Penoplex (0.03 W/m °C) – 1.64x0.03 = 0.0492 m (50 mm).

Next, we include the cost of materials in the calculation and do not forget about the logic. Penoplex, although it shows the most best characteristics, For aerated concrete walls It’s simply not suitable, so you’ll have to choose between mineral wool and polystyrene foam. A cubic meter of inexpensive basalt insulation, which is suitable for insulating a facade, will cost about 2,500 rubles. If we take slabs with a thickness of 70 mm, for this amount it will be possible to cover 14.3 m2.

PSB-S-25f costs 2600 rub/m3. At first glance, the difference is small, but let’s recalculate how much area the slabs will cover if the thermal insulation thickness is 100 mm. It should be explained here that 80 mm sheets do not meet the minimum thermal protection requirements, and 90 mm sheets are not commercially available. So, in fact, for 2,600 rubles you can insulate only 10 square meters. It turns out that the difference in price between polystyrene foam and mineral wool is 4%, and in the insulated area – 43%. However, it's worth doing one more calculation on the calculator. He will show you how much it will cost curtain façade to protect the mineral wool itself, and how the cost will change after plastering and painting the PSB.

For slopes and flat designs Similar calculations are performed, but here you will have to take into account all the working layers in the overall pie. Thus, we obtain the insulation for the roof and its thickness by subtracting the resistances of all other elements from the norm according to SNiP (adjusted by 0.16), after which we simply multiply the difference by its own thermal conductivity coefficient:

S = (R-0.16-S 1 /ʎ 1 -S 2 /ʎ 2 -…-S i /ʎ i)·ʎ (m).

Instead of worrying, you can find recommendations for roof insulation for your region. In Moscow, 200 mm of basalt wool is considered the norm. From here, through the proportion of the thermal conductivity of the materials, we obtain an equivalent replacement: 250 mm of polystyrene foam or 150 mm of Penoplex.

The same calculation rules apply here, but the standard value of R0 changes. If we are talking about floors above a cold basement, in MO they should have a total resistance of 4.12 m2 °C/W, but adjusted for the coefficient of thermal uniformity of the slabs (for reinforced concrete this is 0.8, for wooden floors 0.9). The figure 0.17 is also subtracted from the resulting figure according to SNiP requirements. Then the resistance will be equal to:

R = R 0 ÷ 0.8 – 0.17 = 4.12 ÷ 0.8 – 0.17 = 4.98 m2 °C/W.

Again we subtract the thickness of the ceiling divided by its thermal conductivity, and multiply the finished result by the conductivity of the insulation itself. For example, for Penoplex on a slab with cement screed with a total thickness of 26 cm we get a layer of 160 mm. From here it is already possible to calculate the thickness of mineral wool (215 mm) and foam plastic (265), which could replace it.

How thick should the insulation be, comparison of the thermal conductivity of materials.

January 16, 2006
Published: Construction technologies and materials

The need to use WDVS thermal insulation systems is caused by high economic efficiency.

Following European countries, the Russian Federation adopted new standards for thermal resistance of enclosing and load-bearing structures aimed at reducing operating costs and energy saving. With the release of SNiP II-3-79*, SNiP 02/23/2003 “Thermal protection of buildings”, the previous thermal resistance standards have become outdated. The new standards provide for a sharp increase in the required heat transfer resistance of enclosing structures. Now the previously used approaches in construction do not correspond to the new ones regulatory documents, it is necessary to change the principles of design and construction, introduce modern technologies.

As calculations have shown, single-layer structures do not economically meet the accepted new standards of building heating engineering. For example, in the case of using high load-bearing capacity of reinforced concrete or brickwork, in order for the same material to withstand thermal resistance standards, the thickness of the walls must be increased to 6 and 2.3 meters, respectively, which is contrary to common sense. If you use materials with best performance according to thermal resistance, then their bearing capacity is very limited, for example, as with aerated concrete and expanded clay concrete, and expanded polystyrene and mineral wool, effective insulation materials, are not construction materials at all. On at the moment There is no absolute building material that would have a high load-bearing capacity combined with a high thermal resistance coefficient.

In order to meet all construction and energy saving standards, it is necessary to construct the building according to the principle of multilayer structures, where one part will perform a load-bearing function, the second - the thermal protection of the building. In this case, the thickness of the walls remains reasonable, and the normalized thermal resistance of the walls is observed. In terms of their thermal performance, WDVS systems are the most optimal of all façade systems on the market.

Table of the required insulation thickness to meet the requirements of current standards for thermal resistance in some cities of the Russian Federation:

Table where: 1 - geographical point 2 - average temperature of the heating period 3 - duration of the heating period in days 4 - degree-day of the heating period Dd, °С * day 5 - normalized value of heat transfer resistance Rreq, m2*°C/W of walls 6 - required insulation thickness

Conditions for performing calculations for the table:

1. The calculation is based on the requirements of SNiP 02/23/2003
2. Group of buildings 1 is taken as an example of the calculation - Residential, medical and preventive institutions and children's institutions, schools, boarding schools, hotels and hostels.
3. For load-bearing wall in the table, brickwork with a thickness of 510 mm is taken from ordinary clay bricks on cement-sand mortar l = 0.76 W/(m * °C)
4. The thermal conductivity coefficient is taken for zones A.
5. Estimated indoor air temperature + 21 °C “living room during the cold season” (GOST 30494-96)
6. Rreq is calculated using the formula Rreq=aDd+b for a given geographical location
7. Calculation: Formula for calculating the total heat transfer resistance of multi-layer fencing:
R0= Rв + Rв.п + Rн.к + Ro.к + Rн Rв - heat transfer resistance at the inner surface of the structure
Rн - heat transfer resistance at the outer surface of the structure
Rv.p - thermal conductivity resistance of the air layer (20 mm)
Rн.к - thermal conductivity resistance load-bearing structure
Rо.к - thermal conductivity resistance of the enclosing structure
R = d/l d - thickness of homogeneous material in m,
l - thermal conductivity coefficient of the material, W/(m * °C)
R0 = 0.115 + 0.02/7.3 + 0.51/0.76 + dу/l + 0.043 = 0.832 + dу/l
dу - thickness of thermal insulation
R0 = Rreq
Formula for calculating insulation thickness for given conditions:
dу = l * (Rreq - 0.832)

a) - the average thickness of the air gap between the wall and the thermal insulation is taken to be 20 mm
b) - thermal conductivity coefficient of polystyrene foam PSB-S-25F l = 0.039 W/(m * °C) (based on the test report)
c) - thermal conductivity coefficient of facade mineral wool l = 0.041 W/(m * °C) (based on the test report)

* The table shows the average values for the required thickness of these two types of insulation.

Approximate calculation of the thickness of walls made of a homogeneous material to meet the requirements of SNiP 23-02-2003 “Thermal protection of buildings”.

* For comparative analysis data is used climate zone Moscow and Moscow region.

Conditions for performing calculations for the table:

1. Standardized value of heat transfer resistance Rreq = 3.14
2. Thickness of homogeneous material d= Rreq * l

Thus, from the table it is clear that in order to build a building from a homogeneous material that meets modern requirements thermal resistance, for example, from traditional brickwork, even from perforated brick, the wall thickness should be at least 1.53 meters.

To clearly show what thickness of material is needed to meet the requirements for thermal resistance of walls made of a homogeneous material, a calculation was performed taking into account design features application of materials, the following results were obtained:

This table shows calculated data on thermal conductivity of materials.

According to the table data, for clarity, the following diagram is obtained:

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Insulated Swedish Plate

Insulated Swedish slab (USP) is one of the types of shallow foundation. The technology came from Europe. This type of foundation has two main layers. The bottom, heat-insulating layer prevents the ground from freezing under the house. Top layer...

Film - step-by-step instructions on SFTK technology ("wet facade")

With the support of SIBUR, the Association of Manufacturers and Sellers of Expanded Polystyrene, as well as in cooperation with the companies KREIZEL RUS, TERMOKLIP and ARMAT-TD, a unique educational film was created on the production technology of plaster heat-insulating facades...

In February 2015, another training video on facade systems was released. How to properly make decorative elements for decorating a cottage - this is explained step by step in the video.

With the support of SIBUR, the 1st practical conference “Polymers in Thermal Insulation” was held
On May 27, the 1st practical conference “Polymers in Thermal Insulation” was held in Moscow, organized by the information and analytical center Rupec and the Oil and Gas Vertical magazine with the support of SIBUR. The main topics of the conference were trends in the field of regulatory…
Directory - weight, diameter, width of ferrous metal products (reinforcement, angle, channel, I-beam, pipes)
1. Directory: diameter, weight linear meter reinforcement, section, steel class
The BOLARS TVD-1 and BOLARS TVD-2 systems are absolutely fireproof!
The BOLARS TVD-1 and BOLARS TVD-2 systems are absolutely fireproof! Experts came to this conclusion after conducting fire tests on façade thermal insulation systems TM BOLARS. Systems are assigned a class fire danger K0 – the safest. Huge...

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