Calculation of the thickness of the insulation: the choice of material, the calculation procedure for various surfaces. How to calculate the thickness of the insulation - methods and methods

Online insulation calculator, designed to calculate the amount and volume of insulation for external walls and the side surface of the foundations of buildings. The calculations take into account window and door openings, as well as the cost of insulation and additional materials.

When filling out the data, pay attention to Additional information signed Additional Information

Expanded Polystyrene (EPS) and Extruded Polystyrene (EPS)

I am one of the most accessible and efficient lungs heaters. More than 90% consists of air, which is the best heat insulator. Conventional PPS is used to insulate the external walls of buildings, but since it is a moisture-permeable material, it is not recommended to use it for insulating foundations. For these purposes, EPPS is best suited, which, when insulating foundations, is also a moisture-proof layer.

Mats of stone (basalt) wool

Currently, the most famous manufacturers of stone wool slabs are companies such as Rokwool and TechnoNIKOL.

With the most important advantages of this material are ease of processing, to work with it you will not need any special equipment, a knife or saw with fine teeth is enough. It is worth remembering that the wool slabs must be joined very tightly, but it is forbidden to ram them or compress them. From the inside, the mats are covered vapor barrier membrane, and outside - with a windproof film, this is necessary in order to protect the wool from moisture.

With strong moisture, stone and mineral wool loses its heat-saving characteristics

Sprayed heaters

This method of insulation in our country is still not very widespread. Mainly for wall insulation frame houses using polyurethane foam. It consists of two liquid substances, which turn into foam under air pressure, and after the entire space is filled, its excess is cut off. Working with such material is similar to working with mounting foam.

Ecowool

IN Lately the use of such insulation as cellulose fibers or ecowool has become very popular. It is made from natural material and does not require additional protection, this type of insulation is most suitable for those who want to make their home environmentally friendly.

And there are two ways of laying: it is a dry method and a wet one.

• Dry way

Using a special machine, the wool is blown into an insulated layer until the required density is reached. The disadvantage of this method is that over time it can shrink and begin to let heat into upper layers. Although many manufacturers give a guarantee that there will be no shrinkage for at least 20 years.

• wet way

It can be done with the help of special equipment, ecowool under pressure is “glued” to the walls and to each other, this avoids shrinkage. The main disadvantage is that wet laying of ecowool must be carried out outside before wall cladding.

Further presented full list performed calculations with brief description each item. If you did not find the answer to your question, you can contact us by feedback.

General information on the results of calculations

• The amount of insulation
- The total volume of the necessary insulation
• P area of ​​insulation
- total area insulation, taking into account gables, window and door openings
• Number of dowels "fungi"
- The total number of dowels "fungi" with a consumption of 6 pieces per 1 square meter insulation.
• In the EU of insulation
- The total weight of the insulation of the specified density. Check the density of the material with the sellers.

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

How to do it?

To simplify the task for inexperienced builders, any online calculator calculation program will help. These are easy to find on construction portals or on the official websites of manufacturers. 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 show data for all big 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 insulation layers, as well as exterior finish. We will start from the given figure, not forgetting that we are talking about the minimum allowable indicator.

Here, the thermotechnical calculation of the required insulation thickness begins with an analysis of the selected building material and power 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 scores for cellular concrete and gas blocks - about 0.2-0.3 W / m ° C.
• Wood (depending on the selected species) - 0.10-0.18 W / m ° С.

However, these figures alone 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 real walls.

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

We arm ourselves with a calculator and find that for the thermal protection of the walls of such a house built in Moscow, it is not enough: 3.14-1.5 \u003d 1.64 m ° C / W.

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

• Mineral wool (0.04 W / m ° C) - 1.64x0.04 \u003d 0.0656 m or 66 mm.
• Styrofoam (0.05 W / m ° C) - 1.64x0.05 \u003d 0.082 m (82 mm).
• Penoplex (0.03 W / m ° C) - 1.64x0.03 \u003d 0.0492 m (50 mm).

Further, we include the cost of materials in the calculation and do not forget about logic. Penoplex, although it shows the most best performance, for aerated concrete walls is simply not good, so you have to choose between mineral wool and polystyrene. A cubic meter of inexpensive basalt insulation, which is suitable for facade insulation, will cost about 2,500 rubles. If you take plates with a thickness of 70 mm, for this amount it will be possible to sheathe 14.3 m2.

PSB-S-25f costs 2600 rubles/m3. At first glance, the difference is small, but let's recalculate the area for which the plates will be enough if the thickness of the thermal insulation is 100 mm. It should be explained here that 80 mm sheets do not satisfy minimum requirements thermal protection, and 90 are not available for sale. So in fact, for 2600 rubles you can insulate only 10 squares. It turns out the difference in price for polystyrene foam with mineral wool is 4%, and in the insulated area - 43%. However, it is worth doing one more calculation on the calculator. He'll show you how much hinged facade to protect the mineral wool itself, and how the cost will change after plastering and painting the PSB.

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

S \u003d (R-0.16-S 1 /ʎ 1 -S 2 /ʎ 2 - ... -S i /ʎ i) ʎ (m).

You can not suffer, but 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 thermal conductivity of materials, we obtain an equivalent replacement: 250 mm of foam or 150 of Penoplex.

The same calculation rules apply here, but normative value R 0 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 plates (for reinforced concrete products this is 0.8, for wooden floors 0.9). The indicator 0.17 is also subtracted from the figure obtained according to the requirements of SNiP. Then the resistance will be:

R \u003d R 0 ÷ 0.8 - 0.17 \u003d 4.12 ÷ 0.8 - 0.17 \u003d 4.98 m2 ° C / W.

Again we subtract the thickness of the floor, 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 a cement screed total power 26 cm we get a layer of 160 mm. From here it is already possible to calculate the thickness of the mineral wool (215 mm) and polystyrene (265), which could replace it.

Warming of the house necessary procedure for creatures comfortable conditions residence. Insulation methods can be different - from increasing the thickness of the walls of the building to applying plaster layers, but warming with special heat-insulating materials is always relevant. And here the thickness of the insulation for the walls is important - it is impossible to increase the width of the bearing walls to infinity, but the heat must be kept as large as possible. Therefore, the calculation of the thickness of the insulation should be based on the characteristics of building and insulation materials.

You can insulate the house on the outside or interior walls, you can also combine these solutions, but the most effective is the insulation of exterior walls. Internal insulation shifts the dew point, which necessarily exists between dissimilar materials, inside the house, which means that the walls will gain moisture due to the abundance of accumulated condensate. It will not be possible to remove condensate with ventilation gaps, since the insulation is located inside the house. External insulation, on the contrary, shifts the dew point to the outside of the wall, allowing moisture to evaporate through the wall or ventilation gap, which is often done with external thermal insulation.
Comparison of insulation optionsChoice of material for home insulation

The characteristics of the heat insulator depend on the geographical region and climatic conditions in it, the size of the premises or building, building materials of house building. Also, the thickness of the insulation depends on the functional purpose of the area to be insulated. For residential housing construction, these will be one parameter, and for attic or basement space, others. As such, the thickness of the insulation does not play a primary role - the parameters of the following areas are important here:

1. Weather;
2. Building materials for load-bearing floors and walls;
3. The level of the object above the ground surface;
4. Heat insulator material.

To determine exactly how thick the heat-insulating layer on the walls of the house should be, you need to compare the coefficients of the materials. It is important to pay attention to the fact that the thermal conductivity coefficients of different heaters will always be different.

Comparative information for choosing popular heat-insulating building materials:

1. Expanded polystyrene boards: thermal conductivity coefficient = 0.039 W / m 0 С, material thickness = 120 mm;
2. Mineral, basalt, stone wool: 0.041 W / m 0 C, thickness of the slab or pouring layer = 130 mm;
3. Reinforced concrete, reinforced concrete walls: 1.7 W / m 0 С, thickness H = 533 mm;
4. Silicate brick: 0.76 W / m 0 C, product size = 238 mm;
5. Hollow red brick: 0.5 W / m 0 C, H = 157 mm;
6. Glued profiled timber: 0.16 W / m 0 C, timber thickness = 50 mm;
7. Expanded clay concrete: 0.47 W/m 0 С, H= 148 mm;
8. Gas block: 0.15 W / m 0 C, H = 470 mm;
9. Foam block: 0.3 W / m 0 C, H = 940 mm;
10. Cinder block: 0.6 W/m 0 С, H= 1800 mm.

From the above information, it is clear that the wall insulation must be ≥ 1500 mm thick for a comfortable microclimate in the house. But this is very, very much, so the wall must be made thinner, and the heat insulator layer reduced to 120-130 mm. How to do it? Selection of optimal parameters of building materials and heat-insulating building materials. The table shows the recommended thickness of mineral wool (basalt, stone) for different regions when building a house:

Comparative parameters of thermal insulation coefficients of different heat-insulating materials for these cities and regions:

1. Expanded polystyrene blocks PSB-S-25: thermal conductivity coefficient = 0.042 W / m 0 C, thickness H = 12.4 cm;
2. Mineral wool for insulation of ventilated facades: 0.046 W / m 0 C, H = 13.5 cm;
3. Profiled glued timber with a strength of 500 kg / m³: 0.18 W / m 0 С, H = 53.0 cm;
4. Keramoblocks: 0.17 W/m 0 С, H= 57.5 cm;
5. Gas blocks 600 kg / m³: 0.29 W / m 0 C, H = 98.1 cm;
6. Silicate brick: 0.87 W / m 0 C, H = 256.0 cm.

The use of these heat-insulating materials in the insulation of a house is a direct saving on the thickness of the outer walls.

The thickness of insulation in different climatic regions:

Minimum sub-zero temperature	Area	Material/density
		Stone/1300	Brick/1600	Keramoblock/1200	Concrete/300
		Thickness, mm
-60 0 С	Verkhoyansk	—	900,0	700,0	450,0
-40 0 C	Novosibirsk	—	900,0	700,0	450,0
-30 0 С	Moscow	30,0	640,0	500,0	350,0
-20 0 С	Yerevan	60,0	510,0	300,0	200,0
-10 0 С	Krasnovodsk	45,0	330,0	250,0	160,0

The calculation of the thickness of thermal insulation begins with the selection of material according to the purpose of the room and the average annual outdoor temperature. Common geographic areas:

1. I-th zone: ≥3501 degree-days;
2. II-I: ≈3001-3501 degree-days;
3. III-rd: ≈2501-3000 degree-days;
4. IV-th: ≤2500 degree-days.

The concept of "degree-day" is a parameter that reflects the difference between the air temperatures inside the building and the air temperature outside during the heating period. Its formula:

GSOP = (t v – t 8)z 8 ;

• t v - air temperature inside the building, ° С;
• t 8 - the average temperature of the heating period with average daily temperature air ≤8°С;
• z 8 - the number of days of the heating period with an average daily air temperature ≤ 8 ° С.

As real example the following calculations apply:

Minimum parameters for all four climatic zones: 2.80; 2.50; 2.20; 2.0. Below are the maximum allowable minimum values ​​of heat transfer resistance for different types premises:

1. Floors and wall coverings for buildings and premises without heating: 4.95; 4.50; 3.90; 3.30;
2. Basement and basement rooms without heating: 3.50; 3.30; 3.0; 2.50;
3. Ceilings for unheated basement and basements not below the ground surface: 2.80; 2.60; 2.20; 2.0.
4. Ceiling ceilings of cellars located below the soil surface: 3.70; 3.45; 3.0; 2.70.
5. Balconies and display windows, glazed facades, translucent verandas and terraces: 0.60; 0.56; 0.55; 0.50.
6. Front entrances and living rooms: 0.44; 0.41; 0.39; 0.32.
7. In a private house: hallways, halls and corridors: 0.60; 0.56; 0.54; 0.45.
8. Entrance halls and halls located above the 1st floor: 0.25 for all four zones.

Applying these indicators to a specific building, you can calculate the thickness of the layer of any heat insulator for any object. For an unmistakable choice of heat-insulating material, it is necessary to know and calculate its technical and performance characteristics. The accuracy of the results is influenced by the climatic zone of construction, Construction Materials insulated walls, the functional purpose of the object, the characteristics of each type of heat-insulating materials with approximately the same parameters.

Foreword. For home insulation, 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 the thermal conductivity coefficient and its thickness. In this article, we will tell you how to calculate the thickness of the insulation for the roof, attic, walls and floor in the house so that it is warm and comfortable in winter.

Why is it necessary to calculate the thickness of the insulation

Comfortable living in the house provides maintenance optimal temperature indoors, especially in winter. When erecting a building, you should remember about thermal insulation, you should correctly select and calculate the thickness of insulation for walls, roofs, floors and attics. Any material - brick, wood, foam block or mineral wool has its own value of thermal conductivity and heat resistance.

A warm home is the dream of every owner

Thermal conductivity is the ability of a material to conduct heat. This value is determined in laboratory conditions, and the data obtained are given by the manufacturer on the packaging or. The thermal resistance of a material is the reciprocal of thermal conductivity. A material that conducts heat well has low heat resistance and requires insulation.

When erecting a building, one should remember about high-quality thermal insulation. If mistakes were made in the walls of the house or in other structures during construction, then cold bridges may appear - areas along which heat quickly leaves the house. Condensation may occur in these places, and in the future, mold may form, if not taken during the warming measures.

How to calculate the thickness of insulation for walls

1 . Determine the design and finish of the exterior walls of the house (internal and external). The finishing scheme depends on your preferences, the decision of the exterior and interior of the building. Finishing adds several layers to the wall thickness of the house.

2 . Calculate the thermal resistance of the selected wall (Rpr.) The value can be found by the formula, and you need to know the wall material and its thickness:

Rpr.=(1/α (c))+R1+R2+R3+(1/α (n)),

where R1, R2, R3 are the heat transfer resistance of the layer, α(c) is the heat transfer coefficient inner surface walls, α(n) is the heat transfer coefficient of the outer surface of the wall.

3 . Calculate the minimum heat transfer resistance (Rmin.) for your climate zone according to the formula R=δ/λ, δ, where δ is the thickness of the material layer in meters, λ is the thermal conductivity of the material (W/m*K). Thermal conductivity (the ability of a material to exchange heat with environment) can be found on the packaging of the material or determined from the table of thermal conductivity of mineral wool or other material, for example, for PSB-S 15 foam plastic it is 0.043 W / m, for mineral wool with a density of 200 kg / m3 - 0.08 W / m.

The higher the thermal conductivity, the colder the material. The highest thermal conductivity is in metal, marble, the minimum is in air. Materials based on air are warm, for example, 40 mm of foam plastic is equal in thermal conductivity to 1 meter brickwork. The coefficient has constant value, it can be found in the reference book DBN V.2.6-31:2006 ( Thermal insulation buildings).

4 . Compare Rmin. with Rpr. and find the difference ΔR. If, as a result of your calculation, Rmin. is less than or equal to Rpr., then the insulation of the walls of the house is not necessary, since the existing layers provide the normative thermal insulation of the building. When is Rmin. more than Rpr., then determine the difference between them, for this, subtract from greater value less? R= Rmin.- Rpr.

5 . Choose the thickness of the insulation according to the value of ΔR. The selected insulation must provide the structure with the missing heat transfer resistance. When choosing a material, one should remember about its characteristics: thermal conductivity coefficient, density and combustibility class, water absorption coefficient. Next, let's look at examples of how to calculate the thickness of the insulation for different designs, but you can easily calculate the thermal conductivity of the wall online calculator on our website.

How to calculate insulation for brick walls

Imagine that the house has walls made of foam concrete with a density of 300 (0.3 m), the thermal conductivity of the material is 0.29. Divide 0.3 by 0.29 and we end up with a value of 1.03.

How to calculate the thickness of insulation for walls, allowing you to provide comfortable accommodation in the House? For this you need to know minimum value heat resistance in the city or region where the insulated building is located. Further, the resulting 1.03 must be subtracted from this value, and as a result, the heat resistance that the insulation should have will become known.

If the walls consist of several materials - concrete, brick, a layer of plaster, etc., then their heat resistance indicators should be summed up. The thickness of the wall insulation is calculated taking into account the resistance to heat transfer of the material used (R). To find the parameter, you should find out the value of HOSP (degree day of the heating period) using the formula:

t B reflects the temperature inside the room. According to established standards it is in the range of + 20-22 ° С. The average air temperature is t from, the number of days of the heating period in calendar year– z from. These values ​​are given in "Construction climatology" SNiP 23-01-99. Attention should be paid to the duration and temperature in heating period when the average daily t≤ 8°С.

When the heat resistance of each material is determined, you should find out what should be the thickness of the insulation of the ceiling, floor, walls, roof of the house. Each material of the "multilayer cake" design has its own thermal resistance R and is calculated by the formula:

R TP \u003d R 1 + R 2 + R 3 ... R n,

Where n is understood as the number of layers, while 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

How to calculate the insulation of walls from a foam block

For example, in the construction of a structure, a D600 foam block 30 cm thick is used, URSA basalt wool with a density of 80-125 kg / m3 acts as thermal insulation, and hollow brick with a density of 1000 kg / m3, 12 cm thick acts as a finishing layer.

The thermal conductivity coefficients of the above materials are indicated in the certificates.

Thermal conductivity of concrete 0.26 W/m*0С

Thermal conductivity of the insulation - 0.045 W / m * 0С

The thermal conductivity of a brick is 0.52 W / m * 0С.

Determine R for each material.

Thermal resistance of aerated concrete - R G \u003d δ SG / λ SG \u003d 0.3 / 0.26 \u003d 1.15 m 2 * 0 C / W
Thermal resistance of a brick - R K \u003d δ SK / λ SK \u003d 0.12 / 0.52 \u003d 0.23 m 2 * 0 C / B.

Knowing that the wall consists of 3 layers, we find R TR \u003d R G + R Y + R K, and find the heat resistance of the insulation R Y \u003d R TR - R G - R K.

Imagine that construction is taking place in a region where R TP (22 0 C) is 3.45 m 2 * 0 C / W. We calculate R Y \u003d 3.45 - 1.15 - 0.23 \u003d 2.07 m 2 * 0 C / W. Now we know how much resistance basalt wool or other heater. The thickness of the wall insulation will be determined by the formula:

δ S \u003d R Y x λ S Y \u003d 2.07 x 0.045 \u003d 0.09 m or 9 cm.

If we imagine that R TP (18 0 C) \u003d 3.15 m 2 * 0 C / W, then R U \u003d 1.77 m 2 * 0 C / W, and δ S \u003d 0.08 m or 8 cm.

How to calculate the thickness of the attic insulation

The calculation of this parameter is carried out by analogy with the determination of the thickness of the insulation of the walls of the house. For thermal insulation 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 does not have of great importance. Most often, rolled, matte or slab thermal insulation is used to insulate roof slopes.

The thickness of the insulation for the ceiling is calculated according to the above algorithm. The temperature in the house in winter depends on how correctly the parameters of the insulating material are determined. Experienced builders advise increasing the thickness of the roof insulation up to 50% relative to the design. If used backfill materials, from time to time they need to be loosened.

The thickness of the insulation in the frame house

The role of thermal insulation can be stone wool, ecowool and bulk materials. Calculation of the thickness of the insulation in frame house simple, because its design provides for the presence of insulation. The heat resistance of the walls of a house in Moscow should be R = 3.20 m 2 * 0 C / W. The thermal conductivity of the insulation is presented in the tables or in the product certificate.

For wool, it is λ ut \u003d 0.045 W / m * 0 C. The thickness of the insulation for frame house is determined by the formula:

δ ut \u003d R x λ ut \u003d 3.20 x 0.045 \u003d 0.14 m

Mineral wool slabs are produced with a thickness of 10 cm and 5 cm. In this case, it will be necessary to lay the mineral wool in two layers.

How to calculate the thickness of floor insulation

Before proceeding with the calculations, you should know at what depth the floor is located relative to ground level. You should also have an idea about the temperature of the soil in winter at a depth. Data can be taken from the table of dependence of soil temperature on depth and location:

First you need to determine the GSOP, then calculate the resistance to heat transfer, determine the thickness of the floor layers (for example, reinforced concrete, cement strainer for the heater flooring). Next, we determine the resistance of each of the layers and summarize the obtained values. Thus, we find out the thermal resistance of all layers of the floor, except for the insulation.

To find the thickness of the insulation, we subtract the total resistance of the floor layers from the normative heat resistance, with the exception of the insulating material. The thickness of the insulation for the floor in the house is calculated by multiplying the heat resistance of the insulation by the coefficient of thermal conductivity.

When choosing a material for thermal insulation, a reasonable question arises: “How to calculate the thickness of insulation for walls?”, Especially since there are various sizes of sheets, mats and rolls on sale. The answer depends on many factors.

What determines the thickness

Material

The calculation of the thickness of the insulation for walls is impossible without taking into account many related factors and conditions. It is incorrect to talk about the parameters of some kind of spherical insulation in a vacuum. There are many different materials, each with its own characteristics.

Here is a list of thermal conductivity coefficients of various thermal insulation materials:

• Glass wool URSA - 0.044 W/m×K;
• Stone (basalt) wool Rockwool - 0.039 W / m × K;
• (polystyrene) - 0.037 W/m×K;
• Ecowool - 0.036 W/m×K;
• Polyurethane foam () - 0.03 W / m × K;
• Expanded clay - 0.17 W / m × K;
• Brickwork - 0.520 W/m×K.

• Glass wool URSA - 189 mm;
• Stone (basalt) wool Rockwool - 167 mm;
• Expanded polystyrene (polystyrene) - 159 mm;
• Ecowool - 150 mm;
• Polyurethane foam - 120 mm;
• Expanded clay - 869 mm;
• Brickwork - 1460 mm.

1. Operational density;
2. Load on the wall structure;
3. Environmental safety and composition;
4. Biological stability;
5. Chemical properties and interactions;
6. Corrosion resistance;
7. fire safety;
8. Permeability to air and steam;
9. Condensation formation;
10. The presence of "cold bridges" and heat loss associated with them;
11. Hygroscopicity;
12. Moisture resistance.

The photo shows mineral wool, it has a standard minimum thickness, which satisfies the requirements of the climate of the middle zone

Further, on the basis of these data, one more important quantity should be determined - the resistance to heat transfer or simply thermal resistance. This value is equal to the ratio of the temperature difference along the edges of the material to the value heat flow passing through its thickness.

To calculate the resistance (R), the following formula is adopted:

R = wall thickness / thermal conductivity of the wall.

It becomes obvious that the thickness of the insulation depends not only on the properties of the material of the heat insulator, but also on the properties of the material from which the wall is made, its thickness and finish.

Already at this stage, it is clear that the calculation can only be carried out for a specific insulation, and taking into account a whole bunch of related conditions and factors. For example, the thickness of foam for wall insulation can greatly depend on the type of installation and brand of material, manufacturer, quality of raw materials, and many other parameters.

Advice! When it comes to individual construction, you should not go into the wilds of materials science and heat engineering. Suffice it to consider allowable norms for your region with a margin, the maximum overrun will be insignificant, you are not building up a city.

The thickness of the insulation for external walls must be at least a certain value, it makes no sense to calculate it exactly for many reasons:

• Firstly, you will still be forced to make some assumptions, assumptions and averaging, because you still cannot predict the weather and accurately indicate the movement of heated air masses;
• Secondly, even having received a thickness value accurate to microns, you still cannot find suitable size, since they are standard and rather roughly discrete, with a step of several tens of millimeters;
• Thirdly, as they say, the heat of the bones does not break, too warm is not a problem, just open the window, but when it's cold you have to spend money on heating or endure discomfort;
• Fourthly, a small margin of thickness will increase the total volume of the material not so significantly as to seriously worry about it.

Advice! Insulation thickness for external walls there should be more than one minimum allowable value. At the same time, you can play it safe and make a larger margin, you can save money and set the thickness as close as possible to the permissible minimum, it's up to you.

Climatic conditions

Following important condition, which should be taken into account when calculating the thickness of the foam for wall insulation, these are the climatic conditions of the area where it is supposed to be used. This is an obvious fact, but it is still worth mentioning separately.

Once you have decided on the material, you should find out in what climate zone it will be used. Manufacturers, as a rule, provide information on the recommended insulation parameters for different temperature conditions and zones.

Wall construction

To understand how meaningless universal instruction when calculating the thickness of a particular material, one more thing should be recalled important detail: wall designs. Here the number of layers, their composition, sequence, thickness play a role. As you can see, there are many options.

It is also important where the heat insulator is located - outside, from the side of the room or inside the structure. Equally important are waterproofing, vapor barrier, the presence of drafts and the movement of heated air masses, convection, infrared radiation and wind intensity in the region.

Do not forget about the finish, the thickness of the plaster, the facade coating and the presence of additional insulators. Combinations of heat-insulating materials are often used, such as polystyrene-foam, mineral wool-foam, foam-expanded clay, foam concrete-foam and others. This should also be taken into account.

Other factors

When calculating the parameters of the insulation, factors such as the purpose and functions of the insulation are also taken into account.

For example, it's one thing when you build frame building where the foam will be the main barrier to heat. Here you should play it safe and choose the maximum thickness of the insulation, because the very possibility of living in the house will depend on it.

It is a completely different matter when you are not satisfied with the degree of comfort in a brick house or you want to reduce heating costs. In this case, it would be advisable for you to choose the minimum justified thickness of the material, because the price of such a repair is also important, since we are talking about savings.

Also important role the way of construction plays: if you work with your own hands, it is important for you to control and calculate everything. If you hire a professional performer, your task is to choose the right company, because its specialists will in any case be involved in the calculation of all parameters.

Again, the insulation of a loggia or balcony imposes completely different requirements. These objects have thin walls, are blown with cold air from three sides, and do not have radiators. As you can see, the devil is in the details. universal rules, more often than not, nothing more than a myth.