Comfortable living in a house provides for the creation of conditions for maintaining optimal temperature air especially in winter. In building a house, it is very important to correctly select insulation and calculate its thickness. Any construction material whether it is brick, concrete or foam block has its own thermal conductivity and thermal resistance. Thermal conductivity is understood as the ability of a building material to conduct heat. This value is determined in laboratory conditions, and the obtained data are given by the manufacturer on the packaging or in special tables. Thermal resistance is the reciprocal of thermal conductivity. A material that conducts heat well has a correspondingly low heat resistance.
For the construction and insulation of a house, a material is chosen that has a low thermal conductivity and high resistance. To determine the thermal resistance of a building material, it is enough to know its thickness and thermal conductivity coefficient.
Calculation of the thickness of the insulation of the 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. We divide 0.3 by 0.29 and get 1.03.
How to calculate the thickness of the insulation for the 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 insulated building is located. Further, the obtained 1.03 must be subtracted from this value, as a result, the heat resistance that the insulation should have will become known.
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, the norms of "Thermal protection of buildings" SP50.13330.2012 should be applied. The value of GOSP (degree day of the heating period) is calculated by the formula:
In this case, t B reflects the indoor temperature. According to the established standards, it should vary within + 20-22 ° С. Average air temperature - t from, the number of days of the heating period in a calendar year - z from. These values are given in "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 С.
After the thermal resistance is determined, you should find out what the thickness of the insulation of the ceiling, walls, floor, roof of the house should be.
Each material of the "multilayer pie" of the structure has its own thermal resistance R and is calculated by the formula:
R TP = 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 = δ S / λ S
The thickness of the 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, the role of thermal insulation is basalt wool with a density of 80-125 kg / m 3, as a finishing layer - hollow brick with a density of 1000 kg / m 3, 12 cm thick.The thermal conductivity coefficients of the above materials are indicated in the certificates, they can also be seen in SP50.13330.2012 in Appendix C. So thermal conductivity concrete was 0.26 W / m * 0 C, insulation - 0.045 W / m * 0 C, brick - 0.52 W / m * 0 C. Determine R for each of the materials used.
Knowing the thickness of aerated concrete, we find its thermal resistance R Г = δ SГ / λ SГ = 0.3 / 0.26 = 1.15 m 2 * 0 С / W, the thermal resistance of a brick - R К = δ SК / λ SК = 0.12 / 0.52 = 0.23 m 2 * 0 C / B. Knowing that the wall is made up of 3 layers
R TR = R Г + R У + R К,
we find the thermal resistance of the insulation
R U = R TP - R G - R K.
Let's imagine that construction takes place in a region where R TP (22 0 С) is 3.45 m 2 * 0 С / W. We calculate R Y = 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 the walls will be determined by the formula:
δ S = R У х λ SУ = 2.07 х 0.045 = 0.09 m or 9 cm.
If we imagine that R TP (18 0 C) = 3.15 m 2 * 0 C / W, then R Y = 1.77 m 2 * 0 C / W, and δ S = 0.08 m or 8 cm.
Roof insulation thickness
The calculation of this parameter is made by analogy with determining the thickness of the insulation of the walls of the 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 does not matter much.
Most often, high-performance roll, matte or slab thermal insulation is used to insulate roof slopes, and backfill materials for attic roofs.
The thickness of the insulation for the ceiling is calculated using the above algorithm. From how competently the parameters will be determined insulating material the temperature in the house depends on winter time... Experienced builders advise to increase the thickness of the roof insulation up to 50% relative to the design. If filling or crushing materials are used, they must be loosened from time to time.
The thickness of the insulation in the frame house
Glass wool can act as thermal insulation, stone wool, ecowool, bulk materials. Calculation of the thickness of the insulation in frame house simpler, because its design provides for the presence of the insulation itself and the outer and outer upholstery, as a rule, made of plywood and practically does not affect the degree of thermal protection.
For example, interior walls - plywood 6 mm thick, external - OSB board 9 mm thick, stone wool acts as a heater. The house is being built in Moscow.
The thermal resistance of the walls of a house in Moscow and the region should, on average, be R = 3.20 m 2 * 0 C / W. The thermal conductivity of the insulation is presented in special tables or in the certificate for the product. For stone wool, it is λ ut = 0.045 W / m * 0 С.
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 produced with a thickness of 10 cm and 5 cm. In this case, you will need to lay the mineral wool in two layers.
The thickness of the insulation for the floor on the ground
Before proceeding with the calculations, you should know at what depth the floor of the room is located relative to the ground level. You should also be aware of average temperature soil in winter at this depth. The data can be taken from the table.
First, you need to determine the GSPP, then calculate the resistance to heat transfer, determine the thickness of the floor layers (for example, reinforced concrete, cement strainer on insulation, flooring). Next, we determine the resistance of each of the layers by dividing the thickness by the thermal conductivity coefficient and summarizing the obtained values. Thus, we find out the thermal resistance of all layers of the floor, except for insulation. To find this indicator, we subtract the total thermal resistance of the floor layers from the standard thermal resistance, excluding the thermal conductivity coefficient of the insulating material. The thickness of the insulation for the floor is calculated by multiplying the minimum thermal resistance of the insulation by the thermal conductivity of the selected insulation material.
The correct calculation of thermal insulation will increase the comfort of the house and reduce heating costs. During construction, you cannot do without insulation, whose thickness determined by the climatic conditions of the region and the materials used. For insulation, polystyrene, polystyrene foam, mineral wool or ecowool are used, as well as plaster and other finishing materials.
To calculate how thick the insulation should be, you need to know the value of the minimum thermal resistance... It depends on the characteristics of the climate. When calculating it, the duration of the heating period and the difference between internal and external (average for the same time) temperatures are taken into account. So, for Moscow, the resistance to heat transfer for the outer walls of a residential building should be at least 3.28, in Sochi, 1.79 is enough, and in Yakutsk, 5.28 is required.
The thermal resistance of a wall is defined as the sum of the resistance of all structural layers, bearing and insulating. That's why the thickness of the insulation depends on the material from which the wall is made... For brick and concrete walls, more insulation is required, for wood and foam blocks less. Pay attention to the thickness of the material chosen for the supporting structures, and what is its thermal conductivity. The thinner the supporting structures, the greater the thickness of the insulation should be.
If a thick insulation is required, it is better to insulate the house from the outside. This will provide savings interior space... In addition, external insulation avoids the accumulation of moisture inside the room.
Thermal conductivity
The ability of a material to transmit heat is determined by its thermal conductivity. Wood, brick, concrete, foam blocks conduct heat in different ways. Increased air humidity increases thermal conductivity. The value inverse to thermal conductivity is called thermal resistance. To calculate it, the value of thermal conductivity in a dry state is used, which is indicated in the passport of the material used. You can also find it in the tables.
However, one has to take into account that in corners, joints of load-bearing structures and other special structural elements, thermal conductivity is higher than on a flat surface of walls. Cold bridges may arise through which heat will escape from the house. The walls in these areas will sweat. To prevent this, the value of the thermal resistance in such places is increased by about a quarter compared to the minimum allowable.
Calculation example
It is not difficult to calculate the thickness of the thermal insulation using a simple calculator. To do this, first calculate the resistance to heat transfer for supporting structure... The thickness of the structure is divided by the thermal conductivity of the material used. For example, foam concrete with a density of 300 has a thermal conductivity coefficient of 0.29. With a block thickness of 0.3 meters, the value of the thermal resistance:
The calculated value is subtracted from the minimum value. For Moscow conditions, the insulating layers must have a resistance not less than:
Then, multiplying the thermal conductivity coefficient of the insulation by the required thermal resistance, we obtain the required layer thickness. For example, for mineral wool with a thermal conductivity coefficient of 0.045, the thickness should not be less than:
0.045 * 2.25 = 0.1 m
In addition to thermal resistance, the location of the dew point is taken into account. The dew point is the place in the wall where the temperature can drop so much that condensation will occur - dew. If this place happens to be on inner surface walls, it fogs up and a putrefactive process can begin. The colder it is outside, the closer the dew point is to the room. The warmer and more humid the room, the higher the temperature at the dew point.
The thickness of the insulation in the frame house
As a heater for a frame house, mineral wool or ecowool is most often chosen.
The required thickness is determined by the same formulas as for traditional construction... Additional layers of a multi-layer wall give approximately 10% of its value. The wall thickness of the frame house is less than with traditional technology, and the dew point may be closer to the inner surface. That's why unnecessarily saving on the thickness of the insulation is not worth it.
How to calculate the thickness of roof and attic insulation
The formulas for calculating the resistance for roofs use the same, but the minimum thermal resistance in this case is slightly higher. Unheated attics are covered with bulk insulation. There are no restrictions on thickness, so it is recommended to increase it by 1.5 times relative to the calculated one. V mansard rooms for roof insulation, materials with low thermal conductivity are used.
How to calculate the thickness of floor insulation
Although the greatest heat loss occurs through the walls and roof, it is equally important to correctly calculate the floor insulation. If the basement and foundation are not insulated, it is considered that the temperature in the subfloor is equal to the outside temperature, and the thickness of the insulation is calculated in the same way as for the outer walls. If some insulation of the basement is done, its resistance is subtracted from the value of the minimum required thermal resistance for the construction region.
Calculation of the thickness of the foam
The popularity of foam is determined by its cheapness, low thermal conductivity, light weight and moisture resistance. Styrofoam almost does not allow steam to pass through, so it cannot be used for internal insulation ... It is located outside or in the middle of the wall.
Thermal conductivity of foam, like other materials, depends on density... For example, with a density of 20 kg / m3, the thermal conductivity coefficient is about 0.035. Therefore, a foam thickness of 0.05 m will provide a thermal resistance of 1.5.
For any home, a comfortable and warm atmosphere is important, which will make your stay pleasant and convenient. The correct microclimate will get rid of many troubles, including dampness, heat loss, too high heating costs. To avoid such negative points, it is necessary to select the correct type and thickness of insulation.
For the choice of 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, the construction market offers numerous options for insulation, which differ not only in their size and thickness, but also in the type of raw materials for manufacturing, operational characteristics... When choosing a heat insulator, it is required not only to clarify the thickness, but also to determine for which wall material it will be optimal. Attention should be paid to the climatic region, wind loads. For example, the value of the thickness of the insulation will indicate for which particular 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 heaters
Insulation is chosen based not only on thickness, but also on other indicators. What thickness to take depends on the following:
- climatic region for the construction site;
- the main material of the walls;
- the purpose of the room, its level above the ground;
- material of manufacture.
Manufacturers offer various options... Many people say that aerated concrete or expanded clay concrete is an excellent option for building warm home, here you can save on insulation. But is it really so? It is necessary to compare the coefficients of thermal conductivity. In order for the thickness to be selected correctly, it is necessary to take into account that all heaters differ in their characteristics, the indicators of their thermal conductivity will be different.
As comparative data, you can take:
- Expanded polystyrene heat insulators with a thermal conductivity coefficient of 0.039 W / m * ° C with a thickness of 0.12 m.
- Mineral wool (basalt wool, stone wool) 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 (perforated) brick with data of 0.5 W / m * ° C and 1.57 m.
- Glued wooden beams 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, in which the thermal conductivity coefficient is 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, it can be seen that the thickness of the wall to ensure a normal and comfortable microclimate is from one and a half meters. But this is too much. It is best to make the wall thinner, but at the same time 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 economy, but also the availability of free space in the house and on the site depends 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 the required thickness of 124 mm.
- Rockwool mineral wool for facade insulation: thermal conductivity coefficient - 0.046 W / m * ° С, required thickness -135 mm.
- Glued wooden beams made of spruce or pine with indicators of 500 kg / m³ in accordance with GOST 8486: thermal conductivity coefficient - 0.18 W / m * ° С, required thickness - 530 mm.
- Special warm ceramic blocks with a layer of heat-insulating glue: thermal conductivity coefficient - 0.17 W / m * ° С, required thickness - 575 mm.
- Aerated concrete blocks 600 kg / m³: thermal conductivity coefficient - 0.29 W / m * ° С, required thickness - 981 mm.
- Silicate brick in accordance with GOST 379: thermal conductivity coefficient - 0.87 W / m * ° С, required thickness - 2560 mm.
According to the above data, it can be seen that mineral wool, expanded polystyrene, ordinary timber are in the lead among other materials.
Using them as insulation makes it possible to build brick or concrete walls of lesser thickness. If the house is being built in a warm region, then a 10 cm insulation is enough. For colder regions, 12-13 cm is already required, but taking into account what material the main wall of the house is made of.
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An example of calculating insulation
The choice of thickness for a heat insulator must begin with the fact that the material is selected for its intended purpose for a particular room and for a temperature zone. All zones that are used for calculations can be found in special reference books. Among the frequently used 4:
- 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.
You can give as an example the following calculation options:
- The minimum allowable values for thermal resistance are represented by 4 zones in 2.8; 2.5; 2.2 and 2.
- Ceilings, coverings for unheated, unused attics: 4.95; 4.5; 3.9; 3.3.
- Cold cellars basement floors: 3,5; 3,3; 3; 2,5.
- Overlaps for unheated plinths, basements located at ground level: 2.8; 2.6; 2.2; 2.
- Ceilings for basements that are located below ground level: 3.7; 3.45; 3; 2.7.
- Balcony constructions, display cases and panoramic windows, walls near them, translucent special facades, verandas, covered terraces: 0.6; 0.56; 0.55; 0.5.
- Ceremonial for apartment buildings hallways for large public buildings: 0,44; 0,41; 0,39; 0,32.
- Entrance rooms, 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 such an indicator, you can calculate the thickness of a heat insulator of any structure. For example, the walls of a house are built from sand-lime brick at 51 cm. Insulation was carried out using foam plates of 10 cm. To determine whether the planned thickness of the insulation is suitable, you just need to calculate the coefficient for the thermal resistance of the foam and the wall, then add the obtained values and compare with those presented above.
For walls of 51 cm, the following data are obtained:
- The thermal conductivity coefficient of silicate brick is 0.87.
- The thickness of the wall 51 must be divided by 0.87 in order to obtain the thermal resistance of the brick equal to 0.58.
- They do it differently with foam. Its thickness is divided by the coefficient of thermal conductivity of this material 0.043, the result is 2.32.
- Now we need to add the obtained values, the result is 2.88. This indicator should be compared with the above. If the received data for external walls made of silicate bricks coincide with those required for a specific region ( climatic zone), then 10 cm of foam will be enough.
It must be remembered that if the insulation is used for colder areas, then its thickness should be 12-14 cm to create comfortable conditions living in the house.
In order to choose the right thermal insulation material, it is necessary to carefully approach the determination of its parameters. The influence is exerted by the climatic zone in which the house is being built, what material its walls are made of, for which part of the structure the heat insulator is used. It is important to immediately pay attention to the features of using a certain type of insulation. Usually, mineral wool or foam is purchased, but their characteristics are different, therefore, it is necessary to calculate separately for each material.
Until the second half of the 20th century, few people were interested in environmental problems, only the energy crisis that broke out in the 70s in the West sharply raised the question: how to save heat in a house without heating the street and not overpaying for energy.
There is a solution: wall insulation, but how to determine what should be the thickness of the insulation for the walls so that the structure meets modern requirements for resistance to heat transfer?
The effectiveness of the insulation depends on the characteristics of the insulation and the method of insulation. There are several different ways having their own merits:
- Monolithic construction, can be made of wood or aerated concrete.
- A multi-layer structure in which the insulation occupies an intermediate position between the outer and inner parts of the wall, in this case, during the construction phase, circular masonry is performed with simultaneous insulation.
- External insulation by wet (plastering system) or dry (ventilated facade) method.
- Internal insulation, which is performed when it is impossible to insulate the wall from the outside for some reason.
For insulation of already built and operated buildings, external insulation is used, as the most effective method reducing heat loss.
We calculate the thickness of the insulation
Thermal insulation outer wall gives a decrease in heat loss by a factor of two or more. For a country, most of whose territory belongs to the continental and sharply continental climate with a long period of low negative temperatures, like Russia, the thermal insulation of the building envelope gives a huge economic effect.
Therefore, whether the thickness of the heat insulator for external walls is correctly calculated depends on the durability of the structure and the microclimate in the room: if the thickness of the heat insulator is insufficient, the dew point is inside the wall material or on its inner surface, which causes condensation, high humidity, and, then, the formation of mold and fungal infection.
The method for calculating the thickness of the insulation is prescribed in the Code of Rules “SP 50. 13330. 2012 SNiP 23-02-2003. Thermal protection of buildings ”.
Factors affecting the calculation:
- Characteristics of the wall material - thickness, construction, thermal conductivity, density.
- The climatic characteristics of the structure zone are the air temperature of the coldest five-day period.
- Characteristics of the materials of the additional layers (cladding or plaster of the inner surface of the wall).
Insulation layer that meets regulatory requirements, is calculated by the formula:
In the "ventilated facade" thermal insulation system, the thermal resistance of the curtain wall material and the ventilated gap are not taken into account in the calculation.
Characteristics of various materials
Table 1
The value of the normalized resistance to heat transfer of the outer wall depends on the region of the Russian Federation in which the building is located.
table 2
Required layer thermal insulation material, determined based on the following conditions:
- external building envelope - full-bodied ceramic brick plastic pressing with a thickness of 380 mm;
- interior decoration - cement-lime plaster, 20 mm thick;
- exterior finish - a layer of polymer-cement plaster, layer thickness 0.8 cm;
- the coefficient of thermal engineering homogeneity of the structure is 0.9;
- thermal conductivity coefficient of insulation - λА = 0.040; λB = 0.042.
Calculators for calculating the thickness of the insulation
For the calculation, data will be required:
- wall size;
- wall material;
- thermal conductivity coefficient of the selected insulation;
- finishing layers;
- the city in which the building to be insulated is located.
The calculation will be completed in a matter of seconds.
Since we do not have our own calculator, we want to recommend, in our opinion, a very good online calculator, on which you can calculate the thickness of the heat insulator.
Outcomes
It is advisable to provide for a reduction in the cost of heating a house at the design stage: by laying walls in the project that do not require further insulation, you can save significant funds on operating costs.
In case you need to insulate already finished house, it is not difficult to calculate the required thickness of the insulation. The only disadvantage of such insulation is that its durability is less than the service life of the load-bearing wall.
Thermal insulation of the house must be performed with such a material that has the highest thermal conductivity, but at the same time is able to withstand mechanical stress.
An important parameter is the thermal resistance of the insulation. To calculate it, you need to have data on the coefficient of thermal conductivity, and also take into account the thickness of the material itself. The last parameter must be determined very carefully, since otherwise it will not be possible to provide a comfortable microclimate in the home.
Why is it necessary to calculate the thickness of the insulation
In winter, keep in the house comfortable temperature air is very important. Each material from which a residential building can be erected has its own thermal conductivity and thermal resistance. Thus, they will differ for wood, brick and foam block.
Thermal conductivity is the ability of the material used to transfer thermal energy... To accurately calculate this indicator, laboratory tests are carried out. The results obtained are indicated on the packaging of the material. Accordingly, the thermal resistance becomes a value that is the opposite of the already mentioned thermal conductivity. If the material has low resistance, this means that it conducts heat well and needs additional thermal insulation.
The need for thermal insulation procedures increases if during construction works any mistakes have been made. Then there are bridges of cold through which heat leaves indoor spaces... There is also a threat of condensation in problem areas, which leads to the accumulation of moisture and the development of mold.
How to calculate the thickness of the insulation for the walls
1. First, you need to determine the thermal conductivity of the material that was used to build the house. You will also have to take into account the features exterior decoration... If the latter was carried out with high quality, the same good insulation may no longer be required.
2. The calculation of the thermal resistance of the structure (Rpr.) Is performed. You can determine this parameter using a special formula. But it is also important to know from what material the wall was built and how thick it has. The formula itself looks like this:
R ex. = (1 / α (c)) + R1 + R2 + R3 + (1 / α (n)).
Here, R is understood as the resistance of each layer included in the structure. The parameter α (c) acts as a heat transfer coefficient, which is characteristic for inside walls. Accordingly, α (n) is the level of heat transfer of the wall from the outside.
3. Depending on the specific climatic zone, the minimum heat resistance (Rmin.) Is determined. For this, the formula Rmin. = Δ / λ is taken. As δ is meant the thickness of the material used, expressed in meters. Accordingly, λ is an indicator of the thermal conductivity of the material, which is indicated on the container for the material. Although there are also tables that show these parameters.
With an increase in thermal conductivity, the level of thermal insulation decreases, that is, the material becomes colder. Marble has the highest thermal conductivity. But for air, this figure is the lowest. Accordingly, materials containing air pores in the structure are distinguished by high-quality thermal insulation. It is for this reason that a 4 cm thick foam sheet provides the same insulation as brickwork 100 cm thick.
4. A comparison is made between Rmin. and previously determined Rpr. As a result, the difference ΔR is determined, by which it is judged whether the walls need insulation. This conclusion is reached when Rmin. turns out to be more than Rpr. Otherwise, insulation is not required. To carry out thermal insulation, it is necessary to know the difference between the indicators, if Rpr. less than Rmin.
5. Using the difference ΔR, select the optimal thickness of the thermal insulation material. When choosing, it is necessary to take into account other indicators of the material. Critical have thermal conductivity, density, flammability, water absorption.
How to calculate insulation yourself
Now it's worth considering specific example calculation of the required thickness of the insulation. As a material for the construction of walls, we take foam concrete, the density of which is 0.3 m. For foam concrete, the thermal conductivity is 0.29. Then Rmin. will be equal to 0.3 / 0.29 = 1.03. You will also need to know what R value should be present in a particular climatic zone. By comparing the numbers obtained, you can determine if thermal insulation is needed.
But in addition to the foam concrete itself, other layers may also be present in the wall structure - facing brick, plaster and more. In this case, it will be necessary to add up the thermal resistance coefficients characteristic of each of the layers. At the same time, according to SNiP, the temperature inside the dwelling should be at a level of at least + 22 ° C. Moreover, we are talking about the average temperature throughout the year. That is, you will have to take into account those periods when the air temperature outside is no more than + 8 ° C.
After determining the thermal resistance, it is necessary to calculate what thickness the thermal insulation material should have. Typically, thermal insulation consists of several layers, each of which has its own indicator. Therefore, to determine the total thermal resistance of the "pie", it is necessary to add up all the indicators R. Do not forget that R = δS / λS. That is, to determine it, the thickness of the material is divided by the level of thermal conductivity.
How to calculate the insulation of walls from a foam block
As an example, let's take a foam block D600, from which a wall of 30 cm was erected. To create an insulating layer, basalt wool is used, the density of which is 80-125 kg / m3. Additionally, hollow bricks are used for decoration. Layer thickness - 12 cm, material density - 1000 kg / m3.
To find out the coefficient of thermal conductivity of each of the materials, you need to look at the values indicated in the certificates. For concrete, this parameter is 0.26 W / m * 0С, for a heat insulator - 0.045, for bricks - 0.52. Now you can easily calculate R using the formula R = δS / λS. As a result, it turns out that the R of foam concrete is 1.15, brick - 0.23. To calculate the thermal resistance of the insulation, it is necessary to subtract the previously defined Rg and Rk from the Rtr indicator.
When work is carried out in a region where + 22 ° C is used to calculate Rрт, its value will be 3.45. Accordingly, RU = 3.45-1.15-0.23. Thus, the insulation must have a thermal resistance of 2.07. Knowing this parameter, it is possible to calculate the thickness of the insulation δS = RУ * λSУ, which will turn out to be 0.09 m.As a result, it was possible to determine that to obtain decent insulation, it is enough to use a plate of mineral wool with a thickness of 9 cm.But usually a plate of 10 cm is taken, since this the value is fixed.
How to calculate the thickness of the attic insulation
There is no special specificity in the definition of such a parameter. Here, the same actions are carried out as in the case of calculating the thickness of the thermal insulation material for arranging the walls. It is best if the attic insulation is carried out with a material whose thermal conductivity is 0.04. In the case of an attic, it does not matter how thick the thermal insulation layer... Most often, thermal insulation is carried out with slab or sheet materials, although insulation in the form of rolls is also used. Before using the roll, just roll it out on a flat surface and let it straighten.
However, most professional builders recommend using thicker insulation than is calculated on the project. If the owner wants to get reliable insulation of the attic, it is better for him to take insulation about 50% thicker than the calculated one. When applying bulk materials it should also be borne in mind that periodically it will be necessary to loosen so that the individual granules do not stick to each other.
The thickness of the insulation in the frame house
Usually, the thermal insulation of a frame house is carried out with materials such as stone wool, expanded clay, ecowool. There is nothing complicated in calculating the thickness of the thermal insulation layer for a frame house. The fact is that initially the buildings frame type suggest the presence of insulation. For middle lane the thermal resistance of the walls is at the level of 3.20. To determine the thermal conductivity of a material, it is necessary to look at the indicators presented in the certificate. So, for mineral wool, this parameter is 0.045. Then, to determine the thickness of the insulation, it is necessary to divide the thermal resistance into thermal conductivity. The result is 0.14 m.
The difficulty lies in the fact that mineral wool is produced in slabs, the thickness of which does not exceed 10 cm. Therefore, it is best to take slabs with different thicknesses. First, a layer of mineral wool is laid in 10 cm, and on it - in 5 cm.
How to calculate the thickness of floor insulation
To correctly calculate the thickness of the insulation, you must first take into account how deep the floor is laid in comparison with the ground level. It is also important what temperature the soil has in winter. These indicators are taken from a special table. Based on what air temperature in the room needs to be obtained, the thermal resistance is calculated by summing up the indicators of each of the layers that form the floor. The result will be the level of thermal resistance of the floor as a whole, excluding the insulation.
It remains to subtract the indicator obtained above from the standard thermal resistance. The remainder is multiplied with the coefficient of thermal conductivity of the material that will be used for thermal insulation. The resulting value is the required thickness of the insulation.
The calculation of the thickness of the insulation must be very carefully approached. The comfort of living and the safety of the building itself depends on how well this parameter can be determined.