In recent years, when building a house or repairing it, much attention has been paid to energy efficiency. With the already existing fuel prices, this is very important. And it seems that further savings will become increasingly important. In order to correctly select the composition and thickness of materials in the pie of enclosing structures (walls, floors, ceilings, roofs), it is necessary to know the thermal conductivity of building materials. This characteristic is indicated on the packaging with materials, and it is necessary at the design stage. After all, it is necessary to decide what material to build walls from, how to insulate them, how thick each layer should be.
When choosing building materials for construction, it is necessary to pay attention to the characteristics of the materials. One of the key positions is thermal conductivity. It is displayed by the coefficient of thermal conductivity. This is the amount of heat that a particular material can conduct per unit of time. That is, the smaller this coefficient, the worse the material conducts heat. Conversely, the higher the number, the better the heat is removed.
Materials with low thermal conductivity are used for insulation, with high - for heat transfer or removal. For example, radiators are made of aluminum, copper or steel, as they transfer heat well, that is, they have a high thermal conductivity. For insulation, materials with a low coefficient of thermal conductivity are used - they retain heat better. If an object consists of several layers of material, its thermal conductivity is determined as the sum of the coefficients of all materials. In the calculations, the thermal conductivity of each of the components of the "pie" is calculated, the found values are summarized. In general, we get the heat-insulating ability of the building envelope (walls, floor, ceiling).
There is also such a thing as thermal resistance. It reflects the ability of the material to prevent the passage of heat through it. That is, it is the reciprocal of thermal conductivity. And, if you see a material with high thermal resistance, it can be used for thermal insulation. An example of thermal insulation materials can be popular mineral or basalt wool, polystyrene, etc. Materials with low thermal resistance are needed to remove or transfer heat. For example, aluminum or steel radiators are used for heating, as they give off heat well.
To make it easier for the house to keep warm in winter and cool in summer, the thermal conductivity of walls, floors and roofs must be at least a certain figure, which is calculated for each region. The composition of the "pie" of walls, floor and ceiling, the thickness of the materials are taken in such a way that the total figure is not less (or better - at least a little more) recommended for your region.
When choosing materials, it must be taken into account that some of them (not all) conduct heat much better in conditions of high humidity. If during operation such a situation is likely to occur for a long time, the thermal conductivity for this state is used in the calculations. The thermal conductivity coefficients of the main materials used for insulation are shown in the table.
| Material name | Thermal conductivity W/(m °C) | ||
|---|---|---|---|
| Dry | Under normal humidity | With high humidity | |
| Woolen felt | 0,036-0,041 | 0,038-0,044 | 0,044-0,050 |
| Stone mineral wool 25-50 kg/m3 | 0,036 | 0,042 | 0,045 |
| Stone mineral wool 40-60 kg/m3 | 0,035 | 0,041 | 0,044 |
| Stone mineral wool 80-125 kg/m3 | 0,036 | 0,042 | 0,045 |
| Stone mineral wool 140-175 kg/m3 | 0,037 | 0,043 | 0,0456 |
| Stone mineral wool 180 kg/m3 | 0,038 | 0,045 | 0,048 |
| Glass wool 15 kg/m3 | 0,046 | 0,049 | 0,055 |
| Glass wool 17 kg/m3 | 0,044 | 0,047 | 0,053 |
| Glass wool 20 kg/m3 | 0,04 | 0,043 | 0,048 |
| Glass wool 30 kg/m3 | 0,04 | 0,042 | 0,046 |
| Glass wool 35 kg/m3 | 0,039 | 0,041 | 0,046 |
| Glass wool 45 kg/m3 | 0,039 | 0,041 | 0,045 |
| Glass wool 60 kg/m3 | 0,038 | 0,040 | 0,045 |
| Glass wool 75 kg/m3 | 0,04 | 0,042 | 0,047 |
| Glass wool 85 kg/m3 | 0,044 | 0,046 | 0,050 |
| Expanded polystyrene (polystyrene, PPS) | 0,036-0,041 | 0,038-0,044 | 0,044-0,050 |
| Extruded polystyrene foam (EPS, XPS) | 0,029 | 0,030 | 0,031 |
| Foam concrete, aerated concrete on cement mortar, 600 kg/m3 | 0,14 | 0,22 | 0,26 |
| Foam concrete, aerated concrete on cement mortar, 400 kg/m3 | 0,11 | 0,14 | 0,15 |
| Foam concrete, aerated concrete on lime mortar, 600 kg/m3 | 0,15 | 0,28 | 0,34 |
| Foam concrete, aerated concrete on lime mortar, 400 kg/m3 | 0,13 | 0,22 | 0,28 |
| Foam glass, crumb, 100 - 150 kg/m3 | 0,043-0,06 | ||
| Foam glass, crumb, 151 - 200 kg/m3 | 0,06-0,063 | ||
| Foam glass, crumb, 201 - 250 kg/m3 | 0,066-0,073 | ||
| Foam glass, crumb, 251 - 400 kg/m3 | 0,085-0,1 | ||
| Foam block 100 - 120 kg/m3 | 0,043-0,045 | ||
| Foam block 121- 170 kg/m3 | 0,05-0,062 | ||
| Foam block 171 - 220 kg/m3 | 0,057-0,063 | ||
| Foam block 221 - 270 kg/m3 | 0,073 | ||
| Ecowool | 0,037-0,042 | ||
| Polyurethane foam (PPU) 40 kg/m3 | 0,029 | 0,031 | 0,05 |
| Polyurethane foam (PPU) 60 kg/m3 | 0,035 | 0,036 | 0,041 |
| Polyurethane foam (PPU) 80 kg/m3 | 0,041 | 0,042 | 0,04 |
| Cross-linked polyethylene foam | 0,031-0,038 | ||
| Vacuum | 0 | ||
| Air +27°C. 1 atm | 0,026 | ||
| Xenon | 0,0057 | ||
| Argon | 0,0177 | ||
| Airgel (Aspen aerogels) | 0,014-0,021 | ||
| slag wool | 0,05 | ||
| Vermiculite | 0,064-0,074 | ||
| foamed rubber | 0,033 | ||
| Cork sheets 220 kg/m3 | 0,035 | ||
| Cork sheets 260 kg/m3 | 0,05 | ||
| Basalt mats, canvases | 0,03-0,04 | ||
| Tow | 0,05 | ||
| Perlite, 200 kg/m3 | 0,05 | ||
| Expanded perlite, 100 kg/m3 | 0,06 | ||
| Linen insulating boards, 250 kg/m3 | 0,054 | ||
| Polystyrene concrete, 150-500 kg/m3 | 0,052-0,145 | ||
| Cork granulated, 45 kg/m3 | 0,038 | ||
| Mineral cork on a bitumen basis, 270-350 kg/m3 | 0,076-0,096 | ||
| Cork flooring, 540 kg/m3 | 0,078 | ||
| Technical cork, 50 kg/m3 | 0,037 | ||
Part of the information is taken from the standards that prescribe the characteristics of certain materials (SNiP 23-02-2003, SP 50.13330.2012, SNiP II-3-79 * (Appendix 2)). Those material that are not spelled out in the standards are found on the manufacturers' websites. Since there are no standards, they can vary significantly from manufacturer to manufacturer, so when buying, pay attention to the characteristics of each material you buy.
Walls, floors, floors, can be made from different materials, but it so happened that the thermal conductivity of building materials is usually compared with brickwork. Everyone knows this material, it is easier to make associations with it. The most popular charts, which clearly demonstrate the difference between different materials. One such picture is in the previous paragraph, the second - a comparison of a brick wall and a wall of logs - is given below. That is why thermal insulation materials are chosen for walls made of bricks and other materials with high thermal conductivity. To make it easier to select, the thermal conductivity of the main building materials is tabulated.
| Material name, density | Coefficient of thermal conductivity | ||
|---|---|---|---|
| dry | at normal humidity | at high humidity | |
| CPR (cement-sand mortar) | 0,58 | 0,76 | 0,93 |
| Lime-sand mortar | 0,47 | 0,7 | 0,81 |
| Gypsum plaster | 0,25 | ||
| Foam concrete, aerated concrete on cement, 600 kg/m3 | 0,14 | 0,22 | 0,26 |
| Foam concrete, aerated concrete on cement, 800 kg/m3 | 0,21 | 0,33 | 0,37 |
| Foam concrete, aerated concrete on cement, 1000 kg/m3 | 0,29 | 0,38 | 0,43 |
| Foam concrete, aerated concrete on lime, 600 kg/m3 | 0,15 | 0,28 | 0,34 |
| Foam concrete, aerated concrete on lime, 800 kg/m3 | 0,23 | 0,39 | 0,45 |
| Foam concrete, aerated concrete on lime, 1000 kg/m3 | 0,31 | 0,48 | 0,55 |
| Window glass | 0,76 | ||
| Arbolit | 0,07-0,17 | ||
| Concrete with natural crushed stone, 2400 kg/m3 | 1,51 | ||
| Lightweight concrete with natural pumice, 500-1200 kg/m3 | 0,15-0,44 | ||
| Concrete on granulated slag, 1200-1800 kg/m3 | 0,35-0,58 | ||
| Concrete on boiler slag, 1400 kg/m3 | 0,56 | ||
| Concrete on crushed stone, 2200-2500 kg/m3 | 0,9-1,5 | ||
| Concrete on fuel slag, 1000-1800 kg/m3 | 0,3-0,7 | ||
| Porous ceramic block | 0,2 | ||
| Vermiculite concrete, 300-800 kg/m3 | 0,08-0,21 | ||
| Expanded clay concrete, 500 kg/m3 | 0,14 | ||
| Expanded clay concrete, 600 kg/m3 | 0,16 | ||
| Expanded clay concrete, 800 kg/m3 | 0,21 | ||
| Expanded clay concrete, 1000 kg/m3 | 0,27 | ||
| Expanded clay concrete, 1200 kg/m3 | 0,36 | ||
| Expanded clay concrete, 1400 kg/m3 | 0,47 | ||
| Expanded clay concrete, 1600 kg/m3 | 0,58 | ||
| Expanded clay concrete, 1800 kg/m3 | 0,66 | ||
| Ladder made of ceramic solid bricks at the CPR | 0,56 | 0,7 | 0,81 |
| Masonry of hollow ceramic bricks at the CPR, 1000 kg/m3) | 0,35 | 0,47 | 0,52 |
| Masonry of hollow ceramic bricks at the CPR, 1300 kg/m3) | 0,41 | 0,52 | 0,58 |
| Masonry of hollow ceramic bricks at the CPR, 1400 kg/m3) | 0,47 | 0,58 | 0,64 |
| Masonry of solid silicate bricks at the CPR, 1000 kg/m3) | 0,7 | 0,76 | 0,87 |
| Masonry of hollow silicate bricks at the CPR, 11 voids | 0,64 | 0,7 | 0,81 |
| Masonry of hollow silicate bricks at the CPR, 14 voids | 0,52 | 0,64 | 0,76 |
| Limestone 1400 kg/m3 | 0,49 | 0,56 | 0,58 |
| Limestone 1+600 kg/m3 | 0,58 | 0,73 | 0,81 |
| Limestone 1800 kg/m3 | 0,7 | 0,93 | 1,05 |
| Limestone 2000 kg/m3 | 0,93 | 1,16 | 1,28 |
| Construction sand, 1600 kg/m3 | 0,35 | ||
| Granite | 3,49 | ||
| Marble | 2,91 | ||
| Expanded clay, gravel, 250 kg/m3 | 0,1 | 0,11 | 0,12 |
| Expanded clay, gravel, 300 kg/m3 | 0,108 | 0,12 | 0,13 |
| Expanded clay, gravel, 350 kg/m3 | 0,115-0,12 | 0,125 | 0,14 |
| Expanded clay, gravel, 400 kg/m3 | 0,12 | 0,13 | 0,145 |
| Expanded clay, gravel, 450 kg/m3 | 0,13 | 0,14 | 0,155 |
| Expanded clay, gravel, 500 kg/m3 | 0,14 | 0,15 | 0,165 |
| Expanded clay, gravel, 600 kg/m3 | 0,14 | 0,17 | 0,19 |
| Expanded clay, gravel, 800 kg/m3 | 0,18 | ||
| Gypsum boards, 1100 kg/m3 | 0,35 | 0,50 | 0,56 |
| Gypsum boards, 1350 kg/m3 | 0,23 | 0,35 | 0,41 |
| Clay, 1600-2900 kg/m3 | 0,7-0,9 | ||
| Refractory clay, 1800 kg/m3 | 1,4 | ||
| Expanded clay, 200-800 kg/m3 | 0,1-0,18 | ||
| Expanded clay concrete on quartz sand with porization, 800-1200 kg/m3 | 0,23-0,41 | ||
| Expanded clay concrete, 500-1800 kg/m3 | 0,16-0,66 | ||
| Expanded clay concrete on perlite sand, 800-1000 kg/m3 | 0,22-0,28 | ||
| Clinker brick, 1800 - 2000 kg/m3 | 0,8-0,16 | ||
| Ceramic facing brick, 1800 kg/m3 | 0,93 | ||
| Medium density rubble masonry, 2000 kg/m3 | 1,35 | ||
| Drywall sheets, 800 kg/m3 | 0,15 | 0,19 | 0,21 |
| Drywall sheets, 1050 kg/m3 | 0,15 | 0,34 | 0,36 |
| Plywood | 0,12 | 0,15 | 0,18 |
| Fiberboard, chipboard, 200 kg/m3 | 0,06 | 0,07 | 0,08 |
| Fiberboard, chipboard, 400 kg/m3 | 0,08 | 0,11 | 0,13 |
| Fiberboard, chipboard, 600 kg/m3 | 0,11 | 0,13 | 0,16 |
| Fiberboard, chipboard, 800 kg/m3 | 0,13 | 0,19 | 0,23 |
| Fiberboard, chipboard, 1000 kg/m3 | 0,15 | 0,23 | 0,29 |
| PVC linoleum on a heat-insulating base, 1600 kg/m3 | 0,33 | ||
| PVC linoleum on a heat-insulating base, 1800 kg/m3 | 0,38 | ||
| PVC linoleum on fabric basis, 1400 kg/m3 | 0,2 | 0,29 | 0,29 |
| PVC linoleum on fabric basis, 1600 kg/m3 | 0,29 | 0,35 | 0,35 |
| PVC linoleum on fabric basis, 1800 kg/m3 | 0,35 | ||
| Asbestos-cement flat sheets, 1600-1800 kg/m3 | 0,23-0,35 | ||
| Carpet, 630 kg/m3 | 0,2 | ||
| Polycarbonate (sheets), 1200 kg/m3 | 0,16 | ||
| Polystyrene concrete, 200-500 kg/m3 | 0,075-0,085 | ||
| Shell rock, 1000-1800 kg/m3 | 0,27-0,63 | ||
| Fiberglass, 1800 kg/m3 | 0,23 | ||
| Concrete tile, 2100 kg/m3 | 1,1 | ||
| Ceramic tile, 1900 kg/m3 | 0,85 | ||
| PVC roof tiles, 2000 kg/m3 | 0,85 | ||
| Lime plaster, 1600 kg/m3 | 0,7 | ||
| Cement-sand plaster, 1800 kg/m3 | 1,2 | ||
Wood is one of the building materials with relatively low thermal conductivity. The table provides indicative data for different breeds. When buying, be sure to look at the density and coefficient of thermal conductivity. Not all of them are the same as prescribed in the regulatory documents.
| Name | Coefficient of thermal conductivity | ||
|---|---|---|---|
| Dry | Under normal humidity | With high humidity | |
| Pine, spruce across the grain | 0,09 | 0,14 | 0,18 |
| Pine, spruce along the grain | 0,18 | 0,29 | 0,35 |
| Oak along the grain | 0,23 | 0,35 | 0,41 |
| Oak across the grain | 0,10 | 0,18 | 0,23 |
| Corkwood | 0,035 | ||
| Birch | 0,15 | ||
| Cedar | 0,095 | ||
| Natural rubber | 0,18 | ||
| Maple | 0,19 | ||
| Linden (15% moisture) | 0,15 | ||
| Larch | 0,13 | ||
| Sawdust | 0,07-0,093 | ||
| Tow | 0,05 | ||
| Oak parquet | 0,42 | ||
| Piece parquet | 0,23 | ||
| Panel parquet | 0,17 | ||
| Fir | 0,1-0,26 | ||
| Poplar | 0,17 | ||
Metals conduct heat very well. They are often the bridge of cold in the design. And this must also be taken into account, to exclude direct contact using heat-insulating layers and gaskets, which are called thermal breaks. The thermal conductivity of metals is summarized in another table.
| Name | Coefficient of thermal conductivity | Name | Coefficient of thermal conductivity | |
|---|---|---|---|---|
| Bronze | 22-105 | Aluminum | 202-236 | |
| Copper | 282-390 | Brass | 97-111 | |
| Silver | 429 | Iron | 92 | |
| Tin | 67 | Steel | 47 | |
| Gold | 318 |
In order for the house to be warm in winter and cool in summer, it is necessary that the enclosing structures (walls, floor, ceiling / roof) must have a certain thermal resistance. This value is different for each region. It depends on the average temperature and humidity in a particular area.
Thermal resistance of enclosing
structures for Russian regions
In order for the heating bills not to be too large, it is necessary to select building materials and their thickness so that their total thermal resistance is not less than that indicated in the table.
Modern construction is characterized by a situation where the wall has several layers. In addition to the supporting structure, there is insulation, finishing materials. Each layer has its own thickness. How to determine the thickness of the insulation? The calculation is easy. Based on the formula:
R is thermal resistance;
p is the layer thickness in meters;
k is the thermal conductivity coefficient.
First you need to decide on the materials that you will use in construction. Moreover, you need to know exactly what type of wall material, insulation, finish, etc. will be. After all, each of them contributes to thermal insulation, and the thermal conductivity of building materials is taken into account in the calculation.
First, the thermal resistance of the structural material is considered (from which the wall, ceiling, etc. will be built), then the thickness of the selected insulation is selected according to the "residual" principle. You can also take into account the thermal insulation characteristics of finishing materials, but usually they go "plus" to the main ones. So a certain reserve is laid "just in case". This reserve allows you to save on heating, which subsequently has a positive effect on the budget.
Let's take an example. We are going to build a brick wall - one and a half bricks, we will insulate with mineral wool. According to the table, the thermal resistance of the walls for the region should be at least 3.5. The calculation for this situation is given below.
If the budget is limited, you can take 10 cm of mineral wool, and the missing will be covered with finishing materials. They will be inside and outside. But, if you want the heating bills to be minimal, it is better to start the finish with a “plus” to the calculated value. This is your reserve for the time of the lowest temperatures, since the norms of thermal resistance for enclosing structures are calculated according to the average temperature for several years, and winters are abnormally cold. Because the thermal conductivity of building materials used for decoration is simply not taken into account.
It is better to start the construction of each object with the planning of the project and careful calculation of thermal parameters. Accurate data will allow you to get a table of thermal conductivity of building materials. Proper construction of buildings contributes to optimal climatic parameters in the room. And the table will help you choose the right raw materials that will be used for construction.
The thermal conductivity of materials affects the thickness of the walls
Thermal conductivity is a measure of the transfer of heat energy from heated objects in a room to objects with a lower temperature. The heat exchange process is carried out until the temperature indicators are equalized. To designate thermal energy, a special coefficient of thermal conductivity of building materials is used. The table will help you see all the required values. The parameter indicates how much heat energy is passed through a unit area per unit time. The larger this designation, the better the heat transfer will be. When erecting buildings, it is necessary to use a material with a minimum value of thermal conductivity.
The thermal conductivity coefficient is a value that is equal to the amount of heat passing through a meter of material thickness per hour. The use of such a characteristic is necessary to create the best thermal insulation. Thermal conductivity should be taken into account when selecting additional insulating structures.
Thermal conductivity is determined by such factors:
Materials are represented by structural and heat-insulating varieties. The first type has high thermal conductivity. They are used for the construction of ceilings, fences and walls.
With the help of the table, the possibilities of their heat transfer are determined. In order for this indicator to be low enough for a normal indoor microclimate, walls made of some materials must be especially thick. To avoid this, it is recommended to use additional heat-insulating components.
When creating a project, all methods of heat leakage must be taken into account. It can exit through walls and roofs, as well as through floors and doors. If you do the design calculations incorrectly, you will have to be content with only the thermal energy received from the heating devices. Buildings built from standard raw materials: stone, brick or concrete need to be additionally insulated.
Additional thermal insulation is carried out in frame buildings. At the same time, the wooden frame gives rigidity to the structure, and the insulating material is laid in the space between the uprights. In buildings made of bricks and cinder blocks, insulation is carried out outside the structure.
When choosing heaters, it is necessary to pay attention to such factors as the level of humidity, the effect of elevated temperatures and the type of structure. Consider certain parameters of insulating structures:
The following types are used as heaters:
Bulk types of raw materials can be used for thermal insulation. These are paper granules or perlite. They are resistant to moisture and fire. And from organic varieties, you can consider wood fiber, linen or cork. When choosing, pay special attention to such indicators as environmental friendliness and fire safety.
Note! When designing thermal insulation, it is important to consider the installation of a waterproofing layer. This will avoid high humidity and increase resistance to heat transfer.
The table of thermal conductivity of building materials contains indicators of various types of raw materials that are used in construction. Using this information, you can easily calculate the thickness of the walls and the amount of insulation.
The heat transfer resistance table of materials shows the most popular materials. When choosing a particular option for thermal insulation, it is important to consider not only physical properties, but also such characteristics as durability, price and ease of installation.
Did you know that the easiest way is to install penooizol and polyurethane foam. They are distributed over the surface in the form of foam. Such materials easily fill the cavities of structures. When comparing solid and foam options, it should be noted that the foam does not form joints.
When making calculations, you should know the coefficient of resistance to heat transfer. This value is the ratio of temperatures on both sides to the amount of heat flow. In order to find the thermal resistance of certain walls, a thermal conductivity table is used.
You can do all the calculations yourself. For this, the thickness of the heat insulator layer is divided by the thermal conductivity coefficient. This value is often indicated on the packaging if it is insulation. Household materials are self-measured. This applies to thickness, and the coefficients can be found in special tables.
The resistance coefficient helps to choose a certain type of thermal insulation and the thickness of the material layer. Information on vapor permeability and density can be found in the table.
With the correct use of tabular data, you can choose high-quality material to create a favorable indoor climate.
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The thermal conductivity table of building materials is necessary when designing the protection of a building from heat loss in accordance with the SNiP standards of 2003 under the number 23-02. These measures ensure the reduction of the operating budget, the maintenance of a year-round comfortable microclimate inside the premises. For the convenience of users, all data are summarized in tables, parameters are given for normal operation, conditions of high humidity, since some materials sharply reduce properties with an increase in this parameter.
Thermal conductivity is one of the ways in which heat is lost by living quarters. This characteristic is expressed by the amount of heat that can penetrate a unit area of the material (1 m 2) per second at a standard layer thickness (1 m). Physicists explain the equalization of the temperatures of various bodies, objects through heat conduction by the natural desire for thermodynamic equilibrium of all material substances. Thus, each individual developer, heating the premises in winter, receives losses of thermal energy leaving the dwelling through the outer walls, floors, windows, and roofs. In order to reduce energy consumption for space heating, while maintaining a comfortable microclimate for operation inside them, it is necessary to calculate the thickness of all enclosing structures at the design stage. This will reduce the construction budget. The table of thermal conductivity of building materials allows you to use accurate coefficients for wall structural materials. The SNiP standards regulate the resistance of the facades of the cottage to the transfer of heat to the cold air of the street within 3.2 units. By multiplying these values, you can get the required wall thickness to determine the amount of material. For example, when choosing cellular concrete with a coefficient of 0.12 units, laying in one block 0.4 m long is sufficient. Using cheaper blocks of the same material with a coefficient of 0.16 units, you will need to make the wall thicker - 0.52 m. pine, spruce is 0.18 units. Therefore, in order to comply with the heat transfer resistance condition of 3.2, a 57 cm beam is required, which does not exist in nature. When choosing brickwork with a coefficient of 0.81 unit, the thickness of the outer walls threatens to increase up to 2.6 m, reinforced concrete structures - up to 6.5 m. In practice, walls are made multi-layered, laying a layer of insulation inside or sheathing the outer surface with a heat insulator. These materials have a much lower thermal conductivity coefficient, which makes it possible to reduce the thickness many times over. Structural material ensures the strength of the building, heat insulator reduces heat loss to an acceptable level. Modern facing materials used on facades, internal walls also have resistance to heat loss. Therefore, all layers of future walls are taken into account in the calculations. The above calculations will be inaccurate if you do not take into account the presence of translucent structures in each wall of the cottage. The table of thermal conductivity of building materials in the SNiP standards provides easy access to the thermal conductivity coefficients of these materials. When choosing a typical or individual project, the developer receives a set of documentation necessary for the construction of walls. Power structures are necessarily calculated for strength, taking into account wind, snow, operational, structural loads. The thickness of the walls takes into account the characteristics of the material of each layer, therefore, heat losses are guaranteed to be below the permissible norms of SNiP. In this case, the customer may file a claim with the organization involved in the design, in the absence of the necessary effect during the operation of the dwelling. However, during the construction of a dacha, a garden house, many owners prefer to save on the purchase of project documentation. In this case, wall thickness calculations can be made independently. Experts do not recommend using services on the websites of companies that sell structural materials, insulation. Many of them overestimate the values of thermal conductivity coefficients of standard materials in calculators in order to present their own products in a favorable light. Similar errors in calculations are fraught for the developer with a decrease in the comfort of the interior during the cold period. Self-calculation is not difficult, a limited number of formulas, standard values are used: For example, in order to bring the thickness of a brick wall in line with the normative thermal resistance, it will be necessary to multiply the coefficient for this material, taken from the table, by the normative thermal resistance: 0.76 x 3.5 = 2.66 m Such a fortress is unnecessarily expensive for any developer, therefore, the thickness of the masonry should be reduced to an acceptable 38 cm by adding insulation: The thermal resistance of brickwork in this case will be 0.38 / 0.76 \u003d 0.5 units. Subtracting the result obtained from the standard parameter, we obtain the required thermal resistance of the insulation layer: 3.5 - 0.5 = 3 units When choosing basalt wool with a coefficient of 0.039 units, we get a layer with a thickness of: 3 x 0.039 = 11.7 cm
An example of calculating the wall thickness by thermal conductivity
People also have different thermal conductivity, some warm like fluff, while others take heat like iron.
Yuri Serezhkin
The word "also" in the above statement shows that the concept of "thermal conductivity" is applied to people only conditionally. Although…
Did you know: a fur coat does not heat, it only retains the heat that the human body produces.
This means that the human body has the ability to conduct heat in a literal, and not just figurative sense. This is all poetry, in fact, we will compare heaters in terms of thermal conductivity.
You know better, because you yourself typed in the search engine "thermal conductivity of heaters." What exactly did you want to know? And if without jokes, then it is important to know about this concept, because different materials behave very differently when used. An important, although not a key point in the choice is precisely the ability of the material to conduct thermal energy. If you choose the wrong heat-insulating material, it simply will not perform its function, namely, to keep the heat in the room.
From a school physics course, you most likely remember that there are three types of heat transfer:
So thermal conductivity is a type of heat transfer or movement of thermal energy. It has to do with the internal structure of the bodies. One molecule transfers energy to another. Now would you like a little test?
Which type of substance transmits (transfers) the most energy?
That's right, the crystal lattice of solids transfers energy most of all. Their molecules are closer to each other and therefore can interact more effectively. Gases have the lowest thermal conductivity. Their molecules are at the greatest distance from each other.
We continue our conversation about the thermal conductivity of heaters. All bodies that are nearby tend to equalize the temperature among themselves. A house or apartment, as an object, seeks to equalize the temperature with the street. Are all building materials capable of being insulators? No. For example, concrete allows the heat flow from your house to the street too quickly, so the heating equipment will not have time to maintain the desired temperature in the room. The thermal conductivity coefficient for insulation is calculated by the formula:
Where W is our heat flux, and m2 is the area of \u200b\u200binsulation with a temperature difference of one Kelvin (It is equal to one degree Celsius). For our concrete, this coefficient is 1.5. This means that conditionally, one square meter of concrete with a temperature difference of one degree Celsius is able to pass 1.5 watts of thermal energy per second. But, there are materials with a coefficient of 0.023. It is clear that such materials are much better suited for the role of heaters. Does thickness matter, you ask? Plays. But, here you still can not forget about the heat transfer coefficient. To achieve the same results, you will need a concrete wall 3.2 m thick or a sheet of foam plastic 0.1 m thick. It is clear that although concrete can technically be a heater, it is not economically feasible. That's why:
Insulation can be called a material that conducts the least amount of thermal energy through itself, preventing it from leaving the room and at the same time costing as little as possible.
The best heat insulator is air. Therefore, the task of any insulation is to create a fixed air gap without convection (movement) of air inside it. That is why, for example, foam plastic is 98% air. The most common insulating materials are:
The thermal insulation properties of all the materials listed above lie close to these limits. It is also worth considering: the higher the density of the material, the more it conducts energy through itself. Remember from theory? The closer the molecules are, the more efficiently heat is conducted.
The table shows a comparison of heaters in terms of thermal conductivity declared by manufacturers and corresponding to GOSTs:
Comparative table of thermal conductivity of building materials that are not considered to be heaters:
The heat transfer rate only indicates the rate of heat transfer from one molecule to another. For real life, this indicator is not so important. But you can’t do without a thermal calculation of the wall. Heat transfer resistance is the reciprocal of thermal conductivity. We are talking about the ability of the material (insulation) to retain heat flow. To calculate the resistance to heat transfer, you need to divide the thickness by the coefficient of thermal conductivity. The example below shows the calculation of the thermal resistance of a wall made of a 180 mm thick beam.
As you can see, the thermal resistance of such a wall will be 1.5. Enough? It depends on the region. The example shows the calculation for Krasnoyarsk. For this region, the required coefficient of resistance of enclosing structures is set at 3.62. The answer is clear. Even for Kyiv, which is much further south, this figure is 2.04.
Thermal resistance is the reciprocal of thermal conductivity.
This means that the ability of a wooden house to resist heat loss is not enough. Warming is necessary, and already, with what material - calculate according to the formula.
It is worth saying that all the above indicators are given for DRY materials. If the material gets wet, it will lose its properties by at least half, or even turn into a “rag”. Therefore, it is necessary to protect thermal insulation. Styrofoam is most often insulated under a wet facade, in which the insulation is protected by a layer of plaster. A waterproofing membrane is applied to the mineral wool to prevent moisture from entering.
Another point that deserves attention is wind protection. Heaters have different porosity. For example, let's compare expanded polystyrene boards and mineral wool. If the first one looks solid, the second one clearly shows pores or fibers. Therefore, if you are installing fibrous thermal insulation, such as mineral wool or ecowool, on a wind-blown fence, be sure to take care of the wind protection. Otherwise, the good thermal performance of the insulation will not be useful.
So, we discussed that the thermal conductivity of heaters is their ability to transfer thermal energy. The heat insulator must not release the heat generated by the heating system of the house. The primary task of any material is to keep air inside. It is the gas that has the lowest thermal conductivity. It is also necessary to calculate the thermal resistance of the wall in order to find out the correct coefficient of thermal insulation of the building. If you have any questions about this topic, please leave them in the comments.
As a bonus for the curious who have read to the end an interesting experiment with thermal conductivity:
The process of transferring energy from a hotter part of the body to a less heated one is called thermal conduction. The numerical value of such a process reflects the thermal conductivity of the material. This concept is very important in the construction and repair of buildings. Properly selected materials allow you to create a favorable microclimate in the room and save a significant amount on heating.
Thermal conductivity is the process of thermal energy exchange, which occurs due to the collision of the smallest particles of the body. Moreover, this process will not stop until the moment of temperature equilibrium comes. This takes a certain amount of time. The more time spent on heat exchange, the lower the thermal conductivity.
This indicator is expressed as the coefficient of thermal conductivity of materials. The table contains already measured values for most materials. The calculation is made according to the amount of thermal energy that has passed through a given surface area of the material. The larger the calculated value, the faster the object will give up all its heat.
The thermal conductivity of a material depends on several factors:
When choosing a material for room insulation, it is also important to consider the conditions in which it will be used.
Thermal conductivity is taken into account at the design stage of a building. This takes into account the ability of materials to retain heat. Thanks to their correct selection, residents inside the premises will always be comfortable. During operation, money for heating will be significantly saved.
Insulation at the design stage is optimal, but not the only solution. It is not difficult to insulate an already finished building by carrying out internal or external work. The thickness of the insulation layer will depend on the materials chosen. Some of them (for example, wood, foam concrete) can in some cases be used without an additional layer of thermal insulation. The main thing is that their thickness exceeds 50 centimeters.
Particular attention should be paid to the insulation of the roof, window and door openings, and the floor. Most of the heat escapes through these elements. Visually, this can be seen in the photo at the beginning of the article.
For the construction of buildings, materials with a low coefficient of thermal conductivity are used. The most popular are:
Another popular building material is brick. Depending on the composition, it has the following indicators:
The coefficient of thermal conductivity of the material allows you to use the latter for the construction of garages, sheds, summer houses, baths and other structures. This group includes:
The coefficient of thermal conductivity of thermal insulation materials, the most popular in our time:
For convenience, the coefficient of thermal conductivity of the material is usually entered in the table. In addition to the coefficient itself, such indicators as the degree of humidity, density, and others can be reflected in it. Materials with a high coefficient of thermal conductivity are combined in the table with indicators of low thermal conductivity. An example of this table is shown below:
Using the coefficient of thermal conductivity of the material will allow you to build the desired building. The main thing: to choose a product that meets all the necessary requirements. Then the building will be comfortable for living; it will maintain a favorable microclimate.
Properly selected will reduce due to which it will no longer be necessary to “heat the street”. Thanks to this, financial costs for heating will be significantly reduced. Such savings will soon return all the money that will be spent on the purchase of a heat insulator.
