Comparison of the thermal conductivity of building materials - we study important indicators. Calculation of the thermal conductivity of the wall Dependence of the thermal conductivity on the thickness of the material

Construction business involves the use of any suitable materials. The main criteria are safety for life and health, thermal conductivity, reliability. This is followed by price, aesthetic properties, versatility of use, etc.

Consider one of the most important characteristics of building materials - the coefficient of thermal conductivity, since it is on this property that, for example, the level of comfort in the house largely depends.

Theoretically, and practically too, building materials, as a rule, create two surfaces - external and internal. From the point of view of physics, the warm region always tends to the cold region.

In relation to a building material, heat will tend to move from one surface (warmer) to another surface (less warm). Here, in fact, the ability of the material with respect to such a transition is called the thermal conductivity coefficient or, in the abbreviation, CFT.

Scheme explaining the effect of thermal conductivity: 1 - thermal energy; 2 - coefficient of thermal conductivity; 3 – temperature of the first surface; 4 – temperature of the second surface; 5 - the thickness of the building material

The CHF characteristic is usually built on the basis of tests, when an experimental specimen with dimensions of 100x100 cm is taken and a thermal effect is applied to it, taking into account the temperature difference of two surfaces by 1 degree. Exposure time 1 hour.

Accordingly, thermal conductivity is measured in Watts per meter per degree (W/m°C). The coefficient is denoted by the Greek symbol λ.

By default, the thermal conductivity of various building materials with a value less than 0.175 W/m°C equates these materials to the category of insulating materials.

Modern production has mastered the technology of manufacturing building materials, the level of CFT of which is less than 0.05 W/m°C. Thanks to such products, it is possible to achieve a pronounced economic effect in terms of the consumption of energy resources.

Influence of factors on the level of thermal conductivity

Each individual building material has a certain structure and has a peculiar physical state.

The basis of this are:

• dimension of structure crystals;
• phase state of matter;
• degree of crystallization;
• anisotropy of thermal conductivity of crystals;
• volume of porosity and structure;
• direction of heat flow.

All of these are influencing factors. The chemical composition and impurities also have a certain effect on the level of CHF. The amount of impurities, as practice has shown, has a particularly pronounced effect on the level of thermal conductivity of crystalline components.

Insulating building materials - a class of products for construction, created taking into account the properties of PTS, close to optimal properties. However, it is extremely difficult to achieve ideal thermal conductivity while maintaining other qualities.

In turn, the KTP is influenced by the operating conditions of the building material - temperature, pressure, humidity level, etc.

Building materials with a minimum KTP

According to studies, dry air has a minimum value of thermal conductivity (about 0.023 W / m ° C).

From the point of view of the use of dry air in the structure of a building material, a structure is needed where dry air resides inside numerous enclosed spaces of small volume. Structurally, such a configuration is presented in the form of numerous pores inside the structure.

Hence the logical conclusion: a building material, the internal structure of which is a porous formation, should have a low level of CHF.

Moreover, depending on the maximum allowable porosity of the material, the value of thermal conductivity approaches the value of the CHF of dry air.

The porous structure contributes to the creation of a building material with minimal thermal conductivity. The more pores of different volumes are contained in the structure of the material, the better the CFT can be obtained.

In modern production, several technologies are used to obtain the porosity of a building material.

In particular, technologies are used:

• foaming;
• gas formation;
• water sealing;
• swelling;
• introduction of additives;
• creating fiber scaffolds.

It should be noted: the coefficient of thermal conductivity is directly related to properties such as density, heat capacity, thermal conductivity.

The thermal conductivity value can be calculated using the formula:

λ \u003d Q / S * (T 1 -T 2) * t,

• Q- The amount of heat;
• S is the thickness of the material;
• T1, T2– temperature on both sides of the material;
• t- time.

The average value of density and thermal conductivity is inversely proportional to the value of porosity. Therefore, based on the density of the building material structure, the dependence of thermal conductivity on it can be calculated as follows:

λ \u003d 1.16 √ 0.0196 + 0.22d 2 - 0.16,

Where: d– density value. This is the formula of V.P. Nekrasov, demonstrating the influence of the density of a particular material on the value of its CFT.

The influence of moisture on the thermal conductivity of building materials

Again, judging by the examples of the use of building materials in practice, it turns out the negative effect of moisture on the CTP of building materials. It has been noticed that the more moisture the building material is exposed to, the higher the value of the CFT becomes.

In various ways, they seek to protect the material used in construction from moisture. This measure is fully justified, given the increase in the coefficient for wet building materials

It is easy to justify this point. The impact of moisture on the structure of the building material is accompanied by air humidification in the pores and partial replacement of the air environment.

Considering that the parameter of the thermal conductivity coefficient for water is 0.58 W/m°C, a significant increase in the CTP of the material becomes clear.

A more negative effect should also be noted, when water entering the porous structure is additionally frozen - it turns into ice.

One of the reasons for the refusal of winter construction in favor of summer construction should be considered precisely the factor of possible freezing of some types of building materials and, as a result, an increase in thermal conductivity.

From here, construction requirements regarding the protection of insulating building materials from moisture ingress become apparent. After all, the level of thermal conductivity increases in direct proportion to the quantitative humidity.

Another point is no less significant - the opposite, when the structure of the building material is subjected to significant heating. Excessively high temperature also provokes an increase in thermal conductivity.

This happens due to an increase in the kinematic energy of the molecules that make up the structural basis of the building material.

True, there is a class of materials, the structure of which, on the contrary, acquires the best properties of thermal conductivity in the regime of strong heating. One of these materials is metal.

If, under strong heating, most of the widely used building materials change the thermal conductivity upwards, strong heating of the metal leads to the opposite effect - the CFT of the metal decreases

Methods for determining the coefficient

Different methods are used in this direction, but in fact all measurement technologies are combined by two groups of methods:

1. Stationary measurement mode.
2. Mode of non-stationary measurements.

The stationary technique involves working with parameters that are unchanged over time or vary slightly. This technology, judging by practical applications, makes it possible to count on more accurate results of QFT.

Actions aimed at measuring thermal conductivity, the stationary method can be carried out in a wide temperature range - 20 - 700 ° C. But at the same time, stationary technology is considered a labor-intensive and complex technique that requires a lot of time for execution.

An example of an apparatus designed to perform measurements of the thermal conductivity coefficient. This is one of the modern digital designs that provides fast and accurate results.

Another measurement technology - non-stationary, seems to be more simplified, requiring from 10 to 30 minutes to complete the work. However, in this case, the temperature range is significantly limited. However, the technique has found wide application in the manufacturing sector.

Table of thermal conductivity of building materials

It makes no sense to measure many existing and widely used building materials.

All these products, as a rule, have been tested repeatedly, on the basis of which a table of thermal conductivity of building materials has been compiled, which includes almost all materials needed at a construction site.

One of the options for such a table is presented below, where KTP is the thermal conductivity coefficient:

Material (building material)	Density, m 3	KTP dry, W/mºC	% humidity_1			% humidity_2	KTP at humidity_1, W/mºC		KTP at humidity_2, W/mºC
Roofing bitumen	1400	0,27	0			0	0,27		0,27
Roofing bitumen	1000	0,17	0			0	0,17		0,17
Roofing slate	1800	0,35	2			3	0,47		0,52
Roofing slate	1600	0,23	2			3	0,35		0,41
Roofing bitumen	1200	0,22	0			0	0,22		0,22
Asbestos-cement sheet	1800	0,35	2			3	0,47		0,52
Asbestos-cement sheet	1600	0,23	2			3	0,35		0,41
asphalt concrete	2100	1,05	0			0	1,05		1,05
roofing roofing	600	0,17	0			0	0,17		0,17
Concrete (on a gravel pad)	1600	0,46	4			6	0,46		0,55
Concrete (on a slag pad)	1800	0,46	4			6	0,56		0,67
Concrete (on gravel)	2400	1,51	2			3	1,74		1,86
Concrete (on a sand cushion)	1000	0,28	9			13	0,35		0,41
Concrete (porous structure)	1000	0,29	10			15	0,41		0,47
Concrete (solid structure)	2500	1,89	2			3	1,92		2,04
pumice stone	1600	0,52	4			6	0,62		0,68
Building bitumen	1400	0,27	0			0	0,27		0,27
Building bitumen	1200	0,22	0			0	0,22		0,22
Lightweight mineral wool	50	0,048	2			5	0,052		0,06
Mineral wool heavy	125	0,056	2			5	0,064		0,07
Mineral wool	75	0,052	2			5	0,06		0,064
Vermiculite sheet	200	0,065	1			3	0,08		0,095
Vermiculite sheet	150	0,060	1			3	0,074		0,098
Gas-foam-ash concrete	800	0,17	15			22	0,35		0,41
Gas-foam-ash concrete	1000	0,23	15			22	0,44		0,50
Gas-foam-ash concrete	1200	0,29	15			22	0,52		0,58
	300	0,08	8			12	0,11		0,13
Gas-foam-concrete (foam-silicate)	400	0,11	8			12	0,14		0,15
Gas-foam-concrete (foam-silicate)	600	0,14	8			12	0,22		0,26
Gas-foam-concrete (foam-silicate)	800	0,21	10			15	0,33		0,37
Gas-foam-concrete (foam-silicate)	1000	0,29	10			15	0,41		0,47
Building plaster board	1200	0,35	4			6	0,41		0,46
Expanded clay gravel	600	2,14	2			3	0,21		0,23
Expanded clay gravel	800	0,18	2			3	0,21		0,23
Granite (basalt)	2800	3,49	0			0	3,49		3,49
Expanded clay gravel	400	0,12	2			3	0,13		0,14
Expanded clay gravel	300	0,108	2			3	0,12		0,13
Expanded clay gravel	200	0,099	2			3	0,11		0,12
shungizite gravel	800	0,16	2			4	0,20		0,23
shungizite gravel	600	0,13	2			4	0,16		0,20
shungizite gravel	400	0,11	2			4	0,13		0,14
Pine wood transverse fibers	500	0,09	15			20	0,14		0,18
Plywood	600	0,12	10			13	0,15		0,18
Pine tree along the grain	500	0,18	15			20	0,29		0,35
Oak wood across the grain	700	0,23	10			15	0,18		0,23
Duralumin metal	2600	221	0			0	221		221
Reinforced concrete	2500	1,69	2			3	1,92		2,04
Tuff concrete	1600	0,52	7			10	0,7		0,81
Limestone	2000	0,93	2			3	1,16		1,28
Lime mortar with sand	1700	0,52	2			4	0,70		0,87
Sand for construction work	1600	0,035	1			2	0,47		0,58
Tuff concrete	1800	0,64	7			10	0,87		0,99
Facing cardboard	1000	0,18	5			10	0,21		0,23
Multilayer construction paper	650	0,13	6			12	0,15		0,18
foamed rubber	60-95	0,034	5			15	0,04		0,054
Expanded clay concrete	1400	0,47	5			10	0,56		0,65
Expanded clay concrete	1600	0,58	5			10	0,67		0,78
Expanded clay concrete	1800	0,86	5			10	0,80		0,92
Brick (hollow)	1400	0,41	1			2	0,52		0,58
Brick (ceramic)	1600	0,47	1			2	0,58		0,64
Construction tow	150	0,05	7			12	0,06		0,07
Brick (silicate)	1500	0,64	2			4	0,7		0,81
Brick (solid)	1800	0,88	1			2	0,7		0,81
Brick (slag)	1700	0,52	1,5			3	0,64		0,76
Brick (clay)	1600	0,47	2			4	0,58		0,7
Brick (triple)	1200	0,35	2			4	0,47		0,52
metal copper	8500	407	0			0	407		407
Dry plaster (sheet)	1050	0,15	4			6	0,34		0,36
Mineral wool slabs	350	0,091	2			5	0,09		0,11
Mineral wool slabs	300	0,070	2			5	0,087		0,09
Mineral wool slabs	200	0,070	2			5	0,076		0,08
Mineral wool slabs	100	0,056	2			5	0,06		0,07
Linoleum PVC	1800	0,38	0			0	0,38		0,38
foam concrete	1000	0,29	8			12	0,38		0,43
foam concrete	800	0,21	8			12	0,33		0,37
foam concrete	600	0,14	8			12	0,22		0,26
foam concrete	400	0,11	6			12	0,14		0,15
Foam concrete on limestone	1000	0,31	12			18	0,48		0,55
Foam concrete on cement	1200	0,37	15			22	0,60		0,66
Expanded polystyrene (PSB-S25)	15 – 25	0,029 – 0,033	2			10	0,035 – 0,052		0,040 – 0,059
Expanded polystyrene (PSB-S35)	25 – 35	0,036 – 0,041	2			20	0,034		0,039
Polyurethane foam sheet	80	0,041	2			5	0,05		0,05
Panel polyurethane foam	60	0,035	2			5	0,41		0,41
Lightweight foam glass	200	0,07	1			2	0,08		0,09
Weighted foam glass	400	0,11	1			2	0,12		0,14
glassine	600	0,17	0			0	0,17		0,17
Perlite	400	0,111	1			2	0,12		0,13
Perlite-cement slab	200	0,041	2			3	0,052		0,06
Marble	2800	2,91	0			0	2,91		2,91
tufa	2000	0,76	3			5	0,93		1,05
Ash gravel concrete	1400	0,47	5			8	0,52		0,58
Fiberboard (chipboard)	200	0,06	10			12	0,07		0,08
Fiberboard (chipboard)	400	0,08	10			12	0,11		0,13
Fiberboard (chipboard)	600	0,11	10			12	0,13		0,16
Fiberboard (chipboard)	800	0,13	10			12	0,19		0,23
Fiberboard (chipboard)	1000	0,15	10			12	0,23		0,29
Polystyrene concrete on Portland cement	600	0,14	4			8	0,17		0,20
Vermiculite concrete	800	0,21	8			13	0,23		0,26
Vermiculite concrete	600	0,14	8			13	0,16		0,17
Vermiculite concrete	400	0,09	8			13	0,11		0,13
Vermiculite concrete	300	0,08	8			13	0,09		0,11
Ruberoid	600	0,17	0			0	0,17		0,17
Fiberboard plate	800	0,16	10			15	0,24		0,30
metal steel	7850	58	0			0	58		58
Glass	2500	0,76	0			0	0,76		0,76
glass wool	50	0,048	2			5	0,052		0,06
Fiberglass	50	0,056	2			5	0,06		0,064
Fiberboard plate	600	0,12	10			15	0,18		0,23
Fiberboard plate	400	0,08	10			15	0,13		0,16
Fiberboard plate	300	0,07	10			15	0,09		0,14
Plywood	600	0,12	10			13	0,15		0,18
Reed plate	300	0,07	10			15	0,09		0,14
Cement-sand mortar	1800	0,58	2			4	0,76		0,93
metal cast iron	7200	50	0			0	50		50
Cement-slag mortar	1400	0,41	2			4	0,52		0,64
Complex sand solution	1700	0,52	2			4	0,70		0,87
Dry plaster	800	0,15	4			6	0,19		0,21
Reed plate	200	0,06	10			15	0,07		0,09
cement plaster	1050	0,15	4			6	0,34		0,36
Peat plate	300	0,064	15			20	0,07		0,08
Peat plate	200	0,052	15			20	0,06		0,064

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.

What affects the thermal conductivity?

Thermal conductivity is determined by such factors:

• porosity determines the heterogeneity of the structure. When heat is passed through such materials, the cooling process is negligible;
• an increased density value affects the close contact of the particles, which contributes to faster heat transfer;
• high humidity increases this indicator.

Use of thermal conductivity values ​​in practice

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.

Thermal conductivity indicators for finished buildings. Types of insulation

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 thermal conductivity index affects the quality of the heat-insulating process;
• moisture absorption is of great importance when insulating external elements;
• thickness affects the reliability of insulation. Thin insulation helps to save the useful area of ​​​​the room;
• flammability is important. High-quality raw materials have the ability to self-extinguish;
• thermal stability reflects the ability to withstand temperature changes;
• environmental friendliness and safety;
• soundproofing protects against noise.

The following types are used as heaters:

• mineral wool is fire resistant and environmentally friendly. Important characteristics include low thermal conductivity;

• Styrofoam is a lightweight material with good insulating properties. It is easy to install and is moisture resistant. Recommended for use in non-residential buildings;
• basalt wool, unlike mineral wool, has better resistance to moisture;
• penoplex is resistant to moisture, high temperatures and fire. It has excellent thermal conductivity, easy to install and durable;

• polyurethane foam is known for such qualities as incombustibility, good water-repellent properties and high fire resistance;
• extruded polystyrene foam undergoes additional processing during production. Has a uniform structure;

• penofol is a multilayer insulating layer. Contains polyethylene foam. The surface of the plate is covered with foil to provide reflection.

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.

Table of thermal conductivity of building materials: features of indicators

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.

How to use the table of thermal conductivity of materials and heaters?

The heat transfer resistance table of materials shows the most popular materials. When choosing a particular thermal insulation option, 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.

Values ​​of heat transfer coefficients of materials in the table

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.

Thermal conductivity of building materials (video)

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When building private and multi-apartment buildings, many factors must be taken into account and a large number of norms and standards must be observed. In addition, before construction, a house plan is created, calculations are made for the load on the supporting structures (foundation, walls, ceilings), communications and heat resistance. The calculation of heat transfer resistance is no less important than the others. It not only determines how warm the house will be, and, as a result, energy savings, but also the strength and reliability of the structure. After all, walls and other elements of it can freeze through. The cycles of freezing and thawing destroy the building material and lead to dilapidated and accident-prone buildings.

Thermal conductivity

Any material can conduct heat. This process is carried out due to the movement of particles, which transmit the change in temperature. The closer they are to each other, the faster the heat transfer process. Thus, denser materials and substances cool or heat up much faster. The intensity of heat transfer primarily depends on the density. It is expressed numerically in terms of the thermal conductivity coefficient. It is denoted by the symbol λ and is measured in W/(m*°C). The higher this coefficient, the higher the thermal conductivity of the material. The reciprocal of the thermal conductivity is the thermal resistance. It is measured in (m2*°C)/W and is denoted by the letter R.

Application of concepts in construction

In order to determine the thermal insulation properties of a building material, the heat transfer resistance coefficient is used. Its value for various materials is given in almost all building guides.

Since most modern buildings have a multilayer wall structure, consisting of several layers of different materials (external plaster, insulation, wall, internal plaster), the concept of reduced heat transfer resistance is introduced. It is calculated in the same way, but in the calculations a homogeneous analogue of a multilayer wall is taken, which transmits the same amount of heat over a certain time and with the same temperature difference inside and outside the room.

The reduced resistance is calculated not for 1 square meter, but for the entire structure or some part of it. It summarizes the thermal conductivity of all wall materials.

Thermal resistance of structures

All external walls, doors, windows, roof are enclosing structures. And since they protect the house from the cold in different ways (they have a different coefficient of thermal conductivity), the heat transfer resistance of the enclosing structure is individually calculated for them. Such structures include internal walls, partitions and ceilings, if there is a temperature difference in the premises. This refers to rooms in which the temperature difference is significant. These include the following unheated parts of the house:

• Garage (if it is directly adjacent to the house).
• Hallway.
• Veranda.
• Pantry.
• Attic.
• Basement.

If these rooms are not heated, then the wall between them and the living quarters must also be insulated, like the outer walls.

Thermal resistance of windows

In the air, the particles that participate in heat exchange are located at a considerable distance from each other, and therefore, the air isolated in a sealed space is the best insulator. Therefore, all wooden windows used to be made with two rows of sashes. Due to the air gap between the frames, the heat transfer resistance of the windows increases. The same principle applies to front doors in a private house. To create such an air gap, two doors are placed at some distance from each other or a dressing room is made.

This principle has remained in modern plastic windows. The only difference is that the high heat transfer resistance of double-glazed windows is achieved not due to the air gap, but due to hermetic glass chambers, from which air is pumped out. In such chambers, the air is discharged and there are practically no particles, which means that there is nothing to transfer the temperature to. Therefore, the thermal insulation properties of modern double-glazed windows are much higher than those of old wooden windows. The thermal resistance of such a double-glazed window is 0.4 (m2*°C)/W.

Modern front doors for private houses have a multilayer structure with one or more layers of insulation. In addition, additional heat resistance is provided by the installation of rubber or silicone seals. Thanks to this, the door becomes practically airtight and the installation of a second one is not required.

Calculation of thermal resistance

The calculation of the heat transfer resistance allows you to estimate the heat loss in W and calculate the necessary additional insulation and heat loss. Thanks to this, you can correctly select the required power of heating equipment and avoid unnecessary spending on more powerful equipment or energy carriers.

For clarity, we calculate the thermal resistance of the wall of a house made of red ceramic bricks. Outside, the walls will be insulated with extruded polystyrene foam 10 cm thick. The thickness of the walls will be two bricks - 50 cm.

Heat transfer resistance is calculated using the formula R = d/λ, where d is the thickness of the material and λ is the thermal conductivity of the material. From the building guide it is known that for ceramic bricks λ = 0.56 W / (m * ° C), and for extruded polystyrene foam λ = 0.036 W / (m * ° C). Thus, R (brickwork) \u003d 0.5 / 0.56 \u003d 0.89 (m 2 * ° C) / W, and R (extruded polystyrene foam) \u003d 0.1 / 0.036 \u003d 2.8 (m 2 * °C)/W. In order to find out the total thermal resistance of the wall, you need to add these two values: R \u003d 3.59 (m 2 * ° C) / W.

Table of thermal resistance of building materials

All the necessary information for individual calculations of specific buildings is given by the heat transfer resistance table below. The example of calculations given above, in conjunction with the data in the table, can also be used to estimate the loss of thermal energy. To do this, use the formula Q \u003d S * T / R, where S is the area of ​​\u200b\u200bthe building envelope, and T is the temperature difference between the street and the room. The table shows the data for a wall with a thickness of 1 meter.

Material	R, (m 2 * °C) / W
Reinforced concrete	0,58
Expanded clay blocks	1,5-5,9
ceramic brick	1,8
silicate brick	1,4
Aerated concrete blocks	3,4-12,29
Pine	5,6
Mineral wool	14,3-20,8
Styrofoam	20-32,3
Extruded polystyrene foam	27,8
polyurethane foam	24,4-50

Warm designs, methods, materials

In order to increase the resistance to heat transfer of the entire structure of a private house, as a rule, building materials with a low coefficient of thermal conductivity are used. Thanks to the introduction of new technologies in the construction of such materials is becoming more and more. Among them are the most popular:

• Tree.
• Sandwich panels.
• ceramic block.
• Expanded clay block.
• Aerated concrete block.
• Foam block.
• Polystyrene concrete block, etc.

Wood is a very warm, environmentally friendly material. Therefore, many in the construction of a private house opt for it. It can be either a log house, or a rounded log or a rectangular beam. The material used is mainly pine, spruce or cedar. However, this is a rather capricious material and requires additional measures to protect against weathering and insects.

Sandwich panels are a fairly new product on the domestic building materials market. Nevertheless, its popularity in private construction has increased greatly in recent years. After all, its main advantages are a relatively low cost and good resistance to heat transfer. This is achieved through its structure. On the outside there is a rigid sheet material (OSB boards, plywood, metal profiles), and inside - foamed insulation or mineral wool.

Building blocks

The high resistance to heat transfer of all building blocks is achieved due to the presence of air chambers or a foam structure in their structure. So, for example, some ceramic and other types of blocks have special holes that, when laying the wall, run parallel to it. Thus, closed chambers with air are created, which is a fairly effective measure of preventing heat transfer.

In other building blocks, the high resistance to heat transfer lies in the porous structure. This can be achieved by various methods. In foam concrete aerated concrete blocks, a porous structure is formed due to a chemical reaction. Another way is to add a porous material to the cement mixture. It is used in the manufacture of polystyrene concrete and expanded clay concrete blocks.

The nuances of the use of heaters

If the heat transfer resistance of the wall is insufficient for the given region, then insulation can be used as an additional measure. Wall insulation, as a rule, is carried out outside, but if necessary, it can also be applied on the inside of load-bearing walls.

Today, there are many different heaters, among which the most popular are:

• Mineral wool.
• Polyurethane foam.
• Styrofoam.
• Extruded polystyrene foam.
• Foam glass, etc.

All of them have a very low coefficient of thermal conductivity, therefore, for the insulation of most walls, a thickness of 5-10 mm is usually sufficient. But at the same time, one should take into account such a factor as the vapor permeability of the insulation and wall material. According to the rules, this indicator should increase outwards. Therefore, the insulation of walls made of aerated concrete or foam concrete is possible only with the help of mineral wool. Other heaters can be used for such walls if a special ventilation gap is made between the wall and the heater.

Conclusion

The thermal resistance of materials is an important factor to be considered in construction. But, as a rule, the warmer the wall material, the lower the density and compressive strength. This should be taken into account when planning a house.

Methodological material for self-calculation of the thickness of the walls of the house with examples and a theoretical part.

Part 1. Heat transfer resistance - the primary criterion for determining the thickness of the wall

In order to determine the thickness of the wall, which is necessary to comply with energy efficiency standards, the heat transfer resistance of the designed structure is calculated, in accordance with section 9 "Methodology for designing thermal protection of buildings" SP 23-101-2004.

Heat transfer resistance is a property of a material that indicates how heat is retained by a given material. This is a specific value that shows how slowly heat is lost in watts when a heat flux passes through a unit volume with a temperature difference of 1°C on the walls. The higher the value of this coefficient, the “warmer” the material.

All walls (non-translucent enclosing structures) are considered for thermal resistance according to the formula:

R \u003d δ / λ (m 2 ° C / W), where:

δ is the thickness of the material, m;

λ - specific thermal conductivity, W / (m · ° С) (can be taken from the passport data of the material or from tables).

The resulting value of Rtotal is compared with the tabular value in SP 23-101-2004.

To focus on the regulatory document, it is necessary to calculate the amount of heat required to heat the building. It is performed according to SP 23-101-2004, the resulting value is "degree day". The rules recommend the following ratios.

wall material		Heat transfer resistance (m 2 °C / W) / application area (°C day)
structural	heat-insulating	Double-layer with external thermal insulation	Three-layer with insulation in the middle	With non-ventilated atmospheric layer	With ventilated atmospheric layer
Brickwork	Styrofoam
	Mineral wool
Expanded clay concrete (flexible links, dowels)	Styrofoam
	Mineral wool
Aerated concrete blocks with brick cladding	Cellular concrete
Note. In the numerator (before the line) - the approximate values ​​​​of the reduced resistance to heat transfer of the outer wall, in the denominator (behind the line) - the limiting degrees-days of the heating period, at which this wall structure can be applied.

The results obtained must be verified with the norms of clause 5. SNiP 23-02-2003 "Thermal protection of buildings".

You should also take into account the climatic conditions of the zone where the building is being built: different regions have different requirements due to different temperature and humidity conditions. Those. the thickness of the gas block wall should not be the same for the coastal region, central Russia and the far north. In the first case, it will be necessary to correct the thermal conductivity taking into account the humidity (upward: increased humidity reduces the thermal resistance), in the second case, you can leave it “as is”, in the third case, be sure to take into account that the thermal conductivity of the material will increase due to a larger temperature difference.

Part 2. Thermal conductivity of wall materials

The coefficient of thermal conductivity of wall materials is this value, which shows the specific thermal conductivity of the wall material, i.e. how much heat is lost when a heat flux passes through a conditional unit volume with a temperature difference on its opposite surfaces of 1°C. The lower the value of the coefficient of thermal conductivity of the walls - the warmer the building will turn out, the higher the value - the more power will have to be put into the heating system.

In fact, this is the reciprocal of the thermal resistance discussed in part 1 of this article. But this applies only to specific values ​​for ideal conditions. The real thermal conductivity coefficient for a particular material is affected by a number of conditions: temperature difference on the walls of the material, internal heterogeneous structure, humidity level (which increases the density level of the material, and, accordingly, increases its thermal conductivity) and many other factors. As a rule, tabular thermal conductivity must be reduced by at least 24% to obtain an optimal design for temperate climates.

Part 3. The minimum allowable value of wall resistance for various climatic zones.

The minimum allowable thermal resistance is calculated to analyze the thermal properties of the designed wall for various climatic zones. This is a normalized (basic) value, which shows what the thermal resistance of the wall should be, depending on the region. First, you choose the material for the structure, calculate the thermal resistance of your wall (part 1), and then compare it with the tabular data contained in SNiP 23-02-2003. If the value obtained turns out to be less than that established by the rules, then it is necessary either to increase the thickness of the wall, or to insulate the wall with a heat-insulating layer (for example, mineral wool).

According to paragraph 9.1.2 of SP 23-101-2004, the minimum allowable heat transfer resistance R o (m 2 ° C / W) of the enclosing structure is calculated as

R o \u003d R 1 + R 2 + R 3, where:

R 1 \u003d 1 / α ext, where α ext is the heat transfer coefficient of the inner surface of the enclosing structures, W / (m 2 × ° С), taken according to table 7 of SNiP 23-02-2003;

R 2 \u003d 1 / α ext, where α ext is the heat transfer coefficient of the outer surface of the enclosing structure for the conditions of the cold period, W / (m 2 × ° С), taken according to table 8 of SP 23-101-2004;

R 3 - total thermal resistance, the calculation of which is described in part 1 of this article.

If there is a layer in the enclosing structure ventilated by outside air, the layers of the structure located between the air layer and the outer surface are not taken into account in this calculation. And on the surface of the structure facing towards the layer ventilated from the outside, the heat transfer coefficient α external should be taken equal to 10.8 W / (m 2 · ° С).

Table 2. Normalized values ​​of thermal resistance for walls according to SNiP 23-02-2003.

The updated values ​​of the degree-days of the heating period are indicated in Table 4.1 of the reference manual to SNiP 23-01-99 * Moscow, 2006.

Part 4. Calculation of the minimum allowable wall thickness on the example of aerated concrete for the Moscow region.

When calculating the thickness of the wall structure, we take the same data as indicated in Part 1 of this article, but rebuild the basic formula: δ = λ R, where δ is the wall thickness, λ is the thermal conductivity of the material, and R is the heat resistance norm according to SNiP.

Calculation example the minimum wall thickness of aerated concrete with a thermal conductivity of 0.12 W / m ° C in the Moscow region with an average temperature inside the house during the heating season + 22 ° C.

1. We take the normalized thermal resistance for walls in the Moscow region for a temperature of + 22 ° C: R req \u003d 0.00035 5400 + 1.4 \u003d 3.29 m 2 ° C / W
2. The coefficient of thermal conductivity λ for aerated concrete grade D400 (dimensions 625x400x250 mm) at a humidity of 5% = 0.147 W/m∙°C.
3. Minimum wall thickness of aerated concrete stone D400: R λ = 3.29 0.147 W/m∙°С=0.48 m.

Conclusion: for Moscow and the region, for the construction of walls with a given thermal resistance parameter, an aerated concrete block with a width of at least 500 mm is needed, or a block with a width of 400 mm and subsequent insulation (mineral wool + plastering, for example), to ensure the characteristics and requirements of SNiP in terms of energy efficiency of wall structures.

Table 3. The minimum thickness of walls erected from various materials that meet the standards of thermal resistance according to SNiP.

Material	Wall thickness, m	conductivity,
Expanded clay blocks			For the construction of load-bearing walls, a grade of at least D400 is used.
cinder blocks
silicate brick
Gas silicate blocks d500			I use a brand from D400 and higher for housing construction
Foam block			frame construction only
Cellular concrete			The thermal conductivity of cellular concrete is directly proportional to its density: the “warmer” the stone, the less durable it is.
			Minimum wall size for frame structures
Solid ceramic brick
Sand-concrete blocks			At 2400 kg/m³ under conditions of normal temperature and air humidity.

Part 5. The principle of determining the value of heat transfer resistance in a multilayer wall.

If you plan to build a wall from several types of material (for example, building stone + mineral insulation + plaster), then R is calculated for each type of material separately (using the same formula), and then summed up:

R total \u003d R 1 + R 2 + ... + R n + R a.l where:

R 1 -R n - thermal resistance of various layers

R a.l - resistance of a closed air gap, if it is present in the structure (table values ​​are taken in SP 23-101-2004, p. 9, table 7)

An example of calculating the thickness of a mineral wool insulation for a multilayer wall (cinder block - 400 mm, mineral wool - ? mm, facing brick - 120 mm) with a heat transfer resistance value of 3.4 m 2 * Deg C / W (Orenburg).

R \u003d R cinder block + R brick + R wool \u003d 3.4

R cinder block \u003d δ / λ \u003d 0.4 / 0.45 \u003d 0.89 m 2 × ° C / W

Rbrick \u003d δ / λ \u003d 0.12 / 0.6 \u003d 0.2 m 2 × ° C / W

R cinder block + R brick \u003d 0.89 + 0.2 \u003d 1.09 m 2 × ° C / W (<3,4).

Rwool \u003d R- (R cinder block + R brick) \u003d 3.4-1.09 \u003d 2.31 m 2 × ° C / W

δvata = Rwool λ = 2.31 * 0.045 = 0.1 m = 100 mm (we take λ = 0.045 W / (m × ° С) - the average value of thermal conductivity for mineral wool of various types).

Conclusion: in order to comply with the requirements for heat transfer resistance, expanded clay concrete blocks can be used as the main structure, lined with ceramic bricks and a layer of mineral wool with a thermal conductivity of at least 0.45 and a thickness of 100 mm.

Questions and answers on the topic

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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.

What is thermal conductivity and thermal resistance

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.

Table of thermal conductivity of thermal insulation materials

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 differ significantly from manufacturer to manufacturer, so when buying, pay attention to the characteristics of each material you buy.

Table of thermal conductivity of building materials

Walls, ceilings, 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

How to calculate wall thickness

In order for the house to be warm in winter and cool in summer, it is necessary that the building envelope (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.

Calculation of wall thickness, insulation thickness, finishing layers

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.

An example of calculating the thickness of the insulation

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.

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