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What should be the thermal conductivity of the wall in the house. Thermal insulation thickness calculator online

Everyone who builds a house or is going to carry out repairs asks the question: how thick should the walls be, what thermal insulation and what kind of insulation is best to use.

It is the answers to these questions that will make any house or apartment cozy, comfortable and convenient for living.

Again, the use of low-quality materials and in insufficient quantities, ignoring insulation, as such, can lead to very sad consequences.

In such a house it will simply be difficult to live both in the heat and in the cold. The temperature in the rooms will differ little from the temperature outside.

Therefore, you should find out what thickness the thermal insulation should be specifically for your case.

How best to proceed

Today, you can do it yourself: make the necessary calculations, find out the best materials for work and install them yourself.

You can prefer work to ordering a large company that can, for a fee, make an accurate calculation, select materials and proceed with their installation.

Of course, if you do everything yourself, there will be no one to make claims.

In the case of a company, you can complain about poor quality, dishonest work, or when the required effect from the work performed has not been achieved.

To calculate the thermal conductivity of the wall, you can use special programs, specialized online calculators that will help you get the right numbers.

Or you can do it yourself. Many are mistaken, thinking that they themselves are not able to make calculations, calculate how much thermal insulation for work will be needed for a room, apartment or house. This is extremely simple to do, because it is quite simple to calculate the thickness of the required thermal insulation: manufacturers indicate the coefficient of thermal conductivity on all materials.

Why is it necessary to calculate thermal conductivity and install thermal insulation

As already mentioned, there are a number of reasons for this:

the absence or insufficiency of thermal insulation will lead to freezing of the walls;
there is a possibility of transferring the so-called dew point, which, in turn, will cause condensation on the walls, add excessive humidity to the premises;
in hot weather indoors will be worse than under the bright sun on the street; in such houses it will be hot, stuffy and uncomfortable.

Again, the above reasons will bring you new problems: the same humidity will contribute to spoilage as used indoors. building materials as well as furniture and appliances. This, in turn, will force you to spend money on repairs, upgrades, and the purchase of new things. An example of this can be easily seen below.

So thermal insulation is a guarantee of the safety of your money in the future.

How to calculate the thickness of thermal insulation

To calculate the required thickness, you should know the value of heat resistance, which is constant, the value is different, depending on geographical location, which is different for each individual region. We take the following indicators as a basis: the heat resistance of the walls is 3.5m 2 * K / W, and the ceiling is 6m 2 * K / W. Let's call the first value R1, and the second, respectively, R2.

When calculating walls or a ceiling or a floor consisting of more than one layer, the thermal resistance of each of them should be calculated and then summarized.

R= R+R1+R2 etc.

Accordingly, the required thickness of the thermal insulation, its layer, will be obtained by the following manipulations and using the formulas:

R=p/k, where p is the layer thickness and k is the thermal conductivity of the material, which can be obtained from the manufacturer.

Again, do not forget, if there are several layers, then each should be calculated using this formula, and then the results should be summarized.

An example of such calculations

There is nothing complicated in this process, you can easily carry out the calculation for any material. As an example, we can take the calculation for a brick house.

Let's say the thickness of the measured walls will be 1.5 brick lengths, and we will decide to use mineral wool as thermal insulation.

So, we need a wall thermal resistance of at least 3.5. To start the calculation, we need to know the current thermal resistance of this brick wall.

The thickness is about 38 centimeters, the thermal conductivity coefficient is 0.56.

Accordingly, 0.38 / 0.56 = 0.68. To reach a figure of 3.5, we subtract the result obtained from it (we need 2.85 square meters * K / W).

Now we will calculate the thickness of the thermal insulation, as mentioned above, of mineral wool: 2.85 * 0.045 = 0.128

Let us round the result a little and get the following: if necessary, insulate a brick wall, one and a half bricks thick, we will need a thickness of 130mm of thermal insulation material, provided that we use mineral wool. Considering the upcoming internal and external work, both finishing and decorative, you can afford a 100mm mineral wool layer. As you can see, nothing complicated.

What else will give such a calculation

Using this calculation, you can compare different types of insulation and thermal insulation, you can choose the most effective with the smallest layer.

If you have a problem in space, if you want to save money, then such work will allow you, through simple manipulations, to quickly find out which material will cost you less.

If you are still at the stage of planning a house, you can figure out what will cost you less and less labor. This may be an increase in the thickness of the brickwork, the use of other types thermal insulation materials or the use of other building materials to build a wall, say, instead of bricks, use blocks, etc.

Many are too lazy to do the calculations on their own, in this case, you can easily afford to use the calculators that are offered on the net on many pages.

Here you will find a lot of templates and blanks, almost all the information is collected in reference books, you will only need to substitute the type of building materials, region of residence and thickness indicator. In this case, all calculations will occur very quickly and easily.

But in this case, there is a high probability that this or that site is cheating: they are trying to put the material that is being traded in the best light. In this case, a calculation error is likely, which can cost you dearly.

Do not be afraid of independent calculations, for this you only need a pen, paper and a calculator.

You can easily double-check your calculations at any time or show them to a specialist. Consulting with a familiar builder will be much cheaper than hiring a professional company.

Again, when choosing materials, calculating the required thickness and price for them, consider other beneficial features which you may be interested in.

For example, fire safety, sound insulation, water or moisture resistance. For example, glass wool has sound insulation and thermal insulation.

Yes, unfortunately, such materials will come out a little more expensive, but still, a price difference of 10-20%, given that you get, say, not only thermal insulation, but also sound insulation, should be called a good purchase and a good solution .

Video - calculation of the thermal conductivity of the wall

In this video, you can see firsthand how the thermal conductivity of the wall is calculated using a specialized program.

Instruction

The determination of the thermal conductivity of materials is carried out through the coefficient of thermal conductivity, which is a measure of the ability to transmit heat flow. The lower the value of this indicator, the higher the insulating properties of the material. In this case, the thermal conductivity does not depend on the density.

Numerically, the value of thermal conductivity is equal to the amount of thermal energy that passes through a section of material 1 m thick and 1 sq. m in area in 1 second. In this case, the temperature difference on opposite surfaces is assumed to be 1 Kelvin. The amount of heat is the energy that a material gains or loses when heat is transferred.

The formula for thermal conductivity is as follows: Q = λ*(dT/dx)*S*dτ, where: Q is the thermal conductivity; λ is the thermal conductivity; (dT/dx) is the temperature gradient; S is the cross-sectional area.

When calculating the thermal conductivity of a building structure, it is divided into components and their thermal conductivity is summarized. This allows you to determine the measure of the ability of a house structure (walls, roofs, windows, etc.) to pass heat flow. In fact, the thermal conductivity of a building structure is the combined thermal conductivity of its materials, including air gaps and a film of outside air.

Based on the value of the thermal conductivity of the structure, the volume of heat loss through it is determined. This value is obtained by multiplying the thermal conductivity by the calculated time interval, total area surface, as well as the temperature difference between the outer and inner surfaces of the structure. For example, for a wall with an area of 10 square meters with a thermal conductivity of 0.67 at a temperature difference of 13 °, heat loss in 5 hours will be 0.67 * 5 * 10 * 13 = 435.5 J * m.

Thermal conductivity coefficients various materials are contained in the thermal conductivity table, for example, for vacuum it is 0, and for silver, one of the most thermally conductive materials, 430 W / (m * K).

During construction, along with the thermal conductivity of materials, the phenomenon of convection, which is observed in materials in liquid and gaseous states, should be taken into account. This is especially true when developing a water heating and aeration system. To reduce heat loss in these cases, transverse partitions made of felt, wool and other insulating materials are installed.

When installing heating devices in residential buildings, industrial and office buildings, it is often necessary to know the volume systems heating. It is good when the customer provides such data, but this does not always happen. There are methods for estimating the total volume systems and its individual components depending on the power.

Instruction

To calculate the volume of coolant in heating system when replacing or reconstructing it, use the special calculation tables available in reference books. So, one section of aluminum radiators has a coolant volume of 0.45 liters, a section of new cast iron batteries- 1 l., section of old cast-iron batteries - 1.7 l. In one running meter of a pipe with a diameter of 15 mm - 0.177 liters of coolant, and if pipes with a diameter of, for example, 32 mm are used, then the volume will be 0.8 liters and so on.

One of the common cases when you want to find out the volume systems heating- installation of an expansion tank and make-up pumps. Total volume systems heating at the same time, calculate by adding the volumes of the boiler, heating appliances (radiators) and the pipeline part systems according to the formula: V = (VS x E) / d, where V is the volume of the expansion tank; VS - total volume systems(boiler, radiators, pipes, heat exchangers, etc.); E is the expansion coefficient of the liquid (in percent); d is the efficiency of the expansion tank.

When calculating, take into account such a factor as the expansion of the liquid. For water systems heating it is approximately 4%. If ethylene glycol is used in the system, the expansion coefficient will be approximately 4.4%.

For a less accurate volume calculation systems heating use the power based formula: 1 kW = 15 hp. For such approximate calculations, you need to know the power systems heating, while the need to calculate in detail the volume of pipelines, radiators, the boiler itself and other elements systems disappears. Example: if the heating power for a residential building is 50 kW, then the total volume systems heating VS is calculated as follows: VS \u003d 15 x 50 \u003d 750 liters.

When making calculations, keep in mind that in the case of application in the system heating new and modern radiators and pipes volume systems will be somewhat smaller. Detailed information can be found in the technical documentation of the equipment manufacturer.

Sources:

Calculation of membrane expansion tanks
"Designer's Handbook", I.G. Staroverov, 1990
heating volume

Timber beams are the most economical option for a home. They are very easy to install and to manufacture. Compared to reinforced concrete and steel beams wood has low thermal conductivity. But any beams should be carefully calculated and installed.

You will need

- ruler;
- calculator;
- planer.

Instruction

Calculate the bending strength of the section, given the ratio of 5:7, which means - if you take in height beam 7 measures, then you need to take 5 measures in width. A beam with this ratio will be very strong in both torsion and bending. Be aware: if you take the width more than the height of the beam, then an excessive deflection will appear. If you take it the other way around, then there will be a bend to the side.

Permissible beam deflection floors count based on this ratio - 1/200 or 1/300 of the length of the beam. For example, if you take beam , whose length is 600 meters, then after calculations we get that the deflection is 2 or 3 centimeters.

Sharpen with a planer beam from the side where the beam faces down, by the amount of allowable deflection. That is, give it a kind of arch appearance. So you will ensure that the ceiling will not go down with a “bubble”, because in the middle the beam will become thinner, and everything will remain the same at the edges.

Install beam - you can immediately see that it is arched to the top with an arc. This will not be constant, from the action of beam loads floors straighten up.

Take into account the own weight of the beam, because it also gives a load. For interfloor floors, choose beams with a weight load of 190 kg / m2, but not more than 220 kg / m2, operational (temporary) load - 200 kg / m2. Lay beams floors along the span section, which is shorter. The installation step is equal to the installation step of the frame racks.

Thermal conductivity characteristics
popular building materials

brick houses

A brick house is a reliable, durable home and is popular with our fellow citizens. Its strength and resistance to adverse environmental factors is due to the high density of the material.

brick walls retain heat well, but still require constant heating of the premises. Otherwise, in winter, the brick absorbs moisture and, under the weight of the masonry, begins to collapse. If you keep it for a long time brick house without heating, it will have to warm up to normal temperature for about three days.

Cons of brick buildings:

High heat transfer and the need for additional thermal insulation. Without a heat-insulating layer, the thickness of a brick wall capable of retaining heat must be at least 1.5 m.
Impossibility of periodic (seasonal) use of the building. Brick walls absorb heat and moisture well. In the cold season, a complete warm-up of the house will take at least three days, and it will take at least a month to completely eliminate excess moisture.
A thick cement-sand joint, fastening brickwork, has a three times higher thermal conductivity compared to brick. Accordingly, heat loss through masonry joints is even more significant than through the brick itself.

Technology warm home brick requires additional insulation from the outside of the wall with insulation boards.

wooden houses

A comfortable atmosphere is created faster in a house built of wood. This material practically does not cool and does not heat up, so the temperature inside the room quickly stabilizes. With sufficient wall thickness, houses made of timber or logs can not be insulated, since the tree itself can serve as thermal insulation.

However, in order to wooden house was warm, the thickness of the outer walls of solid wood should be more than 40 cm, of glued laminated timber 35-40 cm, and of logs of more than 50 cm. The cost of building such housing is very high. It remains either to ignore modern requirements and build a house, for example, from glued laminated timber with a minimum thickness of 20-22 cm or from logs with a diameter of 24-28 cm (while understanding that heating costs will be quite high, especially if there is no main gas in the house) or walls wooden house still need to be warmed up.

For people who put comfort and expediency in the first place, it is better to think about warming a wooden house. Then the tree will create in the house optimal microclimate, and insulation will save on heating. Compared to brick, the heat loss of a wooden house is much less. But still, in order to warm house made of wood was also economical, it required additional thermal insulation.

Frame houses

According to its characteristics, the frame construction technology looks much better than a brick or wooden house and does not require additional insulation. If in the climate zone where construction is planned country house, there are low temperatures in winter, then frame technology is the most ideal option.

The technology of frame housing construction implies a layer of thermal insulation inside the walls, which allows you to protect the premises from the outside cold. The big advantage of building a frame house, in comparison with a wooden or brick one, is high energy efficiency with a very small wall thickness.

This technology allows you to build objects that are completely different in their functional purpose:

Frame houses for seasonal living.
For example, frame-panel houses, houses from SIP panels and other "economy" options, used mainly
like summer cottages.

Warm frame houses for permanent residence.
For example, buildings on a monolithic foundation, with wall insulation of at least 200 mm, with internal utilities.

Buildings using 3D frame technology are not only the warmest frame houses for permanent residence, but are also leaders in energy efficiency. In this, the opinions of many experts agree: The 3D frame has an exceptional ability to retain heat, has the parameters of a "passive house" and is recommended for use throughout our country as energy-efficient housing.

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 navigate to normative 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
Brickwork	Mineral wool
Expanded clay concrete (flexible links, dowels)	Styrofoam
Expanded clay concrete (flexible links, dowels)	Mineral wool
Aerated concrete blocks with brick cladding	Cellular concrete
Note. In the numerator (before the line) - indicative values 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 design 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: for different regions different requirements due to different temperature and humidity conditions. Those. the thickness of the gas block wall should not be the same for the seaside area, middle lane 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 average temperature inside the house during the heating season + 22 ° С.

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
The coefficient of thermal conductivity λ for aerated concrete grade D400 (dimensions 625x400x250 mm) at a humidity of 5% = 0.147 W/m∙°C.
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 size walls 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

δwool = Rwool λ = 2.31 * 0.045 = 0.1 m = 100 mm (we take λ = 0.045 W / (m × ° C) - 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

No questions have been asked for the material yet, you have the opportunity to be the first to do so

To build a warm house - a heater is required. Nobody is objecting to this. In modern conditions, it is impossible to build a house that meets the requirements of SNiP without the use of insulation.

That is, a wooden or brick house, of course, is possible to build. And they build everything. However, in order to comply with the requirements of Building Codes and Rules, its coefficient of resistance to heat transfer of walls R must be at least 3.2. And this is 150 cm.

Why, one wonders, to build a "fortress wall" of one and a half meters, when it is possible to use only 15 cm of highly efficient insulation - basalt wool or foam plastic to obtain the same indicator R = 3.2?

And if you do not live in the Moscow region, but in the Novosibirsk region or in the Khanty-Mansiysk Autonomous Okrug? Then for you the heat transfer resistance coefficient for the walls will be different. What? See table.

Table 4. Rated resistance to heat transfer SNiP 23-02-2003 (document text):

We carefully watch and comment. If something is not clear, ask questions through or write to the site editor - the answer will be in your e-mail or in the NEWS section.

So, in this table we are interested in two types of premises - residential and domestic. Residential premises, this, of course, is in a residential building, which must comply with the requirements of SNiP. And household premises are insulated and heated baths, a boiler room and a garage. Sheds, pantries and other outbuildings are not subject to insulation, which means that there are no indicators for the thermal resistance of walls and ceilings for them.

All requirements governing the reduced resistance to heat transfer according to SNiP are divided by region. The regions differ from each other in the duration of the heating season in the cold season and the extreme negative temperatures.

A table showing the degree-days of the heating season for all the main cities of Russia can be seen at the end of the material (Appendix 1).

For example, the Moscow region belongs to the region with D = 4000 degree-days of the heating period. For this region, the following indicators of SNiP heat transfer resistance (R) are established:

Walls = 2.8
Ceilings (floor of the 1st floor, attic or attic ceiling) = 3.7
Windows and doors = 0.35

To make, we use the calculation formula and used in construction. All these materials are available on our website - available by clicking on the links.

With calculations for the cost of insulation, everything is extremely simple. We take the resistance of the wall to heat transfer and select such a heater that, with its minimum thickness, will suit us according to the budget and fit into the requirements of SNiP 23-02-2003.

We are now looking at the degree-days of the heating season for your city where you live. If you live not in the city, but nearby, you can use values 2-3 degrees higher, since the actual winter temperature in large cities is 2-3 degrees higher than in the region. This is facilitated by large heat losses in heating mains and the release of heat into the atmosphere by thermal power plants.

Table 4.1. Degree-days of the heating season for the main cities of the Russian Federation (Appendix 1):

To use this table in calculations where the normalized resistance to heat transfer appears, you can take the average values of the internal temperature of the premises at + 22C.

But here, as they say, the taste and color - someone likes to be warm and sets the regulator in his air to + 24C. And someone is used to living in a cooler house and keeps the room temperature at + 19C. As you can see, the cooler the constant temperature in the room, the less gas or wood you use to heat your home.

By the way, doctors tell us that living in a house at a temperature of +19C is much more beneficial than at +24C.

What should be the thermal conductivity of the wall in the house. Thermal insulation thickness calculator online

How best to proceed

Why is it necessary to calculate thermal conductivity and install thermal insulation

How to calculate the thickness of thermal insulation

An example of such calculations

What else will give such a calculation

Video - calculation of the thermal conductivity of the wall

Thermal conductivity characteristics popular building materials

brick houses

wooden houses

Frame houses

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

Part 2. Thermal conductivity of wall materials

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

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

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

Questions and answers on the topic

Thermal conductivity characteristics
popular building materials