Good day!

I decided to post here the results of calculations for floor insulation on the ground. The calculations were carried out in the Therm 6.3 program.

Floor on the ground - concrete slab thickness 250mm with a coefficient of thermal conductivity of 1.2
Walls - 310 mm with a thermal conductivity coefficient of 0.15 (aerated concrete or wood)
For simplicity, wall to ground. There can be many options for insulation and cold bridges of the node, for simplicity we omit them.
Soil - with a coefficient of thermal conductivity 1. Wet clay or wet sand. Dry ones are more heat-shielding.

Warming. Here are 4 options:
1. There is no insulation. Just a slab on the ground.
2. Insulated blind area 1m wide, 10cm thick. Thermal insulation with EPS. The top layer of the blind area itself was not taken into account, since it does not play a big role.
3. The basement tape is insulated at a depth of 1m. Insulation is also 10cm, EPS. The concrete is not drawn as it is close to the ground in terms of thermal conductivity.
4. The slab under the house is insulated. 10cm, EPS.

The thermal conductivity coefficient of EPS was taken equal to 0.029.
The slab width is taken as 5.85 m.

Initial data on temperatures:
- inside +21;
- outside -3;
- at a depth of 6m +3.

6m here is the GWL estimate. I took 6m because it is the closest to my house option, although I do not have flooring on the ground, but the results also apply to my warm underground.

You can see the results graphically. Attached in two versions - with isotherms and "IK".

The digital data for the floor surface is obtained in the form of U-factor, the value inverse to our resistance to heat transfer ([R] = K * m2 / W).

Converted, the results are as follows (on average by gender):

1.R = 2.86
2.R = 3.31
3. R ​​= 3.52
4.R = 5.59

These are very interesting results for me. In particular a sufficiently high value for the 1st option indicates that it is not so necessary to insulate the slab along the floor in any way. It is necessary to insulate the soil when it is near groundwater and then we have option 4, with the soil partially cut off from the heat circuit. Moreover, with a close GWL we will not get 5.59. since the 6m of soil taken in the calculation does not participate in insulation. You should wait for R ~ 3 in this case or so.

It is also very significant that the edge of the slab in the design version is rather warm 17.5oC according to the first non-insulated version, therefore, freezing, condensation and mold are not expected there, even if the temperature gradient doubles (-27 outside). Moreover, it should be understood that in such calculations, peak temperatures do not play any role, since the system is very heat-intensive and the soil freezes for weeks or months.

Options 1,2,3. And especially option 2 is the most inertial. Here, the soil is involved in the thermal circuit, not only the one directly under the house, but also under the blind area. The time to establish the temperature regime as in the figure is years and in fact temperature regime will be the average for the year. A period of about 3 months only manages to involve 2-3 m of soil in heat exchange. But this is a separate story, so for now I will finish it, just note that the characteristic time is proportional to the thickness of the layer squared. Those. if 2m is 3 months, then 4m is already 9 months.

I also note that in practice, probably, with a relatively small GWL (type 4.5 m and below), one should expect worse results of the thermal insulation properties of the soil due to the evaporation of water from it. Unfortunately, I am not familiar with the tool that could carry out the calculation under conditions of evaporation in the soil. And there is a big problem with the initial data.

The assessment with the influence of evaporation in the ground was carried out as follows.
Dug up data that water in loams rises by capillary forces from the GWL by 4-5m

Well, I will use this figure as the initial data.
I will impudently assume that the same 5m remain in my calculation under any circumstances.
In 1m of soil, steam diffuses to the floor, and the value of the vapor permeability coefficient can be dug. Vapor permeability coefficient of sand 0.17, adobe 0.1. Well, for reliability, I'll take 0.2 mg / m / h / Pa.
At a depth of one meter in the calculated variants, except for variant 4, about 15 degrees.
The total water vapor pressure there is 1700 Pa (100% rel).
Indoors we take 21 degrees 40% (rel.) => 1000 Pa
In total, we have a 700Pa vapor pressure gradient per 1m of clay with Mu = 0.2 and 0.25m of concrete with Mu = 0.09
The final vapor permeability of the two-layer is 1 / (1 / 0.2 + 0.25 / 0.09) = 0.13
As a result, we have a steam flow from the soil 0.13 * 700 = 90 mg / m2 / h = 2.5e-8 kg / m2 / s
We multiply by the heat of evaporation of water 2.3 MJ / kg and we get additional heat loss for evaporation => 0.06 W / m2. It's little things. If we speak in the language of R (resistance to heat transfer), then such accounting for moisture leads to a decrease in R by about 0.003, i.e. unimportant.

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Despite the fact that heat losses through the floor of most one-story industrial, administrative and residential buildings rarely exceed 15% of the total heat losses, and with an increase in the number of stories, sometimes they do not even reach 5%, the importance of solving the problem correctly ...

The definition of heat loss from the air of the first floor or basement to the ground does not lose its relevance.

This article discusses two options for solving the problem posed in the title. Conclusions - at the end of the article.

Considering heat loss, one should always distinguish between the concepts of “building” and “premises”.

When performing a calculation for the entire building, the goal is to find the power of the source and the entire heat supply system.

When calculating the heat losses of each individual room of the building, the problem is solved to determine the power and the number of heating devices (batteries, convectors, etc.) required for installation in each specific room in order to maintain a given temperature of internal air.

The air in the building is heated by receiving thermal energy from the Sun, external sources heat supply through the heating system and from a variety of internal sources - from people, animals, office equipment, household appliances, lighting lamps, hot water supply systems.

The indoor air cools down due to the loss of thermal energy through the building envelope, which are characterized by thermal resistances measured in m 2 ° C / W:

R = Σ (δ i /λ i )

δ i- thickness of the material layer of the enclosing structure in meters;

λ i- coefficient of thermal conductivity of the material in W / (m · ° С).

Protect the house from external environment ceiling (ceiling) of the upper floor, exterior walls, windows, doors, gates and the floor of the lower floor (possibly a basement).

The external environment is the outside air and soil.

The calculation of heat loss by a building is performed at the design temperature of the outside air for the coldest five-day period of the year in the area where the facility is built (or will be built)!

But, of course, no one forbids you to make a calculation for any other time of the year.

Calculation inExcelheat loss through the floor and walls adjacent to the ground according to the generally accepted zonal technique of V.D. Machinsky.

The temperature of the soil under the building depends primarily on the thermal conductivity and heat capacity of the soil itself and on the temperature of the ambient air in a given area during the year. Since the outside temperature varies significantly in different climatic zones, then the soil has different temperatures at different times of the year at different depths in different areas.

To simplify the solution of the complex problem of determining heat loss through the floor and walls of the basement into the ground, the method of dividing the area of ​​enclosing structures into 4 zones has been successfully used for more than 80 years.

Each of the four zones has its own fixed resistance to heat transfer in m 2 ° C / W:

R 1 = 2.1 R 2 = 4.3 R 3 = 8.6 R 4 = 14.2

Zone 1 is a strip on the floor (in the absence of deepening the soil under the building) 2 meters wide, measured from the inner surface of the outer walls along the entire perimeter or (in the case of a subfloor or basement) a strip of the same width, measured down internal surfaces outside walls from the edge of the ground.

Zones 2 and 3 are also 2 meters wide and are located behind zone 1 closer to the center of the building.

Zone 4 covers the entire remaining central square.

In the figure below, zone 1 is located entirely on the walls of the basement, zone 2 is partially on the walls and partially on the floor, zones 3 and 4 are completely on the basement floor.

If the building is narrow, then zones 4 and 3 (and sometimes 2) may simply not exist.

Square sex zone 1 in the corners is counted twice in the calculation!

If the entire zone 1 is located on vertical walls, then the area is considered in fact without any additions.

If part of zone 1 is on the walls and part on the floor, then only the corner parts of the floor are counted twice.

If the entire zone 1 is located on the floor, then the calculated area should be increased by 2 × 2x4 = 16 m 2 when calculating (for a rectangular house in the plan, i.e. with four corners).

If the building is not buried in the ground, then this means that H =0.

Below is a screenshot of the program for calculating the heat loss through the floor and recessed walls in Excel for rectangular buildings.

Areas of zones F 1 , F 2 , F 3 , F 4 calculated according to the rules of ordinary geometry. The task is cumbersome and often requires sketching. The program greatly facilitates the solution of this task.

The total heat loss to the surrounding soil is determined by the formula in kW:

Q Σ =((F 1 + F1y )/ R 1 + F 2 / R 2 + F 3 / R 3 + F 4 / R 4 ) * (t vr -t nr) / 1000

The user only needs to fill in the first 5 lines in the Excel table and read the result below.

To determine heat loss to soil premises areas of zones will have to be counted manually and then substitute in the above formula.

The following screenshot shows, as an example, the calculation in Excel of heat loss through a floor and recessed walls. for the lower right (according to the picture) basement room.

The sum of heat losses to the ground by each room is equal to the total heat losses to the ground of the entire building!

The figure below shows simplified diagrams of typical floor and wall structures.

The floor and walls are considered non-insulated if the thermal conductivity coefficients of the materials ( λ i), of which they are composed, is more than 1.2 W / (m · ° C).

If the floor and / or walls are insulated, that is, they contain layers with λ <1,2 W / (m ° C), then the resistance is calculated for each zone separately according to the formula:

Rinsulatedi = Rnot warmi + Σ (δ j /λ j )

Here δ j- the thickness of the insulation layer in meters.

For floors on logs, the resistance to heat transfer is also calculated for each zone, but using a different formula:

Ron lagsi =1,18*(Rnot warmi + Σ (δ j /λ j ) )

Calculation of heat losses inMS Excelthrough the floor and walls adjacent to the ground according to the method of Professor A.G. Sotnikov.

A very interesting technique for buildings buried in the ground is described in the article "Thermophysical calculation of heat loss in the underground part of buildings". The article was published in 2010 in No. 8 of the "AVOK" magazine in the "Discussion club" section.

Those who want to understand the meaning of what is written below should first study the above.

A.G. Sotnikov, relying mainly on the conclusions and experience of other scientists-predecessors, is one of the few who, in almost 100 years, tried to move the topic that worries many heat engineers off the ground. I am very impressed by his approach from the point of view of fundamental heating technology. But the difficulty of correctly assessing the temperature of the soil and its thermal conductivity coefficient in the absence of appropriate survey work somewhat shifts the method of A.G. Sotnikov into the theoretical plane, moving away from practical calculations. Although at the same time, continuing to rely on the zonal method of V.D. Machinsky, everyone simply blindly believes the results and, understanding the general physical meaning of their occurrence, they cannot definitely be sure of the numerical values ​​obtained.

What is the meaning of Professor A.G. Sotnikov? He suggests that all heat losses through the floor of a buried building "go" into the interior of the planet, and all heat losses through walls in contact with the ground are eventually transferred to the surface and "dissolve" in the ambient air.

This is somewhat similar to the truth (without mathematical justification) if there is sufficient deepening of the floor of the lower floor, but if the depth is less than 1.5 ... 2.0 meters, doubts arise about the correctness of the postulates ...

Despite all the critical remarks made in the previous paragraphs, it is the development of the algorithm of Professor A.G. Sotnikov looks very promising.

Let's calculate in Excel the heat loss through the floor and walls into the ground for the same building as in the previous example.

We write in the block of initial data the dimensions of the basement of the building and the calculated air temperatures.

Next, you need to fill in the soil characteristics. As an example, we will take sandy soil and enter into the initial data its thermal conductivity coefficient and temperature at a depth of 2.5 meters in January. The temperature and thermal conductivity of the soil for your area can be found on the Internet.

We will make the walls and floor of reinforced concrete ( λ = 1.7 W / (m ° C)) 300mm thick ( δ =0,3 m) with thermal resistance R = δ / λ = 0.176 m 2 ° C / W.

And, finally, we add to the initial data the values ​​of the heat transfer coefficients on the inner surfaces of the floor and walls and on the outer surface of the soil in contact with the outside air.

The program performs the calculation in Excel according to the formulas below.

Floor area:

F pl =B * A

Wall area:

F st = 2 *h *(B + A )

Conditional thickness of the soil layer behind the walls:

δ conv = f(h / H )

Thermal resistance of soil under the floor:

R 17 = (1 / (4 * λ gr) * (π / Fpl ) 0,5

Heat loss through the floor:

Qpl = Fpl *(tv — tgr )/(R 17 + Rpl + 1 / α c)

Thermal resistance of soil behind walls:

R 27 = δ conv / λ gr

Heat loss through walls:

Qst = Fst *(tv — tn ) / (1 / α n +R 27 + Rst + 1 / α c)

General heat loss to the ground:

Q Σ = Qpl + Qst

Remarks and conclusions.

The heat loss of a building through the floor and walls into the ground, obtained by two different methods, is significantly different. According to the algorithm of A.G. Sotnikov value Q Σ =16,146 KW, which is almost 5 times more than the value according to the generally accepted "zonal" algorithm - Q Σ =3,353 KW!

The fact is that the reduced thermal resistance of the soil between the buried walls and the outside air R 27 =0,122 m 2 ° C / W is clearly small and hardly corresponds to reality. This means that the conditional soil thickness δ conv not quite correct!

In addition, the "bare" reinforced concrete of the walls, which I have chosen in the example, is also a completely unrealistic option for our time.

An attentive reader of A.G. Sotnikova will find a number of errors, not copyright errors, but those that arose when typing. Then in formula (3) the factor 2 appears λ , then disappears later. In the example, when calculating R 17 there is no division sign after the unit. In the same example, when calculating heat losses through the walls of the underground part of the building, for some reason, the area is divided by 2 in the formula, but then it is not divided when writing the values ​​... What are these non-insulated walls and floor in the example with Rst = Rpl =2 m 2 ° C / W? In this case, their thickness must be at least 2.4 m! And if the walls and floor are insulated, then, it seems, it is incorrect to compare these heat losses with the option of calculating by zones for an uninsulated floor.

R 27 = δ conv / (2 * λ gr) = K (cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

About the question regarding the presence of a factor of 2 in λ gr has already been said above.

I have divided the complete elliptic integrals by each other. As a result, it turned out that the graph in the article shows the function for λ gr = 1:

δ conv = (½) *TO(cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

But it should be mathematically correct:

δ conv = 2 *TO(cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

or, if the factor is 2 y λ gr not needed:

δ conv = 1 *TO(cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

This means that the graph for determining δ conv gives erroneous 2 or 4 times lower values ​​...

It turns out that while everyone has no choice but to continue either "counting" or "determining" heat loss through the floor and walls into the ground by zones? No other decent method has been invented in 80 years. Or came up with, but not finalized ?!

I invite blog readers to test both calculation options in real projects and present the results in the comments for comparison and analysis.

Everything that is said in the last part of this article is solely the opinion of the author and does not claim to be the ultimate truth. I would be glad to hear in the comments the opinion of experts on this topic. I would like to understand to the end the algorithm of A.G. Sotnikov, because he actually has a more rigorous thermophysical substantiation than the generally accepted method.

I beg respecting the author's work to download the file with the calculation programs after subscribing to article announcements!

P. S. (25.02.2016)

Almost a year after writing the article, we managed to sort out the issues mentioned a little higher.

First, a program for calculating heat loss in Excel according to the method of A.G. Sotnikova thinks everything is correct - exactly according to the formulas of A.I. Pekhovich!

Secondly, formula (3) from the article by A.G. Sotnikova shouldn't look like this:

R 27 = δ conv / (2 * λ gr) = K (cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

In the article by A.G. Sotnikov is not a correct record! But then the graph is built, and the example is calculated using the correct formulas !!!

So it should be, according to A.I. Pekhovich (page 110, additional task to item 27):

R 27 = δ conv / λ gr= 1 / (2 * λ gr) * K (cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

δ conv = R27 * λ gr = (½) * K (cos((h / H ) * (π / 2))) / K (sin((h / H ) * (π / 2)))

Previously, we calculated the heat loss of the floor over the ground for a 6m wide house with a groundwater level of 6m and +3 degrees in depth.
Results and problem statement here -
We also took into account the heat loss to the street air and deep into the earth. Now I will separate the flies from the cutlets, namely, I will carry out the calculation purely into the ground, excluding the heat transfer to the outside air.

I will carry out calculations for option 1 from the previous calculation (without insulation). and the following data combinations
1.GLV 6m, +3 on GWL
2.GLV 6m, +6 on GWL
3. GWL 4m, +3 on GWL
4. GWL 10m, +3 on GWL.
5. GWL 20m, +3 on GWL.
Thus, we will close the issues related to the influence of the depth of the groundwater level and the effect of temperature on the groundwater level.
The calculation, as before, is stationary, does not take into account seasonal fluctuations and does not take into account the outside air at all
The conditions are the same. The soil has Lambda = 1, walls 310mm Lambda = 0.15, floor 250mm Lambda = 1.2.

The results, as before, are two pictures each (isotherms and "IK"), and numerical - the resistance to heat transfer to the ground.

Numerical results:
1.R = 4.01
2.R = 4.01 (Everything is normalized for the difference, otherwise it should not have been)
3.R = 3.12
4.R = 5.68
5.R = 6.14

About the values. If we correlate them with the GWL depth, we get the following
4m. R / L = 0.78
6m. R / L = 0.67
10m. R / L = 0.57
20m. R / L = 0.31
R / L would be equal to one (or rather, the inverse coefficient of soil thermal conductivity) for an infinitely large house, but our house dimensions are comparable to the depth to which heat losses are carried out and the smaller the house is in comparison with the depth, the less this ratio should be.

The resulting R / L dependence should depend on the ratio of the width of the house to the GWL (B / L), plus, as already mentioned, at B / L-> infinity R / L-> 1 / Lambda.
In total, there are the following points for an infinitely long house:
L / B | R * Lambda / L
0 | 1
0,67 | 0,78
1 | 0,67
1,67 | 0,57
3,33 | 0,31
This dependence is well approximated by the exponential one (see the graph in the commentary).
Moreover, the exponent can be written in a simpler manner without much loss of accuracy, namely
R * Lambda / L = EXP (-L / (3B))
This formula at the same points gives the following results:
0 | 1
0,67 | 0,80
1 | 0,72
1,67 | 0,58
3,33 | 0,33
Those. error within 10%, i.e. very satisfactory.

Hence, for an infinite house of any width and for any GWL in the considered range, we have a formula for calculating the resistance to heat transfer in GWL:
R = (L / Lambda) * EXP (-L / (3B))
here L is the depth of the groundwater level, Lambda is the thermal conductivity of the soil, B is the width of the house.
The formula is applicable in the L / 3B range from 1.5 to about infinity (high GWL).

If we use the formula for deeper groundwater levels, then the formula gives a significant error, for example, for 50m depth and 6m width of the house, we have: R = (50/1) * exp (-50/18) = 3.1, which is obviously too small.

Have a great day everyone!

Conclusions:
1. An increase in the depth of the groundwater level does not lead to a corresponding decrease in heat loss to groundwater, since everything is involved large quantity soil.
2. At the same time, systems with a GWL of type 20m and more may never go to the hospital received in the calculation during the "life" of the house.
3. R ​​into the ground is not so great, it is at the level of 3-6, so the heat loss deep into the floor along the ground is very significant. This is consistent with the previously obtained result about the absence of a large reduction in heat loss when insulating a tape or blind area.
4. A formula has been derived from the results, use it for health (at your own peril and risk, naturally, I ask you to know in advance that I am not responsible for the reliability of the formula and other results and their applicability in practice).
5. Follows from a little research carried out below in the commentary. Heat loss outside reduces heat loss to the ground. Those. It is incorrect to consider the two heat transfer processes separately. And by increasing heat protection from the street, we increase heat loss to the ground and thus it becomes clear why the effect of warming the house contour previously obtained is not so significant.

Usually, the heat loss of the floor in comparison with similar indicators of other building envelopes (external walls, window and door openings) is a priori assumed to be insignificant and taken into account in the calculations of heating systems in a simplified form. Such calculations are based on a simplified system of accounting and correction coefficients of heat transfer resistance of various building materials.

If we take into account that the theoretical substantiation and methodology for calculating the heat loss of a ground floor was developed quite a long time ago (i.e., with a large design margin), we can safely speak of the practical applicability of these empirical approaches in modern conditions... Coefficients of thermal conductivity and heat transfer of various building materials, insulation and floor coverings well known and others physical characteristics for calculating heat loss through the floor is not required. According to their thermal characteristics, floors are usually divided into insulated and non-insulated, structurally - floors on the ground and logs.

Calculation of heat loss through an uninsulated floor on the ground is based on the general formula for assessing heat loss through the building envelope:

where Q- main and additional heat loss, W;

A- total area of ​​the enclosing structure, m2;

tv , tн- temperature inside the room and outside air, оС;

β - the share of additional heat losses in the total;

n- correction factor, the value of which is determined by the location of the enclosing structure;

Ro- resistance to heat transfer, m2 ° С / W.

Note that in the case of a homogeneous single-layer floor overlap, the heat transfer resistance Rо is inversely proportional to the heat transfer coefficient of the non-insulated floor material on the ground.

When calculating heat loss through a non-insulated floor, a simplified approach is used, in which the value (1+ β) n = 1. It is customary to produce heat loss through the floor by zoning the heat transfer area. This is due to the natural heterogeneity of the temperature fields of the soil under the floor.

The heat loss of the non-insulated floor is determined separately for each two-meter zone, the numbering of which starts from the outer wall of the building. In total, it is customary to take into account four such strips with a width of 2 m, considering the temperature of the soil in each zone to be constant. The fourth zone includes the entire surface of the non-insulated floor within the boundaries of the first three strips. Heat transfer resistance is taken: for the 1st zone R1 = 2.1; for the 2nd R2 = 4.3; respectively for the third and fourth R3 = 8.6, R4 = 14.2 m2 * оС / W.

Fig. 1. Zoning of the floor surface on the ground and adjacent recessed walls when calculating heat loss

In the case of recessed rooms with an unpaved base of the floor: the area of ​​the first zone adjacent to the wall surface is taken into account in the calculations twice. This is quite understandable, since the heat losses of the floor are summed up with the heat losses in the adjacent vertical enclosing structures of the building.

The calculation of heat loss through the floor is carried out for each zone separately, and the results obtained are summed up and used for the heat engineering substantiation of the building project. The calculation for the temperature zones of the outer walls of recessed rooms is made according to formulas similar to those given above.

In the calculations of heat loss through an insulated floor (and it is considered as such if its structure contains layers of material with a thermal conductivity of less than 1.2 W / (m ° C)), the value of the heat transfer resistance of an uninsulated floor on the ground increases in each case by the heat transfer resistance of the insulating layer:

Ru.s = δs / λs,

where δу.с- thickness of the insulating layer, m; λw.s- thermal conductivity of the insulating layer material, W / (m ° C).

The essence of thermal calculations of premises, in varying degrees, located in the ground, is reduced to determining the influence of atmospheric "cold" on their thermal regime, or rather, to what extent a certain soil insulates a given room from atmospheric temperature effects. Because thermal insulation properties soil depend on too a large number factors, the so-called 4-zone technique was adopted. It is based on the simple assumption that the thicker the soil layer, the higher its thermal insulation properties (in to a greater extent the influence of the atmosphere decreases). The shortest distance (vertically or horizontally) to the atmosphere is divided into 4 zones, 3 of which have a width (if it is a floor along the ground) or a depth (if these are walls along the ground) of 2 meters, and the fourth has these characteristics equal to infinity. Each of the 4 zones is assigned its own permanent heat-insulating properties according to the principle - the farther the zone (the larger its serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can make a simple conclusion that the farther a point in the room is from the atmosphere (with a multiplicity of 2 m), the more favorable conditions(from the point of view of the influence of the atmosphere) it will be located.

Thus, the counting of conditional zones begins along the wall from the ground level, provided there are walls along the ground. If there are no walls on the ground, then the first zone will be the floor strip closest to outside wall... Further, zones 2 and 3 are numbered 2 meters wide. The remaining zone is zone 4.

It is important to consider that the zone can start on the wall and end on the floor. In this case, you should be especially careful when making calculations.

If the floor is not insulated, then the values ​​of the heat transfer resistances of the non-insulated floor by zones are:

zone 1 - R n.p. = 2.1 m2 * C / W

zone 2 - R n.p. = 4.3 m2 * C / W

zone 3 - R n.p. = 8.6 m2 * C / W

zone 4 - R n.p. = 14.2 m2 * C / W

To calculate the resistance to heat transfer for insulated floors, you can use the following formula:

- resistance to heat transfer of each zone of the non-insulated floor, m2 * C / W;

- insulation thickness, m;

- coefficient of thermal conductivity of the insulation, W / (m * C);