Heat transfer through the fences of the house is a complex process. In order to maximize these difficulties, the measurement of the premises during the calculations of the heat loss is made by certain rulesSuppose of conditional increase or decrease in area. Below are the main provisions of these rules.

Rules of the area of \u200b\u200bthe area of \u200b\u200benclosing structures: A - section of a building with an attic overlap; b - section of the building with a combined coating; in the building plan; 1 - gender above the basement; 2 - floor on lags; 3 - Paul on the ground;

The area of \u200b\u200bwindows, doors and other openings is measured by the smallest building effect.

The ceiling area (PT) and gender (PL) (except for the floor on the soil) is measured between the axes of the inner walls and the inner surface outdoor Wall.

The size of the outer walls are taken horizontally along the outer perimeter between the axes of the inner walls and the outer angle of the wall, and in height - on all floors, except for the lower: on the level of clean floor to the floor of the next floor. On the last floor The top of the outer wall coincides with the top of the coating or attic overlap. On the lower floor, depending on the design of the floor: a) from the inner surface of the floor on the ground; b) from the surface of the preparation under the structure of the floor on the lags; c) from the lower verge of overlapping over the unheated underground or basement.

When determining heat loss through interior walls They are measured by the inner perimeter. Warm losses through the inner fences of the premises can not be taken into account if the difference in air temperatures in these rooms is 3 ° C and less.

Flooring the surface of the floor (a) and beugoned parts of the outer walls (b) to the calculated zones I-IV

The transfer of heat from the room through the structure of the floor or walls and the thickness of the soil with which they come into contact, obeys complex patterns. To calculate the resistance of heat transfer structures located on the ground, use a simplified technique. The surface of the floor and walls (at the same time the floor is considered as a continuation of the wall) on the ground is divided into strips with a width of 2 m, parallel to the joint of the outer wall and the ground surface.

The countdown of zones begins on the wall from the ground level, and if there are no walls on the soil, the zone I is the floor strip closest to the outer wall. The following two bands will have numbers II and III, and the rest of the floor will be zone IV. Moreover, one zone can begin on the wall, but continue on the floor.

Floor or walls that do not contain insulation layers from materials with a thermal conductivity coefficient of less than 1.2 W / (M · ° C) are called displeasted. The resistance of the heat transfer of such sex is taken to denote R NP, m 2 · ° C / W. For each zone of the laptile, the regulatory values \u200b\u200bof heat transfer resistance are provided:

• zone i - ri \u003d 2.1 m 2 · ° C / W
• zone II - RII \u003d 4.3 m 2 · ° C / W;
• zone III - RIII \u003d 8.6 m 2 · ° C / W
• zone IV - Riv \u003d 14.2 m 2 · ° C / W.

If in the structure of the floor, located on the ground, there are insulation layers, it is called insulated, and its heat transfer resistance R supplies, m 2 · ° C / W, is determined by the formula:

R pack \u003d R np + r Us1 + R Us2 ... + R USN

Where R np is the heat transfer resistance of the zone under consideration of the radiated gender, m 2 · ° C / W
R mustache resistance to the heat transfer of the insulation layer, m 2 · ° C / W;

For the floor on the lags, the resistance of heat transfer RL, M 2 · ° C / W is calculated by the formula.

The essence of the thermal calculations of the premises, in one degree or another in the ground, is reduced to the determination of the influence of the atmospheric "cold" on their thermal regime, or rather, to what a certain soil isolates this room from atmospheric temperature effects. Because heat insulating properties Soil depend on too big number factors, then the so-called method of 4 zones was adopted. It is based on a simple assumption that the thickness of the soil layer, the higher its thermal insulation properties (in more than The effect of the atmosphere is reduced). The shortest distance (vertical or horizontal) to the atmosphere is divided into 4 zones, 3 of which have a width (if it is the floor via the soil) or depth (if it is the walls of the soil) 2 meters, and the fourth of these characteristics are equal to infinity. Each of the 4 zones is assigned their constant thermal insulation properties on the principle - the further zone (the more its serial number), the influence of the atmosphere is less. Updating formalized approach, it is possible to make a simple conclusion that the further a certain point in the room is from the atmosphere (with multiplicity of 2 m), the more favorable conditions (From the point of view of the influence of the atmosphere) it will be.

Thus, the countdown of the conditional zones is starting along the wall from the level of the Earth, subject to the presence of walls on the soil. If there are no walls on the soil, the first zone will be the floor strip closest to the outer wall. Next, zones 2 and 3 2 meters width are numbered. The remaining zone is zone 4.

It is important to take into account that the zone can begin on the wall and end on the floor. In this case, one should be particularly attentive when calculating.

If the floor is radiant, the resistance values \u200b\u200bof the heat transfer of the radiated sex in the zones are equal:

zone 1 - R N.P. \u003d 2.1 sq. M. * C / W

zone 2 - R N.P. \u003d 4.3 sq. M. * C / W

zone 3 - R N.P. \u003d 8.6 sq. M. * C / W

zone 4 - R N.P. \u003d 14.2 sq. M. * C / W

To calculate the heat transfer resistance for insulated floors, you can take advantage of the following formula:

- resistance to heat transfer of each zone of the radiated floor, sq. M. * C / W;

- the thickness of the insulation, m;

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

Typically, the heat of the floor in comparison with the same indicators of other enclosing buildings of the building (exterior walls, window and doorways) a priori are taken insignificant and taken into account in the calculations of the heating systems in a simplified form. The basis of such calculations is laid by a simplified system of accounting and correction coefficients of resistance of heat transfer various building materials.

If we consider that the theoretical substantiation and methodology for calculating the heat transfer of the soil floor was developed sufficiently long ago (i.e. with a large design reserve), one can safely talk about the practical applicability of these empirical approaches in modern conditions. The coefficients of thermal conductivity and heat transfer of various building materials, insulation and outdoor coatings well known and other physical characteristics To calculate heat loss through the floor is not required. In terms of its heat engineering characteristics, the floors are accepted on insulated and unheated, structurally - floors on the ground and lags.

Calculation of heat loss through the dispelled floor on the ground is based on the general formula of the evaluation of heat loss through the enclosing building structures:

where Q. - main and additional heat loss, W;

BUT - the total area of \u200b\u200bthe enclosing structure, m2;

tB , tN - temperature indoors and outdoor air, OS;

β - the proportion of additional heat loss in total;

n. - correction coefficient whose value is determined by the location of the enclosing structure;

RO - heat transfer resistance, m2 ° C / W.

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

When calculating heat loss through the dispelled gender, a simplified approach is used, in which the value (1+ β) n \u003d 1. Heat loss through the floor is made by zoning the heat transfer area. This is due to the natural heterogeneity of soil temperature fields under the overlap.

The heat loss of the radiated gender is determined separately for each two-meter zone, the numbering of which begins on the outer wall of the building. In total, 2 m wide should be taken into account four, counting the temperature of the soil in each area of \u200b\u200bconstant. The fourth zone includes the entire surface of the laptile floor within the borders of the first three bands. The heat transfer resistance is accepted: for the 1st zone R1 \u003d 2.1; for 2nd R2 \u003d 4.3; Accordingly, for the third and fourth R3 \u003d 8.6, R4 \u003d 14.2 m2 * OS / W.

Fig.1. Zoning the surface of the floor on the ground and the adjoining swallowed walls when calculating the Heat Pottery

In the case of beaten premises with the ground base: the area of \u200b\u200bthe first zone adjacent to the wall surface is taken into account in the calculations twice. This is quite understandable, since the heat loss of the floor is summed up with heat loss in adjacent vertical enclosing building structures.

Calculation of heat loss across the floor is made for each zone separately, and the results obtained are summed up and used for the heat engineering justification of the building project. The calculation for the temperature zones of the outer walls of the plated rooms is made by formulas similar to the above.

In the calculations of heat loss through the insulated floor (and that it is considered if there is a layer of material with a thermal conductivity of less than 1.2 W / (M ° C)) in its design, the magnitude of the heat transfer heat transfer in the soil increases in each case to the heat transfer resistance of the insulation layer:

Row \u003d Δu.c / λu.s,

where Δu.s. - the thickness of the insulation layer, m; Λu.s. - heat conduction material of the insulation layer, W / (M ° C).

To perform the calculation of heat loss through the floor and the ceiling, the following data will be required:

• Dimensions of house 6 x 6 meters.
• Floors - a cutting board, whipped with a thickness of 32 mm, are trimmed with a chipboard with a thickness of 0.01 m, insulated with a mineral wool insulation with a thickness of 0.05 m. Under the house is arranged underground for storing vegetables and preservation. In winter, the temperature in the underground on average is + 8 ° C.
• Ceiling overlap - ceilings are made of wooden shields, ceilings are insulated on the side of the attic room with a mineral wool insulation of a layer thickness of 0.15 meters, with a pair-waterproofing layer device. Attic room displeased.

Calculation of heat loss through the floor

R Board \u003d b / k \u003d 0.032 m / 0.15 W / MK \u003d 0.21 m² ° C / W, where b is the thickness of the material, K is the coefficient of heat resistance.

R chip \u003d b / k \u003d 0,01m / 0,15W / mk \u003d 0.07m² ° C / W

R Heat \u003d b / k \u003d 0.05 m / 0.039 W / mk \u003d 1.28 m² ° C / W

The total value of R floor \u003d 0.21 + 0.07 + 1.28 \u003d 1.56 m² of ° C / W

Considering that in the underground, the temperature in the winter is constantly holding around + 8 ° C, then the DT needed to calculate the heat loss is 22-8 \u003d 14 degrees. Now there is all the data for calculating heat loss through the floor:

Q floor \u003d SXDT / R \u003d 36 m²14 degrees / 1,56 m² ° C / W \u003d 323.07 W.Ch (0.32 kWh)

Calculation of heat loss through the ceiling

The ceiling area is the same as the floor S ceiling \u003d 36 m 2

When calculating the heat resistance of the ceiling we do not take into account wooden shieldsbecause They do not have a dense compound among themselves and do not perform the role of heat insulator. Therefore, the thermal resistance of the ceiling:

R ceiling \u003d R insulation \u003d insulation thickness 0.15 m / thermal conductivity of insulation 0.039 W / MK \u003d 3.84 m²s ° C / W

We make the calculation of heat loss through the ceiling:

Q ceiling \u003d SXDT / R \u003d 36 m²52 degrees / 3.84 m²s ° C / W \u003d 487.5 W.Ch (0.49 kWh)

Despite the fact that Heatlopotieri in most of the majority of single-storey industrial, administrative and residential buildings rarely exceed 15% of the total heat loss, and with increasing floors, sometimes do not reach 5%, the importance of the proper solution to the problem ...

Definitions of heat loss from the air of the first floor or basement into the ground does not lose its relevance.

This article discusses two options for solving the title task. Conclusions - at the end of the article.

Considering the loss of heat, it should always be distinguished by the concepts of "building" and "room".

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

When calculating the thermal loss of each individual building, the task of determining the power and the number of heat devices (batteries, convectors, etc.) is solved, which is necessary for installation in each particular room in order to maintain the specified temperature of the inner air.

Air in the building heats up due to heat 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 systems.

The air indoors cools out due to the loss of thermal energy through the enclosing structures of the structure, which are characterized by thermal resistances measured in m 2 · ° C / W:

R. = Σ (δ I. /λ I. )

δ I. - The thickness of the layer of the material of the enclosing structure in meters;

λ I. - coefficient of thermal conductivity of material in W / (M · ° C).

Fencing House Ot external environment The ceiling (overlapping) of the upper floor, exterior walls, windows, doors, gates and gender of the lower floor (possibly - basement).

External environment is the outer air and soil.

The calculation of the heat loss by the structure is performed at the calculated temperature of the outdoor air for the coldest five days per year in the area, where it is built (or will be built) object!

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

Calculation B.Excel Heat loss through the floor and walls adjacent to the soil according to the generally accepted zonal method 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 ambient temperature in this area during the year. Since the outdoor temperature is significantly different in different climatic zones, then the soil has different temperature At different periods of the year at different depths in various districts.

To simplify the solution of a complex problem of determining the heat loss through the floor and the walls of the basement in the ground, for more than 80 years, the method of splitting the area of \u200b\u200benclosing structures on 4 zones has been successfully used.

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

R 1 \u003d 2.1 R 2 \u003d 4.3 R 3 \u003d 8.6 R 4 \u003d 14.2

Zone 1 is a strip on the floor (in the absence of a soil under the structure) 2 meters width, measured from the inner surface of the outer walls along the entire perimeter or (in the case of the presence of an underground or basement) the strip of the same width measured down inland surfaces Outdoor walls from the edge of the soil.

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

Zone 4 occupies the rest of the central square.

The figure presented by just below zone 1 is located entirely on the walls of the basement, zone 2 - partially on the walls and partially on the floor, zone 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 be.

Area floor Zones 1 in the corners are taken into account when calculating twice!

If the whole zone 1 is located on vertical WallsThe area is considered to be in fact without any additives.

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

If the entire zone 1 is located on the floor, then the calculated area should be increased by 2 × 2x4 \u003d 16 m 2 (for home rectangular in the plan, i.e. with four angles).

If the blocking of the building is not in the soil, then this means that H. =0.

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

Square Zone F. 1 , F. 2 , F. 3 , F. 4 Calculated according to the rules of ordinary geometry. The task is cumbersome, requires often drawing sketch. The program greatly facilitates the solution of this task.

Common heat loss into the surrounding soil are determined by the formula in kW:

Q Σ. =((F. 1 + F. 1U. )/ R. 1 + F. 2 / R. 2 + F. 3 / R. 3 + F. 4 / R. 4 ) * (T BP -T HP) / 1000

The user needs only to fill in the Excel table with values \u200b\u200bof the first 5 lines and read the result at the bottom.

To determine heat losses in the ground premises Square Zone we'll have to be considered manually And then substitute to the above formula.

The following screenshot is shown as an example calculation in Excel heat loss through Paul and Blowered Walls for the right lower (in drawing) basement.

The amount of heat loss in the ground by each room is equal to the general thermal loss in the ground of the whole building!

The figure below shows simplified schemes of typical structures of floors and walls.

The floor and walls are considered displeasted if the thermal conductivity coefficients of the materials ( λ I. ), from which they consist, more than 1.2 W / (m · ° C).

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

R. instext I. = R. unfortunate I. + Σ (δ J. /λ J. )

Here δ J. - The thickness of the insulation layer in meters.

For floors on the lags, heat transfer resistance is also calculated for each zone, but on another formula:

R. on lags I. =1,18*(R. unfortunate I. + Σ (δ J. /λ J. ) )

Calculation of thermal losses inMS. Excel Through the floor and walls, adjacent to the soil according to the method of professor A.G. Sotnikova.

A very interesting technique for the buildings are placed in the soil in the soil is set forth in the article "Thermophysical calculation of the heat loss of the underground part of the buildings." The article was published in 2010 in No. 8 of the magazine "Avok" in the heading "Discussion Club".

Those who want to understand the meaning written further should be previously learned to learn the above.

A.G. Sotnikov, based mainly to the conclusions and experience of other predecessor scientists, is one of the few who almost over 100 years old tried to move the topic with a dead point that exciting many heat engineers. Very impresses his approach from the point of view of fundamental heat engineering. But the complexity of proper estimation of the temperature of the soil and its coefficient of thermal conductivity in the absence of the corresponding survey works somewhat shifts the AG technique Sotnikova in the theoretical plane, giving away from practical calculations. Although, while continuing to rely on the zonal method of V.D. Maachinsky, everyone just blindly believe the results and, understanding the general physical meaning of their occurrence, cannot definitely be confident in the numerical values \u200b\u200bobtained.

What is the meaning of the methodology of Professor A.G. Sotnikova? It suggests that all heat loss through the floor of a plated building "go" into the depths of the planet, and all the weight losses through the walls in contact with the soil are transmitted as a result on the surface and "dissolve" in the air of the environment.

It looks like a partly for the truth (without mathematical justifications) in the presence of sufficiently bullet down the floor of the lower floor, but with a gloss of less than 1.5 ... 2.0 meters there are doubts about the correctness of the postulates ...

Despite all critical comments made in previous paragraphs, it is the development of the algorithm of Professor A.G. Sotnikova seems very promising.

Perform the calculation in Excel heat loss through the floor and walls in the ground for the same building as in the previous example.

We write in the source data unit sizes of the basement of the building and the calculated air temperatures.

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

Wall and floor from reinforced concrete ( λ \u003d 1.7 W / (m · ° C)) 300mm thick ( δ =0,3 m) with thermal resistance R. = δ / λ \u003d 0.176. m 2 · ° C / W.

And finally, we add to the initial data the values \u200b\u200bof heat transfer coefficients on the inner surfaces of the floor and walls and on the outer surface of the soil in contact with the outer air.

The program calculates in Excel according to the formulas below.

Floor area:

F pl \u003dB * A.

Square walls:

F Art \u003d 2 *h. *(B. + A. )

The conditional thickness of the soil layer behind the walls:

δ SL = f.(h. / H. )

Thermo resistance of the soil under the floor:

R. 17 \u003d (1 / (4 * λ gr) * (π / F. PL ) 0,5

Teplockotieri through the floor:

Q. PL = F. PL *(t. in — t. G. )/(R. 17 + R. PL + 1 / α c)

Thermo resistance of the soil behind the walls:

R. 27 = δ SL / λ gr

Heat loss through the walls:

Q. Art = F. Art *(t. in — t. N. ) / (1 / α n +R. 27 + R. Art + 1 / α c)

General heat loss in ground:

Q. Σ = Q. PL + Q. Art

Comments and conclusions.

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

The fact is that the reduced thermal resistance of the soil between the swelled walls and the outer air R. 27 =0,122 M 2 · ° C / W is clearly little and is unlikely to be reality. And this means that the conditional thickness of the soil δ SL Defined not entirely correct!

In addition, the "naked" reinforced concrete walls chosen by me in the example is also completely unreal for our time option.

Attentive reader Article A.G. Sotnikova will find a number of errors, rather not copyright, but arising when typing. That in the formula (3) there is a multiplier of 2 λ , in the future disappears. In the example when calculating R. 17 No after a unit of division sign. In the same example, when calculating the heat loss through the walls of the underground part of the building, the area for some reason is divided into 2 in the formula, but then it is not divided when recording values \u200b\u200b... what is this laptile wall and gender in the example with R. Art = R. PL =2 m 2 · ° C / W? Their thickness should be in this case a minimum of 2.4 m! And if the walls and the floor are insulated, then it seems to incorrectly compare these heat loss with a calculation option for zones for a laptile floor.

R. 27 = δ SL / (2 * λ gr) \u003d K (cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

About the question regarding the presence of a multiplier 2 λ G. It was already said above.

I shared full elliptical integrals on each other. As a result, it turned out that the graph shows the function when λ G \u003d 1:

δ SL = (½) *TO(cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

But it should be mathematically:

δ SL = 2 *TO(cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

or, if the multiplier 2 λ G. not needed:

δ SL = 1 *TO(cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

This means that the schedule for determining δ SL Gives erroneous at 2 or 4 times the value ...

It turns out anything else anything else remains, how to continue not that "count", not that "define" heat loss through the floor and walls in the soil in zones? There were no other decent method for 80 years. Or invented, but did not finalize?!

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

Everything that has been said in the last part of this article is solely the author's opinion and does not claim the truth in the last instance. I will be glad to listen to the opinion of specialists on this topic in the comments. I would like to figure out to the end with the algorithm A.G. Sotnikova, because it really has a stricter thermophysical justification than the generally accepted technique.

ask respectful the work of the author download the file with the calculation programs after subscribing to the announcements of articles!

P. S. (02/25/2016)

Altern a year after writing the article managed to deal with questions voiced a little higher.

First, the program for calculating heat loss in Excel according to the AG methodology Sotnikova considers everything correctly - exactly according to the formulas A.I. Pekhovich!

Secondly, which brought the Sumyatitsa into my arguments of Formula (3) from the article A.G. Sotnikova should not look like this:

R. 27 = δ SL / (2 * λ gr) \u003d K (cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

Article A.G. Sotnikova - not the right record! But then the schedule is built, and the example is designed for the right formulas !!!

So it must be according to A.I. Panchovich (p. 110, additional task to paragraph 27):

R. 27 = δ SL / λ gr\u003d 1 / (2 * λ gr) * K (cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))

δ SL \u003d R. 27 * λ GR \u003d (½) * K (cos.((h. / H. ) * (π / 2))) / k (sin.((h. / H. ) * (π / 2)))