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. Such calculations are based on a simplified system of accounting and correction coefficients of resistance of heat transfer of various building materials.
If we consider that the theoretical substantiation and methodology for calculating the heat lifting of the soil floor was developed sufficiently long ago (that is, 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 floor coatings are well known, and other physical characteristics are not required to calculate heat loss through the floor. 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).
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.
The air in the building is heated due to the production of heat energy from the Sun, external sources of 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 the house from the external environment 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 air temperature varies significantly in different climatic zones, then the soil has a 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, 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 the inner surfaces of the outer walls from The edges 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.
Square floor Zones 1 in the corners are taken into account when calculating twice!
If the entire zone 1 is located on the vertical walls, then the area is considered 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. at — 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. at — 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)))
Heat loss through the floor, located on the ground, are calculated on zones according to. For this purpose, the surface of the floor is divided into strips with a width of 2 m, parallel to the outer walls. The strip close to the outer wall is denoted by the first zone, the following two bands - the second and third zone, and the rest of the floor is the fourth zone.
When calculating the heat loss of basements, the breakdown on the bands-zone in this case is made from the ground level over the surface of the underground part of the walls and then on the floor. The conditional resistances of heat transfer for zones in this case are accepted and calculated in the same way as for a warmed floor in the presence of insulation layers, which in this case are the layers of the wall design.
The heat transfer coefficient K, W / (m 2 ∙ ° C) for each zone of the insulated floor on the ground is determined by the formula:
where - the resistance of the heat transfer of the insulated gender on the ground, m 2 ∙ ° C / W is calculated by the formula:
\u003d + Σ, (2.2)
where - the resistance of the heat transfer of the laptile sex of the i-that zone;
Δ j - the thickness of the J-that layer of the insulation structure;
λ j is the coefficient of thermal conductivity of the material from which the layer consists of.
For all zones of the displeased floor there is data on heat transfer resistance that are accepted by:
2.15 m 2 ∙ ° C / W - for the first zone;
4.3 m 2 ∙ ° C / W - for the second zone;
8.6 m 2 ∙ ° C / W - for the third zone;
14.2 m 2 ∙ ° C / W - for the fourth zone.
In this project, the floors on the ground have 4 layers. The floor design is shown in Figure 1.2, the wall design is shown in Figure 1.1.
An example of a heat engineering calculation of floors located on the ground for the room 002 of the Ventscamera:
1. The division into zones in the venger room is conditionally presented in Figure 2.3.
Figure 2.3. Ventcamera
The figure shows that part of the wall and part of the floor enters the second zone. Therefore, the coefficient of resistance to the heat transfer of this zone is calculated twice.
2. Determine the resistance of the heat transfer of the insulated floor on the ground ,, 2 ∙ ° C / W:
2,15 + 
\u003d 4.04 m 2 ∙ ° C / W,
4,3 + \u003d 7.1 m 2 ∙ ° C / W,
4,3 + 
\u003d 7.49 m 2 ∙ ° C / W,
8,6 + \u003d 11.79 m 2 ∙ ° C / W,
14,2 + \u003d 17.39 m 2 ∙ ° C / W.
Good day!
I decided to post here the results of calculations for the insulation of sex on the soil. Calculations were carried out in the program THERM 6.3.
Paul on the soil - a concrete plate with a thickness of 250mm with a thermal conductivity coefficient of 1.2
Walls - 310 mm with thermal conductivity coefficient 0.15 (aerated concrete or tree)
For simplicity of the wall to the soil. There may be many options for insulation and the bridges of the hod of the node, for simplicity they are lowered.
Soil - with thermal conductivity coefficient 1. Wet clay or wet sand. Dry - more heat-stash.
Insulation. Here are 4 options:
1. There is no insulation. Just a stove on the soil.
2. The gesture is insulated with 1m width, 10 cm thick. Heat EPPS. The top layer itself was not taken into account, since it does not have a big role.
3. The foundation tape is insulated on 1m depth. Warming also 10cm, EPPS. Concrete is not drawn as close to the soil by thermal conductivity.
4. Heated plate under the house. 10cm, EPPS.
The thermal conductivity coefficient of EPPS was taken equal to 0.029.
Plate width is taken 5.85m.
Initial data on temperature:
- inside +21;
- Outside -3;
- At the depth 6m +3.
6m here is an assessment of the corner. He took 6m because it is closest to the option with my home, although I have no floors on the soil, but the results are also applicable for my warm underground.
The results in graphical form you see. Applied in two versions - with isotherms and "IC".
In digital data obtained for the floor surface in the form of U-Factor, the inverse of our heat transfer resistance ([R] \u003d K * m2 / W).
In terms of the following results (on average on the floor):
1. R \u003d 2.86
2. R \u003d 3.31
3. R \u200b\u200b\u003d 3.52
4. R \u003d 5.59
For me, so very interesting results. In particular A sufficient high value according to the 1st option indicates that it is not so necessary to insulate the slab on the floor in any way.It is necessary to warm the soil when the groundwater is near and then we have option 4, with partially cut-off ground from the thermal circuit. With a close corner, we do not get 5.59. Since the 6m soil adopted in the calculation is not involved in insulation. It should be waiting for R ~ 3 in this case or so.
It is also quite essential that The edge of the plate in the calculated version is pretty warm 17.5 ° C for the first laptile optionIt was not expected to be freezing, condensate and mold, even with an increase in temperature gradient twice as well (-27 on the street). In fact, it should be understood that with such calculations, peak temperatures do not have any role, since the system is very heathed and the soil freezes the weeks-months.
Options 1,2,3. And especially option 2 - the most inertial. In the heat circuit, the ground is involved not only by the one that is directly under the house, but also under the slaughterhouse. The time of setting the temperature regime as in the figure is the years and actually the temperature regime will be average per year. The period of 3 months has time to engage in heat exchange only 2-3m soil. But this is a separate story, so I will complete until I note that the characteristic time is proportional to the thickness of the layer in the square. Those. If 2m - 3 months, then 4m is already 9 months.
I also note that in practice, it is likely that with a relatively small AGB (type 4.5m and below), it is necessary to wait for the worst results of the heat-insulating properties of the soil due to the evaporation of water from it. Unfortunately, the instrument that could carry out the calculation in the conditions of evaporation in the soil I am not familiar with me. Yes, and with the source data here is a big problem.
Evaluation with the effect of evaporation in the ground carried out as follows.
Data out the data that water in the loams rises with the capital forces from the corner at 4-5m
Well, as the source data of this number and use.
It is brazen that the same 5m is stored in my calculation under any circumstances.
In the 1m soil to the floor, the pairs diffuse, and the magnitude of the vapor permeability coefficient can be nosed. Parry permeability coefficient of sand 0.17, globckers 0.1. Well, for reliability, take 0.2 mg / m / h / pa.
At the depth of the meter in the calculated versions, in addition to the option 4, about 15 grades.
Total there is water vapor pressure 1700pa (100% rel).
In the room, take 21grad 40% (rel.) \u003d\u003e 1000pa
TOTAL 700PA Course pressure gradient for 1M clay with Mu \u003d 0.2 and 0.25m concrete with MU \u003d 0.09
The final vapor permeability of the two-layer 1 / (1 / 0.2 + 0.25 / 0.09) \u003d 0.13
As a result, we have a stream of a pair of soil 0.13 * 700 \u003d 90 mg / m2 / h \u003d 2,5E-8 kg / m2 / s
We multiply on the warmth of the evaporation of water 2,3mge / kg and we obtain additional heat loss to evaporation \u003d\u003e 0.06W / m2. Trivia is. If we speak in the language R (heat transfer), then such accounting of moisture leads to a decrease in r about 0.003, i.e. Unnecessarily.
Attachments:
Comments
Methods for calculating the heat loss of the premises and the order of its execution (see SP 50.13330.2012, thermal protection of buildings, paragraph 5).
The house loses heat through the enclosing structures (walls, overlappings, windows, roof, foundation), ventilation and sewage. The main weight losses go through the enclosing structures - 60-90% of all heat loss.
In any case, keeping the heat loss to produce for all structures of the enclosing type, which are present in the heated room.
At the same time, it is not necessary to take into account the loss of heat, which are carried out through the internal structures, if the difference between their temperature with the temperature in the neighboring rooms does not exceed 3 degrees Celsius.
Heat loss through fencing structures
Thermal losses of the premises are mainly dependent on:
1 temperature differences in the house and on the street (the difference more, the loss above),
2 heat protection properties of walls, windows, doors, coatings, gender (so-called enclosing room structures).
Fencing structures are mainly not homogeneous by structure. And usually consist of several layers. Example: wall of shell \u003d plaster + shell + outer decoration. This design may include closed air interlayers (example: cavities inside bricks or blocks). The above materials have heat engineering characteristics different from each other. The main characteristic for the design layer is its heat transfer resistance R.
Where q is the amount of heat that loses the square meter of the enclosing surface (is usually measured in W / M.KV.)
ΔT is the difference between the temperature inside the cleaned room and the outer air temperature (the temperature of the coldest five days of ° C for the Climatic region in which the building is calculated).
Basically the inner temperature in the premises is accepted. Residential premises 22 OS. Non-residential 18 OS. Water treatment zones 33 OS.
When it comes to a multi-layered structure, the resistance of the layer of the design is folded.
Δ - layer thickness, m;
λ is the estimated coefficient of thermal conductivity of the material of the design layer, taking into account the conditions of operation of the enclosing structures, W / (M2 OS).
Well, here with the basic data required for the calculation figured out.
So, to calculate heat losses through the enclosing structures, we need:
1. Resistance to the heat transfer structures (if the design is multilayer, then σ r layers)
2. The difference between the temperature in the calculated room and on the street (the temperature of the coldest five days of ° C.). Δt.
3. Plaza of fences F (separate walls, windows, doors, ceiling, floor)
4. The orientation of the building in relation to the parties is useful.
The formula for calculating heat loss fence looks like this:
Qogra \u003d (Δt / Rogr) * Fogr * n * (1 + σB)
QGR - Heat loss through fencing structures, W
Rogr - heat transfer resistance, M.KV. ° C / W; (If several layers then σ Rogr layers)
Fogr - area of \u200b\u200bthe enclosing structure, m;
n is the coefficient of contact of the enclosing construction with the outer air.
| Walling | N. coefficient |
| 1. Outdoor walls and coatings (including ventilated by outer air), the overlap of attic (with roofing materials) and over drives; Overlapping over cold (without enclosing walls) underground in the northern construction and climatic zone | |
| 2. Overlapping over cold basements communicating with the outer air; Ceurface overlaps (with roofing of roll materials); Overlapping over cold (with enclosing walls) underground and cold floors in the northern construction and climatic zone | 0,9 |
| 3. Cleansing over not heated basements with light openings in the walls | 0,75 |
| 4. Cleansing over not heated basements without light openings in the walls located above the ground level | 0,6 |
| 5. Overlapping over not heated technical undergrounds located below ground level | 0,4 |
The heat loss of each enclosing structure is considered separately. The magnitude of heat loss through the enclosing designs of the entire room will be the amount of heat loss through each enclosing room design
Calculation of heat loss through the floors
Unheated floor on the ground
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. Such calculations are based on a simplified system of accounting and correction coefficients of resistance of heat transfer of various building materials.
If we consider that the theoretical substantiation and methodology for calculating the heat lifting of the soil floor was developed sufficiently long ago (that is, 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 floor coatings are well known, and other physical characteristics are not required to calculate heat loss through the floor. 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).