Methodology for calculating heat loss in premises and the procedure for its implementation (see SP 50.13330.2012 Thermal protection buildings, point 5).
The house loses heat through enclosing structures (walls, ceilings, windows, roof, foundation), ventilation and sewerage. The main heat losses occur through the enclosing structures - 60–90% of all heat losses.
In any case, heat loss must be taken into account for all enclosing structures that are present in the heated room.
In this case, it is not necessary to take into account heat losses that occur through internal structures if the difference in their temperature with the temperature in adjacent rooms does not exceed 3 degrees Celsius.
Heat loss through building envelopes
Heat losses in premises mainly depend on:
1 Temperature differences in the house and outside (the greater the difference, the higher the losses),
2 Thermal insulation properties of walls, windows, doors, coatings, floors (the so-called enclosing structures of the room).
Enclosing structures are generally not homogeneous in structure. And they usually consist of several layers. Example: shell wall = plaster + shell + exterior decoration. This design may also include closed air gaps (example: cavities inside bricks or blocks). The above materials have thermal characteristics that differ from each other. The main characteristic for a structural layer is its heat transfer resistance R.
Where q is the amount of heat that is lost square meter enclosing surface (usually measured in W/sq.m.)
ΔT is the difference between the temperature inside the calculated room and the outside air temperature (the coldest five-day temperature °C for the climatic region in which the calculated building is located).
Basically, the internal temperature in the rooms is taken. Living quarters 22 oC. Non-residential 18 oC. Water treatment areas 33 °C.
When it comes to a multilayer structure, the resistances of the layers of the structure add up.
δ - layer thickness, m;
λ is the calculated thermal conductivity coefficient of the material of the construction layer, taking into account the operating conditions of the enclosing structures, W / (m2 oC).
Well, we’ve sorted out the basic data required for the calculation.
So, to calculate heat losses through building envelopes, we need:
1. Heat transfer resistance of structures (if the structure is multilayer, then Σ R layers)
2. The difference between the temperature in settlement room and outside (the temperature of the coldest five-day period is °C.). ΔT
3. Fencing areas F (separately walls, windows, doors, ceiling, floor)
4. The orientation of the building in relation to the cardinal directions is also useful.
The formula for calculating heat loss by a fence looks like this:
Qlimit=(ΔT / Rolim)* Folim * n *(1+∑b)
Qlim - heat loss through enclosing structures, W
Rogr – heat transfer resistance, m2°C/W; (If there are several layers then ∑ Rogr layers)
Fogr – area of the enclosing structure, m;
n is the coefficient of contact between the enclosing structure and the outside air.
| Walling | Coefficient n |
| 1. External walls and coverings (including those ventilated by outside air), attic floors (with roofing made of piece materials) and over driveways; ceilings over cold (without enclosing walls) undergrounds in the Northern construction-climatic zone | |
| 2. Ceilings over cold basements communicating with outside air; attic floors (with a roof made of roll materials); ceilings above cold (with enclosing walls) undergrounds and cold floors in the Northern construction-climatic zone | 0,9 |
| 3. Ceilings over unheated basements with light openings in the walls | 0,75 |
| 4. Ceilings over unheated basements without light openings in the walls, located above ground level | 0,6 |
| 5. Ceilings over unheated technical undergrounds located below ground level | 0,4 |
The heat loss of each enclosing structure is calculated separately. The amount of heat loss through the enclosing structures of the entire room will be the sum of heat losses through each enclosing structure of the room
Calculation of heat loss through floors
Uninsulated floor on the ground
Typically, 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 is taken into account in the calculations of heating systems in a simplified form. The basis for such calculations is a simplified system of accounting and correction coefficients for heat transfer resistance of various building materials.
Considering that theoretical basis and the methodology for calculating heat loss from a ground floor was developed quite a long time ago (i.e., with a large design margin), we can safely talk about the practical applicability of these empirical approaches in modern conditions. Thermal conductivity and heat transfer coefficients of various building materials, insulation materials and floor coverings are well known, and no other physical characteristics are required to calculate heat loss through the floor. According to their thermal characteristics, floors are usually divided into insulated and non-insulated, and structurally - floors on the ground and on joists.
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 losses, W;
A– total area of the enclosing structure, m2;
tв , tн– indoor and outdoor air temperature, °C;
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– heat transfer resistance, m2 °C/W.
Note that in the case of a homogeneous single-layer floor covering, the heat transfer resistance Ro is inversely proportional to the heat transfer coefficient of the non-insulated floor material on the ground.
When calculating heat loss through an uninsulated floor, a simplified approach is used, in which the value (1+ β) n = 1. Heat loss through the floor is usually carried out by zoning the heat transfer area. This is due to the natural heterogeneity of the temperature fields of the soil under the ceiling.
Heat loss from an uninsulated floor is determined separately for each two-meter zone, numbered starting from outer wall building. A total of four such strips 2 m wide are usually taken into account, considering the ground temperature in each zone to be constant. The fourth zone includes the entire surface of the uninsulated floor within the boundaries of the first three stripes. Heat transfer resistance is assumed: 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 the floor surface on the ground and adjacent recessed walls when calculating heat loss
In the case of recessed rooms with a soil base floor: the area of the first zone adjacent to the wall surface is taken into account twice in the calculations. This is quite understandable, since the heat loss of the floor is summed up with the heat loss in the adjacent vertical enclosing structures of the building.
Calculation of heat loss through the floor is carried out for each zone separately, and the results obtained are summarized and used for the thermal engineering justification of the building design. Calculation for temperature zones of external walls of recessed rooms is carried out using formulas similar to those given above.
In calculations of heat loss through an insulated floor (and it is considered such if its design contains layers of material with a thermal conductivity of less than 1.2 W/(m °C)), the value of the heat transfer resistance of a non-insulated floor on the ground increases in each case by the heat transfer resistance of the insulating layer:
Rу.с = δу.с / λу.с,
Where δу.с– thickness of the insulating layer, m; λу.с– thermal conductivity of the insulating layer material, W/(m °C).
Despite the fact that heat loss through the floor of most one-story industrial, administrative and residential buildings rarely exceeds 15% of the total heat loss, and with an increase in the number of storeys sometimes does not reach 5%, the importance the right decision tasks...
Determining 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 problem posed in the title. Conclusions are at the end of the article.
When calculating heat loss, you should always distinguish between the concepts of “building” and “room”.
When performing calculations 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 separate room building, the problem of determining the power and number of thermal devices (batteries, convectors, etc.) required for installation in each specific room in order to maintain a given internal air temperature is solved.
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 due to thermal energy losses 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 layer of material of the enclosing structure in meters;
λ i– coefficient of thermal conductivity of the material in W/(m °C).
Protect the house from external environment ceiling (floor) top floor, external walls, windows, doors, gates and the floor of the lower floor (possibly a basement).
The external environment is the outside air and soil.
Calculation of heat loss from a building is carried out at the calculated outside air temperature for the coldest five-day period of the year in the area where the facility was built (or will be built)!
But, of course, no one forbids you to make calculations 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 method V.D. Machinsky.
The temperature of the soil under a building depends primarily on the thermal conductivity and heat capacity of the soil itself and on the ambient air temperature in the area throughout the year. Since the outside air temperature varies significantly in different climatic zones, the soil also has different temperatures in different periods years at different depths in different areas.
To simplify the solution difficult task To determine heat loss through the floor and walls of the basement into the ground, the technique 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 heat transfer resistance 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 of the soil under the building) 2 meters wide, measured from the inner surface of the external walls along the entire perimeter or (in the case of an underground or basement) a strip of the same width, measured downwards internal surfaces external 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 occupies the entire remaining central area.
In the figure presented just 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 located entirely on the basement floor.
If the building is narrow, then zones 4 and 3 (and sometimes 2) may simply not exist.
Square gender Zone 1 in the corners is taken into account 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 in the calculation by 2 × 2 x 4 = 16 m 2 (for a house with a rectangular plan, i.e. with four corners).
If the structure is not buried in the ground, this means that H =0.
Below is a screenshot of the calculation program in Excel Heat Loss through floors and recessed walls for rectangular buildings.
Zone areas F 1 , F 2 , F 3 , F 4 are calculated according to the rules of ordinary geometry. The task is cumbersome and requires frequent sketching. The program greatly simplifies solving this problem.
The total heat loss to the surrounding soil is determined by the formula in kW:
Q Σ =((F 1 + F1у )/ 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 with values and read the result below.
To determine heat losses into the ground premises zone areas will have to count manually and then substitute into the above formula.
The following screenshot shows, as an example, the calculation in Excel of heat loss through the floor and recessed walls for the lower right (as shown in the picture) basement room.
The amount of heat loss into the ground by each room is equal to the total heat loss into the ground of the entire building!
The figure below shows simplified diagrams standard designs floors and walls.
The floor and walls are considered uninsulated if the thermal conductivity coefficients of the materials ( λ i) of which they consist 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 using the formula:
Rinsulationi = Rinsulatedi + Σ (δ j /λ j )
Here δ j– thickness of the insulation layer in meters.
For floors on joists, the heat transfer resistance is also calculated for each zone, but using a different formula:
Ron the joistsi =1,18*(Rinsulatedi + 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. Sotnikova.
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 issue No. 8 of the ABOK 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 predecessor scientists, is one of the few who, in almost 100 years, tried to move the needle on a topic that worries many heating engineers. I am very impressed by his approach from the point of view of fundamental thermal engineering. But the difficulty of correctly assessing the soil temperature and its thermal conductivity coefficient in the absence of appropriate survey work somewhat shifts A.G.’s methodology. Sotnikov into a 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, cannot be definitely confident in the obtained numerical values.
What is the meaning of Professor A.G.’s methodology? Sotnikova? He suggests that all heat losses through the floor of a buried building “go” deep into the planet, and all heat losses through walls in contact with the ground are ultimately transferred to the surface and “dissolve” in the ambient air.
This seems partly true (without mathematical justification) if there is sufficient depth 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 criticisms made in the previous paragraphs, it was the development of the algorithm of Professor A.G. Sotnikova seems 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 record the dimensions of the basement of the building and the calculated air temperatures in the source data block.
Next, you need to fill in the soil characteristics. As an example, let’s take sandy soil and enter its thermal conductivity coefficient and temperature at a depth of 2.5 meters in January into the initial data. The temperature and thermal conductivity of the soil for your area can be found on the Internet.
The walls and floor will be made of reinforced concrete ( λ =1.7 W/(m°C)) thickness 300mm ( 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 internal surfaces of the floor and walls and on the external surface of the soil in contact with the outside air.
The program performs calculations in Excel using 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 the 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 in )
Thermal resistance of the soil behind the walls:
R 27 = δ conv /λ gr
Heat loss through walls:
Qst = Fst *(tV — tn )/(1/α n +R 27 + Rst +1/α in )
Total heat loss into the ground:
Q Σ = Qpl + Qst
Comments and conclusions.
The heat loss of a building through the floor and walls into the ground, obtained using two different methods, differs significantly. According to the algorithm of A.G. Sotnikov meaning 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 unlikely to correspond to reality. This means that the conditional thickness of the soil δ conv is not defined quite correctly!
In addition, the “bare” reinforced concrete walls that I chose in the example are also a completely unrealistic option for our time.
An attentive reader of the article by A.G. Sotnikova will find a number of errors, most likely not the author’s, but those that arose during 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 loss 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 recording the values... What are these uninsulated walls and floors in the example with Rst = Rpl =2 m 2 °C/W? Their thickness should then be at least 2.4 m! And if the walls and floor are insulated, then it seems incorrect to compare these heat losses with the option of calculating by zone for an uninsulated floor.
R 27 = δ conv /(2*λ gr)=K(cos((h / H )*(π/2)))/K(sin((h / H )*(π/2)))
Regarding the question regarding the presence of a multiplier of 2 λ gr has already been said above.
I divided the complete elliptic integrals by each other. As a result, it turned out that the graph in the article shows the function at λ gr =1:
δ conv = (½) *TO(cos((h / H )*(π/2)))/K(sin((h / H )*(π/2)))
But mathematically it should be correct:
δ conv = 2 *TO(cos((h / H )*(π/2)))/K(sin((h / H )*(π/2)))
or, if the multiplier is 2 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 values that are underestimated by 2 or 4 times...
It turns out that everyone has no choice but to continue to either “count” or “determine” heat loss through the floor and walls into the ground by zone? No other worthy method has been invented in 80 years. Or did they come up with it, but didn’t finalize it?!
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 will be glad to hear the opinions of experts on this topic in the comments. I would like to fully understand A.G.’s algorithm. Sotnikov, because it actually has a more rigorous thermophysical justification than the generally accepted method.
I beg respectful author's work download a file with calculation programs after subscribing to article announcements!
P.S. (02/25/2016)
Almost a year after writing the article, we managed to sort out the questions raised just above.
Firstly, a program for calculating heat loss in Excel using the method of A.G. Sotnikova believes everything is correct - exactly according to the formulas of A.I. Pekhovich!
Secondly, formula (3) from the article by A.G., which brought confusion into my reasoning. Sotnikova should not 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. Sotnikova is not a correct entry! But then the graph was built, and the example was calculated using the correct formulas!!!
This is how it should be according to A.I. Pekhovich (page 110, additional task to paragraph 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)))
The essence of thermal calculations of premises, to one degree or another located in the ground, comes down to determining the influence of atmospheric “cold” on their thermal regime, or more precisely, to what extent a certain soil insulates a given room from atmospheric temperature effects. Because Since the thermal insulation properties of soil depend on too many 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 (the influence of the atmosphere is reduced to a greater extent). 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 on the ground) or a depth (if it is walls on 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 further away the zone (the higher its serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can draw a simple conclusion that the further a certain 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.
Thus, the counting of conditional zones begins along the wall from ground level, provided that there are walls on the ground. If there are no ground walls, then the first zone will be the floor strip closest to the outer wall. Next, zones 2 and 3 are numbered, each 2 meters wide. The remaining zone is zone 4.
It is important to consider that the zone can begin 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 heat transfer resistance values of the non-insulated floor by zone are equal to:
zone 1 - R n.p. =2.1 sq.m*S/W
zone 2 - R n.p. =4.3 sq.m*S/W
zone 3 - R n.p. =8.6 sq.m*S/W
zone 4 - R n.p. =14.2 sq.m*S/W
To calculate the heat transfer resistance for insulated floors, you can use the following formula:
— heat transfer resistance of each zone of the non-insulated floor, sq.m*S/W;
— insulation thickness, m;
— thermal conductivity coefficient of insulation, W/(m*C);
To calculate heat loss through the floor and ceiling, the following data will be required:
- house dimensions 6 x 6 meters.
- The floors are edged boards, tongue-and-groove 32 mm thick, covered with chipboard 0.01 m thick, insulated with 0.05 m thick mineral wool insulation. There is an underground space under the house for storing vegetables and canning. In winter, the temperature in the underground averages +8°C.
- Ceiling - the ceilings are made of wooden panels, the ceilings are insulated on the attic side with mineral wool insulation, layer thickness 0.15 meters, with a vapor-waterproofing layer. The attic space is not insulated.
Calculation of heat loss through the floor
R boards =B/K=0.032 m/0.15 W/mK =0.21 m²x°C/W, where B is the thickness of the material, K is the thermal conductivity coefficient.
R chipboard =B/K=0.01m/0.15W/mK=0.07m²x°C/W
R insulation =B/K=0.05 m/0.039 W/mK=1.28 m²x°C/W
Total value R of the floor =0.21+0.07+1.28=1.56 m²x°C/W
Considering that the underground temperature in winter is constantly around +8°C, the dT required for calculating heat loss is 22-8 = 14 degrees. Now we have all the data to calculate heat loss through the floor:
Q floor = SxdT/R=36 m²x14 degrees/1.56 m²x°C/W=323.07 Wh (0.32 kWh)
Calculation of heat loss through the ceiling
Ceiling area is the same as the floor S ceiling = 36 m2
When calculating the thermal resistance of the ceiling, we do not take into account wooden panels, because they do not have a tight connection with each other and do not act as a heat insulator. Therefore, the thermal resistance of the ceiling is:
R ceiling = R insulation = insulation thickness 0.15 m/thermal conductivity of insulation 0.039 W/mK=3.84 m²x°C/W
We calculate heat loss through the ceiling:
Ceiling Q =SхdT/R=36 m²х52 degrees/3.84 m²х°С/W=487.5 Wh (0.49 kWh)
Previously, we calculated the heat loss of the floor along the ground for a house 6 m wide with a ground water level of 6 m and +3 degrees in depth.
Results and problem statement here -
Heat loss to the street air and deep into the ground was also taken into account. Now I will separate the flies from the cutlets, namely, I will carry out the calculation purely into the ground, excluding 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. GWL 6m, +3 at GWL
2. GWL 6m, +6 at GWL
3. GWL 4m, +3 at GWL
4. GWL 10m, +3 at GWL.
5. GWL 20m, +3 at GWL.
Thus, we will close the questions related to the influence of groundwater depth and the influence of temperature on groundwater.
The calculation is, as before, stationary, not taking into account seasonal fluctuations and generally not taking into account outside air
The conditions are the same. The ground has Lyamda=1, walls 310mm Lyamda=0.15, floor 250mm Lyamda=1.2.
The results, as before, are two pictures (isotherms and “IR”), and numerical ones - resistance to heat transfer into the soil.
Numerical results:
1. R=4.01
2. R=4.01 (Everything is normalized for the difference, it shouldn’t have been otherwise)
3. R=3.12
4. R=5.68
5. R=6.14
Regarding the sizes. If we correlate them with the depth of the groundwater level, 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 unity (or rather the inverse coefficient of thermal conductivity of the soil) for an infinitely large house, but in our case the dimensions of the house are comparable to the depth to which heat loss occurs, and the smaller the house compared to the depth, the smaller this ratio should be.
The resulting R/L relationship should depend on the ratio of the width of the house to the ground level (B/L), plus, as already said, for B/L->infinity R/L->1/Lamda.
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 an exponential one (see graph in the comments).
Moreover, the exponent can be written more simply 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 groundwater level in the considered range, we have a formula for calculating the resistance to heat transfer in the groundwater level:
R=(L/Lamda)*EXP(-L/(3B))
here L is the depth of the groundwater level, Lyamda is the coefficient of 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 approximately infinity (high GWL).
If we use the formula for deeper groundwater levels, the formula gives a significant error, for example, for a 50m depth and 6m width of a house we have: R=(50/1)*exp(-50/18)=3.1, which is obviously too small.
Have a nice day everyone!
Conclusions:
1. An increase in the depth of the groundwater level does not lead to a corresponding reduction in heat loss in groundwater, since everything is involved large quantity soil.
2. At the same time, systems with a ground water level of 20 m or more may never reach the stationary level 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 the tape or blind area.
4. A formula is derived from the results, use it to your health (at your own peril and risk, of course, please know in advance that I am in no way responsible for the reliability of the formula and other results and their applicability in practice).
5. It follows from a small study carried out below in the commentary. Heat loss to the street reduces heat loss to the ground. Those. It is incorrect to consider the two heat transfer processes separately. And by increasing thermal protection from the street, we increase heat loss into the ground and thus it becomes clear why the effect of insulating the outline of the house obtained earlier is not so significant.
