Calculation of heat loss from the floor to the ground in angular units. Thermal engineering calculation of floors located on the ground Floor area by zone example
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 set temperature internal air.
The air in the building is heated by receiving thermal 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 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 the ceiling (floor) of the upper 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, then the soil has different temperatures at different times of the year 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 buried 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 down the internal surfaces of the external walls from edges of the soil.
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 calculated 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 a program for calculating heat loss through the floor and recessed walls in Excel 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)))
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, as more and more soil is involved.
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.
Heat transfer through the enclosure of a house is complex process. To take into account these difficulties as much as possible, measurements of rooms when calculating heat loss are done according to certain rules, which provide for a conditional increase or decrease in area. Below are the main provisions of these rules.
Rules for measuring areas of enclosing structures: a - section of a building with an attic floor; b - section of a building with a combined covering; c - building plan; 1 - floor above the basement; 2 - floor on joists; 3 - floor on the ground;
The area of windows, doors and other openings is measured by the smallest construction opening.
The area of the ceiling (pt) and floor (pl) (except for the floor on the ground) is measured between the axes of the internal walls and the inner surface of the external wall.
The dimensions of the external walls are taken horizontally along the outer perimeter between the axes of the internal walls and the outer corner of the wall, and in height - on all floors except the bottom: from the level of the finished floor to the floor of the next floor. On top floor the top of the outer wall coincides with the top of the covering or attic floor. On the lower floor, depending on the floor design: a) from the inner surface of the floor along the ground; b) from the preparation surface for the floor structure on the joists; c) from the bottom edge of the ceiling above an unheated underground or basement.
When determining heat loss through interior walls their areas are measured along the internal perimeter. Heat losses through the internal enclosures of rooms can be ignored if the difference in air temperatures in these rooms is 3 °C or less.
Breakdown of the floor surface (a) and recessed parts of external walls (b) into design zones I-IV
The transfer of heat from a room through the structure of the floor or wall and the thickness of the soil with which they come into contact is subject to complex laws. To calculate the heat transfer resistance of structures located on the ground, a simplified method is used. The surface of the floor and walls (where the floor is considered as a continuation of the wall) is divided along the ground into strips 2 m wide, parallel to the junction of the outer wall and the ground surface.
The zones are counted along the wall from the ground level, and if there are no walls on the ground, then zone I is the floor strip closest to external wall. The next two stripes will be numbered II and III, and the rest of the floor will be zone IV. Moreover, one zone can begin on the wall and continue on the floor.
A floor or wall that does not contain insulating layers made of materials with a thermal conductivity coefficient of less than 1.2 W/(m °C) is called uninsulated. The heat transfer resistance of such a floor is usually denoted by R np, m 2 °C/W. For each zone of the non-insulated floor there are standard values heat transfer resistance:
- zone I - RI = 2.1 m 2 °C/W;
- zone II - RII = 4.3 m 2 °C/W;
- zone III - RIII = 8.6 m 2 °C/W;
- zone IV - RIV = 14.2 m 2 °C/W.
If the structure of a floor located on the ground has insulating layers, it is called insulated, and its heat transfer resistance R unit, m 2 °C/W, is determined by the formula:
R up = R np + R us1 + R us2 ... + R usn
Where R np is the heat transfer resistance of the considered zone of the non-insulated floor, m 2 °C/W;
R us - heat transfer resistance of the insulating layer, m 2 °C/W;
For a floor on joists, the heat transfer resistance Rl, m 2 °C/W, is calculated using the formula.
Heat loss through a floor located on the ground is calculated by zone according to. To do this, the floor surface is divided into strips 2 m wide, parallel to the outer walls. The strip closest to the outer wall is designated the first zone, the next two strips are the second and third zones, and the rest of the floor surface is the fourth zone.
When calculating heat loss basements in this case, the division into strip-zones is made from ground level along the surface of the underground part of the walls and further along the floor. Conditional heat transfer resistances for zones in this case are accepted and calculated in the same way as for an insulated floor in the presence of insulating layers, which in this case are layers of the wall structure.
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 is the heat transfer resistance of an insulated floor on the ground, m 2 ∙°C/W, calculated by the formula:
= + Σ , (2.2)
where is the heat transfer resistance of the uninsulated floor of the i-th zone;
δ j – thickness of the j-th layer of the insulating structure;
λ j is the thermal conductivity coefficient of the material the layer consists of.
For all areas of non-insulated floors there is data on heat transfer resistance, which is accepted according to:
2.15 m 2 ∙°С/W – for the first zone;
4.3 m 2 ∙°С/W – for the second zone;
8.6 m 2 ∙°С/W – for the third zone;
14.2 m 2 ∙°С/W – for the fourth zone.
In this project, the floors on the ground have 4 layers. The floor structure is shown in Figure 1.2, the wall structure is shown in Figure 1.1.
An example of thermal engineering calculation of floors located on the ground for room 002 ventilation chamber:
1. The division into zones in the ventilation chamber is conventionally presented in Figure 2.3.
Figure 2.3. Division of the ventilation chamber into zones
The figure shows that the second zone includes part of the wall and part of the floor. Therefore, the heat transfer resistance coefficient of this zone is calculated twice.
2. Let’s determine the heat transfer resistance of an insulated floor on the ground, , m 2 ∙°C/W:
2,15 + 
= 4.04 m 2 ∙°С/W,
4,3 + = 7.1 m 2 ∙°С/W,
4,3 + 
= 7.49 m 2 ∙°С/W,
8,6 + = 11.79 m 2 ∙°С/W,
14,2 + = 17.39 m 2 ∙°C/W.
According to SNiP 41-01-2003, the floors of the building floors, located on the ground and joists, are delimited into four zone-strips 2 m wide parallel to the outer walls (Fig. 2.1). When calculating heat loss through floors located on the ground or joists, the surface of the floor areas near the corner of the external walls ( in zone I ) is entered into the calculation twice (square 2x2 m).
Heat transfer resistance should be determined:
a) for uninsulated floors on the ground and walls located below ground level, with thermal conductivity l ³ 1.2 W/(m×°C) in zones 2 m wide, parallel to the external walls, taking R n.p. . , (m 2 ×°C)/W, equal to:
2.1 – for zone I;
4.3 – for zone II;
8.6 – for zone III;
14.2 – for zone IV (for the remaining floor area);
b) for insulated floors on the ground and walls located below ground level, with thermal conductivity l.s.< 1,2 Вт/(м×°С) утепляющего слоя толщиной d у.с. , м, принимая R u.p. , (m 2 ×°C)/W, according to the formula
c) thermal resistance to heat transfer of individual floor zones on joists R l, (m 2 ×°C)/W, determined by the formulas:
I zone – 
;
II zone – 
;
III zone – 
;
IV zone – 
,
where , , , are the values of thermal resistance to heat transfer of individual zones of non-insulated floors, (m 2 × ° C)/W, respectively numerically equal to 2.1; 4.3; 8.6; 14.2; – the sum of the values of thermal resistance to heat transfer of the insulating layer of floors on joists, (m 2 × ° C)/W.
The value is calculated by the expression:
, (2.4)
here is the thermal resistance of closed air gaps
(Table 2.1); δ d – thickness of the layer of boards, m; λ d – thermal conductivity of wood material, W/(m °C).
Heat loss through a floor located on the ground, W:
, (2.5)
where , , , are the areas of zones I, II, III, IV, respectively, m 2 .
Heat loss through the floor located on the joists, W:
, (2.6)
Example 2.2.
Initial data:
– first floor;
– external walls – two;
– floor construction: concrete floors covered with linoleum;
– estimated internal air temperature °C;
Calculation procedure.
Rice. 2.2. Fragment of the plan and location of floor areas in living room No. 1
(for examples 2.2 and 2.3)
2. In living room No. 1 only the first and part of the second zone are located.
I-th zone: 2.0´5.0 m and 2.0´3.0 m;
II zone: 1.0´3.0 m.
3. The areas of each zone are equal:
4. Determine the heat transfer resistance of each zone using formula (2.2):
(m 2 ×°C)/W,
(m 2 ×°C)/W.
5. Using formula (2.5), we determine the heat loss through the floor located on the ground:
Example 2.3.
Initial data:
– floor construction: wooden floors on joists;
– external walls – two (Fig. 2.2);
– first floor;
– construction area – Lipetsk;
– estimated internal air temperature °C; °C.
Calculation procedure.
1. We draw a plan of the first floor to scale indicating the main dimensions and divide the floor into four zones-strips 2 m wide parallel to the external walls.
2. In living room No. 1 only the first and part of the second zone are located.
We determine the dimensions of each zone-strip: