Heat transfer through the enclosure of a house is complex process. In order to take into account these difficulties as much as possible, measurements of premises 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 internal surface outer 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 inner surface floors on 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 counting of zones begins along the wall from ground level, and if there are no walls along the ground, then zone I is the floor strip closest to the outer 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 an uninsulated floor, standard heat transfer resistance values are provided:
- 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.
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 a 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 layers
(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;
– design temperature internal air°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:
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 infinitely big house, in our case the dimensions of the house are comparable to the depth to which heat loss occurs and what smaller 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.
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, the numbering of which starts from the outer wall of the 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).
To calculate heat loss through the floor and ceiling, the following data will be required:
- house dimensions 6 x 6 meters.
- Floors - edged boards, tongue and groove 32 mm thick, covered with chipboard 0.01 m thick, insulated mineral wool insulation 0.05 m thick. Under the house there is an underground space 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. Attic space uninsulated.
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 boards, 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)
