Good afternoon!
I decided to post here the results of calculations for floor insulation on the ground. The calculations were carried out using the Therm 6.3 program.
Ground floor - concrete slab 250mm thick with a thermal conductivity coefficient of 1.2
Walls - 310 mm with a thermal conductivity coefficient of 0.15 (aerated concrete or wood)
For simplicity, the wall to the ground. There can be many options for warming and cold bridges of the node, for simplicity we omit them.
Soil - with a coefficient of thermal conductivity of 1. Wet clay or wet sand. Dry - more heat-shielding.
Warming. There are 4 options here:
1. There is no insulation. Just a slab on the ground.
2. The blind area is insulated with a width of 1m, a thickness of 10cm. EPPS insulation. The top layer of the blind area itself was not taken into account, since it does not play a big role.
3. The foundation tape is insulated at a depth of 1m. Insulation is also 10cm, EPS. Concrete is not traced as it is close to the ground in terms of thermal conductivity.
4. The stove under the house is insulated. 10cm, EPS.
The EPSS thermal conductivity coefficient was taken equal to 0.029.
The width of the slab is taken as 5.85 m.
Initial data on temperatures:
- inside +21;
- outside -3;
- at a depth of 6m +3.
6m here is the GWL estimate. I took 6m because it's the closest to my house, although I don't have ground floors, the results also apply to my warm underground.
You can see the results in graphical form. Attached in two versions - with isotherms and "IR".
Digitally obtained data for the floor surface in the form of U-factor, the reciprocal of our resistance to heat transfer ([R]=K*m2/W).
In terms of results, the results are as follows (on average by sex):
1.R=2.86
2.R=3.31
3.R=3.52
4.R=5.59
For me, the results are very interesting. In particular a sufficiently 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 insulate the soil when there is groundwater nearby, and then we have option 4, with partially cut off soil from thermal circuit. Moreover, with a close GWL, we will not get 5.59. since the 6 m of soil taken into account do not participate in the insulation. One should expect R~3 in this case or so.
It is also very significant that the edge of the slab in the calculated version is quite warm 17.5oC according to the first non-insulated version, therefore, freezing, condensate and mold are not expected there, even if the temperature gradient doubles (-27 outside). Moreover, it should be understood that peak temperatures do not play any role in such calculations, since the system is very heat-intensive and the soil freezes for weeks or months.
Options 1,2,3. And especially option 2 - the most inertial. Here, the soil is involved in the thermal circuit, not only the one that is directly under the house, but also under the blind area. The time for establishing the temperature regime as in the figure is years and in fact temperature regime will be the average for the year. A period of about 3 months manages to involve only 2-3 m of soil in heat exchange. But this is a separate story, so for now I will conclude, I will only note that the characteristic time is proportional to the thickness of the layer squared. Those. if 2m is 3 months, then 4m is already 9 months.
I also note that in practice, probably, with a relatively small groundwater level (such as 4.5m and below), worse results should be expected. thermal insulation properties soil due to evaporation of water from it. Unfortunately, I am not familiar with the tool that could carry out the calculation under the conditions of evaporation in the soil. Yes, and with the original data there is a big problem.
Evaluation with the influence of evaporation in the soil was carried out as follows.
I dug up the data that the water in the loams rises by capillary forces from the groundwater level by 4-5m
Well, I will use this figure as the initial data.
I will brazenly assume that the same 5m are saved in my calculation under any circumstances.
In 1 m of soil, steam diffuses to the floor, and the value of the vapor permeability coefficient can be dug. The coefficient of vapor permeability of sand is 0.17, adobe 0.1. Well, for reliability, I'll take 0.2 mg / m / h / Pa.
At a depth of a meter in the design options, except for option 4, about 15 degrees.
The total water vapor pressure there is 1700 Pa (100% rel).
Indoors, we take 21 degrees 40% (rel.) => 1000Pa
In total, we have a 700Pa vapor pressure gradient per 1m of clay with Mu=0.2 and 0.25m of concrete with Mu=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 steam flow from the soil 0.13*700=90 mg/m2/h=2.5e-8 kg/m2/s
We multiply by the heat of evaporation of water 2.3 MJ / kg and get additional heat loss for evaporation => 0.06 W / m2. Little things it is. If we speak in the language of R (heat transfer resistance), then such an allowance for moisture leads to a decrease in R by about 0.003, i.e. insignificant.
Attachments:
Comments
Example 1
It is required to determine the thickness of the concrete underlay in the passage of the warehouse. Floor covering, concrete, thickness h 1 = 2.5 cm. Floor load - from MAZ-205 cars; base soil - loam. Ground water is absent.
For the MAZ-205 vehicle, which has two axles with a wheel load of 42 kN, the calculated wheel load according to the formula ( 6 ):
R p \u003d 1.2 42 \u003d 50.4 kN
The wheel track area of the MAZ-205 is 700 cm 2
According to the formula ( 5 ) we calculate:
r = D/2 = 30/2 = 15 cm
According to the formula ( 3 ) r p \u003d 15 + 2.5 \u003d 17.5 cm
2. For loamy soil of the base in the absence of groundwater according to Table. 2.2
To 0 \u003d 65 N / cm 3:
For the underlying layer, we will take concrete in terms of compressive strength B22.5. Then in the travel area in warehouse where no stationary floor is installed on the floors technological equipment(according to par. 2.2 group I), when loaded from trackless Vehicle according to the table 2.1 Rδt = 1.25 MPa, E b = 28500 MPa.
3. σ R. The load from the car, according to par. 2.4 , is the load simple form and is transmitted along the trace of a round shape. Therefore, the calculated bending moment is determined by the formula ( 11 ). According to par. 2.13 let's ask approximately h\u003d 10 cm. Then according to p. 2.10 accept l= 44.2 cm. For ρ = r R / l\u003d 17.5 / 44.2 \u003d 0.395 according to the table. 2.6 find K 3 = 103.12. According to the formula ( 11 ): M p = To 3 · R p \u003d 103.12 50.4 \u003d 5197 N cm / cm. According to the formula ( 7 ) calculate the stresses in the plate:
Tension in slab thickness h= 10 cm exceeds design resistance Rδt = 1.25 MPa. In accordance with par. 2.13 Let's repeat the calculation by asking great value h= 12 cm, then l= 50.7 cm; p = r R / l = 17,5/50,7 = 0,345; To 3 = 105,2; M R= 105.2 50.4 = 5302 N cm / cm
Received σ R= 1.29 MPa differs from the design resistance Rδt = 1.25 MPa (see tab. 2.1 ) by less than 5%, therefore, we accept the underlying layer of concrete in terms of compressive strength class B22.5 with a thickness of 12 cm.
Example 2
For mechanical workshops, it is required to determine the thickness of the concrete sub-base used as a floor without covering ( h 1 = 0 cm). Floor load - from the machine weighing P p= 180 kN, standing directly on the underlying layer, is evenly distributed along the track in the form of a rectangle measuring 220×120 cm. There are no special requirements for the deformation of the base. The base soil is fine sand, located in the zone of capillary rise of groundwater.
1. Let's determine the design parameters.
Estimated track length according to par. 2.5 and according to the formula ( 1 ) a p \u003d a \u003d 220 cm. Estimated track width according to the formula ( 2 ) b p = b = 120 cm. For the base soil of fine sand located in the zone of capillary rise of groundwater, according to Table. 2.2 K 0 \u003d 45 N / cm 3. For the underlying layer, we will take concrete in terms of compressive strength class B22.5. Then in mechanical workshops, where stationary technological equipment is installed on the floors without special requirements for the deformation of the base (according to paragraph 1 of Art. 2.2 group II), with a fixed load according to table. 2.1 Rδt = 1.5 MPa, E b = 28500 MPa.
2. Determine the tensile stress in the concrete of the slab during bending σ R. The load is transferred along the trail rectangular shape and, according to par. 2.5 , is a load of simple form.
Therefore, the calculated bending moment is determined by the formula ( 9 ). According to par. 2.13 let's ask approximately h\u003d 10 cm. Then according to p. 2.10 accept l= 48.5 cm.
Taking into account α = a p / l= 220/48.5 = 4.53 and β = b p / l\u003d 120 / 48.5 \u003d 2.47 according to the table. 2.4 find To 1 = 20,92.
According to the formula ( 9 ): M p = To one · R p \u003d 20.92 5180 \u003d 3765.6 N cm / cm.
According to the formula ( 7 ) calculate the stress in the plate:
Tension in slab thickness h= 10 cm significantly smaller Rδt = 1.5 MPa. In accordance with par. 2.13 Let's recalculate and keep h\u003d 10 cm, we find a lower brand of concrete of the underlying layer slab, at which σ R » Rδt. Let's take concrete of class B15 for compressive strength, for which Rδt = 1.2 MPa, E b = 23000 MPa.
Then l= 46.2 cm; α = a p / l= 220/46.2 = 4.76 and β = b p / l= 120/46.2 = 2.60; according to the table 2.4 To 1 = 18,63;. M R\u003d 18.63 180 \u003d 3353.4 N cm / cm.
The resulting tensile stress in a slab of concrete of compressive strength class B15 is less Rδt = 1.2 MPa. Let's take the underlying layer of concrete of the compressive strength class B15 with a thickness h= 10 cm.
Example 3
It is required to determine the thickness of the concrete underlayment of the floor in the machine-building shop under loads from automated line machines and ZIL-164 vehicles. The layout of the loads is shown in fig. 1 in", 1 in"", 1 at """. The center of the vehicle wheel track is 50 cm from the edge of the machine track. Weight of the machine in working condition R R= 150 kN is distributed evenly over the area of a rectangular track 260 cm long and 140 cm wide.
The floor covering is the hardened surface of the underlying layer. The base soil is sandy loam. The base is located in the zone of capillary rise of groundwater
Let's define the calculated parameters.
For the ZIL-164 car, which has two axles with a wheel load of 30.8 kN, the calculated wheel load according to the formula ( 6 ):
R R= 1.2 30.8 = 36.96 kN
The wheel track area of the ZIL-164 is 720 cm 2
According to par. 2.5
r R = r = D/2 = 30/2 = 15 cm
For sandy soil base, located in the zone of capillary rise of groundwater, according to Table. 2.2 To 0 \u003d 30 N / cm 3. For the underlying layer, we will take concrete of the compressive strength class B22.5. Then for the machine-building shop, where an automated line is installed on the floors (according to paragraph 2.2 group IV), with the simultaneous action of fixed and dynamic loads according to Table. 2.1 Rδt = 0.675 MPa, E b= 28500 MPa.
Let's ask approximately h\u003d 10 cm, then according to p. 2.10 accept l= 53.6 cm. In this case, the distance from the center of gravity of the car wheel track to the edge of the machine track is 50 cm l = 321.6 cm, i.e. according to par. 2.4 The loads acting on the floor are complex loads.
In accordance with par. 2.17 set the position of the calculation centers in the centers of gravity of the trace of the machine (O 1) and the wheel of the car (O 2). From the load layout (Fig. 1 c") it follows that for the calculation center O 1 it is not clear which direction of the OS axis should be set. Therefore, we define the bending moment as with the direction of the OS axis parallel to the long side of the machine trace (Fig. 1 c"), and perpendicular to this side (Fig. 1 in""). For the calculation center O 2, we will take the direction of the OS through the centers of gravity of the traces of the machine and the wheel of the car (Fig. 1 in""").
Calculation 1 Determine the tensile stress in the concrete of the slab during bending σ R for the calculation center O 1 when the OS is directed parallel to the long side of the machine track (Fig. 1 c"). In this case, the load from the machine with a rectangular track refers to a load of a simple type. For the machine track according to p. 2.5 without floor covering h 1 \u003d 0 cm) a p \u003d a \u003d 260 cm; b p \u003d b \u003d 140 cm.
Taking into account the values α = a р / l= 260/53.6 = 4.85 and β = b p / l\u003d 140 / 53.6 \u003d 2.61 according to the table. 2.4 find K 1 = 18,37.
For machine R 0 = R R= 150 kN in accordance with p. 2.14 determined by the formula ( 9 ):
M p = To one · R p \u003d 18.37 150 \u003d 27555.5 N cm / cm.
Coordinates of the center of gravity of the car wheel track: x i= 120 cm and y i= 0 cm.
Taking into account the ratios x i /l= 120/53.6 = 2.24 and y i /l\u003d 0 / 53.6 \u003d 0 according to the table. 2.7 find To 4 = -20,51.
Bending moment in calculation center O 1 from the car wheel according to the formula ( 14 ):
M i\u003d -20.51 36.96 \u003d -758.05 N cm / cm.
13 ):
M p I = M 0 + Σ M i= 2755.5 - 758.05 = 1997.45 N cm/cm
7 ):
Calculation 2 Determine the tensile stress in the concrete of the slab during bending σ R II for settlement center O 1 when the OS is directed perpendicular to the long side of the machine trace (Fig. 1 in""). We divide the area of the machine footprint into elementary areas according to paragraph 1. 2.18 . Compatible with clearing house O 1 center of gravity of elementary area square shape with side length a p \u003d b p \u003d 140 cm.
Let's define loads R i per elementary area according to the formula ( 15 ), for which we first determine the area of the machine footprint F\u003d 260 140 \u003d 36400 cm 2;
To determine the bending moment M 0 from load R 0 is calculated for an elementary square-shaped platform with the center of gravity in the calculation center O 1 values α = β = a p / l= b p / l\u003d 140 / 53.6 \u003d 2.61 and taking them into account according to table. 2.4 find K 1=36.0; according to the instructions of 2.14 and formula ( 9 ) we calculate:
M 0 = To one · R 0 \u003d 36.0 80.8 \u003d 2908.8 N cm / cm.
M i, from loads located outside the calculation center O 1 . The calculated data are given in table. 2.10 .
Table 2.10
Calculated data with the calculation center O 1 and the direction of the y-axis perpendicular to the long side of the machine trace
| I | x i | y i | x i /l | y i /l | To 4 according to the table. 2.7 | P i, kN | n i number of loads | M i = n i · To four · P i |
|
| 1 | 0 | 120 | 0 | 2,24 | 9,33 | 36,96 | 1 | 363,3 |
|
| 2 | 120 | 35 | 1,86 | 0,65 | -17,22 | 17,31 | 4 | -1192,3 |
|
| Σ M i= -829.0 Ncm/cm |
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Estimated bending moment from the wheel of the car and the machine according to the formula ( 13 ):
M p II = M 0 + Σ M i= 2908.8 - 829.0 = 2079.8 N cm / cm
Tensile stress in the plate during bending according to the formula ( 7 ):
Calculation 3 Determine the tensile stress in the concrete of the slab during bending σ R III for the settlement center O 2 (Fig. 1 in """). Divide the area of the machine footprint into elementary areas according to p. 2.18 . Let's define loads R i per elementary area, according to the formula ( 15 ).
Let us determine the bending moment from the load created by the pressure of the car wheel, for which we find ρ = r R / l= 15/53.6 = 0.28; according to the table 2.6 find To 3 = 112.1. According to the formula ( 11 ):M 0 = To 3 · R p \u003d 112.1 36.96 \u003d 4143.22 N cm / cm.
Let us determine the total bending moment Σ M i from loads located outside the settlement center O 2 . The calculated data are given in table. 2.11 .
Table 2.11
Design data with settlement center O 2
| I | x i | y i | x i /l | y i /l | To 4 according to the table. 2.7 | P i, kN | n i number of loads | M i = n i · To four · P i |
|
| 1 | 0 | 65 | 0 | 1,21 | 40,97 | 4,9 | 1 | 200,75 |
|
| 2 | 0 | 100 | 0 | 1,87 | 16,36 | 6,6 | 1 | 107,98 |
|
| 3 | 0 | 155 | 0 | 2,89 | 2,89 | 11,5 | 1 | 33,24 |
|
| 4 | 40 | 65 | 0,75 | 1,21 | 19,1 | 4,9 | 2 | 187,18 |
|
| 5 | 40 | 100 | 0,75 | 1,87 | 8,44 | 6,6 | 2 | 111,41 |
|
| 6 | 40 | 155 | 0,75 | 2,89 | 1,25 | 11,5 | 2 | 28,75 |
|
| 7 | 95 | 65 | 1,77 | 1,21 | -10,78 | 8,7 | 2 | -187,57 |
|
| 8 | 95 | 100 | 1,77 | 1,87 | -5,89 | 11,5 | 2 | -135,47 |
|
| 9 | 95 | 155 | 1,77 | 2,89 | -2,39 | 20,2 | 2 | -96,56 |
|
| Σ M i= 249.7 N cm/cm |
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Estimated bending moment from the wheel of the car and the machine according to the formula ( 13 ):
M p III = M 0 + Σ M i= 4143.22 + 249.7 = 4392.92 N cm/cm
Tensile stress in the plate during bending according to the formula ( 7 ):
more Rδt = 0.675 MPa, as a result of which we repeat the calculation, setting a large value h. We will carry out the calculation only according to the loading scheme with the calculation center O 2 , for which the value σ R III in the first calculation turned out to be the largest.
For the recalculation, we tentatively set h\u003d 19 cm, then according to p. 2.10 accept l= 86.8 cm; p = r R / l =15/86,8 = 0,1728; To 3 = 124,7; M 0 = To 3 · R p\u003d 124.7 36.96 \u003d 4608.9 N cm / cm.
Let us determine the total bending moment from loads located outside the calculation center O 2 . The calculated data are given in table. 2.12 .
Table 2.12
Calculated data for recalculation
| I | x i | y i | x i /l | y i /l | To 4 according to the table. 2.7 | P i, kN | n i number of loads | M i = n i · To four · P i |
|
| 1 | 0 | 65 | 0 | 0,75 | 76,17 | 4,9 | 1 | 373,23 |
|
| 2 | 0 | 100 | 0 | 1,15 | 44,45 | 6,6 | 1 | 293,37 |
|
| 3 | 0 | 155 | 0 | 1,79 | 18,33 | 11,5 | 1 | 210,79 |
|
| 4 | 40 | 65 | 0,46 | 0,75 | 48,36 | 4,9 | 2 | 473,93 |
|
| 5 | 40 | 100 | 0,46 | 1,15 | 32,39 | 6,6 | 2 | 427,55 |
|
| 6 | 40 | 155 | 0,46 | 1,79 | 14,49 | 11,5 | 2 | 333,27 |
|
| 7 | 95 | 65 | 1,09 | 0,75 | 1,84 | 8,7 | 2 | 32,02 |
|
| 8 | 95 | 100 | 1,09 | 1,15 | 3,92 | 11,5 | 2 | 90,16 |
|
| 9 | 95 | 155 | 1,09 | 1,79 | 2,81 | 20,2 | 2 | 113,52 |
|
| Σ M i= 2347.84 N cm/cm. |
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M p= M 0 + Σ M i= 4608.9 + 2347.84 = 6956.82 Ncm/cm
Tensile stress in the plate during bending according to the formula ( 7 ):
Received value σ R= 0.67 MPa differs from Rδt = 0.675 MPa by less than 5%. We accept the underlying layer of concrete of the compressive strength class B22.5 with a thickness h= 19 cm.
Heat losses through the floor located on the ground are calculated by zones 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 - the second and third zones, and the rest of the floor surface - the fourth zone.
When calculating heat loss basements breakdown into strip-zones in this case is made from the 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 the layers of the wall structure.
The heat transfer coefficient K, W / (m 2 ∙ ° С) for each zone of the insulated floor on the ground is determined by the formula:
where - heat transfer resistance of the insulated floor on the ground, m 2 ∙ ° С / W, is calculated by the formula:
= + Σ , (2.2)
where is the heat transfer resistance of the uninsulated floor of the i-th zone;
δ j is the thickness of the jth layer of the insulating structure;
λ j is the coefficient of thermal conductivity of the material of which the layer consists.
For all areas of an uninsulated floor, there is data on heat transfer resistance, which are taken 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 a thermal calculation of floors located on the ground for room 002 ventilation chamber:
1. The division into zones in the ventilation chamber is conventionally shown in Figure 2.3.
Figure 2.3. Division into zones of the ventilation chamber
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 the insulated floor on the ground, m 2 ∙ ° С / W:
2,15 + 
\u003d 4.04 m 2 ∙ ° С / W,
4,3 + \u003d 7.1 m 2 ∙ ° С / W,
4,3 + 
\u003d 7.49 m 2 ∙ ° С / W,
8,6 + \u003d 11.79 m 2 ∙ ° С / W,
14,2 + \u003d 17.39 m 2 ∙ ° С / W.
Usually, floor heat losses in comparison with similar indicators of other building envelopes (external walls, window and door openings) are a priori assumed to be insignificant and are taken into account in the calculations of heating systems in a simplified form. Such calculations are based on a simplified system of accounting and correction coefficients of resistance to heat transfer of various building materials.
Considering that theoretical background and the methodology for calculating the heat loss of the 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. The coefficients of thermal conductivity and heat transfer of various building materials, insulation and floor coverings well known, and others physical characteristics to calculate heat loss through the floor is not required. According to their thermal characteristics, floors are usually divided into insulated and non-insulated, structurally - floors on the ground and logs.
The calculation of heat loss through an uninsulated floor on the ground is based on the general formula for estimating heat loss through the building envelope:
where Q are the main and additional heat losses, W;
BUT is the total area of the enclosing structure, m2;
tv , tn- temperature inside the room and outside air, °C;
β - share of additional heat losses in total;
n- correction factor, the value of which is determined by the location of the enclosing structure;
Ro– resistance to heat transfer, m2 °С/W.
Note that in the case of a homogeneous single-layer floor slab, the heat transfer resistance Ro is inversely proportional to the heat transfer coefficient of the uninsulated 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 floor.
The heat loss of an uninsulated floor is determined separately for each two-meter zone, the numbering of which starts from outer wall building. In total, four such strips 2 m wide are taken into account, considering the soil 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 strips. Heat transfer resistance is accepted: 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 of the floor surface on the ground and adjacent recessed walls when calculating heat losses
In the case of recessed rooms with ground 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 losses of the floor are added to the heat losses in the vertical enclosing structures of the building adjacent to it.
Calculation of heat loss through the floor is made for each zone separately, and the results obtained are summed up and used for the thermal engineering justification of the building design. The calculation for the temperature zones of the outer walls of recessed rooms is carried out according to formulas similar to those given above.
In calculations of heat loss through an insulated floor (and it is considered as such if its structure contains layers of material with a thermal conductivity of less than 1.2 W / (m ° C)) the value of the heat transfer resistance of an uninsulated floor on the ground increases in each case by the heat transfer resistance of the insulating layer:
Ru.s = δy.s / λy.s,
where δy.s– thickness of the insulating layer, m; λu.s- thermal conductivity of the material of the insulating layer, W / (m ° C).
