Heat loss of insulated floors in corner rooms calculation. Thermotechnical calculation of floors located on the ground. Remarks and Conclusions

The essence of thermal calculations of premises, to some extent located in the ground, is to determine the influence of atmospheric "cold" on their thermal regime, or rather, to what extent a certain soil isolates a given room from atmospheric temperature effects. Because Since the thermal insulation properties of the 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 more the influence of the atmosphere is reduced). 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 a wall 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 farther the zone (the more it serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can make a simple conclusion that the farther a certain point in the room is from the atmosphere (with a multiplicity of 2 m), the more favorable conditions(in terms of the influence of the atmosphere) it will be.

Thus, the countdown of conditional zones starts along the wall from the ground level, provided that there are walls along 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 start 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 values of heat transfer resistance of the uninsulated floor by zones are equal to:

zone 1 - R n.p. \u003d 2.1 sq.m * C / W

zone 2 - R n.p. \u003d 4.3 sq.m * C / W

zone 3 - R n.p. \u003d 8.6 sq.m * C / W

zone 4 - R n.p. \u003d 14.2 sq. m * C / W

To calculate the heat transfer resistance for insulated floors, you can use the following formula:

- resistance to heat transfer of each zone of an uninsulated floor, sq.m * C / W;

— insulation thickness, m;

- coefficient of thermal conductivity of the insulation, W / (m * C);

To perform the calculation of heat loss through the floor and ceiling, the following data will be required:

The dimensions of the house are 6 x 6 meters.
Floors - edged board, grooved 32 mm thick, sheathed with chipboard 0.01 m thick, insulated with mineral wool insulation 0.05 m thick. Under the house there is an underground for storing vegetables and preserving. In winter, the temperature in the underground averages + 8 ° С.
Ceiling - the ceilings are made of wooden panels, the ceilings are insulated from the attic side with mineral wool insulation, layer thickness is 0.15 meters, with a vapor-waterproofing layer. The attic is uninsulated.

Calculation of heat loss through the floor

R boards \u003d B / K \u003d 0.032 m / 0.15 W / mK \u003d 0.21 m²x ° C / W, where B is the thickness of the material, K is the thermal conductivity coefficient.

R chipboard \u003d B / K \u003d 0.01m / 0.15W / mK \u003d 0.07m²x ° C / W

R insulation \u003d B / K \u003d 0.05 m / 0.039 W / mK \u003d 1.28 m²x ° C / W

The total value of R floor \u003d 0.21 + 0.07 + 1.28 \u003d 1.56 m²x ° C / W

Given that in the underground the temperature in winter is constantly kept at about + 8 ° C, then dT required for calculating heat loss is 22-8 = 14 degrees. Now there is all the data for calculating heat loss through the floor:

Q floor \u003d SxdT / R \u003d 36 m²x14 degrees / 1.56 m²x ° C / W \u003d 323.07 Wh (0.32 kWh)

Calculation of heat loss through the ceiling

The ceiling area is the same as the floor S ceiling = 36 m 2

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 play the role of a heat insulator. Therefore, the thermal resistance of the ceiling:

R ceiling \u003d R insulation \u003d insulation thickness 0.15 m / thermal conductivity of insulation 0.039 W / mK \u003d 3.84 m² x ° C / W

We calculate the heat loss through the ceiling:

Ceiling Q \u003d SхdT / R \u003d 36 m² x 52 degrees / 3.84 m² x ° C / W \u003d 487.5 Wh (0.49 kWh)

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 the theoretical justification and 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 are well known, and no other physical characteristics are required to calculate the heat loss through the floor. By their own thermal performance 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 general formula assessment of heat losses through the building envelope:

where Q are the main and additional heat losses, W;

A 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 a soil base of the 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 added to the heat loss 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 project. 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).

Previously, we calculated the heat loss of the floor on the ground for a house 6m wide with a groundwater level of 6m and +3 degrees in depth.
Results and problem statement here -
The heat losses to the outdoor air and deep into the earth were 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. UGV 6m, +3 on UGV
2. UGV 6m, +6 on UGV
3. UGV 4m, +3 on UGV
4. UGV 10m, +3 on UGV.
5. UGV 20m, +3 on UGV.
Thus, we will close the issues related to the influence of the GWL depth and the influence of temperature on the GWL.
The calculation, as before, is stationary, not taking into account seasonal fluctuations, and generally not taking into account the outside air
The conditions are the same. The ground has Lamda=1, walls 310mm Lamda=0.15, floor 250mm Lamda=1.2.

The results, as before, in two pictures (isotherms and "IR"), and numerical - resistance to heat transfer into the soil.

Numerical results:
1.R=4.01
2. R = 4.01 (Everything is normalized for the difference, otherwise it should not have been)
3.R=3.12
4.R=5.68
5.R=6.14

About the sizes. If we correlate them with the GWL depth, 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 one (more precisely, the reciprocal of the 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 losses are carried out and how smaller house compared with the depth, the smaller this ratio should be.

The resulting dependence R / L should depend on the ratio of the width of the house to the groundwater level (B / L), plus, as already mentioned, with B / L-> infinity R / L-> 1 / Lamda.
In total, there are the following points for an infinitely long house:
L/B | R*lamda/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 the graph in the comments).
Moreover, the exponent can be written in a simpler way 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 GWL in the considered range, we have a formula for calculating the resistance to heat transfer in the GWL:
R=(L/lamda)*EXP(-L/(3B))
here L is the depth of the GWL, Lamda is the 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 you use the formula for deeper groundwater levels, then 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 good day everyone!

Conclusions:
1. An increase in the GWL depth does not lead to a consistent decrease in heat loss in ground water, since everything is involved large quantity soil.
2. At the same time, systems with a GWL of the type of 20m or more may never reach the hospital, which is calculated 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 the tape or blind area is insulated.
4. A formula has been derived from the results, use it to your health (at your own peril and risk, of course, I ask you to 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. Follows from a small study conducted below in the commentary. Heat loss to the street reduces heat loss to the ground. Those. It is incorrect to consider two heat transfer processes separately. And by increasing the thermal protection from the street, we increase heat loss to the ground and thus it becomes clear why the effect of warming the contour of the house, obtained earlier, is not so significant.

The method for calculating the heat loss of premises and the procedure for its implementation (see SP 50.13330.2012 Thermal protection buildings, point 5).

The house loses heat through the building envelope (walls, ceilings, windows, roof, foundation), ventilation and sewerage. The main heat losses go through the building envelope - 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.

At the same time, it is not necessary to take into account heat losses that occur through internal structures if the difference between their temperature and the temperature in neighboring rooms does not exceed 3 degrees Celsius.

Heat loss through building envelopes

Heat loss premises mainly depend on:
1 Temperature differences in the house and on the street (the greater the difference, the higher the losses),
2 Heat-shielding properties of walls, windows, doors, coatings, floors (the so-called enclosing structures of the room).

Enclosing structures are generally not homogeneous in structure. And usually consist of several layers. Example: shell wall = plaster + shell + exterior finish. This design may also include closed air gaps (example: cavities inside bricks or blocks). The above materials have different thermal characteristics from each other. The main such characteristic for a structural layer is its heat transfer resistance R.

Where q is the amount of heat lost square meter enclosing surface (usually measured in W/m2)

ΔT is the difference between the temperature inside the calculated room and outdoor temperature air (the temperature of the coldest five-day period °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. Zones of water procedures 33 °C.

When it comes to a multilayer structure, the resistances of the layers of the structure add up.

δ - layer thickness, m;

λ - calculated factor thermal conductivity of the material of the structure layer, taking into account the operating conditions of the enclosing structures, W / (m2 °C).

Well, now we figured 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 outdoors (the temperature of the coldest five-day period is °C.). ∆T

3. Square fences F (Separate walls, windows, doors, ceiling, floor)

4. Another useful orientation of the building in relation to the cardinal points.

The formula for calculating the heat loss of a fence looks like this:

Qlimit=(ΔT / Rlimit)* Flimit * n *(1+∑b)

Qlimit - heat loss through the building envelope, W

Rogr - resistance to heat transfer, m.sq. ° C / W; (If there are several layers, then ∑ Rlimit of layers)

Fogr – area of the enclosing structure, m;

n is the coefficient of contact of the building envelope with the outside air.

Walling	Coefficient n
1. External walls and coverings (including those ventilated by outside air), attic ceilings (with a roof made of piece materials) and over driveways; ceilings over cold (without enclosing walls) undergrounds in the Northern building-climatic zone
2. Ceilings over cold cellars communicating with outside air; attic ceilings (with a roof made of roll materials); ceilings over cold (with enclosing walls) undergrounds and cold floors in the Northern building-climatic zone	0,9
3. Ceilings over unheated basements with skylights in the walls	0,75
4. Ceilings above 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 considered 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

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 for the resistance to heat transfer of various building materials.

Considering that the theoretical justification and 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 say that these empirical approaches are practically applicable in modern conditions. The coefficients of thermal conductivity and heat transfer of various building materials, insulation and floor coverings are well known, and other physical characteristics are not required to calculate heat loss through the floor. 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;

A 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.

The heat loss of an uninsulated floor is determined separately for each two-meter zone, the numbering of which starts from the outer wall of the 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

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).