The essence of thermal calculations of premises, to one degree or another located in the ground, is to determine the effect of atmospheric "cold" on their thermal regime, or rather, to what extent a certain soil isolates a given room from atmospheric temperature effects. Because thermal insulation properties ground depend too a large number 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 larger its 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 (by a factor of 2 m), the more favorable conditions (from the point of view 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 walls on the ground, then the first zone will be the floor strip closest to 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. Attic space 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
Considering that in the underground the temperature in winter is constantly kept at about + 8 ° С, 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 shields, because they do not have a tight connection with each other and do not play the role of a heat insulator. That's why thermal resistance 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 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 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
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).
Heat transfer through the fences of a house is a complex process. In order to take into account these difficulties as much as possible, the measurement of premises when calculating heat losses is 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 the areas of enclosing structures: a - section of a building with an attic floor; b - section of a building with a combined coating; c - building plan; 1 - floor above the basement; 2 - floor on logs; 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 inner walls and the inner surface of the outer wall.
The dimensions of the outer walls are taken horizontally along the outer perimeter between the axes of the inner walls and the outer corner of the wall, and in height - on all floors except the lower one: from the level of the finished floor to the floor of the next floor. On the last floor the top of the outer wall coincides with the top of the cladding, or attic floor. On the lower floor, depending on the floor design: a) from the inner surface of the floor on the ground; b) from the preparation surface for the floor structure on the logs; c) from the lower edge of the ceiling over an unheated underground or basement.
When determining heat loss through internal walls their areas are measured along the inner perimeter. Heat losses through the internal enclosures of the premises can be ignored if the air temperature difference in these premises is 3 °C or less.
Breakdown of the floor surface (a) and recessed parts of the outer walls (b) into design zones I-IV
The transfer of heat from the 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 resistance to heat transfer of structures located on the ground, a simplified method is used. The surface of the floor and walls (in this case, 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 starts along the wall from the 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 strips 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 non-insulated. The resistance to heat transfer of such a floor is usually denoted as R np, m 2 ° C / W. For each zone of an uninsulated floor, 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 \u003d 8.6 m 2 ° C / W;
- zone IV - RIV \u003d 14.2 m 2 ° C / W.
If there are insulating layers in the floor structure located on the ground, it is called insulated, and its resistance to heat transfer R unit, m 2 ° C / W, is determined by the formula:
R pack \u003d R np + R us1 + R us2 ... + R usn
Where R np - resistance to heat transfer of the considered zone of an uninsulated floor, m 2 · ° С / W;
R us - heat transfer resistance of the insulating layer, m 2 · ° C / W;
For a floor on logs, the heat transfer resistance Rl, m 2 · ° С / W, is calculated by the formula.
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.
