The essence of thermal calculations of premises, to one degree or another located in the ground, comes down to determining the influence of atmospheric “cold” on their thermal regime, or more precisely, to what extent a certain soil insulates a given room from atmospheric temperature effects. Because thermal insulation properties 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 (in to a greater extent 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 walls 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 further away the zone (the higher its serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can draw a simple conclusion that the further a certain point in the room is from the atmosphere (with a multiplicity of 2 m), the more favorable conditions(from the point of view of the influence of the atmosphere) it will be located.
Thus, the counting of conditional zones begins along the wall from 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 external 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 begin 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 heat transfer resistance values of the non-insulated floor by zone are equal to:
zone 1 - R n.p. =2.1 sq.m*S/W
zone 2 - R n.p. =4.3 sq.m*S/W
zone 3 - R n.p. =8.6 sq.m*S/W
zone 4 - R n.p. =14.2 sq.m*S/W
To calculate the heat transfer resistance for insulated floors, you can use the following formula:
— heat transfer resistance of each zone of the non-insulated floor, sq.m*S/W;
— insulation thickness, m;
— thermal conductivity coefficient of insulation, W/(m*C);
Methodology for calculating heat loss in premises and the procedure for its implementation (see SP 50.13330.2012 Thermal protection of buildings, paragraph 5).
The house loses heat through enclosing structures (walls, ceilings, windows, roof, foundation), ventilation and sewerage. The main heat losses occur through the enclosing structures - 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.
In this case, 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 adjacent rooms does not exceed 3 degrees Celsius.
Heat loss through building envelopes
Heat losses in premises mainly depend on:
1 Temperature differences in the house and outside (the greater the difference, the higher the losses),
2 Thermal insulation properties of walls, windows, doors, coatings, floors (the so-called enclosing structures of the room).
Enclosing structures are generally not homogeneous in structure. And they usually consist of several layers. Example: shell wall = plaster + shell + exterior decoration. This design may also include closed air gaps(example: cavities inside bricks or blocks). The above materials have thermal characteristics that differ from each other. The main characteristic for a structural layer is its heat transfer resistance R.
Where q is the amount of heat that is lost square meter enclosing surface (usually measured in W/sq.m.)
ΔT - the difference between the temperature inside the calculated room and outside temperature air (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 water procedures 33 oC.
When it comes to multilayer construction, then the resistances of the layers of the structure add up.
δ - layer thickness, m;
λ is the calculated thermal conductivity coefficient of the material of the construction layer, taking into account the operating conditions of the enclosing structures, W / (m2 oC).
Well, we’ve sorted 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 the calculation room and outside (temperature of the coldest five-day period °C). ΔT
3. Fencing areas F (separately walls, windows, doors, ceiling, floor)
4. The orientation of the building in relation to the cardinal directions is also useful.
The formula for calculating heat loss by a fence looks like this:
Qlimit=(ΔT / Rolim)* Folim * n *(1+∑b)
Qlim - heat loss through enclosing structures, W
Rogr – heat transfer resistance, m2°C/W; (If there are several layers then ∑ Rogr layers)
Fogr – area of the enclosing structure, m;
n is the coefficient of contact between the enclosing structure and the outside air.
| Walling | Coefficient n |
| 1. External walls and coverings (including those ventilated by outside air), attic floors (with a roof made of piece materials) and over driveways; ceilings over cold (without enclosing walls) undergrounds in the Northern construction-climatic zone | |
| 2. Ceilings over cold basements communicating with outside air; attic floors (with a roof made of roll materials); ceilings above cold (with enclosing walls) undergrounds and cold floors in the Northern construction-climatic zone | 0,9 |
| 3. Ceilings over unheated basements with light openings in the walls | 0,75 |
| 4. Ceilings over 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 calculated 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
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.
If we take into account that the theoretical justification and methodology for calculating the heat loss of 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 own thermal characteristics Floors are usually divided into insulated and non-insulated, structurally - floors on the ground and logs.
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 soil foundation 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).
Heat transfer through the enclosure of a home is a 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 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 the 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 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.
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 floor R value =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. That's why thermal resistance ceiling:
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)
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.