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.
The given thermal resistance to heat transfer of a floor structure located directly on the ground is taken using a simplified method, according to which the floor surface is divided into four strips 2 m wide, parallel to the outer walls.
1. For the first zone = 2.1.
,
2. For the second zone = 4.3.
The heat transfer coefficient is equal to:
,
3. For the third zone = 8.6.
The heat transfer coefficient is equal to:
,
4. For the fourth zone = 14.2.
The heat transfer coefficient is equal to:
.
Thermal engineering calculation of external doors.
1. Determine the required heat transfer resistance for the wall:
where: n – correction factor for the calculated temperature difference
t in – design temperature of internal air
t n B – design temperature of outside air
Δt n – normalized temperature difference between temperatures internal air and temperature of the inner surface of the fence
α in – heat absorption coefficient of the inner surface of the fence = 8.7 W/(m 2 /ºС)
2. Determine the heat transfer resistance of the front door:
R odd = 0.6 · R ons tr = 0.6 · 1.4 =0.84 , (2.5),
3. Doors with known R req 0 =2.24 are accepted for installation,
4. Determine the heat transfer coefficient of the front door:
,
(2.6),
5. Determine the adjusted heat transfer coefficient of the entrance door:
2.2. Determination of heat loss through building envelopes.
In buildings, structures and premises with a constant thermal regime during the heating season, in order to maintain the temperature at a given level, heat loss and heat gain are compared in the calculated steady state, when the greatest heat deficit is possible.
Heat loss in rooms general view consist of heat loss through the enclosing structures Q ogp, heat consumption for heating the external infiltrating air entering through opened doors and other openings and cracks in the fences.
Heat loss through fences is determined by the formula:
where: A is the estimated area of the enclosing structure or part thereof, m 2 ;
K is the heat transfer coefficient of the enclosing structure, ;
t int - internal air temperature, 0 C;
t ext - outside air temperature according to parameter B, 0 C;
β – additional heat loss, determined as a fraction of the main heat loss. Additional heat losses are taken according to;
n – coefficient taking into account the dependence of the position of the outer surface of the enclosing structures in relation to the outside air, is taken according to Table 6.
According to the requirements of clause 6.3.4, the design did not take into account heat loss through the internal enclosing structures, with a temperature difference in them of 3°C or more.
When calculating heat loss in basements, the distance from the finished floor of the first floor to the ground level is taken as the height of the above-ground part. The underground parts of the external walls are considered floors on the ground. Heat loss through floors on the ground is calculated by dividing the floor area into 4 zones (I-III zones 2 m wide, IV zone the remaining area). The division into zones starts from ground level along the outer wall and is transferred to the floor. The heat transfer resistance coefficients of each zone are taken according to .
Heat consumption Qi, W, for heating the infiltrating air is determined by the formula:
Q i = 0.28G i c(t in – t ext)k , (2.9),
where: G i is the flow rate of infiltrated air, kg/h, through the enclosing structures of the room;
C- specific heat air, equal to 1 kJ/kg°C;
k is the coefficient for taking into account the influence of oncoming heat flow in structures, equal to 0.7 for windows with triple sashes;
There is no flow rate of infiltrated air in the room G i , kg/h, through leaks in the external enclosing structures, due to the fact that fiberglass sealed structures are installed in the room, preventing the penetration of outside air into the room, and infiltration through panel joints is taken into account only for residential buildings .
Calculation of heat loss through the building envelope was carried out in the Potok program, the results are given in Appendix 1.
To calculate heat loss through the floor and ceiling, the following data will be required:
- house dimensions 6 x 6 meters.
- The floors are edged boards, tongue-and-groove 32 mm thick, covered with chipboard 0.01 m thick, insulated with 0.05 m thick mineral wool insulation. There is an underground space under the house 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. 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)
Methodology for calculating heat loss in premises and the procedure for its implementation (see SP 50.13330.2012 Thermal protection buildings, point 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 in their temperature with the temperature in adjacent 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 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 a multilayer structure, the resistances of the layers of the structure add up.
δ - layer thickness, m;
λ - calculated coefficient thermal conductivity 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 with outside air), attic floors (with a roof made of piece materials) and over passages; 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 underground 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 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 general formula assessment of 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, numbered starting from outer wall 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).
Basements are often home to gyms, saunas, billiard rooms, not to mention sanitary standards Many countries even allow bedrooms to be placed in basements. In this regard, the question arises about heat loss through basements.
Basement floors are in conditions where the average temperature fluctuations are very small and range from 11 to 9°C. Thus, heat loss through the floor, although not very large, is constant throughout the year. According to computer analysis, heat loss through an uninsulated concrete floor is 1.2 W/m2.
Heat losses occur along stress lines in the ground to a depth of 10 to 20 m from the surface of the earth or from the base of the building. Applying polystyrene insulation with a thickness of about 25 mm can reduce heat loss by approximately 5%, which is no more than 1% total number heat loss from the building.
The installation of the same roof insulation allows reducing heat loss in winter time by 20% or improve the overall thermal efficiency of the building by 11%. Thus, in order to save energy, roof insulation is significantly more efficient than basement floor insulation.
This position is confirmed by an analysis of the microclimate inside the building in the summer. In the case when the lower part of the foundation walls of the building is not insulated, the incoming air heats the room, but the thermal inertia of the soil begins to affect heat loss, creating a stable temperature regime; At the same time, heat loss increases, and the temperature inside the basement decreases.
Thus, free heat exchange through structures helps maintain summer indoor air temperatures at a comfortable level. The installation of thermal insulation under the floor significantly disrupts the conditions of heat exchange between the concrete floor and the ground.
The installation of floor (internal) thermal insulation from an energy point of view leads to unproductive costs, but at the same time it is necessary to take into account moisture condensation on cold surfaces and, in addition, the need to create comfortable conditions for a person.
To mitigate the feeling of cold, you can apply thermal insulation by placing it under the floor, which will bring the floor temperature closer to the air temperature in the room and isolate the floor from the underlying layer of earth, which has a relatively low temperature. Although such insulation may increase the temperature of the floor, in this case the temperature usually does not exceed 23°C, which is 14°C below human body temperature.
Thus, to reduce the feeling of cold from the floor in order to provide the most comfortable conditions, it is best to use carpets or install a wooden floor on a concrete base.
The last aspect to be considered in this energy analysis concerns the heat loss at the junction of the floor and the wall that is not protected by backfill. This type of knot is found in buildings located on a slope.
As the analysis of heat losses shows, significant heat losses are possible in this zone in winter. Therefore, to reduce the influence of weather conditions, it is recommended to insulate the foundation along the outer surface.
