Every building, regardless of design features, skips thermal energy through the fences. Heat loss in environment needs to be restored using a heating system. The sum of heat losses with a normalized reserve is the required power of the heat source that heats the house. To create in a home comfortable conditions, the calculation of heat loss is carried out taking into account various factors: the structure of the building and the layout of the premises, orientation to the cardinal points, wind direction and the average mildness of the climate in cold period, physical qualities of building and thermal insulation materials.
According to the results thermotechnical calculation choose a heating boiler, specify the number of battery sections, calculate the power and length of underfloor heating pipes, select a heat generator for the room - in general, any unit that compensates for heat loss. By by and large, it is necessary to determine heat losses in order to heat the house economically - without excess power reserves of the heating system. Calculations are performed manually or choose a suitable computer program into which the data is inserted.
How to perform the calculation?
First, it’s worth understanding the manual technique to understand the essence of the process. To find out how much heat a house loses, the losses through each building envelope are determined separately and then added up. The calculation is performed in stages.
1. Form a base of initial data for each room, preferably in the form of a table. The first column records the pre-calculated area of door and window blocks, external walls, ceilings, and floors. The thickness of the structure is entered in the second column (this is design data or measurement results). In the third - the thermal conductivity coefficients of the corresponding materials. Table 1 contains standard values that will be needed in further calculations:
The higher λ, the more heat goes through the meter thickness of this surface.
2. Determine the thermal resistance of each layer: R = v/ λ, where v is the thickness of the building or thermal insulation material.
3. Calculate the heat loss of each structural element using the formula: Q = S*(T in -T n)/R, where:
- Tn – outside temperature, °C;
- T in – indoor temperature, °C;
- S – area, m2.
Of course, during the heating season the weather varies (for example, the temperature ranges from 0 to -25°C), and the house is heated to the required level comfort (say, up to +20°C). Then the difference (T in -T n) varies from 25 to 45.
To make the calculation you need average difference temperatures for the entire heating season. To do this, in SNiP 23-01-99 “Building climatology and geophysics” (Table 1), the average temperature of the heating period for a particular city is found. For example, for Moscow this figure is -26°. In this case the average difference is 46°C. To determine the heat consumption through each structure, the heat losses of all its layers are added up. So, for walls, plaster is taken into account, masonry material, external thermal insulation, cladding.
4. Calculate the total heat loss, defining it as the sum Q external walls, floors, doors, windows, ceilings.
5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the addition result. If you install high-quality double-glazed windows in your house and do not abuse ventilation, the infiltration coefficient can be taken as 0.1. Some sources indicate that the building does not lose heat at all, since leaks are compensated by solar radiation and household heat emissions.
Manual counting
Initial data. Cottage area 8x10 m, height 2.5 m. The walls are 38 cm thick and made of ceramic bricks, the inside is finished with a layer of plaster (thickness 20 mm). The floor is made of 30mm edged boards, insulated with mineral wool (50 mm), sheathed chipboard sheets(8 mm). The building has a basement, the temperature in which in winter is 8°C. The ceiling is covered with wooden panels and insulated with mineral wool (thickness 150 mm). The house has 4 windows 1.2x1 m, an oak entrance door 0.9x2x0.05 m.
Assignment: determine the total heat loss of a house based on the assumption that it is located in the Moscow region. The average temperature difference during the heating season is 46°C (as mentioned earlier). The room and the basement have a temperature difference: 20 – 8 = 12°C.
1. Heat loss through external walls.
Total area (minus windows and doors): S = (8+10)*2*2.5 – 4*1.2*1 – 0.9*2 = 83.4 m2.
Thermal resistance is determined brickwork and plaster layer:
- R clade. = 0.38/0.52 = 0.73 m2*°C/W.
- R pieces = 0.02/0.35 = 0.06 m2*°C/W.
- R total = 0.73 + 0.06 = 0.79 m2*°C/W.
- Heat loss through the walls: Q st = 83.4 * 46/0.79 = 4856.20 W.
2. Heat loss through the floor.
Total area: S = 8*10 = 80 m2.
The thermal resistance of a three-layer floor is calculated.
- R boards = 0.03/0.14 = 0.21 m2*°C/W.
- R chipboard = 0.008/0.15 = 0.05 m2*°C/W.
- R insulation = 0.05/0.041 = 1.22 m2*°C/W.
- R total = 0.03 + 0.05 + 1.22 = 1.3 m2*°C/W.
We substitute the values of the quantities into the formula for finding heat loss: Q floor = 80*12/1.3 = 738.46 W.
3. Heat loss through the ceiling.
Square ceiling surface equal to floor area S = 80 m2.
When determining the thermal resistance of the ceiling, in this case they do not take into account wooden boards: They are secured with gaps and do not act as a barrier to the cold. The thermal resistance of the ceiling coincides with the corresponding insulation parameter: R sweat. = R insulation = 0.15/0.041 = 3.766 m2*°C/W.
Amount of heat loss through the ceiling: Q sweat. = 80*46/3.66 = 1005.46 W.
4. Heat loss through windows.
Glazing area: S = 4*1.2*1 = 4.8 m2.
For the manufacture of windows, a three-chamber PVC profile(occupies 10% of the window area), as well as a double-chamber double-glazed window with a glass thickness of 4 mm and a distance between glasses of 16 mm. Among technical characteristics the manufacturer indicated the thermal resistance of the glass unit (R st.p. = 0.4 m2*°C/W) and profile (R prof. = 0.6 m2*°C/W). Taking into account the dimensional fraction of each structural element, the average thermal resistance of the window is determined:
- R approx. = (R st.p.*90 + R prof.*10)/100 = (0.4*90 + 0.6*10)/100 = 0.42 m2*°C/W.
- Based on the calculated result, heat loss through the windows is calculated: Q approx. = 4.8*46/0.42 = 525.71 W.
Door area S = 0.9*2 = 1.8 m2. Thermal resistance R dv. = 0.05/0.14 = 0.36 m2*°C/W, and Q dv. = 1.8*46/0.36 = 230 W.
The total amount of heat loss at home is: Q = 4856.20 W + 738.46 W + 1005.46 W + 525.71 W + 230 W = 7355.83 W. Taking into account infiltration (10%), losses increase: 7355.83 * 1.1 = 8091.41 W.
To accurately calculate how much heat a building loses, they use online calculator heat loss This is a computer program into which not only the data listed above is entered, but also various additional factors, influencing the result. The advantage of the calculator is not only the accuracy of the calculations, but also an extensive reference data base.
Of course, the main sources of heat loss in a house are doors and windows, but when viewing the picture through a thermal imager screen, it is easy to see that these are not the only sources of leakage. Heat is also lost through poorly installed roofs, cold floors, and uninsulated walls. Heat loss at home today is calculated using a special calculator. This allows you to select best option heating and holding additional work for building insulation. It is interesting that for each type of building (made of timber, logs, the level of heat loss will be different. Let's talk about this in more detail.
Basics of calculating heat loss
Control of heat loss is systematically carried out only for rooms heated in accordance with the season. Premises not intended for seasonal living do not fall under the category of buildings amenable to thermal analysis. The home heat loss program in this case will have no practical significance.
To spend full analysis, calculate thermal insulation materials and choose a heating system with optimal power, you need to have knowledge of the real heat loss of your home. Walls, roof, windows and floors are not the only sources of energy leakage from a home. Most of heat leaves the room through improperly installed ventilation systems.
Factors influencing heat loss
The main factors influencing the level of heat loss are:
- High level of temperature difference between the internal microclimate of the room and the temperature outside.
- Character thermal insulation properties enclosing structures, which include walls, ceilings, windows, etc.
Heat loss measurement values
Enclosing structures perform a barrier function for heat and do not allow it to freely escape outside. This effect is explained by the thermal insulation properties of the products. The quantity used to measure thermal insulation properties is called heat transfer resistance. This indicator is responsible for reflecting the temperature difference when the nth amount of heat passes through a section of fencing structures with an area of 1 m2. So, let’s figure out how to calculate the heat loss of a house.
The main quantities necessary to calculate the heat loss of a house include:
- q is a value indicating the amount of heat leaving the room to the outside through 1 m 2 of the barrier structure. Measured in W/m2.
- ∆T is the difference between the temperature in the house and outside. It is measured in degrees (o C).
- R - heat transfer resistance. It is measured in °C/W/m² or °C·m²/W.
- S is the area of the building or surface (used as needed).
Formula for calculating heat loss
The home heat loss program is calculated using a special formula:
When making calculations, remember that for structures consisting of several layers, the resistance of each layer is summed up. So, how to calculate heat loss frame house lined with brick on the outside? The resistance to heat loss will be equal to the sum of the resistance of brick and wood, taking into account the air gap between the layers.
Important! Please note that the resistance calculation is carried out for the coldest time of the year, when the temperature difference reaches its peak. Reference books and manuals always indicate exactly this reference value, which is used for further calculations.
Features of calculating heat loss of a wooden house
The calculation of heat loss in a house, the features of which must be taken into account when calculating, is carried out in several stages. The process requires special attention and concentration. You can calculate heat loss in a private house using a simple scheme like this:
- Determined through the walls.
- Calculated through window structures.
- Through doorways.
- Calculations are made through the floors.
- Calculate heat loss wooden house through the floor covering.
- Add the previously obtained values.
- Taking into account thermal resistance and energy loss through ventilation: from 10 to 360%.
For the results of points 1-5, the standard formula for calculating the heat loss of a house (made of timber, brick, wood) is used.
Important! Thermal resistance for window designs taken from SNIP II-3-79.
Construction reference books often contain information in a simplified form, that is, the results of calculating the heat loss of a house made of timber are given for different types walls and ceilings. For example, they calculate the resistance at a temperature difference for atypical rooms: corner and not corner rooms, single- and multi-storey buildings.
The need to calculate heat loss
Arrangement comfortable home requires strict process control at each stage of the work. Therefore, the organization of the heating system, which is preceded by the choice of the method of heating the room itself, should not be overlooked. When working on building a house, you will have to devote a lot of time not only project documentation, but also the calculation of heat loss at home. If in the future you are going to work in the field of design, then the engineering skills of calculating heat loss will definitely be useful to you. So why not practice doing this work experimentally and make a detailed calculation of heat loss for your own home.
Important! The choice of method and power of the heating system directly depends on the calculations you have made. If you calculate the heat loss indicator incorrectly, you risk freezing in cold weather or sweltering from the heat due to excessive heating of the room. It is necessary not only to choose the right device, but also to determine the number of batteries or radiators that can heat one room.
Estimation of heat loss using a calculated example
If you do not need to study the calculation of heat loss at home in detail, we will focus on the evaluation analysis and determination of heat loss. Sometimes errors occur during the calculation process, so it is better to add minimum value to estimated power heating system. In order to begin calculations, you need to know the resistance indicator of the walls. It differs depending on the type of material from which the building is made.
Resistance (R) for houses made of ceramic bricks (with a masonry thickness of two bricks - 51 cm) is 0.73 °C m²/W. The minimum thickness with this value should be 138 cm. When using expanded clay concrete as a base material (with a wall thickness of 30 cm), R is 0.58 °C m²/W with a minimum thickness of 102 cm. wooden house or a timber building with a wall thickness of 15 cm and a resistance level of 0.83 °C m²/W is required minimum thickness at 36 cm.
Building materials and their resistance to heat transfer
Based on these parameters, you can easily carry out calculations. You can find resistance values in the reference book. In construction, bricks, timber or log frames, foam concrete, wooden floor, ceilings.
Heat transfer resistance values for:
- brick wall (2 bricks thick) - 0.4;
- timber frame (200 mm thick) - 0.81;
- log house (diameter 200 mm) - 0.45;
- foam concrete (thickness 300 mm) - 0.71;
- wooden floor - 1.86;
- ceiling overlap - 1.44.
Based on the information provided above, we can conclude that for correct calculation heat loss requires only two values: the temperature difference and the level of heat transfer resistance. For example, a house is made of wood (logs) 200 mm thick. Then the resistance is 0.45 °C m²/W. Knowing this data, you can calculate the percentage of heat loss. To do this, a division operation is carried out: 50/0.45 = 111.11 W/m².
Calculation of heat loss by area is performed as follows: heat loss is multiplied by 100 (111.11*100=11111 W). Taking into account the decoding of the value (1 W=3600), we multiply the resulting number by 3600 J/hour: 11111*3600=39.999 MJ/hour. By carrying out such simple mathematical operations, any owner can find out about the heat loss of his home in an hour.
Calculation of heat loss in a room online
There are many sites on the Internet offering the service of online calculation of heat loss of a building in real time. The calculator is a program with a special form to fill out, where you enter your data and after automatic calculation you will see the result - a figure that will indicate the amount of heat released from the living space.
A residential building is a building in which people live throughout the entire heating season. As a rule, country houses, where the heating system operates periodically and as needed, do not fall into the category of residential buildings. To retool and achieve optimal mode heat supply, you will have to carry out a number of works and, if necessary, increase the power of the heating system. Such re-equipment may take a long period. In general, the whole process depends on the design features of the house and the indicators of increasing the power of the heating system.
Many have not even heard of the existence of such a thing as “heat loss at home,” and subsequently, having made constructive correct installation heating system, suffer all their lives from a lack or excess of heat in the house, without even realizing the real reason. That is why it is so important to take into account every detail when designing a home, to personally control and build it in order to ultimately get a high-quality result. In any case, a home, regardless of what material it is built from, should be comfortable. And such an indicator as the heat loss of a residential building will help make staying at home even more pleasant.
During the cold period, when the indoor air temperature is much higher than the outdoor air temperature, heat flows (heat loss) occur through the building enclosure.
Heat loss in premises consists of two main components: transmission heat loss and heat consumption for heating the air infiltrated through leaks.
Transmission heat loss is the loss of heat through external enclosures due to heat transfer.
Transmission heat losses are found using the formulas:
where is heat loss, W;
Thermal resistance of the fence ()/W, determined thermotechnical calculation;
K - heat transfer coefficient of the fence W / (),
F is the surface area of the fence,
– design air temperature in the room, °C, table 2
Estimated outside air temperature equal to the average temperature of the coldest five-day period, °C, Table 3
N – correction factor to the calculated temperature difference;
Additional heat loss, W.
To calculate the surface areas F in formulas (1.24) and (1.25.) we are guided by generally accepted method determining the linear dimensions of the enclosing structure.
Rice. 2. Measurement of fences:
a – vertically; b – in plan; 1 – floor on the ground; 2- floor along joists; 3 – floor above the basement; O – windows; NS – external wall; Pl – floor; Fri – ceiling.
It is customary to determine the heat loss of the floor lying on the ground by zones. Each zone has its own thermal resistance.
; 4.3()/W;
The amount of heat loss through the i-th zone is found by the formula:
where is the resistance of the i-th zone, ()/ W;
– area of the i-th zone (area of a 2 m wide ring strip along the building’s contour). The area of zone I in the corners of the building is multiplied by 2.
Rice. 3. Heat flows from floors along the ground and buried walls:
a – through the floor; b – through a recessed wall; c – division of the floor into zones 1,2,3,4; d – division of the recessed shade and floor into zones 1,2,3,4.
Heat loss through the floors is obtained by summing the heat loss by zone
If the floors are laid on joists or on insulating material (have an air gap) and the thermal resistance of these additional elements The calculation method remains the same (in this case, the resistance of each zone increases by the amount of resistance of the underlying layers.)
The same technique is used to calculate heat loss through the walls of a building buried in the ground (heated basements).
The division into zones begins from the ground surface outside the building, the floors are considered as a continuation of the walls.
Additional heat loss is determined as follows:
1. Additions for orientation to the cardinal points are made to all vertical fences or vertical projections of inclined fences as follows:
N, N-W, N-E, E-10%; W, SE – 5%; S, S-W – 0%.
2. For the rush of cold air through external doors when they are opened briefly at a building height of N, m:
Double doors with vestibules – 27% of H;
The same without a vestibule - 34% of H;
Single doors – 22% of N.
3. For the floors of the first floor above the cold basements of buildings in areas with an estimated outside air temperature (five days) of minus 40 ° C and below, it is assumed to be equal to 5%.
By summing the transmission heat losses across all enclosures, we find the heat losses of the entire room.
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Let's look at how to calculate the heat loss of a house through the building envelope. The calculation is given using the example of a one-story residential building. This calculation can also be used to calculate the heat loss of a separate room, an entire house or an individual apartment.
An example of a technical specification for calculating heat loss
First, we draw up a simple house plan indicating the area of the premises, the size and location of windows and front door. This is necessary to determine the surface area of the house through which heat loss occurs.
Formula for calculating heat loss
To calculate heat loss, we use the following formulas:
R=B/K- this is a formula for calculating the thermal resistance of the building envelope.
- R - thermal resistance, (m2*K)/W;
- K - coefficient of thermal conductivity of the material, W/(m*K);
- B - material thickness, m.
Q =S. dT/R- this is the formula for calculating heat loss.
- Q - heat loss, W;
- S - area of the building envelope, m2;
- dT - temperature difference between interior space and street, K;
- R - value of thermal resistance of the structure, m2.K/W
To calculate the temperature inside the house, we take +21..+23°C - this mode is the most comfortable for a person. The minimum street temperature for calculating heat loss was taken to be -30°C, since in winter period in the region: where the house was built (Yaroslavl region, Russia), such a temperature can last more than one week and it is recommended to include the lowest temperature indicator in the calculations, while the temperature difference is dT = 51..53, on average - 52 degrees.
The total heat loss of a house consists of the heat loss of all enclosing structures, therefore, using these formulas, we perform:
After the calculation we received the following data:
- Q walls - 0.49 kWh,
- Q ceiling- 0.49 kWh,
- Floor Q - 0.32 kWh,
- Q windows - 0.38 kWh.
- Q entrance door - 0.16 kWh.
Total: the total result of heat loss through the enclosing structures was 1.84 kWh.
CHAPTER 3. HEAT BALANCE OF ROOMS AND HEAT CONSUMES FOR HEATING BUILDINGS
Design power of heating systems
Thermal conditions can be constant or variable.
Constant - supported around the clock in residential, industrial and continuous mode work in buildings, children's and medical institutions, hotels, sanatoriums.
Variable - in industrial buildings with one- and two-shift work, administrative, commercial, educational buildings, service enterprises. During non-working hours, use the existing heating system, or standby heating - low temperature.
The heat balance is compiled into a form (Table 3.1).
Table 3.1. Heat balance form
If heat loss is greater than heat release, then heating is required.
Calculated thermal power heating systems:
Q с,о = ∑Q sweat - ∑Q post, (3.1)
If in industrial building ∑Q post >∑Q sweat, then supply ventilation is arranged.
Heat loss through building envelopes
To determine heat loss you must have:
Floor plans with all building dimensions;
Copy from the general plan with the designation of the cardinal points and the wind rose;
The purpose of each room;
Geographical location of the building's construction;
Designs of all external fencing.
All rooms on the plans indicate:
Numbered from left to right, staircases are designated by letters or Roman numerals regardless of the floor and are considered as one room.
Heat loss in premises through enclosing structures, rounded up to 10 W:
Q limit = (F/R o)(t in – t n B)(1 + ∑β)n = kF(t in – t n B)(1 - ∑β)n,(3.2)
Where F, k, R o- design area, heat transfer coefficient, heat transfer resistance of the enclosing structure, m2, W/(m2 oC), (m2 oC)/W; t in- estimated room air temperature, o C; t n B- estimated outside air temperature (B) or air temperature in a colder room; P- coefficient taking into account the position of the outer surface of the enclosing structures in relation to the outside air (Table 2.4); β - additional heat losses in fractions of the main losses.
Heat exchange through fences between adjacent heated rooms is taken into account if the temperature difference in them is more than 3°C.
Squares F, m2, fences (external walls (NS), windows (O), doors (D), lanterns (F), ceiling (Pt), floor (P)) are measured according to plans and sections of the building (Fig. 3.1).
1. Height of the walls of the first floor: if the floor is on the ground, between the floor levels of the first and second floors ( h 1); if the floor is on joists - from the external level of preparation of the floor on joists to the floor level of the second floor ( h 1 1); for an unheated basement or underground - from the level of the lower surface of the floor structure of the first floor to the level of the finished floor of the second floor ( h 1 11), and in one-story buildings with an attic floor, the height is measured from the floor to the top of the insulating layer of the floor.
2. The height of the walls of the intermediate floor is between the levels of the finished floors of this and the overlying floors ( h 2), and the upper floor - from the level of its clean floor to the top of the insulating layer attic floor (h 3) or roofless roofing.
3. The length of external walls in corner rooms - from the edge of the outer corner to the axes interior walls (l 1 And l 2l 3).
4. The length of the internal walls - from the internal surfaces of the external walls to the axes of the internal walls ( m 1) or between the axes of internal walls (T).
5. Areas of windows, doors and lanterns - according to smallest sizes construction openings in the light ( A And b).
6. The areas of ceilings and floors above basements and underground spaces in corner rooms - from the inner surface of the external walls to the axes of opposite walls ( m 1 And P), and in non-corner ones - between the axes of the internal walls ( T) and from the inner surface of the outer wall to the axis of the opposite wall ( P).
The error of linear dimensions is ±0.1 m, area error is ±0.1 m2.
Rice. 3.1. Measuring diagram for heat transfer fencing
Figure 3.2. Scheme for determining heat loss through floors and walls buried below ground level
1 - first zone; 2 – second zone; 3 – third zone; 4 – fourth zone (last).
Heat loss through the floors is determined by zone-strips 2 m wide, parallel to the external walls (Fig. 5.2).
Reduced heat transfer resistance R n.p., m 2 K/W, areas of uninsulated floors on the ground and walls below ground level, with thermal conductivity λ > 1.2 W/(m o C): for the 1st zone - 2.1; for zone 2 - 4.3; for the 3rd zone - 8.6; for the 4th zone (remaining floor area) - 14.2.
Formula (3.2) when calculating heat losses Q pl, W, through the floor located on the ground, takes the form:
Q pl = (F 1 / R 1n.p +F 2 / R 2n.p +F 3 / R 3n.p +F 4 / R 4n.p)(t in – t n B)(1 + ∑β) n,(3.3)
Where F 1 - F 4- area of 1 - 4 zone-strips, m2; R 1, n.p. - R 4, n.p.- heat transfer resistance of floor zones, m 2 K/W; n =1.
Heat transfer resistance of insulated floors on the ground and walls below ground level (λ< 1,2 Вт/(м· оС)) R y .п, m 2 o C/W, also determined for zones using the formula
R u.p = R n.p +∑(δ u.s. /λ u.s.),(3.4)
Where R n.a.- heat transfer resistance of non-insulated floor zones (Fig. 3.2), m 2 o C/W; sum of fraction- the sum of the thermal resistances of the insulating layers, m 2 o C/W; δ у.с- thickness of the insulating layer, m.
Heat transfer resistance of floors on joists R l, m 2 o C/W:
R l.p = 1.18 (R n.p +∑(δ u.s. /λ u.s.)),(3.5)
Insulating layers - air gap and a plank floor on joists.
When calculating heat losses, floor areas in the corners of external walls (in the first two-meter zone) are entered into the calculation twice in the direction of the walls.
Heat loss through the underground part of the external walls and the floors of the heated basement is also calculated in zones 2 m wide, counting them from ground level (see Fig. 3.2). Then the floors (when counting zones) are considered as a continuation of the underground part of the external walls. Heat transfer resistance is determined in the same way as for uninsulated or insulated floors.
Additional heat loss through fences. In (3.2) the term (1+∑β) takes into account additional heat loss as a share of the main heat loss:
1. On orientation in relation to the cardinal points. β external vertical and inclined (vertical projection) walls, windows and doors.
Rice. 3.3. Addition to the main heat loss depending on the orientation of the fences in relation to the cardinal points
2. For ventilation of rooms with two or more external walls. IN standard projects through walls, doors and windows facing all countries of the world β = 0.08 with one external wall and 0.13 for corner rooms and in all residential premises.
3. At the design temperature of the outside air. For unheated floors of the first floor above cold underground areas of buildings in areas with t n B minus 40°C and below - β = 0,05.
4. To heat the rushing cold air. For external doors, without air or air-thermal curtains, at building height N, m:
- β = 0,2N- for triple doors with two vestibules between them;
- β = 0,27 N - For double doors with a vestibule between them;
- β = 0,34 N - for double doors without vestibule;
- β = 0,22 N - for single doors.
For external non-equipped gates β =3 without vestibule and β = 1 - with a vestibule at the gate. For summer and emergency external doors and gates β = 0.
Heat losses through the building envelopes are entered in the form (Table 3.2).
Table 3.2. Form (form) for calculating heat loss
The area of the walls in the calculation is measured with the area of the windows, thus the area of the windows is taken into account twice, therefore in column 10 the coefficient k windows is taken as the difference between its values for windows and walls.
Heat loss calculations are carried out by room, floor, building.