Reducing heat loss through windows by installing double and triple glazing. Calculation of heat loss of PVC windows Window, balcony door
To date heat saving is an important parameter that is taken into account when constructing a residential or office space. In accordance with SNiP 23-02-2003 “Thermal protection of buildings”, heat transfer resistance is calculated using one of two alternative approaches:
- Prescriptive;
- Consumer.
To calculate home heating systems, you can use the calculator for calculating heating and home heat loss.
Prescriptive Approach- these are the standards for individual elements of thermal protection of a building: external walls, floors above unheated spaces, coverings and attic floors, windows, entrance doors, etc.
Consumer approach(heat transfer resistance can be reduced in relation to the prescribed level, provided that the design specific heat energy consumption for space heating is lower than the standard one).
Sanitary and hygienic requirements:
- The difference between indoor and outdoor air temperatures should not exceed certain permissible values. Maximum valid values temperature difference for the outer wall is 4°C. for roofing and attic flooring 3°C and for ceilings over basements and crawl spaces 2°C.
- The temperature on the inner surface of the fence must be above the dew point temperature.
Eg: for Moscow and the Moscow region, the required thermal resistance of the wall according to the consumer approach is 1.97 °C m 2 /W, and according to the prescriptive approach:
- for home permanent residence 3.13 °C m 2 / W.
- for administrative and other public buildings, including structures for seasonal residence 2.55 °C m 2 / W.
For this reason, when choosing a boiler or other heating devices solely according to those specified in their technical documentation parameters. You must ask yourself whether your house was built with strict regard to the requirements of SNiP 02/23/2003.
Therefore, for the right choice heating boiler power or heating devices, it is necessary to calculate real heat loss from your home. As a rule, a residential building loses heat through the walls, roof, windows, and ground; significant heat losses can also occur through ventilation.
Heat loss mainly depends on:
- temperature differences in the house and outside (the higher the difference, the higher the losses).
- heat-protective characteristics of walls, windows, ceilings, coatings.
Walls, windows, ceilings have a certain resistance to heat leakage, the heat-shielding properties of materials are assessed by a value called heat transfer resistance.
Heat transfer resistance will show how much heat will leak through square meter structures at a given temperature difference. This question can be formulated differently: what temperature difference will occur when a certain amount of heat passes through a square meter of fencing.
R = ΔT/q.
- q is the amount of heat that escapes through a square meter of wall or window surface. This amount of heat is measured in watts per square meter (W/m2);
- ΔT is the difference between the temperature outside and in the room (°C);
- R is the heat transfer resistance (°C/W/m2 or °C m2/W).
In cases where we are talking about a multilayer structure, the resistance of the layers is simply summed up. For example, the resistance of a wall made of wood, which is lined with brick, is the sum of three resistances: brick and wooden wall and the air gap between them:
R(total)= R(wood) + R(air) + R(brick)
Temperature distribution and air boundary layers during heat transfer through a wall.
Heat loss calculation performed for the coldest period of the year, which is the coldest and windiest week of the year. In construction literature, the thermal resistance of materials is often indicated based on this condition and the climatic region (or outside temperature) where your home is located.
Heat transfer resistance table various materials
at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)
|
Wall material and thickness |
Heat transfer resistance Rm.
|
|
Brick wall |
0.592 |
|
Log house Ø 25 |
0.550 |
|
Log house made of timber Thickness 20 centimeters |
0.806 |
|
Frame wall (board + |
|
|
Foam concrete wall 20 centimeters |
0.476 |
|
Plastering on brick, concrete. |
|
|
Ceiling (attic) floor |
|
|
Double wooden doors |
Table of heat losses of windows of various designs at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)
|
Note
. Even numbers in symbol double glazed windows indicate air
gap in millimeters;
. The letters Ar mean that the gap is filled not with air, but with argon;
. The letter K means that the outer glass has a special transparent
heat-protective coating.
As can be seen from the above table, modern double-glazed windows make it possible reduce heat loss windows almost doubled. For example, for 10 windows measuring 1.0 m x 1.6 m, savings can reach up to 720 kilowatt-hours per month.
To correctly select materials and wall thickness, apply this information to a specific example.
Two quantities are involved in calculating heat losses per m2:
- temperature difference ΔT.
- heat transfer resistance R.
Let's say the room temperature is 20 °C. and the outside temperature will be -30 °C. In this case, the temperature difference ΔT will be equal to 50 °C. The walls are made of timber 20 centimeters thick, then R = 0.806 °C m 2 / W.
Heat losses will be 50 / 0.806 = 62 (W/m2).
To simplify calculations of heat loss in construction reference books indicate heat loss various type of walls, floors, etc. for some values winter temperature air. Typically, different numbers are given for corner rooms(the turbulence of the air that swells the house influences this) and non-angular, and also takes into account the difference in temperatures for the rooms of the first and upper floors.
Table of specific heat loss of building enclosure elements (per 1 m 2 inner contour walls) depending on the average temperature of the coldest week of the year.
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Note. In the case when there is an external unheated room behind the wall (canopy, glazed veranda, etc.), then the heat loss through it will be 70% of the calculated value, and if behind this unheated room If there is another outdoor room, then the heat loss will be 40% of the calculated value.
Table of specific heat loss of building enclosure elements (per 1 m2 along the internal contour) depending on the average temperature of the coldest week of the year.
Example 1.
Corner room(1st floor)
Room characteristics:
- 1st floor.
- room area - 16 m2 (5x3.2).
- ceiling height - 2.75 m.
- There are two external walls.
- material and thickness of the external walls - timber 18 centimeters thick, covered with plasterboard and covered with wallpaper.
- windows - two (height 1.6 m, width 1.0 m) with double glazing.
- floors - wooden insulated. basement below.
- above the attic floor.
- estimated outside temperature -30 °C.
- required room temperature +20 °C.
- Area of external walls minus windows: S walls (5+3.2)x2.7-2x1.0x1.6 = 18.94 m2.
- Window area: S windows = 2x1.0x1.6 = 3.2 m2
- Floor area: S floor = 5x3.2 = 16 m2
- Ceiling area: Ceiling S = 5x3.2 = 16 m2
Square internal partitions does not participate in the calculation, since the temperature on both sides of the partition is the same, therefore heat does not escape through the partitions.
Now let's calculate the heat loss of each surface:
- Q walls = 18.94x89 = 1686 W.
- Q windows = 3.2x135 = 432 W.
- Floor Q = 16x26 = 416 W.
- Ceiling Q = 16x35 = 560 W.
Total heat loss rooms will be: Q total = 3094 W.
It should be borne in mind that much more heat escapes through walls than through windows, floors and ceilings.
Example 2
Room under the roof (attic)
Room characteristics:
- top floor.
- area 16 m2 (3.8x4.2).
- ceiling height 2.4 m.
- exterior walls; two roof slopes (slate, continuous sheathing, 10 centimeters of mineral wool, lining). pediments (beams 10 centimeters thick covered with clapboard) and side partitions ( frame wall with expanded clay filling 10 centimeters).
- windows - 4 (two on each gable), 1.6 m high and 1.0 m wide with double glazing.
- estimated outside temperature -30°C.
- required room temperature +20°C.
- Area of the end external walls minus windows: S end walls = 2x(2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 m2
- Area of roof slopes bordering the room: S sloped walls = 2x1.0x4.2 = 8.4 m2
- Area of the side partitions: S side partition = 2x1.5x4.2 = 12.6 m 2
- Window area: S windows = 4x1.6x1.0 = 6.4 m2
- Ceiling area: Ceiling S = 2.6x4.2 = 10.92 m2
Next we calculate heat losses these surfaces, it is necessary to take into account that in this case the heat will not escape through the floor, since there is a warm room. Heat loss for walls We calculate as for corner rooms, and for the ceiling and side partitions we enter a 70 percent coefficient, since unheated rooms are located behind them.
- Q end walls = 12x89 = 1068 W.
- Q pitched walls = 8.4x142 = 1193 W.
- Q side burnout = 12.6x126x0.7 = 1111 W.
- Q windows = 6.4x135 = 864 W.
- Ceiling Q = 10.92x35x0.7 = 268 W.
The total heat loss of the room will be: Q total = 4504 W.
As we see, warm room 1st floor loses (or consumes) significantly less heat, how attic room with thin walls and a large glazing area.
To make this room suitable for winter accommodation, it is necessary first of all to insulate the walls, side partitions and windows.
Any enclosing surface can be represented in the form multilayer wall, each layer of which has its own thermal resistance and its own resistance to air passage. By summing the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also, if you sum up the resistance to the passage of air of all layers, you can understand how the wall breathes. The most best wall made of timber should be equivalent to a wall made of timber with a thickness of 15 - 20 centimeters. The table below will help with this.
Table of resistance to heat transfer and air passage of various materials ΔT = 40 ° C (T external = -20 ° C. T internal = 20 ° C.)
|
|
Thickness |
Resistance |
Resistance |
|
|
Equivalent |
||||
|
Brickwork from the usual 12 centimeters |
12 |
0.15 |
12 |
6 |
|
Masonry made of expanded clay concrete blocks 1000 kg/m3 |
1.0 |
75 |
17 |
|
|
Foam aerated concrete 30 cm thick 300 kg/m3 |
2.5 |
190 |
7 |
|
|
Thick timbered wall (pine) 10 centimeters |
10 |
0.6 |
45 |
10 |
To get a complete picture of the heat loss of the entire room, you need to take into account
- Heat loss through the contact of the foundation with frozen ground, as a rule, take 15% of heat loss through the walls of the first floor (taking into account the complexity of the calculation).
- Heat losses associated with ventilation. These losses are calculated taking into account building codes(SNiP). A residential building requires about one air change per hour, that is, during this time it is necessary to supply the same volume fresh air. Thus, the losses associated with ventilation will be slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation 50%. In European standards for ventilation and wall insulation, the heat loss ratio is 30% and 60%.
- If the wall “breathes”, like a wall made of timber or logs 15 - 20 centimeters thick, then heat returns. This allows you to reduce heat losses by 30%. therefore, the value of the thermal resistance of the wall obtained during the calculation must be multiplied by 1.3 (or, accordingly reduce heat loss).
By summing up all the heat loss at home, you can understand what power the boiler has and heating devices necessary for comfortable heating of the house on the coldest and windiest days. Also, such calculations will show where the “weak link” is and how to eliminate it using additional insulation.
You can also calculate heat consumption using aggregated indicators. So, in 1-2 storey houses that are not very insulated at an outside temperature of -25 ° C, 213 W per 1 m2 is required total area, and at -30 °C - 230 W. For well-insulated houses, this figure will be: at -25 °C - 173 W per m 2 of total area, and at -30 °C - 177 W.
The first step in organizing the heating of a private home is calculating heat loss. The purpose of this calculation is to find out how much heat escapes out through walls, floors, roofing and windows (commonly known as building envelopes) during the most severe frosts in a given area. Knowing how to calculate heat loss according to the rules, you can get a fairly accurate result and begin selecting a heat source based on power.
Basic formulas
To get a more or less accurate result, you need to perform calculations according to all the rules; a simplified method (100 W of heat per 1 m² of area) will not work here. The total heat loss of a building during the cold season consists of 2 parts:
- heat loss through enclosing structures;
- loss of energy used to heat ventilation air.
The basic formula for calculating the thermal energy consumption through external fences is as follows:
Q = 1/R x (t in - t n) x S x (1+ ∑β). Here:
- Q is the amount of heat lost by a structure of one type, W;
- R - thermal resistance of the construction material, m²°C / W;
- S—external fence area, m²;
- t in — internal air temperature, °C;
- t n - most low temperature environment, °C;
- β - additional heat loss, depending on the orientation of the building.
The thermal resistance of the walls or roof of a building is determined based on the properties of the material from which they are made and the thickness of the structure. To do this, use the formula R = δ / λ, where:
- λ—reference value of the thermal conductivity of the wall material, W/(m°C);
- δ is the thickness of the layer of this material, m.
If a wall is built from 2 materials (for example, brick with mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. Outdoor temperature is selected according to regulatory documents, and according to personal observations, internal - as necessary. Additional heat losses are coefficients determined by the standards:
- When a wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
- If the structure faces southeast or west, β = 0.05.
- β = 0 when the outer fence faces the south or southwest.
Calculation order
To take into account all the heat leaving the house, it is necessary to calculate the heat loss of the room, each separately. To do this, measurements are taken of all fences adjacent to the environment: walls, windows, roof, floor and doors.
Important point: measurements should be taken according to outside, capturing the corners of the building, otherwise the calculation of the heat loss of the house will give an underestimated heat consumption.
Windows and doors are measured by the opening they fill.
Based on the measurement results, the area of each structure is calculated and substituted into the first formula (S, m²). The value R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity coefficient of the building material. In the case of new windows made of metal-plastic, the R value will be told to you by a representative of the installer.
As an example, it is worth calculating heat loss through enclosing walls made of brick 25 cm thick, with an area of 5 m² at an ambient temperature of -25°C. It is assumed that the temperature inside will be +20°C, and the plane of the structure faces north (β = 0.1). First you need to take the thermal conductivity coefficient of brick (λ) from the reference literature; it is equal to 0.44 W/(m°C). Then, using the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:
R = 0.25 / 0.44 = 0.57 m²°C / W
To determine the heat loss of a room with this wall, all initial data must be substituted into the first formula:
Q = 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) = 434 W = 4.3 kW
If the room has a window, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same actions are repeated regarding the floors, roofing and front door. At the end, all the results are summed up, after which you can move on to the next room.
Heat metering for air heating
When calculating the heat loss of a building, it is important to take into account the amount of thermal energy consumed by the heating system to heat the ventilation air. The share of this energy reaches 30% of total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss of a house through the heat capacity of the air using a popular formula from a physics course:
Q air = cm (t in - t n). In it:
- Q air - heat consumed by the heating system for heating supply air, W;
- t in and t n - the same as in the first formula, °C;
- m is the mass flow of air entering the house from outside, kg;
- c is the heat capacity of the air mixture, equal to 0.28 W / (kg °C).
Here all quantities are known, except for the mass air flow rate during ventilation of premises. In order not to complicate your task, you should agree to the condition that the air environment in the entire house is renewed once an hour. Then the volumetric air flow rate can be easily calculated by adding the volumes of all rooms, and then you need to convert it into mass air flow through density. Since the density of the air mixture changes depending on its temperature, you need to take the appropriate value from the table:
m = 500 x 1.422 = 711 kg/h
Heating such a mass of air by 45°C will require the following amount of heat:
Q air = 0.28 x 711 x 45 = 8957 W, which is approximately equal to 9 kW.
At the end of the calculations, the results of heat losses through external fences are summed up with ventilation heat losses, which gives the total heat load on the building’s heating system.
The presented calculation methods can be simplified if the formulas are entered into Excel in the form of tables with data, this will significantly speed up the calculation.
The choice of thermal insulation, options for insulating walls, ceilings and other enclosing structures is a difficult task for most customer-developers. There are too many conflicting problems to solve at once. This page will help you figure it all out.
Currently, heat conservation of energy resources has become of great importance. According to SNiP 23-02-2003 “Thermal protection of buildings”, heat transfer resistance is determined using one of two alternative approaches:
prescriptive ( regulatory requirements apply to individual elements of the building’s thermal protection: external walls, floors above unheated spaces, coverings and attic floors, windows, entrance doors, etc.)
consumer (the heat transfer resistance of the fence can be reduced in relation to the prescriptive level, provided that the design specific heat energy consumption for heating the building is lower than the standard one).
Hygiene requirements must be met at all times.
These include
The requirement that the difference between the temperatures of the internal air and on the surface of the enclosing structures does not exceed permissible values. The maximum permissible drop values for an external wall are 4°C, for roofing and attic flooring 3°C, and for ceilings above basements and crawl spaces 2°C.
The requirement is that the temperature on the inner surface of the fence be above the dew point temperature.
For Moscow and its region, the required thermal resistance of the wall according to the consumer approach is 1.97 °C m. sq./W, and according to the prescriptive approach:
for a permanent home 3.13 °C m. sq./W,
for administrative and other public buildings, incl. buildings for seasonal residence 2.55 °С m. sq./W.
Table of thicknesses and thermal resistance of materials for the conditions of Moscow and its region.
|
Name of wall material |
Wall thickness and corresponding thermal resistance |
Required thickness according to the consumer approach (R=1.97 °C sq.m/W) and according to the prescriptive approach (R=3.13 °C sq.m/W) |
|
Solid solid clay brick (density 1600 kg/m3) |
510 mm (two bricks), R=0.73 °С m. sq./W |
1380 mm 2190 mm |
|
Expanded clay concrete (density 1200 kg/m3) |
300 mm, R=0.58 °С m. sq./W |
1025 mm 1630 mm |
|
Wooden beam |
150 mm, R=0.83 °С m. sq./W |
355 mm 565 mm |
|
Wooden shield with filling mineral wool(the thickness of the internal and external cladding of boards is 25 mm) |
150 mm, R=1.84 °С m. sq./W |
160 mm 235 mm |
Table of required heat transfer resistance of enclosing structures in houses in the Moscow region.
|
Exterior wall |
Window, balcony door |
Covering and floors |
Attic floors and floors over unheated basements |
Entrance door |
|
According to the prescriptive approach |
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|
According to consumer approach |
||||
From these tables it is clear that the majority of suburban housing in the Moscow region does not meet the requirements for heat conservation, while even the consumer approach is not observed in many newly constructed buildings.
Therefore, by selecting a boiler or heating devices only according to the ability to heat a certain area indicated in their documentation, you claim that your house was built with strict regard to the requirements of SNiP 02/23/2003.
The conclusion follows from the above material. To correctly select the power of the boiler and heating devices, it is necessary to calculate the actual heat loss of the premises of your home.
Below we will show a simple method for calculating the heat loss of your home.
The house loses heat through the wall, roof, strong emissions of heat come through the windows, heat also goes into the ground, significant heat losses can occur through ventilation.
Heat losses mainly depend on:
temperature differences in the house and outside (the greater the difference, the higher the losses),
heat-insulating properties of walls, windows, ceilings, coatings (or, as they say, enclosing structures).
Enclosing structures resist heat leakage, therefore their heat-protective properties are assessed by a value called heat transfer resistance.
Heat transfer resistance shows how much heat will be lost through a square meter of the building envelope for a given temperature difference. We can also say, conversely, what temperature difference will occur when a certain amount of heat passes through a square meter of fencing.
where q is the amount of heat lost per square meter of the enclosing surface. It is measured in watts per square meter (W/m2); ΔT is the difference between the temperature outside and in the room (°C) and R is the heat transfer resistance (°C/W/m2 or °C·m2/W).
When it comes to a multilayer structure, the resistance of the layers simply adds up. For example, the resistance of a wall made of wood lined with brick is the sum of three resistances: the brick and wooden walls and the air gap between them:
R(total)= R(wood) + R(air) + R(brick).
Temperature distribution and air boundary layers during heat transfer through a wall
Calculation of heat loss is carried out for the most unfavorable period, which is the coldest and windiest week of the year.
Construction reference books, as a rule, indicate the thermal resistance of materials based on this condition and the climatic region (or outside temperature) where your home is located.
Table – Heat transfer resistance of various materials at ΔT = 50 °C (T adv. = –30 °С, T internal = 20 °C.)
|
Wall material and thickness |
Heat transfer resistanceR m , |
|
Brick wall 3 bricks thick (79 cm) 2.5 bricks thick (67 cm) 2 bricks thick (54 cm) 1 brick thick (25 cm) |
0,592 0,502 0,405 0,187 |
|
Log house Ø 25 Ø 20 |
|
|
Log house made of timber 20 cm thick 10 cm thick |
|
|
Frame wall (board + mineral wool + board) 20 cm |
|
|
Foam concrete wall 20 cm 30 cm |
|
|
Plaster on brick, concrete, foam concrete (2-3 cm) |
|
|
Ceiling (attic) floor |
|
|
Wooden floors |
|
|
Double wooden doors |
Table – Heat losses of windows of various designs at ΔT = 50 °C (T adv. = –30 °С, T internal = 20 °C.)
Note Even numbers in the designation of a double-glazed window indicate an air gap in mm; The symbol Ar means that the gap is filled not with air, but with argon; The letter K means that the outer glass has a special transparent heat-protective coating. |
As can be seen from the previous table, modern double-glazed windows can reduce the heat loss of a window by almost half. For example, for ten windows measuring 1.0 m x 1.6 m, the savings will reach a kilowatt, which gives 720 kilowatt-hours per month.
To correctly select materials and thicknesses of enclosing structures, we will apply this information to a specific example.
When calculating heat losses per sq. meter there are two quantities involved:
temperature difference ΔT,
heat transfer resistance R.
Let's define the room temperature as 20 °C, and take the outside temperature to be –30 °C. Then the temperature difference ΔT will be equal to 50 °C. The walls are made of timber 20 cm thick, then R = 0.806 °C m. sq./W.
Heat losses will be 50 / 0.806 = 62 (W/m2).
To simplify calculations of heat loss, heat loss is given in construction reference books different types walls, ceilings, etc. for some values of winter air temperature. In particular, different figures are given for corner rooms (the turbulence of the air that swells the house is affected there) and non-corner rooms, and the different thermal picture for the rooms of the first and upper floors is also taken into account.
Table – Specific heat loss of building enclosure elements (per 1 sq.m. along the internal contour of the walls) depending on the average temperature of the coldest week of the year.
Note If behind the wall there is an external unheated room (canopy, glass veranda etc.), then the heat loss through it is 70% of the calculated value, and if behind this unheated room there is not a street, but another room outside (for example, a canopy opening onto the veranda), then 40% of the calculated value. |
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Table – Specific heat loss of building enclosure elements (per 1 sq.m. along the internal contour) depending on the average temperature of the coldest week of the year.
|
Characteristics of the fence |
Outside temperature, °C |
Heat loss, kW |
|
Double glazed window |
||
|
Solid wooden doors (double) |
||
|
Attic floor |
||
|
Wood floors above basement |
Let's consider an example of calculating heat losses of two different rooms one area using tables.
Example 1.
Corner room (ground floor)
Room characteristics:
first floor,
room area – 16 sq.m. (5x3.2),
ceiling height – 2.75 m,
external walls - two,
material and thickness of the external walls - timber 18 cm thick, covered with plasterboard and covered with wallpaper,
windows – two (height 1.6 m, width 1.0 m) with double glazing,
floors – wooden insulated, basement below,
above the attic floor,
estimated outside temperature –30 °С,
required room temperature +20 °C.
Area of external walls excluding windows:
S walls (5+3.2)x2.7-2x1.0x1.6 = 18.94 sq. m.
Window area:
S windows = 2x1.0x1.6 = 3.2 sq. m.
Floor area:
S floor = 5x3.2 = 16 sq. m.
Ceiling area:
Ceiling S = 5x3.2 = 16 sq. m.
The area of the internal partitions is not included in the calculation, since heat does not escape through them - after all, the temperature is the same on both sides of the partition. The same applies to the inner door.
Now let's calculate the heat loss of each surface:
Q total = 3094 W.
Note that more heat escapes through walls than through windows, floors and ceilings.
The calculation result shows the heat loss of the room on the coldest (T ambient = –30 °C) days of the year. Naturally, the warmer it is outside, the less heat will leave the room.
Example 2
Room under the roof (attic)
Room characteristics:
top floor,
area 16 sq.m. (3.8x4.2),
ceiling height 2.4 m,
exterior walls; two roof slopes (slate, solid sheathing, 10 cm mineral wool, lining), gables (10 cm thick timber, covered with lining) and side partitions (frame wall with 10 cm expanded clay filling),
windows – four (two on each gable), 1.6 m high and 1.0 m wide with double glazing,
estimated outside temperature –30°С,
required room temperature +20°C.
Let's calculate the areas of heat-transfer surfaces.
Area of the end external walls excluding windows:
S end wall = 2x(2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 sq. m.
Area of roof slopes bordering the room:
S sloped walls = 2x1.0x4.2 = 8.4 sq. m.
Area of side partitions:
S side burner = 2x1.5x4.2 = 12.6 sq. m.
Window area:
S windows = 4x1.6x1.0 = 6.4 sq. m.
Ceiling area:
Ceiling S = 2.6x4.2 = 10.92 sq. m.
Now let’s calculate the heat losses of these surfaces, taking into account that heat does not escape through the floor (the room is warm there). We calculate heat loss for walls and ceilings as for corner rooms, and for the ceiling and side partitions we introduce a 70 percent coefficient, since behind them there are unheated rooms.
The total heat loss of the room will be:
Q total = 4504 W.
As you can see, a warm room on the first floor loses (or consumes) significantly less heat than an attic room with thin walls and a large glazing area.
To make such a room suitable for winter living, you must first insulate the walls, side partitions and windows.
Any enclosing structure can be presented in the form of a multilayer wall, each layer of which has its own thermal resistance and its own resistance to air passage. Adding up the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also, by summing up the resistance to the passage of air of all layers, we will understand how the wall breathes. Perfect wall made of timber should be equivalent to a wall made of timber with a thickness of 15 - 20 cm. The table below will help with this.
Table – Resistance to heat transfer and air passage of various materials ΔT=40 °C (T adv. =–20 °С, T internal =20 °C.)
|
Wall Layer |
Wall layer thickness (cm) |
Heat transfer resistance of the wall layer |
Resistance air permeability equivalent to timber wall thickness (cm) |
|
|
Equivalent brickwork thickness (cm) |
||||
|
Brickwork made of ordinary clay bricks with a thickness of: 12 cm 25 cm 50 cm 75 cm |
0,15 0,3 0,65 1,0 |
|||
|
Masonry made of expanded clay concrete blocks 39 cm thick with density: 1000 kg / cubic m 1400 kg / cubic m 1800 kg / cubic m |
||||
|
Foam aerated concrete 30 cm thick, density: 300 kg / cubic m 500 kg / cubic m 800 kg / cubic m |
||||
|
Thick timbered wall (pine) 10 cm 15 cm 20 cm |
||||
For an objective picture of the heat loss of the entire house, it is necessary to take into account
Heat loss through the contact of the foundation with frozen soil is usually assumed to be 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
Heat losses associated with ventilation. These losses are calculated taking into account building codes (SNiP). A residential building requires about one air change per hour, that is, during this time it is necessary to supply the same volume of fresh air. Thus, losses associated with ventilation are slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss through ventilation is 50%. In European standards for ventilation and wall insulation, the ratio of heat losses is 30% and 60%.
If the wall “breathes”, like a wall made of timber or logs 15–20 cm thick, then heat returns. This allows you to reduce heat losses by 30%, so the value of the wall’s thermal resistance obtained in the calculation should be multiplied by 1.3 (or heat losses should be reduced accordingly).
By summing up all the heat loss in the house, you will determine what power the heat generator (boiler) and heating appliances are needed to comfortably heat the house on the coldest and windiest days. Also, calculations of this kind will show where the “weak link” is and how to eliminate it using additional insulation.
Heat consumption can also be calculated using aggregated indicators. Thus, in one- and two-story houses that are not heavily insulated, at an outside temperature of –25 °C, 213 W per square meter of total area is required, and at –30 °C – 230 W. For well-insulated houses this is: at –25 °C – 173 W per sq.m. total area, and at –30 °C – 177 W.
The cost of thermal insulation relative to the cost of the entire house is significantly small, but during the operation of the building the main costs are for heating. In no case should you skimp on thermal insulation, especially when living comfortably in large areas. Energy prices around the world are constantly rising.
Modern Construction Materials have higher thermal resistance than traditional materials. This allows you to make walls thinner, which means cheaper and lighter. All this is good, but thin walls less heat capacity, that is, they store heat worse. You have to constantly heat it - the walls heat up quickly and cool down quickly. In old houses with thick walls, it is cool on a hot summer day; the walls, which cooled down overnight, “accumulated cold.”
Insulation must be considered in conjunction with the air permeability of the walls. If an increase in the thermal resistance of walls is associated with a significant decrease in air permeability, then it should not be used. An ideal wall in terms of breathability is equivalent to a wall made of timber 15...20 cm thick.
Very often, improper use of vapor barrier leads to deterioration of the sanitary and hygienic properties of housing. With properly organized ventilation and “breathable” walls, it is unnecessary, and with poorly breathable walls it is unnecessary. Its main purpose is to prevent infiltration of walls and protect insulation from wind.
Insulating walls from the outside is much more effective than internal insulation.
You should not endlessly insulate the walls. The effectiveness of this approach to energy saving is not high.
Ventilation is the main source of energy saving.
By applying modern systems glazing (double-glazed windows, heat-insulating glass, etc.), low-temperature heating systems, effective thermal insulation enclosing structures, you can reduce heating costs by 3 times.
Options additional insulation building structures based on building thermal insulation of the “ISOVER” type, in the presence of air exchange and ventilation systems in the premises.
Insulation of tiled roofs using ISOVER thermal insulation
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Insulation of a wall made of lightweight concrete blocks |
Insulation of a brick wall with a ventilated gap |
Insulation of a log wall |
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How effective is double glazing than single glazing? Does it make sense to install K and i-glasses? Does the thickness of the air gap and argon filling play a role? And what is the difference between all this?
All answers in one simple table.
For ease of comparison, a basic level of An ordinary single-chamber double-glazed window with four-millimeter glass panes and an inter-glass distance of 16 mm was taken. Also added to the table are comparative values of sound insulation of double-glazed windows and the difference in cost.
Comparative table of double-glazed windows efficiency
| Double-glazed window formula (“k” - K-glass, “a” - argon) |
Thickness, mm | How much “warmer”, % | How much “quieter”, % | How much more expensive, % | Resistance heat transfer, m 2 *C/W | Soundproofing, dBA |
| 4 — 6 — 4 | 14 | -15% | -16% | 0,308 | 30 | |
| 4 — 8 — 4 | 16 | -9% | -13% | 0,33 | 30 | |
| 4 — 10 — 4 | 18 | -4% | -10% | 0,347 | 30 | |
| 4 — 12 — 4 | 20 | -1% | -6% | 0,358 | 30 | |
| 4 — 16 — 4 | 24 | 0,361 | 30 | |||
| 4 — 14 — 4 | 22 | 0% | -3% | 0,362 | 30 | |
| 4 - 6 - 4k | 14 | 7% | 46% | 0,386 | 30 | |
| 4k - 6 - 4k | 14 | 11% | 107% | 0,4 | 30 | |
| 4 - 8 - 4k | 16 | 24% | 49% | 0,446 | 30 | |
| 4 — 6 — 4 — 6 — 4 | 24 | 25% | 32% | 39% | 0,452 | 34 |
| 4k - 8 - 4k | 16 | 30% | 111% | 0,469 | 30 | |
| 4 - 6a - 4k | 14 | 31% | 66% | 0,472 | 30 | |
| 4 — 8 — 4 — 8 — 4 | 28 | 37% | 41% | 46% | 0,495 | 35 |
| 4 - 10 - 4k | 18 | 38% | 52% | 0,498 | 30 | |
| 4k - 6a - 4k | 14 | 39% | 127% | 0,5 | 30 | |
| 4 — 9 — 4 — 9 — 4 | 30 | 42% | 41% | 49% | 0,512 | 35 |
| 4 - 16 - 4k | 24 | 45% | 62% | 0,524 | 30 | |
| 4 - 12 - 4k | 20 | 46% | 55% | 0,526 | 30 | |
| 4 - 6 - 4 - 6 - 4k | 24 | 46% | 32% | 101% | 0,526 | 34 |
| 4 — 10 — 4 — 10 — 4 | 32 | 47% | 52% | 52% | 0,529 | 36 |
| 4 - 14 - 4k | 22 | 47% | 59% | 0,529 | 30 | |
| 4k - 10 - 4k | 18 | 47% | 114% | 0,532 | 30 | |
| 4 - 8a - 4k | 16 | 51% | 69% | 0,546 | 30 | |
| 4 — 12 — 4 — 12 — 4 | 36 | 54% | 62% | 59% | 0,555 | 37 |
| 4k - 16 - 4k | 24 | 55% | 124% | 0,559 | 30 | |
| 4 — 14 — 4 — 14 — 4 | 40 | 55% | 74% | 65% | 0,561 | 38 |
| 4k - 12 - 4k | 20 | 57% | 117% | 0,565 | 30 | |
| 4k - 14 - 4k | 22 | 57% | 120% | 0,565 | 30 | |
| 4k - 8a - 4k | 16 | 64% | 131% | 0,592 | 30 | |
| 4 - 10a - 4k | 18 | 67% | 72% | 0,602 | 30 | |
| 4 - 8 - 4 - 8 - 4k | 28 | 68% | 41% | 108% | 0,606 | 35 |
| 4 - 6 - 4k - 6 - 4k | 24 | 68% | 32% | 163% | 0,606 | 34 |
| 4 - 16a - 4k | 24 | 69% | 82% | 0,61 | 30 | |
| 4 - 14a - 4k | 22 | 71% | 79% | 0,617 | 30 | |
| 4 - 12a - 4k | 20 | 72% | 75% | 0,621 | 30 | |
| 4 - 9 - 4 - 9 - 4k | 30 | 78% | 41% | 111% | 0,641 | 35 |
| 4 - 6a - 4 - 6a - 4k | 24 | 78% | 32% | 121% | 0,641 | 34 |
| 4k - 10a - 4k | 18 | 85% | 134% | 0,667 | 30 | |
| 4k - 16a - 4k | 24 | 85% | 143% | 0,667 | 30 | |
| 4 - 10 - 4 - 10 - 4k | 32 | 87% | 52% | 114% | 0,676 | 36 |
| 4k - 14a - 4k | 22 | 88% | 140% | 0,68 | 30 | |
| 4k - 12a - 4k | 20 | 90% | 137% | 0,685 | 30 | |
| 4 - 12 - 4 - 12 - 4k | 36 | 101% | 62% | 120% | 0,725 | 37 |
| 4 - 8 - 4k - 8 - 4k | 28 | 101% | 41% | 169% | 0,725 | 35 |
| 4 - 8a - 4 - 8a - 4k | 28 | 104% | 41% | 127% | 0,735 | 35 |
| 4 - 9a - 4 - 9a - 4k | 30 | 115% | 41% | 131% | 0,775 | 35 |
| 4 - 6a - 4k - 6a - 4k | 24 | 115% | 32% | 203% | 0,775 | 34 |
| 4 - 10a - 4 - 10a - 4k | 32 | 125% | 52% | 134% | 0,813 | 36 |
| 4 - 10 - 4k - 10 - 4k | 32 | 131% | 52% | 176% | 0,833 | 36 |
| 4 - 12a - 4 - 12a - 4k | 36 | 137% | 62% | 140% | 0,855 | 37 |
| 4 - 12 - 4k - 12 - 4k | 36 | 154% | 62% | 182% | 0,917 | 37 |
| 4 - 8a - 4k - 8a - 4k | 28 | 157% | 41% | 209% | 0,926 | 35 |
| 4 - 10a - 4k - 10a - 4k | 32 | 192% | 52% | 216% | 1,053 | 36 |
| 4 - 12a - 4k - 12a - 4k | 36 | 218% | 62% | 222% | 1,149 | 37 |
Explanations and symbols:
In the column “glass unit formula” the thickness in millimeters of its “components” is indicated, where 4 mm glasses are separated from each other air gaps(chambers) filled with ordinary air or argon (where the letter “a” is indicated).
K-glass is energy-saving low-emissivity glass, which differs from ordinary glass by a special transparent coating made of metal oxides InSnO2. This coating reflects long-wave thermal radiation back into the room. If the emissivity value plain glass is 0.84, then K-glass is usually about 0.2. This means that K-glass returns approximately 70% of the thermal radiation that hits it into the room. At the same time, K-glass can protect the room from heating up in hot weather. sunny weather, also reflecting most heat waves.
There is even more efficient low-emissivity i-glass (they are not in the table). It's about one and a half times more efficient than K-glass and has an emissivity value of up to 0.04.
The article uses information from the OT-inform private enterprise.