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", the 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, heat loss at home.
Prescriptive approach are the standards for individual elements thermal protection of the building: external walls, floors above unheated spaces, coverings and attic ceilings, windows, entrance doors, etc.
consumer approach(heat transfer resistance may be reduced from the prescriptive level, provided that the design specific consumption thermal energy for space heating below the standard).
Sanitary and hygienic requirements:
- The difference between the air temperatures inside and outside the room should not exceed certain permissible values. Maximum allowed values temperature difference for outer wall 4°C. to cover and attic floor 3°С and for ceilings over cellars and undergrounds 2°С.
- The temperature on the inner surface of the enclosure must be above the dew point temperature.
For example: for Moscow and the Moscow region, the required thermal resistance of the wall according to the consumer approach is 1.97 ° С m 2 /W, and according to the prescriptive approach:
- for home permanent residence 3.13 °С m 2 / W.
- for administrative and other public buildings, including facilities seasonal residence 2.55 °С m 2 / W.
For this reason, choosing a boiler or other heating devices solely according to those indicated in their technical documentation parameters. You should ask yourself if your house was built with strict adherence to the requirements of SNiP 23-02-2003.
Therefore, for right choice power of the heating boiler or heating devices, it is necessary to calculate the real heat loss in your home. As a rule, a residential building loses heat through the walls, roof, windows, ground, as well as significant heat losses can occur through ventilation.
Heat loss mainly depends on:
- temperature difference in the house and on the street (the higher the difference, the higher the loss).
- heat-shielding characteristics of walls, windows, ceilings, coatings.
Walls, windows, floors, have a certain resistance to heat leakage, the heat-shielding properties of materials are evaluated by a value called heat transfer resistance.
Heat transfer resistance will show how much heat will seep through square meter structures for 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 fences.
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 / m 2);
- ΔT is the difference between the temperature in the street and in the room (°C);
- R is the heat transfer resistance (°C / W / m 2 or ° C m 2 / 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 air gap between them:
R(sum)= R(wood) + R(car) + R(brick)
Temperature distribution and boundary layers of air during heat transfer through a wall.
Heat loss calculation is performed for the coldest period of the year of the period, which is the coldest and windiest week of the year. In building literature, the thermal resistance of materials is often indicated based on the given conditions and the climatic area (or outside temperature) where your house is located.
Heat transfer resistance table various materials
at ΔT = 50 °С (T external = -30 °С. Т internal = 20 °С.)
|
Wall material and thickness |
Heat transfer resistance Rm.
|
|
Brick wall |
0.592 |
|
Log cabin Ø 25 |
0.550 |
|
Log cabin Thickness 20 centimeters |
0.806 |
|
Frame wall (board + |
|
|
Foam concrete wall 20 centimeters |
0.476 |
|
Plastering on brick, concrete. |
|
|
Ceiling (attic) ceiling |
|
|
Double wooden doors |
Table of heat losses of windows of various designs at ΔT = 50 °C (T out = -30 °C. T int. = 20 °C.)
|
Note
. Even numbers in symbol double glazing 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 protection 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, the savings can reach up to 720 kilowatt-hours per month.
For the correct choice of materials and wall thicknesses, we apply this information to a specific example.
Two quantities are involved in the calculation of heat losses per m 2:
- 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 loss will be 50 / 0.806 = 62 (W / m 2).
To simplify the calculation of heat losses in construction guides indicate heat loss different kind walls, floors, etc. for some values winter temperature air. Usually given various numbers for corner rooms(the swirl of air flowing through the house affects it) and non-angular, and also takes into account the difference in temperatures for the premises of the first and upper floors.
Table of specific heat losses of building fencing elements (per 1 m 2 along the inner contour of the 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, and if behind this unheated room there is another outdoor room, then the heat loss will be 40% of the calculated value.
Table of specific heat losses of building fencing elements (per 1 m 2 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 m 2 (5x3.2).
- ceiling height - 2.75 m.
- outer walls - two.
- the material and thickness of the outer walls - a timber 18 centimeters thick is sheathed 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.
- design outside temperature -30 °С.
- the required temperature in the room is +20 °С.
- The area of the outer walls minus the windows: S walls (5+3.2)x2.7-2x1.0x1.6 = 18.94 m2.
- Windows area: S windows \u003d 2x1.0x1.6 \u003d 3.2 m 2
- Floor area: S floor \u003d 5x3.2 \u003d 16 m 2
- Ceiling area: S ceiling \u003d 5x3.2 \u003d 16 m 2
Square internal partitions does not participate in the calculation, since the temperature is the same on both sides of the partition, therefore, heat does not escape through the partitions.
Now Let's calculate the heat loss of each of the surfaces:
- Q walls \u003d 18.94x89 \u003d 1686 watts.
- Q windows \u003d 3.2x135 \u003d 432 watts.
- Q floor \u003d 16x26 \u003d 416 watts.
- Q ceiling \u003d 16x35 \u003d 560 watts.
The total heat loss of the room will be: Q total \u003d 3094 W.
It should be borne in mind that much more heat escapes through walls than through windows, floors and ceilings.
Example 2
Roof room (attic)
Room characteristics:
- upper floor.
- area 16 m 2 (3.8x4.2).
- ceiling height 2.4 m.
- exterior walls; two roof slopes (slate, continuous crate. 10 cm of mineral wool, lining). pediments (beam 10 cm thick sheathed with clapboard) and side partitions ( frame wall with expanded clay filling 10 cm).
- windows - 4 (two on each gable), 1.6 m high and 1.0 m wide with double glazing.
- design outside temperature -30°С.
- required room temperature +20°C.
- The area of the end external walls minus the windows: S end walls = 2x (2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 m 2
- The area of \u200b\u200bthe roof slopes that bound the room: S slopes. walls \u003d 2x1.0x4.2 \u003d 8.4 m 2
- The area of the side partitions: S side partition = 2x1.5x4.2 = 12.6 m 2
- Windows area: S windows \u003d 4x1.6x1.0 \u003d 6.4 m 2
- Ceiling area: S ceiling \u003d 2.6x4.2 \u003d 10.92 m 2
Next, we calculate heat loss these surfaces, while 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 both for corner rooms, and for the ceiling and side partitions we introduce a 70 percent coefficient, since unheated rooms are located behind them.
- Q end walls \u003d 12x89 \u003d 1068 W.
- Q slope walls \u003d 8.4x142 \u003d 1193 W.
- Q side burner = 12.6x126x0.7 = 1111 W.
- Q windows \u003d 6.4x135 \u003d 864 watts.
- Q ceiling \u003d 10.92x35x0.7 \u003d 268 watts.
The total heat loss of the room will be: Q total \u003d 4504 W.
As we see warm room 1 floor loses (or consumes) significantly less heat than attic room with thin walls and a large glass area.
To make this space suitable for winter residence, it is necessary first of all to insulate walls, side partitions and windows.
Any enclosing surface can be represented as a multilayer wall, each layer of which has its own thermal resistance and its own resistance to the passage of air. Summing up 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 from a bar should be equivalent to a wall from a bar with a thickness of 15 - 20 centimeters. The table below will help you with this.
Table of resistance to heat transfer and air passage of various materials ΔT=40 °C (T ext. = -20 °C. T int. =20 °C.)
|
|
Thickness |
Resistance |
Resist. |
|
|
Equivalent |
||||
|
Brickwork out of the ordinary 12 centimeters |
12 |
0.15 |
12 |
6 |
|
Claydite-concrete block masonry 1000 kg / m 3 |
1.0 |
75 |
17 |
|
|
Foam aerated concrete 30 cm thick 300 kg / m 3 |
2.5 |
190 |
7 |
|
|
Brusoval wall thick (pine) 10 centimeters |
10 |
0.6 |
45 |
10 |
For a complete picture of the heat loss of the entire room, it is necessary to take into account
- Heat loss through foundation contact with frozen ground, as a rule, take 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
- Heat loss associated with ventilation. These losses are calculated taking into account building codes(SNiP). For a residential building, about one air exchange per hour is required, 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 sum of heat losses attributable to the building envelope. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation fifty%. In European standards for ventilation and wall insulation, the ratio of heat loss is 30% and 60%.
- If the wall "breathes", like a wall made of timber or logs 15 - 20 centimeters thick, then heat is returned. This reduces heat loss by 30%. therefore, the value obtained in the calculation thermal resistance walls must be multiplied by 1.3 (or, respectively, reduce heat loss).
Summing up all the heat losses at home, you can understand what power the boiler and heating appliances are necessary for comfortable heating of the house in the coldest and windy days. Also, such calculations will show where the “weak link” is and how to eliminate it with the help of additional insulation.
You can also calculate the heat consumption using aggregated indicators. So, in 1-2 storey not very insulated houses with outdoor temperature-25 °С 213 W required per 1 m2 total area, and at -30 ° С - 230 W. For well-insulated houses, this figure will be: at -25 ° C - 173 W per m 2 of the total area, and at -30 ° C - 177 W.
The calculation of the heating of a private house can be done independently by taking some measurements and substituting your values into necessary formulas. Let's tell you how it's done.
We calculate the heat loss of the house
Several critical parameters of the heating system depend on the calculation of heat loss at home, and first of all, the power of the boiler.
The calculation sequence is as follows:
We calculate and write down in a column the area of windows, doors, external walls, floors, ceilings of each room. Opposite each value we write down the coefficient from which our house is built.
If you didn't find desired material in, then look in the extended version of the table, which is called so - the coefficients of thermal conductivity of materials (soon on our website). Further, according to the formula below, we calculate the heat loss of each structural element of our house.
Q=S*ΔT/R,where Q– heat loss, W
S— construction area, m2
Δ T— temperature difference between indoors and outdoors for the coldest days °C
R— the value of thermal resistance of the structure, m2 °C/W
R layer = V / λwhere V— layer thickness in m,
λ - coefficient of thermal conductivity (see table for materials).
We summarize the thermal resistance of all layers. Those. for walls, both plaster and wall material and external insulation (if any) are taken into account.
Putting it all together Q for windows, doors, exterior walls, floors, ceilings
We add 10-40% of ventilation losses to the amount received. They can also be calculated by the formula, but with good windows and moderate ventilation, you can safely set 10%.
The result is divided by the total area of the house. It is the general, because heat will indirectly be spent on corridors where there are no radiators. The calculated value of specific heat loss can vary within 50-150 W/m2. The highest heat losses are in the rooms of the upper floors, the lowest in the middle ones.
After graduation installation work, conduct walls, ceilings and other structural elements to make sure that there are no heat leaks anywhere.
The table below will help you more accurately determine the indicators of materials.
Determining the temperature
This stage is directly related to the choice of the boiler and the method of space heating. If you intend to install " warm floors", Maybe, the best solution– condensing boiler and low-temperature regime of 55C in the supply and 45C in the “return”. This mode ensures the maximum efficiency of the boiler and, accordingly, the best economy gas. In the future, if you want to use high-tech heating methods, ( , solar collectors) you do not have to remake the heating system for new equipment, because it is specifically designed for low temperature regimes. Additional pluses - the air in the room does not dry out, the flow rate is lower, less dust is collected.
In the case of choosing a traditional boiler, it is better to choose the temperature regime as close as possible to European standards 75C - at the outlet of the boiler, 65C - return flow, 20C - room temperature. This mode is provided in the settings of almost all imported boilers. In addition to choosing a boiler, the temperature regime affects the calculation of the power of radiators.
Selection of power radiators
For the calculation of heating radiators for a private house, the material of the product does not play a role. This is a matter of taste of the owner of the house. Only the power of the radiator indicated in the product passport is important. Often manufacturers indicate inflated figures, so the result of the calculations will be rounded up. The calculation is made for each room separately. Simplifying somewhat the calculations for a room with ceilings of 2.7 m, we give a simple formula:
Where To- desired number of radiator sections
S- area of the room
P- power indicated in the product passport
Calculation example: For a room with an area of 30 m2 and a power of one section of 180 W, we get: K = 30 x 100/180
K=16.67 rounded 17 sections
The same calculation can be applied to cast iron batteries, assuming that
1 rib(60 cm) = 1 section.
Hydraulic calculation of the heating system
The meaning of this calculation is to choose the right pipe diameter and characteristics. Due to the complexity of the calculation formulas, it is easier for a private house to select pipe parameters from the table.
Here is the total power of the radiators for which the pipe supplies heat.
| Pipe diameter | Min. radiator power kW | Max. radiator power kW |
| Metal-plastic pipe 16 mm | 2,8 | 4,5 |
| Metal-plastic pipe 20 mm | 5 | 8 |
| Metal-plastic pipe 25 mm | 8 | 13 |
| Metal-plastic pipe 32 mm | 13 | 21 |
| Polypropylene pipe 20 mm | 4 | 7 |
| Polypropylene pipe 25 mm | 6 | 11 |
| Polypropylene pipe 32 mm | 10 | 18 |
| Polypropylene pipe 40 mm | 16 | 28 |
We calculate the volume of the heating system
This value is necessary to select the correct volume expansion tank. It is calculated as the sum of the volumes in the radiators, pipelines and boiler. reference Information for radiators and pipelines is given below, for the boiler - indicated in his passport.
The volume of coolant in the radiator:
- aluminum section - 0.450 liters
- bimetallic section - 0.250 liters
- new cast iron section- 1,000 liters
- old cast iron section - 1,700 liters
The volume of the coolant in 1 l.m. pipes:
- ø15 (G ½") - 0.177 liters
- ø20 (G ¾") - 0.310 liters
- ø25 (G 1.0″) - 0.490 liters
- ø32 (G 1¼") - 0.800 liters
- ø15 (G 1½") - 1.250 liters
- ø15 (G 2.0″) - 1.960 liters
Installation of the heating system of a private house - the choice of pipes
It is carried out with pipes from different materials:
Steel
- They have a lot of weight.
- They require proper skill, special tools and equipment for installation.
- Corrosion resistant
- May accumulate static electricity.
Copper
- Withstand temperatures up to 2000 C, pressure up to 200 atm. (in a private house, completely unnecessary dignity)
- Reliable and durable
- Have a high cost
- Mounted with special equipment, silver solder
Plastic
- Antistatic
- Corrosion resistant
- Inexpensive
- Have minimal hydraulic resistance
- Requires no special skills for installation
Summarize
Correctly made calculation of the heating system of a private house provides:
- Comfortable warmth in the rooms.
- Sufficient amount of hot water.
- Silence in the pipes (no gurgling or growling).
- Optimal boiler operating modes
- Correct load on the circulation pump.
- Minimum installation costs
The choice of thermal insulation, options for insulating walls, ceilings and other building envelopes is a difficult task for most building developers. Too many conflicting problems need to be solved at the same time. This page will help you figure it all out.
At present, the heat saving of energy resources has acquired 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 are applied to individual elements of the thermal protection of the building: external walls, floors above unheated spaces, coverings and attic ceilings, windows, entrance doors, etc.)
- consumer (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 below the standard).
Sanitary and hygienic requirements must be observed 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 the permissible values. The maximum allowable differential values for the outer wall are 4°C, for roofing and attic floors 3°C and for ceilings above basements and undergrounds 2°C.
The requirement that the temperature on the inner surface of the enclosure 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 °C m. sq./ W.
Table of thicknesses and thermal resistance of materials for the conditions of Moscow and its region.
| Wall material name | Wall thickness and corresponding thermal resistance | Required thickness according to consumer approach (R=1.97 °C m/W) and prescriptive approach (R=3.13 °C m/W) |
|---|---|---|
| Solid solid clay brick (density 1600 kg/m3) | 510 mm (two-brick masonry), 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 infill mineral wool(thickness of inner and outer skin from boards of 25 mm) | 150 mm, R=1.84 °С m. sq./W | 160 mm 235 mm |
Table of required resistance to heat transfer of enclosing structures in houses in the Moscow region.
| outer wall | Window, balcony door | Coating and overlays | Ceiling attic and ceilings over unheated basements | front door |
|---|---|---|---|---|
| Byprescriptive approach | ||||
| 3,13 | 0,54 | 3,74 | 3,30 | 0,83 |
| By consumer approach | ||||
| 1,97 | 0,51 | 4,67 | 4,12 | 0,79 |
These tables show that the majority of suburban housing in the Moscow region does not meet the requirements for heat saving, while even the consumer approach is not observed in many newly built buildings.
Therefore, when selecting a boiler or heaters only according to their ability to heat certain area, You claim that your house was built with strict compliance with the requirements of SNiP 23-02-2003.
The conclusion follows from the above material. For the correct choice of the power of the boiler and heating devices, it is necessary to calculate the actual heat loss of the premises of your house.
Below we will show a simple method for calculating the heat loss of your home.
The house loses heat through the wall, roof, strong heat emissions go through the windows, heat also goes into the ground, significant heat losses can occur through ventilation.
Heat losses mainly depend on:
- temperature difference in the house and on the street (the greater the difference, the higher the losses),
- heat-shielding properties of walls, windows, ceilings, coatings (or, as they say, enclosing structures).
Enclosing structures resist heat leakage, so their heat-shielding properties are evaluated by a value called heat transfer resistance.
The heat transfer resistance shows how much heat will go through a square meter of the building envelope at a given temperature difference. It can be said, and vice versa, what temperature difference will occur when a certain amount of heat passes through a square meter of fences.
where q is the amount of heat that a square meter of enclosing surface loses. It is measured in watts per square meter (W/m2); ΔT is the difference between the temperature in the street and in the room (°C) and, R is the heat transfer resistance (°C / W / m2 or °C m2 / W).
When it comes to multi-layer construction, the resistance of the layers simply add up. For example, the resistance of a wall made of wood lined with bricks is the sum of three resistances: a brick and wooden wall and an air gap between them:
R(sum)= R(wood) + R(cart) + R(brick).
Temperature distribution and boundary layers of air during heat transfer through a wall
Calculation of heat loss is carried out for the most unfavorable period, which is the most frosty and windy week of the year.
Building guides usually indicate the thermal resistance of materials based on this condition and the climatic area (or outside temperature) where your house is located.
Table- Heat transfer resistance of various materials at ΔT = 50 °C (T out = -30 °C, T int = 20 °C.)
| Wall material and thickness | Heat transfer resistance Rm, |
|---|---|
| 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 cabin Ø 25 Ø 20 |
0,550 0,440 |
| Log cabin 20 cm thick |
0,806 0,353 |
| Frame wall (board + mineral wool + board) 20 cm |
0,703 |
| Foam concrete wall 20 cm 30 cm |
0,476 0,709 |
| Plastering on brick, concrete, foam concrete (2-3 cm) |
0,035 |
| Ceiling (attic) ceiling | 1,43 |
| wooden floors | 1,85 |
| Double wooden doors | 0,21 |
Table- Thermal losses of windows various designs at ΔT = 50 °С (T external = -30 °С, Т internal = 20 °С.)
Note |
As can be seen from the previous table, modern double-glazed windows can reduce window heat loss 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.
For the correct choice of materials and thicknesses of enclosing structures, we apply this information to a specific example.
In the calculation of heat losses per square. meter involved two quantities:
- temperature difference ΔT,
- heat transfer resistance R.
Let's define the indoor temperature as 20 °C, and take the outside temperature as -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 / sq.m.).
To simplify the calculations of heat loss in construction guides, heat losses are given different kind walls, floors, etc. for some values of winter air temperature. In particular, different figures are given for corner rooms (where the swirl of air flowing through the house affects) and non-corner ones, and different thermal patterns are taken into account for rooms on the first and upper floors.
Table- Specific heat loss of building fencing elements (per 1 sq.m. along the inner contour of the walls) depending on the average temperature of the coldest week of the year.
Note |
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Table- Specific heat losses of building fencing elements (per 1 sq.m. along the internal contour) depending on the average temperature of the coldest week of the year.
| Fence characteristic | outdoor temperature, °С | heat loss, kW |
|---|---|---|
| double glazed window | -24 -26 -28 -30 |
117 126 131 135 |
| Solid wood doors (double) | -24 -26 -28 -30 |
204 219 228 234 |
| Attic floor | -24 -26 -28 -30 |
30 33 34 35 |
| Wooden floors above basement | -24 -26 -28 -30 |
22 25 26 26 |
Consider an example of calculating the heat losses of two different rooms one area using tables.
Example 1
Corner room (first floor)
Room characteristics:
- first floor,
- room area - 16 sq.m. (5x3.2),
- ceiling height - 2.75 m,
- outer walls - two,
- material and thickness of the outer walls - timber 18 cm thick, sheathed with plasterboard and covered with wallpaper,
- windows - two (height 1.6 m, width 1.0 m) with double glazing,
- floors - wooden insulated, basement below,
- higher attic floor,
- design outside temperature -30 °С,
- the required temperature in the room is +20 °С.
External wall area excluding windows:
S walls (5 + 3.2) x2.7-2x1.0x1.6 \u003d 18.94 square meters. m.
window area:
S windows \u003d 2x1.0x1.6 \u003d 3.2 square meters. m.
Floor area:
S floor \u003d 5x3.2 \u003d 16 square meters. m.
Ceiling area:
S ceiling \u003d 5x3.2 \u003d 16 square meters. 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 we calculate the heat loss of each of the surfaces:
Q total = 3094 watts.
Note that more heat escapes through walls than through windows, floors and ceilings.
The result of the calculation shows the heat loss of the room in the most frosty (T outdoor = -30 ° C) days of the year. Naturally, the warmer it is outside, the less heat will leave the room.
Example 2
Roof room (attic)
Room characteristics:
- top floor,
- area 16 sq.m. (3.8x4.2),
- ceiling height 2.4 m,
- exterior walls; two roof slopes (slate, solid lathing, 10 cm mineral wool, lining), gables (10 cm thick timber, sheathed with lining) and side partitions (frame wall with expanded clay filling 10 cm),
- windows - four (two on each gable), 1.6 m high and 1.0 m wide with double glazing,
- design outside temperature -30°С,
- required room temperature +20°C.
Calculate the area of heat transfer surfaces.
The area of the end external walls minus the windows:
S end walls \u003d 2x (2.4x3.8-0.9x0.6-2x1.6x0.8) \u003d 12 square meters. m.
The area of the roof slopes that bound the room:
S slope walls \u003d 2x1.0x4.2 \u003d 8.4 square meters. m.
The area of the side partitions:
S side cut = 2x1.5x4.2 = 12.6 sq. m.
window area:
S windows \u003d 4x1.6x1.0 \u003d 6.4 square meters. m.
Ceiling area:
S ceiling \u003d 2.6x4.2 \u003d 10.92 square meters. m.
Now we calculate the heat losses of these surfaces, while taking into account that heat does not escape through the floor (there is a warm room). We consider heat losses for walls and ceilings as for corner rooms, and for the ceiling and side partitions we introduce a 70% coefficient, since unheated rooms are located behind them.
The total heat loss of the room will be:
Q total = 4504 watts.
As you can see, a warm room on the first floor loses (or consumes) much less heat than an attic room with thin walls and a large glass area.
In order to make such a room suitable for winter living, it is first necessary to insulate the walls, side partitions and windows.
Any enclosing structure can be represented as a multilayer wall, each layer of which has its own thermal resistance and its own resistance to the passage of air. Adding the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also summing up the resistance to the passage of air of all layers, we will understand how the wall breathes. Perfect Wall from a bar should be equivalent to a wall from a bar 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 external = -20 °С, T internal =20 °С.)
| wall layer | Thickness layer walls | Resistance heat transfer wall layer | Resist. air duct permeability equivalent to timber wall thick (cm) |
|
|---|---|---|---|---|
| Ro, | Equivalent brick masonry thick (cm) |
|||
| Brickwork from ordinary clay brick thickness: 12 cm |
12 25 50 75 |
0,15 0,3 0,65 1,0 |
12 25 50 75 |
6 12 24 36 |
| Claydite-concrete block masonry 39 cm thick with density: 1000 kg / m3 |
39 |
1,0 0,65 0,45 |
75 50 34 |
17 23 26 |
| Foam aerated concrete 30 cm thick density: 300 kg / m3 |
30 |
2,5 1,5 0,9 |
190 110 70 |
7 10 13 |
| Brusoval wall thick (pine) 10 cm |
10 15 20 |
0,6 0,9 1,2 |
45 68 90 |
10 15 20 |
For an objective picture of the heat loss of the whole house, it is necessary to take into account
- Heat loss through the contact of the foundation with frozen ground usually takes 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
- Heat loss associated with ventilation. These losses are calculated taking into account building codes (SNiP). For a residential building, about one air exchange per hour is required, that is, during this time it is necessary to supply the same volume of fresh air. Thus, the losses associated with ventilation are slightly less than the sum of heat losses attributable to the building envelope. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation is 50%. In European norms 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 is returned. This allows you to reduce heat losses by 30%, therefore, the value of the thermal resistance of the wall obtained during the calculation should be multiplied by 1.3 (or, accordingly, heat losses should be reduced).
Summing up all the heat losses at home, you will determine what power the heat generator (boiler) and heaters are needed for comfortable heating of 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 with the help of additional insulation.
You can also calculate the heat consumption by aggregated indicators. So, in one- and two-story houses that are not very insulated at an outside temperature of -25 ° C, 213 W are required per square meter of total area, 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 low, but during the operation of the building, the main costs are for heating. In no case should you save on thermal insulation, especially when comfortable living over large areas. Energy prices around the world are constantly rising.
- Modern building materials have a higher thermal resistance than traditional materials. This allows you to make the 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 heat constantly - 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 that have cooled down during the night have “accumulated cold”.
- Insulation must be considered in conjunction with the air permeability of the walls. If an increase in the thermal resistance of the walls is associated with a significant decrease in air permeability, then it should not be used. An ideal wall in terms of air permeability is equivalent to a wall made of timber with a thickness of 15 ... 20 cm.
- Very often, improper use of vapor barrier leads to a deterioration in the sanitary and hygienic properties of housing. With properly organized ventilation and “breathing” walls, it is unnecessary, and with poorly breathable walls, this is unnecessary. Its main purpose is to prevent wall infiltration and protect insulation from wind.
- Wall insulation from the outside is much more effective than internal insulation.
- Do not endlessly insulate walls. The effectiveness of this approach to energy saving is not high.
- Ventilation - these are the main reserves of energy saving.
- Applying modern systems glazing (double-glazed windows, heat-shielding glass, etc.), low-temperature heating systems, effective thermal insulation of enclosing structures, it is possible to 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.
Any construction of the house, begins with drawing up the project of the house. Already at this stage, you should think about warming your home, because. there are no buildings and houses with zero heat loss that we pay for cold winter, in heating season. Therefore, it is necessary to carry out the insulation of the house outside and inside, taking into account the recommendations of the designers.
What and why to insulate?
During the construction of houses, many do not know, and do not even realize that in a private house built, during the heating season, up to 70% of the heat will go to heat the street.
Concerned about saving family budget and the problem of home insulation, many are wondering: what and how to insulate ?
This question is very easy to answer. It is enough to look at the screen of the thermal imager in winter, and you will immediately notice through which structural elements the heat escapes into the atmosphere.
If you do not have such a device, then it does not matter, below we will describe the statistics that show where and in what percentage the heat leaves the house, as well as post a video of the thermal imager from a real project.
When insulating a house it is important to understand that heat escapes not only through floors and roofs, walls and foundations, but also through old windows and doors that will need to be replaced or insulated during the cold season.
Distribution of heat losses in the house
All experts recommend insulation of private houses
, apartments and industrial premises not only from the outside, but also from the inside. If this is not done, then the warmth that is “dear” to us, in the cold season, will simply quickly disappear into nowhere.
Based on statistics and data from specialists, according to which, if the main heat leaks are identified and eliminated, it will already be possible to save 30% or more percent on heating in winter.
So, let's analyze in what directions, and in what percentage our heat leaves the house.
Most big losses heat occur through:
Heat loss through the roof and floors
As is known, warm air always rises to the top, so it heats the uninsulated roof of the house and ceilings, through which 25% of our heat leaks.
To produce house roof insulation and reduce heat loss to a minimum, you need to use roof insulation with a total thickness of 200mm to 400mm. The technology of insulating the roof of the house can be seen by enlarging the picture on the right.
Heat loss through walls
Many will probably wonder: why is the heat loss through the uninsulated walls of the house (about 35%) more than through the uninsulated roof of the house, because all the warm air rises to the top?
Everything is very simple. First, the wall area is much more area roofs, and secondly, different materials have different thermal conductivity. Therefore, during construction country houses, you need to take care of the house wall insulation. For this, insulation for walls with a total thickness of 100 to 200 mm is suitable.
For proper insulation walls of the house you need to have knowledge of technology and special tool. The technology of insulating the walls of a brick house can be seen by enlarging the picture on the right.
Heat loss through floors
Strange as it may seem, but not insulated floors in the house take from 10 to 15% of the heat (the figure may be more if your house is built on piles). This is due to ventilation under the house during the cold period of winter.
To minimize heat loss through insulated floors in the house, you can use insulation for floors with a thickness of 50 to 100mm. This will be enough to walk barefoot on the floor in the cold winter season. The technology of home floor insulation can be seen by enlarging the picture on the right.
Heat loss through windows
Window- perhaps this is the very element that is almost impossible to insulate, because. then the house will become like a dungeon. The only thing that can be done to reduce heat loss by up to 10% is to reduce the number of windows in the design, insulate the slopes and install at least double-glazed windows.
Heat loss through doors
The last element in the design of the house, through which up to 15% of heat escapes, is the doors. This is due to the constant opening of the entrance doors through which heat constantly escapes. For reducing heat loss through doors to a minimum, it is recommended to install double doors, seal them with sealing rubber and install thermal curtains.
Benefits of an insulated home
- Payback in the first heating season
- Savings on air conditioning and heating at home
- Cool indoors in summer
- Excellent additional soundproofing walls and floors, ceilings and floors
- Protection of house structures from destruction
- Increased indoor comfort
- It will be possible to turn on the heating much later
The results of the insulation of a private house
It is very profitable to warm the house , and in most cases even necessary, because this is due large quantity advantages over non-insulated houses, and allows you to save your family budget.
Having carried out external and internal insulation at home, your a private house becomes like a thermos. Heat will not fly away from it in winter and heat will not come in in summer, and all costs for the complete insulation of the facade and roof, basement and foundation will pay off within one heating season.
For optimal choice heater for home , we recommend that you read our article: The main types of insulation for the house, which discusses in detail the main types of insulation used in the insulation of a private house outside and inside, their pros and cons.
Video: Real project - where does the heat go in the house
The first step in organizing the heating of a private house is the calculation of heat loss. The purpose of this calculation is to find out how much heat escapes outside through walls, floors, roofs and windows (common name - building envelope) 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 start selecting a heat source by power.
Basic Formulas
To get a more or less accurate result, it is necessary 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;
- energy loss for heating ventilation air.
The basic formula for calculating the consumption of thermal energy through external fences is as follows:
Q \u003d 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 is the thermal resistance of the construction material, m²°C / W;
- S is the area of the outer fence, m²;
- t in - internal air temperature, ° С;
- t n - most low temperature environment, °С;
- 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. For this, the formula R = δ / λ is used, where:
- λ is the 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 the wall is built from 2 materials (for example, a brick with a mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summarized. The outdoor temperature is selected as regulatory documents, and according to personal observations, internal - by necessity. Additional heat losses are the coefficients defined by the standards:
- When the wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
- If the structure is facing southeast or west, β = 0.05.
- β = 0 when the outer fence faces 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 made of all fences adjacent to the environment: walls, windows, roofs, floors 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 results of measurements, the area of \u200b\u200beach structure is calculated and substituted into the first formula (S, m²). The value of R is also inserted there, obtained by dividing the thickness of the fence by the coefficient of thermal conductivity building material. In the case of new metal-plastic windows, the value of R will be prompted by a representative of the installer.
As an example, it is worthwhile to calculate the heat loss through the enclosing walls made of bricks 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 is facing north (β = 0.1). First you need to take from the reference literature the coefficient of thermal conductivity of the brick (λ), it is equal to 0.44 W / (m ° C). Then, according to the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:
R \u003d 0.25 / 0.44 \u003d 0.57 m² ° C / W
To determine the heat loss of a room with this wall, all the initial data must be substituted into the first formula:
Q \u003d 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) \u003d 434 W \u003d 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 for floors, roofs and front door. At the end, all the results are summarized, 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 heat energy consumed by the heating system for heating the ventilation air. The share of this energy reaches 30% of the total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss at home through the heat capacity of the air using the popular formula from the physics course:
Q air \u003d 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, ° С;
- m is the mass flow rate of air entering the house from the outside, kg;
- c is the heat capacity of the air mixture, equal to 0.28 W / (kg ° С).
Here, all quantities are known except mass flow air for ventilation. In order not to complicate your task, you should agree with the condition that air environment is updated throughout the house 1 time per hour. Then it is not difficult to calculate the volumetric air flow by adding the volumes of all rooms, and then you need to convert it into mass air through density. Since the density of the air mixture varies with its temperature, you need to take 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 \u003d 0.28 x 711 x 45 \u003d 8957 W, which is approximately equal to 9 kW.
Upon completion of the calculations, the results of heat losses through the external enclosures are added to the ventilation heat losses, which gives the total heat load to the heating system of the building.
The presented calculation methods can be simplified if the formulas are entered into the Excel program in the form of tables with data, this will significantly speed up the calculation.
