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 quite exact result and proceed to the selection of 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;
- loss of energy used to heat the 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. outdoor temperature is chosen 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 measurement results, 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 thermal conductivity of the 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 heat transfer resistance of a brick wall of 0.25 m is calculated:
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 the floors, roof 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 the 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 for the mass air flow during ventilation of rooms. In order not to complicate your task, it is worth agreeing 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 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 \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.
Infiltration is associated with:Your gag is due to two things
- Air penetration through walls. Let's assume it is 0.
- Air penetration through openings
- Window
- self-ventilation
- Special devices (climatic valves, special channels). If you do not have them in the window design (do not confuse them with drainage ones) - do not take them into account in your task.
- Partially breathable seals. If you do not have them in the window construct, do not take them into account in your task
- Adjacency leaks. If your double-glazed windows are not defective - do not take into account in your task.
- doors
- Ventilation - you opened the window to ventilate. Do not take into account in your task.
- Self-ventilating - suppose you don't have a wind blowing through your door. Do not take into account in your task.
In any case, as part of the questions about PVC windows:
- In Valtek, ventilation was classified as infiltration (heat consumption for heating the air required by sanitary standards for living quarters and kitchens). This is clearer, but, strictly speaking, it is not true.
- You think that "you will breathe less, the air will be fresher, so you need to heat less."
- You can accept that "we will plug all hoods and ventilate by hand when necessary". In this case, you do not need 14 kW of heat to compensate for heat loss through infiltration / ventilation, but along the way you will solve the issue of humidity.
- You can accept that "fuck it, I'll heat it up more." In this case, you will need to provide additional generation, distribution and supply of 14 kW. But it's up to you to generate/distribute/serve. That is, it is more powerful to mount the boiler, there are more warm floors - but do not heat so much.
Regarding "accept heat loss / insolation through openings at the worst estimate":
- You do not take into account infiltration through openings, since it is negligible with ventilation through the ventilation system.
- To the thesis “but when I put my hand to the window - it feels like it’s blowing,” I immediately say - most likely it’s not blowing and the sensations are related to the fact that the window is cold, and this is a consequence of installing windows / slope insulation / quarter insulation / etc.
- Infiltration in Valtec is shown crookedly, because it is not so much infiltration as ventilation. And compensation for ventilation heat loss is a separate topic of the forum and a complex issue.
- I tried for a very long time to find out the heat resistance of my 40 m2 glazing.
- When I realized that the numbers on the Internet are different, take into account the influence of different distance frames- it is practically impossible, and the heat resistance of the 4-14Ar-4-16Ar-4I package in seven different offices on paper is shown by 8 different numbers.
- Scored and took the worst estimate. - because the price of a mistake is high.
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 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 presented to individual elements 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 consumption thermal energy for heating the building 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 outer wall 4°C, for covering and attic flooring 3°C and for covering over 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 home permanent residence 3.13 °С m. sq./W,
for administrative and other public buildings, incl. buildings seasonal residence 2.55 °С 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 the consumer approach (R=1.97 °C.m.sq./W) and prescriptive approach (R=3.13 °C.m.sq./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 cladding 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 |
|
Prescriptive approach |
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By consumer approach |
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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, by selecting a boiler or heaters only according to the ability to heat a certain area indicated in their documentation, you confirm that your house was built with strict consideration of the requirements of SNiP 23-02-2003.
The conclusion follows from the above material. For right choice 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. Conversely, one can also say what temperature difference will occur when a certain amount of heat passes through square meter fences.
where q is the amount of heat lost per square meter of enclosing surface. It is measured in watts per square meter (W/m2); ΔT is the difference between the temperature in the street and in the room (°С) and, R is the heat transfer resistance (°С/W/m2 or °С·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 a wooden wall and 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 various materials at ΔT = 50 °С (Т nar. = -30 °C, T internal = 20 °C.)
|
Wall material and thickness |
Heat transfer resistanceR m , |
|
Brick wall 3 bricks (79 cm) thick 2.5 bricks (67 cm) thick 2 bricks (54 cm) thick 1 brick (25 cm) thick |
0,592 0,502 0,405 0,187 |
|
Log cabin Ø 25 Ø 20 |
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Log cabin 20 cm thick 10 cm thick |
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|
Frame wall (board + mineral wool + board) 20 cm |
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|
Foam concrete wall 20 cm 30 cm |
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Plaster on brick, concrete, foam concrete (2-3 cm) |
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Ceiling (attic) ceiling |
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wooden floors |
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Double wooden doors |
Table – Heat losses of windows of various designs at ΔT = 50 °С (Т nar. = -30 °C, T internal = 20 °C.)
Note Even numbers in symbol double glazing mean air gap in mm; The symbol Ar means that the gap is not filled with air, but with argon; The letter K means that the outer glass has a special transparent heat-shielding coating. |
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.
We 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 calculation of heat losses in construction guides, heat losses of different type of 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 is affected) and non-corner rooms, and different thermal patterns are taken into account for rooms on the first and upper floors.
Table – Specific heat losses of the 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 If there is an external unheated room behind the wall (canopy, glazed porch, etc.), then the heat loss through it is 70% of the calculated one, and if behind this unheated room there is not a street, but one more room outside (for example, a canopy overlooking to the veranda), then 40% of the calculated value. |
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Table – Specific heat loss 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 |
Outside temperature, °C |
Heat loss, kW |
|
Window with double glazed |
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Solid wood doors (double) |
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Attic floor |
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Wooden floors above basement |
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 °C.
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.
Square internal partitions does not participate 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 outside = -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 sheathing, 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 we see, warm room the first floor loses (or consumes) significantly less heat than attic room with thin walls and large area glazing.
In order to make such a room suitable for winter residence, you must first 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 °С (Т nar. =–20 °C, T internal =20 °C.)
|
wall layer |
Wall layer thickness (cm) |
Heat transfer resistance of the wall layer |
Resist. air permeability equivalent to timber wall thickness (cm) |
|
|
Equivalent masonry thickness (cm) |
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|
Brickwork of ordinary clay brick thickness: 12 cm 25 cm 50 cm 75 cm |
0,15 0,3 0,65 1,0 |
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Masonry of expanded clay concrete blocks 39 cm thick with a density of: 1000 kg/cu m 1400 kg/cu m 1800 kg/cu m |
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Foam-aerated concrete 30 cm thick density: 300 kg/cu m 500 kg/cu m 800 kg/cu m |
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Brusoval wall thick (pine) 10 cm 15 cm 20 cm |
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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 heating appliances are necessary for comfortable heating of the house in the coldest and windy 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 heavily 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, these are: 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 Construction Materials have 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 is the main reserve 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.
Tiled roof insulation using ISOVER thermal insulation
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Wall insulation made of lightweight concrete blocks |
Insulation of a brick wall with a ventilated gap |
Log wall insulation |
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To date heat saving is important parameter, which is taken into account when constructing a residential or office space. In accordance with SNiP 23-02-2003 " Thermal protection 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, heat loss at home.
Prescriptive approach- these are the standards for individual elements of the thermal protection of a building: external walls, floors above unheated spaces, coatings and attic ceilings, windows, entrance doors, etc.
consumer approach(heat transfer resistance can be reduced in relation to the prescriptive level, provided that the design specific heat energy consumption for space heating is below the standard).
Sanitary and hygienic requirements:
- The difference between the air temperatures inside and outside the room should not exceed certain allowable values. The maximum allowable temperature difference for the outer wall is 4°C. for covering and attic flooring 3°С and for covering over basements and undergrounds 2°С.
- The temperature on the inner surface of the enclosure 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 ° С m 2 / W, and according to the prescriptive approach:
- for a permanent home 3.13 ° С m 2 / W.
- for administrative and other public buildings, including structures for seasonal residence 2.55 °C 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, in order to correctly select the power of the heating boiler, either heating appliances, 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 a square meter of construction 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 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 wooden wall lined with brick is the sum of three resistances: a brick and wooden wall and an 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.
Table of heat transfer resistance of 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 |
|
|
wooden floors |
|
|
Double wooden doors |
Window heat loss table various designs at ΔT = 50 °С (T external = -30 °С. Т internal = 20 °С.)
|
Note
. Even numbers in the symbol of a double-glazed window 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 loss in building reference books indicate heat loss different kind walls, floors, etc. for some values of winter air temperature. Usually given various numbers For corner rooms(the swirl of air flowing through the house affects the house) 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 outdoor unheated room behind the wall (canopy, glazed veranda, etc.), then the heat loss through it will be 70% of the calculated one, and if there is another outdoor room behind this unheated room, then the heat loss will be 40 % of 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
The area of the internal partitions is not included 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.
Total heat loss rooms will be: Q total \u003d 3094 watts.
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, solid sheathing. 10 cm of mineral wool, lining). gables (beam 10 cm thick lined 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 the heat losses of these surfaces, while it is necessary to take into account that through the floor in this case the heat will not go away, as it is located below 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 can see, a warm room on the 1st floor loses (or consumes) much less heat than an attic room with thin walls and a large glazing area.
To this room make it suitable for winter living, it is necessary first of all to insulate the 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 the contact of the foundation with frozen ground, as a rule, 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 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 50%. 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 of the thermal resistance of the wall obtained in the calculation 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 heaters need to comfortably heat the house on the coldest and windiest 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 at an outside temperature of -25 ° C, 213 W per 1 m 2 of the total area is needed, and at -30 ° C - 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.
Before you start building a house, you need to buy a house project - that's what the architects say. It is necessary to buy the services of professionals - so the builders say. It is necessary to buy high-quality building materials - this is what sellers and manufacturers of building materials and insulation say.
And you know, in some ways they are all a little bit right. However, no one but you will be so interested in your housing to take into account all the points and bring together all the issues of its construction.
One of the most important issues that should be resolved at the stage is the heat loss of the house. The design of the house, its construction, and what building materials and insulation you will purchase will depend on the calculation of heat loss.
There are no houses with zero heat loss. To do this, the house would have to float in a vacuum with walls 100 meters high. effective insulation. We do not live in a vacuum, and we do not want to invest in 100 meters of insulation. So, our house will have heat loss. Let them be, as long as they are reasonable.
Heat loss through walls
Heat loss through the walls - all the owners think about it at once. The thermal resistance of the building envelope is considered, they are insulated until the standard indicator R is reached, and this completes their work on the insulation of the house. Of course, heat loss through the walls of the house must be considered - the walls have the maximum area of all the enclosing structures of the house. But they are not the only way for heat to get out.
Home insulation is the only way to reduce heat loss through the walls.
In order to limit heat loss through the walls, it is enough to insulate the house 150 mm for the European part of Russia or 200-250 mm of the same insulation for Siberia and the northern regions. And on this you can leave this indicator alone and move on to others, no less important.
Floor heat loss
The cold floor in the house is a disaster. The heat loss of the floor, relative to the same indicator for walls, is about 1.5 times more important. And it is exactly the same amount that the thickness of the insulation in the floor should be greater than the thickness of the insulation in the walls.
Floor heat loss becomes significant when you have a cold basement or just outside air under the floor of the first floor, for example, with screw piles.
Insulate the walls and insulate the floor.
If you lay 200 mm into the walls basalt wool or polystyrene, then you will have to lay 300 millimeters of equally effective insulation in the floor. Only in this case it will be possible to walk barefoot on the floor of the first floor to any, even the most fierce,.
If you have a heated basement under the floor of the first floor or a well-insulated basement with a well-insulated wide blind area, then the insulation of the floor of the first floor can be neglected.
Moreover, it is worth pumping heated air into such a basement or basement from the first floor, and preferably from the second. But the walls of the basement, its slab should be insulated as much as possible so as not to "heat" the ground. Of course, the constant temperature of the soil is +4C, but this is at a depth. And in winter, around the walls of the basement are the same -30C, as well as on the surface of the soil.
Heat loss through the ceiling
All heat goes up. And there it seeks to go outside, that is, to leave the room. Heat loss through the ceiling in your house is one of the largest values that characterizes the heat loss to the street.
The thickness of the insulation on the ceiling should be 2 times the thickness of the insulation in the walls. Mount 200 mm into walls - mount 400 mm into the ceiling. In this case, you will be guaranteed the maximum thermal resistance of your thermal circuit.
What do we get? Walls 200 mm, floor 300 mm, ceiling 400 mm. Consider that you will save money with which you will heat your home.
Windows heat loss
What is completely impossible to insulate are the windows. The heat loss of windows is the most great value, which describes the amount of heat leaving your home. Whatever you make your double-glazed windows - two-chamber, three-chamber or five-chamber, the heat loss of windows will still be gigantic.
How to reduce heat loss through windows? First, it is worth reducing the area of glazing throughout the house. Of course, with large glazing, the house looks chic, and its facade reminds you of France or California. But there is already one thing - either half-wall stained-glass windows or good heat resistance of your house.
If you want to reduce the heat loss of windows, do not plan a large area of them.
Secondly, it should be well insulated window slopes- places where the bindings adhere to the walls.
And, thirdly, it is worth using novelties in the construction industry for additional heat conservation. For example, automatic night heat-saving shutters. Or films that reflect heat radiation back into the house, but freely transmit the visible spectrum.
Where does the heat from the house go?
The walls are insulated, the ceiling and the floor too, the shutters are put on five-chamber double-glazed windows, with might and main it is fired up. But the house is still cold. Where does the heat from the house continue to go?
It's time to look for cracks, cracks and cracks, where the heat leaves the house.
First, the ventilation system. Cold air comes in supply ventilation into the house, warm air leaves the house exhaust ventilation. To reduce heat loss through ventilation, you can install a heat exchanger - a heat exchanger that takes heat from the outgoing warm air and heating the incoming cold air.
One way to reduce heat loss at home through the ventilation system is to install a heat exchanger.
Secondly, the entrance doors. To prevent heat loss through the doors, it is necessary to mount cold vestibule, which will buffer between entrance doors and street air. The tambour should be relatively airtight and unheated.
Thirdly, it is worth at least once to look at your house in the cold with a thermal imager. Departure of experts costs not so big money. But you will have a “map of facades and ceilings” on hand, and you will clearly know what other measures to take in order to reduce heat loss at home during the cold season.
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