Norm of heat loss of the outer wall Tue. Calculation of heat loss at home: online calculator. Video: Real project - where does the heat go in the house
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, coatings 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 of thermal energy 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. Maximum allowed values differential for outer wall 4°C, for coating and attic floor 3°С and for ceilings over cellars and undergrounds 2°С.
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 including 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 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, 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 boiler power and heating appliances, it is necessary to calculate real heat loss premises of your home.
Below we will show a simple method for calculating the heat loss of your home.
The house loses heat through the wall, roof, strong 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 measures how much heat is lost through square meter 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 we are talking about a multi-layer design, the resistance layers simply add up. For example, the resistance of a wall made of wood lined with bricks is the sum of three resistances: brick and 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 of various materials at ΔT = 50 °C (T out = -30 °C, T int = 20 °C.)
| Wall material and thickness | Heat transfer resistance R m, |
|---|---|
| Brick wall 3 bricks thick (79 cm) 2.5 bricks thick (67 cm) 2 bricks thick (54 cm) 1 brick thick (25 cm) |
0,592 0,502 0,405 0,187 |
| Log 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 building reference books, heat losses are given different kind walls, floors, etc. for some values winter temperature air. 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 loss 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, °C | 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 °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 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 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 let's calculate heat loss 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) much less heat than an attic room with thin walls and a large glazing area.
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. An ideal timber wall should be equivalent to a 15 - 20 cm thick timber wall. The table below will help you 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 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%, so the value obtained in the calculation thermal resistance walls should be multiplied by 1.3 (or reduce heat loss accordingly).
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 not very insulated houses with outdoor temperature-25 °C requires 213 W 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 can you save on thermal insulation, especially with a comfortable stay on 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 have 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. When correct organized ventilation and "breathing" walls, it is unnecessary, and with poorly breathable walls it 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 for additional insulation of building structures based on building thermal insulation type "ISOVER", in the presence of air exchange and ventilation systems in the premises.
Today, many families choose for themselves Vacation home as a place of permanent residence or year-round recreation. However, its content, and in particular the payment utilities, are quite costly, while most homeowners are not oligarchs at all. One of the most significant expenses for any homeowner is the cost of heating. To minimize them, it is necessary to think about energy saving even at the stage of building a cottage. Let's consider this question in more detail.
« The problems of energy efficiency of housing are usually remembered from the perspective of urban housing and communal services, however, the owners individual houses this topic is sometimes much closer,- considers Sergey Yakubov , deputy director of sales and marketing, a leading manufacturer of roofing and facade systems in Russia. - The cost of heating a house can be much more than half the cost of maintaining it in the cold season and sometimes reach tens of thousands of rubles. However, with a competent approach to the thermal insulation of a residential building, this amount can be significantly reduced.».
Actually, you need to heat the house in order to constantly maintain in it comfortable temperature no matter what's going on outside. In this case, it is necessary to take into account heat losses both through the building envelope and through ventilation, because. heat leaves with heated air, which is replaced by cooled air, as well as the fact that a certain amount of heat is emitted by people in the house, Appliances, incandescent lamps, etc.
To understand how much heat we need to get from our heating system and how much money we have to spend on it, let's try to evaluate the contribution of each of the other factors to the heat balance using the example of a brick building located in the Moscow region two-story house With with total area premises 150 m2 (to simplify the calculations, we considered that the dimensions of the cottage in terms of approximately 8.7x8.7 m and it has 2 floors 2.5 m high).
Heat loss through building envelope (roof, walls, floor)
The intensity of heat loss is determined by two factors: the temperature difference inside and outside the house and the resistance of its enclosing structures to heat transfer. By dividing the temperature difference Δt by the heat transfer resistance coefficient Ro of walls, roofs, floors, windows and doors and multiplying by their surface area S, we can calculate the intensity of heat loss Q:
Q \u003d (Δt / R o) * S
The temperature difference Δt is not constant, it changes from season to season, during the day, depending on the weather, etc. However, our task is simplified by the fact that we need to estimate the need for heat in total for the year. Therefore, for an approximate calculation, we may well use such an indicator as the average annual air temperature for the selected area. For the Moscow region it is +5.8°C. If we take +23°C as a comfortable temperature in the house, then our average difference will be
Δt = 23°C - 5.8°C = 17.2°C
Walls. The area of the walls of our house (2 square floors 8.7x8.7 m high 2.5 m) will be approximately equal to
S \u003d 8.7 * 8.7 * 2.5 * 2 \u003d 175 m 2
However, the area of windows and doors must be subtracted from this, for which we will calculate the heat loss separately. Suppose we have one front door, standard size 900x2000 mm, i.e. area
S doors \u003d 0.9 * 2 \u003d 1.8 m 2,
and windows - 16 pieces (2 on each side of the house on both floors) with a size of 1500x1500 mm, the total area of \u200b\u200bwhich will be
S windows \u003d 1.5 * 1.5 * 16 \u003d 36 m 2.
Total - 37.8 m 2. Remaining area brick walls -
S walls \u003d 175 - 37.8 \u003d 137.2 m 2.
The heat transfer resistance coefficient of a 2-brick wall is 0.405 m2°C/W. For simplicity, we will neglect the resistance to heat transfer of the layer of plaster covering the walls of the house from the inside. Thus, the heat dissipation of all the walls of the house will be:
Q walls \u003d (17.2 ° C / 0.405 m 2 ° C / W) * 137.2 m 2 \u003d 5.83 kW
Roof. For simplicity of calculations, we will assume that the resistance to heat transfer roofing cake equal to the heat transfer resistance of the insulation layer. For light mineral wool insulation 50-100 mm thick, most often used for roof insulation, it is approximately equal to 1.7 m 2 °C / W. We will neglect the heat transfer resistance of the attic floor: let's assume that the house has an attic, which communicates with other rooms and heat is distributed evenly between all of them.
Square gable roof with a slope of 30 ° will be
Roof S \u003d 2 * 8.7 * 8.7 / Cos30 ° \u003d 87 m 2.
Thus, its heat dissipation will be:
Roof Q \u003d (17.2 ° C / 1.7 m 2 ° C / W) * 87 m 2 \u003d 0.88 kW
Floor. The heat transfer resistance of a wooden floor is approximately 1.85 m2°C/W. Having made similar calculations, we obtain heat dissipation:
Q floor = (17.2°C / 1.85m 2 °C/W) * 75 2 = 0.7 kW
Doors and windows. Their resistance to heat transfer is approximately equal to 0.21 m 2 °C / W, respectively (double wooden door) and 0.5 m 2 °C / W (ordinary double-glazed window, without additional energy-efficient "gadgets"). As a result, we get heat dissipation:
Q door = (17.2°C / 0.21W/m 2 °C) * 1.8m 2 = 0.15 kW
Q windows \u003d (17.2 ° C / 0.5 m 2 ° C / W) * 36 m 2 \u003d 1.25 kW
Ventilation. According to building codes, the air exchange coefficient for a dwelling should be at least 0.5, and preferably 1, i.e. in an hour, the air in the room should be completely updated. Thus, with a ceiling height of 2.5 m, this is approximately 2.5 m 3 of air per hour per square meter. This air must be heated from outdoor temperature (+5.8°C) to room temperature (+23°C).
The specific heat capacity of air is the amount of heat required to raise the temperature of 1 kg of a substance by 1 ° C - approximately 1.01 kJ / kg ° C. At the same time, the air density in the temperature range of interest to us is approximately 1.25 kg/m3, i.e. the mass of 1 cubic meter of it is 1.25 kg. Thus, to heat the air by 23-5.8 = 17.2 ° C for each square meter of area, you will need:
1.01 kJ / kg ° C * 1.25 kg / m 3 * 2.5 m 3 / hour * 17.2 ° C = 54.3 kJ / hour
For a house of 150 m2, this will be:
54.3 * 150 \u003d 8145 kJ / h \u003d 2.26 kW
| Heat loss through | Temperature difference, °C | Area, m2 |
Heat transfer resistance, m2°C/W |
Heat loss, kW |
| Walls |
17,2 |
175 |
0,41 |
5,83 |
| Roof |
17,2 |
87 |
1,7 |
0,88 |
| Floor |
17,2 |
75 |
1,85 |
0,7 |
| doors |
17,2 |
1,8 |
0,21 |
0,15 |
| Window |
17,2 |
36 |
0,5 |
0,24 |
| Ventilation |
17,2 |
- |
- |
2,26 |
|
Total: |
|
|
|
11,06 |
Let's breathe now!
Suppose a family of two adults with two children lives in a house. The nutritional norm for an adult is 2600-3000 calories per day, which is equivalent to a heat dissipation power of 126 watts. The heat dissipation of a child will be estimated at half the heat dissipation of an adult. If everyone who lived at home is in it 2/3 of the time, then we get:
(2*126 + 2*126/2)*2/3 = 252W
Let's say that there are 5 rooms in the house, lit by ordinary incandescent lamps with a power of 60 W (not energy-saving), 3 per room, which are turned on for an average of 6 hours a day (i.e. 1/4 of the total time). Approximately 85% of the power consumed by the lamp is converted into heat. In total we get:
5*60*3*0.85*1/4=191W
The refrigerator is a very efficient heating device. Its heat dissipation is 30% of the maximum power consumption, i.e. 750 W.
Other household appliances (let it be washing and dishwasher) releases about 30% of the maximum power input as heat. The average power of these devices is 2.5 kW, they work for about 2 hours a day. Total we get 125 watts.
A standard electric stove with an oven has a power of approximately 11 kW, however, the built-in limiter regulates the operation of the heating elements so that their simultaneous consumption does not exceed 6 kW. However, it is unlikely that we will ever use more than half of the burners at the same time or all the heating elements of the oven at once. Therefore, we will proceed from the fact that the average operating power of the stove is approximately 3 kW. If she works 3 hours a day, then we get 375 watts of heat.
Each computer (and there are 2 in the house) emits approximately 300 W of heat and works 4 hours a day. Total - 100 watts.
TV is 200 W and 6 hours a day, i.e. per circle - 50 watts.
In total we get: 1.84 kW.
Now we calculate the required thermal power heating systems:
Heating Q = 11.06 - 1.84 = 9.22 kW
heating costs
Actually, above we calculated the power that will be needed to heat the coolant. And we will heat it, of course, with the help of a boiler. Thus, heating costs are fuel costs for this boiler. Since we are considering the most general case, we will make a calculation for the most universal liquid (diesel) fuel, since gas pipelines are far from being everywhere (and the cost of their summing up is a figure with 6 zeros), but solid fuel it is necessary, firstly, to bring it somehow, and secondly, to throw it into the boiler furnace every 2-3 hours.
To find out what volume V of diesel fuel per hour we have to burn to heat the house, we need specific heat its combustion q (the amount of heat released during the combustion of a unit mass or volume of fuel, for diesel fuel - approximately 13.95 kWh / l) multiplied by the boiler efficiency η (approximately 0.93 for diesel) and then the required power of the heating system Qheating ( 9.22 kW) divided by the resulting figure:
V = heating Q / (q * η) = 9.22 kW / (13.95 kW * h / l) * 0.93) = 0.71 l / h
With an average cost of diesel fuel for the Moscow Region of 30 rubles per liter per year, it will take us
0.71 * 30 rub. * 24 hours * 365 days = 187 thousand rubles. (rounded).
How to save?
The natural desire of any homeowner is to reduce heating costs even at the construction stage. Where does it make sense to invest money?
First of all, you should think about the insulation of the facade, which, as we saw earlier, accounts for the bulk of all heat loss at home. In the general case, external or internal additional insulation can be used for this. However internal insulation much less efficient: when installing thermal insulation from the inside, the boundary between the warm and cold areas “moves” inside the house, i.e. moisture will condense in the thickness of the walls.
There are two ways to insulate facades: “wet” (plaster) and by installing a hinged ventilated facade. Practice shows that due to the need for constant repairs, “wet” insulation, taking into account operating costs, ends up being almost twice as expensive as a ventilated facade. The main disadvantage of the plaster facade is high price its service and content. " The initial costs for the arrangement of such a facade are lower than for a hinged ventilated one, by only 20-25%, a maximum of 30%,- explains Sergey Yakubov ("Metal Profile"). - However, considering the cost of Maintenance, which must be done at least once every 5 years, after the first five years, the plaster facade will be equal in cost to the ventilated one, and in 50 years (the service life of the ventilated facade) it will be 4-5 times more expensive».
What is a hinged ventilated facade? This is an external "screen" attached to a light metal frame, which is attached to the wall with special brackets. A light insulation is placed between the wall of the house and the screen (for example, Isover "VentFacade Bottom" with a thickness of 50 to 200 mm), as well as a wind and hydroprotective membrane (for example, Tyvek Housewrap). Can be used as exterior cladding various materials, but in individual construction most commonly used steel siding. " The use of modern high-tech materials in the production of siding, such as steel coated with Colorcoat Prisma ™, allows you to choose almost any design decision, - says Sergey Yakubov. - This material has excellent resistance to both corrosion and mechanical stress. The warranty period for it is 20 years with a real life of 50 years or more. Those. provided that steel siding is used, the entire facade structure will last 50 years without repair».
An additional layer of facade insulation made of mineral wool has a heat transfer resistance of approximately 1.7 m2°C/W (see above). In construction, to calculate the heat transfer resistance of a multi-layer wall, add up the corresponding values for each of the layers. As we remember, our main bearing wall in 2 bricks has a heat transfer resistance of 0.405 m2°C/W. Therefore, for a wall with a ventilated facade, we get:
0.405 + 1.7 = 2.105 m 2 °C / W
Thus, after insulation, the heat dissipation of our walls will be
Q facade \u003d (17.2 ° C / 2.105 m 2 ° C / W) * 137.2 m 2 \u003d 1.12 kW,
which is 5.2 times less than the same indicator for an uninsulated facade. Impressive, isn't it?
Again we calculate the required heat output of the heating system:
Q heating-1 = 6.35 - 1.84 = 4.51 kW
Diesel fuel consumption:
V 1 \u003d 4.51 kW / (13.95 kW * h / l) * 0.93) \u003d 0.35 l / h
Amount for heating:
0.35 * 30 rub. * 24 hours * 365 days = 92 thousand rubles.
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 solved 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 heat 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.
Warming of the house the only way reduce heat loss through 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 home is one of the largest values, which characterizes the escape of heat 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. Window heat loss is the largest measure of 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 exclude heat loss through the doors, a cold vestibule should be mounted, which will be a 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.
Reinforced concrete Concrete on gravel or crushed stone natural stone Dense silicate concrete Expanded clay concrete. sand and expanded clay foam concrete Р=1800 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=1600 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=1400 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=1200 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1000 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=800 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=600 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete Р=500 Expanded clay concrete on quartz sand with porization Р=1200 Expanded clay concrete on quartz sand with porization Р=1000 Expanded clay concrete on quartz sand with porization Р=800 Perlite concrete Р=1200 Perlite concrete Р=1000 Perlite concrete Р=800 Perlite concrete Р=60 0 Agloporite concrete and concretes on fuel slag P=1800 Aggloporite concrete and concretes on fuel slag Р=1600 Aggloporite concrete and concrete on fuel slag Р=1400 Aggloporite concrete and concrete on fuel slag Р=1200 Aggloporite concrete and concrete on fuel slag Р=1000 Concrete on ash gravel Р= 1400 Concrete on ash gravel Р=1200 Concrete on ash gravel Р=1000 Polystyrene concrete Р=600 Polystyrene concrete Р=500 Gas and foam concrete. gas and foam silicate Р=1000 Gas and foam concrete. gas and foam silicate Р=900 Gas and foam concrete. gas and foam silicate Р=800 Gas and foam concrete. gas and foam silicate Р=700 Gas and foam concrete. gas and foam silicate Р=600 Gas and foam concrete. gas and foam silicate Р=500 Gas and foam concrete. gas and foam silicate Р=400 Gas and foam concrete. gas and foam silicate Р=300 Gas and foam ash concrete Р=1200 Gas and foam ash concrete Р=100 Gas and foam ash concrete Р=800 Cement-sand mortar Complex (sand, lime, cement) mortar Lime-sand mortar Cement-slag mortar P=1400 Cement-slag mortar P=1200 Cement-perlite mortar P=1000 Cement-perlite mortar P=800 Gypsum-perlite mortar Porous gypsum-perlite mortar P=500 Porous gypsum-perlite mortar P=400 Gypsum slabs P=1200 Gypsum slabs P=1000 Gypsum sheathing sheets (dry plaster) Ordinary clay brick silicate brick P=2000 Sand-lime brick P=1900 Sand-lime brick P=1800 Sand-lime brick P=1700 Sand-lime brick P=1600 ceramic brick Р=1600 Ceramic brick Р=1400 Ceramic stone Р=1700 Thickened silicate brick Р=1600 Thickened silicate brick Р=1400 Silicate stone Р=1400 Silicate stone Р=1300 Granite. gneiss and basalt Marble Limestone Р=2000 Limestone Р=1800 Limestone Р=1600 Limestone Р=1400 Tuff Р=2000 Tuff Р=1800 Tuff Р=1600 Tuff Р=1400 Tuff Р=1200 Tuff Р=1000 Pine and spruce across the grain Pine and spruce along the fibers Oak across the fibers Oak along the fibers Glued plywood Facing cardboard Building multi-layer cardboard Wood fiber boards. and wood chip., skopodrevovovolok. Р=1000 Wood fiber boards. and wood chip., skopodrevovovolok. Р=800 Wood fiber boards. and wood chip., skopodrevovovolok. Р=400 Wood fiber boards. and wood chip., skopodrevovovolok. P=200 Fibrolite slabs and arbolite on Portland cement P=800 Fibrolite slabs and arbolite on Portland cement P=600 Fibrolite slabs and arbolite on Portland cement P=400 Fibrolite slabs and arbolite on Portland cement P=300 Fibrous heat-insulating slabs from faux fur waste P=175 Plates Fibrous heat-insulating boards from artificial fur waste P=150 Fibrous heat-insulating boards from artificial fur waste P=125 Flax-boned insulating boards Peat heat-insulating boards P=300 Heat-insulating peat boards P=200 Tow Mineral wool mats stitched P=125 Mineral wool stitched mats P=100 Mineral wool mats stitched nye R=75 Mineral wool pierced mats R=50 Mineral wool slabs on a synthetic binder R=250 Mineral wool slabs on a synthetic binder R=200 Mineral wool slabs on a synthetic binder R=175 Mineral wool slabs on a synthetic binder R=125 Mineral wool slabs on a synthetic binder R=75 Plates Polystyrene foam p = 50 plate polystyrene foam p = 35 plate polystyrene foam p = 25 slabs Polistyle polystyrene p = 15 polyurethane foam P = 80 Popenoliuretan P = 60 Popeniopoliac P = 40 stoves from resollain -formaldehydine foam p = 100 slabs from resolve -phynoldehydal foam p = 75 plate of stoves from rezin phenol -formaldehyde foam p =50 Slabs of resole-phenol-formaldehyde foam P=40 Polystyrene-concrete heat-insulating slabs P=300 Polystyrene-concrete heat-insulating slabs P=260 Polystyrene-concrete heat-insulating slabs P=230 Expanded clay gravel P=800 Expanded clay gravel P=600 Expanded clay gravel P=400 Expanded clay gravel new Р=300 Expanded clay gravel Р =200 Crushed stone and sand from expanded perlite P=600 Crushed stone and sand from expanded perlite P=400 Crushed stone and sand from expanded perlite P=200 Sand for construction works Foam glass and gas glass Р=200 Foam glass and gas glass Р=180 Foam glass and gas glass Р=160 Flat asbestos-cement sheets Р=1800 Flat asbestos-cement sheets Р=1600 Petroleum construction and roofing bitumen Р=1400 Petroleum construction and roofing bitumen Р=1200 Petroleum construction bitumen and roofing P=1000 Asphalt concrete Products from expanded perlite on a bituminous binder P=400 Products from expanded perlite on a bitumen binder P=300 Roofing material. glassine. roofing Polyvinylchloride multilayer linoleum R=1800 Polyvinylchloride multilayer linoleum R=1600 Polyvinylchloride linoleum on a fabric base R=1800 Polyvinylchloride linoleum on a fabric base P=1600 Polyvinylchloride linoleum on a fabric base P=1400 Rebar steel Cast iron Aluminum Copper Window glassThe exact calculation of heat loss at home is a painstaking and slow task. For its production, initial data are required, including the dimensions of all enclosing structures of the house (walls, doors, windows, ceilings, floors).
For single-layer and / or multi-layer walls, as well as floors, the heat transfer coefficient is easy to calculate by dividing the thermal conductivity of the material by the thickness of its layer in meters. For a multilayer structure, the overall heat transfer coefficient will be equal to the reciprocal of the sum of the heat resistances of all layers. For windows, you can use the table of thermal characteristics of windows.
Walls and floors lying on the ground are calculated by zones, so in the table it is necessary to create separate lines for each of them and indicate the corresponding heat transfer coefficient. The division into zones and the values of the coefficients are indicated in the rules for measuring premises.
Column 11. Basic heat loss. Here, the main heat losses are automatically calculated based on the data entered in the previous cells of the line. Specifically, Temperature Difference, Area, Heat Transfer Coefficient and Position Coefficient are used. Formula in cell:
Column 12. Orientation addition. In this column, the additive for orientation is automatically calculated. Depending on the contents of the Orientation cell, the appropriate coefficient is inserted. The formula for calculating a cell looks like this:
IF(H9="E",0.1,IF(H9="SE",0.05,IF(H9="S",0,IF(H9="SW",0,IF(H9="W ";0.05;IF(H9="SW";0.1;IF(H9="S";0.1;IF(H9="SW";0.1;0))))))) )
This formula inserts a factor into a cell as follows:
- East - 0.1
- Southeast - 0.05
- South - 0
- Southwest - 0
- West - 0.05
- Northwest - 0.1
- North - 0.1
- Northeast - 0.1
Column 13. Other additive. Here you enter the addition factor when calculating the floor or doors in accordance with the conditions in the table:
Column 14. Heat loss. Here is the final calculation of the heat loss of the fence according to the line. Cell formula:
As the calculations progress, you can create cells with formulas for summing heat losses by rooms and deriving the sum of heat losses from all the fences of the house.
There are also heat losses due to air infiltration. They can be neglected, since they are compensated to some extent by household heat emissions and heat gains from solar radiation. For a more complete, exhaustive calculation of heat loss, you can use the methodology described in the reference manual.
As a result, to calculate the power of the heating system, we increase the resulting amount of heat loss of all the fences of the house by 15 - 30%.
Others, more simple ways heat loss calculation:
- quick calculation in the mind approximate method of calculation;
- somewhat more complex calculation using coefficients;
- the most accurate way to calculate heat loss in real time;
