Simple calculation of heat losses of buildings. House insulation

I figured out the loss of overlap (floors on the ground without insulation) even STRONGLY a lot
with a thermal conductivity of concrete of 1.8, it turns out 61491 kWh season
I think the average temperature difference should not be taken as 4033 * 24, since the earth is still warmer than atmospheric air

For floors, the temperature difference will be less, the air outside is -20 degrees and the ground under the floors can be +10 degrees. That is, at a temperature in the house of 22 degrees, to calculate the heat loss in the walls, the temperature difference will be 42 degrees, and for the floors at the same time it will be only 12 degrees.

I also made such a calculation for myself last year in order to choose an economically justified thickness of insulation. But I made a more complex calculation. I found in the internet for my city statistics on temperatures for the previous year, and in increments of every four hours. Ie I consider that during four hours the temperature is constant. For each temperature, he determined how many hours a year this temperature had and calculated the losses for each temperature for the season, of course, divided into articles, walls, attic, floor, windows, ventilation. For the floor, I took a temperature difference of a constant 15 degrees, like (I have a basement). I did it all in an excel spreadsheet. I set the thickness of the insulation and immediately see the result.

My walls are silicate brick 38 cm. The house is two-story plus a basement, the area with a basement is 200 sq. m. The results are as follows:
Styrofoam 5 cm. Savings for the season will be 25919 rubles, a simple payback period (without inflation) is 12.8 years.
Styrofoam 10 cm. Savings for the season will be 30,017 rubles, a simple payback period (without inflation) is 12.1 years.
Styrofoam 15 cm. Savings per season will be 31,690 rubles, a simple payback period (without inflation) is 12.5 years.

Now let's think of a slightly different number. compare 10 cm and the payback to them of an additional 5 cm (up to 15)
So, additional savings at +5 cm is about 1,700 rubles per season. and the additional costs for warming are approximately 31,500 rubles, that is, these additional. 5 cm of insulation will pay off only after 19 years. It is not worth it, although before the calculations I was determined to make 15 cm in order to reduce the operating costs of gas, but now I see that the sheepskin skin is not worth the candle, add. saving 1700 rubles a year, it's not serious

For comparison, to the first five cm, we additionally add another 5 cm, then add. savings will be 4100 per year, add. costs 31500, payback 7.7 years, this is already normal. I will do 10 cm thinner, but I don’t want to, not seriously like that.

Yes, according to my calculations, I got the following results
brick wall 38 cm plus 10 cm foam.
energy saving windows.
Ceiling 20 cm min. wool (I didn’t count the boards, plus two films and an air gap of 5 cm, and there will also be an air gap between the ceiling and the finishing ceiling, the losses will mean even less, but so far I don’t take it into account), the floor of foam boards or whatever another 10 cm plus ventilation.

The total losses for the year are 41,245 kW. h, it is approximately 4,700 cubic meters of gas per year or so 17500 rub/ year (1460 rubles / month) It seems to me that it turned out okay. I also want to make a self-made heat exchanger for ventilation, otherwise I estimated 30-33% of all heat losses, these are ventilation losses, something needs to be decided with this., I don’t want to sit in a corked box.

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, which we pay for in the cold winter, during the 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.

Having asked the question of saving the 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.

The largest heat loss occurs through:

Heat loss through the roof and floors

As you know, warm air always rises to the top, so it heats the non-insulated 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. Firstly, the wall area is much larger than the roof area, and secondly, different materials have different thermal conductivity. Therefore, when building country houses, first of all, you need to take care of house wall insulation. For this, insulation for walls with a total thickness of 100 to 200 mm is suitable.

For proper insulation of the walls of the house, you must have knowledge of technology and a 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 sound insulation of walls and 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 to the large number of advantages over non-insulated houses, and allows you to save your family budget.

Having carried out the external and internal insulation of the house, your private house will become 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 the optimal choice of insulation for the 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

Not all materials that are used in construction are able to provide the proper level of heat saving for a private house. Through the walls, roof, floor, window openings there is a constant leakage of heat. Having determined with the help of a thermal imager which structural elements of the building act as “weak links”, it is possible to significantly reduce heat loss in a private house by means of complex or fragmentary insulation.

Insulation of windows at home is most often performed according to Swedish technology, for which all window sashes are removed from the frames, then a groove is selected along the perimeter of the frame with a cutter, into which a tubular sealant made of silicone (with a diameter of 2 to 7 mm) is filled - this allows you to reliably seal the window porches. Small gaps in the frames, the gaps between the double-glazed window and the frame are filled with sealant after preliminary washing, cleaning and drying the windows.

Window insulation can also be performed using a heat-saving film, which is fixed to the window frame with a self-adhesive strip. Letting light into the room, the film reliably shields heat flows due to metallized sputtering, returning about 60% of heat back into the room. Significant heat losses through windows are often associated with a violation of the geometry of the frame, gaps between the frame and slopes, sagging and skewed sashes, poor functioning of fittings - to eliminate these problems, qualified adjustment or repair of windows is required.

The most significant heat loss - about 40% - occurs through the walls of buildings, so thoughtful insulation of the main walls of a private house will drastically improve its heat-saving parameters. Wall insulation can be done from the inside or/and outside - the method of insulation depends on the material used in the construction of the house. Brick and foam concrete houses are most often insulated from the outside, but a heat insulator can also be laid from the inside of these buildings. Wooden houses are almost never insulated from the inside, in order to avoid the greenhouse effect in the rooms. Outside, houses are insulated from a bar, sometimes from a log house.

Insulation of the walls of the house can be performed using the technology of a "wet" or hinged facade - the main difference between these methods lies in the principle of mounting the facade cladding. When arranging a “wet” facade, a dense heat insulator (expanded polystyrene, polystyrene) is attached to the wall, and then a decorative finish is performed using adhesive mixtures. When installing a hinged facade, after installing a heater (mineral or glass wool), a crate is mounted, and then facing modules are fixed in its profiles. An obligatory element of the "pie" of the walls is a vapor barrier film, which removes condensate from the insulation layer, protects it from getting wet and prevents the loss of insulating properties.

The roof of the house is another surface through which heat constantly escapes from the house. Depending on the material used in the construction of the roof deck, the roof may be more or less warm. Capital insulation, as a rule, requires metal roofing from corrugated board and metal tiles. Roofs made of ondulin, flexible and ceramic tiles have low thermal conductivity, so for them the insulating "pie" can be thinner than in the case of metal. Similar to the technology for insulating other surfaces of the house, a vapor barrier must be included in the “pie” of the roof, and one or two ventilation gaps are provided for effective ventilation of the under-roof space.

Unlike walls and window openings, heat leakage through the floor of a private house is small - approximately 10%, and subject to the arrangement of insulation, it will be reduced to a minimum. The same polystyrene foam, polystyrene or mineral wool is used as insulation for floors, but it is also possible to use expanded clay, foamed concrete, cement-bonded mixtures and peat mats. An additional insulation measure in a country house can be the installation of underfloor heating: water, cable or infrared.

Similar to the device for insulating walls and roofs, a vapor barrier membrane acts as an obligatory component of the “pie” of the floor, which shields moisture-saturated steam that seeps out of the interior of the house. Thus, the heat-insulating layer is reliably protected from getting wet.

So that your house does not turn out to be a bottomless pit for heating costs, we suggest studying the basic directions of thermal engineering research and calculation methodology. Without a preliminary calculation of thermal permeability and moisture accumulation, the whole essence of housing construction is lost.

Physics of thermal processes

Different areas of physics have much in common in describing the phenomena they study. So it is in heat engineering: the principles describing thermodynamic systems clearly echo the foundations of electromagnetism, hydrodynamics and classical mechanics. After all, we are talking about the description of the same world, so it is not surprising that models of physical processes are characterized by some common features in many areas of research.

The essence of thermal phenomena is easy to understand. The temperature of a body or the degree of its heating is nothing but a measure of the intensity of oscillations of the elementary particles of which this body is composed. Obviously, when two particles collide, the one with the higher energy level will transfer energy to the particle with lower energy, but never vice versa. However, this is not the only way of energy exchange; transfer is also possible through thermal radiation quanta. At the same time, the basic principle is necessarily preserved: a quantum emitted by a less heated atom is not able to transfer energy to a hotter elementary particle. It is simply reflected from it and either disappears without a trace, or transfers its energy to another atom with less energy.

Thermodynamics is good because the processes occurring in it are absolutely clear and can be interpreted under the guise of various models. The main thing is to follow the basic postulates, such as the law of energy transfer and thermodynamic equilibrium. So if your presentation complies with these rules, you will easily understand the method of thermal engineering calculations from and to.

The concept of resistance to heat transfer

The ability of a material to transfer heat is called thermal conductivity. In the general case, it is always higher, the greater the density of the substance and the better its structure is adapted to transmit kinetic vibrations.

The quantity inversely proportional to thermal conductivity is thermal resistance. For each material, this property takes on unique values ​​depending on the structure, shape, and a number of other factors. For example, the efficiency of heat transfer in the thickness of materials and in the zone of their contact with other media may differ, especially if there is at least a minimal layer of matter between the materials in a different state of aggregation. Quantitatively, thermal resistance is expressed as the temperature difference divided by the heat flow rate:

R t \u003d (T 2 - T 1) / P

• R t - thermal resistance of the section, K / W;
• T 2 - temperature of the beginning of the section, K;
• T 1 - temperature of the end of the section, K;
• P is the heat flux, W.

In the context of calculating heat losses, thermal resistance plays a decisive role. Any enclosing structure can be represented as a plane-parallel barrier to the heat flow. Its total thermal resistance is the sum of the resistances of each layer, while all the partitions are folded into a spatial structure, which is, in fact, a building.

R t \u003d l / (λ S)

• R t - thermal resistance of the circuit section, K / W;
• l is the length of the section of the thermal circuit, m;
• λ is the thermal conductivity of the material, W/(m K);
• S is the cross-sectional area of ​​the site, m 2.

Factors affecting heat loss

Thermal processes correlate well with electrical ones: the temperature difference acts as a voltage, the heat flux can be considered as a current strength, but you don’t even need to come up with your own term for resistance. The concept of least resistance, which appears in heat engineering as cold bridges, is also fully true.

If we consider an arbitrary material in a section, it is quite easy to establish the path of the heat flow both at the micro- and at the macro level. As the first model, we will take a concrete wall, in which, due to technological necessity, through fastenings are made with steel rods of arbitrary section. Steel conducts heat somewhat better than concrete, so we can distinguish three main heat flows:

• through the concrete
• through steel bars
• from steel bars to concrete

The last heat flow model is the most interesting. Since the steel rod warms up faster, the temperature difference between the two materials will be observed closer to the outer part of the wall. Thus, steel not only "pumps" heat outward by itself, it also increases the thermal conductivity of adjacent masses of concrete.

In porous media, thermal processes proceed in a similar way. Almost all building materials consist of a branched web of solid matter, the space between which is filled with air. Thus, a solid, dense material serves as the main conductor of heat, but due to the complex structure, the path along which heat spreads turns out to be larger than the cross section. Thus, the second factor that determines thermal resistance is the heterogeneity of each layer and the building envelope as a whole.

The third factor affecting the thermal conductivity, we can name the accumulation of moisture in the pores. Water has a thermal resistance 20-25 times lower than that of air, so if it fills the pores, the overall thermal conductivity of the material becomes even higher than if there were no pores at all. When water freezes, the situation becomes even worse: thermal conductivity can increase up to 80 times. The source of moisture, as a rule, is room air and atmospheric precipitation. Accordingly, the three main methods of combating this phenomenon are external waterproofing of walls, the use of vapor protection and the calculation of moisture accumulation, which is necessarily carried out in parallel with predicting heat loss.

Differentiated calculation schemes

The simplest way to determine the amount of heat loss in a building is to sum up the values ​​of heat flow through the structures that form this building. This technique fully takes into account the difference in the structure of different materials, as well as the specifics of the heat flow through them and at the junctions of one plane to another. Such a dichotomous approach greatly simplifies the task, because different enclosing structures can differ significantly in the design of thermal protection systems. Accordingly, in a separate study, it is easier to determine the amount of heat loss, because various calculation methods are provided for this:

• For walls, heat leakage is quantitatively equal to the total area multiplied by the ratio of the temperature difference to the thermal resistance. At the same time, the orientation of the walls to the cardinal points is necessarily taken into account to take into account their heating in the daytime, as well as the ventilation of building structures.
• For floors, the methodology is the same, but the presence of an attic space and its mode of operation are taken into account. Also, a value of 3-5 ° C higher is taken as room temperature, the calculated humidity is also increased by 5-10%.
• Heat losses through the floor are calculated zonal, describing the belts along the perimeter of the building. This is due to the fact that the temperature of the soil under the floor is higher near the center of the building compared to the foundation part.
• The heat flux through the glazing is determined by the nameplate data of the windows; the type of adjoining windows to the walls and the depth of the slopes must also be taken into account.

Q = S (ΔT / Rt)

• Q is heat loss, W;
• S - wall area, m 2;
• ΔT - temperature difference inside and outside the room, ° С;
• R t - resistance to heat transfer, m 2 ° C / W.

Calculation example

Before moving on to the demo, let's answer the last question: how to correctly calculate the integral thermal resistance of complex multilayer structures? This, of course, can be done manually, since not many types of load-bearing bases and insulation systems are used in modern construction. However, it is quite difficult to take into account the presence of decorative finishes, interior and facade plaster, as well as the influence of all transient processes and other factors, it is better to use automated calculations. One of the best online resources for such tasks is smartcalc.ru, which additionally plots the dew point shift depending on climatic conditions.

For example, let's take an arbitrary building, having studied the description of which the reader will be able to judge the set of initial data necessary for the calculation. There is a one-story house of a regular rectangular shape with dimensions of 8.5x10 m and a ceiling height of 3.1 m, located in the Leningrad region. The house has an uninsulated floor on the ground with boards on logs with an air gap, the floor height is 0.15 m higher than the ground planning mark on the site. The wall material is a slag monolith 42 cm thick with internal cement-lime plaster up to 30 mm thick and external slag-cement plaster of the "fur coat" type up to 50 mm thick. The total area of ​​glazing is 9.5 m 2 , the windows used are a double-glazed window in a heat-saving profile with an average thermal resistance of 0.32 m 2 °C / W. The ceiling is made on wooden beams: the bottom is plastered on shingles, filled with blast-furnace slag and covered with clay screed on top, above the ceiling there is a cold-type attic. The task of calculating heat loss is the formation of a thermal protection system for walls.

First of all, heat losses through the floor are determined. Since their share in the total heat outflow is the smallest, and also due to the large number of variables (density and type of soil, freezing depth, massiveness of the foundation, etc.), heat loss is calculated using a simplified method using the reduced heat transfer resistance. Along the perimeter of the building, starting from the line of contact with the ground, four zones are described - encircling strips 2 meters wide. For each of the zones, its own value of the reduced resistance to heat transfer is taken. In our case, there are three zones with an area of ​​74, 26 and 1 m 2. Do not be confused by the total area of ​​the zones, which is 16 m 2 more than the area of ​​the building, the reason for this is the double recalculation of the intersecting strips of the first zone in the corners, where heat losses are much higher compared to sections along the walls. Applying heat transfer resistance values ​​of 2.1, 4.3 and 8.6 m 2 °C / W for zones one through three, we determine the heat flow through each zone: 1.23, 0.21 and 0.05 kW respectively.

Walls

Using the terrain data, as well as the materials and thickness of the layers that form the walls, you need to fill in the appropriate fields on the smartcalc.ru service mentioned above. According to the results of the calculation, the resistance to heat transfer is equal to 1.13 m 2 ° C / W, and the heat flux through the wall is 18.48 W per square meter. With a total wall area (excluding glazing) of 105.2 m 2, the total heat loss through the walls is 1.95 kW / h. In this case, heat loss through the windows will be 1.05 kW.

Covering and roofing

The calculation of heat loss through the attic floor can also be performed in the online calculator by selecting the desired type of enclosing structures. As a result, the resistance of the overlap to heat transfer is 0.66 m 2 · ° С / W, and the heat loss is 31.6 W per square meter, that is, 2.7 kW from the entire area of ​​the building envelope.

The total total heat loss according to the calculations is 7.2 kWh. Given the rather low quality of building structures, this indicator is obviously much lower than the real one. In fact, such a calculation is idealized, it does not take into account special coefficients, blowing, the convection component of heat transfer, losses through ventilation and entrance doors. In fact, due to poor-quality installation of windows, lack of protection at the junction of the roof to the Mauerlat and poor waterproofing of the walls from the foundation, real heat losses can be 2 or even 3 times more than the calculated ones. Nevertheless, even basic thermal engineering studies help to determine whether the structures of the house under construction will comply with sanitary standards, at least in the first approximation.

Finally, we give one important recommendation: if you really want to get a complete understanding of the thermal physics of a particular building, you must use an understanding of the principles described in this overview and special literature. For example, a reference manual by Elena Malyavina “Heat Loss of a Building” can be a very good help in this matter, where the specifics of heat engineering processes are explained in great detail, links are given to the necessary regulatory documents, as well as examples of calculations and all the necessary background information.

The choice of thermal insulation, options for insulating walls, ceilings and other enclosing structures 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 are imposed on 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 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.

Table of required resistance to heat transfer of enclosing structures in houses in the Moscow region.

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, choosing 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 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.)

Table- Heat losses of windows of various designs at ΔT = 50 °C (T outside = -30 °C, T inside = 20 °C.)

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 losses in building reference books, heat losses of various types of walls, ceilings, etc. are given. for some values ​​of winter air temperature. In particular, different numbers are given for corner rooms (where the swirl of air flowing through the house affects) and non-corner rooms, 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.

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.

Consider an example of calculating the heat loss of two different rooms of the same 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.

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. 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.)

For an objective picture of the heat loss of the whole house, it is necessary to take into account

1. 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).
2. 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%.
3. 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.

1. 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 comfortable living in large areas. Energy prices around the world are constantly rising.
2. 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 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”.
3. 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.
4. 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.
5. Wall insulation from the outside is much more effective than internal insulation.
6. Do not endlessly insulate walls. The effectiveness of this approach to energy saving is not high.
7. Ventilation - these are the main reserves of energy saving.
8. By applying modern glazing systems (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 of the "ISOVER" type, in the presence of air exchange and ventilation systems in the premises.