Heat loss at home, calculation of heat loss. Reducing heat loss through windows by installing double and triple glazing Heat loss depends on proper installation

In the energy saving program during the construction and operation of buildings, translucent barriers play an important role, since the current level of their thermal protection is not inferior to the thermal protection of the building envelope (wall) structures (up to 40% of all building losses).

Heat losses through the window occur through several channels: losses through the window block and bindings (cold bridges, leaks), losses due to the thermal conductivity of air and convective flows between the panes, as well as heat losses through thermal radiation.

Currently, the following main methods are used in Russia to improve the energy efficiency of translucent structures:

Transition from one- and two-chamber double-glazed windows to three- and more-chamber ones;
- use of thermal film (heat-absorbing glazing);
- filling double-glazed windows with inert gases.

In modern translucent designs of heat-shielding windows, one- or two-chamber double-glazed windows are used, and wooden, aluminum, fiberglass, plastic (PVC) profiles or their combinations are used to make window sashes and frames. In the manufacture of double-glazed windows using float glass, the windows provide a calculated reduced heat transfer resistance of not more than 0.56 m 2 ∙ºС / W or more.

Another way to increase the energy efficiency of translucent structures is heat-absorbing glazing. The thermal conductivity of glazing depends on the angle of incidence of sunlight and the thickness of the glass. Heat-reflecting glasses are covered with metal or polymer films. The heat transmission coefficient of such glasses is 0.2÷0.6.

Another energy-efficient method is the method of filling double-glazed windows with inert gases. At the same time, convection currents inside the double-glazed window are reduced, which leads to a decrease in heat loss.

In order to add description of energy saving technology to the Catalog, fill out the questionnaire and send it to marked "to Catalog".

As practice shows, a very large proportion of the heat from the house escapes through the windows. Since many houses have plastic windows that practically reduce drafts and cooling of the premises due to the influx of cold air, this has an advantage over ordinary windows. And yet, plastic windows can lose heat from 20 to 40% of the total heat loss at home, let's figure out the reasons for this and how to prevent heat loss through windows.

Heat loss through double glazing

They are able to retain heat very well and this figure is higher, the thicker the double-glazed window. As practice shows, it is not so important for how many chambers your double-glazed window consists of. Two, or three cameras, or one - it's not so important. Heat leakage occurs through the entire glass area. This radiation lies in the infrared region of the spectrum.

Modern technologists cope with this task in the following way: the so-called energy-saving double-glazed windows have been invented. They differ from the usual ones in that a special layer of low-emission sputtering is applied to its glass. Thanks to this layer, heat is reflected back into the room. Thanks to this double-glazed window, it is possible to prevent heat leakage through the window by 50%. At the same time, the glass does not lose its transparency and aesthetic appearance at all. At the same time, solar radiation also does not penetrate such glass, which is very good for regions with a hot climate.

Double glazing will provide you with the required window thickness for better heat savings. And at the same time, it must be remembered that such a double-glazed window is noticeably heavier than usual, which can lead to sagging of the wings over time. Among other things, it has been noticed that from street noise such a double-glazed window can begin to emit low-frequency sounds. This is due to the fact that a standing sound wave can occur between the glasses, which can contribute to the occurrence of resonance and the appearance of a characteristic bounce.

In some double-glazed windows, neutral gas is pumped instead of air. However, after two or three years there is no trace of this advantage, since this gas escapes and is replaced by ordinary air.

Another unpleasant moment is the freezing of windows in winter, as well as the appearance of ice on the double-glazed window. Most often, this is an indicator that the window sealant has become unusable. This happens because of its destruction. In order for the foam sealant not to collapse, it must be covered with moisture-proof mastic during installation.

Also check the tightness of the protective sealing rubber window. In order for the rubber to retain its insulating function, it must be lubricated at least twice with a special lubricant from the plastic window maintenance kit. You will be surprised how much dirt can accumulate on rubber in six months, when you finally decide to wash it with detergent. If this is not done, the rubber will crack and lose its elasticity. Silicone grease will help extend the life of the plastic window sealing rubber. If, nevertheless, the rubber has lost its qualities and is not able to perform its functions, replace it.

To date heat saving is an important parameter that is taken into account when constructing a residential or office space. In accordance with SNiP 23-02-2003 "Thermal protection of buildings", heat transfer resistance is calculated using one of two alternative approaches:

• prescriptive;
• Consumer.

To calculate home heating systems, you can use the calculator for calculating heating, 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.

For example: for Moscow and the Moscow region, the required thermal resistance of the wall according to the consumer approach is 1.97 ° С m 2 / W, and according to the prescriptive approach:

• for a permanent home 3.13 ° С m 2 / W.
• for administrative and other public buildings, including structures for seasonal residence 2.55 ° С m 2 / W.

For this reason, choosing a boiler or other heating devices solely according to the parameters indicated in their technical documentation. You should ask yourself if your house was built with strict adherence to the requirements of SNiP 23-02-2003.

Therefore, for the correct choice of the power of the heating boiler or heating devices, it is necessary to calculate the real heat loss in your home. As a rule, a residential building loses heat through the walls, roof, windows, ground, as well as significant heat losses can occur through ventilation.

Heat loss mainly depends on:

• temperature difference in the house and on the street (the higher the difference, the higher the loss).
• heat-shielding characteristics of walls, windows, ceilings, coatings.

Walls, windows, floors, have a certain resistance to heat leakage, the heat-shielding properties of materials are evaluated by a value called heat transfer resistance.

Heat transfer resistance will show how much heat will seep through 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.)

Table of heat losses of windows of various designs at ΔT = 50 °C (T out = -30 °C. T int. = 20 °C.)

Note
. Even numbers in 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 various types of walls, ceilings, etc. for some values ​​of winter air temperature. As a rule, different figures are given for corner rooms(the swirl of air flowing through the house affects it) and non-angular, and also takes into account the difference in temperatures for the premises of the first and upper floors.

Table of specific heat losses of building fencing elements (per 1 m 2 along the inner contour of the walls) depending on the average temperature of the coldest week of the year.

Note. In the case when there is an outdoor unheated room behind the wall (canopy, glazed porch, 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 °C.

• 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.

The total heat loss of the room will be: Q total \u003d 3094 W.

It should be borne in mind that much more heat escapes through walls than through windows, floors and ceilings.

Example 2

Roof room (attic)

Room characteristics:

• upper floor.
• area 16 m 2 (3.8x4.2).
• ceiling height 2.4 m.
• exterior walls; two roof slopes (slate, 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 taking into account that in this case, heat will not escape through the floor, since a warm room is located below. 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.

In order to make this room 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 best timber wall should be equivalent to a timber wall 15 to 20 inches thick. 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.)

For a complete picture of the heat loss of the entire room, it is necessary to take into account

1. 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).
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 will be slightly less than the sum of heat losses attributable to the building envelope. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation fifty%. In European standards for ventilation and wall insulation, the ratio of heat loss is 30% and 60%.
3. 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.

Let's use a simple example to analyze the option of calculating the heat loss of a house through the windows and the front door of a house, for which insulation can be used ecowool extra . For the calculation, we take two windows on different walls of the house measuring 100x120 cm (1x1.2 m), another smaller window which is 60x120 cm (0.6x1.2 m).

To calculate the heat loss of a house through the front door, we take the following door parameters 80x120x5 cm (door width - 0.8 m, door height - 2 m, door leaf thickness - 0.05 m). the structure of the door leaf is solid pine. The door from the street side is protected from direct exposure to atmospheric phenomena by an unheated terrace, therefore, according to the rules for calculating heat losses, it is necessary to apply a reduction factor equal to 0.7.

Calculation of heat loss through windows

To start calculating the heat loss of the house through the windows, it is necessary to calculate the total area of ​​​​all previously agreed windows. We will calculate according to the formula:

S windows = 1 ∙ 1.2 ∙ 2 + 0.6 ∙ 1.2 = 3.12 m2

Now, to continue calculating the heat loss of the house through the windows, we find out their characteristics. For example, take the following technical indicators:

• Windows made of three-chamber PVC profile
• The windows have a double-glazed window (4-16-4-16-4, where 4 is the thickness of the glass, 16 is the distance between the glass panes of each window).

Now you can proceed to further calculations and find out the thermal resistance of the installed windows. Thermal resistance of a double-glazed window and a three-chamber profile of such a window design:

• R st-a \u003d 0.4 m² ∙ ° С / W - thermal resistance of a double-glazed window
• R profile \u003d 0.6 m² ∙ ° С / W - thermal resistance of a three-chamber profile

Most of the window - 90%, is occupied by a double-glazed window and 10% - by a PVC profile. The thermal resistance of the window is calculated by the formula:

R windows \u003d (R st-a ∙ 90 + R profile ∙ 10) / 100 \u003d 0.42 m² ∙ ° C / W.

Having data on the area of ​​windows and their thermal resistance, we calculate the heat loss through the windows:

Q windows \u003d S ∙ dT ∙ / R \u003d 3.1 m² ∙ 52 degrees / 0.42 m² ∙ ° С / W = 383.8 W (0.38 kW), this is you and I get the heat loss of the house through the windows, now calculate the heat loss of the house through the front door.

Each building, regardless of design features, passes thermal energy through the fences. Heat loss to the environment must be restored using the heating system. The sum of heat losses with a normalized margin is the required power of the heat source that heats the house. In order to create comfortable conditions in a dwelling, heat loss is calculated taking into account various factors: building design and layout of premises, orientation to the cardinal points, wind direction and average mildness of the climate during the cold period, physical qualities of building and heat-insulating materials.

Based on the results of the heat engineering calculation, a heating boiler is selected, the number of battery sections is specified, the power and length of the underfloor heating pipes are considered, a heat generator is selected for the room - in general, any unit that compensates for heat loss. By and large, it is necessary to determine heat losses in order to heat the house economically - without an extra supply of power from the heating system. Calculations are performed manually or a suitable computer program is selected into which data are substituted.

How to make a calculation?

First, you should deal with the manual technique - to understand the essence of the process. To find out how much heat a house loses, determine the losses through each building envelope separately, and then add them up. The calculation is carried out in stages.

1. Form a base of initial data for each room, preferably in the form of a table. In the first column, the pre-calculated area of ​​door and window blocks, external walls, ceilings, and floors is recorded. The thickness of the structure is entered in the second column (these are design data or measurement results). In the third - the coefficients of thermal conductivity of the corresponding materials. Table 1 contains the normative values ​​that will be needed in the further calculation:

The higher λ, the more heat escapes through the meter thickness of the given surface.

2. The heat resistance of each layer is determined: R = v/ λ, where v is the thickness of the building or heat-insulating material.

3. Calculate the heat loss of each structural element according to the formula: Q \u003d S * (T in -T n) / R, where:

• T n - outdoor temperature, ° C;
• T in - indoor temperature, ° C;
• S is the area, m2.

Of course, during the heating period, the weather varies (for example, the temperature ranges from 0 to -25°C), and the house is heated to the desired level of comfort (for example, up to +20°C). Then the difference (T in -T n) varies from 25 to 45.

To make a calculation, you need the average temperature difference for the entire heating season. To do this, in SNiP 23-01-99 "Construction climatology and geophysics" (table 1) find the average temperature of the heating period for a particular city. For example, for Moscow this figure is -26°. In this case, the average difference is 46°C. To determine the heat consumption through each structure, the heat losses of all its layers are added. So, for walls, plaster, masonry material, external thermal insulation, and cladding are taken into account.

4. Calculate the total heat loss, defining them as the sum of Q external walls, floors, doors, windows, ceilings.

5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the result of addition. If high-quality double-glazed windows are installed in the house, and ventilation is not abused, the infiltration coefficient can be taken as 0.1. In some sources, it is indicated that the building does not lose heat at all, since leakages are compensated for by solar radiation and domestic heat emissions.

Counting by hand

Initial data. A one-story house with an area of ​​8x10 m, a height of 2.5 m. The walls, 38 cm thick, are made of ceramic bricks, finished with a layer of plaster from the inside (thickness 20 mm). The floor is made of 30 mm edged board, insulated with mineral wool (50 mm), sheathed with chipboard sheets (8 mm). The building has a cellar, where the temperature in winter is 8°C. The ceiling is covered with wooden panels, insulated with mineral wool (thickness 150 mm). The house has 4 windows 1.2x1 m, an entrance oak door 0.9x2x0.05 m.

Task: determine the total heat loss of the house based on the fact that it is located in the Moscow region. The average temperature difference in the heating season is 46°C (as mentioned earlier). The room and basement have a difference in temperature: 20 – 8 = 12°C.

1. Heat loss through external walls.

Total area (excluding windows and doors): S \u003d (8 + 10) * 2 * 2.5 - 4 * 1.2 * 1 - 0.9 * 2 \u003d 83.4 m2.

The heat resistance of the brickwork and the plaster layer is determined:

• R clade. = 0.38/0.52 = 0.73 m2*°C/W.
• R pieces. = 0.02/0.35 = 0.06 m2*°C/W.
• R total = 0.73 + 0.06 = 0.79 m2*°C/W.
• Heat loss through walls: Q st \u003d 83.4 * 46 / 0.79 \u003d 4856.20 W.

2. Heat loss through the floor.

Total area: S = 8*10 = 80 m2.

The heat resistance of a three-layer floor is calculated.

• R boards = 0.03 / 0.14 = 0.21 m2 * ° C / W.
• R chipboard = 0.008/0.15 = 0.05 m2*°C/W.
• R insulation = 0.05/0.041 = 1.22 m2*°C/W.
• R total = 0.03 + 0.05 + 1.22 = 1.3 m2*°C/W.

We substitute the values ​​\u200b\u200bof the quantities into the formula for finding heat losses: Q floor \u003d 80 * 12 / 1.3 \u003d 738.46 W.

3. Heat loss through the ceiling.

The area of ​​the ceiling surface is equal to the area of ​​the floor S = 80 m2.

When determining the thermal resistance of the ceiling, in this case, wooden panels are not taken into account: they are fixed with gaps and are not a barrier to cold. The thermal resistance of the ceiling coincides with the corresponding parameter of the insulation: R pot. = R ins. = 0.15/0.041 = 3.766 m2*°C/W.

The amount of heat loss through the ceiling: Q sweat. \u003d 80 * 46 / 3.66 \u003d 1005.46 W.

4. Heat loss through windows.

Glazing area: S = 4*1.2*1 = 4.8 m2.

For the manufacture of windows, a three-chamber PVC profile was used (occupies 10% of the window area), as well as a two-chamber double-glazed window with a glass thickness of 4 mm and a distance between glasses of 16 mm. Among the technical characteristics, the manufacturer indicated the thermal resistance of a double-glazed window (R st.p. = 0.4 m2 * ° C / W) and a profile (R prof. = 0.6 m2 * ° C / W). Taking into account the dimensional fraction of each structural element, the average heat resistance of the window is determined:

• R ok. \u003d (R st.p. * 90 + R prof. * 10) / 100 \u003d (0.4 * 90 + 0.6 * 10) / 100 \u003d 0.42 m2 * ° C / W.
• Based on the calculated result, the heat losses through the windows are calculated: Q approx. \u003d 4.8 * 46 / 0.42 \u003d 525.71 W.

Door area S = 0.9 * 2 = 1.8 m2. Thermal resistance R dv. \u003d 0.05 / 0.14 \u003d 0.36 m2 * ° C / W, and Q ext. \u003d 1.8 * 46 / 0.36 \u003d 230 W.

The total amount of heat loss at home is: Q = 4856.20 W + 738.46 W + 1005.46 W + 525.71 W + 230 W = 7355.83 W. Taking into account infiltration (10%), the losses increase: 7355.83 * 1.1 = 8091.41 W.

To accurately calculate how much heat a building loses, use an online heat loss calculator. This is a computer program into which not only the data listed above are entered, but also various additional factors that affect the result. The advantage of the calculator is not only the accuracy of calculations, but also an extensive database of reference data.