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 the house, for which insulation can be used extra ecowool ... 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 the direct influence of 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 a house through the windows, it is necessary to calculate the total area of all previously agreed windows. The calculation will be carried out 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 a house through the windows, we will find out their characteristics. Let's take the following technical indicators as an example:
- Windows are made of three-chamber PVC profile
- The windows have a double-glazed unit (4-16-4-16-4, where 4 is the thickness of the glass, 16 is the distance between the glass units of each window).
Now you can proceed to further calculations and find out the thermal resistance of the installed windows. Thermal resistance of a two-chamber glass unit and a three-chamber profile of such a window design:
- R st-a = 0.4 m2 ∙ ° С / W - thermal resistance of a glass unit
- R profile = 0.6 m2 ∙ ° С / W - thermal resistance of a three-chamber profile
Most of the window - 90%, is a double-glazed unit and 10% - a PVC profile. We calculate the thermal resistance of the window using the formula:
R window = (R st-a ∙ 90 + R profile ∙ 10) / 100 = 0.42 m2 ∙ ° С / W.
Having data on the area of the windows and their thermal resistance, we calculate the heat loss through the windows:
Q windows = S ∙ dT ∙ / R = 3.1 m² ∙ 52 degrees / 0.42 m² ∙ ° С / W = 383.8 W (0.38 kW), we get the heat loss at home through the windows, now we will calculate the heat loss of the house through the front door.
In the energy saving program during the construction and operation of buildings, translucent barriers play an important role, since the modern 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 sashes (cold bridges, leaks), losses due to heat conduction of air and convective flows between glass, as well as heat loss through thermal radiation.
Currently, the following main methods of increasing the energy efficiency of translucent structures are used in Russia:
Transition from one- and two-chamber double-glazed windows to three or more chamber ones;
- the use of thermal film (heat absorbing glazing);
- filling double-glazed windows with inert gases.
In modern translucent structures of heat-shielding windows, one- or two-chamber double-glazed windows are used, and for the manufacture of window sashes and frames - wooden, aluminum, fiberglass, plastic (PVC) profiles or their combinations. In the manufacture of double-glazed windows using float glass, the windows provide the calculated reduced resistance to heat transfer no more than 0.56 m 2 ∙ ºС / W and more.
Another way to increase the energy efficiency of translucent structures is heat-absorbing glazing. The heat transmission capacity of the 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 glass units with inert gases. At the same time, convection currents inside the glass unit decrease, which leads to a decrease in heat loss.
To add a description of the energy-saving technology to the Catalog, fill out the questionnaire and send it to marked "to the Catalog".
The choice of thermal insulation, insulation options for walls, ceilings and other enclosing structures is a difficult task for most building customers. Too many conflicting problems need to be solved at the same time. This page will help you figure it out.
Nowadays, heat conservation of energy resources has acquired great importance. According to SNiP 23-02-2003 "Thermal protection of buildings", the resistance to heat transfer is determined by one of two alternative approaches:
prescriptive (regulatory requirements apply to individual elements of building thermal protection: external walls, floors over unheated spaces, coatings and attic ceilings, windows, entrance doors, etc.)
consumer (the resistance to heat transfer of the fence can be reduced in relation to the prescriptive level, provided that the design specific heat consumption for heating the building is lower than the standard).
Sanitary and hygienic requirements must always be met.
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 permissible differential values for the outer wall are 4 ° C, for the covering and attic ceiling 3 ° C and for the ceiling above basements and underground 2 ° C.
The requirement that the temperature on the inner surface of the fence 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 ° С · m. sq / W, and according to the prescriptive approach:
for a house of permanent residence 3.13 ° C · m. sq. / W,
for administrative and other public buildings incl. buildings with seasonal residence 2.55 ° С · m. sq. / W.
Table of thicknesses and thermal resistance of materials for the conditions of Moscow and its region.
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Wall material name |
Wall thickness and corresponding thermal resistance |
The required thickness according to the consumer approach (R = 1.97 ° С · m2 / W) and according to the prescriptive approach (R = 3.13 ° С · m2 / W) |
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Solid solid clay brick (density 1600 kg / m3) |
510 mm (laying in two bricks), R = 0.73 ° С m. sq. / W |
1380 mm 2190 mm |
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Expanded clay concrete (density 1200 kg / m3) |
300 mm, R = 0.58 ° С m. sq. / W |
1025 mm 1630 mm |
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Wooden beams |
150 mm, R = 0.83 ° С m. sq. / W |
355 mm 565 mm |
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Wooden board filled with mineral wool (thickness of the inner and outer sheathing from boards of 25 mm) |
150 mm, R = 1.84 ° C m. sq. / W |
160 mm 235 mm |
Table of required resistances to heat transfer of enclosing structures in houses of the Moscow region.
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Outer wall |
Window, balcony door |
Covering and slabs |
Attic slab and ceilings over unheated basements |
Front door |
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By prescriptive approach |
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According to the consumer approach |
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These tables show that the majority of suburban housing in the Moscow region does not meet the requirements for heat conservation, while even the consumer approach is not followed in many newly constructed buildings.
Therefore, choosing a boiler or heating devices only according to the ability to heat a certain area indicated in their documentation, you claim that your house was built in strict accordance with the requirements of SNiP 23-02-2003.
The conclusion follows from the above material. For the correct choice of the power of the boiler and heating devices, it is necessary to calculate the real 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 be due to ventilation.
Heat loss mainly depends on:
temperature differences in the house and outside (the greater the difference, the higher the losses),
heat-shielding properties of walls, windows, floors, coatings (or, as they say, enclosing structures).
The enclosing structures resist heat leakage, therefore their heat-shielding properties are assessed by a value called heat transfer resistance.
Heat transfer resistance shows how much heat will go through a square meter of the enclosing structure 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 lost per square meter of the enclosing surface. It is measured in watts per square meter (W / m. Sq.); ΔT is the difference between the temperature outside and in the room (° С) and, R is the heat transfer resistance (° С / W / m2 or ° С · m2 / W).
When it comes to multi-layer construction, the resistance of the layers just adds up. For example, the resistance of a wall made of wood lined with brick is the sum of three resistances: a brick and a wooden wall and an air gap between them:
R (sum) = R (wood) + R (carriage) + R (brick).
Temperature distribution and boundary layers of air during heat transfer through the wall
Calculation for heat loss is carried out for the most unfavorable period, which is the most frosty and windy week of the year.
In building directories, as a rule, they indicate the thermal resistance of materials based on this condition and the climatic region (or outside temperature) where your house is located.
table - Resistance to heat transfer of various materials at ΔT = 50 ° C (T bunk bed = -30 ° C, T int. = 20 ° C.)
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Wall material and thickness |
Heat transfer resistanceR m , |
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Brick wall 3 bricks (79 cm) thick 2.5 bricks (67 cm) thick 2 bricks (54 cm) thick 1 brick (25 cm) thick |
0,592 0,502 0,405 0,187 |
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Log cabin Ø 25 Ø 20 |
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Log house 20 cm thick 10 cm thick |
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Frame wall (board + mineral wool + board) 20 cm |
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Foam concrete wall 20 cm 30 cm |
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Plaster on brick, concrete, foam concrete (2-3 cm) |
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Ceiling (attic) overlap |
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Wood floors |
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Double wooden doors |
table - Heat losses of windows of various designs at ΔT = 50 ° C (T bunk bed = -30 ° C, T int. = 20 ° C.)
Note Even numbers in the symbol of a double-glazed unit mean the air gap in mm; Ar symbol means that the gap is filled not with air, but with argon; Letter K means that the outer glass has a special transparent heat-shielding coating. |
As can be seen from the previous table, modern double-glazed windows can reduce the heat loss of a window by almost half. For example, for ten windows measuring 1.0 mx 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 the enclosing structures, we will apply this information to a specific example.
In calculating heat losses per sq. meter, two quantities are involved:
temperature difference ΔT,
heat transfer resistance R.
The temperature in the room is determined at 20 ° С, and the outside temperature is assumed to be –30 ° С. Then the temperature difference ΔT will be equal to 50 ° С. The walls are made of 20 cm thick timber, then R = 0.806 ° С · m. sq. / W.
Heat losses will be 50 / 0.806 = 62 (W / m2).
To simplify the calculations of heat losses in construction reference books, heat losses of various types of walls, ceilings, etc. are given. for some values of winter air temperature. In particular, different figures are given for corner rooms (where the swirl of the air swelling the house affects) and non-corner rooms, and different thermal patterns for rooms on the first and upper floors are taken into account.
table - Specific heat loss of building fencing elements (per 1 sq. M. Along the inner contour of the walls) depending on the average temperature of the coldest week of the year.
Note If there is an outside unheated room behind the wall (a canopy, a glazed veranda, etc.), then the heat loss through it is 70% of the calculated, and if behind this unheated room there is not a street, but another room outside (for example, a canopy overlooking on the veranda), then 40% of the calculated value. |
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table - Specific heat loss of building fencing elements (per 1 sq. M. Along the inner contour) depending on the average temperature of the coldest week of the year.
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Fence characteristic |
Outside temperature, ° С |
Heat loss, kW |
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Double glazed window |
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Solid wood doors (double) |
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Attic floor |
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Wooden floors above the basement |
Let's consider an example of calculating heat losses of two different rooms of the same area using tables.
Example 1.
Corner room (ground floor)
Room characteristics:
first floor,
room area - 16 sq.m. (5x3.2),
ceiling height - 2.75 m,
external walls - two,
material and thickness of external 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,
required temperature in the room +20 ° С.
Exterior wall area minus windows:
S walls (5 + 3.2) x2.7-2x1.0x1.6 = 18.94 sq. m.
Window area:
S windows = 2x1.0x1.6 = 3.2 sq. m.
Floor area:
S floor = 5x3.2 = 16 sq. m.
Ceiling area:
S ceiling = 5x3.2 = 16 sq. m.
The area of the internal partitions is not involved 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 let's calculate the heat loss for each of the surfaces:
Q total = 3094 W.
Note that more heat escapes through the walls than through windows, floors and ceilings.
The result of the calculation shows the heat loss of the room on the coldest (T out. = –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), pediments (10 cm thick timber sheathed with clapboard) and side partitions (frame wall with expanded clay filling 10 cm),
windows - four (two on each pediment), 1.6 m high and 1.0 m wide with double glazing,
design outside temperature -30 ° С,
required temperature in the room + 20 ° С.
Let's calculate the area of the heat-transfer surfaces.
The area of the end external walls minus the windows:
S end walls = 2x (2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 sq. m.
The area of the roof slopes bounding the room:
S sloped walls = 2x1.0x4.2 = 8.4 sq. m.
Side partitions area:
S side overburden = 2x1.5x4.2 = 12.6 sq. m.
Window area:
S windows = 4x1.6x1.0 = 6.4 sq. m.
Ceiling area:
S ceiling = 2.6x4.2 = 10.92 sq. m.
Now we will 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 calculate heat loss for walls and ceiling as for corner rooms, and for the ceiling and side partitions we enter a 70% coefficient, since unheated rooms are located behind them.
The total heat loss of the room will be:
Q total = 4504 W.
As you can see, a warm room on the ground floor loses (or consumes) significantly less heat than an attic room with thin walls and a large glazing area.
To make such a room suitable for winter living, you first need to insulate the walls, side partitions and windows.
Any enclosing structure can be represented as a multi-layer wall, each layer of which has its own thermal resistance and its own resistance to air passage. Adding the thermal resistance of all layers, we obtain 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 bunk bed = –20 ° С, Т int. = 20 ° C.)
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Wall layer |
Wall layer thickness (cm) |
Heat transfer resistance of the wall layer |
Resistance air permeability is equivalent to a timber wall thickness (cm) |
|
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Equivalent to masonry thickness (cm) |
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Ordinary clay brick masonry thickness: 12 cm 25 cm 50 cm 75 cm |
0,15 0,3 0,65 1,0 |
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Masonry made of lightweight aggregate concrete blocks 39 cm thick with a density: 1000 kg / cubic meter 1400 kg / cubic meter 1800 kg / cubic meter |
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Foam-aerated concrete with a thickness of 30 cm with a density: 300 kg / cubic meter 500 kg / cubic meter 800 kg / cubic meter |
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Brushed wall thick (pine) 10 cm 15 cm 20 cm |
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For an objective picture of the heat loss of the whole house, it is necessary to take into account
Heat losses through the contact of the foundation with frozen soil usually take 15% of the heat losses through the walls of the first floor (taking into account the complexity of the calculation).
Heat loss due to ventilation. These losses are calculated taking into account building codes (SNiP). A residential building requires about one air change per hour, 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 enclosing structures. It turns out that the heat loss through the walls and glazing is only 40%, and the heat loss for ventilation is 50%. In European ventilation and wall insulation standards, the ratio of heat losses is 30% and 60%.
If the wall "breathes" like a wall made of timber or logs 15 - 20 cm thick, then heat is returned. This allows you to reduce heat losses by 30%, therefore, the value of the thermal resistance of the wall obtained in the calculation should be multiplied by 1.3 (or, accordingly, reduce the heat loss).
Summing up all the heat losses at home, you will determine what power the heat generator (boiler) and heating devices are needed to comfortably heat 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 additional insulation.
It is possible to calculate the heat consumption according to the enlarged indicators. So, in one- and two-story, not very insulated houses at an external temperature of -25 ° C, 213 W is required per one square meter of the total area, and at -30 ° C - 230 W. For well-insulated houses it is: at –25 ° С - 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, however, during the operation of the building, the main costs are spent on heating. In no case should you save on thermal insulation, especially when living comfortably in large areas. Energy prices are constantly increasing all over the world.
Modern building materials have a higher thermal resistance than traditional materials. This allows the walls to be made 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 all the time - the walls quickly heat up and cool down quickly. It is cool in old houses with thick walls on a hot summer day, the walls that cooled down during the night “accumulated cold”.
Insulation must be considered in conjunction with the air permeability of the walls. If the increase in the thermal resistance of the walls is associated with a significant decrease in air permeability, then it should not be used. An ideal wall in terms of air permeability is equivalent to a wall made of timber with a thickness of 15 ... 20 cm.
Very often, improper use of vapor barrier leads to a deterioration in the sanitary and hygienic properties of housing. With properly organized ventilation and "breathing" walls, it is superfluous, and with poorly air-permeable walls, it is unnecessary. Its main purpose is to prevent wall infiltration and to protect the insulation from the wind.
Insulation of walls from the outside is much more effective than internal insulation.
You should not endlessly insulate the walls. The effectiveness of this approach to energy conservation is not high.
Ventilation is the main reserves of energy saving.
Using modern glazing systems (double-glazed windows, heat-insulating glass, etc.), low-temperature heating systems, effective thermal insulation of enclosing structures, you can reduce heating costs by 3 times.
Options for additional thermal insulation of building structures based on construction thermal insulation of the "ISOVER" type, in the presence of air exchange and ventilation systems in the premises.
Thermal insulation of tiled roofs using ISOVER thermal insulation
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Wall insulation made of lightweight concrete blocks |
Thermal insulation of a brick wall with a ventilated gap |
Insulation of a log wall |
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The first step in organizing heating of a private house is the calculation of heat loss. The purpose of this calculation is to find out how much heat goes out through walls, floors, roofs and windows (the common name is enclosing structures) during the most severe frosts in a given area. Knowing how to calculate heat loss according to the rules, you can get a fairly accurate result and start selecting a heat source by power.
Basic formulas
To get a more or less accurate result, it is necessary to perform calculations according to all the rules, a simplified method (100 W of heat per 1 m2 of area) will not work here. The total loss of heat by the building during the cold season consists of 2 parts:
- heat loss through enclosing structures;
- energy losses used to heat the ventilation air.
The basic formula for calculating the thermal energy consumption through outdoor fences is as follows:
Q = 1 / R x (t in - t n) x S x (1+ ∑β). Here:
- Q is the amount of heat lost by a structure of one type, W;
- R - thermal resistance of the construction material, m² ° С / W;
- S is the area of the outer fence, m²;
- t in - internal air temperature, ° С;
- t n - the lowest ambient temperature, ° С;
- β - additional heat loss, depending on the orientation of the building.
The thermal resistance of the walls or roof of a building is determined based on the properties of the material from which they are made and the thickness of the structure. For this, the formula R = δ / λ is used, where:
- λ - reference value of the thermal conductivity of the wall material, W / (m ° C);
- δ is the thickness of the layer of this material, m.
If the wall is built of 2 materials (for example, a brick with mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. The outdoor temperature is selected both according to regulatory documents and personal observations, the indoor temperature is selected as needed. Additional heat losses are coefficients determined by the norms:
- When the wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
- If the structure is facing southeast or west, β = 0.05.
- β = 0 when the outside railing faces south or southwest.
Calculation order
To take into account all the heat leaving the house, it is necessary to calculate the heat loss of the room, each separately. For this, measurements are made of all fences adjacent to the environment: walls, windows, roofs, floors and doors.
An important point: measurements should be carried out on the outside, capturing the corners of the structure, otherwise the calculation of the heat loss of the house will give an underestimated heat consumption.
Windows and doors are measured by the opening they fill.
Based on the results of measurements, the area of each structure is calculated and substituted into the first formula (S, m²). The R value obtained by dividing the thickness of the fence by the thermal conductivity of the building material is also inserted there. In the case of new windows made of metal-plastic, the R value will be advised by the representative of the installer.
As an example, it is worth calculating the heat loss through the enclosing walls made of bricks 25 cm thick, with an area of 5 m² at an ambient temperature of -25 ° C. It is assumed that the temperature inside will be + 20 ° С, and the plane of the structure is facing north (β = 0.1). First, you need to take the thermal conductivity of a brick (λ) from the reference literature, it is equal to 0.44 W / (m ° C). Then, using the second formula, the heat transfer resistance of a brick wall of 0.25 m is calculated:
R = 0.25 / 0.44 = 0.57 m2 ° C / W
To determine the heat loss of a room with this wall, all the initial data must be substituted into the first formula:
Q = 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) = 434 W = 4.3 kW
If there is a window in the room, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same steps are repeated for floors, roof and front door. At the end, all the results are summed up, after which you can move on to the next room.
Heat metering for air heating
When calculating the heat loss of a building, it is important to take into account the amount of heat energy consumed by the heating system to heat the ventilation air. The share of this energy reaches 30% of the total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss at home through the heat capacity of the air using the popular formula from the physics course:
Q air = cm (t in - t n). In it:
- Q air - heat consumed by the heating system to warm up the supply air, W;
- t in and t n - the same as in the first formula, ° С;
- m is the mass flow rate of air entering the house from the outside, kg;
- с - heat capacity of the air mixture, equal to 0.28 W / (kg ° C).
Here, all values are known, except for the mass air flow rate for ventilation of premises. In order not to complicate the task for yourself, it is worth agreeing with the condition that the air environment is renewed throughout the house once an hour. Then the volumetric air flow can be easily calculated by adding the volumes of all rooms, and then you need to convert it to mass through the density. Since the density of the air mixture changes depending on its temperature, you need to take a suitable value from the table:
m = 500 x 1.422 = 711 kg / h
Heating such a mass of air by 45 ° C will require such an amount of heat:
Q air = 0.28 x 711 x 45 = 8957 W, which is approximately equal to 9 kW.
At the end of the calculations, the results of heat losses through the outer fences are summed up with ventilation heat losses, which gives the total heat load on the building's heating system.
The presented calculation methods can be simplified if the formulas are entered into the Excel program in the form of tables with data, this will significantly speed up the calculation.
Heating bills and hot ox bills make up a significant part of the splits in the dwelling and, to a certain extent, reflect the level of thermal energy consumption. In the past, energy was cheap. Now its price has increased and is unlikely to decrease in the foreseeable future. But you can reduce the cost of heating and hot water. This is done using thermomolerenesiacin. It will reduce heat leakage through house structures and increase the efficiency of heating and hot water systems. Of course, thermal modernization will require considerable financial costs, but if it is done correctly, then the costs will be reimbursed due to the funds saved on heating.
Where does the heat go?
Let's consider the main reasons for the high level of thermal energy consumption in private houses. The heat goes away:
☰ through ventilation. In modern houses of traditional designs, 30-40% of the heat is removed in this way;
☰ windows and doors. They usually account for up to 25% of the total heat loss at home.
☰ In some houses, the size of the windows is determined not by rational norms of natural lighting, but by the architectural fashion that has come to us from countries with a warmer climate;
☰ external walls. 15-20% of the heat escapes through the structure of the walls. The building codes of the past years did not require high thermal insulation capacity from the wall structure; moreover, they were often violated as it is;
☰ roof. Up to 15% of the heat leaves through it;
☰ floor on the ground. A common solution in houses without a basement, with insufficient thermal insulation, can lead to losses of 5-10% of heat;
☰ cold bridges, or thermal bridges. They cause the loss of about 5% of the heat.
Insulation of external walls
It consists in creating an additional layer of thermal insulation on the outside or inside of the outer wall of the house. At the same time, heat loss decreases, and the temperature of the inner surface of the step increases, which makes living in the house more comfortable and eliminates the cause of increased humidity and mold formation. After additional insulation, the thermal insulation properties of the wall are improved three to four times.
Insulation from the outside is much more convenient and effective, therefore it is used in the overwhelming majority of cases. It provides:
☰ uniformity of thermal insulation over the entire surface of the outer wall;
☰ an increase in the thermostaticity of the wall, that is, the latter becomes a heat accumulator. During the day from the sunlight, it heats up, and at night, cooling down, it gives off heat to the room;
☰ elimination of unevenness of the wall and creation of a new, more aesthetic facade of the house;
☰ performance of work without inconvenience for residents.
Insulation of a house from the inside is used only in exceptional cases, for example, in houses with richly decorated facades or when only some rooms are insulated.
Insulation of ceilings and roofs
Overlappings in an unheated attic are insulated by laying a layer of slabs, mats or bulk materials. If the attic is planned to be used, then a layer of boards or a cement screed is laid over the insulation. Installing an extra layer of insulation in an easily accessible attic is actually simple and inexpensive.
The situation with the so-called ventilated combined roof is more complicated, where there is a space of several tens of centimeters above the ceiling of the last floor, to which there is no direct access. Then a special insulation is blown into this space so that, having hardened, it forms a thick heat-insulating layer on the ceiling.
It is possible to insulate the combined roof (this is usually arranged above the attic floors) by laying an additional layer of thermal insulation on it and making a new roof covering. Ceilings over basements are most easily insulated by gluing or hanging the insulation with anchors and steel mesh. The insulation layer can be left open or covered with aluminum foil, wallpaper, plaster, etc.
Reducing heat loss through windows
There are several ways to reduce heat loss through the window "joinery".
Here are the SIMPLEST ones:
☰ reduce windows;
☰ notice shutters and blinds;
☰ change windows.
The most radical way to reduce heat loss is the latter. Windows with higher thermal insulation properties are installed instead of the old ones. The market offers various types of energy-saving trenches: wooden, plastic, aluminum, with two- and three-chamber double-glazed windows, with special low-emission glass. It is not cheap to change windows, but new ones are easier to care for (plastic windows do not need to be painted), their high density prevents dust from entering, sound and thermal insulation improves.
Some houses have too many windows, significantly more than are necessary for natural lighting of the premises. Therefore, you can reduce their area by filling some of the openings with wall material.
The lowest temperatures outside the house are usually removed at night, when there is no daylight. Therefore, heat loss can be reduced by using shutters or blinds.
Heating and hot water supply system
If the heat supply of the house is carried out with the help of a boiler room, which has been used for 10-15 years, then it requires thermal modernization. The biggest disadvantage of older boilers is their low performance. In addition, such coal-fired devices emit a lot of combustion products. Therefore, it is advisable to replace them with modern gas or liquid fuel boilers: they have more performance, and they pollute the air less.
You can also upgrade the heating system itself in the house. For this, they arrange thermal insulation on heating and hot water pipes that pass through unheated rooms. In addition, thermostatic valves are installed on all radiators. This allows you to set the required temperature and not heat non-residential premises. You can also arrange air heating or "warm floor". Modernization of the hot water network is the replacement of leaking pipelines and thermal insulation of new ones, optimization of the operation of the system that prepares hot water, and the inclusion of a circulation pump in it.
Ventilation system
To reduce heat loss through this system, you can install a recuperator - a device that allows you to use the heat of the air leaving the house. In addition, supply air heating can be used. The simplest devices that reduce heat loss through dense modern windows are ventilation pockets that supply air to the premises.
Unconventional energy sources
For heating the house, you can use renewable energy. For example, heat from burning wood, wood waste (sawdust) and straw. For this, special boilers are used. The cost of heating in this way is significantly lower than systems that run on traditional fuels.
To use solar heat for heating, solar collectors are used, located on the roof or on the wall of the house. For maximum efficiency of their work, the collectors should be placed on the southern slope of the roof with a slope of about 45 °. In our climatic conditions, collectors are usually combined with another source of heat, for example, a convection gas boiler or a solid fuel boiler.
For heating and hot water supply, you can use heat pumps that use the heat of the earth or groundwater. However, they require electricity to operate. The cost of heat generated by heat pumps is low, but the cost of the pump and the heating system is quite high. The annual heat demand for individual houses is 120-160 kWh / m2. It is easy to calculate that 24000-32000 kWh will be required to heat a dwelling with an area of 200 m2 throughout the year. By applying a number of technical measures, this value can be reduced by almost half.