Heat loss during the heating season. Calculation of heat loss at home according to standards

Accurately calculating heat loss at home is a painstaking and slow task. For its production, initial data is required, including the dimensions of all enclosing structures of the house (walls, doors, windows, ceilings, floors).

For single-layer and/or multi-layer walls, as well as floors, the heat transfer coefficient can be easily calculated by dividing the thermal conductivity coefficient of the material by the thickness of its layer in meters. For multilayer construction the overall heat transfer coefficient will be equal to the value, the reciprocal of the sum of the thermal resistances of all layers. For windows, you can use the table of thermal characteristics of windows.

Walls and floors lying on the ground are calculated by zone, so it is necessary to create separate rows in the table for each of them and indicate the corresponding heat transfer coefficient. The division into zones and the values ​​of the coefficients are indicated in the rules for measuring premises.

Box 11. Main heat losses. Here, the main heat losses are automatically calculated based on the data entered in the previous cells of the line. Specifically, Temperature Difference, Area, Heat Transfer Coefficient and Position Coefficient are used. Formula in cell:

Column 12. Additive for orientation. In this column, the additive for orientation is automatically calculated. Depending on the contents of the Orientation cell, the appropriate coefficient is inserted. The cell calculation formula looks like this:

IF(H9="E";0.1;IF(H9="SE";0.05;IF(H9="S";0;IF(H9="SW";0;IF(H9="W ";0.05;IF(H9="NW";0.1;IF(H9="N";0.1;IF(H9="NW";0.1;0))))))) )

This formula inserts a coefficient into a cell as follows:

• East - 0.1
• Southeast - 0.05
• South - 0
• Southwest - 0
• West - 0.05
• North-west - 0.1
• North - 0.1
• Northeast - 0.1

Box 13. Other additive. Here you enter the additive factor when calculating the floor or doors in accordance with the conditions in the table:

Box 14. Heat loss. Here is the final calculation of the heat loss of the fence based on the line data. Cell formula:

As the calculations progress, you can create cells with formulas for summing up heat loss by room and deriving the sum of heat loss from all the fences of the house.

There are also heat losses due to air infiltration. They can be neglected, since they are to some extent compensated by household heat emissions and heat gains from solar radiation. For a more complete, comprehensive calculation of heat loss, you can use the methodology described in the reference manual.

As a result, to calculate the power of the heating system, we increase the resulting amount of heat loss from all the fences of the house by 15 - 30%.

Other, simpler ways to calculate heat loss:

• quick mental calculation; approximate method of calculation;
• a slightly more complex calculation using coefficients;
• the most accurate way to calculate heat loss in real time;

Calculation of heat loss at home is the basis of the heating system. It is needed, at a minimum, to choose the right boiler. You can also estimate how much money will be spent on heating in the planned house and conduct an analysis financial efficiency insulation i.e. understand whether the costs of installing insulation will be recouped by fuel savings over the service life of the insulation. Very often, when choosing the power of a room’s heating system, people are guided by the average value of 100 W per 1 m 2 of area at standard height ceilings up to three meters. However, this power is not always sufficient to completely replenish heat loss. Buildings differ in the composition of building materials, their volume, location in different climatic zones etc. To correctly calculate thermal insulation and select the power of heating systems, you need to know about the real heat loss of the house. We will tell you how to calculate them in this article.

Basic parameters for calculating heat loss

Heat loss in any room depends on three basic parameters:

• volume of the room - we are interested in the volume of air that needs to be heated
• difference in temperature inside and outside the room - than more difference the faster heat exchange occurs and the air loses heat
• thermal conductivity of enclosing structures - the ability of walls and windows to retain heat

The simplest calculation of heat loss

Qt (kW/hour)=(100 W/m2 x S (m2) x K1 x K2 x K3 x K4 x K5 x K6 x K7)/1000

This formula calculation of heat loss using aggregated indicators, which are based on average conditions of 100 W per 1 square meter. Where the main calculation indicators for calculating the heating system are the following values:

Qt- thermal power proposed waste oil heater, kW/hour.

100 W/m2- specific value of heat loss (65-80 watt/m2). It includes leakage of thermal energy through its absorption by windows, walls, ceilings and floors; leaks through ventilation and room leaks and other leaks.

S- area of ​​the room;

K1- heat loss coefficient of windows:

• conventional glazing K1=1.27
• double glazing K1=1.0
• triple glazing K1=0.85;

K2- wall heat loss coefficient:

• poor thermal insulation K2=1.27
• wall of 2 bricks or insulation 150 mm thick K2=1.0
• good thermal insulation K2=0.854

K3 window to floor area ratio:

• 10% K3=0.8
• 20% K3=0.9
• 30% K3=1.0
• 40% K3=1.1
• 50% K3=1.2;

K4- outside temperature coefficient:

• -10oC K4=0.7
• -15oC K4=0.9
• -20oC K4=1.1
• -25oC K4=1.3
• -35oC K4=1.5;

K5- number of walls facing outside:

• one - K5=1.1
• two K5=1.2
• three K5=1.3
• four K5=1.4;

K6- type of room that is located above the calculated one:

• cold attic K6=1.0
• warm attic K6=0.9
• heated room K6-0.8;

K7- room height:

• 2.5 m K7=1.0
• 3.0 m K7=1.05
• 3.5 m K7=1.1
• 4.0 m K7=1.15
• 4.5 m K7=1.2.

Simplified calculation of heat loss at home

Qt = (V x ∆t x k)/860; (kW)

V- room volume (cub.m)
∆t- temperature delta (outdoor and indoor)
k- dissipation coefficient

• k= 3.0-4.0 – without thermal insulation. (Simplified wooden structure or corrugated sheet metal construction).
• k= 2.0-2.9 – low thermal insulation. (Simplified building structure, single brickwork, simplified window and roof structure).
• k= 1.0-1.9 – average thermal insulation. (Standard construction, double brickwork, few windows, standard roof).
• k= 0.6-0.9 – high thermal insulation. (Improved construction, double insulated brick walls, few double glazed windows, thick floor base, high quality insulated roof).

This formula very conditionally takes into account the dispersion coefficient and it is not entirely clear which coefficients to use. In the classics there is a rare modern one, made of modern materials taking into account current standards, the room has enclosing structures with a dispersion coefficient of more than one. For a more detailed understanding of the calculation methodology, we offer the following more accurate methods.

I would like to immediately draw your attention to the fact that enclosing structures are generally not homogeneous in structure, but usually consist of several layers. Example: shell wall = plaster + shell + exterior decoration. This design may also include closed air gaps (example: cavities inside bricks or blocks). The above materials have thermal characteristics that differ from each other. The main characteristic for a structural layer is its heat transfer resistance R.

q– this is the amount of heat that is lost per square meter of the enclosing surface (usually measured in W/sq.m.)

ΔT- the difference between the temperature inside the calculated premises and the external air temperature (the temperature of the coldest five-day period °C for the climatic region in which the calculated building is located).

Basically, the internal temperature in the premises is taken:

• Living quarters 22C
• Non-residential 18C
• Zones water procedures 33C

When it comes to a multilayer structure, the resistances of the layers of the structure add up. Separately, I would like to draw your attention to the calculated coefficient thermal conductivity of the layer material λ W/(m°C). Since material manufacturers most often indicate it. Having the calculated thermal conductivity coefficient of the construction layer material, we can easily obtain layer heat transfer resistance:

δ - layer thickness, m;

λ - calculated coefficient of thermal conductivity of the material of the construction layer, taking into account the operating conditions of the enclosing structures, W / (m2 oC).

So, to calculate heat losses through building envelopes, we need:

1. Heat transfer resistance of structures (if the structure is multilayer then Σ R layers)R
2. The difference between the temperature in the calculation room and outside (temperature of the coldest five-day period °C). ΔT
3. Fencing areas F (separate walls, windows, doors, ceiling, floor)
4. Orientation of the building in relation to the cardinal directions.

The formula for calculating heat loss by a fence looks like this:

Qlimit=(ΔT / Rolim)* Folim * n *(1+∑b)

Qlimit- heat loss through enclosing structures, W
Rogr– heat transfer resistance, m2°C/W; (If there are several layers then ∑ Rogr layers)
Folim– area of ​​the enclosing structure, m;
n– coefficient of contact between the enclosing structure and the outside air.

Type of enclosing structure	Coefficient n
1. External walls and coverings (including those ventilated by outside air), attic floors (with a roof made of piece materials) and over driveways; ceilings over cold (without enclosing walls) undergrounds in the Northern construction-climatic zone
2. Ceilings over cold basements communicating with outside air; attic floors (with a roof made of roll materials); ceilings above cold (with enclosing walls) undergrounds and cold floors in the Northern construction-climatic zone
3. Ceilings over unheated basements with light openings in the walls
4. Ceilings over unheated basements without light openings in the walls, located above ground level
5. Ceilings over unheated technical underground located below ground level

(1+∑b) – additional heat losses in fractions of the main losses. Additional heat losses b through the enclosing structures should be taken as a proportion of the main losses:

a) in premises of any purpose through external vertical and inclined (vertical projection) walls, doors and windows facing north, east, northeast and northwest - in the amount of 0.1, to the southeast and west - in the amount 0.05; in corner rooms additionally - 0.05 for each wall, door and window, if one of the fences faces the north, east, north-east and north-west and 0.1 - in other cases;

b) in rooms developed for standard design, through walls, doors and windows facing any of the cardinal directions, in the amount of 0.08 for one external wall and 0.13 for corner rooms (except residential), and in all residential premises - 0.13;

c) through unheated floors of the first floor above the cold undergrounds of buildings in areas with an estimated outside air temperature of minus 40 °C and below (parameters B) - in the amount of 0.05,

d) through external doors not equipped with air or air-heat curtains, with a building height of N, m, from the average level of the ground to the top of the cornice, the center of the exhaust openings of the lantern or the mouth of the shaft in the amount of: 0.2 N - for triple doors with two vestibules between them; 0.27 H - for double doors with vestibules between them; 0.34 H - for double doors without vestibule; 0.22 H - for single doors;

e) through external gates not equipped with air and air-thermal curtains - in size 3 if there is no vestibule and in size 1 - if there is a vestibule at the gate.

For summer and emergency external doors and gates, additional heat losses under subparagraphs “d” and “e” should not be taken into account.

Separately, let’s take such an element as a floor on the ground or on joists. There are some peculiarities here. A floor or wall that does not contain insulating layers made of materials with a thermal conductivity coefficient λ less than or equal to 1.2 W/(m °C) is called not insulated. The heat transfer resistance of such a floor is usually denoted Rn.p, (m2 oC) / W. For each zone of a non-insulated floor, standard heat transfer resistance values ​​are provided:

• zone I - RI = 2.1 (m2 oC) / W;
• zone II - RII = 4.3 (m2 oC) / W;
• zone III - RIII = 8.6 (m2 oC) / W;
• zone IV - RIV = 14.2 (m2 oC) / W;

The first three zones are strips located parallel to the perimeter of the external walls. The remaining area is classified as the fourth zone. The width of each zone is 2 m. The beginning of the first zone is where the floor adjoins the outer wall. If the non-insulated floor is adjacent to a wall buried in the ground, then the beginning is transferred to the upper boundary of the wall’s burial. If the structure of a floor located on the ground has insulating layers, it is called insulated, and its heat transfer resistance Rу.п, (m2 оС) / W, is determined by the formula:

Rу.п. = Rn.p. + Σ (γу.с. / λу.с.)

Rn.p- heat transfer resistance of the considered zone of the non-insulated floor, (m2 oC) / W;
γу.с- thickness of the insulating layer, m;
λу.с- thermal conductivity coefficient of the insulating layer material, W/(m °C).

For a floor on joists, the heat transfer resistance Rl, (m2 oC) / W, is calculated using the formula:

Rl = 1.18 * Rу.п

The heat loss of each enclosing structure is calculated separately. The amount of heat loss through the enclosing structures of the entire room will be the sum of heat losses through each enclosing structure of the room. It is important not to get confused in measurements. If instead of (W) (kW) appears, or even (kcal), you will get the wrong result. You can also inadvertently specify Kelvins (K) instead of degrees Celsius (°C).

Advanced calculation of heat loss at home

Heating in civil and residential buildings, heat loss of premises consists of heat loss through various enclosing structures, such as windows, walls, ceilings, floors, as well as heat consumption for heating air, which is infiltrated through leaks in the protective structures (enclosing structures) of a given room. IN industrial buildings There are other types of heat loss. Calculation of heat loss in a room is made for all enclosing structures of all heated rooms. Heat loss through internal structures may not be taken into account when the temperature difference in them with the temperature of neighboring rooms is up to 3C. Heat loss through the building envelope is calculated using the following formula, W:

Qlimit = F (tin – tnB) (1 + Σ β) n / Rо

tnB– outside air temperature, °C;
tvn– room temperature, °C;
F– area of ​​the protective structure, m2;
n– coefficient that takes into account the position of the fence or protective structure (its outer surface) relative to the outside air;
β – additional heat losses, fractions of the main ones;
Ro– heat transfer resistance, m2 °C / W, which is determined by the following formula:

Rо = 1/ αв + Σ (δі / λі) + 1/ αн + Rв.п., where

αв – heat absorption coefficient of the fence (its inner surface), W/ m2 o C;
λі and δі – calculated thermal conductivity coefficient for the material of a given structural layer and the thickness of this layer;
αн – heat transfer coefficient of the fence (its outer surface), W/ m2 o C;
Rв.n – if there is a closed air gap in the structure, its thermal resistance, m2 o C / W (see Table 2).
The coefficients αн and αв are accepted according to SNiP and for some cases are given in Table 1;
δі - usually assigned according to the specifications or determined from the drawings of enclosing structures;
λі – accepted from reference books.

Table 1. Heat absorption coefficients αв and heat transfer coefficients αн

Surface of the building envelope	αv, W/ m2 o C	αn, W/ m2 o C
Internal surface of floors, walls, smooth ceilings
External surface of walls, roofless ceilings
Attic floors and ceilings over unheated basements with light openings
Ceilings over unheated basements without light openings

Table 2. Thermal resistance of closed air layers Rв.n, m2 o C / W

Air layer thickness, mm	Horizontal and vertical layers when heat flow from bottom to top		Horizontal layer with heat flow from top to bottom
	At the temperature in the air gap space

For doors and windows, heat transfer resistance is calculated very rarely, and is more often taken depending on their design according to reference data and SNiPs. The areas of fences for calculations are determined, as a rule, according to construction drawings. Temperature tin for residential buildings selected from Appendix I, tnB – from Appendix 2 of SNiP depending on the location of the construction site. Additional heat loss is indicated in Table 3, coefficient n - in Table 4.

Table 3. Additional heat loss

Fencing, its type	Terms	Additional heat loss β
Windows, doors and exterior vertical walls:	orientation northwest east, north and northeast
	west and southeast
External doors, doors with vestibules 0.2 N without air curtain at building height N, m	triple doors with two vestibules
	double doors with vestibule
Corner rooms additionally for windows, doors and walls	one of the fences is oriented to the east, north, northwest or northeast
	other cases

Table 4. The value of the coefficient n, which takes into account the position of the fence (its outer surface)

The heat consumption for heating the external infiltrating air in public and residential buildings for all types of premises is determined by two calculations. The first calculation determines the consumption of thermal energy Qi for heating the outside air, which enters the i-th room as a result of the action of natural exhaust ventilation. The second calculation determines the consumption of thermal energy Qi for heating the outside air, which penetrates into a given room through leaks in the fences as a result of wind and (or) thermal pressure. For the calculation, the largest value of heat loss determined by the following equations (1) and (or) (2) is taken.

Qі = 0.28 L ρн s (tin – tnB) (1)

L, m3/hour c – the flow rate of air removed from the premises; for residential buildings, 3 m3/hour per 1 m2 of residential area, including kitchens;
With – specific heat air (1 kJ /(kg °C));
ρн– air density outside the room, kg/m3.

The specific gravity of air γ, N/m3, its density ρ, kg/m3, are determined according to the formulas:

γ = 3463/ (273 +t), ρ = γ / g, where g = 9.81 m/s2, t, ° C – air temperature.

The heat consumption for heating the air that enters the room through various leaks of protective structures (fences) as a result of wind and thermal pressure is determined according to the formula:

Qi = 0.28 Gi s (tin – tnB) k, (2)

where k is a coefficient taking into account the counter-current heat flow, for separate-binding balcony doors and windows, 0.8 is accepted, for single and double-sash windows – 1.0;
Gi – flow rate of air penetrating (infiltrating) through protective structures (enclosing structures), kg/h.

For balcony doors and windows, the Gi value is determined:

Gi = 0.216 Σ F Δ Рі 0.67 / Ri, kg/h

where Δ Рi is the difference in air pressure on the internal Рвн and external Рн surfaces of doors or windows, Pa;
Σ F, m2 – estimated areas of all building fences;
Ri, m2·h/kg – air permeability resistance of this fence, which can be accepted in accordance with Appendix 3 of SNiP. IN panel buildings, in addition, the additional flow of air infiltrated through leaks in the joints of the panels is determined.

The value of Δ Рi is determined from the equation, Pa:

Δ Рі= (H – hі) (γн – γвн) + 0.5 ρн V2 (се,n – се,р) k1 – ріnt,
where H, m is the height of the building from the zero level to the mouth of the ventilation shaft (in buildings without attics the mouth is usually located 1 m above the roof, and in buildings with an attic - 4–5 m above the attic floor);
hі, m – height from zero level to the top of balcony doors or windows for which air flow is calculated;
γн, γвн – specific weights of external and internal air;
ce, pu ce, n – aerodynamic coefficients for the leeward and windward surfaces of the building, respectively. For rectangular buildings se,r= –0.6, ce,n= 0.8;

V, m/s – wind speed, which is taken for calculation according to Appendix 2;
k1 – coefficient that takes into account the dependence of wind speed pressure and building height;
рінт, Pa – conditionally constant air pressure that occurs during forced ventilation; when calculating residential buildings, ріnt can be ignored, since it is equal to zero.

For fences with a height of up to 5.0 m, the coefficient k1 is 0.5, for a height of up to 10 m it is 0.65, for a height of up to 20 m it is 0.85, and for fences of 20 m and above it is taken to be 1.1.

Total estimated heat loss in the room, W:

Qcalc = Σ Qlim + Qunf – Qbyt

where Σ Qlim – total heat loss through all safety fences premises;
Qinf – maximum flow heat for heating the air that is infiltrated, taken from calculations according to formulas (2) u (1);
Qhousehold – all heat emissions from household electrical appliances, lighting, and other possible heat sources, which are accepted for kitchens and living quarters in the amount of 21 W per 1 m2 of calculated area.

Vladivostok -24.
Vladimir -28.
Volgograd -25.
Vologda -31.
Voronezh -26.
Ekaterinburg -35.
Irkutsk -37.
Kazan -32.
Kaliningrad -18
Krasnodar -19.
Krasnoyarsk -40.
Moscow -28.
Murmansk -27.
Nizhny Novgorod -30.
Novgorod -27.
Novorossiysk -13.
Novosibirsk -39.
Omsk -37.
Orenburg -31.
Eagle -26.
Penza -29.
Perm -35.
Pskov -26.
Rostov -22.
Ryazan -27.
Samara -30.
St. Petersburg -26.
Smolensk -26.
Tver -29.
Tula -27.
Tyumen -37.
Ulyanovsk -31.

The choice of thermal insulation, options for insulating walls, ceilings and other enclosing structures is a difficult task for most customer-developers. There are too many conflicting problems to solve at once. This page will help you figure it all out.

Currently, heat conservation of energy resources has become great value. According to SNiP 23-02-2003 “Thermal protection of buildings”, heat transfer resistance is determined using one of two alternative approaches:

• prescriptive ( regulatory requirements presented to individual elements thermal protection of the building: external walls, floors above unheated spaces, coverings and attic floors, windows, entrance doors, etc.)
• consumer (the heat transfer resistance of the fence can be reduced in relation to the prescriptive level, provided that the design specific consumption thermal energy for heating the building is below standard).

Hygiene requirements must be met 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 permissible values. Maximum valid values difference for outer wall 4°C, for roofing and attic flooring 3°C and for ceilings over basements and crawl spaces 2°C.

The requirement is 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 °C m. sq./W, and according to the prescriptive approach:

• for home permanent residence 3.13 °С m. sq./W,
• for administrative and other public buildings incl. buildings for seasonal residence 2.55 °С m. sq./W.

Table of thicknesses and thermal resistance of materials for the conditions of Moscow and its region.

Name of wall material	Wall thickness and corresponding thermal resistance	Required thickness according to consumer approach (R=1.97 °C m2/W) and a prescriptive approach (R=3.13 °C m2/W)
Solid solid clay brick (density 1600 kg/m3)	510 mm (two bricks), R=0.73 °С m. sq./W	1380 mm 2190 mm
Expanded clay concrete (density 1200 kg/m3)	300 mm, R=0.58 °С m. sq./W	1025 mm 1630 mm
Wooden beam	150 mm, R=0.83 °С m. sq./W	355 mm 565 mm
Wooden shield with filling mineral wool(thickness of internal and external cladding from 25 mm boards)	150 mm, R=1.84 °С m. sq./W	160 mm 235 mm

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

Exterior wall	Window, balcony door	Covering and floors	Attic floors and floors over unheated basements	Entrance door
Byprescriptive approach
3,13	0,54	3,74	3,30	0,83
According to consumer approach
1,97	0,51	4,67	4,12	0,79

From these tables it is clear 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 observed in many newly constructed buildings.

Therefore, when selecting a boiler or heating devices only according to the heating capabilities indicated in their documentation certain area, You claim that your house was built with strict regard to the requirements of SNiP 02/23/2003.

The conclusion follows from the above material. For the right choice power of the boiler and heating devices, it is necessary to calculate the real heat loss of the premises of your home.

Below we will show a simple method for calculating the heat loss of your home.

The house loses heat through the wall, roof, strong emissions of heat come through the windows, heat also goes into the ground, significant heat losses can occur through ventilation.

Heat losses mainly depend on:

• temperature differences in the house and outside (the greater the difference, the higher the losses),
• heat-insulating properties of walls, windows, ceilings, coatings (or, as they say, enclosing structures).

Enclosing structures resist heat leakage, therefore their heat-protective properties are assessed by a value called heat transfer resistance.

Heat transfer resistance shows how much heat will be lost through a square meter of the building envelope for a given temperature difference. We can also say, conversely, what temperature difference will occur when a certain amount of heat passes through a square meter of fencing.

where q is the amount of heat lost per square meter of the enclosing surface. It is measured in watts per square meter (W/m2); ΔT is the difference between the temperature outside and in the room (°C) and R is the heat transfer resistance (°C/W/m2 or °C·m2/W).

When we're talking about On a multilayer structure, the resistance of the layers simply adds up. For example, the resistance of a wall made of wood lined with brick is the sum of three resistances: the brick and wooden walls and the air gap between them:

R(total)= R(wood) + R(air) + R(brick).

Temperature distribution and air boundary layers during heat transfer through a wall

Heat loss calculations are carried out for the most unfavorable period, which is the coldest and windiest week of the year.

Construction reference books, as a rule, indicate the thermal resistance of materials based on this condition and the climatic region (or outside temperature) where your home is located.

Table- Heat transfer resistance various materials at ΔT = 50 °C (T external = -30 °C, T internal = 20 °C.)

Wall material and thickness	Heat transfer resistance R m,
Brick wall 3 bricks thick (79 cm) 2.5 bricks thick (67 cm) 2 bricks thick (54 cm) 1 brick thick (25 cm)	0,592 0,502 0,405 0,187
Log house Ø 25 Ø 20	0,550 0,440
Log house made of timber 20 cm thick 10 cm thick	0,806 0,353
Frame wall (board + mineral wool + board) 20 cm	0,703
Foam concrete wall 20 cm 30 cm	0,476 0,709
Plastering on brick, concrete, foam concrete (2-3 cm)	0,035
Ceiling (attic) floor	1,43
Wooden floors	1,85
Double wooden doors	0,21

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

Window type R T q, W/m2 Q, W Regular window with double frames 0,37 135 216 Double-glazed window (glass thickness 4 mm) 4-16-4 4-Ar16-4 4-16-4K 4-Ar16-4K 0,32 0,34 0,53 0,59 156 147 94 85 250 235 151 136 Double-glazed window 4-6-4-6-4 4-Ar6-4-Ar6-4 4-6-4-6-4K 4-Ar6-4-Ar6-4K 4-8-4-8-4 4-Ar8-4-Ar8-4 4-8-4-8-4K 4-Ar8-4-Ar8-4K 4-10-4-10-4 4-Ar10-4-Ar10-4 4-10-4-10-4K 4-Ar10-4-Ar10-4K 4-12-4-12-4 4-Ar12-4-Ar12-4 4-12-4-12-4K 4-Ar12-4-Ar12-4K 4-16-4-16-4 4-Ar16-4-Ar16-4 4-16-4-16-4K 4-Ar16-4-Ar16-4K 0,42 0,44 0,53 0,60 0,45 0,47 0,55 0,67 0,47 0,49 0,58 0,65 0,49 0,52 0,61 0,68 0,52 0,55 0,65 0,72 119 114 94 83 111 106 91 81 106 102 86 77 102 96 82 73 96 91 77 69 190 182 151 133 178 170 146 131 170 163 138 123 163 154 131 117 154 146 123 111 Note . Even numbers in symbol double glazing means air clearance in mm; . The symbol Ar means that the gap is filled not with air, but with argon; . The letter K means that the outer glass has a special transparent heat-protective coating.

As can be seen from the previous table, modern double glazed windows allow you to 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.

To correctly select materials and thicknesses of enclosing structures, we will apply this information to a specific example.

When calculating heat losses per sq. meter there are two quantities involved:

• temperature difference ΔT,
• heat transfer resistance R.

Let's define the room temperature as 20 °C, and take the outside temperature to be -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/m2).

To simplify calculations of heat loss, heat loss is given in construction reference books different types walls, ceilings, etc. for some values ​​of winter air temperature. In particular, different figures are given for corner rooms (the turbulence of the air that swells the house is affected there) and non-corner rooms, and the different thermal picture for the rooms of the first and upper floors is also taken into account.

Table- Specific heat loss of building enclosure elements (per 1 sq.m. along the internal contour of the walls) depending on average temperature the coldest week of the year.

Characteristic fencing Outdoor temperature, °C Heat loss, W First floor Top floor Corner room Unangle room Corner room Unangle room Wall 2.5 bricks (67 cm) with internal plaster -24 -26 -28 -30 76 83 87 89 75 81 83 85 70 75 78 80 66 71 75 76 Wall of 2 bricks (54 cm) with internal plaster -24 -26 -28 -30 91 97 102 104 90 96 101 102 82 87 91 94 79 87 89 91 Chopped wall(25 cm) with internal sheathing -24 -26 -28 -30 61 65 67 70 60 63 66 67 55 58 61 62 52 56 58 60 Chopped wall (20 cm) with internal sheathing -24 -26 -28 -30 76 83 87 89 76 81 84 87 69 75 78 80 66 72 75 77 Wall made of timber (18 cm) with internal sheathing -24 -26 -28 -30 76 83 87 89 76 81 84 87 69 75 78 80 66 72 75 77 Wall made of timber (10 cm) with internal sheathing -24 -26 -28 -30 87 94 98 101 85 91 96 98 78 83 87 89 76 82 85 87 Frame wall (20 cm) with expanded clay filling -24 -26 -28 -30 62 65 68 71 60 63 66 69 55 58 61 63 54 56 59 62 Foam concrete wall (20 cm) with internal plaster -24 -26 -28 -30 92 97 101 105 89 94 98 102 87 87 90 94 80 84 88 91 Note If behind the wall there is an external unheated room (canopy, glass veranda etc.), then the heat loss through it is 70% of the calculated value, and if beyond this unheated room not a street, but another room outside (for example, a canopy opening onto a veranda), then 40% of the calculated value.

Table- Specific heat loss of building enclosure elements (per 1 sq.m. along the internal contour) depending on the average temperature of the coldest week of the year.

Characteristics of the fence	Outdoor temperature, °C	Heat loss kW
Double glazed window	-24 -26 -28 -30	117 126 131 135
Solid wooden doors (double)	-24 -26 -28 -30	204 219 228 234
Attic floor	-24 -26 -28 -30	30 33 34 35
Wood floors above basement	-24 -26 -28 -30	22 25 26 26

Let's consider an example of calculating heat losses of two different rooms one area using tables.

Example 1.

Corner room(first floor)

Room characteristics:

• first floor,
• room area - 16 sq.m. (5x3.2),
• ceiling height - 2.75 m,
• external walls - two,
• material and thickness of the external walls - timber 18 cm thick, covered 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,
• estimated outside temperature -30 °C,
• required room temperature +20 °C.

Area of ​​external walls excluding 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:

Ceiling S = 5x3.2 = 16 sq. m.

Square internal partitions does not participate in the calculation, since heat does not escape through them - after all, the temperature is the same on both sides of the partition. The same applies to the inner door.

Now let's calculate the heat loss of each surface:

Q total = 3094 W.

Note that more heat escapes through the walls than through the windows, floors and ceiling.

The calculation result shows the heat loss of the room on the coldest (T ambient = -30 °C) days of the year. Naturally, the warmer it is outside, the less heat will leave the room.

Example 2

Room under the roof (attic)

Room characteristics:

• top floor,
• area 16 sq.m. (3.8x4.2),
• ceiling height 2.4 m,
• external walls; two roof slopes (slate, continuous sheathing, 10 cm mineral wool, lining), gables (10 cm thick timber, covered 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,
• estimated outside temperature -30°С,
• required room temperature +20°C.

Let's calculate the areas of heat-transfer surfaces.

Area of ​​the end external walls excluding windows:

S end wall = 2x(2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 sq. m.

Area of ​​roof slopes bordering the room:

S sloped walls = 2x1.0x4.2 = 8.4 sq. m.

Area of ​​side partitions:

S side burner = 2x1.5x4.2 = 12.6 sq. m.

Window area:

S windows = 4x1.6x1.0 = 6.4 sq. m.

Ceiling area:

Ceiling S = 2.6x4.2 = 10.92 sq. m.

Now let's calculate heat losses these surfaces, while taking into account that heat does not escape through the floor (there warm room). We calculate heat loss for walls and ceilings as for corner rooms, and for the ceiling and side partitions we introduce a 70 percent coefficient, since behind them there are unheated rooms.

The total heat loss of the room will be:

Q total = 4504 W.

As you can see, a warm room on the first floor loses (or consumes) significantly less heat than attic room with thin walls and a large glazing area.

To make such a room suitable for winter accommodation, you must first insulate the walls, side partitions and windows.

Any enclosing structure can be represented in the form multilayer wall, each layer of which has its own thermal resistance and its own resistance to air passage. Adding up the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also, by 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 with this.

Table- Resistance to heat transfer and air passage of various materials ΔT = 40 ° C (T external = -20 ° C, T internal = 20 ° C.)

Wall Layer	Thickness layer walls	Resistance heat transfer of the wall layer		Resistance air-flow worthlessness equivalent timber wall thick (cm)
		Ro,	Equivalent brick masonry thick (cm)
Brickwork from the usual clay brick thickness: 12 cm 25 cm 50 cm 75 cm	12 25 50 75	0,15 0,3 0,65 1,0	12 25 50 75	6 12 24 36
Masonry made of expanded clay concrete blocks 39 cm thick with density: 1000 kg/cu m 1400 kg/cu m 1800 kg/cu m	39	1,0 0,65 0,45	75 50 34	17 23 26
Foam aerated concrete 30 cm thick density: 300 kg/cu m 500 kg/cu m 800 kg/cu m	30	2,5 1,5 0,9	190 110 70	7 10 13
Thick timbered wall (pine) 10 cm 15 cm 20 cm	10 15 20	0,6 0,9 1,2	45 68 90	10 15 20

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

1. Heat loss through the contact of the foundation with frozen soil is usually assumed to be 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
2. Heat losses associated with 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 fresh air. Thus, losses associated with ventilation are slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss through ventilation is 50%. In European standards 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 returns. This allows you to reduce heat losses by 30%, so the value of the wall’s thermal resistance obtained in the calculation should be multiplied by 1.3 (or heat losses should be reduced accordingly).

By summing up all the heat loss at home, you will determine the power of the heat generator (boiler) and heating devices necessary 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 using additional insulation.

Heat consumption can also be calculated using aggregated indicators. So, in one- and two-story houses that are not very insulated with outside temperature-25 °C requires 213 W per square meter of total area, and at -30 °C - 230 W. For well-insulated houses this is: at -25 °C - 173 W per sq.m. total area, and at -30 °C - 177 W.

1. The cost of thermal insulation relative to the cost of the entire house is significantly small, but during the operation of the building the main costs are for heating. In no case should you skimp on thermal insulation, especially when living comfortably in large areas. Energy prices around the world are constantly rising.
2. Modern building materials have higher thermal resistance than traditional materials. This allows you to make walls thinner, which means cheaper and lighter. All this is good, but thin walls have less heat capacity, that is, they store heat less well. You have to constantly heat it - 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, which cooled down overnight, “accumulated cold.”
3. Insulation must be considered in conjunction with the air permeability of the walls. If an increase in the thermal resistance of walls is associated with a significant decrease in air permeability, then it should not be used. An ideal wall in terms of breathability is equivalent to a wall made of timber 15...20 cm thick.
4. Very often, improper use of vapor barrier leads to deterioration of the sanitary and hygienic properties of housing. When correct organized ventilation and “breathing” walls it is unnecessary, and with poorly breathable walls it is unnecessary. Its main purpose is to prevent infiltration of walls and protect insulation from wind.
5. Insulating walls from the outside is much more effective than internal insulation.
6. You should not endlessly insulate the walls. The effectiveness of this approach to energy saving is not high.
7. Ventilation is the main source of energy saving.
8. By applying modern systems glazing (double glazing, thermal insulation glass, etc.), low-temperature heating systems, effective thermal insulation of building envelopes, heating costs can be reduced by 3 times.

Options additional insulation building structures based on building thermal insulation of the “ISOVER” type, in the presence of air exchange and ventilation systems in the premises.

How to properly arrange heating appliances and increase their efficiency

Heat loss at home

To date heat saving is important parameter, which is taken into account when constructing a residential or office space. In accordance with SNiP 23-02-2003 “Thermal protection 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 and home heat loss.

Prescriptive Approach- these are the standards for individual elements of thermal protection of a building: external walls, floors above unheated spaces, coverings and attic floors, windows, entrance doors, etc.

Consumer approach(heat transfer resistance can be reduced in relation to the prescribed level, provided that the design specific heat energy consumption for space heating is lower than the standard one).

Sanitary and hygienic requirements:

• The difference between indoor and outdoor air temperatures should not exceed certain permissible values. The maximum permissible temperature difference for an external wall is 4°C. for roofing and attic flooring 3°C and for ceilings over basements and crawl spaces 2°C.
• The temperature on the inner surface of the fence 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 °C m 2 /W, and according to the prescriptive approach:

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

For this reason, when choosing a boiler or other heating devices solely according to those specified in their technical documentation parameters. You must ask yourself whether your house was built with strict regard to the requirements of SNiP 02/23/2003.

Therefore, to correctly select the power of the heating boiler, either heating devices, it is necessary to calculate real heat loss from your home. As a rule, a residential building loses heat through the walls, roof, windows, and ground; significant heat losses can also occur through ventilation.

Heat loss mainly depends on:

• temperature differences in the house and outside (the higher the difference, the higher the losses).
• heat-protective characteristics of walls, windows, ceilings, coatings.

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

Heat transfer resistance will show how much heat will leak through a square meter of structure 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/m2);
• ΔT is the difference between the temperature outside and in the room (°C);
• R is the heat transfer resistance (°C/W/m2 or °C m2/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 wall made of wood, which is lined with brick, is the sum of three resistances: the brick and wooden walls and the air gap between them:

R(total)= R(wood) + R(air) + R(brick)

Temperature distribution and air boundary layers during heat transfer through a wall.

Heat loss calculation performed for the coldest period of the year, which is the coldest and windiest week of the year. In construction literature, the thermal resistance of materials is often indicated based on this condition and the climatic region (or outside temperature) where your home is located.

Table of heat transfer resistance of various materials

at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)

Wall material and thickness	Heat transfer resistance R m.
Brick wall thickness in 3 bricks. (79 centimeters) thickness in 2.5 bricks. (67 centimeters) thickness in 2 bricks. (54 centimeters) thickness in 1 brick. (25 centimeters)	0.592 0.502 0.405 0.187
Log house Ø 25 Ø 20	0.550 0.440
Log house made of timber Thickness 20 centimeters Thickness 10 centimeters	0.806 0.353
Frame wall (board + mineral wool + board) 20 centimeters
Foam concrete wall 20 centimeters 30 cm	0.476 0.709
Plastering on brick, concrete. foam concrete (2-3 cm)
Ceiling (attic) floor
Wooden floors
Double wooden doors

Window heat loss table various designs at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)

Window type R T q . W/m2 Q . W Regular double glazed window Double-glazed window (glass thickness 4 mm) 4-16-4 4-Ar16-4 4-16-4K 4-Ar16-4K 0.32 0.34 0.53 0.59 156 147 94 85 250 235 151 136 Double-glazed window 4-6-4-6-4 4-Ar6-4-Ar6-4 4-6-4-6-4K 4-Ar6-4-Ar6-4K 4-8-4-8-4 4-Ar8-4-Ar8-4 4-8-4-8-4K 4-Ar8-4-Ar8-4K 4-10-4-10-4 4-Ar10-4-Ar10-4 4-10-4-10-4K 4-Ar10-4-Ar10-4К 4-12-4-12-4 4-Ar12-4-Ar12-4 4-12-4-12-4K 4-Ar12-4-Ar12-4К 4-16-4-16-4 4-Ar16-4-Ar16-4 4-16-4-16-4K 4-Ar16-4-Ar16-4K 0.42 0.44 0.53 0.60 0.45 0.47 0.55 0.67 0.47 0.49 0.58 0.65 0.49 0.52 0.61 0.68 0.52 0.55 0.65 0.72 119 114 94 83 111 106 91 81 106 102 86 77 102 96 82 73 96 91 77 69 190 182 151 133 178 170 146 131 170 163 138 123 163 154 131 117 154 146 123 111

Note
. Even numbers in the designation 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-protective coating.

As can be seen from the above table, modern double-glazed windows make it possible reduce heat loss windows almost 2 times. For example, for 10 windows measuring 1.0 m x 1.6 m, savings can reach up to 720 kilowatt-hours per month.

To correctly select materials and wall thickness, apply this information to a specific example.

Two quantities are involved in calculating heat losses per m2:

• 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 losses will be 50 / 0.806 = 62 (W/m2).

To simplify calculations of heat loss in construction reference books indicate heat loss various types walls, ceilings, etc. for some values ​​of winter air temperature. Typically, different numbers are given for corner rooms(the turbulence of the air that swells the house influences this) and non-angular, and also takes into account the difference in temperatures for the rooms of the first and upper floors.

Table of specific heat loss of building enclosure elements (per 1 m2 along the internal contour of the walls) depending on the average temperature of the coldest week of the year.

Characteristic fencing Outdoor temperature. °C Heat loss. W 1st floor 2nd floor Corner room Unangle room Corner room Unangle room Wall 2.5 bricks (67 cm) with internal plaster 24 -26 -28 -30 76 83 87 89 75 81 83 85 70 75 78 80 66 71 75 76 Wall of 2 bricks (54 cm) with internal plaster 24 -26 -28 -30 91 97 102 104 90 96 101 102 82 87 91 94 79 87 89 91 Chopped wall (25 cm) with internal sheathing 24 -26 -28 -30 61 65 67 70 60 63 66 67 55 58 61 62 52 56 58 60 Chopped wall (20 cm) with internal sheathing 24 -26 -28 -30 76 83 87 89 76 81 84 87 69 75 78 80 66 72 75 77 Wall made of timber (18 cm) with internal sheathing 24 -26 -28 -30 76 83 87 89 76 81 84 87 69 75 78 80 66 72 75 77 Wall made of timber (10 cm) with internal sheathing 24 -26 -28 -30 87 94 98 101 85 91 96 98 78 83 87 89 76 82 85 87 Frame wall (20 cm) with expanded clay filling 24 -26 -28 -30 62 65 68 71 60 63 66 69 55 58 61 63 54 56 59 62 Foam concrete wall (20 cm) with internal plaster 24 -26 -28 -30 92 97 101 105 89 94 98 102 87 87 90 94 80 84 88 91

Note. In the case when there is an external unheated room behind the wall (canopy, glazed veranda etc.), then the heat loss through it will be 70% of the calculated value, and if there is another outdoor room behind this unheated room, then the heat loss will be 40% of the calculated value.

Table of specific heat loss of building enclosure elements (per 1 m2 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 m2 (5x3.2).
• ceiling height - 2.75 m.
• There are two external walls.
• material and thickness of the external walls - timber 18 centimeters thick, covered 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.
• estimated outside temperature -30 °C.
• required room temperature +20 °C.

• Area of ​​external walls minus windows: S walls (5+3.2)x2.7-2x1.0x1.6 = 18.94 m2.
• Window area: S windows = 2x1.0x1.6 = 3.2 m2
• Floor area: S floor = 5x3.2 = 16 m2
• Ceiling area: Ceiling S = 5x3.2 = 16 m2

The area of ​​the internal partitions is not included in the calculation, since the temperature on both sides of the partition is the same, therefore heat does not escape through the partitions.

Now let's calculate the heat loss of each surface:

• Q walls = 18.94x89 = 1686 W.
• Q windows = 3.2x135 = 432 W.
• Floor Q = 16x26 = 416 W.
• Ceiling Q = 16x35 = 560 W.

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

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

Example 2

Room under the roof (attic)

Room characteristics:

• top floor.
• area 16 m2 (3.8x4.2).
• ceiling height 2.4 m.
• external walls; two roof slopes (slate, continuous sheathing, 10 centimeters of mineral wool, lining). gables (beams 10 centimeters thick covered with clapboard) and side partitions (frame wall with expanded clay filling 10 centimeters).
• windows - 4 (two on each gable), 1.6 m high and 1.0 m wide with double glazing.
• estimated outside temperature -30°C.
• required room temperature +20°C.

• Area of ​​the end external walls minus windows: S end walls = 2x(2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 m2
• Area of ​​roof slopes bordering the room: S sloped walls = 2x1.0x4.2 = 8.4 m2
• Area of ​​the side partitions: S side partition = 2x1.5x4.2 = 12.6 m 2
• Window area: S windows = 4x1.6x1.0 = 6.4 m2
• Ceiling area: Ceiling S = 2.6x4.2 = 10.92 m2

Next, we will calculate the heat losses of these surfaces, while it is necessary to take into account that through the floor in in this case the heat will not escape, since there is a warm room below. Heat loss for walls We calculate as for corner rooms, and for the ceiling and side partitions we enter a 70 percent coefficient, since unheated rooms are located behind them.

• Q end walls = 12x89 = 1068 W.
• Q sloped walls = 8.4x142 = 1193 W.
• Q side burnout = 12.6x126x0.7 = 1111 W.
• Q windows = 6.4x135 = 864 W.
• Ceiling Q = 10.92x35x0.7 = 268 W.

The total heat loss of the room will be: Q total = 4504 W.

As we can see, a warm room on the 1st floor loses (or consumes) significantly less heat than an attic room with thin walls and a large glazing area.

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 presented in the form of a multilayer wall, each layer of which has its own thermal resistance and its own resistance to air passage. By summing the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also, if you sum up the resistance to the passage of air of all layers, you can understand how the wall breathes. The most best wall made of timber should be equivalent to a wall made of timber with a thickness of 15 - 20 centimeters. The table below will help with this.

Table of resistance to heat transfer and air passage of various materials ΔT = 40 ° C (T external = -20 ° C. T internal = 20 ° C.)

Wall Layer	Thickness layer walls	Resistance heat transfer of the wall layer		Resistance Air flow worthlessness equivalent timber wall thick (cm)
			Equivalent brick masonry thick (cm)
Ordinary brickwork clay brick thickness: 12 centimeters 25 centimeters 50 centimeters 75 centimeters	12 25 50 75	0.15 0.3 0.65 1.0	12 25 50 75	6 12 24 36
Masonry made of expanded clay concrete blocks 39 cm thick with density: 1000 kg/m3 1400 kg/m3 1800 kg/m3		1.0 0.65 0.45	75 50 34	17 23 26
Foam aerated concrete 30 cm thick density: 300 kg/m3 500 kg/m3 800 kg/m3		2.5 1.5 0.9	190 110 70	7 10 13
Thick timbered wall (pine) 10 centimeters 15 centimeters 20 centimeters	10 15 20	0.6 0.9 1.2	45 68 90	10 15 20

To get a complete picture of the heat loss of the entire room, you need to take into account

1. Heat loss through the contact of the foundation with frozen soil is usually assumed to be 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
2. Heat losses associated with 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 will be slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation 50%. In European standards for ventilation and wall insulation, the heat loss ratio is 30% and 60%.
3. If the wall “breathes”, like a wall made of timber or logs 15 - 20 centimeters thick, then heat returns. This allows you to reduce heat losses by 30%. therefore, the value of the thermal resistance of the wall obtained during the calculation must be multiplied by 1.3 (or, accordingly reduce heat loss).

By summing up all the heat loss in the house, you can understand what power the boiler and heating appliances are needed 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 using additional insulation.

You can also calculate heat consumption using aggregated indicators. So, in 1-2 storey houses that are not very insulated, at an outside temperature of -25 ° C, 213 W per 1 m 2 of total area is required, and at -30 ° C - 230 W. For well-insulated houses, this figure will be: at -25 °C - 173 W per m 2 of total area, and at -30 °C - 177 W.

Before you start building a house, you need to buy a house plan - that’s what the architects say. You need to buy the services of professionals - that’s what the builders say. It is necessary to buy high-quality building materials - this is what sellers and manufacturers of building materials and insulation materials say.

And you know, in some ways they are all a little right. However, no one except you will be so interested in your home as to take into account all the points and bring together all the issues regarding its construction.

One of the most important issues, which should be solved at the stage, is the heat loss of the house. The design of the house, its construction, and what building materials and insulation materials you will purchase will depend on the calculation of heat loss.

There are no houses with zero heat loss. To do this, the house would have to float in a vacuum with walls of 100 meters of highly efficient insulation. We don’t live in a vacuum, and we don’t want to invest in 100 meters of insulation. This means that our house will experience heat loss. Let them be, as long as they are reasonable.

Heat loss through walls

Heat loss through the walls - all owners immediately think about this. Calculate the thermal resistance of the enclosing structures and insulate them until they reach standard indicator R and this is where they finish their work on insulating the house. Of course, heat loss through the walls of the house must be considered - the walls have maximum area from all enclosing structures of the house. But they are not the only way for heat to escape.

Insulation of the house - the only way reduce heat loss through walls.

In order to limit heat loss through the walls, it is enough to insulate the house with 150 mm for the European part of Russia or 200-250 mm of the same insulation for Siberia and northern regions. And with that, you can leave this indicator alone and move on to others that are no less important.

Floor heat loss

A cold floor in a house is a disaster. Heat loss from the floor, relative to the same indicator for the walls, is approximately 1.5 times more important. And the thickness of the insulation in the floor should be exactly the same amount greater than the thickness of the insulation in the walls.

Heat loss from the floor becomes significant when you have a cold base or just street air under the floor of the first floor, for example, with screw piles.

If you insulate the walls, insulate the floor too.

If you put 200 mm of basalt wool or polystyrene in the walls, then you will have to put 300 mm of equally effective insulation in the floor. Only in this case will it be possible to walk on the floor of the first floor barefoot in any, even the most severe, conditions.

If you have a heated basement under the floor of the first floor or a well-insulated basement with a well-insulated wide blind area, then the insulation of the first floor floor can be neglected.

Moreover, such a basement or basement should be pumped with heated air from the first floor, or better yet, from the second. But the walls of the basement and its slab should be insulated as much as possible so as not to “heat” the soil. Certainly, constant temperature soil +4C, but this is at depth. And in winter around the basement walls it’s still the same -30C as on the ground surface.

Heat loss through the ceiling

All the heat goes up. And there it strives to go outside, that is, to leave the room. Heat loss through the ceiling in your home is one of the largest quantities that characterizes the loss of heat to the street.

The thickness of the insulation on the ceiling should be 2 times the thickness of the insulation in the walls. If you mount 200 mm in the walls, mount 400 mm on the ceiling. In this case, you will be guaranteed the maximum thermal resistance of your thermal circuit.

What are we doing? Walls 200 mm, floor 300 mm, ceiling 400 mm. Consider the savings you will use to heat your home.

Heat loss from windows

What is completely impossible to insulate are windows. Window heat loss is the largest quantity that describes the amount of heat leaving your home. No matter what you make your double-glazed windows - two-chamber, three-chamber or five-chamber, the heat loss of the windows will still be gigantic.

How to reduce heat loss through windows? Firstly, it is worth reducing the glass area throughout the house. Of course, with large glazing, the house looks chic, and its facade reminds you of France or California. But there is only one thing here - either stained glass windows in half the wall or good thermal resistance of your home.

If you want to reduce heat loss from windows, do not plan a large area.

Secondly, the window slopes – the places where the sashes meet the walls – should be well insulated.

And thirdly, it is worth using new products from the construction industry for additional heat conservation. For example, automatic night heat-saving shutters. Or films that reflect thermal radiation back into the house, but freely transmit the visible spectrum.

Where does the heat leave the house?

The walls are insulated, the ceiling and floor too, the shutters are installed on five-chamber double-glazed windows, the fire is in full swing. But the house is still cool. Where does the heat continue to go from the house?

Now is the time to look for cracks, crevices and crevices where heat is escaping from your home.

Firstly, the ventilation system. Cold air comes through supply ventilation into the house, warm air leaves the house through exhaust ventilation. To reduce heat loss through ventilation, you can install a recuperator - a heat exchanger that takes heat from the outlet warm air and heating the incoming cold air.

One way to reduce heat loss at home through the ventilation system is to install a recuperator.

Secondly, the entrance doors. To eliminate heat loss through the doors, you should install a cold vestibule, which will serve as a buffer between entrance doors and street air. The vestibule should be relatively sealed and unheated.

Thirdly, it’s worth looking at your house at least once in cold weather with a thermal imager. Visiting specialists does not cost that much money. But you will have a “map of facades and ceilings” in your hands, and you will clearly know what other measures to take in order to reduce heat loss at home during the cold period.