Heat losses during the heating season. How to calculate heat loss at home: features, recommendations and program

Below is a pretty simple one heat loss calculation buildings, which, nevertheless, will help to accurately determine the power required to heat your warehouse, shopping center or another similar building. This will make it possible, even at the design stage, to preliminarily estimate the cost of heating equipment and subsequent heating costs, and, if necessary, adjust the project.

Where does the heat go? Heat escapes through walls, floors, roofs and windows. In addition, heat is lost during ventilation of the premises. To calculate heat loss through the enclosing structures, use the formula:

Q - heat loss, W

S - construction area, m2

T - temperature difference between indoor and outdoor air, ° C

R is the value of the thermal resistance of the structure, m2 ° C / W

The calculation scheme is as follows - we calculate heat loss individual elements, summarize and add the heat loss during ventilation. Everything.

Suppose we want to calculate the heat loss for the object shown in the figure. The height of the building is 5 ... 6 m, width - 20 m, length - 40 m, and thirty windows measuring 1.5 x 1.4 meters. Indoor temperature 20 ° С, external temperature -20 ° С.

We consider the area of the enclosing structures:

floor: 20 m * 40 m = 800 m2

roof: 20.2 m * 40 m = 808 m2

window: 1.5 m * 1.4 m * 30 pcs = 63 m2

walls:(20 m + 40 m + 20 m + 40 m) * 5 m = 600 m2 + 20 m2 (accounting pitched roof) = 620 m2 - 63 m2 (windows) = 557 m2

Now let's see the thermal resistance of the materials used.

The value of thermal resistance can be taken from the table of thermal resistances or calculated based on the value of the thermal conductivity coefficient using the formula:

R - thermal resistance, (m2 * K) / W

? - coefficient of thermal conductivity of the material, W / (m2 * K)

d - material thickness, m

The value of the thermal conductivity coefficients for different materials can be viewed.

floor: concrete screed 10 cm and mineral wool with a density of 150 kg / m3. 10 cm thick.

R (concrete) = 0.1 / 1.75 = 0.057 (m2 * K) / W

R (mineral wool) = 0.1 / 0.037 = 2.7 (m2 * K) / W

R (floor) = R (concrete) + R (mineral wool) = 0.057 + 2.7 = 2.76 (m2 * K) / W

roof:

R (roof) = 0.15 / 0.037 = 4.05 (m2 * K) / W

window: the value of the thermal resistance of windows depends on the type of glass unit used
R (windows) = 0.40 (m2 * K) / W for a single-chamber glass unit 4–16–4 at ΔT = 40 ° С

walls: mineral wool panels 15 cm thick
R (walls) = 0.15 / 0.037 = 4.05 (m2 * K) / W

Let's calculate heat losses:

Q (floor) = 800 m2 * 20 ° C / 2.76 (m2 * K) / W = 5797 W = 5.8 kW

Q (roof) = 808 m2 * 40 ° C / 4.05 (m2 * K) / W = 7980 W = 8.0 kW

Q (windows) = 63 m2 * 40 ° C / 0.40 (m2 * K) / W = 6300 W = 6.3 kW

Q (walls) = 557 m2 * 40 ° C / 4.05 (m2 * K) / W = 5500 W = 5.5 kW

We get that the total heat loss through the enclosing structures will be:

Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kWh

Now about ventilation losses.

To heat 1 m3 of air from a temperature of - 20 ° С to + 20 ° С, 15.5 watts are required.

Q (1 m3 of air) = 1.4 * 1.0 * 40 / 3.6 = 15.5 W, here 1.4 is the air density (kg / m3), 1.0 is the specific heat capacity of air (kJ / ( kg K)), 3.6 is the conversion factor to watts.

It remains to decide on the amount of required air. It is believed that with normal breathing, a person needs 7 m3 of air per hour. If you use a building as a warehouse and 40 people work on it, then you need to heat 7 m3 * 40 people = 280 m3 of air per hour, this will require 280 m3 * 15.5 W = 4340 W = 4.3 kW. And if you have a supermarket and on average there are 400 people on the territory, then heating the air will require 43 kW.

The final result:

Heating the proposed building requires a heating system of about 30 kW / h, and a ventilation system with a capacity of 3000 m3 / h with a heater with a capacity of 45 kW / h.

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 material of construction, m² ° С / W;
S is the area of the outer fence, m²;
t in - internal air temperature, ° С;
t n - the most low temperature environment, ° 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 according to regulatory documents, and on personal observation, internal - 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 according to outside, capturing the corners of the building, 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 for heating 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 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 introduced into Excel program in the form of tables with data, this will significantly speed up the calculation.

Calculation of heat loss at home is the basis of the heating system. It is needed, at least, in order to choose the right boiler. You can also estimate how much money will be spent on heating in the planned house, analyze the financial efficiency of insulation, i.e. to understand whether the cost of installing insulation will pay off with fuel economy over the life of the insulation. Very often, when choosing the power of the heating system of a room, people are guided by an 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 fully replenish the heat loss. Buildings vary in composition building materials, their volume, location in different climatic zones, etc. For competent calculation of thermal insulation and power selection heating systems you need to know about the real heat loss at home. How to calculate them - we will tell 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
the difference in temperature inside and outside the room - the greater the difference, the faster heat exchange occurs and the air loses heat
thermal conductivity of enclosing structures - the ability of walls, windows to retain heat

The easiest calculation of heat loss

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

This formula for calculating heat loss according to enlarged indicators, which are based on the 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 of the proposed waste oil heater, kW / hour.

100 W / m2- specific value of heat losses (65-80 watts / m2). It includes thermal energy leaks by absorbing it by windows, walls, ceiling and floor; leaks through ventilation and leaks in the premises and other leaks.

S- the area of the room;

K1- coefficient of heat loss of windows:

ordinary glazing K1 = 1.27
double glazing K1 = 1.0
triple glass unit K1 = 0.85;

K2- coefficient of heat loss of walls:

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 ratio of the area of windows and floor:

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

K4- outdoor temperature coefficient:

-10oC K4 = 0.7
-15oC K4 = 0.9
-20oC K4 = 1.1
-25oC K4 = 1.3
-35 ° C 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- the 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 (cubic meters)
∆t- delta temperatures (outdoor and indoor)
k- dissipation factor

k = 3.0-4.0 - without thermal insulation. (Simplified wood structure or corrugated sheet metal structure).
k = 2.0-2.9 - small thermal insulation. (Simplified building structure, single brickwork, simplified construction of windows and roof).
k = 1.0-1.9 - average thermal insulation. ( Standard design, double brickwork, few windows, roof with standard roof).
k = 0.6-0.9 - high thermal insulation. (Improved construction, double insulated brick walls, few double glazed windows, thick subfloor, high quality insulating roof).

In this formula, the scattering coefficient is very conditionally taken into account and it is not entirely clear which coefficients to use. In the classics, a rare contemporary made of modern materials taking into account the 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.

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

q Is the amount of heat that is lost square meter enclosing surface (usually measured in W / m2)

ΔT- the difference between the temperature inside the calculated room and the outside 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 rooms is taken:

Living quarters 22C
Non-residential 18С
Water treatment zones 33C

When it comes to a multi-layer structure, the resistances of the layers of the structure add up. I would also 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 coefficient of thermal conductivity of the material of the structure layer, we can easily obtain layer heat transfer resistance:

δ - layer thickness, m;

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

So, to calculate heat losses through enclosing structures, 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. Fence area F (Separate walls, windows, doors, ceiling, floor)
4. The orientation of the building in relation to the cardinal points.

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

Qlim = (ΔT / Rlim) * Flim * n * (1 + ∑b)

Qlim- heat loss through enclosing structures, W
Rogr- resistance to heat transfer, m2 ° C / W; (If there are several layers then ∑ Rlim layers)
Flim- area of the enclosing structure, m;
n- the coefficient of contact of the enclosing structure with the outside air.

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

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

a) in rooms 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 size 0.05; in corner rooms additionally - 0.05 for each wall, door and window, if one of the fences faces 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 points, in the amount of 0.08 with one outer wall and 0.13 for corner rooms (except for residential), and in all living quarters - 0.13;

c) through the 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 that are not equipped with air or air-thermal curtains, at the height of buildings H, m, from the average planning level of the earth to the top of the cornice, the center of the exhaust openings of the lantern or the mouth of the mine in the size: 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 a vestibule; 0.22 H - for single doors;

e) through the outer gates not equipped with air and air-thermal curtains - in size 3 in the absence of a vestibule and in size 1 if there is a vestibule at the gate.

For summer and emergency external doors and gates, additional heat losses according to subparagraphs "d" and "d" should not be taken into account.

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

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

The first three zones are stripes parallel to the perimeter of the outer walls. The rest of the area belongs to the fourth zone. The width of each zone is 2 m. The beginning of the first zone is at the point where the floor joins the outer wall. If the non-insulated floor adjoins a wall buried in the ground, then the beginning is transferred to the upper boundary of the wall deepening. If there are insulating layers in the floor structure located on the ground, it is called insulated, and its resistance to heat transfer Rу.п, (m2 оС) / W, is determined by the formula:

Ru.p. = Rn.p. + Σ (γv.s. / λ.s.)

Rn.p- resistance to heat transfer of the considered area of the non-insulated floor, (m2 oC) / W;
γu.s- thickness of the insulating layer, m;
λw.s- coefficient of thermal conductivity of the material of the insulating layer, W / (m · ° С).

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

Rl = 1.18 * Ru.p

Heat losses of each enclosing structure are counted 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 confuse the measurements. If instead of (W) appears (kW) or in general (kcal), you will get an incorrect result. You can also inadvertently specify Kelvin (K) instead of Celsius (° C).

Advanced home heat loss calculation

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 the air, which infiltrates through leaks in the protective structures (enclosing structures) of the given premises. There are other types of heat loss in industrial buildings. The calculation of the heat loss of the premises is carried out for all enclosing structures of all heated premises. Heat losses through internal structures may not be taken into account, if the temperature difference in them with the temperature of neighboring rooms is up to 3C. Heat loss through the enclosing structures is calculated according to the following formula, W:

Qlim = F (tvn - tnB) (1 + Σ β) n / Rо

tnB- outdoor air temperature, оС;
tvn- indoor temperature, оС;
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 loss, shares of the main;
Ro- resistance to heat transfer, m2 oC / W, which is determined by the following formula:

Rо = 1 / αv + Σ (δі / λі) + 1 / αн + Rv.p., where

αw - coefficient of heat absorption of the fence (its inner surface), W / m2 · о С;
λі and δі - the calculated coefficient of thermal conductivity for the material of this layer of the structure and the thickness of this layer;
αн - coefficient of heat transfer of the enclosure (its outer surface), W / m2 · о С;
Rv.n - if there is a closed air gap in the structure, its thermal resistance, m2 o C / W (see Table 2).
Coefficients αн and αв are accepted according to SNiP and for some cases are given in table 1;
δі - usually assigned according to the assignment or determined according to the drawings of the enclosing structures;
λі - is taken according to reference books.

Table 1. Coefficients of heat absorption αw and heat transfer αн

The surface of the enclosing structure	αw, W / m2 o С	αн, W / m2 о С
Internal surface of floors, walls, smooth ceilings
Surface outer wall, attic ceilings
Attic ceilings and ceilings over unheated basements with light openings
Ceilings over unheated basements without skylights

Table 2. Thermal resistance of closed air layers Rv.n, m2 · о С / W

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

For doors and windows, resistance to heat transfer is calculated very rarely, and more often it is 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. The temperature tvn for residential buildings is selected from Appendix i, tnB - from Appendix 2 to SNiP, depending on the location of the construction site. Additional heat losses are shown in Table 3, coefficient n - in Table 4.

Table 3. Additional heat losses

Fencing, its type	Conditions	Additional heat loss β
Windows, doors and external vertical walls:	orientation northwest east, north and northeast
Windows, doors and external vertical walls:	west and southeast
External doors, doors with vestibules 0.2 N without air curtain at a 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
Corner rooms additionally for windows, doors and walls	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 outside infiltrated air in public and residential buildings for all types of premises is determined by two calculations. The first calculation determines the consumption of heat energy Qі for heating the outside air, which enters the і-th room as a result of the action of natural exhaust ventilation... The second calculation determines the consumption of thermal energy Qі for heating the outside air, which penetrates into the given room through leaks in the enclosures as a result of wind and (or) thermal pressure. For the calculation, take the largest amount of heat loss determined by the following equations (1) and (or) (2).

Qі = 0.28 L ρн с (tвн - tнБ) (1)

L, m3 / h c - the flow rate of air removed from the premises, for residential buildings, take 3 m3 / hour per 1 m2 of the area of residential premises, including the kitchen;
with- specific heat capacity of air (1 kJ / (kg oC));
ρн- air density outside the premises, kg / m3.

Specific gravity 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 is the air temperature.

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

Qі = 0.28 Gі s (tvn - tnB) k, (2)

where k is the coefficient taking into account the occasional heat flow, for single-binding balcony doors and windows are taken as 0.8, for single and double-bound windows - 1.0;
Gі - flow rate of air penetrating (infiltrating) through protective structures (enclosing structures), kg / h.

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

Gі = 0.216 Σ F Δ Pi 0.67 / Ri, kg / h

where Δ Рі is the difference in air pressures on the internal Рвн and external Рн surfaces of doors or windows, Pa;
Σ F, m2 - calculated areas of all building fences;
Ri, m2 · h / kg - resistance to air permeability of this fence, which can be taken in accordance with Appendix 3 of SNiP. In panel buildings, in addition, it is determined additional expense air infiltrating through leaks in the joints of the panels.

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

Δ Рі = (H - hі) (γн - γвн) + 0.5 ρн V2 (ce, n - ce, р) k1 - ріnt,
where H, m - building height from zero level to the mouth of the ventshakhty (in non-attic buildings, the mouth is usually located 1 m above the roof, and in buildings with an attic - 4–5 m above the ceiling of the attic);
hі, m - the height from the zero level to the top of the balcony doors or windows, for which the air flow rate is calculated;
γн, γвн - specific weights of external and internal air;
se, pu se, n - aerodynamic coefficients for the leeward and windward surfaces of the building, respectively. For rectangular buildings ce, r= –0.6, ce, n = 0.8;

V, m / s - wind speed, which is taken for calculation in accordance with Appendix 2;
k1 is a coefficient that takes into account the dependence of the high-speed wind pressure and the height of the building;
pint, Pa - conditionally constant air pressure, which occurs when ventilation is forced to operate, when calculating residential buildings, pint 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, a height of up to 10 m is 0.65, with a height of up to 20 m - 0.85, and for fences of 20 m and above, 1.1 is taken.

General calculated heat loss in the room, W:

Qcalculated = Σ Qlim + Qunf - Qbyt

where Σ Qlim - total heat losses through all protective enclosures of the room;
Qinf - the maximum heat consumption for heating the air, which is infiltrated, taken from the calculations according to formulas (2) u (1);
Qbyt - all heat emissions from household electrical appliances, lighting, other possible sources of heat, which are accepted for kitchens and living quarters in the amount of 21 W per 1 m2 of the estimated area.

Vladivostok -24.
Vladimir -28.
Volgograd -25.
Vologda -31.
Voronezh -26.
Yekaterinburg -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.

Today many families choose for themselves Vacation home as a place permanent residence or year-round recreation. However, its maintenance, and especially the payment of utilities, is quite costly, and most homeowners are not oligarchs at all. One of the most significant expense items for any homeowner is heating costs. To minimize them, it is necessary to think about energy saving at the stage of building a cottage. Let's consider this issue in more detail.

« The problems of energy efficiency of housing are usually remembered from the perspective of urban housing and communal services, but the owners individual houses this topic is sometimes much closer,- considers Sergey Yakubov , Deputy Director for Sales and Marketing, a leading manufacturer of roofing and facade systems in Russia. - The cost of heating a house can be much more than half of the cost of its maintenance in the cold season and sometimes reach tens of thousands of rubles. However, with a competent approach to the thermal insulation of a residential building, this amount can be significantly reduced.».

Actually, you need to heat the house in order to constantly maintain a comfortable temperature in it, regardless of what is happening on the street. In this case, it is necessary to take into account the heat loss both through the enclosing structures and through the ventilation, because heat leaves along with heated air, which is replaced by cooled air, as well as the fact that some of the heat is emitted by people in the house, Appliances, incandescent lamps, etc.

To understand how much heat we should get from our heating system and how much money will have to be spent on it, we will try to assess the contribution of each of the other factors to the heat balance using the example of a brick building located in the Moscow region. two-story house with a total area of 150 m2 (to simplify calculations, we assumed that the dimensions of the cottage in terms of about 8.7x8.7 m and it has 2 floors with a height of 2.5 m each).

Heat loss through enclosing structures (roof, walls, floor)

The intensity of heat loss is determined by two factors: the difference in temperatures inside and outside the house and the resistance of its enclosing structures to heat transfer. Dividing the temperature difference Δt by the coefficient of resistance to heat transfer Ro of walls, roofs, floors, windows and doors and multiplying by the area S of their surface, we can calculate the intensity of heat loss Q:

Q = (Δt / R o) * S

The temperature difference Δt is not a constant value, it changes from season to season, during the day, depending on the weather, etc. However, our task is simplified by the fact that we need to estimate the demand for heat in total for the year. Therefore, for an approximate calculation, we can easily use such an indicator as the average annual air temperature for the selected area. For the Moscow region it is + 5.8 ° C. If we take + 23 ° C as a comfortable temperature in the house, then our average difference will be

Δt = 23 ° C - 5.8 ° C = 17.2 ° C

Walls. The area of the walls of our house (2 square floors 8.7x8.7 m 2.5 m high) will be approximately equal to

S = 8.7 * 8.7 * 2.5 * 2 = 175 m 2

However, from this you need to subtract the area of windows and doors, for which we will calculate the heat loss separately. Let's pretend that Entrance door we have one standard size 900x2000 mm, i.e. area

S door = 0.9 * 2 = 1.8 m 2,

and windows - 16 pieces (2 on each side of the house on both floors) measuring 1500x1500 mm, the total area of which will be

S windows = 1.5 * 1.5 * 16 = 36 m 2.

Total - 37.8 m 2. Remaining area brick walls -

S walls = 175 - 37.8 = 137.2 m 2.

The coefficient of resistance to heat transfer of a wall in 2 bricks is 0.405 m2 ° C / W. For simplicity, we will neglect the heat transfer resistance of the plaster layer that covers the walls of the house from the inside. Thus, the heat dissipation of all walls of the house will be:

Q walls = (17.2 ° C / 0.405m 2 ° C / W) * 137.2 m 2 = 5.83 kW

Roof. For simplicity of calculations, we will assume that the heat transfer resistance roofing cake equal to the heat transfer resistance of the insulation layer. For lightweight mineral wool insulation with a thickness of 50-100 mm, most often used for insulating roofs, it is approximately equal to 1.7 m 2 ° C / W. We will neglect the heat transfer resistance of the attic floor: let's say that there is an attic in the house, which communicates with other rooms and the heat is evenly distributed between all of them.

Square gable roof with a slope of 30 ° will be

S roof = 2 * 8.7 * 8.7 / Cos30 ° = 87 m 2.

Thus, its heat release will be:

Q of the roof = (17.2 ° C / 1.7m 2 ° C / W) * 87 m 2 = 0.88 kW

Floor. The heat transfer resistance of a wooden floor is approximately 1.85 m2 ° C / W. Making similar calculations, we get the heat release:

Floor Q = (17.2 ° C / 1.85m 2 ° C / W) * 75 2 = 0.7 kW

Doors and windows. Their resistance to heat transfer is approximately equal to respectively 0.21 m 2 ° C / W (double wooden door) and 0.5 m 2 ° C / W (ordinary double-glazed window, without additional energy-efficient "gadgets"). As a result, we get heat dissipation:

Q door = (17.2 ° C / 0.21W / m 2 ° C) * 1.8m 2 = 0.15 kW

Q window = (17.2 ° C / 0.5m 2 ° C / W) * 36m 2 = 1.25 kW

Ventilation. By building codes the air exchange coefficient for a dwelling should be at least 0.5, and better - 1, i.e. in an hour, the air in the room must be completely renewed. Thus, with a ceiling height of 2.5 m, this is approximately 2.5 m 3 of air per hour per square meter of area. This air must be heated from outside temperature(+ 5.8 ° C) to room temperature (+ 23 ° C).

The specific heat capacity of air is the amount of heat required to raise the temperature of 1 kg of a substance by 1 ° C - it is approximately 1.01 kJ / kg ° C. In this case, the air density in the temperature range of interest to us is approximately 1.25 kg / m 3, i.e. the mass of 1 cubic meter is 1.25 kg. Thus, to heat the air by 23-5.8 = 17.2 ° C for each square meter of area, you will need:

1.01 kJ / kg ° C * 1.25 kg / m 3 * 2.5 m 3 / hour * 17.2 ° C = 54.3 kJ / hour

For a house with an area of 150 m2, this will be:

54.3 * 150 = 8145 kJ / hour = 2.26 kW

Summarize

Heat loss through	Temperature difference, ° C	Area, m2	Heat transfer resistance, m2 ° C / W	Heat loss, kW
Walls	17,2	175	0,41	5,83
Roof	17,2	87	1,7	0,88
Floor	17,2	75	1,85	0,7
Doors	17,2	1,8	0,21	0,15
Window	17,2	36	0,5	0,24
Ventilation	17,2	-	-	2,26
Total:				11,06

Let's breathe now!

Suppose there is a family of two adults with two children living in a house. The nutritional norm of an adult is 2,600-3,000 calories per day, which is equivalent to a heat dissipation power of 126 watts. The heat release of a child will be estimated at half the heat release of an adult. If everyone lived at home are in it 2/3 of the time, then we get:

(2 * 126 + 2 * 126/2) * 2/3 = 252 W

Let's say that there are 5 rooms in a house, illuminated with ordinary 60 W incandescent lamps (not energy-saving), 3 per room, which are turned on on average 6 hours a day (i.e. 1/4 of the total time). Approximately 85% of the power consumed by the lamp is converted into heat. Total we get:

5 * 60 * 3 * 0.85 * 1/4 = 191 W

Refrigerator - very efficient heating device... Its heat dissipation is 30% of the maximum power consumption, i.e. 750 watts

Other household appliances (let it be washing and dishwasher) generates in the form of heat about 30% of the maximum power consumption. The average power of these devices is 2.5 kW, they work for about 2 hours a day. This gives us a total of 125 watts.

A standard electric stove with an oven has a power of about 11 kW, but a built-in limiter regulates the operation heating elements so that their simultaneous consumption does not exceed 6 kW. However, it is unlikely that we ever use more than half of the burners at the same time or all oven heating elements at once. Therefore, we will proceed from the fact that the average working power of the stove is approximately 3 kW. If it works 3 hours a day, then we get 375 watts of heat.

Each computer (and there are 2 of them in the house) emits about 300 watts of heat and works 4 hours a day. Total - 100 watts.

TV is 200 W and 6 hours a day, i.e. per circle - 50 watts.

In total we get: 1.84 kW.

Now we calculate the required heat output heating systems:

Heating Q = 11.06 - 1.84 = 9.22 kW

Heating costs

Actually, above we calculated the power that will be needed to heat the coolant. And we will heat it, of course, with the help of a boiler. Thus, heating costs are the fuel costs for this boiler. Since we are considering the most general case, we will make a calculation for the most universal liquid (diesel) fuel, since gas mains are far from everywhere (and the cost of supplying them is a figure with 6 zeros), and solid fuel must, firstly, be brought in somehow, and secondly, every 2-3 hours it must be thrown into the boiler furnace.

To find out what volume V of diesel fuel per hour we have to burn to heat the house, we need to multiply the specific heat of its combustion q (the amount of heat released when burning a unit of mass or volume of fuel, for diesel fuel - about 13.95 kW * h / l) multiplied by The boiler efficiency η (about 0.93 for diesel) and then the required heating system power Q heating (9.22 kW) divided by the resulting figure:

V = Q heating / (q * η) = 9.22 kW / (13.95 kW * h / l) * 0.93) = 0.71 l / h

With an average cost of diesel fuel for the Moscow region of 30 rubles / liter per year, we will spend on heating a house

0.71 * 30 rubles * 24 hours * 365 days = 187 thousand rubles. (rounded off).

How to save money?

The natural desire of any homeowner is to reduce heating costs even during the construction phase. Where does it make sense to invest money?

First of all, one should think about the insulation of the facade, which, as we saw earlier, accounts for the bulk of all heat losses at home. In general, external or internal additional insulation can be used for this. However, internal insulation is much less effective: when installing thermal insulation from the inside, the boundary between the warm and cold areas "moves" inside the house, i.e. moisture will condense in the walls.

There are two ways to insulate facades: "wet" (plaster) and by installing a hinged ventilated facade. Practice shows that due to the need for constant repair, "wet" insulation, taking into account operating costs, turns out to be almost twice as expensive as a ventilated facade. The main disadvantage of a plaster facade is the high cost of its maintenance and maintenance. " The initial cost of arranging such a facade is lower than for a hinged ventilated one, by only 20-25%, by a maximum of 30%,- explains Sergey Yakubov (Metal Profile). - However, taking into account the costs of Maintenance, which must be done at least once every 5 years, after the end of the first five years plaster facade will be equal in cost to the ventilated one, and in 50 years (the service life of the ventilated facade) it will be 4-5 times more expensive».

What is a hinged ventilated facade? This is an external "screen" attached to a light metal frame, which is attached to the wall with special brackets. A light insulation is placed between the wall of the house and the screen (for example, Isover "VentFasad Niz" with a thickness of 50 to 200 mm), as well as a wind-waterproof membrane (for example, Tyvek Housewrap). Can be used as external cladding various materials, but in individual construction, steel siding is most often used. " The use of modern high-tech materials in the production of siding, such as steel coated with Colorcoat Prisma ™, allows you to choose almost any design solution,- says Sergey Yakubov. - This material has excellent resistance to both corrosion and mechanical stress. The warranty period for it is 20 years with real time operation for 50 years or more. Those. subject to the use of steel siding all facade structure will last 50 years without repair».

Additional layer facade insulation mineral wool has a heat transfer resistance of approximately 1.7 m2 ° C / W (see above). In construction, to calculate the heat transfer resistance of a multi-layer wall, add the corresponding values for each of the layers. As we remember, our main bearing wall in 2 bricks has a heat transfer resistance of 0.405 m2 ° C / W. Therefore, for a wall with a ventilated facade we get:

0.405 + 1.7 = 2.105 m 2 ° C / W

Thus, after insulation, the heat dissipation of our walls will be

Q facade = (17.2 ° C / 2.105m 2 ° C / W) * 137.2 m 2 = 1.12 kW,

which is 5.2 times less than the same indicator for a non-insulated facade. Impressive, isn't it?

Let's calculate the required heat output of the heating system again:

Q heating-1 = 6.35 - 1.84 = 4.51 kW

Diesel fuel consumption:

V 1 = 4.51 kW / (13.95 kW * h / l) * 0.93) = 0.35 l / h

Heating amount:

0.35 * 30 rubles. * 24 hours * 365 days = 92 thousand rubles.

Energy efficient building renovation will help to save heat energy and increase the comfort of living. The greatest savings potential lies in the good thermal insulation of the outer walls and roof. The easiest way to assess opportunities effective repair Is the consumption of thermal energy. If more than 100 kWh of electricity is consumed per year (10 m³ natural gas) per square meter of heated area, including wall area, then energy-efficient renovation can be beneficial.

Heat loss through the outer shell

The basic concept of an energy efficient building is a continuous layer of thermal insulation over the heated surface of the house circuit.

Roof. With a thick layer of thermal insulation, heat loss through the roof can be reduced;

Important! In timber structures, heat-shielding roof sealing is difficult, as the timber swells and can be damaged by high humidity.

Walls. As with a roof, heat loss is reduced with the use of a special coating. In the case of internal wall insulation, there is a risk that condensation will collect behind the insulation if the humidity in the room is too high;

Floor or basement. For practical reasons, thermal insulation is done from the inside of the building;
Thermal bridges. Thermal bridges are unwanted cooling fins (heat conductors) outside the building. For example, a concrete floor, which is also a balcony floor. Many thermal bridges are in the soil, parapets, window and door frames... There are also temporary thermal bridges if the wall parts are secured metal elements... Thermal bridges can account for a significant portion of heat loss;
Window. Over the past 15 years, thermal insulation window glass improved 3 times. Today's windows have a special reflective layer on the glass, which reduces radiation losses, this is one thing - and double-glazed windows;
Ventilation. A typical building has air leaks, especially in the area of windows, doors and on the roof, which provides the necessary air exchange. However, in the cold season, this causes significant heat loss at home from the outgoing heated air. Good ones modern buildings are sufficiently airtight, and it is necessary to regularly ventilate the premises by opening the windows for a few minutes. In order to reduce heat loss due to ventilation, more and more comfortable ventilation systems... This type of heat loss is estimated at 10-40%.

Thermographic footage of a poorly insulated building gives an idea of how much heat is lost. This is very good tool for quality control of renovations or new construction.

Methods for assessing heat loss at home

There are complex calculation methods that take into account various physical processes: convection exchange, radiation, but they are often unnecessary. Usually, simplified formulas are used, and if necessary, you can add 1-5% to the result. The orientation of the building is taken into account in new buildings, but solar radiation also does not significantly affect the calculation of heat loss.

Important! When applying formulas for calculating heat energy losses, the time spent by people in a particular room is always taken into account. The smaller it is, the lower the temperature indicators should be taken as a basis.

Average values. The most approximate method is not accurate enough. There are tables compiled for individual regions, taking into account climatic conditions and average building parameters. For example, for a specific area, the power value in kilowatts is indicated, which is required to heat 10 m² of the area of a room with 3 m high ceilings and one window. If the ceilings are lower or higher and there are 2 windows in the room, the power readings are adjusted. This method does not take into account the degree of thermal insulation of the house at all and will not save thermal energy;
Calculation of heat loss of the building envelope. The area is summed up outer walls minus the dimensions of the areas of windows and doors. Additionally, there is a roof area with a floor. Further calculations are carried out according to the formula:

Q = S x ΔT / R, where:

S is the area found;
ΔT is the difference between indoor and outdoor temperatures;
R is the resistance to heat transfer.

The results obtained for the walls, floor and roof are combined. Then ventilation losses are added.

Important! Such calculation of heat loss will help determine the boiler capacity for the building, but will not allow calculating the number of radiators per room.

Calculation of heat loss by room. Using a similar formula, the losses are calculated for all rooms in the building separately. Then the heat loss for ventilation is found by determining the volume of the air mass and the approximate number of times a day it changes in the room.

Important! When calculating ventilation losses, it is imperative to take into account the purpose of the room. Enhanced ventilation is required for the kitchen and bathroom.

An example of calculating the heat loss of a residential building

The second method of calculation is used, only for the external structures of the house. Up to 90 percent of thermal energy goes through them. Accurate results are important in order to select the correct boiler for efficient heat output without overheating the premises. It is also an indicator of the economic efficiency of the selected materials for thermal protection, showing how quickly you can recoup the cost of their purchase. Simplified calculations for a building without a multilayer thermal insulation layer.

The house has an area of 10 x 12 m and a height of 6 m. The walls are 2.5 bricks (67 cm) thick, covered with plaster, with a layer of 3 cm. The house has 10 windows 0.9 x 1 m and a door 1 x 2 m.

Calculation of the resistance to heat transfer of walls:

R = n / λ, where:

n - wall thickness,
λ - specific thermal conductivity (W / (m ° C).

This value is searched for in the table for its material.

For bricks:

Rkir = 0.67 / 0.38 = 1.76 m2 ° C / W.

For plastering:

Rpc = 0.03 / 0.35 = 0.086 m2 ° C / W;

Total value:

Rst = Rkir + Rsht = 1.76 + 0.086 = 1.846 m2 ° C / W;

Calculating the area of external walls:

Total area of external walls:

S = (10 + 12) x 2 x 6 = 264 sq.m.

Window and doorway area:

S1 = ((0.9 x 1) x 10) + (1 x 2) = 11 sq.m.

Adjusted wall area:

S2 = S - S1 = 264 - 11 = 253 sq.m.

Heat losses for walls will be determined by:

Q = S x ΔT / R = 253 x 40 / 1.846 = 6810.22 W.

Important! The ΔT value is taken arbitrarily. For each region in the tables, you can find the average value of this value.

At the next stage, heat losses through the foundation, windows, roof, door are calculated in an identical way. When calculating the heat loss index for the foundation, a smaller temperature difference is taken. Then you need to add up all the numbers obtained and get the final one.

To determine the possible consumption of electricity for heating, you can present this figure in kWh and calculate it for heating season.

If you use only the number for the walls, it turns out:

per day:

6810.22 x 24 = 163.4 kWh;

per month:

163.4 x 30 = 4903.4 kWh;

for the heating season 7 months:

4903.4 x 7 = 34 323.5 kWh.

When heating is gas, the gas consumption is determined based on its calorific value and coefficient useful action boiler.

Heat losses for ventilation

Find the air volume of a house:

10 x 12 x 6 = 720 m³;

Air mass is found by the formula:

M = ρ x V, where ρ is the air density (taken from the table).

M = 1.205 x 720 = 867.4 kg.

It is necessary to determine the number of times the air changes in the whole house per day (for example, 6 times), and calculate the heat loss for ventilation:

Qw = nxΔT xmx С, where С is the specific heat capacity for air, n is the number of times the air is replaced.

Qw = 6 x 40 x 867.4 x 1.005 = 209217 kJ;

Now it is necessary to convert to kWh. Since there are 3600 kilojoules in one kilowatt-hour, then 209217 kJ = 58.11 kWh

Some calculation methods suggest taking heat losses for ventilation from 10 to 40 percent of the total heat losses, without calculating them using formulas.

To make it easier to calculate the heat loss at home, there are online calculators where you can calculate the result for each room or the whole house. In the proposed fields, you simply enter your data.