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

Below is a pretty simple heat loss calculation buildings, which, nevertheless, will help to accurately determine the power required for heating your warehouse, shopping center or other 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 building envelope, 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 the heat loss individual elements, summarize and add the heat loss during ventilation. All.

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 °C, outside temperature -20 °C.

We consider the area of 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 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) \u003d 0.1 / 0.037 \u003d 2.7 (m2 * K) / W

R (floor) \u003d R (concrete) + R (mineral wool) \u003d 0.057 + 2.7 \u003d 2.76 (m2 * K) / W

roof:

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

window: the value of thermal resistance of windows depends on the type of double-glazed window used
R (windows) \u003d 0.40 (m2 * K) / W for single-chamber glass wool 4–16–4 at? T \u003d 40 ° С

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

Let's calculate the heat loss:

Q (floor) \u003d 800 m2 * 20 ° C / 2.76 (m2 * K) / W \u003d 5797 W \u003d 5.8 kW

Q (roof) \u003d 808 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 7980 W \u003d 8.0 kW

Q (windows) \u003d 63 m2 * 40 ° C / 0.40 (m2 * K) / W \u003d 6300 W \u003d 6.3 kW

Q (walls) \u003d 557 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 5500 W \u003d 5.5 kW

We get that the total heat loss through the building envelope 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 °C to +20 °C, 15.5 W will be required.

Q (1 m3 of air) \u003d 1.4 * 1.0 * 40 / 3.6 \u003d 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 determine the amount of air needed. 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 air heating will require 43 kW.

Final result:

For heating the proposed building, a heating system of the order of 30 kWh is required, and a ventilation system with a capacity of 3000 m3 / h with a heater with a power of 45 kW / h.

The first step in organizing the heating of a private house is the calculation of heat loss. The purpose of this calculation is to find out how much heat escapes outside through walls, floors, roofs and windows (common name - building envelope) 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 heat loss of a building during the cold season consists of 2 parts:

heat loss through enclosing structures;
loss of energy used to heat the ventilation air.

The basic formula for calculating the consumption of thermal energy through external fences is as follows:

Q \u003d 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 construction material, m²°С / W;
S is the area of the outer fence, m²;
t in - internal air temperature, ° С;
t n - 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 is the 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 from 2 materials (for example, a brick with a mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summarized. The outdoor temperature is selected as regulatory documents, and according to personal observations, internal - by necessity. Additional heat losses are the coefficients defined by the standards:

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 outer fence 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. To do this, measurements are made of all fences adjacent to the environment: walls, windows, roofs, floors and doors.

Important point: measurements should be taken 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 \u200b\u200beach structure is calculated and substituted into the first formula (S, m²). The value of R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity of the building material. In the case of new metal-plastic windows, the value of R will be prompted by a representative of the installer.

As an example, it is worthwhile to calculate 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°C, and the plane of the structure is facing north (β = 0.1). First you need to take from the reference literature the coefficient of thermal conductivity of the brick (λ), it is equal to 0.44 W / (m ° C). Then, according to the second formula, the heat transfer resistance of a brick wall of 0.25 m is calculated:

R \u003d 0.25 / 0.44 \u003d 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 \u003d 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) \u003d 434 W \u003d 4.3 kW

If the room has a window, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same actions are repeated for the floors, roof and front door. At the end, all the results are summarized, 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 for heating 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 \u003d 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;
c is the heat capacity of the air mixture, equal to 0.28 W / (kg ° С).

Here, all quantities are known, except for the mass air flow during ventilation of rooms. In order not to complicate your task, you should agree with the condition that the air environment is updated throughout the house 1 time per hour. Then it is not difficult to calculate the volumetric air flow by adding the volumes of all rooms, and then you need to convert it into mass air through density. Since the density of the air mixture varies with its temperature, you need to take appropriate value from the table:

m = 500 x 1.422 = 711 kg/h

Heating such a mass of air by 45°C will require the following amount of heat:

Q air \u003d 0.28 x 711 x 45 \u003d 8957 W, which is approximately equal to 9 kW.

Upon completion of the calculations, the results of heat losses through the external fences are added to the ventilation heat losses, which gives the total heat load on the building's heating system.

The presented calculation methods can be simplified if the formulas are entered in Excel program in the form of tables with data, this will significantly speed up the calculation.

Calculation of heat loss at home - the basis of the heating system. It is needed, at least, 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. understand whether the cost of installing insulation will pay off with fuel savings 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 with standard height ceilings up to three meters. However, this power is not always sufficient to fully replenish heat losses. 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 actual heat loss of the house. How to calculate them - we will tell in this article.

Basic parameters for calculating heat loss

The heat loss of any room depends on three basic parameters:

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

The simplest calculation of heat loss

Qt (kWh)=(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 aggregated indicators, which are based on averaged conditions of 100 W per 1 sq. meter. Where the main calculated indicators for calculating the heating system are the following values:

Qt- thermal power of the proposed heater on waste oil, kW / h.

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

S- area of the room;

K1- window heat loss coefficient:

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

K2- coefficient of heat loss of walls:

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

K3 the ratio of the areas 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
-35oC K4=1.5;

K5- the number of walls facing the outside:

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

K6- type of room, which 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- temperature delta (outdoor and indoor)
k- dispersion coefficient

k= 3.0-4.0 - without thermal insulation. (Simplified wooden structure or corrugated metal sheet structure).
k \u003d 2.0-2.9 - small thermal insulation. (Simplified building design, single brickwork, simplified design of windows and roof).
k \u003d 1.0-1.9 - average thermal insulation. ( Standard design, double brickwork, few windows, standard roof).
k \u003d 0.6-0.9 - high thermal insulation. (Improved construction, double-insulated brick walls, few double-pane windows, thick subfloor, high quality insulating material roof).

In this formula, the dispersion coefficient is taken into account very conditionally and it is not entirely clear which coefficients to use. In the classics, a rare modern, 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.

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

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

∆T- 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 the premises is taken:

Residential premises 22C
Non-residential 18C
Zones of water procedures 33С

When it comes to a multilayer structure, the resistances of the layers of the structure add up. Separately, I want to focus your attention on 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 construction layer, we can easily obtain layer heat transfer resistance:

δ - layer thickness, m;

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

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 calculated room and on the street (the temperature of the coldest five-day period is °C.). ∆T
3. Fencing area F (Separate walls, windows, doors, ceiling, floor)
4. The orientation of the building in relation to the cardinal points.

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

Qlimit=(ΔT / Rlimit)* Flimit * n *(1+∑b)

Qlimit- heat loss through the building envelope, W
Rogr– resistance to heat transfer, m.sq.°C/W; (If there are several layers, then ∑ Rlimit of layers)
Fogr– area of the enclosing structure, m;
n- the coefficient of contact of the building envelope with the outside air.

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

(1+∑b) – additional heat losses as a share of the main losses. Additional heat losses b through the building envelope should be taken as a fraction 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, 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 is facing north, east, northeast and northwest, and 0.1 - in other cases;

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

c) through the unheated floors of the first floor above the cold undergrounds of buildings in areas with an estimated outdoor 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 curtains, at the height of buildings H, m, from the average planning mark of the earth to the top of the eaves, the center of the exhaust holes of the lantern or the mouth of the mine 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 a vestibule; 0.22 H - for single doors;

e) through external gates not equipped with air and air-thermal curtains - in the amount of 3 in the absence of a vestibule and in the amount of 1 - in the presence of a vestibule at the gate.

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

Separately, we take such an element as a floor on the ground or on logs. There are features 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 as Rn.p, (m2 °C) / W. For each zone of an uninsulated floor, standard values of resistance to heat transfer are provided:

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

The first three zones are strips located 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 located at the junction of the floor to the outer wall. If an uninsulated floor adjoins a wall buried in the ground, then the beginning is transferred to the upper boundary of the wall penetration. If there are insulating layers in the structure of the floor located on the ground, it is called insulated, and its resistance to heat transfer Rу.p, (m2 оС) / W, is determined by the formula:

Ru.p. = Rn.p. + Σ (γc.s. / λc.s)

Rn.p- resistance to heat transfer of the considered zone of the non-insulated floor, (m2 °C) / W;
γy.s- thickness of the insulating layer, m;
λu.s- coefficient of thermal conductivity of the material of the insulating layer, W / (m ° C).

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

Rl \u003d 1.18 * Ry.p

The heat loss of each enclosing structure is considered 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 the measurements. If instead of (W) appears (kW) or in general (kcal), you will get an incorrect result. You can also inadvertently indicate Kelvins (K) instead of degrees Celsius (°C).

Advanced home heat loss calculation

Heating in civil and residential buildings heat losses of premises consist of heat losses through various enclosing structures, such as windows, walls, ceilings, floors, as well as heat consumption for heating air, which infiltrates 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 of the room is made for all enclosing structures of all heated rooms. 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 losses through the building envelope are calculated according to the following formula, W:

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

tnB- outdoor air temperature, °C;
tvn- temperature in the room, °C;
F is the 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, shares from the main ones;
Ro- resistance to heat transfer, m2 °C / W, which is determined by the following formula:

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

αv - coefficient of heat absorption of the fence (its inner surface), W / m2 o C;
λі and δі are the design coefficient of thermal conductivity for the material of a given layer of the structure and the thickness of this layer;
αn - heat transfer coefficient of the fence (its outer surface), W/ m2 o C;
Rv.n - in the case of 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 task or determined from the drawings of enclosing structures;
λі - taken from directories.

Table 1. Heat absorption coefficients αv and heat transfer coefficients αn

The surface of the building envelope	αw, W/ m2 o C	αn, W/ m2 o C
Interior surface of floors, walls, smooth ceilings
Surface exterior walls, bare floors
Attic ceilings and ceilings over unheated basements with light openings
Ceilings over unheated basements without light openings

Table 2. Thermal resistance of closed air spaces Rv.n, m2 o C / 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 a temperature in the space of the air gap

For doors and windows, heat transfer resistance is calculated very rarely, but 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 of SNiP, depending on the location of the construction site. Additional heat losses are indicated in Table 3, the 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 building height H, m	triple doors with two vestibules
	double doors with vestibule
Corner rooms optional for windows, doors and walls	one of the fences is oriented to the east, north, northwest or northeast
Corner rooms optional 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 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 Qі 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 Qі for heating the outside air, which penetrates into this room through the leaks of the fences as a result of wind and (or) thermal pressure. For calculation, the largest heat loss is taken from those determined by the following equations (1) and (or) (2).

Qi = 0.28 L ρn s (tin – tnB) (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 kitchens;
With– specific heat capacity of air (1 kJ /(kg oC));
ρn– air density outside the room, 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 , ° s 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 pressure is determined according to the formula:

Qі = 0.28 Gі s (tin – tnB) k, (2)

where k is a coefficient that takes into account the counter heat flux, for separate-binding balcony doors and windows is taken 0.8, for single and double-binding windows - 1.0;
Gі is the flow rate of air penetrating (infiltrating) through protective structures (enclosing structures), kg/h.

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

Gi = 0.216 Σ F Δ Рі 0.67 / Ri, kg/h

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

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

Δ Рі= (H - hі) (γн - γin) + 0.5 ρн V2 (сe,n - ce,р) k1 - ріnt,
where H, m is the height of the building from zero level to the mouth of the ventilation shaft (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 attic floor);
hі, m - height from zero level to the top of balcony doors or windows for which the air flow rate is calculated;
γn, γin – specific weights of outdoor and indoor air;
ce, ru ce, n - aerodynamic coefficients for the leeward and windward surfaces of the building, respectively. For rectangular buildings se, p= –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 wind pressure and the height of the building;
ріnt, Pa - conditionally constant air pressure, which occurs when ventilation is operated with forced impulse, 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, with a height of up to 10 m it 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.

Total calculated heat loss in the room, W:

Qcalc \u003d Σ Qlimit + Qunf - Qlife

where Σ Qlimit - total heat loss through all protective enclosures of the room;
Qinf is the maximum heat consumption for heating the air that is infiltrated, taken from calculations according to formulas (2) u (1);
Qlife - all heat generation from domestic electrical appliances, lighting, other possible heat sources that are accepted for kitchens and living quarters in the amount of 21 W per 1 m2 of the calculated 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 like a place permanent residence or year-round vacation. However, its maintenance, and especially the payment of utilities, is quite expensive, while most homeowners are not oligarchs at all. One of the most significant expenses for any homeowner is the cost of heating. To minimize them, it is necessary to think about energy saving even at the stage of building a cottage. Let's consider this question in more detail.

« The problems of energy efficiency of housing are usually remembered from the perspective of urban housing and communal services, however, 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 the cost of maintaining it 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 heat losses both through the building envelope and through ventilation, because. heat leaves with heated air, which is replaced by cooled air, as well as the fact that a certain amount of heat is emitted by people in the house, Appliances, incandescent lamps, etc.

To understand how much heat we need to get from our heating system and how much money we have to spend on it, let's try to evaluate 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 the calculations, we assumed that the dimensions of the cottage in terms of approximately 8.7x8.7 m and it has 2 floors 2.5 m high).

Heat loss through building envelope (roof, walls, floor)

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

Q \u003d (Δt / R o) * S

The temperature difference Δt is not constant, 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 need for heat in total for the year. Therefore, for an approximate calculation, we may well 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 high 2.5 m) will be approximately equal to

S \u003d 8.7 * 8.7 * 2.5 * 2 \u003d 175 m 2

However, the area of windows and doors must be subtracted from this, 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 doors \u003d 0.9 * 2 \u003d 1.8 m 2,

and windows - 16 pieces (2 on each side of the house on both floors) with a size of 1500x1500 mm, the total area of \u200b\u200bwhich will be

S windows \u003d 1.5 * 1.5 * 16 \u003d 36 m 2.

Total - 37.8 m 2. Remaining area brick walls -

S walls \u003d 175 - 37.8 \u003d 137.2 m 2.

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

Q walls \u003d (17.2 ° C / 0.405 m 2 ° C / W) * 137.2 m 2 \u003d 5.83 kW

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

Square gable roof with a slope of 30 ° will be

Roof S \u003d 2 * 8.7 * 8.7 / Cos30 ° \u003d 87 m 2.

Thus, its heat dissipation will be:

Roof Q \u003d (17.2 ° C / 1.7 m 2 ° C / W) * 87 m 2 \u003d 0.88 kW

Floor. The heat transfer resistance of a wooden floor is approximately 1.85 m2°C/W. Having made similar calculations, we obtain heat dissipation:

Q floor = (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 0.21 m 2 °C / W, respectively (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 windows \u003d (17.2 ° C / 0.5 m 2 ° C / W) * 36 m 2 \u003d 1.25 kW

Ventilation. By building codes the air exchange coefficient for a dwelling should be at least 0.5, and preferably 1, i.e. in an hour, the air in the room should be completely updated. Thus, with a ceiling height of 2.5 m, this is approximately 2.5 m 3 of air per hour per square meter. This air must be heated from outdoor 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 - approximately 1.01 kJ / kg ° C. At the same time, the air density in the temperature range of interest to us is approximately 1.25 kg/m3, i.e. the mass of 1 cubic meter of it 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 of 150 m2, this will be:

54.3 * 150 \u003d 8145 kJ / h \u003d 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 a family of two adults with two children lives in a house. The nutritional norm for an adult is 2600-3000 calories per day, which is equivalent to a heat dissipation power of 126 watts. The heat dissipation of a child will be estimated at half the heat dissipation of an adult. If everyone who lived at home is in it 2/3 of the time, then we get:

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

Let's say that there are 5 rooms in the house, lit by ordinary incandescent lamps with a power of 60 W (not energy-saving), 3 per room, which are turned on for an average of 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. In total we get:

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

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

Other household appliances (let it be washing and dishwasher) releases about 30% of the maximum power input as heat. The average power of these devices is 2.5 kW, they work for about 2 hours a day. Total we get 125 watts.

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

Each computer (and there are 2 in the house) emits approximately 300 W 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 thermal power 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 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 pipelines are far from being everywhere (and the cost of their summing up is a figure with 6 zeros), and solid fuel must, firstly, be brought somehow, and secondly, thrown into the boiler furnace every 2-3 hours.

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

V = heating Q / (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 per liter per year, it will take us

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

How to save?

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

First of all, you should think about the insulation of the facade, which, as we saw earlier, accounts for the bulk of all heat loss at home. In the general case, 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 thickness of 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 repairs, “wet” insulation, taking into account operating costs, ends up being almost twice as expensive as a ventilated facade. The main disadvantage of the plaster facade is the high cost of its maintenance and upkeep. " The initial costs for the arrangement of such a facade are lower than for a hinged ventilated one, by only 20-25%, a maximum of 30%,- explains Sergey Yakubov ("Metal Profile"). - However, considering the cost of Maintenance, which must be done at least once every 5 years, already after the first five years plaster facade will be equal in cost to a ventilated one, and in 50 years (the service life of a ventilated facade) it will be 4-5 times more expensive than it».

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 "VentFacade Bottom" with a thickness of 50 to 200 mm), as well as a wind and hydroprotective membrane (for example, Tyvek Housewrap). Can be used as exterior 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 real time operation for 50 years or more. Those. subject to the use of steel siding all facade construction will last 50 years without repair».

Extra layer facade insulation from mineral wool has a resistance to heat transfer of approximately 1.7 m2 ° C / W (see above). In construction, to calculate the heat transfer resistance of a multi-layer wall, add up 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 \u003d (17.2 ° C / 2.105 m 2 ° C / W) * 137.2 m 2 \u003d 1.12 kW,

which is 5.2 times less than the same indicator for an uninsulated facade. Impressive, isn't it?

Again we calculate the required heat output of the heating system:

Q heating-1 = 6.35 - 1.84 = 4.51 kW

Diesel fuel consumption:

V 1 \u003d 4.51 kW / (13.95 kW * h / l) * 0.93) \u003d 0.35 l / h

Amount for heating:

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

Energy-efficient reconstruction of the building will help to save thermal energy and increase the comfort of life. 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 (10 m³ natural gas) per square meter of heated area, including wall area, then energy saving renovations can be beneficial.

Heat loss through the outer shell

The basic concept of an energy-saving building is a continuous layer of thermal insulation over the heated surface of the house contour.

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

Important! In wooden structures, the thermal seal of the roof is difficult, as the wood swells and can be damaged by high humidity.

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

Floor or basement. For practical reasons, thermal insulation is made from inside the building;
thermal bridges. Thermal bridges are unwanted cooling fins (heat conductors) on the outside of a building. For example, a concrete floor, which is also a balcony floor. Many thermal bridges are found in soil areas, parapets, window and door frames. There are also temporary thermal bridges if the wall details are fixed 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, these are single- and double-glazed windows;
Ventilation. A typical building has air leaks, especially around windows, doors and on the roof, which provides the necessary air exchange. However, during the cold season, this causes significant heat loss from the house from the outgoing heated air. Good modern buildings are sufficiently airtight, and it is necessary to regularly ventilate the premises by opening windows for a few minutes. To reduce heat loss through ventilation, comfort panels are increasingly being installed. ventilation systems. This type of heat loss is estimated at 10-40%.

Thermographic surveys in a poorly insulated building give an idea of how much heat is being wasted. This is very good tool for quality control of repair or new construction.

Ways to assess heat loss at home

There are complex calculation methods that take into account various physical processes: convection exchange, radiation, but they are often redundant. Simplified formulas are usually used, and if necessary, 1-5% can be added to the result. Building orientation 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 losses, the time spent by people in a particular room is always taken into account. The smaller it is, the lower temperature indicators should be taken as a basis.

Average values. The most approximate method does not have sufficient accuracy. 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 room area 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 indicators 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 enclosing contour of the building. Summarized area external 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 found area;
ΔT is the difference between indoor and outdoor temperatures;
R is the resistance to heat transfer.

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

Important! Such a calculation of heat losses will help determine the boiler capacity for the building, but will not allow you to calculate the number of radiators per room.

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

Important! When calculating ventilation losses, it is necessary to take into account the purpose of the room. The kitchen and bathroom need enhanced ventilation.

An example of calculating the heat loss of a residential building

The second calculation method is used, only for the external structures of the house. Through them, up to 90 percent of thermal energy is lost. Accurate results are important in order to select the right boiler to deliver efficient heat without overheating the rooms. 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. The calculations are simplified, 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 resistance to heat transfer of walls:

R = n/λ, where:

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

This value is looked up in the table for its material.

For brick:

Rkir \u003d 0.67 / 0.38 \u003d 1.76 sq.m ° C / W.

For plaster coating:

Rpcs \u003d 0.03 / 0.35 \u003d 0.086 sq.m ° C / W;

Total value:

Rst \u003d Rkir + Rsht \u003d 1.76 + 0.086 \u003d 1.846 sq.m ° C / W;

Calculation of the area of external walls:

Total area of external walls:

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

Area of windows and doorway:

S1 \u003d ((0.9 x 1) x 10) + (1 x 2) \u003d 11 sq.m.

Adjusted wall area:

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

Heat losses for walls will be determined by:

Q \u003d S x ΔT / R \u003d 253 x 40 / 1.846 \u003d 6810.22 W.

Important! The value of ΔT 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, and 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 sum up all the received numbers and get the final one.

To determine the possible consumption of electricity for heating, you can represent 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 of 7 months:

4903.4 x 7 \u003d 34,323.5 kWh.

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

Heat losses for ventilation

Find the air volume of the house:

10 x 12 x 6 = 720 m³;

The mass of air is found by the formula:

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

M \u003d 1, 205 x 720 \u003d 867.4 kg.

It is necessary to determine the figure, how many times the air in the whole house is replaced per day (for example, 6 times), and calculate the heat loss for ventilation:

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

Qv \u003d 6 x 40 x 867.4 x 1.005 \u003d 209217 kJ;

Now we need 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 facilitate the calculation of heat loss at home, there are online calculators where you can calculate the result for each room or the entire house. You simply enter your data in the proposed fields.