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

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

Q – heat loss, W

S – structure area, m2

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

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

The calculation scheme is as follows: we calculate the heat loss of individual elements, sum it up and add 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. Room temperature 20 °C, outside temperature-20 °C.

We calculate the areas 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 look at 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 you can see .

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

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

Let's do the math 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 find that the total heat loss through the enclosing structures will be:

Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kW/h

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) = 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 – conversion factor to watts.

It remains to decide on the quantity required air. It is believed that during normal breathing a person needs 7 m3 of air per hour. If you use the 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.

Final result:

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

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

Basic parameters for calculating heat loss

Heat loss in any room depends on three basic parameters:

• volume of the room - we are interested in the volume of air that needs to be heated
• 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 and windows to retain heat

The simplest calculation of heat loss

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

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

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

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

S- area of ​​the room;

K1- heat loss coefficient of windows:

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

K2- wall heat loss coefficient:

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

K3 window to floor area ratio:

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

K4- outside temperature coefficient:

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

K5- number of walls facing outside:

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

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

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

K7- room height:

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

Simplified calculation of heat loss at home

Qt = (V x ∆t x k)/860; (kW)

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

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

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

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

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

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

Basically, the internal temperature in the premises is taken:

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

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

δ - layer thickness, m;

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

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

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

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

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

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

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

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

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

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

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

d) through external doors not equipped with air or air-thermal curtains, with a building height of H, m, from the average planning level of the ground to the top of the eaves, center exhaust openings lantern or shaft mouth in 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 vestibule; 0.22 H - for single doors;

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

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

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

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

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

Rу.п. = Rn.p. + Σ (γу.с. / λу.с.)

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

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

Rl = 1.18 * Rу.п

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

Advanced calculation of heat loss at home

Heating in civil and residential buildings, heat loss of premises consists of heat loss through various enclosing structures, such as windows, walls, ceilings, floors, as well as heat consumption for heating air, which is infiltrated through leaks in the protective structures (enclosing structures) of a given room. There are other types of heat loss in industrial buildings. Calculation of heat loss of the room is carried out for all enclosing structures of all heated rooms. Heat loss through internal structures, when the temperature difference in them with the temperature of neighboring rooms is up to 3C. Heat loss through the enclosing structures is calculated using the following formula, W:

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

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

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

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

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

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

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

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

For doors and windows, heat transfer resistance is calculated very rarely, and is more often taken depending on their design according to reference data and SNiPs. The areas of fences for calculations are determined, as a rule, according to construction drawings. Temperature 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 loss is indicated in Table 3, coefficient n - in Table 4.

Table 3. Additional heat loss

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

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

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

Qі = 0.28 L ρн s (tin – tnB) (1)

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

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

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

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

Qi = 0.28 Gi s (tin – tnB) k, (2)

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

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

Gi = 0.216 Σ F Δ Рі 0.67 / Ri, kg/h

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

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

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

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

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

Total estimated heat loss in the room, W:

Qcalc = Σ Qlim + Qunf – Qbyt

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

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

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

• Prescriptive;
• Consumer.

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

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

Consumer approach(heat transfer resistance can be reduced relative to the prescribed level, provided that the design specific consumption thermal energy for space heating is below standard).

Sanitary and hygienic requirements:

• The difference between indoor and outdoor air temperatures should not exceed certain permissible values. Maximum permissible temperature differences for outer wall 4°C. for roofing and attic flooring 3°C and for ceilings over basements and crawl spaces 2°C.
• The temperature on the inner surface of the fence must be above the dew point temperature.

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

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

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

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

Heat loss mainly depends on:

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

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

Heat transfer resistance will show how much heat will leak through a square meter of structure at a given temperature difference. This question can be formulated differently: what temperature difference will occur when a certain amount of heat passes through a square meter of fencing.

R = ΔT/q.

• q is the amount of heat that escapes through a square meter of wall or window surface. This amount of heat is measured in watts per square meter (W/m2);
• ΔT is the difference between the temperature outside and in the room (°C);
• R is the heat transfer resistance (°C/W/m2 or °C m2/W).

In cases where we are talking about a multilayer structure, the resistance of the layers is simply summed up. For example, the resistance of a wall made of wood, which is lined with brick, is the sum of three resistances: the brick and wooden walls and the air gap between them:

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

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

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

Heat transfer resistance table various materials

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

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

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

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

Note
. Even numbers in symbol double glazed windows indicate air
gap in millimeters;
. The letters Ar mean that the gap is filled not with air, but with argon;
. The letter K means that the outer glass has a special transparent
heat-protective coating.

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

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

Two quantities are involved in calculating heat losses per m2:

• temperature difference ΔT.
• heat transfer resistance R.

Let's say the room temperature is 20 °C. and the outside temperature will be -30 °C. In this case, the temperature difference ΔT will be equal to 50 °C. The walls are made of timber 20 centimeters thick, then R = 0.806 °C m 2 / W.

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

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

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

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

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

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

Example 1.

Corner room (1st floor)

Room characteristics:

• 1st floor.
• room area - 16 m2 (5x3.2).
• ceiling height - 2.75 m.
• There are two external walls.
• material and thickness of the external walls - timber 18 centimeters thick, covered with plasterboard and covered with wallpaper.
• windows - two (height 1.6 m, width 1.0 m) with double glazing.
• floors - wooden insulated. basement below.
• above the attic floor.
• estimated outside temperature -30 °C.
• required room temperature +20 °C.

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

Square internal partitions does not participate in the calculation, since the temperature on both sides of the partition is the same, therefore heat does not escape through the partitions.

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

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

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

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

Example 2

Room under the roof (attic)

Room characteristics:

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

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

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

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

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

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

To make this room suitable for winter living, it is necessary first of all to insulate the walls, side partitions and windows.

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

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

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

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

1. Heat loss through the contact of the foundation with frozen soil is usually assumed to be 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
2. Heat losses associated with ventilation. These losses are calculated taking into account building codes (SNiP). A residential building requires about one air change per hour, that is, during this time it is necessary to supply the same volume of fresh air. Thus, the losses associated with ventilation will be slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation 50%. In European standards for ventilation and wall insulation, the heat loss ratio is 30% and 60%.
3. If the wall “breathes”, like a wall made of timber or logs 15 - 20 centimeters thick, then heat returns. This allows you to reduce heat losses by 30%. therefore, the value of the thermal resistance of the wall obtained during the calculation must be multiplied by 1.3 (or, accordingly reduce heat loss).

By summing up all the heat loss in the house, you can understand what power the boiler and heating appliances are needed to comfortably heat the house on the coldest and windiest days. Also, such calculations will show where the “weak link” is and how to eliminate it using additional insulation.

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

The first step in organizing the heating of a private home is calculating heat loss. The purpose of this calculation is to find out how much heat escapes out through walls, floors, roofing and windows (commonly known as building envelopes) 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 begin selecting a heat source based on power.

Basic formulas

To get a more or less accurate result, you need 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 ventilation air.

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

Q = 1/R x (t in - t n) x S x (1+ ∑β). Here:

• Q is the amount of heat lost by a structure of one type, W;
• R - thermal resistance of the construction material, m²°C / W;
• S—external fence area, m²;
• t in — internal air temperature, °C;
• t n - lowest temperature environment, °C;
• 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. To do this, use the formula R = δ / λ, 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 a wall is built from 2 materials (for example, brick with mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. Outdoor temperature is selected according to regulatory documents, and according to personal observations, internal - as necessary. Additional heat losses are coefficients determined by the standards:

1. When a wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
2. If the structure faces southeast or west, β = 0.05.
3. β = 0 when the outer fence faces the 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 taken of all fences adjacent to the environment: walls, windows, roof, floor 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 measurement results, the area of ​​each structure is calculated and substituted into the first formula (S, m²). The value R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity coefficient building material. In the case of new windows made of metal-plastic, the R value will be told to you by a representative of the installer.

As an example, it is worth calculating heat loss through enclosing walls made of brick 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 faces north (β = 0.1). First you need to take the thermal conductivity coefficient of brick (λ) from the reference literature; it is equal to 0.44 W/(m°C). Then, using the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:

R = 0.25 / 0.44 = 0.57 m²°C / W

To determine the heat loss of a room with this wall, all 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 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 regarding floors, roofing 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 thermal energy consumed by the heating system to heat the ventilation air. The share of this energy reaches 30% of total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss of a house through the heat capacity of the air using a popular formula from a physics course:

Q air = cm (t in - t n). In it:

• Q air - heat consumed by the heating system to warm up the supply air, W;
• t in and t n - the same as in the first formula, °C;
• m is the mass flow of air entering the house from outside, kg;
• c is the heat capacity of the air mixture, equal to 0.28 W / (kg °C).

Here all quantities are known, except mass flow air during room ventilation. In order not to complicate your task, you should agree to the condition that the air environment in the entire house is renewed once an hour. Then the volumetric air flow rate can be easily calculated by adding the volumes of all rooms, and then you need to convert it into mass air flow through density. Since the density of the air mixture changes depending on its temperature, you need to take the 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 = 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 external fences are summed up with ventilation heat losses, which gives the total heat load on the building’s heating system.

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

Calculation of heat loss at home is a necessary step in designing a heating system. Performed using complex formulas. Incorrectly leads to insufficient heating of the room (if heat loss indicators are underestimated) or to overpayments for the system and for heating (if the indicators are overestimated).

Heat supply calculations must be performed on top level

Initial data for calculating heat loss at home

To carry out the calculation correctly, you need to have a basic set of data. Only with them is it possible to work.

1. Heated area (you will also need it in the future to calculate the volume of heated air);
2. Floor plan of the building (used, among other things, when determining the installation locations of heating units);
3. Section of the building (sometimes not required);
4. The type of climate of the area is taken into account in the calculation. You can find out from SNB – 2. 04. 02 – 2000 “Building climatology”. The resulting coefficient is taken into account in the calculation;
5. Geographical position of the building, location of the heated volume relative to the north, south, west and east;
6. Building materials from which walls and floors are made;
7. Structure of enclosing structures (walls, floors). You need a profile listing the layers of materials, their location and thickness;
8. each type of building material, etc.;
9. Type and design of doors from the premises, their profile, section;
10. The materials from which the doors are made, determining the specific density of each, the location and thickness of the layers and the thermal conductivity coefficient. Those. the same information is required as for wall materials;
11. Calculation of the thermal power of a heating system is impossible without information on windows, if any. It is necessary to take into account their dimensions, geometry, type of glass unit, and sometimes materials. A profile and data similar to doors may also be required;
12. Roof data: structure, type, height, profile listing the type of materials and thickness, position of layers. Characteristics of building materials - thermal conductivity, quantity, etc.;
13. Window sill height. It is calculated as the distance from the surface of the top layer of the floor (not the cladding, but the clean layer) to the underside of the board;
14. Presence or absence of heating batteries;
15. If there is a “warm floor” - its profile, the building material of the covering above the communications, listing the thickness of the layers, their location, thermal conductivity coefficient, etc.;
16. Construction material and type of pipeline.

Determined data for the walls of a residential building

Think about what the future functions of the room are, based on this, draw a conclusion about the desired temperature regime (for example, in warehouses the temperature may be lower than in those where staff are constantly located; in greenhouses, flower shops there are even more specific requirements for heating ).

The next step is to determine temperature regime premises. It is carried out by periodically measuring temperatures. Determined desired temperatures that need to be supported. The heating scheme and the intended (or desired) installation locations for the risers are selected. The source of heat supply is determined.

When calculating heat loss, the architecture of the building, in particular its shape and geometry, also plays an important role. Since 2003, SNiP has taken into account the indicator of the shape of a structure. It is calculated as the ratio of the area of ​​the shell (walls, floor and ceiling) to the volume that it surrounds. Until 2003, this parameter was not taken into account, which led to the fact that energy was significantly overused.

Progress of work: calculation of the percentage of permissible heat loss for a country house made of timber, logs, bricks, panels

Before starting work directly, the contractor conducts some field surveys at the site. The premises are examined and measured, the wishes and information from the customer are taken into account. This process involves certain actions:

1. Full-scale measurements of premises;
2. Their specification according to the customer;
3. Study of the heating system, if available;
4. Ideas for improving or correcting heating errors (in the existing system);
5. Study of the hot water supply system;
6. Developing ideas for using it for heating or reducing heat loss (for example, using Valtec equipment);
7. Calculation of heat loss and others necessary to develop a heating system plan.

After these stages, the contractor provides the necessary technical documentation. It includes floor plans, profiles, where each heating device And general device systems, materials according to the specifics and type of equipment used.

Calculations: where the greatest heat losses in an insulated frame house come from and how to reduce them using the device

The most important process in heating design is the calculations of the future system. The heat loss through the enclosing structures is calculated, additional heat losses and gains are determined, the required number of heating devices of the selected type is determined, etc. The calculation of the heat loss coefficient of a house should be done by an experienced person.

The heat balance equation plays an important role in determining heat losses and developing methods for compensating them. is given below:

V is the volume of the room, calculated taking into account the area of ​​the room and the height of the ceilings. T is the difference between the external and internal temperatures of the building. K – heat loss coefficient.

The heat balance formula does not give the most exact indicators, therefore it is rarely used.

The main value used in the calculation is thermal load for heaters. To determine it, the values ​​of heat loss and are used. allows you to calculate the amount of heat that the heating system will produce, has the form:

Volume heat loss () is multiplied by 1.2. This is a reserve thermal coefficient - a constant that helps compensate for some heat losses that are random in nature (prolonged opening of doors or windows, etc.).

Calculating heat loss is quite difficult. On average, different building envelopes contribute to different amounts of energy loss. 10% is lost through the roof, 10% - through the floor, foundation, 40% - walls, 20% each - windows and poor insulation, ventilation system, etc. Specific thermal performance different materials are not the same. Therefore, the formula contains coefficients that allow you to take into account all the nuances. The table below shows the coefficient values ​​needed to calculate the amount of heat.

The heat loss formula is as follows:

In the formula, the specific heat loss is equal to 100 watts per square meter. m. Pl – area of ​​the room, also involved in the determination. Now a formula can be applied to calculate the amount of heat required to be released by the boiler.

Do the math correctly and your home will be warm

An example of calculating the heat loss coefficient in a private house: formula for success

The formula for calculating heat for space heating is easily applicable to any building. As an example, consider a hypothetical building with simple glazing, wooden walls and a window-to-floor ratio of 20%. It is located in a temperate climate zone, where the minimum outside temperature is 25 degrees. It has 4 walls, 3 m high. Above the heated room there is a cold attic. The value of the coefficients is determined from the table K1 - 1.27, K2 - 1.25, K3 - 1, K4 - 1.1, K5 - 1.33, K6 - 1, K7 - 1.05. The area of ​​the premises is 100 sq.m. The formula for the heat balance equation is not complicated and can be done by anyone.

Since the formula is known, the amount of heat required to heat a room can be calculated as follows:

Tp = 100*100*1.27*1.25*1*1.1*1.33*1*1.05 = 24386.38 W = 24.386 kW

And to calculate the thermal energy for heating, the boiler power formula is used as follows:

Mk = 1.2*24.386 = 29.2632 kW.

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At further stages, the number of required heating elements and the load on each of them, as well as energy consumption for heating, are determined. Calculating the heat loss of a house in our time of saving is very relevant.