Calculation of heating by heat load. Calculation of the heat load for heating the building

To find out how much power the heat-power equipment of a private house should have, it is necessary to determine the total load on the heating system, for which a thermal calculation is performed. In this article, we will not talk about an enlarged method for calculating the area or volume of a building, but we will present a more accurate method used by designers, only in a simplified form for better perception. So, 3 types of loads fall on the heating system of the house:

• compensation for losses of thermal energy leaving through building structures (walls, floors, roofs);
• heating the air required for ventilation of the premises;
• heating water for DHW needs (when a boiler is involved in this, and not a separate heater).

Determination of heat loss through external fences

First, let's present the formula from SNiP, which calculates the heat energy lost through building structures that separate the interior of the house from the street:

Q \u003d 1 / R x (tv - tn) x S, where:

• Q is the consumption of heat leaving through the structure, W;
• R - resistance to heat transfer through the material of the fence, m2ºС / W;
• S is the area of ​​this structure, m2;
• tv - the temperature that should be inside the house, ºС;
• tn is the average outdoor temperature for the 5 coldest days, ºС.

For reference. According to the methodology, heat loss calculation is performed separately for each room. In order to simplify the task, it is proposed to take the building as a whole, assuming an acceptable average temperature of 20-21 ºС.

The area for each type of external fencing is calculated separately, for which windows, doors, walls and floors with a roof are measured. This is done because they are made from different materials different thickness. So the calculation will have to be done separately for all types of structures, and then the results will be summed up. You probably know the coldest street temperature in your area of ​​​​residence from practice. But the parameter R will have to be calculated separately according to the formula:

R = δ / λ, where:

• λ is the coefficient of thermal conductivity of the fence material, W/(mºС);
• δ is the thickness of the material in meters.

Note. The value of λ is a reference value, it is easy to find it in any reference literature, and for plastic windows, manufacturers will tell you this coefficient. Below is a table with the coefficients of thermal conductivity of some building materials, and for calculations it is necessary to take the operational values ​​of λ.

As an example, let's calculate how much heat will be lost by 10 m2 of a brick wall 250 mm thick (2 bricks) with a temperature difference between outside and inside the house of 45 ºС:

R = 0.25 m / 0.44 W / (m ºС) = 0.57 m2 ºС / W.

Q \u003d 1 / 0.57 m2 ºС / W x 45 ºС x 10 m2 \u003d 789 W or 0.79 kW.

If the wall consists of different materials (structural material plus insulation), then they must also be calculated separately according to the above formulas, and the results summarized. Windows and roofing are calculated in the same way, but the situation is different with floors. First of all, you need to draw a building plan and divide it into zones 2 m wide, as is done in the figure:

Now you should calculate the area of ​​\u200b\u200beach zone and alternately substitute it into the main formula. Instead of parameter R, you need to take the standard values ​​​​for zone I, II, III and IV, indicated in the table below. At the end of the calculations, the results are added up and we get the total heat loss through the floors.

Ventilation air heating consumption

Uninformed people often do not take into account that the supply air in the house also needs to be heated, and this heat load also falls on the heating system. Cold air still enters the house from the outside, whether we like it or not, and it takes energy to heat it. Moreover, a full-fledged supply and exhaust ventilation usually with a natural urge. Air exchange is created due to the presence of draft in the ventilation ducts and the boiler chimney.

The method for determining the heat load from ventilation proposed in the regulatory documentation is rather complicated. Pretty accurate results can be obtained if this load is calculated using the well-known formula through the heat capacity of the substance:

Qvent = cmΔt, here:

• Qvent - the amount of heat required to heat the supply air, W;
• Δt - temperature difference in the street and inside the house, ºС;
• m is the mass of the air mixture coming from outside, kg;
• c is the heat capacity of air, assumed to be 0.28 W / (kg ºС).

The complexity of calculating this type of heat load lies in the correct determination of the mass of heated air. It is difficult to find out how much it gets inside the house with natural ventilation. Therefore, it is worth referring to the standards, because buildings are built according to projects where the required air exchanges are laid down. And the regulations say that in most rooms the air environment should change 1 time per hour. Then we take the volumes of all rooms and add to them the air consumption rates for each bathroom - 25 m3 / h and the kitchen gas stove - 100 m3 / h.

To calculate the heat load on heating from ventilation, the resulting volume of air must be converted into mass, knowing its density at different temperatures from the table:

Let us assume that the total amount of supply air is 350 m3/h, the outside temperature is minus 20 ºС, and the inside temperature is plus 20 ºС. Then its mass will be 350 m3 x 1.394 kg / m3 = 488 kg, and the heat load on the heating system will be Qvent = 0.28 W / (kg ºС) x 488 kg x 40 ºС = 5465.6 W or 5.5 kW.

Heat load from DHW heating

To determine this load, you can use the same simple formula, only now you need to calculate the thermal energy spent on heating water. Its heat capacity is known and amounts to 4.187 kJ/kg °С or 1.16 W/kg °С. Considering that a family of 4 people needs 100 liters of water for 1 day, heated to 55 ° C, for all needs, we substitute these numbers into the formula and get:

QDHW \u003d 1.16 W / kg ° С x 100 kg x (55 - 10) ° С \u003d 5220 W or 5.2 kW of heat per day.

Note. By default, it is assumed that 1 liter of water is equal to 1 kg, and the temperature of cold tap water is 10 °C.

The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But it is impossible to divide this figure by 24, because we want to receive hot water as soon as possible, and for this the boiler must have a power reserve. That is, this load must be added to the rest as is.

Conclusion

This calculation of home heating loads will give much more accurate results than the traditional method by area, although you will have to work hard. The final result must be multiplied by the safety factor - 1.2, or even 1.4, and selected according to the calculated value boiler equipment. Another way to enlarge the calculation of thermal loads according to the standards is shown in the video:

Creating a heating system in your own home or even in a city apartment is an extremely responsible task. At the same time, it would be completely unreasonable to purchase boiler equipment, as they say, “by eye”, that is, without taking into account all the features of housing. In this, it is quite possible to fall into two extremes: either the power of the boiler will not be enough - the equipment will work “to its fullest”, without pauses, but will not give the expected result, or, conversely, an overly expensive device will be purchased, the capabilities of which will remain completely unclaimed.

But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place heat exchange devices in the premises - radiators, convectors or "warm floors". And again, relying only on your intuition or the "good advice" of your neighbors is not the most reasonable option. In a word, certain calculations are indispensable.

Of course, ideally, such heat engineering calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it interesting to try to do it yourself? This publication will show in detail how heating is calculated by the area of ​​\u200b\u200bthe room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, will help you perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to get a result with a completely acceptable degree of accuracy.

The simplest methods of calculation

In order for the heating system to create comfortable living conditions during the cold season, it must cope with two main tasks. These functions are closely related, and their separation is very conditional.

• The first is maintaining an optimal level of air temperature in the entire volume of the heated room. Of course, the temperature level may vary slightly with altitude, but this difference should not be significant. Quite comfortable conditions are considered to be an average of +20 ° C - it is this temperature that, as a rule, is taken as the initial temperature in thermal calculations.

In other words, the heating system must be able to heat a certain volume of air.

If we approach with complete accuracy, then for individual rooms in residential buildings the standards for the required microclimate have been established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:

Purpose of the premises	Air temperature, °С		Relative humidity, %		Air speed, m/s
	optimal	admissible	optimal	admissible, max	optimal, max	admissible, max
For the cold season
Living room	20÷22	18÷24 (20÷24)	45÷30	60	0.15	0.2
The same, but for living rooms in regions with minimum temperatures from -31 ° C and below	21÷23	20÷24 (22÷24)	45÷30	60	0.15	0.2
Kitchen	19:21	18:26	N/N	N/N	0.15	0.2
Toilet	19:21	18:26	N/N	N/N	0.15	0.2
Bathroom, combined bathroom	24÷26	18:26	N/N	N/N	0.15	0.2
Premises for rest and study	20÷22	18:24	45÷30	60	0.15	0.2
Inter-apartment corridor	18:20	16:22	45÷30	60	N/N	N/N
lobby, stairwell	16÷18	14:20	N/N	N/N	N/N	N/N
Storerooms	16÷18	12÷22	N/N	N/N	N/N	N/N
For the warm season (The standard is only for residential premises. For the rest - it is not standardized)
Living room	22÷25	20÷28	60÷30	65	0.2	0.3

• The second is the compensation of heat losses through the structural elements of the building.

The main "enemy" of the heating system is heat loss through building structures.

Alas, heat loss is the most serious "rival" of any heating system. They can be reduced to a certain minimum, but even with the highest quality thermal insulation, it is not yet possible to completely get rid of them. Thermal energy leaks go in all directions - their approximate distribution is shown in the table:

Building element	Approximate value of heat loss
Foundation, floors on the ground or over unheated basement (basement) premises	from 5 to 10%
"Cold bridges" through poorly insulated joints of building structures	from 5 to 10%
Entry places engineering communications(sewerage, water supply, gas pipes, electrical cables, etc.)	up to 5%
External walls, depending on the degree of insulation	from 20 to 30%
Poor quality windows and exterior doors	about 20÷25%, of which about 10% - through non-sealed joints between the boxes and the wall, and due to ventilation
Roof	up to 20%
Ventilation and chimney	up to 25 ÷30%

Naturally, in order to cope with such tasks, the heating system must have a certain thermal power, and this potential must not only meet the general needs of the building (apartment), but also be correctly distributed over the premises, in accordance with their area and a number of other important factors.

Usually the calculation is carried out in the direction "from small to large". Simply put, the required amount of thermal energy for each heated room is calculated, the obtained values ​​​​are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler needs. And the values ​​​​for each room will be the starting point for calculating the required number of radiators.

The most simplified and most commonly used method in a non-professional environment is to accept the norm of 100 W of thermal energy per square meter of area:

The most primitive way of counting is the ratio of 100 W / m²

Q = S× 100

Q- the required thermal power for the room;

S– area of ​​the room (m²);

100 — specific power per unit area (W/m²).

For example, room 3.2 × 5.5 m

S= 3.2 × 5.5 = 17.6 m²

Q= 17.6 × 100 = 1760 W ≈ 1.8 kW

The method is obviously very simple, but very imperfect. It is worth mentioning right away that it is conditionally applicable only with a standard ceiling height - approximately 2.7 m (permissible - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.

It is clear that in this case the value of specific power is calculated per cubic meter. It is taken equal to 41 W / m³ for a reinforced concrete panel house, or 34 W / m³ - in brick or made of other materials.

Q = S × h× 41 (or 34)

h- ceiling height (m);

41 or 34 - specific power per unit volume (W / m³).

For example, the same room, in a panel house, with a ceiling height of 3.2 m:

Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW

The result is more accurate, since it already takes into account not only all linear dimensions rooms, but even, to a certain extent, the features of the walls.

But still, it is still far from real accuracy - many nuances are “outside the brackets”. How to perform calculations closer to real conditions - in the next section of the publication.

You may be interested in information about what they are

Carrying out calculations of the required thermal power, taking into account the characteristics of the premises

The calculation algorithms discussed above are useful for the initial “estimate”, but you should still rely on them completely with very great care. Even to a person who does not understand anything in building heat engineering, the indicated average values ​​\u200b\u200bmay seem doubtful - they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk Region. In addition, the room - the room is different: one is located on the corner of the house, that is, it has two external walls ki, and the other on three sides is protected from heat loss by other rooms. In addition, the room may have one or more windows, both small and very large, sometimes even panoramic. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - just such features are visible even to the "naked eye".

In a word, there are a lot of nuances that affect the heat loss of each particular room, and it is better not to be too lazy, but to carry out a more thorough calculation. Believe me, according to the method proposed in the article, this will not be so difficult to do.

General principles and calculation formula

The calculations will be based on the same ratio: 100 W per 1 square meter. But only the formula itself is “overgrown” with a considerable number of various correction factors.

Q = (S × 100) × a × b × c × d × e × f × g × h × i × j × k × l × m

The Latin letters denoting the coefficients are taken quite arbitrarily, in alphabetical order, and are not related to any standard quantities accepted in physics. The meaning of each coefficient will be discussed separately.

• "a" - a coefficient that takes into account the number of external walls in a particular room.

Obviously, the more external walls in the room, the larger the area through which heat loss occurs. In addition, the presence of two or more external walls also means corners - extremely vulnerable places in terms of the formation of "cold bridges". The coefficient "a" will correct for this specific feature of the room.

The coefficient is taken equal to:

- external walls No (interior): a = 0.8;

- outer wall one: a = 1.0;

- external walls two: a = 1.2;

- external walls three: a = 1.4.

• "b" is a coefficient that takes into account the location of the external walls of the room relative to the cardinal points.

You may be interested in information about what are

Even on the coldest winter days, solar energy still has an effect on the temperature balance in a building. It is quite natural that the side of the house that is facing south receives some heating from the sun's rays, and heat loss through it is lower.

But the walls and windows facing north never “see” the Sun. The eastern part of the house, although it "grabs" the morning sun's rays, still does not receive any effective heating from them.

Based on this, we introduce the coefficient "b":

- the outer walls of the room look at North or East: b = 1.1;

- the outer walls of the room are oriented towards South or West: b = 1.0.

• "c" - coefficient taking into account the location of the room relative to the winter "wind rose"

Perhaps this amendment is not so necessary for houses located in areas protected from the winds. But sometimes the prevailing winter winds can make their own “hard adjustments” to the thermal balance of the building. Naturally, the windward side, that is, "substituted" to the wind, will lose much more body, compared to the leeward, opposite.

Based on the results of long-term meteorological observations in any region, the so-called "wind rose" is compiled - a graphic diagram showing the prevailing wind directions in winter and summer. This information can be obtained from the local hydrometeorological service. However, many residents themselves, without meteorologists, know perfectly well where the winds mainly blow from in winter, and from which side of the house the deepest snowdrifts usually sweep.

If there is a desire to carry out calculations with higher accuracy, then the correction factor “c” can also be included in the formula, taking it equal to:

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

- wall located parallel to the direction of the wind: c = 1.1.

• "d" - a correction factor that takes into account the peculiarities of the climatic conditions of the region where the house was built

Naturally, the amount of heat loss through all the building structures of the building will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer indicators “dance” in a certain range, but for each region there is an average indicator of the lowest temperatures characteristic of the coldest five-day period of the year (usually this is characteristic of January). For example, below is a map-scheme of the territory of Russia, on which approximate values ​​​​are shown in colors.

Usually this value is easy to check with the regional meteorological service, but you can, in principle, rely on your own observations.

So, the coefficient "d", taking into account the peculiarities of the climate of the region, for our calculations in we take equal to:

— from – 35 °С and below: d=1.5;

— from – 30 °С to – 34 °С: d=1.3;

— from – 25 °С to – 29 °С: d=1.2;

— from – 20 °С to – 24 °С: d=1.1;

— from – 15 °С to – 19 °С: d=1.0;

— from – 10 °С to – 14 °С: d=0.9;

- not colder - 10 ° С: d=0.7.

• "e" - coefficient taking into account the degree of insulation of external walls.

The total value of the heat loss of the building is directly related to the degree of insulation of all building structures. One of the "leaders" in terms of heat loss are walls. Therefore, the value of thermal power required to maintain comfortable conditions living indoors depends on the quality of their thermal insulation.

The value of the coefficient for our calculations can be taken as follows:

- external walls are not insulated: e = 1.27;

- medium degree of insulation - walls in two bricks or their surface thermal insulation with other heaters is provided: e = 1.0;

– the insulation was carried out qualitatively, on the basis of the thermotechnical calculations: e = 0.85.

Later in the course of this publication, recommendations will be given on how to determine the degree of insulation of walls and other building structures.

• coefficient "f" - correction for ceiling height

Ceilings, especially in private homes, can have different heights. Therefore, the thermal power for heating one or another room of the same area will also differ in this parameter.

It will not be a big mistake to accept the following values ​​​​of the correction factor "f":

– ceiling height up to 2.7 m: f = 1.0;

— flow height from 2.8 to 3.0 m: f = 1.05;

– ceiling height from 3.1 to 3.5 m: f = 1.1;

– ceiling height from 3.6 to 4.0 m: f = 1.15;

– ceiling height over 4.1 m: f = 1.2.

• « g "- coefficient taking into account the type of floor or room located under the ceiling.

As shown above, the floor is one of the significant sources of heat loss. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:

- cold floor on the ground or over an unheated room (for example, basement or basement): g= 1,4 ;

- insulated floor on the ground or over an unheated room: g= 1,2 ;

- a heated room is located below: g= 1,0 .

• « h "- coefficient taking into account the type of room located above.

The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which takes into account this feature of the calculated room:

- a "cold" attic is located on top: h = 1,0 ;

- an insulated attic or other insulated room is located on top: h = 0,9 ;

- any heated room is located above: h = 0,8 .

• « i "- coefficient taking into account the design features of windows

Windows are one of the "main routes" of heat leaks. Naturally, much in this matter depends on the quality of the window structure itself. Old wooden frames, which were previously installed everywhere in all houses, are significantly inferior to modern multi-chamber systems with double-glazed windows in terms of their thermal insulation.

Without words, it is clear that the thermal insulation qualities of these windows are significantly different.

But even between PVC-windows there is no complete uniformity. For example, a two-chamber double-glazed window (with three glasses) will be much warmer than a single-chamber one.

This means that it is necessary to enter a certain coefficient "i", taking into account the type of windows installed in the room:

- standard wooden windows with conventional double glazing: i = 1,27 ;

– modern window systems with single-chamber double-glazed windows: i = 1,0 ;

– modern window systems with two-chamber or three-chamber double-glazed windows, including those with argon filling: i = 0,85 .

• « j" - correction factor for the total glazing area of ​​the room

Whatever quality windows however they were, it will still not be possible to completely avoid heat loss through them. But it is quite clear that it is impossible to compare a small window with panoramic glazing almost on the entire wall.

First you need to find the ratio of the areas of all the windows in the room and the room itself:

x = ∑SOK /SP

∑ SOK- the total area of ​​windows in the room;

SP- area of ​​the room.

Depending on the value obtained and the correction factor "j" is determined:

- x \u003d 0 ÷ 0.1 →j = 0,8 ;

- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;

- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;

- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;

- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;

• « k" - coefficient that corrects for the presence of an entrance door

The door to the street or to an unheated balcony is always an additional "loophole" for the cold

The door to the street or to an open balcony is able to make its own adjustments to the heat balance of the room - each opening of it is accompanied by the penetration of a considerable amount of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient "k", which we take equal to:

- no door k = 1,0 ;

- one door to the street or balcony: k = 1,3 ;

- two doors to the street or to the balcony: k = 1,7 .

• « l "- possible amendments to the connection diagram of heating radiators

Perhaps this will seem like an insignificant trifle to some, but still - why not immediately take into account the planned scheme for connecting heating radiators. The fact is that their heat transfer, and hence their participation in maintaining a certain temperature balance in the room, changes quite noticeably with different types of insertion of supply and return pipes.

Illustration	Radiator insert type	The value of the coefficient "l"
	Diagonal connection: supply from above, "return" from below	l = 1.0
	Connection on one side: supply from above, "return" from below	l = 1.03
	Two-way connection: both supply and return from the bottom	l = 1.13
	Diagonal connection: supply from below, "return" from above	l = 1.25
	Connection on one side: supply from below, "return" from above	l = 1.28
	One-way connection, both supply and return from below	l = 1.28

• « m "- correction factor for the features of the installation site of heating radiators

And finally, the last coefficient, which is also associated with the features of connecting heating radiators. It is probably clear that if the battery is installed openly, is not obstructed by anything from above and from the front, then it will give maximum heat transfer. However, such an installation is far from always possible - more often, radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating priors into the created interior ensemble, hide them completely or partially with decorative screens - this also significantly affects the heat output.

If there are certain “baskets” on how and where the radiators will be mounted, this can also be taken into account when making calculations by entering a special coefficient “m”:

Illustration	Features of installing radiators	The value of the coefficient "m"
	The radiator is located on the wall openly or is not covered from above by a window sill	m = 0.9
	The radiator is covered from above by a window sill or a shelf	m = 1.0
	The radiator is blocked from above by a protruding wall niche	m = 1.07
	The radiator is covered from above with a window sill (niche), and from the front - with a decorative screen	m = 1.12
	The radiator is completely enclosed in a decorative casing	m = 1.2

So, there is clarity with the calculation formula. Surely, some of the readers will immediately take up their heads - they say, it's too complicated and cumbersome. However, if the matter is approached systematically, in an orderly manner, then there is no difficulty at all.

Any good homeowner must have a detailed graphical plan of their "possessions" with dimensions, and usually oriented to the cardinal points. It is not difficult to specify the climatic features of the region. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "vertical neighborhood" from above and below, the location of the entrance doors, the proposed or existing scheme for installing heating radiators - no one except the owners knows better.

It is recommended to immediately draw up a worksheet, where you enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will help to carry out the built-in calculator, in which all the coefficients and ratios mentioned above are already “laid”.

If some data could not be obtained, then, of course, they can not be taken into account, but in this case, the “default” calculator will calculate the result, taking into account the least favorable conditions.

It can be seen with an example. We have a house plan (taken completely arbitrary).

The region with the level of minimum temperatures in the range of -20 ÷ 25 °С. Predominance of winter winds = northeasterly. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal diagonal connection of radiators, which will be installed under the window sills, has been selected.

Let's create a table like this:

The room, its area, ceiling height. Floor insulation and "neighborhood" from above and below	The number of external walls and their main location relative to the cardinal points and the "wind rose". Degree of wall insulation	Number, type and size of windows	Existence of entrance doors (to the street or to the balcony)	Required heat output (including 10% reserve)
Area 78.5 m²				10.87 kW ≈ 11 kW
1. Hallway. 3.18 m². Ceiling 2.8 m. Warmed floor on the ground. Above is an insulated attic.	One, South, the average degree of insulation. Leeward side	Not	One	0.52 kW
2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic	Not	Not	Not	0.62 kW
3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well insulated floor on the ground. Svehu - insulated attic	Two. South, west. Average degree of insulation. Leeward side	Two, single-chamber double-glazed window, 1200 × 900 mm	Not	2.22 kW
4. Children's room. 18.3 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic	Two, North - West. High degree of insulation. windward	Two, double glazing, 1400 × 1000 mm	Not	2.6 kW
5. Bedroom. 13.8 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic	Two, North, East. High degree of insulation. windward side	One, double-glazed window, 1400 × 1000 mm	Not	1.73 kW
6. Living room. 18.0 m². Ceiling 2.8 m. Well insulated floor. Top - insulated attic	Two, East, South. High degree of insulation. Parallel to wind direction	Four, double glazing, 1500 × 1200 mm	Not	2.59 kW
7. Bathroom combined. 4.12 m². Ceiling 2.8 m. Well insulated floor. Above is an insulated attic.	One, North. High degree of insulation. windward side	One. Wooden frame with double glazing. 400 × 500 mm	Not	0.59 kW
TOTAL:

Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values ​​\u200b\u200bfor each room - this will be the required total power of the heating system.

The result for each room, by the way, will help you choose the right number of heating radiators - it remains only to divide by the specific heat output of one section and round up.

Hello dear readers! Today a small post about the calculation of the amount of heat for heating according to aggregated indicators. In general, the heating load is taken according to the project, that is, the data that the designer calculated are entered into the heat supply contract.

But often there is simply no such data, especially if the building is small, such as a garage, or some kind of utility room. In this case, the heating load in Gcal / h is calculated according to the so-called aggregated indicators. I wrote about this. And already this figure is included in the contract as the estimated heating load. How is this number calculated? And it is calculated according to the formula:

Qot \u003d α * qo * V * (tv-tn.r) * (1 + Kn.r) * 0.000001; where

α is a correction factor that takes into account the climatic conditions of the area, it is applied in cases where the calculated outdoor air temperature differs from -30 ° С;

qo is the specific heating characteristic of the building at tn.r = -30 °С, kcal/m3*С;

V - the volume of the building according to the external measurement, m³;

tv is the design temperature inside the heated building, °С;

tn.r - design outdoor air temperature for heating design, °C;

Kn.r is the infiltration coefficient, which is due to thermal and wind pressure, that is, the ratio of heat losses from the building with infiltration and heat transfer through external fences at the outdoor air temperature, which is calculated for heating design.

So, in one formula, you can calculate the heat load on the heating of any building. Of course, this calculation is largely approximate, but it is recommended in the technical literature on heat supply. Heat supply organizations also enter this figure of the heating load Qfrom, in Gcal / h, into heat supply contracts. So the calculation is correct. This calculation is well presented in the book - V.I. Manyuk, Ya.I. Kaplinsky, E.B. Khizh and others. This book is one of my desktop books, a very good book.

Also, this calculation of the heat load on the heating of the building can be done according to the "Methodology for determining the amount of thermal energy and heat carrier in public water supply systems" of RAO Roskommunenergo of the Gosstroy of Russia. True, there is an inaccuracy in the calculation in this method (in formula 2 in Appendix No. 1, 10 to the minus third power is indicated, but it should be 10 to the minus sixth power, this must be taken into account in the calculations), you can read more about this in the comments to this article.

I fully automated this calculation, added reference tables, including a table of climatic parameters for all regions former USSR(from SNiP 23.01.99 "Construction climatology"). You can buy a calculation in the form of a program for 100 rubles by writing to me by e-mail [email protected]

I will be glad to comments on the article.

1. Heating

1.1. The estimated hourly heat load of heating should be taken according to standard or individual building designs.

If the value of the calculated outdoor air temperature adopted in the project for designing heating differs from the current standard value for a particular area, it is necessary to recalculate the estimated hourly heat load of the heated building given in the project according to the formula:

where Qo max is the calculated hourly heat load of the heating of the building, Gcal/h;

Qo max pr - the same, according to a standard or individual project, Gcal / h;

tj - design air temperature in the heated building, °С; taken in accordance with Table 1;

to - design outdoor air temperature for designing heating in the area where the building is located, according to SNiP 23-01-99, ° С;

to.pr - the same, according to a standard or individual project, ° С.

Table 1. Estimated air temperature in heated buildings

In areas with an estimated outdoor air temperature for heating design of -31 °С and below, the value of the calculated air temperature inside heated residential buildings should be taken in accordance with the chapter SNiP 2.08.01-85 equal to 20 °С.

1.2. In the absence of design information, the estimated hourly heat load of heating an individual building can be determined by aggregated indicators:

where  is a correction factor that takes into account the difference in the calculated outdoor temperature for heating design to from to = -30 °С, at which the corresponding qo value is determined; taken according to table 2;

V is the volume of the building according to the external measurement, m3;

qo - specific heating characteristic of the building at to = -30 °С, kcal/m3 h°С; taken according to tables 3 and 4;

Ki.r - calculated coefficient of infiltration due to thermal and wind pressure, i.e. the ratio of heat losses from a building with infiltration and heat transfer through external fences at an outside air temperature calculated for heating design.

Table 2. Correction factor  for residential buildings

Table 3. Specific heating characteristic of residential buildings

External building volume V, m3	Specific heating characteristic qo, kcal/m3 h °C
building before 1958	building after 1958

Table 3a. Specific heating characteristic of buildings built before 1930

Table 4. Specific thermal characteristic of administrative, medical, cultural and educational buildings, children's institutions

Name of buildings	Volume of buildings V, m3	Specific thermal characteristics
for heating qo, kcal/m3 h °C	for ventilation qv, kcal/m3 h °C
Administrative buildings, offices
	over 15000
	over 10000
Cinemas
	over 10000
	over 30000
The shops
	over 10000
Kindergartens and nurseries
Schools and higher education institutions
	over 10000
Hospitals
	over 15000
	over 10000
Laundries
	over 10000
Catering establishments, canteens, kitchen factories
	over 10000
Laboratories
	over 10000
fire stations

The value of V, m3, should be taken according to the information of a typical or individual design of a building or a technical inventory bureau (BTI).

If the building has an attic floor, the value V, m3, is determined as the product of the horizontal cross-sectional area of ​​the building at the level of its first floor (above the basement floor) and the free height of the building - from the level of the finished floor of the first floor to the upper plane of the attic floor heat-insulating layer, with roofs, combined with attic ceilings - up to the average mark of the top of the roof. Architectural details protruding beyond the surface of the walls and niches in the walls of the building, as well as unheated loggias, are not taken into account when determining the calculated hourly heat load of heating.

If there is a heated basement in the building, 40% of the volume of this basement must be added to the resulting volume of the heated building. The construction volume of the underground part of the building (basement, ground floor) is defined as the product of the horizontal section of the building at the level of its first floor by the height of the basement (ground floor).

The calculated infiltration coefficient Ki.r is determined by the formula:

where g - free fall acceleration, m/s2;

L - free height of the building, m;

w0 - calculated wind speed for the given area during the heating season, m/s; accepted according to SNiP 23-01-99.

It is not necessary to enter into the calculation of the calculated hourly heat load of the heating of the building the so-called correction for the effect of wind, because this quantity has already been taken into account in formula (3.3).

In areas where the design value of the outdoor air temperature for heating design is up to  -40 °С, for buildings with unheated basements, additional heat losses through the unheated floors of the first floor in the amount of 5% should be taken into account.

For buildings completed by construction, the calculated hourly heat load of heating should be increased for the first heating period for stone buildings built:

In May-June - by 12%;

In July-August - by 20%;

In September - by 25%;

In the heating period - by 30%.

1.3. The specific heating characteristic of a building qo, kcal/m3 h °C, in the absence of a qo value corresponding to its construction volume in Tables 3 and 4, can be determined by the formula:

where a \u003d 1.6 kcal / m 2.83 h ° С; n = 6 - for buildings under construction before 1958;

a \u003d 1.3 kcal / m 2.875 h ° C; n = 8 - for buildings under construction after 1958

1.4. If a part of a residential building is occupied by a public institution (office, shop, pharmacy, laundry collection point, etc.), the estimated hourly heating load must be determined according to the project. If the calculated hourly heat load in the project is indicated only for the whole building, or is determined by aggregated indicators, the heat load of individual rooms can be determined from the heat exchange surface area of ​​the installed heating devices using the general equation describing their heat transfer:

Q = k F t, (3.5)

where k is the heat transfer coefficient of the heating device, kcal/m3 h °C;

F - heat exchange surface area of ​​the heating device, m2;

t - temperature difference of the heating device, °С, defined as the difference between the average temperature of the convective-radiative heating device and the air temperature in the heated building.

The methodology for determining the calculated hourly heat load of heating on the surface of installed heating devices of heating systems is given in.

1.5. When heated towel rails are connected to the heating system, the calculated hourly heat load of these heaters can be determined as the heat transfer of uninsulated pipes in a room with an estimated air temperature tj \u003d 25 ° C according to the method given in.

1.6. In the absence of design data and the determination of the estimated hourly heat load for heating industrial, public, agricultural and other non-standard buildings (garages, heated underground passages, swimming pools, shops, kiosks, pharmacies, etc.) according to aggregated indicators, the values ​​of this load should be refined according to the heat exchange surface area of ​​the installed heating devices of heating systems in accordance with the methodology given in. The initial information for calculations is revealed by a representative of the heat supply organization in the presence of a representative of the subscriber with the preparation of an appropriate act.

1.7. The consumption of thermal energy for the technological needs of greenhouses and conservatories, Gcal/h, is determined from the expression:

, (3.6)

where Qcxi - heat energy consumption for i-e technological operations, Gcal/h;

n is the number of technological operations.

In its turn,

Qcxi \u003d 1.05 (Qtp + Qv) + Qfloor + Qprop, (3.7)

where Qtp and Qv are heat losses through the building envelope and during air exchange, Gcal/h;

Qpol + Qprop - consumption of thermal energy for heating irrigation water and steaming the soil, Gcal/h;

1.05 - coefficient taking into account the consumption of thermal energy for heating domestic premises.

1.7.1. Heat loss through building envelope, Gcal/h, can be determined by the formula:

Qtp = FK (tj - to) 10-6, (3.8)

where F is the surface area of ​​the building envelope, m2;

K is the heat transfer coefficient of the enclosing structure, kcal/m2 h °C; for single glazing, K = 5.5 can be taken, for a single-layer film fence K = 7.0 kcal / m2 h ° C;

tj and to are the process temperature in the room and the calculated outdoor air for the design of the corresponding agricultural facility, °C.

1.7.2. Heat losses during air exchange for greenhouses with glass coatings, Gcal/h, are determined by the formula:

Qv \u003d 22.8 Finv S (tj - to) 10-6, (3.9)

where Finv is the inventory area of ​​the greenhouse, m2;

S - volume coefficient, which is the ratio of the volume of the greenhouse and its inventory area, m; can be taken in the range from 0.24 to 0.5 for small greenhouses and 3 or more m - for hangars.

Heat losses during air exchange for film-coated greenhouses, Gcal/h, are determined by the formula:

Qv \u003d 11.4 Finv S (tj - to) 10-6. (3.9a)

1.7.3. The consumption of thermal energy for heating irrigation water, Gcal/h, is determined from the expression:

, (3.10)

where Fcreep - effective area greenhouses, m2;

n - duration of watering, h.

1.7.4. The consumption of thermal energy for steaming the soil, Gcal/h, is determined from the expression:

2. Supply ventilation

2.1. If there is a standard or individual design of the building and the compliance of the installed equipment of the supply ventilation system with the project, the calculated hourly heat load of ventilation can be taken according to the project, taking into account the difference in the values ​​of the calculated outdoor temperature for designing ventilation adopted in the project, and the current standard value for the area where the considered building.

Recalculation is carried out according to a formula similar to formula (3.1):

, (3.1a)

Qv.pr - the same, according to the project, Gcal / h;

tv.pr is the calculated outdoor air temperature at which the heat load of supply ventilation in the project is determined, °С;

tv is the calculated outdoor air temperature for designing supply ventilation in the area where the building is located, °С; accepted according to the instructions of SNiP 23-01-99.

2.2. In the absence of projects or inconsistency of the installed equipment with the project, the calculated hourly heat load of supply ventilation must be determined from the characteristics of the equipment actually installed, in accordance with the general formula describing the heat transfer of air heaters:

Q = Lc (2 + 1) 10-6, (3.12)

where L is the volumetric flow rate of heated air, m3/h;

 - density of heated air, kg/m3;

c is the heat capacity of the heated air, kcal/kg;

2 and 1 - calculated values ​​of air temperature at the inlet and outlet of the calorific unit, °C.

The methodology for determining the estimated hourly heat load of supply air heaters is set out in.

It is permissible to determine the calculated hourly heat load of the supply ventilation of public buildings according to aggregated indicators according to the formula:

Qv \u003d Vqv (tj - tv) 10-6, (3.2a)

where qv is the specific thermal ventilation characteristic of the building, depending on the purpose and construction volume of the ventilated building, kcal/m3 h °C; can be taken from Table 4.

3. Hot water supply

3.1. The average hourly heat load of hot water supply of a consumer of thermal energy Qhm, Gcal/h, during the heating period is determined by the formula:

where a is the rate of water consumption for hot water supply of the subscriber, l / unit. measurements per day; must be approved by the local government; in the absence of approved norms, it is adopted according to the table of Appendix 3 (mandatory) SNiP 2.04.01-85;

N - the number of units of measurement, referred to the day, - the number of residents, students in educational institutions, etc.;

tc - tap water temperature during the heating season, °С; in the absence of reliable information, tc = 5 °С is accepted;

T - the duration of the operation of the hot water supply system of the subscriber per day, h;

Qt.p - heat losses in local system hot water supply, in the supply and circulation pipelines of the external hot water supply network, Gcal / h.

3.2. The average hourly heat load of hot water supply in the non-heating period, Gcal, can be determined from the expression:

, (3.13a)

where Qhm is the average hourly heat load of hot water supply during the heating period, Gcal/h;

 - coefficient taking into account the decrease in the average hourly load of hot water supply in the non-heating period compared to the load in the heating period; if the value of  is not approved by the local government,  is taken equal to 0.8 for the housing and communal sector of cities in central Russia, 1.2-1.5 - for resorts, southern cities and towns, for enterprises - 1.0;

ths, th - hot water temperature in non-heating and heating periods, °С;

tcs, tc - tap water temperature during the non-heating and heating period, °C; in the absence of reliable information, tcs = 15 °С, tc = 5 °С are accepted.

3.3. Heat losses by pipelines of the hot water supply system can be determined by the formula:

where Ki is the heat transfer coefficient of a section of an uninsulated pipeline, kcal/m2 h °C; you can take Ki = 10 kcal/m2 h °C;

di and li - diameter of the pipeline in the section and its length, m;

tн and tк ​​- temperature of hot water at the beginning and end of the calculated section of the pipeline, ° С;

tamb - ambient temperature, °C; take the form of laying pipelines:

In furrows, vertical channels, communication shafts of sanitary cabins tacr = 23 °С;

In bathrooms tamb = 25 °С;

In kitchens and toilets tamb = 21 °С;

On stairwells tocr = 16 °С;

In the underground laying channels of the external hot water supply network tcr = tgr;

In tunnels tcr = 40 °С;

In unheated basements tocr = 5 °С;

In attics tambi = -9 °С (at the average outdoor temperature of the coldest month of the heating period tн = -11 ... -20 °С);

 - efficiency of thermal insulation of pipelines; accepted for pipelines with a diameter of up to 32 mm  = 0.6; 40-70 mm  = 0.74; 80-200 mm  = 0.81.

Table 5. Specific heat losses of pipelines of hot water supply systems (according to the place and method of laying)

Place and method of laying	Thermal losses of the pipeline, kcal / hm, with a nominal diameter, mm
Main supply riser in a ditch or communication shaft, insulated
Riser without heated towel rails, insulated, in the sanitary cabin shaft, furrow or utility shaft
Same with towel rails.
Riser uninsulated in the sanitary cabin shaft, furrow or communication shaft or open in the bathroom, kitchen
Distribution insulated pipelines (supply):
in the basement, on the stairwell
in a cold attic
in a warm attic
Circulation pipelines isolated:
in the basement
in a warm attic
in a cold attic
Circulation pipelines uninsulated:
in apartments
on the stairwell
Circulation risers in the duct of a sanitary cabin or bathroom:
isolated
uninsulated

Note. In the numerator - specific heat losses of pipelines of hot water supply systems without direct water intake in heat supply systems, in the denominator - with direct water intake.

Table 6. Specific heat losses of pipelines of hot water supply systems (by temperature difference)

Temperature drop, °С

Thermal losses of the pipeline, kcal / h m, with a nominal diameter, mm

Note. If the hot water temperature drop is different from its given values, the specific heat losses should be determined by interpolation.

3.4. In the absence of the initial information necessary for calculating heat losses by hot water supply pipelines, heat losses, Gcal / h, can be determined using a special coefficient Kt.p, taking into account the heat losses of these pipelines, according to the expression:

Qt.p = Qhm Kt.p. (3.15)

The heat flow to hot water supply, taking into account heat losses, can be determined from the expression:

Qg = Qhm (1 + Kt.p). (3.16)

Table 7 can be used to determine the values ​​of the coefficient Kt.p.

Table 7. Coefficient taking into account heat losses by pipelines of hot water supply systems

studfiles.net

How to calculate the heat load for heating a building

In houses that have been put into operation in recent years, these rules are usually met, so the calculation of the heating power of the equipment is based on standard coefficients. An individual calculation can be carried out at the initiative of the owner of the housing or the communal structure involved in the supply of heat. This happens when spontaneous replacement of heating radiators, windows and other parameters.

See also: How to calculate the power of a heating boiler by area of ​​\u200b\u200bthe house

Calculation of norms for heating in an apartment

In an apartment served by a utility company, the calculation of the heat load can only be carried out upon transfer of the house in order to track the parameters of SNIP in the premises taken on balance. Otherwise, the owner of the apartment does this in order to calculate his heat losses in the cold season and eliminate the shortcomings of insulation - use heat-insulating plaster, glue the insulation, mount penofol on the ceilings and install metal-plastic windows with a five-chamber profile.

The calculation of heat leaks for the public utility in order to open a dispute, as a rule, does not give a result. The reason is that there are heat loss standards. If the house is put into operation, then the requirements are met. At the same time, heating devices comply with the requirements of SNIP. Replacing batteries and extracting more heat is prohibited, as the radiators are installed according to approved building standards.

The method of calculating the norms for heating in a private house

Private houses are heated by autonomous systems, which at the same time calculates the load is carried out to comply with the requirements of SNIP, and the correction of heating power is carried out in conjunction with work to reduce heat loss.

Calculations can be done manually using a simple formula or a calculator on the site. The program helps to calculate required power heating systems and heat leakage typical of the winter period. Calculations are carried out for a certain thermal zone.

Basic principles

The methodology includes a number of indicators that together allow us to assess the level of insulation of the house, compliance with SNIP standards, as well as the power of the heating boiler. How it works:

• depending on the parameters of walls, windows, insulation of the ceiling and foundation, you calculate heat leakage. For example, your wall consists of a single layer of clinker bricks and frame bricks with insulation, depending on the thickness of the walls, they have a certain thermal conductivity in combination and prevent heat from escaping in winter. Your task is to ensure that this parameter is not less than recommended in SNIP. The same is true for the foundation, ceilings and windows;
• find out where heat is lost, bring the parameters to standard ones;
• calculate the power of the boiler based on the total volume of rooms - for every 1 cubic meter. m of the room takes 41 W of heat (for example, a hallway of 10 m² with a ceiling height of 2.7 m requires 1107 W of heating, two 600 W batteries are needed);
• you can calculate from the opposite, that is, from the number of batteries. Each section of the aluminum battery gives 170 W of heat and heats 2-2.5 m of the room. If your house requires 30 battery sections, then the boiler that can heat the room must be at least 6 kW.

The worse the house is insulated, the higher the heat consumption from the heating system

An individual or average calculation is carried out for the object. The main purpose of such a survey is to good insulation and small heat leaks in winter, 3 kW can be used. In a building of the same area, but without insulation, at low winter temperatures, the power consumption will be up to 12 kW. Thus, the thermal power and load are estimated not only by area, but also by heat loss.

The main heat loss of a private house:

• windows - 10-55%;
• walls - 20-25%;
• chimney - up to 25%;
• roof and ceiling - up to 30%;
• low floors - 7-10%;
• temperature bridge in the corners - up to 10%

These indicators can vary for better and worse. They are rated according to the types installed windows, thickness of walls and materials, degree of insulation of the ceiling. For example, in poorly insulated buildings, heat loss through walls can reach 45% percent, in which case the expression “we drown the street” is applicable to the heating system. Methodology and The calculator will help you evaluate the nominal and calculated values.

Specificity of calculations

This technique can still be found under the name "thermal calculation". The simplified formula looks like this:

Qt = V × ∆T × K / 860, where

V is the volume of the room, m³;

∆T is the maximum difference between indoors and outdoors, °С;

K is the estimated heat loss coefficient;

860 is the conversion factor in kWh.

The heat loss coefficient K depends on the building structure, thickness and thermal conductivity of the walls. For simplified calculations, you can use the following parameters:

• K \u003d 3.0-4.0 - without thermal insulation (non-insulated frame or metal structure);
• K \u003d 2.0-2.9 - low thermal insulation (laying in one brick);
• K \u003d 1.0-1.9 - average thermal insulation (brickwork in two bricks);
• K \u003d 0.6-0.9 - good thermal insulation according to the standard.

These coefficients are averaged and do not allow estimating heat loss and heat load on the room, so we recommend using the online calculator.

gidpopechi.ru

Calculation of the heat load on the heating of a building: formula, examples

When designing a heating system, whether it is an industrial building or a residential building, it is necessary to carry out competent calculations and draw up a diagram of the heating system circuit. At this stage, experts recommend paying special attention to the calculation of the possible heat load on the heating circuit, as well as the amount of fuel consumed and heat generated.

This term refers to the amount of heat given off by heating devices. The carried out preliminary calculation of the heat load will allow avoiding unnecessary costs for the purchase of components of the heating system and for their installation. Also, this calculation will help to correctly distribute the amount of heat generated economically and evenly throughout the building.

There are many nuances in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Experts try to take into account as many factors and characteristics as possible to obtain a more accurate result.

The calculation of the heat load with errors and inaccuracies leads to inefficient operation of the heating system. It even happens that you have to redo sections of an already working structure, which inevitably leads to unplanned expenses. Yes, and housing and communal organizations calculate the cost of services based on data on heat load.

Main Factors

An ideally calculated and designed heating system must maintain the set temperature in the room and compensate for the resulting heat losses. When calculating the indicator of the heat load on the heating system in the building, you need to take into account:

Purpose of the building: residential or industrial.

Characteristics of the structural elements of the structure. These are windows, walls, doors, roof and ventilation system.

Housing dimensions. The larger it is, the more powerful the heating system should be. Be sure to take into account the area of ​​window openings, doors, exterior walls and the volume of each interior space.

Availability of rooms special purpose(bath, sauna, etc.).

Degree of equipment with technical devices. That is, the presence of hot water supply, ventilation systems, air conditioning and the type of heating system.

Temperature regime for a single room. For example, in rooms intended for storage, it is not necessary to maintain a comfortable temperature for a person.

Number of points with hot water supply. The more of them, the more the system is loaded.

Area of ​​glazed surfaces. Rooms with French windows lose a significant amount of heat.

Additional terms. In residential buildings, this can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, technological chain production process etc.

Climatic conditions of the region. When calculating heat losses, street temperatures are taken into account. If the differences are insignificant, then a small amount of energy will be spent on compensation. While at -40 ° C outside the window it will require significant expenses.

Features of existing methods

The parameters included in the calculation of the heat load are in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system, digital characteristics are taken regarding a specific heating radiator, boiler, etc. And also traditionally:

The heat consumption, taken to the maximum for one hour of operation of the heating system,

The maximum heat flow from one radiator,

Total heat costs in a certain period (most often - a season); if an hourly calculation of the load on the heating network is required, then the calculation must be carried out taking into account the temperature difference during the day.

The calculations made are compared with the heat transfer area of ​​the entire system. The index is quite accurate. Some deviations happen. For example, for industrial buildings, it will be necessary to take into account the reduction in heat energy consumption on weekends and holidays, and in residential buildings - at night.

Methods for calculating heating systems have several degrees of accuracy. To reduce the error to a minimum, it is necessary to use rather complex calculations. Less accurate schemes are used if the goal is not to optimize the costs of the heating system.

Basic calculation methods

To date, the calculation of the heat load on the heating of a building can be carried out in one of the following ways.

Three main

• Aggregated indicators are taken for calculation.
• The indicators of the structural elements of the building are taken as the base. Here, it will be important to calculate the heat loss used to heat the internal volume of air.
• All objects included in the heating system are calculated and summarized.

One exemplary

There is also a fourth option. It has a fairly large error, because the indicators are taken very average, or they are not enough. Here is the formula - Qot \u003d q0 * a * VH * (tEN - tHRO), where:

• q0 - specific thermal characteristic of the building (most often determined by the coldest period),
• a - correction factor (depends on the region and is taken from ready-made tables),
• VH is the volume calculated from the outer planes.

Example of a simple calculation

For a building with standard parameters (ceiling heights, room sizes and good thermal insulation characteristics), a simple ratio of parameters can be applied, adjusted for a coefficient depending on the region.

Suppose that a residential building is located in the Arkhangelsk region, and its area is 170 square meters. m. The heat load will be equal to 17 * 1.6 \u003d 27.2 kW / h.

Such a definition of thermal loads does not take into account many important factors. For example, the design features of the structure, temperature, the number of walls, the ratio of the areas of walls and window openings, etc. Therefore, such calculations are not suitable for serious heating system projects.

Calculation of a heating radiator by area

It depends on the material from which they are made. Most often today, bimetallic, aluminum, steel are used, much less often cast iron radiators. Each of them has its own heat transfer index (thermal power). Bimetal radiators with a distance between the axes of 500 mm, on average they have 180 - 190 watts. Aluminum radiators have almost the same performance.

The heat transfer of the described radiators is calculated for one section. Steel plate radiators are non-separable. Therefore, their heat transfer is determined based on the size of the entire device. For example, the thermal power of a two-row radiator 1,100 mm wide and 200 mm high will be 1,010 W, and a steel panel radiator 500 mm wide and 220 mm high will be 1,644 W.

The calculation of the heating radiator by area includes the following basic parameters:

Ceiling height (standard - 2.7 m),

Thermal power (per sq. m - 100 W),

One outer wall.

These calculations show that for every 10 sq. m requires 1,000 W of thermal power. This result is divided by the heat output of one section. The answer is the required number of radiator sections.

For the southern regions of our country, as well as for the northern ones, decreasing and increasing coefficients have been developed.

Average calculation and exact

Given the factors described, the average calculation is carried out according to the following scheme. If for 1 sq. m required 100W heat flow, then a room of 20 sq. m should receive 2,000 watts. A radiator (popular bimetallic or aluminum) of eight sections emits about 150 watts. We divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.

The exact one looks a little intimidating. Actually, nothing complicated. Here is the formula:

Qt = 100 W/m2 × S(room)m2 × q1 × q2 × q3 × q4 × q5 × q6 × q7, where:

• q1 - type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
• q2 – wall insulation (weak or absent = 1.27, 2-brick wall = 1.0, modern, high = 0.85);
• q3 - the ratio of the total area of ​​window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
• q4 - outdoor temperature (the minimum value is taken: -35оС = 1.5, -25оС = 1.3, -20оС = 1.1, -15оС = 0.9, -10оС = 0.7);
• q5 - the number of external walls in the room (all four = 1.4, three = 1.3, corner room = 1.2, one = 1.2);
• q6 - type of design room above the design room (cold attic = 1.0, warm attic = 0.9, residential heated room = 0.8);
• q7 - ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.1, 3.0 m = 1.05, 2.5 m = 1.3).

Using any of the methods described, it is possible to calculate the heat load of an apartment building.

Approximate calculation

These are the conditions. The minimum temperature in the cold season is -20°C. Room 25 sq. m with triple glazing, double-leaf windows, ceiling height of 3.0 m, two-brick walls and an unheated attic. The calculation will be as follows:

Q = 100 W/m2 × 25 m2 × 0.85 × 1 × 0.8(12%) × 1.1 × 1.2 × 1 × 1.05.

The result, 2 356.20, is divided by 150. As a result, it turns out that 16 sections need to be installed in a room with the specified parameters.

If calculation is required in gigacalories

In the absence of a heat energy meter on the open heating circuit calculation of the heat load for heating the building is calculated by the formula Q \u003d V * (T1 - T2) / 1000, where:

• V - the amount of water consumed by the heating system, calculated in tons or m3,
• T1 - a number indicating the temperature of hot water, measured in ° C, and for calculations, the temperature corresponding to a certain pressure in the system is taken. This indicator has its own name - enthalpy. If it is not possible to remove temperature indicators in a practical way, they resort to an average indicator. It is in the range of 60-65oC.
• T2 is the temperature of cold water. It is quite difficult to measure it in the system, so constant indicators have been developed that depend on the temperature regime on the street. For example, in one of the regions, in the cold season, this indicator is taken equal to 5, in summer - 15.
• 1,000 is the coefficient for obtaining the result immediately in gigacalories.

In the case of a closed circuit, the heat load (gcal/h) is calculated differently:

Qot \u003d α * qo * V * (tin - tn.r) * (1 + Kn.r) * 0.000001, where

• α is a coefficient designed to correct climatic conditions. It is taken into account if the street temperature differs from -30 ° C;
• V - the volume of the building according to external measurements;
• qo - specific heating index of the structure at a given tn.r = -30 ° C, measured in kcal / m3 * C;
• tv is the calculated internal temperature in the building;
• tn.r - estimated street temperature for drafting a heating system;
• Kn.r – infiltration coefficient. It is due to the ratio of heat losses of the calculated building with infiltration and heat transfer through external structural elements at the street temperature, which is set within the framework of the project being drawn up.

The calculation of the heat load turns out to be somewhat enlarged, but it is this formula that is given in the technical literature.

Inspection with a thermal imager

Increasingly, in order to increase the efficiency of the heating system, they resort to thermal imaging surveys of the building.

These works are carried out at night. For a more accurate result, you must observe the temperature difference between the room and the street: it must be at least 15 °. Fluorescent and incandescent lamps are switched off. It is advisable to remove carpets and furniture to the maximum, they knock down the device, giving some error.

The survey is carried out slowly, the data are recorded carefully. The scheme is simple.

The first stage of work takes place indoors. The device is moved gradually from doors to windows, paying special attention to corners and other joints.

The second stage is the examination of the external walls of the building with a thermal imager. The joints are still carefully examined, especially the connection with the roof.

The third stage is data processing. First, the device does this, then the readings are transferred to a computer, where the corresponding programs complete the processing and give the result.

If the survey was conducted by a licensed organization, then it will issue a report with mandatory recommendations based on the results of the work. If the work was carried out personally, then you need to rely on your knowledge and, possibly, the help of the Internet.

highlogistic.ru

Calculation of the heat load for heating: how to correctly perform?

The first and most important stage in the difficult process of organizing the heating of any real estate object (whether it be a country house or an industrial facility) is the competent design and calculation. In particular, it is necessary to calculate thermal loads on the heating system, as well as the volume of heat and fuel consumption.

Thermal loads

Performing preliminary calculations is necessary not only in order to obtain the entire range of documentation for organizing the heating of a property, but also to understand the volumes of fuel and heat, the selection of one or another type of heat generators.

Thermal loads of the heating system: characteristics, definitions

The definition of “heat load on heating” should be understood as the amount of heat that is collectively given off by heating devices installed in a house or other facility. It should be noted that before installing all the equipment, this calculation is made to exclude any troubles, unnecessary financial costs and work.

Calculation of heat loads for heating will help to organize uninterrupted and efficient work real estate heating systems. Thanks to this calculation, you can quickly complete absolutely all the tasks of heat supply, ensure their compliance with the norms and requirements of SNiP.

A set of instruments for performing calculations

The cost of an error in the calculation can be quite significant. The thing is that, depending on the calculated data received, the maximum expenditure parameters will be allocated in the housing and communal services department of the city, limits and other characteristics will be set, from which they are repelled when calculating the cost of services.

The total heat load on a modern heating system consists of several main load parameters:

• On the common system central heating;
• per system floor heating(if it is available in the house) - underfloor heating;
• Ventilation system (natural and forced);
• Hot water supply system;
• For all kinds of technological needs: swimming pools, baths and other similar structures.

Calculation and components of thermal systems at home

The main characteristics of the object, important to take into account when calculating the heat load

The most correctly and competently calculated heat load on heating will be determined only when absolutely everything, even the smallest details and parameters, is taken into account.

This list is quite large and can include:

• Type and purpose of real estate objects. A residential or non-residential building, an apartment or an administrative building - all this is very important for obtaining reliable thermal calculation data.

Also, the load rate, which is determined by heat supplier companies and, accordingly, heating costs, depends on the type of building;

• Architectural part. The dimensions of all kinds of external fences (walls, floors, roofs), the dimensions of openings (balconies, loggias, doors and windows) are taken into account. The number of storeys of the building, the presence of basements, attics and their features are important;
• Temperature requirements for each of the premises of the building. This parameter should be understood as temperature regimes for each room of a residential building or zone of an administrative building;
• The design and features of external fences, including the type of materials, thickness, the presence of insulating layers;

Physical indicators of room cooling - data for calculating the heat load

• The nature of the premises. As a rule, it is inherent in industrial buildings, where for a workshop or site it is necessary to create some specific thermal conditions and modes;
• Availability and parameters of special premises. The presence of the same baths, pools and other similar structures;
• Degree Maintenance- the presence of hot water supply, such as central heating, ventilation and air conditioning systems;
• The total number of points from which hot water is drawn. It is on this characteristic that special attention should be paid, because what more number points - the greater the heat load on the entire heating system as a whole;
• The number of people living in the home or in the facility. The requirements for humidity and temperature depend on this - factors that are included in the formula for calculating the heat load;

Equipment that can affect thermal loads

• Other data. For an industrial facility, such factors include, for example, the number of shifts, the number of workers per shift, and working days per year.

As for a private house, you need to take into account the number of people living, the number of bathrooms, rooms, etc.

Calculation of heat loads: what is included in the process

Do-it-yourself calculation of the heating load itself is carried out even at the design stage of a country cottage or other real estate object - this is due to simplicity and the absence of extra cash costs. At the same time, the requirements of various norms and standards, TCP, SNB and GOST are taken into account.

The following factors are mandatory for determination during the calculation of thermal power:

• Heat losses of external protections. Includes the desired temperature conditions in each of the rooms;
• The power required to heat the water in the room;
• The amount of heat required to heat the air ventilation (in the case when forced ventilation is required);
• The heat needed to heat the water in the pool or bath;

Gcal/hour - a unit of measurement of thermal loads of objects

• Possible developments of the further existence of the heating system. It implies the possibility of outputting heating to the attic, to the basement, as well as all kinds of buildings and extensions;

Heat loss in a standard residential building

Advice. With a "margin", thermal loads are calculated in order to exclude the possibility of unnecessary financial costs. It is especially important for a country house, where additional connection of heating elements without preliminary study and preparation will be prohibitively expensive.

Features of calculating the heat load

As already mentioned earlier, the design parameters of indoor air are selected from the relevant literature. At the same time, heat transfer coefficients are selected from the same sources (passport data of heating units are also taken into account).

The traditional calculation of heat loads for heating requires a consistent determination of the maximum heat flow from heating devices (all heating batteries actually located in the building), the maximum hourly heat energy consumption, as well as the total heat power consumption for a certain period, for example, the heating season.

Distribution of heat fluxes from various types heaters

The above instructions for calculating thermal loads, taking into account the surface area of ​​​​heat exchange, can be applied to various real estate objects. It should be noted that this method allows you to competently and most correctly develop a justification for the use of efficient heating, as well as energy inspection of houses and buildings.

An ideal calculation method for the standby heating of an industrial facility, when temperatures are expected to drop during non-working hours (holidays and weekends are also taken into account).

Methods for determining thermal loads

Currently, thermal loads are calculated in several main ways:

1. Calculation of heat losses by means of enlarged indicators;
2. Determination of parameters through various elements of enclosing structures, additional losses for air heating;
3. Calculation of heat transfer of all heating and ventilation equipment installed in the building.

Enlarged method for calculating heating loads

Another method for calculating the loads on the heating system is the so-called enlarged method. As a rule, such a scheme is used in the case when there is no information about projects or such data does not correspond to the actual characteristics.

Examples of heat loads for residential apartment buildings and their dependence on the number of people living and area

For an enlarged calculation of the heat load of heating, a rather simple and uncomplicated formula is used:

Qmax from.=α*V*q0*(tv-tn.r.)*10-6

The following coefficients are used in the formula: α is a correction factor that takes into account the climatic conditions in the region where the building was built (applied when the design temperature is different from -30C); q0 specific heating characteristic, selected depending on the temperature of the coldest week of the year (the so-called "five days"); V is the outer volume of the building.

Types of thermal loads to be taken into account in the calculation

In the course of calculations (as well as in the selection of equipment), it is taken into account a large number of a wide variety of thermal loads:

1. seasonal loads. As a rule, they have the following features:

• Throughout the year, there is a change in thermal loads depending on the air temperature outside the premises;
• Annual heat consumption, which is determined by the meteorological features of the region where the facility is located, for which heat loads are calculated;

Thermal load regulator for boiler equipment

• Changing the load on the heating system depending on the time of day. Due to the heat resistance of the building's external enclosures, such values ​​are accepted as insignificant;
• Thermal energy costs ventilation system by hours of the day.

1. Year-round thermal loads. It should be noted that for heating and hot water supply systems, most domestic facilities have heat consumption throughout the year, which changes quite a bit. So, for example, in summer the cost of thermal energy in comparison with winter is reduced by almost 30-35%;
2. Dry heat - convection heat transfer and thermal radiation from other similar devices. Determined by dry bulb temperature.

This factor depends on the mass of parameters, including all kinds of windows and doors, equipment, ventilation systems and even air exchange through cracks in the walls and ceilings. It also takes into account the number of people who can be in the room;

1. Latent heat is evaporation and condensation. Based on wet bulb temperature. The amount of latent heat of humidity and its sources in the room is determined.

Heat loss of a country house

In any room, humidity is affected by:

• People and their number who are simultaneously in the room;
• Technological and other equipment;
• Air flows that pass through cracks and crevices in building structures.

Thermal load regulators as a way out of difficult situations

As you can see in many photos and videos of modern industrial and domestic heating boilers and other boiler equipment, they come with special heat load regulators. The technique of this category is designed to provide support for a certain level of loads, to exclude all kinds of jumps and dips.

It should be noted that RTN can significantly save on heating bills, because in many cases (and especially for industrial enterprises) certain limits are set that cannot be exceeded. Otherwise, if jumps and excesses of thermal loads are recorded, fines and similar sanctions are possible.

An example of the total heat load for a certain area of ​​the city

Advice. Loads on heating, ventilation and air conditioning systems are an important point in designing a house. If it is impossible to carry out the design work on your own, then it is best to entrust it to specialists. At the same time, all formulas are simple and uncomplicated, and therefore it is not so difficult to calculate all the parameters by yourself.

Loads on ventilation and hot water supply - one of the factors of thermal systems

Thermal loads for heating, as a rule, are calculated in combination with ventilation. This is a seasonal load, it is designed to replace the exhaust air with clean air, as well as heat it up to the set temperature.

Hourly heat consumption for ventilation systems is calculated according to a certain formula:

Qv.=qv.V(tn.-tv.), where

Measurement of heat loss in a practical way

In addition to, in fact, ventilation, thermal loads are also calculated on the hot water supply system. The reasons for such calculations are similar to ventilation, and the formula is somewhat similar:

Qgvs.=0.042rv(tg.-tx.)Pgav, where

r, in, tg., tx. - the calculated temperature of hot and cold water, the density of water, as well as the coefficient in which the values ​​\u200b\u200bare taken into account maximum load hot water supply to the average value established by GOST;

Comprehensive calculation of thermal loads

In addition to, in fact, theoretical issues of calculation, some practical work. So, for example, comprehensive thermal surveys include mandatory thermography of all structures - walls, ceilings, doors and windows. It should be noted that such works make it possible to determine and fix the factors that have a significant impact on the heat loss of the building.

Device for calculations and energy audit

Thermal imaging diagnostics will show what the real temperature difference will be when a certain strictly defined amount of heat passes through 1m2 of enclosing structures. Also, it will help to find out the heat consumption at a certain temperature difference.

Practical measurements are an indispensable component of various computational works. In combination, such processes will help to obtain the most reliable data on thermal loads and heat losses that will be observed in a particular building over a certain period of time. A practical calculation will help to achieve what the theory does not show, namely the "bottlenecks" of each structure.

Conclusion

Calculation of thermal loads, as well as hydraulic calculation of the heating system - important factor, which must be calculated before starting the organization of the heating system. If all the work is done correctly and the process is approached wisely, you can guarantee trouble-free operation of heating, as well as save money on overheating and other unnecessary costs.

Page 2

Heating boilers

One of the main components of comfortable housing is the presence of a well-thought-out heating system. At the same time, the choice of the type of heating and the required equipment is one of the main questions that need to be answered at the design stage of the house. An objective calculation of the heating boiler power by area will eventually allow you to get a completely efficient heating system.

We will now tell you about the competent conduct of this work. In doing so, consider the features inherent in different types heating. After all, they must be taken into account when carrying out calculations and the subsequent decision to install one or another type of heating.

Basic calculation rules

• room area (S);
• specific power of the heater per 10 m² of heated area - (W sp.). This value is determined adjusted for the climatic conditions of a particular region.

This value (W beats) is:

• for the Moscow region - from 1.2 kW to 1.5 kW;
• for the southern regions of the country - from 0.7 kW to 0.9 kW;
• for the northern regions of the country - from 1.5 kW to 2.0 kW.

Let's do the calculations

The power calculation is carried out as follows:

W cat. \u003d (S * Wsp.): 10

Advice! For simplicity, a simplified version of this calculation can be used. In it Wud.=1. Therefore, the heat output of the boiler is defined as 10kW per 100m² of heated area. But with such calculations, at least 15% must be added to the obtained value in order to get a more objective figure.

Calculation example

As you can see, the instructions for calculating the heat transfer intensity are simple. But, nevertheless, we will accompany it with a concrete example.

The conditions will be as follows. The area of ​​heated premises in the house is 100m². Specific power for the Moscow region is 1.2 kW. Substituting the available values ​​into the formula, we get the following:

W boiler \u003d (100x1.2) / 10 \u003d 12 kilowatts.

Calculation for different types of heating boilers

The degree of efficiency of the heating system depends primarily on right choice her type. And of course, from the accuracy of the calculation of the required performance of the heating boiler. If the calculation of the thermal power of the heating system was not carried out accurately enough, then negative consequences will inevitably arise.

If the heat output of the boiler is less than required, it will be cold in the rooms in winter. In the case of excess performance, there will be an overexpenditure of energy and, accordingly, the money spent on heating the building.

House heating system

To avoid these and other problems, it is not enough just to know how to calculate the power of a heating boiler.

It is also necessary to take into account the features inherent in systems using different types of heaters (you can see a photo of each of them later in the text):

• solid fuel;
• electric;
• liquid fuel;
• gas.

The choice of one or another type largely depends on the region of residence and the level of infrastructure development. Equally important is the availability of the possibility of acquiring a certain type of fuel. And, of course, its cost.

Solid fuel boilers

The calculation of the power of a solid fuel boiler must be made taking into account the features characterized by the following features of such heaters:

• low popularity;
• relative accessibility;
• the possibility of autonomous operation - it is provided for in a number of modern models of these devices;
• economy during operation;
• the need for additional fuel storage space.

solid fuel heater

Another characteristic feature that should be taken into account when calculating the heating power of a solid fuel boiler is the cyclicity of the temperature obtained. That is, in rooms heated with its help, the daily temperature will fluctuate within 5ºС.

Therefore, such a system is far from the best. And if possible, it should be abandoned. But, if this is not possible, there are two ways to smooth out the existing shortcomings:

1. Using a bulb, which is needed to adjust the air supply. This will increase the burning time and reduce the number of furnaces;
2. The use of water heat accumulators with a capacity of 2 to 10 m². They are included in the heating system, allowing you to reduce energy costs and, thereby, save fuel.

All this will reduce the required performance of a solid fuel boiler for heating a private house. Therefore, the effect of the application of these measures must be taken into account when calculating the power of the heating system.

Electric boilers

Electric boilers for home heating are characterized by the following features:

• high cost of fuel - electricity;
• possible problems due to network interruptions;
• environmental friendliness;
• ease of management;
• compactness.

electric boiler

All these parameters should be taken into account when calculating the power of an electric heating boiler. After all, it is not purchased for one year.

Oil boilers

They have the following characteristic features:

• not eco-friendly;
• convenient in operation;
• require additional storage space for fuel;
• have an increased fire hazard;
• use fuel, the price of which is quite high.

Oil heater

gas boilers

In most cases, they are the best option for organizing a heating system. Household gas heating boilers have the following characteristic features that must be taken into account when calculating the power of the heating boiler:

• ease of operation;
• do not require a place to store fuel;
• safe in operation;
• low cost of fuel;
• economy.

A gas boiler

Calculation for heating radiators

Let's say you decide to install a heating radiator with your own hands. But first you need to buy it. And choose exactly the one that suits the power.

• First, we determine the volume of the room. To do this, multiply the area of ​​​​the room by its height. As a result, we get 42m³.
• Further, you should know that it takes 41 watts to heat 1m³ of a room in central Russia. Therefore, to find out the desired performance of the radiator, we multiply this figure (41 W) by the volume of the room. As a result, we get 1722W.
• Now let's calculate how many sections our radiator should have. Make it simple. Each element of a bimetallic or aluminum radiator has a heat transfer of 150W.
• Therefore, we divide the performance we obtained (1722W) by 150. We get 11.48. Round up to 11.
• Now you need to add another 15% to the resulting figure. This will help smooth out the increase in required heat transfer during the most severe winters. 15% of 11 is 1.68. Round up to 2.
• As a result, we add 2 more to the existing figure (11). We get 13. So, to heat a room with an area of ​​​​14m², we need a radiator with a power of 1722W, which has 13 sections.

Now you know how to calculate the desired performance of the boiler, as well as the heating radiator. Take advantage of our advice and provide yourself with an efficient and at the same time not wasteful heating system. If you need more detailed information, then you can easily find it in the corresponding video on our website.

Page 3

All this equipment, indeed, requires a very respectful, prudent attitude - mistakes lead not only to financial losses, but to losses in health and attitude to life.

When we decide to build our own private house, we are primarily guided by largely emotional criteria - we want to have our own separate housing, independent of city utilities, much larger in size and made according to our own ideas. But somewhere in the soul, of course, there is an understanding that you will have to count a lot. The calculations relate not so much to the financial component of all work, but to the technical one. One of the most important types of calculations will be the calculation of the mandatory heating system, without which there is no escape.

First, of course, you need to take up the calculations - a calculator, a piece of paper and a pen will be the first tools

To begin with, decide what is called, in principle, about the methods of heating your home. After all, you have several options for providing heat at your disposal:

• Autonomous heating electrical appliances. It is possible that such devices are good, and even popular, as auxiliary means of heating, but they cannot be considered as the main ones.

• Electric heating floors. But this method of heating may well be used as the main one for a single living room. But there is no question of providing all the rooms in the house with such floors.

• Heating fireplaces. A brilliant option, it warms not only the air in the room, but also the soul, creates an unforgettable atmosphere of comfort. But then again, no one considers fireplaces as a means of providing heat throughout the house - only in the living room, only in the bedroom, and nothing more.

• Centralized water heating. Having “torn off” yourself from the high-rise building, you, nevertheless, can bring its “spirit” into your home by connecting to centralized system heating. Is it worth it!? Is it worth it again to rush "out of the fire, but into the frying pan." This should not be done, even if such a possibility exists.

• Autonomous water heating. But this method of providing heat is the most efficient, which can be called the main one for private houses.

You can not do without a detailed plan of the house with a layout of equipment and wiring of all communications

After resolving the issue in principle

When the solution to the fundamental question of how to provide heat in the house using an autonomous water system has taken place, you need to move on and understand that it will be incomplete if you do not think about

• Installation of reliable window systems that will not just “lower” all your successes in heating to the street;

• Additional insulation of both external and internal walls of the house. The task is very important and requires a separate serious approach, although it is not directly related to the future installation of the heating system itself;

• Fireplace installation. Recently, this auxiliary heating method has been increasingly used. It may not replace general heating, but it is such an excellent support for it that in any case it helps to significantly reduce heating costs.

The next step is to create a very accurate diagram of your building with all the elements of the heating system integrated into it. Calculation and installation of heating systems without such a scheme is impossible. The elements of this scheme will be:

• Heating boiler, as the main element of the entire system;
• A circulation pump that provides the coolant current in the system;
• Pipelines, as a kind of "blood vessels" of the entire system;
• Heating batteries are those devices that have long been known to everyone and which are the final elements of the system and are responsible in our eyes for the quality of its work;
• Devices for monitoring the state of the system. An accurate calculation of the volume of the heating system is unthinkable without the presence of such devices that provide information about the actual temperature in the system and the volume of the passing coolant;
• Locking and adjusting devices. Without these devices, the work will be incomplete, it is they who will allow you to regulate the operation of the system and adjust according to the readings of the control devices;
• Various fitting systems. These systems could well be attributed to pipelines, but their influence on the successful operation of the entire system is so great that fittings and connectors are separated into a separate group of elements for the design and calculation of heating systems. Some experts call electronics the science of contacts. It is possible, without fear of making a big mistake, to call the heating system - in many respects, the science of the quality of the compounds that provide the elements of this group.

The heart of the entire hot water heating system is the heating boiler. Modern boilers are entire systems for providing the entire system with hot coolant

Helpful advice! When it comes to the heating system, this word “coolant” often appears in the conversation. It is possible, with some degree of approximation, to consider ordinary “water” as the medium that is intended to move through the pipes and radiators of the heating system. But there are some nuances that are associated with the way water is supplied to the system. There are two ways - internal and external. External - from an external cold water supply. In this situation, indeed, the coolant will be ordinary water, with all its shortcomings. Firstly, in general availability, and, secondly, purity. When choosing this method of introducing water from the heating system, we highly recommend installing a filter at the inlet, otherwise severe contamination of the system cannot be avoided in just one season of operation. If a completely autonomous filling of water into the heating system is chosen, then do not forget to “flavor” it with all kinds of additives against solidification and corrosion. It is water with such additives that is already called a coolant.

Types of heating boilers

Among the heating boilers available for your choice are the following:

• Solid fuel - can be very good in remote areas, in the mountains, in the Far North, where there are problems with external communications. But if access to such communications is not difficult, solid fuel boilers are not used, they lose in the convenience of working with them, if it is still necessary to keep one level of heat in the house;

• Electric - and where now without electricity. But you need to understand that the cost of this type of energy in your house when using electric heating boilers will be so high that the solution to the question “how to calculate the heating system” in your house will lose any meaning - everything will go into electric wires;

• Liquid fuel. Such boilers on gasoline, solarium, suggest themselves, but they, due to their non-environmental friendliness, are very unloved by many, and rightly so;

• Domestic gas heating boilers are the most common types of boilers, very easy to operate and do not require a supply of fuel. The efficiency of such boilers is the highest of all available on the market and reaches 95%.

Pay special attention to the quality of all materials used, there is no time for savings, the quality of each component of the system, including pipes, must be perfect

Boiler calculation

When they talk about the calculation of an autonomous heating system, they first of all mean the calculation of a heating gas boiler. Any example of calculating the heating system includes the following formula for calculating the boiler power:

W \u003d S * Wsp / 10,

• S is the total area of ​​the heated premises in square meters;
• Wsp - specific power of the boiler per 10 sq.m. premises.

The specific power of the boiler is set depending on the climatic conditions of the region of its use:

• for the Middle band, it ranges from 1.2 to 1.5 kW;
• for areas of the level of Pskov and above - from 1.5 to 2.0 kW;
• for Volgograd and below - from 0.7 - 0.9 kW.

But, after all, our climate of the XXI century has become so unpredictable that, by and large, the only criterion when choosing a boiler is your acquaintance with the experience of other heating systems. Perhaps, understanding this unpredictability, for simplicity, it has long been accepted in this formula to always take the specific power as a unit. Although do not forget about the recommended values.

Calculation and design of heating systems, to a large extent - the calculation of all junction points, the latest connecting systems, of which there are a huge number on the market, will help here

Helpful advice! This is the desire - to get acquainted with the existing, already working, autonomous heating systems will be very important. If you decide to establish such a system at home, and even with your own hands, then be sure to get acquainted with the heating methods used by your neighbors. Getting a "heating system calculation calculator" first hand will be very important. You will kill two birds with one stone - you will get a good adviser, and maybe in the future a good neighbor, and even a friend, and avoid mistakes that your neighbor may have made at one time.

Circulation pump

The method of supplying the coolant to the system largely depends on the heated area - natural or forced. Natural does not require any additional equipment and involves the movement of the coolant through the system due to the principles of gravity and heat transfer. Such a heating system can also be called passive.

Much more widespread are active heating systems, in which heat carrier is used to move circulation pump. It is more common to install such pumps on the line from radiators to the boiler, when the water temperature has already subsided and will not be able to adversely affect the operation of the pump.

There are certain requirements for pumps:

• they must be quiet, because they work constantly;
• they should consume little, again because of their constant work;
• they must be very reliable, and this is the most important requirement for pumps in a heating system.

Piping and radiators

The most important component of the entire heating system, which any user constantly encounters, is pipes and radiators.

When it comes to pipes, we have three types of pipes at our disposal:

• steel;
• copper;
• polymeric.

Steel - the patriarchs of heating systems, used from time immemorial. Now steel pipes are gradually disappearing "from the scene", they are inconvenient to use, and, in addition, require welding and are subject to corrosion.

Copper - very popular pipes, especially if carried out hidden wiring. Such pipes are extremely resistant to external influences, but, unfortunately, are very expensive, which is the main brake on their widespread use.

Polymer - as a solution to the problems of copper pipes. It is polymer pipes that are the hit of use in modern heating systems. High reliability, resistance to external influences, a huge selection of additional auxiliary equipment specifically for use in heating systems with polymer pipes.

The heating of the house is largely ensured by the precise selection of the piping system and the laying of pipes.

Calculation of radiators

The thermotechnical calculation of the heating system necessarily includes the calculation of such an indispensable element of the network as a radiator.

The purpose of calculating the radiator is to obtain the number of its sections for heating a room of a given area.

Thus, the formula for calculating the number of sections in a radiator is:

K = S / (W / 100),

• S - the area of ​​​​the heated room in square meters (we heat, of course, not the area, but the volume, but the standard height of the room is 2.7 m);
• W - heat transfer of one section in Watts, characteristic of the radiator;
• K is the number of sections in the radiator.

Providing heat in the house is a solution to a whole range of tasks, often not related to each other, but serving the same purpose. Installing a fireplace can be one of these standalone tasks.

In addition to the calculation, radiators also require compliance with certain requirements during their installation:

• installation must be carried out strictly under the windows, in the center, a long and generally accepted rule, but some manage to break it (such an installation prevents the movement of cold air from the window);
• The "ribs" of the radiator must be aligned vertically - but this requirement, somehow no one particularly claims to violate it, is obvious;
• something else is not obvious - if there are several radiators in the room, they should be located on the same level;
• it is necessary to provide at least 5 cm gaps from the top to the window sill and from the bottom to the floor from the radiator, ease of maintenance plays an important role here.

Skillful and accurate placement of radiators ensures the success of the whole end result - here you can not do without diagrams and modeling of the location depending on the size of the radiators themselves

Calculation of water in the system

The calculation of the volume of water in the heating system depends on the following factors:

• the volume of the heating boiler - this characteristic is known;
• pump performance - this characteristic is also known, but it should, in any case, provide the recommended speed of movement of the coolant through the system of 1 m / s;
• the volume of the entire pipeline system - this must already be calculated in fact, after the installation of the system;
• the total volume of radiators.

The ideal, of course, is to hide all communications behind a plasterboard wall, but this is not always possible, and it raises questions from the point of view of the convenience of future maintenance of the system.

Helpful advice! It is often impossible to accurately calculate the required volume of water in the system with mathematical accuracy. So they act a little differently. First, the system is filled, presumably by 90% of the volume, and its performance is checked. As you work, vent excess air and continue filling. Hence, there is a need for an additional reservoir with a coolant in the system. As the system operates, a natural decrease in the coolant occurs as a result of evaporation and convection processes, therefore, the calculation of the replenishment of the heating system consists in monitoring the loss of water from the additional reservoir.

Definitely turn to the experts.

Many repair work Of course, you can also do housework by yourself. But creating a heating system requires too much knowledge and skills. Therefore, even having studied all the photo and video materials on our website, even having familiarized yourself with such indispensable attributes of each element of the system as an “instruction”, we still recommend that you contact professionals for installing a heating system.

As the top of the entire heating system - the creation of warm heated floors. But the feasibility of installing such floors should be very carefully calculated.

The cost of errors when installing an autonomous heating system is very high. It's not worth the risk in this situation. The only thing left for you is the smart maintenance of the entire system and the call of the masters for its maintenance.

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Competently made calculations of the heating system for any building - a residential building, workshop, office, shop, etc., will guarantee its stable, correct, reliable and silent operation. In addition, you will avoid misunderstandings with housing and communal services workers, unnecessary financial costs and energy losses. Heating can be calculated in several stages.

When calculating heating, many factors must be taken into account.

Calculation stages

• First you need to know the heat loss of the building. This is necessary to determine the power of the boiler, as well as each of the radiators. Heat losses are calculated for each room with an external wall.

Note! The next step is to check the data. Divide the resulting numbers by the quadrature of the room. Thus, you will get specific heat losses (W/m²). As a rule, this is 50/150 W / m². If the received data is very different from those indicated, then you made a mistake. Therefore, the price of assembling the heating system will be too high.

• Next, you need to choose temperature regime. It is advisable to take the following parameters for calculations: 75-65-20 ° (boiler-radiators-room). Such a temperature regime, when calculating heat, complies with the European heating standard EN 442.

Heating scheme.

• Then you need to select the power of the heating batteries, based on the data on heat losses in the rooms.
• After that, a hydraulic calculation is carried out - heating without it will not be effective. It is needed to determine the diameter of the pipes and technical properties circulation pump. If the house is private, then the pipe section can be selected according to the table, which will be given below.
• Next, you need to decide on a heating boiler (domestic or industrial).
• Then the volume of the heating system is found. You need to know its capacity in order to choose expansion tank or make sure that the volume of the water tank already built into the heat generator is enough. Any online calculator will help you get the necessary data.

Thermal calculation

To carry out the heat engineering stage of designing a heating system, you will need initial data.

What you need to get started

House project.

1. First of all, you will need a building project. It should indicate the external and internal dimensions of each of the rooms, as well as windows and external doorways.
2. Next, find out the data on the location of the building in relation to the cardinal points, as well as the climatic conditions in your area.
3. Gather information about the height and composition of the exterior walls.
4. You will also need to know the parameters of the floor materials (from the room to the ground), as well as the ceiling (from the premises to the street).

After collecting all the data, you can start calculating the heat consumption for heating. As a result of the work, you will collect information on the basis of which you can carry out hydraulic calculations.

Required formula

Building heat loss.

Calculation of thermal loads on the system should determine the heat losses and boiler output. In the latter case, the formula for calculating heating is as follows:

Mk = 1.2 ∙ Tp, where:

• Mk is the power of the heat generator, in kW;
• Tp - heat loss of the building;
• 1.2 is a margin equal to 20%.

Note! This safety factor takes into account the possibility of a pressure drop in the gas pipeline system in winter, in addition to unforeseen heat losses. For example, as the photo shows, due to a broken window, poor thermal insulation of doors, severe frosts. Such a margin allows you to widely regulate the temperature regime.

It should be noted that when the amount of thermal energy is calculated, its losses throughout the building are not evenly distributed, on average, the figures are as follows:

• external walls lose about 40% of the total figure;
• 20% goes through the windows;
• floors give about 10%;
• 10% escapes through the roof;
• 20% leave through ventilation and doors.

Material coefficients

Thermal conductivity coefficients of some materials.

• K1 - type of windows;
• K2 - thermal insulation of walls;
• K3 - means the ratio of the area of ​​\u200b\u200bwindows and floors;
• K4 - the minimum temperature regime outside;
• K5 - the number of external walls of the building;
• K6 - number of storeys of the structure;
• K7 - the height of the room.

As for windows, their heat loss coefficients are:

• traditional glazing - 1.27;
• double-glazed windows - 1;
• three-chamber analogues - 0.85.

The larger the windows are relative to the floors, the more heat the building loses.

When calculating the consumption of thermal energy for heating, keep in mind that the material of the walls has the following coefficient values:

• concrete blocks or panels - 1.25 / 1.5;
• timber or logs - 1.25;
• masonry in 1.5 bricks - 1.5;
• masonry in 2.5 bricks - 1.1;
• foam concrete blocks - 1.

At negative temperatures, heat leakage also increases.

1. Up to -10°, the coefficient will be equal to 0.7.
2. From -10° it will be 0.8.
3. At -15 °, you need to operate with a figure of 0.9.
4. Up to -20° - 1.
5. From -25° the value of the coefficient will be 1.1.
6. At -30° it will be 1.2.
7. Up to -35°, this value is 1.3.

When you calculate thermal energy, keep in mind that its loss also depends on how many external walls are in the building:

• one external wall - 1%;
• 2 walls - 1.2;
• 3 outer walls - 1.22;
• 4 walls - 1.33.

The greater the number of floors, the more difficult the calculations.

The number of floors or the type of premises located above the living room affect the coefficient K6. When the house has two floors or more, the calculation of heat energy for heating takes into account the coefficient 0.82. If at the same time the building has a warm attic, the figure changes to 0.91, if this room is not insulated, then to 1.

The height of the walls affects the level of the coefficient as follows:

• 2.5 m - 1;
• 3 m - 1.05;
• 3.5 m - 1.1;
• 4 m - 1.15;
• 4.5 m - 1.2.

Among other things, the methodology for calculating the need for thermal energy for heating takes into account the area of ​​\u200b\u200bthe room - Pk, as well as the specific value of heat losses - UDtp.

The final formula for the necessary calculation of the heat loss coefficient looks like this:

Tp \u003d UDtp ∙ Pl ∙ K1 ∙ K2 ∙ K3 ∙ K4 ∙ K5 ∙ K6 ∙ K7. In this case, UDtp is 100 W/m².

Calculation example

The building for which we will find the load on the heating system will have the following parameters.

1. Windows with double glazing, i.e. K1 is 1.
2. External walls - foam concrete, the coefficient is the same. 3 of them are external, in other words K5 is 1.22.
3. The square of the windows is 23% of the same indicator of the floor - K3 is 1.1.
4. Outside temperature is -15°, K4 is 0.9.
5. The attic of the building is not insulated, in other words, K6 will be 1.
6. The height of the ceilings is three meters, i.е. K7 is 1.05.
7. The area of ​​the premises is 135 m².

Knowing all the numbers, we substitute them into the formula:

Fri = 135 ∙ 100 ∙ 1 ∙ 1 ∙ 1.1 ∙ 0.9 ∙ 1.22 ∙ 1 ∙ 1.05 = 17120.565 W (17.1206 kW).

Mk = 1.2 ∙ 17.1206 = 20.54472 kW.

Hydraulic calculation for heating system

An example of a hydraulic calculation scheme.

This design stage will help you choose the right length and diameter of pipes, as well as correctly balance the heating system using radiator valves. This calculation will give you the opportunity to choose the power of the electric circulation pump.

High quality circulation pump.

According to the results of hydraulic calculations, you need to find out the following numbers:

• M is the amount of water flow in the system (kg/s);
• DP - head loss;
• DP1, DP2… DPn, - loss of pressure, from the heat generator to each battery.

The flow rate of the coolant for the heating system is found by the formula:

M = Q/Cp ∙ DPt

1. Q means the total heating power, taken taking into account the heat losses of the house.
2. Cp is the specific heat capacity of water. To simplify the calculations, it can be taken as 4.19 kJ.
3. DPt is the temperature difference at the inlet and outlet of the boiler.

In the same way, it is possible to calculate the consumption of water (coolant) in any section of the pipeline. Select sections so that the fluid velocity is the same. According to the standard, division into sections must be carried out before reduction or tee. Next, sum up the power of all batteries to which water is supplied through each pipe interval. Then substitute the value in the above formula. These calculations must be made for the pipes in front of each of the batteries.

• V is the speed of advancement of the coolant (m/s);
• M - water consumption in the pipe section (kg / s);
• P is its density (1 t/m³);
  • F is the cross-sectional area of ​​the pipes (m²), it is found by the formula: π ∙ r / 2, where the letter r means the inner diameter.

DPptr = R ∙ L,

• R means specific friction loss in the pipe (Pa/m);
• L is the length of the section (m);

After that, calculate the pressure loss on the resistances (fittings, fittings), the action formula:

Dms = Σξ ∙ V²/2 ∙ P

• Σξ denotes the sum of the coefficients of local resistance in a given section;
• V - water velocity in the system
• P is the density of the coolant.

Note! In order for the circulation pump to sufficiently provide heat to all batteries, the pressure loss on the long branches of the system should not be more than 20,000 Pa. The coolant flow rate should be from 0.25 to 1.5 m/s.

If the speed is above the specified value, noise will appear in the system. The minimum speed value of 0.25 m / s is recommended by snip No. 2.04.05-91 so that the pipes do not air.

Pipes made of different materials have different properties.

In order to comply with all the voiced conditions, it is necessary to choose the right diameter of the pipes. You can do this according to the table below, which shows the total power of the batteries.

At the end of the article, you can watch a tutorial video on its topic.

Page 5

For installation, heating design standards must be observed

Numerous companies, as well as individuals, offer the population heating design with its subsequent installation. But do you really, if you are managing a construction site, do you definitely need a specialist in the field of calculation and installation of heating systems and appliances? The fact is that the price of such work is quite high, but with some effort, you can do it yourself.

How to heat your house

It is impossible to consider the installation and design of heating systems of all types in one article - it is better to pay attention to the most popular ones. Therefore, let's dwell on the calculations of water radiator heating and some features of boilers for heating water circuits.

Calculation of the number of radiator sections and installation location

Sections can be added and removed by hand

• Some Internet users have an obsessive desire to find SNiP for heating calculations in the Russian Federation, but such installations simply do not exist. Such rules are possible for a very small region or country, but not for a country with the most diverse climate. The only thing that can be advised to lovers of printed standards is to refer to the tutorial on designing water heating systems for the universities of Zaitsev and Lyubarets.
• The only standard that deserves attention is the amount of heat energy that should be released by a radiator per 1m2 of the room, with an average ceiling height of 270 cm (but not more than 300 cm). The heat transfer power should be 100W, therefore, the formula is suitable for calculations:

Knumber of sections \u003d S room area * 100 / P power of one section

• For example, you can calculate how many sections you need for a room of 30m2 with a specific power of one section of 180W. In this case, K=S*100/P=30*100/180=16.66. Round this number up for the margin and get 17 sections.

Panel radiators

• But what if the design and installation of heating systems is carried out by panel radiators, where it is impossible to add or remove part of the heater. In this case, it is necessary to select the battery power according to the cubic capacity of the heated room. Now we need to apply the formula:

P panel radiator power = V volume of heated room * 41 required amount of W per 1 cu.

• Let's take a room of the same size with a height of 270 cm and get V=a*b*h=5*6*2?7=81m3. Let's substitute the initial data to the formula: P=V*41=81*41=3.321kW. But such radiators do not exist, so let's go up and get a device with a power reserve of 4 kW.

The radiator must be hung under the window

• Whatever metal the radiators are made of, the rules for designing heating systems provide for their location under the window. The battery heats the air enveloping it, and as it heats up, it becomes lighter and rises. These warm streams create a natural barrier to cold streams moving from window panes, thus increasing the efficiency of the appliance.
• Therefore, if you have calculated the number of sections or calculated the required radiator power, this does not mean at all that you can limit yourself to one device if there are several windows in the room (for some panel radiators, the instructions mention this). If the battery consists of sections, then they can be divided, leaving the same amount under each window, and you just need to purchase several pieces of water for panel heaters, but of lesser power.

Boiler selection for the project

Covtion gas boiler Bosch Gaz 3000W

• The terms of reference for the design of the heating system also include the choice of a domestic heating boiler, and if it runs on gas, then in addition to the difference in design power, it may turn out to be convection or condensing. The first system is quite simple - in this case, thermal energy arises only from gas combustion, but the second is more complex, because water vapor is also involved there, as a result of which fuel consumption is reduced by 25-30%.
• It is also possible to choose between an open or closed combustion chamber. In the first situation, you need a chimney and natural ventilation - this is a cheaper way. The second case involves the forced supply of air into the chamber by a fan and the same removal of combustion products through a coaxial chimney.

gas boiler

• If the design and installation of heating provides for a solid fuel boiler for heating a private house, then it is better to give preference to a gas generating device. The fact is that such systems are much more economical than conventional units, because the combustion of fuel in them occurs almost without a trace, and even that evaporates in the form of carbon dioxide and soot. When burning wood or coal from the lower chamber, the pyrolysis gas falls into another chamber, where it burns to the end, which justifies the very high efficiency.

Recommendations. There are other types of boilers, but about them now more briefly. So, if you opted for a liquid fuel heater, you can give preference to a unit with a multi-stage burner, thereby increasing the efficiency of the entire system.

Electrode boiler "Galan"

If you prefer electric boilers, then instead of a heating element it is better to purchase an electrode heater (see photo above). This is a relatively new invention in which the coolant itself serves as a conductor of electricity. But, nevertheless, it is completely safe and very economical.

Fireplace for heating a country house

The thermal calculation of the heating system seems to most to be an easy task that does not require special attention. A huge number of people believe that the same radiators should be chosen based only on the area of ​​\u200b\u200bthe room: 100 W per 1 sq. m. Everything is simple. But this is the biggest misconception. You cannot limit yourself to such a formula. What matters is the thickness of the walls, their height, material and much more. Of course, you need to set aside an hour or two to get the numbers you need, but everyone can do it.

Initial data for designing a heating system

To calculate the heat consumption for heating, you need, firstly, a house project.

The plan of the house allows you to get almost all the initial data that is needed to determine the heat loss and load on the heating system

Secondly, data on the location of the house in relation to the cardinal points and the construction area will be needed - the climatic conditions in each region are different, and what is suitable for Sochi cannot be applied to Anadyr.

Thirdly, we collect information about the composition and height of the outer walls and the materials from which the floor (from the room to the ground) and the ceiling (from the rooms and outward) are made.

After collecting all the data, you can get to work. Calculation of heat for heating can be performed using formulas in one to two hours. You can, of course, use a special program from Valtec.

To calculate the heat loss of heated rooms, the load on the heating system and heat transfer from heating devices, it is enough to enter only the initial data into the program. A huge number of functions make it an indispensable assistant for both the foreman and the private developer.

It greatly simplifies everything and allows you to get all the data on heat losses and hydraulic calculation heating systems.

Formulas for calculations and reference data

The calculation of the heat load for heating involves the determination of heat losses (Tp) and boiler power (Mk). The latter is calculated by the formula:

Mk \u003d 1.2 * Tp, where:

• Mk - thermal performance of the heating system, kW;
• Tp - heat loss at home;
• 1.2 - safety factor (20%).

A 20% safety factor makes it possible to take into account the possible pressure drop in the gas pipeline during the cold season and unforeseen heat losses (for example, broken window, poor-quality thermal insulation of entrance doors or unprecedented frosts). It allows you to insure against a number of troubles, and also makes it possible to widely regulate the temperature regime.

As can be seen from this formula, the power of the boiler directly depends on the heat loss. They are not evenly distributed throughout the house: the outer walls account for about 40% of the total value, the windows - 20%, the floor gives 10%, the roof 10%. The remaining 20% ​​disappear through the doors, ventilation.

Poorly insulated walls and floors, a cold attic, ordinary glazing on windows - all this leads to large heat losses, and, consequently, to an increase in the load on the heating system. When building a house, it is important to pay attention to all the elements, because even ill-conceived ventilation in the house will release heat into the street.

The materials from which the house is built have the most direct impact on the amount of heat lost. Therefore, when calculating, you need to analyze what the walls, and the floor, and everything else consist of.

In the calculations, to take into account the influence of each of these factors, the appropriate coefficients are used:

• K1 - type of windows;
• K2 - wall insulation;
• K3 - the ratio of floor area and windows;
• K4 - minimum temperature on the street;
• K5 - the number of external walls of the house;
• K6 - number of storeys;
• K7 - the height of the room.

For windows, the heat loss coefficient is:

• ordinary glazing - 1.27;
• double-glazed window - 1;
• three-chamber double-glazed window - 0.85.

Naturally, the last option will keep the heat in the house much better than the previous two.

Properly executed wall insulation is the key not only to a long life of the house, but also to a comfortable temperature in the rooms. Depending on the material, the value of the coefficient also changes:

• concrete panels, blocks - 1.25-1.5;
• logs, timber - 1.25;
• brick (1.5 bricks) - 1.5;
• brick (2.5 bricks) - 1.1;
• foam concrete with increased thermal insulation - 1.

The larger the window area relative to the floor, the more heat the house loses:

The temperature outside the window also makes its own adjustments. At low rates of heat loss increase:

• Up to -10С - 0.7;
• -10C - 0.8;
• -15C - 0.90;
• -20C - 1.00;
• -25C - 1.10;
• -30C - 1.20;
• -35C - 1.30.

Heat loss also depends on how many external walls the house has:

• four walls - 1.33;%
• three walls - 1.22;
• two walls - 1.2;
• one wall - 1.

It’s good if a garage, a bathhouse or something else is attached to it. But if it is blown from all sides by winds, then you will have to buy a more powerful boiler.

The number of floors or the type of room that is located above the room determine the K6 coefficient as follows: if the house has two or more floors above, then for calculations we take the value 0.82, but if it is an attic, then for warm - 0.91 and 1 for cold .

As for the height of the walls, the values ​​\u200b\u200bwill be as follows:

• 4.5 m - 1.2;
• 4.0 m - 1.15;
• 3.5 m - 1.1;
• 3.0 m - 1.05;
• 2.5 m - 1.

In addition to the above coefficients, the area of ​​\u200b\u200bthe room (Pl) and the specific value of heat loss (UDtp) are also taken into account.

The final formula for calculating the heat loss coefficient:

Tp \u003d UDtp * Pl * K1 * K2 * K3 * K4 * K5 * K6 * K7.

The UDtp coefficient is 100 W/m2.

Analysis of calculations on a specific example

The house for which we will determine the load on the heating system has double-glazed windows (K1 \u003d 1), foam concrete walls with increased thermal insulation (K2 \u003d 1), three of which go outside (K5 \u003d 1.22). The area of ​​windows is 23% of the floor area (K3=1.1), on the street about 15C frost (K4=0.9). The attic of the house is cold (K6=1), the height of the premises is 3 meters (K7=1.05). The total area is 135m2.

Fri \u003d 135 * 100 * 1 * 1 * 1.1 * 0.9 * 1.22 * 1 * 1.05 \u003d 17120.565 (Watts) or Fri \u003d 17.1206 kW

Mk \u003d 1.2 * 17.1206 \u003d 20.54472 (kW).

Calculation of load and heat loss can be done independently and quickly enough. You just need to spend a couple of hours putting the source data in order, and then just substitute the values ​​​​into the formulas. The numbers that you will receive as a result will help you decide on the choice of a boiler and radiators.