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Calculation of heating boiler power, in particular a gas boiler, is necessary not only for choosing a boiler and heating equipment, but also to ensure comfortable functioning heating system in general and eliminating unnecessary operating costs.

From a physics point of view, only four parameters are involved in calculating thermal power: the air temperature outside, the required temperature inside, the total volume of the premises and the degree of thermal insulation of the house, on which heat loss depends. But in reality, everything is not so simple. Outdoor temperature changes depending on the time of year, the requirements for internal temperature are determined by the mode of residence, the total volume of the premises must first be calculated, and heat loss depends on the materials and design of the house, as well as on the size, number and quality of windows.

Calculator for gas boiler power and gas consumption per year

Power calculator shown here gas boiler and gas consumption for the year can significantly facilitate your task of choosing a gas boiler - just select the appropriate field values, and you will get the required values.

Please note that the calculator calculates not only optimal power gas boiler for heating the house, but also the average annual gas consumption. That is why the “number of residents” parameter was introduced into the calculator. It is necessary to take into account average consumption gas for cooking and hot water for domestic needs.

This parameter is relevant only if you also use gas for your stove and water heater. If you use other appliances for this, for example, electric ones, or even don’t cook at home and do without hot water, put zero in the “number of residents” field.

The following data is used in the calculation:

• duration of the heating season - 5256 hours;
• duration of temporary residence (summer and weekends 130 days) - 3120 hours;
• the average temperature during the heating period is minus 2.2°C;
• the air temperature of the coldest five-day period in St. Petersburg is minus 26°C;
• ground temperature under the house during the heating season - 5°C;
• reduced room temperature in the absence of a person - 8.0°C;
• insulation attic floor— a layer of mineral wool with a density of 50 kg/m³ and a thickness of 200 mm.

Coziness and comfort of housing does not begin with the choice of furniture, decoration and appearance generally. They start with the heat that heating provides. And simply purchasing an expensive heating boiler () and high-quality radiators for this purpose is not enough - first you need to design a system that will maintain the optimal temperature in the house. But to get good result, you need to understand what should be done and how, what nuances exist and how they affect the process. In this article you will become familiar with the basic knowledge about this matter - what heating systems are, how it is carried out and what factors influence it.

Why is thermal calculation necessary?

Some owners of private houses or those who are just planning to build them are interested in whether there is any point in the thermal calculation of the heating system? After all we're talking about about simple country cottage, and not about an apartment building or an industrial enterprise. It would seem that it would be enough just to buy a boiler, install radiators and run pipes to them. On the one hand, they are partially right - for private households, the calculation of the heating system is not as critical an issue as for production premises or multi-apartment residential complexes. On the other hand, there are three reasons why such an event is worth holding. , you can read in our article.

1. Thermal calculation significantly simplifies the bureaucratic processes associated with gasification of a private home.
2. Determining the power required for heating a home allows you to select a heating boiler with optimal characteristics. You will not overpay for excessive product characteristics and will not experience inconvenience due to the fact that the boiler is not powerful enough for your home.
3. Thermal calculation allows you to more accurately select pipes, shut-off valves and other equipment for the heating system of a private home. And in the end, all these rather expensive products will work for as long as is included in their design and characteristics.

Initial data for thermal calculation of the heating system

Before you begin to calculate and work with data, you need to obtain it. Here for those owners of country houses who have not previously dealt with project activities, the first problem arises - what characteristics should you pay attention to. For your convenience, they are summarized in a short list below.

1. Building area, ceiling height and internal volume.
2. Type of building, presence of adjacent buildings.
3. Materials used in the construction of the building - what and how the floor, walls and roof are made of.
4. The number of windows and doors, how they are equipped, how well they are insulated.
5. For what purposes will these or those parts of the building be used - where the kitchen, bathroom, living room, bedrooms will be located, and where - non-residential and technical premises.
6. Duration of the heating season, average minimum temperature during this period.
7. “Wind rose”, the presence of other buildings nearby.
8. An area where a house has already been built or is about to be built.
9. Preferred temperature for residents in certain rooms.
10. Location of points for connecting to water supply, gas and electricity.

Calculation of heating system power based on housing area

One of the fastest and easiest to understand ways to determine the power of a heating system is to calculate the area of ​​the room. This method is widely used by sellers of heating boilers and radiators. Calculating the power of a heating system by area occurs in a few simple steps.

Step 1. Based on the plan or already erected building, the internal area of ​​the building in square meters is determined.

Step 2. The resulting figure is multiplied by 100-150 - this is exactly how many watts of the total power of the heating system are needed for each m 2 of housing.

Step 3. Then the result is multiplied by 1.2 or 1.25 - this is necessary to create a power reserve so that the heating system is able to maintain a comfortable temperature in the house even in the event of the most severe frosts.

Step 4. The final figure is calculated and recorded - the power of the heating system in watts required to heat a particular home. As an example, to maintain a comfortable temperature in a private house with an area of ​​120 m2, approximately 15,000 W will be required.

Advice! In some cases, cottage owners share internal area housing for that part that requires serious heating, and that for which this is unnecessary. Accordingly, different coefficients are used for them - for example, for living rooms this is 100, and for technical rooms - 50-75.

Step 5. Based on the already determined calculation data, a specific model of the heating boiler and radiators is selected.

It should be understood that the only advantage of this method of thermal calculation of a heating system is speed and simplicity. However, the method has many disadvantages.

1. Lack of consideration of the climate in the area where housing is being built - for Krasnodar, a heating system with a power of 100 W per square meter will be clearly excessive. But for the Far North it may not be sufficient.
2. Failure to take into account the height of the premises, the type of walls and floors from which they are built - all these characteristics seriously affect the level of possible heat losses and, consequently, the required power heating system for the home.
3. The very method of calculating the heating system by power was originally developed for large industrial premises and apartment buildings. Therefore, for separate cottage it is not correct.
4. Lack of accounting for the number of windows and doors facing the street, and yet each of these objects is a kind of “cold bridge”.

So does it make sense to use a heating system calculation based on area? Yes, but only as preliminary estimates that allow us to get at least some idea of ​​the issue. To achieve better and more accurate results, you should turn to more complex techniques.

Let's imagine the following method for calculating the power of a heating system - it is also quite simple and understandable, but at the same time it differs more high accuracy the final result. IN in this case The basis for calculations is not the area of ​​the room, but its volume. In addition, the calculation takes into account the number of windows and doors in the building and the average level of frost outside. Let's imagine small example application of a similar method - there is a house with a total area of ​​80 m2, the rooms in which have a height of 3 m. The building is located in the Moscow region. There are a total of 6 windows and 2 doors facing outside. The calculation of the power of the thermal system will look like this. "How to make , You can read in our article.”

Step 1. The volume of the building is determined. This can be the sum of each individual room or the total figure. In this case, the volume is calculated as follows - 80 * 3 = 240 m 3.

Step 2. The number of windows and the number of doors facing the street are counted. Let's take the data from the example - 6 and 2, respectively.

Step 3. A coefficient is determined depending on the area in which the house is located and how severe the frost is there.

Table. Values ​​of regional coefficients for calculating heating power by volume.

Since the example is about a house built in the Moscow region, the regional coefficient will have a value of 1.2.

Step 4. For detached private cottages, the value of the volume of the building determined in the first operation is multiplied by 60. We do the calculation - 240 * 60 = 14,400.

Step 5. Then the calculation result of the previous step is multiplied by the regional coefficient: 14,400 * 1.2 = 17,280.

Step 6. The number of windows in the house is multiplied by 100, the number of doors facing outside is multiplied by 200. The results are summed up. The calculations in the example look like this – 6*100 + 2*200 = 1000.

Step 7 The numbers obtained from the fifth and sixth steps are summed up: 17,280 + 1000 = 18,280 W. This is the power of the heating system required to maintain optimal temperature in the building under the conditions specified above.

It is worth understanding that the calculation of the heating system by volume is also not absolutely accurate - the calculations do not pay attention to the material of the walls and floor of the building and their thermal insulation properties. Also no correction is made for natural ventilation characteristic of any home.

To ensure a comfortable temperature throughout the winter, the heating boiler must produce the amount of thermal energy that is necessary to replenish all heat losses of the building/room. Plus, it is also necessary to have a small power reserve in case of abnormal cold weather or expansion of the area. We’ll talk about how to calculate the required power in this article.

To determine the performance of heating equipment, you must first determine the heat loss of the building/room. This calculation is called thermotechnical. This is one of the most complex calculations in the industry as there are many components to consider.

Of course, the amount of heat loss is influenced by the materials used in the construction of the house. Therefore, the building materials from which the foundation, walls, floor, ceiling, floors, attic, roof, window and door openings are made are taken into account. The type of system wiring and the presence of heated floors. In some cases, they even consider the presence household appliances, which generates heat during operation. But such precision is not always required. There are methods that allow you to quickly estimate the required performance of a heating boiler without plunging into the jungle of heating engineering.

Calculation of heating boiler power by area

For a rough estimate of the required performance of a heating unit, the area of ​​the premises is sufficient. In the very simple version for central Russia, it is believed that 1 kW of power can heat 10 m 2 of area. If you have a house with an area of ​​160 m2, the boiler power for heating it is 16 kW.

These calculations are approximate, because neither ceiling height nor climate are taken into account. For this purpose, there are coefficients derived experimentally, with the help of which appropriate adjustments are made.

The specified norm is 1 kW per 10 m2, suitable for ceilings of 2.5-2.7 m. If you have higher ceilings in the room, you need to calculate the coefficients and recalculate. To do this, divide the height of your premises by the standard 2.7 m and obtain a correction factor.

Calculating the power of a heating boiler by area is the easiest way

For example, the ceiling height is 3.2 m. We calculate the coefficient: 3.2m/2.7m=1.18, round it up, we get 1.2. It turns out that to heat a room of 160 m 2 with a ceiling height of 3.2 m, a heating boiler with a capacity of 16 kW * 1.2 = 19.2 kW is required. They usually round up, so 20 kW.

To take into account climatic features, there are ready-made coefficients. For Russia they are:

• 1.5-2.0 for northern regions;
• 1.2-1.5 for Moscow region regions;
• 1.0-1.2 for the middle band;
• 0.7-0.9 for the southern regions.

If the house is in middle lane, just south of Moscow, a coefficient of 1.2 is used (20 kW * 1.2 = 24 kW), if in the south of Russia in the Krasnodar Territory, for example, the coefficient is 0.8, that is, less power is required (20 kW * 0.8 = 16 kW).

Heating calculation and boiler selection - important stage. Find the power incorrectly and you can get the following result...

These are the main factors that need to be taken into account. But the values ​​found are valid if the boiler operates only for heating. If you also need to heat water, you need to add 20-25% of the calculated figure. Then you need to add a “reserve” for peak winter temperatures. That's another 10%. In total we get:

• For heating a house and hot water in the middle zone 24 kW + 20% = 28.8 kW. Then the reserve for cold weather is 28.8 kW + 10% = 31.68 kW. We round up and get 32 ​​kW. If we compare it with the original figure of 16 kW, the difference is twofold.
• House in Krasnodar region. We add power to heat hot water: 16 kW + 20% = 19.2 kW. Now the “reserve” for cold weather is 19.2+10%=21.12 kW. Round up: 22 kW. The difference is not so striking, but still quite significant.

From the examples it is clear that at least these values ​​must be taken into account. But it is obvious that when calculating the boiler power for a house and an apartment, there should be a difference. You can go the same way and use coefficients for each factor. But there is an easier way that allows you to make corrections in one go.

When calculating a heating boiler for a home, a coefficient of 1.5 is used. It takes into account the presence of heat loss through the roof, floor, and foundation. Valid for an average (normal) degree of wall insulation - masonry with two bricks or building materials with similar characteristics.

For apartments, different coefficients apply. If there is a heated room on top (another apartment) the coefficient is 0.7, if there is a heated attic - 0.9, if there is an unheated attic - 1.0. You need to multiply the boiler power found using the method described above by one of these coefficients and get a fairly reliable value.

To demonstrate the progress of the calculations, we will calculate the power of a gas heating boiler for an apartment of 65 m2 with 3 m ceilings, which is located in central Russia.

1. We determine the required power by area: 65m 2 /10m 2 = 6.5 kW.
2. We make an adjustment for the region: 6.5 kW * 1.2 = 7.8 kW.
3. The boiler will heat the water, so we add 25% (we like it hot) 7.8 kW * 1.25 = 9.75 kW.
4. Add 10% for cold weather: 7.95 kW * 1.1 = 10.725 kW.

Now we round the result and get: 11KW.

The specified algorithm is valid for the selection of heating boilers using any type of fuel. Calculating the power of an electric heating boiler will be no different from calculating a solid fuel, gas or liquid fuel boiler. The main thing is the productivity and efficiency of the boiler, and heat loss does not change depending on the type of boiler. The whole question is how to spend less energy. And this is the area of ​​insulation.

Boiler power for apartments

When calculating heating equipment for apartments, you can use SNiP standards. The use of these standards is also called calculating boiler power by volume. SNiP sets the required amount of heat to heat one cubic meter of air in typical buildings:

• for heating 1m 3 v panel house 41W required;
• V brick house per m 3 there are 34W.

Knowing the area of ​​the apartment and the height of the ceilings, you will find the volume, then, multiplying by the norm, you will find out the power of the boiler.

For example, let’s calculate the required boiler power for premises in a brick house with an area of ​​74 m2 with ceilings of 2.7 m.

1. We calculate the volume: 74m2 *2.7m=199.8m3
2. We calculate according to the norm how much heat will be needed: 199.8*34W=6793W. We round and convert to kilowatts, we get 7 kW. This will be the required power that the thermal unit must produce.

It is easy to calculate the power for the same room, but in a panel house: 199.8*41W=8191W. In principle, in heating engineering they always round up, but you can take into account the glazing of your windows. If the windows have energy-saving double-glazed windows, you can round down. We believe that the double-glazed windows are good and get 8 kW.

The choice of boiler power depends on the type of building - brick buildings require less heat to heat than panel ones

Next, you need, just as in the calculation for a house, to take into account the region and the need to prepare hot water. Corrections for abnormal cold weather are also relevant. But in apartments, the location of the rooms and the number of floors play a big role. Walls facing the street need to be taken into account:

• One outer wall — 1,1
• Two - 1.2
• Three - 1.3

After taking into account all the coefficients, you will get enough exact value, which you can rely on when choosing heating equipment. If you want to get an accurate thermal calculation, you need to order it from a specialized organization.

There is another method: determine real losses using a thermal imager - modern device, which will also show the places through which heat leaks more intensely. At the same time, you can eliminate these problems and improve thermal insulation. And the third option is to use a calculator program that will calculate everything for you. You just need to select and/or enter the required data. At the output you will receive the calculated power of the boiler. True, there is a certain amount of risk here: it is not clear how correct the algorithms are at the basis of such a program. So you still have to at least roughly calculate it to compare the results.

We hope you now have an idea of ​​how to calculate the boiler power. And you don’t get confused about what it is and not solid fuel, or vice versa.

You may be interested in articles about and. In order to have general idea Watch the video about mistakes that are often encountered when planning a heating system.

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

But that's not all. It is not enough to correctly purchase the necessary heating boiler - it is very important to optimally select and correctly arrange 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, it is impossible to do without certain calculations.

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

The simplest calculation methods

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 to each other, and their division is very conditional.

• The first is maintaining an optimal level of air temperature throughout the entire volume of the heated room. Of course, the temperature level may vary somewhat with altitude, but this difference should not be significant. An average of +20 °C is considered quite comfortable conditions - this is the temperature that is usually taken as the initial one in thermal calculations.

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

If we approach it with complete accuracy, then for individual rooms in residential buildings 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:

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

The most important “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 occur in all directions - their approximate distribution is shown in the table:

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 among the rooms, 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 is calculated for each heated room, 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 is needed. And the values ​​​​for each room will become the starting point for calculating the required number of radiators.

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

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

Q = S× 100

Q– required heating power for the room;

S– room area (m²);

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

For example, a 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 when standard height ceilings - approximately 2.7 m (acceptable - 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 power density is calculated at cubic meter. It is taken equal to 41 W/m³ for 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 the linear dimensions of the room, 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 is 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 can be useful for an initial “estimate,” but you should still rely on them completely with great caution. Even to a person who does not understand anything about building heating engineering, the indicated average values ​​may certainly seem dubious - they cannot be equal, say, for Krasnodar region and for the Arkhangelsk region. In addition, the room is different: one is located on the corner of the house, that is, it has two external walls, and the other is protected from heat loss by other rooms on three sides. 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 - it’s just that such features are visible even to the naked eye.

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

General principles and calculation formula

The calculations will be based on the same ratio: 100 W per 1 square meter. But 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 completely 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” is a coefficient that takes into account the number of external walls in a particular room.

Obviously, the more external walls there are in a room, the larger area, through which it occurs heat losses. In addition, the presence of two or more external walls also means corners - extremely vulnerable places from the point of view of the formation of “cold bridges”. Coefficient “a” will correct for this specific feature of the room.

The coefficient is taken equal to:

— external walls No (interior space): a = 0.8;

- external 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 directions.

You might be interested in information about what types of

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

But walls and windows facing north “never see” the Sun. East End at home, although he “grabs” the morning Sun rays, still does not receive any effective heating from them.

Based on this, we introduce the coefficient “b”:

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

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

• “c” is a coefficient that takes into account the location of the room relative to the winter “wind rose”

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

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

If you want to carry out calculations with higher accuracy, you can include the correction factor “c” in the formula, taking it equal to:

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

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

• “d” is a correction factor taking into account the peculiarities climatic conditions region where the house was built

Naturally, the amount of heat loss through all building structures will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer readings “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 typical for January). For example, below is a map diagram of the territory of Russia, on which approximate values ​​are shown in colors.

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

So, the coefficient “d”, which takes into account the climate characteristics of the region, for our calculations is taken equal to:

— from – 35 °C 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;

- no colder - 10 °C: d = 0.7.

• “e” is a coefficient that takes into account the degree of insulation of external walls.

The total value of heat losses of a building is directly related to the degree of insulation of all building structures. One of the “leaders” in 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 do not have insulation: e = 1.27;

- average degree of insulation - walls made of two bricks or their surface thermal insulation is provided with other insulation materials: e = 1.0;

— insulation was carried out with high quality, based on thermal engineering calculations: e = 0.85.

Below 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 heights

Ceilings, especially in private homes, can have different heights. Therefore, the thermal power to warm up a particular room of the same area will also differ in this parameter.

It would not be a big mistake to accept the following values ​​for the correction factor “f”:

- ceiling heights up to 2.7 m: f = 1.0;

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

- ceiling heights from 3.1 to 3.5 m: f = 1.1;

- ceiling heights from 3.6 to 4.0 m: f = 1.15;

- ceiling height more than 4.1 m: f = 1.2.

• « g" is a coefficient that takes 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. This means that it is necessary to make some adjustments to account for this feature of a particular room. The correction factor “g” can be taken equal to:

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

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

— the heated room is located below: g= 1,0 .

• « h" is a coefficient that takes 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 loss is inevitable, which will require an increase in the required thermal power. Let us introduce the coefficient “h”, which takes into account this feature of the calculated room:

— the “cold” attic is located on top: h = 1,0 ;

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

— any heated room is located on top: h = 0,8 .

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

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

Without words it is clear that the thermal insulation qualities of these windows differ significantly

But there is no complete uniformity between PVH windows. For example, 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 glass: 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 total area room glazing

No matter how high-quality the windows are, it will still not be possible to completely avoid heat loss through them. But it is quite clear that one cannot compare a small window with panoramic glazing almost 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– total area of ​​windows in the room;

SP– area of ​​the room.

Depending on the obtained value, the correction factor “j” is determined:

— x = 0 ÷ 0.1 →j = 0,8 ;

— x = 0.11 ÷ 0.2 →j = 0,9 ;

— x = 0.21 ÷ 0.3 →j = 1,0 ;

— x = 0.31 ÷ 0.4 →j = 1,1 ;

— x = 0.41 ÷ 0.5 →j = 1,2 ;

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

A door to the street or to an unheated balcony is always an additional “loophole” for the cold

Door to the street or open balcony is capable of making adjustments to the thermal balance of the room - each opening of it is accompanied by the penetration of a considerable volume 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 to the balcony: k = 1,3 ;

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

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

Perhaps this may seem like an insignificant detail to some, but still, why not immediately take into account the planned connection diagram for the heating radiators. The fact is that their heat transfer, and therefore their participation in maintaining a certain temperature balance in the room, changes quite noticeably when different types 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 below	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 peculiarities of the installation location of heating radiators

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

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

So, the calculation formula is clear. Surely, some of the readers will immediately grab their head - they say, it’s too complicated and cumbersome. However, if you approach the matter systematically and in an orderly manner, then there is no trace of complexity.

Any good homeowner must have a detailed graphic plan of his “possessions” with dimensions indicated, and usually oriented to the cardinal points. The climatic features of the region are easy to clarify. All that remains is to walk through all the rooms with a tape measure and clarify some of the nuances for each room. Features of housing - “vertical proximity” above and below, location entrance doors, the proposed or existing installation scheme for heating radiators - no one except the owners knows better.

It is recommended to immediately create a worksheet where you can enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will be helped by the built-in calculator, which already contains all the coefficients and ratios mentioned above.

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

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

A region with minimum temperatures ranging from -20 ÷ 25 °C. Predominance of winter winds = northeast. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal one has been selected diagonal connection radiators that will be installed under window sills.

Let's create a table something like this:

Then, using the calculator below, we make calculations for each room (already taking into account the 10% reserve). It won't take much time using the recommended app. After this, all that remains is to sum up the obtained values ​​for 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 - all that remains is to divide by the specific thermal power of one section and round up.

In the process of building any house, sooner or later the question arises - how to correctly calculate the heating system? This urgent problem will never exhaust its resource, because if you buy a boiler of less power than necessary, you will have to spend a lot of effort to create secondary heating with oil and infrared radiators, heat guns, and electric fireplaces.

In addition, monthly maintenance, due to expensive electricity, will cost you a pretty penny.

The same thing will happen if you buy a boiler with increased power, which will work at half power and consume no less fuel. Our calculator for calculating the heating of a private house will help you prevent typical mistakes

novice builders. You will receive the value of heat loss and the required heat output of the boiler as close as possible to reality according to the current data of SNiPs and SPs (codes of rules).

The main advantage of the calculator on the website is the reliability of the calculated data and the absence of manual calculations, the entire process is automated, the initial parameters are as generalized as possible, you can easily see their values ​​in the plan of your house or fill them in based on your own experience.

Calculation of a boiler for heating a private house

Using our heating calculation calculator for a private home, you can easily find out the required boiler power to heat your cozy “nest”. As you remember, in order to calculate the heat loss rate, you need to know several values ​​​​of the main components of the house, which together account for more than 90% of the total losses. For your convenience, we have added to the calculator only those fields that you can fill in:

• without special knowledge
• glazing;
• thermal insulation;
• window to floor area ratio;
• outside temperature;
• number of walls facing outside;
• what room is above the one being calculated;
• room height;

room area.

After you receive the value of heat loss at home, to calculate the required boiler power, a correction factor of 1.2 is taken.

How to use the calculator

Remember that the thicker the glazing and the better the thermal insulation, the less heating power will be required.

1. Choose one of the proposed types of glazing (triple or double glazing, regular double-chamber glass).
2. How are your walls insulated? Good thick insulation made from a couple of layers of mineral wool, polystyrene foam, EPS for the north and Siberia. Maybe you live in Central Russia and one layer of insulation is enough for you. Or you are one of those who is building a house in the southern regions and double hollow bricks will suit him.
3. What is your window to floor area ratio, in %. If you don’t know this value, it is calculated very simply: divide the floor area by the window area and multiply by 100%.
4. Specify minimum temperature V winter period over a couple of seasons and round up. No need to use average temperature in winter, otherwise you risk getting a boiler of lower power, and the house will not be heated enough.
5. Are we calculating for the whole house or just one wall?
6. What is above our premises? If you have cottage, select the type of attic (cold or warm), if the second floor, then a heated room.
7. The height of the ceilings and the area of ​​the room are necessary to calculate the volume of the apartment, which in turn is the basis for all calculations.

Calculation example:

• one-story house in the Kaliningrad region;
• the length of the walls is 15 and 10 m, insulated with one layer of mineral wool;
• ceiling height 3 m;
• 6 windows of 5 m2 each from double-glazed windows;
• the minimum temperature over the last 10 years is 26 degrees;
• we calculate for all 4 walls;
• a warm heated attic above;

The area of ​​our house is 150 m2, and the window area is 30 m2. 30/150*100=20% ratio between windows and floor.

We know everything else, select the appropriate fields in the calculator and get that our house will lose 26.79 kW of heat.

26.79*1.2=32.15 kW - the required heating output of the boiler.

DIY heating system

It is impossible to calculate the heating circuit of a private house without assessing the heat loss of surrounding structures.

Russia typically has long, cold winters and buildings lose heat due to temperature changes inside and outside the premises. The larger the area of ​​the house, enclosing and through structures (roof, windows, doors), the higher value heat loss comes out. The material and thickness of the walls, the presence or absence of thermal insulation have a significant influence.

For example, walls made of wood and aerated concrete have a much lower thermal conductivity than brick. Materials with maximum thermal resistance are used as insulation (mineral wool, polystyrene foam).

Before creating a heating system for a house, you need to carefully consider all the organizational and technical aspects, so that immediately after building the “box”, you can begin the final phase of construction, and not postpone the long-awaited occupancy for many months.

Heating in a private house is based on “three elephants”:

• heating element (boiler);
• pipe system;
• radiators.

Which boiler is better to choose for your home?

Heating boilers are the main component of the entire system. They are the ones who will provide warmth to your home, so you need to be especially careful when choosing them. Based on the type of food they are divided into:

• electrical;
• solid fuel;
• liquid fuel;
• gas.

Each of them has a number of significant advantages and disadvantages.

1. Electric boilershave not gained much popularity, primarily due to their relatively high cost and high maintenance costs. Electricity tariffs leave much to be desired, and there is a possibility of broken power lines, which could leave your home without heating.
2. Solid fuelboilersoften used in remote villages and towns where there are no centralized communication networks. They heat water using wood, briquettes and coal. An important disadvantage is the need for constant monitoring of fuel; if the fuel burns out and you do not have time to replenish supplies, the house will stop heating. IN modern models this problem has been solved due to an automatic feeder, but the price of such devices is incredibly high.
3. Liquid fuel boilers, in the vast majority of cases, work for diesel fuel. They have excellent performance due to high efficiency fuel, but high price on raw materials and the need for diesel tanks limits many buyers.
4. The most optimal solution for country house are gas boilers . Due to their small size, low gas price and high heat output, they have won the trust of most of the population.

How to choose heating pipes?

Heating mains supply all heating devices in the house. Depending on the material of manufacture, they are divided into:

• metal;
• metal-plastic;
• plastic.

Metal pipes the most difficult to install (due to the need to weld seams), are susceptible to corrosion, are heavy and expensive. The advantages are high strength, resistance to temperature changes and the ability to withstand high pressure. They are used in apartment buildings, in private construction it is inappropriate to use them.

Polymer pipes made of metal-plastic and polypropylene are very similar in their parameters. Lightness of the material, ductility, lack of corrosion, noise suppression and, of course, low price. The only difference between the former is the presence of an aluminum layer between two layers of plastic, due to which the thermal conductivity increases. Therefore, metal-plastic pipes are used for heating, and plastic pipes for water supply.

Choosing radiators for the home

Last element classical system heating - radiators. They are also divided according to material into the following groups:

• cast iron;
• steel;
• aluminum.

Cast iron batteries are familiar to everyone since childhood, because they were installed in almost all apartment buildings. They have high heat capacity (they take a long time to cool down) and are resistant to temperature and pressure changes in the system. The downside is the high price, fragility and complexity of installation.

They were replaced steel radiators. A wide variety of shapes and sizes, low cost and ease of installation have contributed to their widespread use. However, they also have their drawbacks. Due to their low thermal capacity, the batteries cool down quickly, and their thin body does not allow them to be used in high-pressure networks.

IN Lately heaters from aluminum. Their main advantage is high heat transfer, which allows you to warm the room to an acceptable temperature in 10-15 minutes. However, they are demanding on the coolant if inside the system there is large quantities contains alkali or acid, the service life of the radiator is significantly reduced.

Use the proposed tools to calculate the heating of a private home and design a heating system that will heat your home efficiently, reliably and for a long time, even in the harshest winters.